Apr 22, 2017 - ... essential public services such as maintenance of law and order, defence services, judiciary .... but eventually started buying smaller cars. 24 ...
Course Material
Subject Code/Name: MG 6863 – Engineering Economics Semester: VIII Regulation: 2013 Academic year: 2016-2017
Prepared by Mr. Sathish.D, Assistant Professor, Department of Mechanical Engineering, Sri Ramakrishna Institute of Technology, Coimbatore - 10.
1
What is Economics?
2
3
- Robbins
4
5
Needs and Wants
• NEEDS – stuff e must ha e to sur i e, generally: food, shelter, clothing
• WANTS – stuff e ould really like to ha e (Fancy food, shelter, clothing, big screen TVs, jewelry, conveniences . . . Also known as LUXURIES. 6
VS.
7
ECONOMICS
MICRO
MACRO
8
The Study of Economics • Macroeconomics – The big picture: growth, employment, etc. – Choices made by large groups (like countries)
• Microeconomics – How do individuals make economic decisions.
9
10
11
The Factors of Production
Land
Labour
Capital
Organization
Product
12
13
ECONOMY
MARKET
COMMAND
MIXED
14
15 Contd….
16
Contd….
17
18
19 Contd…
• • • •
Unlimited Wants Scarce Resources – Land, Labour, Capital Many Uses of Resources Choices
20
Circular flow of income????? • The term circular flow of income or circular flow of economic activity refers to “a simple economic model which describes the circulation/flow of income between producers and consumers”. • In the circular flow model, producer and consumer are referred to as "firms" and "households" respectively. 1
Circular Flow Concepts Product Market – where goods and services are exchanged
Households – suppliers of the factors of production & demanders of goods and services
Government – providers of public goods and services & demanders of both private goods and services and the factors of production Businesses / Firms – suppliers of goods and services & demanders of the factors of production
Factor Market – where the factors of production are exchanged
2
The circular-flow diagram is a model that represents the transactions in an economy by flows around a circle. Two sectors models a.) savings economy b.) non-savings economy
Three sectors models Four sectors models 3
Two sectors model (no savings economy) Factors of payment (rent, wages, interest, profit)
Factors of Production (land, labor, capital, entrepreneur)
Household Sector
Business Firm
Output of G/S
Consumption of G/S 4
(savings economy) Factors of payment (rent, wages, interest, profit) Factors of Production (land, labor, capital, entrepreneur)
Business Firm
Household Sector
Output of G/S Consumption of G/S Household savings
Expansion of business 5
Three sectors models It includes household sector, producing sector and government sector. It will study a circular flow income in these sectors excluding rest of the world i.e. closed economy income. Here flows from household sector and producing sector to government sector are in the form of taxes. The income received from the government sector flows to producing and household sector in the form of payments for government purchases of goods and services as well as payment of subsides and transfer payments. Every payment has a receipt in response of it by which aggregate expenditure of an economy becomes identical to aggregate income and makes this circular flow unending. 6
Factors of payment (rent, wages, interest, profit)
Taxes
Factors of Production (land, labor, capital, entrepreneur)
Business Firm
Household Sector
Government (BIR)
Output of G/S Consumption of G/S Household savings
Expansion of business Government expenditure
7
Four sectors models • A modern monetary economy comprises a network of four sector economy these are- 1.Household sector 2. Firms or Producing sector 3. Government sector 4. Rest of the world sector. Each of the above sectors receives some payments from the other in lieu of goods and services which makes a regular flow of goods and physical services. Money facilitates such an exchange smoothly. A residual of each market comes in capital market as saving which in turn is invested in firms and government sector. Technically speaking, so long as lending is equal to the borrowing i.e. leakage is equal to injections, the circular flow will continue indefinitely. However this job is done by financial institutions in the economy. 8
Factors of payment (rent, wages, interest, profit)
Taxes
Import
Government (BIR)
Other countries
Factors of Production (land, labor, capital, entrepreneur)
Business Firm
Household Sector
Output of G/S Consumption of G/S Household savings
Expansion of business
Export
Government expenditure 9
Significance of Study of Circular Flow of Income
Measurement of National Income- National income is an estimation of aggregation of any of economic activity of the circular flow. It is either the income of all the factors of production or the expenditure of various sectors of economy. Knowledge of Interdependence- Circular flow of income signifies the interdependence of each of activity upon one another
Unending Nature of Economic Activities- It signifies that production, income and expenditure are of unending nature, therefore, economic activities in an economy can never come to a halt. National income is also bound to rise in future. To understand about the leakages in economy and injections
10
Foreign sector
Household
Determinants
Firms/ busines s
Financial institutions
11
HOUSE HOLD
12
Households: It is a person or a group of people that share their income. The members of households have two functions:
they supply different factors of production members of household also work as consumers
13
FIRMS
14
Firms: An organization that produces goods and services for sale. main objective is to maximize profit in the production process. The two main functions are as follows: Produce goods and services and supply them in the market.
Firms purchase inputs or raw materials from households to use them in the production process
15
GOVERNMENT
16
Government Just like households and firms the government also earns incomes and makes expenses. Two major functions played are: Government earns revenue either from tax or non-tax sources both from households and firms.
Government provides essential public services such as maintenance of law and order, defence services, judiciary etc. 17
FINANCIAL INSTITUTIONS
18
Financial institutions Financial Institution : consists of banks and non-bank intermediaries who engage in the borrowing (savings from households)and lending of money. the leakage that financial institutions provide in the economy is the option for households to save their money.
19
Foreign sector
20
Foreign Market: It consists of two kinds of international economic transactions i.e. export and import of goods and services inflow and outflow of capital. 21
Have a glance on Two sector model
Three sector model Four sector model
Five sector model 22
TWO SECTOR MODEL
23
24
25
26
27
28
Three sector model Three sector model is created by adding the Government sector to the Two sector model Three kinds of monetary flows between the government and the rest of the economy i.e.1)direct taxes on both households and firms2)government expenditure3)transfer payments and subsidies
Government spends a part of its tax revenue as factor payments to the households and a part in the form of transfer payments as pension and food subsidy etc. 29
30
Govt borrowings
31
32
This circular flow of model shows the four macro economic sectors of the economy i.e. household, business firm, government, and financial institutions. These four sectors capture four fundamental macroeconomic functions and their expenditures are combined together to purchase the economy's total production.
33
To introduce the financial market, it is assumed that household saves in the financial market.
There are no inter-households borrowing If the households save a part of their income in the financial market (such as banks, insurance companies, stock market etc), this reduces the expenditure of household on goods and services Ultimately it reduces the flow of money/income of the economy. So saving known as the leakage of the economy 34
35
FIVE SECTOR MODEL
36
In five sector circular flow of income model, the state of equilibrium occurs when the total leakages are equal to the total injections that occur in the economy. This can be shown as: Savings + Taxes + Imports = Investment + Government
Spending + Exports
S + T + M = I + G + X. 37
CIRCULAR FLOW OF INCOME IN DIFFERENT SECTORS 1.
TWO SECTOR MODEL
It studies the circular flow of income between • Household sector & • Producing sector On the assumption that there are only two sectors in the economy
38
2. THREE SECTOR MODEL It refers to the study of the circular flow of income among • Household sector • Producing sector • Govt. Sector On the assumption that there are three sectors It is a closed economy 2. FOUR SECTOR MODEL It implies the study of the circular flow of income among • Household sector • Producing sector • Govt. Sector • Foreign sector It is a open economy 39
IMPORTANCE OF CIRCULAR FLOW IN ECONOMIC ACTIVITIES 1.
2.
IT HIGHLIGHTS THE STATE OF INTERDEPENDENCE IN THE ECONOMY All sectors household sector, business sector, government sector & foreign sector are mutually interdependent & complementary to one another in such a way that economic activity remains intact
IT SHOWS THE STATE OF EQUILIBRIUM IN THE ECONOMY The circular flow demonstrate the position of equilibrium in the economy. If there is any distortion in the circular flow , the economy will be in disequilibrium & the smooth functioning of the economy will be disturbed
40
3. ESTIMATION OF NATIONAL INCOME. National product or national expenditure can be estimated with the help of circular flow. GDP can be estimated by aggregating the market value of the final goods & services produced in a year. By adding to it net factor income from abroad one can estimate gross national product. By adding all the factors of production in the household sector one can calculate national income
41
4. TRIPLE IDENTITY OF PRODUCTION , INCOME & EXPENDITURE The aggregate of the mkt. value of final goods & services produced in a year by the producing sector is known as national product. This product is received by the households in the form of factor income. This income is spent by he household sector on consumption goods, by the producing sector on investment, by the govt. Sector on goods & factor services , in the rest of the world sector , the flow of income is reflected in terms of net exports of goods & factors as well as nonfactor services. Its aggregates constitutes national expenditure. In this way, circular flow establishes that PRODUCTION = INCOME = EXPENDITURE 42
What is Demand? • Quantity demanded of a product or service is the number that would be bought by the public at a given price
1
The Law of Demand • When a good’s price is lower, consumers will buy more of it • When a good’s price is higher, consumers will buy less of it
2
The Law of Demand • The Law of Demand is affected by two behavior patterns –The Substitution Effect –The Income Effect
3
The Substitution Effect • As the price for one good rises compared to a similar good, consumers will substitute the similar good for their purchases.
4
The Income Effect • As prices go up, your money becomes worth less than it was worth before • People are less likely to buy the good now
5
Demand Schedule • A demand schedule shows the likely number of purchases based on a series of arbitrarily chosen prices
6
Demand Schedule
7
Demand Schedule › Demand Curve
8
A Change (Shift) in Demand • If one of 5 other factors changes, the entire demand curve will shift to the left or right • The curve does NOT shift if the price of the good is the only change
9
A Change in Demand (Graph)
10
A Change in Quantity Demanded (Graph)
11
Income • When people’s income changes, demand shifts accordingly –Normal Goods – • Higher income = higher demand • Lower income = lower demand
12
Income • When people’s income changes, demand shifts accordingly –Inferior Goods – • Higher income = lower demand • Lower income = higher demand
13
Consumer Expectations • If consumers expect a price to rise in the future, current demand increases • If consumers expect a price to fall in the future, current demand decreases 14
Population • When one sector of the population grows, demand increases for products that sector uses • Fastest growing sector of the population today?
15
Consumer Tastes and Advertising
• Increased advertising can increase consumer demand • Bad news about a product can decrease demand
16
Price of Related Goods • Complimentary Goods – goods that are bought and used together –Higher Complementary Price = decrease in demand –Lower Complementary Price = increase in demand
17
Price of Related Goods • Substitute Goods – goods that are used in place of one another –Higher Substitute Price = increase in demand –Lower Substitute Price = decrease in demand
18
Elasticity of Demand
• Elasticity refers to how responsive the quantity demanded is to a change in prices 19
Elasticity of Demand
• An inelastic good will still sell about the same quantity even if the price goes up or down
20
Elasticity of Demand • An elastic good will have a higher change in Qd when there is a price change
21
Calculating Elasticity • Elasticity = % change in quantity demanded __________________________ % change in price
22
Calculating Elasticity • If Elasticity is < 1, the good is inelastic • If Elasticity is > 1, the good is elastic • If Elasticity = 1, the good has a unitary elastic demand 23
Factors Affecting Elasticity • • •
•
Availability of Substitutes – if you have no other options, demand is inelastic. Availability of Substitutes – if you have equally appealing options, demand is highly elastic Relative Importance – what percentage of your budget is spent on the good? – If it is low, price changes will not alter demand – If it is high, even small price changes can greatly affect demand Necessities vs. Luxuries – consumption of milk might stay the same with price changes, while consumption of lobster would greatly change with price changes Change over time – price changes may produce inelastic demand in the short term, but elastic demand long term – 1970s fuel crisis – people still bought the same amount of gas at first, but eventually started buying smaller cars 24
Elasticity and Revenue • Total Revenue – the amount of money a company receives by selling its good or service • With elastic demand, revenue will decrease greatly with price increases 25
Elasticity and Revenue • With Inelastic demand, price increases will increase revenue
26
Determinants of Demand
Number of buyers Income Tastes
Prices of related goods
Expectations
27
Number Of Buyers Consumer Income
Price of Related Goods
Demand Tastes And Preferences
Expectations
Demographics
28
Supply • The quantity supplied is the number of units that sellers want to sell over a specified period of time at a particular price.
29
The Law of Supply • There is a direct relationship between price and quantity supplied. – Quantity supplied rises as price rises, other things constant. – Quantity supplied falls as price falls, other things constant.
30
Law of Supply • Law of Supply – As the price of a product rises, producers will be willing to supply more. – The height of the supply curve at any quantity shows the minimum price necessary to induce producers to supply that next unit to market. – The height of the supply curve at any quantity also shows the opportunity cost of producing the next unit of the good.
31
The Law of Supply • The law of supply is accounted for by two factors: – When prices rise, firms substitute production of one good for another. – Assuming firms’ costs are constant, a higher price means higher profits.
32
The Supply Curve • The supply curve is the graphic representation of the law of supply. • The supply curve slopes upward to the right. • The slope tells us that the quantity supplied varies directly – in the same direction – with the price.
33
Price (per unit)
A Sample Supply Curve S
PA
0
A
QA Quantity supplied (per unit of time) 34
Supply Curve DVDs
35
Shifts in Supply Versus Movements Along a Supply Curve
• Supply refers to a schedule of quantities a seller is willing to sell per unit of time at various prices, other things constant.
36
Shifts in Supply Versus Movements Along a Supply Curve • Quantity supplied refers to a specific amount that will be supplied at a specific price.
37
Shifts in Supply Versus Movements Along a Supply Curve • Changes in price causes changes in quantity supplied represented by a movement along a supply curve.
38
Shifts in Supply Versus Movements Along a Supply Curve
• A movement along a supply curve – the graphic representation of the effect of a change in price on the quantity supplied.
39
Shifts in Supply Versus Movements Along a Supply Curve
• If the amount supplied is affected by anything other than a change in price, there will be a shift in supply.
40
Shifts in Supply Versus Movements Along a Supply Curve • Shift in supply – the graphic representation of the effect of a change in a factor other than price on supply.
41
Change in Quantity Supplied
Price (per unit)
S0
B
$15
A
Change in quantity supplied (a movement along the curve)
1,250 1,500 Quantity supplied (per unit of time) 42
Shift in Supply S0
Price (per unit)
S1
$15
A
B Shift in Supply (a shift of the curve)
1,250 1,500 Quantity supplied (per unit of time) 43
Shift Factors of Supply • Other factors besides price affect how much will be supplied: – Prices of inputs used in the production of a good. – Technology. – Suppliers’ expectations. – Taxes and subsidies.
44
Factors that Shift Supply Resource Prices
Prices of Related Goods and Services
Supply
Number Of Producers
Technology And Productivity
Expectations Of Producers 45
Price of Inputs (Resource Prices) • When costs go up, profits go down, so that the incentive to supply also goes down.
46
Technology • Advances in technology reduce the number of inputs needed to produce a given supply of goods. • Costs go down, profits go up, leading to increased supply.
47
Expectations • If suppliers expect prices to rise in the future, they may store today's supply to reap higher profits later.
48
Number of Suppliers • As more people decide to supply a good the market supply increases (Rightward Shift).
49
Individual and Market Supply Curves • The market supply curve is derived by horizontally adding the individual supply curves of each supplier.
50
From Individual Supplies to a Market Supply (1) (2) (3) (4) (5) Quantities Price Ann's Barry's Charlie's Market Supplied (per DVD) Supply Supply Supply Supply A B C D E F G H I
$0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00
0 1 2 3 4 5 6 7 8
0 0 1 2 3 4 5 5 5
0 0 0 0 0 0 0 2 2
0 1 3 5 7 9 11 14 15 51
From Individual Supplies to a Market Supply $4.00
Charlie
Barry
Ann
Market Supply
Price per DVD
3.50
H
3.00
G
2.50
F
2.00
E
1.50
D
1.00
0.50 0 A
I
C B
CA
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Quantity of DVDs supplied (per week) 52
Aggregation of Supply (I)
53
Aggregation of Supply (II)
54
Price of Related Goods or Services • The opportunity cost of producing and selling any good is the forgone opportunity to produce another good. • If the price of alternate good changes then the opportunity cost of producing changes too! • Example Mc Don selling Hamburgers vs. Salads. 55
Taxes and Subsidies • When taxes go up, costs go up, and profits go down, leading suppliers to reduce output. • When government subsidies go up, costs go down, and profits go up, leading suppliers to increase output.
56
Decrease in Supply
57
Increase in Supply
58
Change in Supply vs. a Change in the Quantity Supplied
59
Types of Elasticity
• There are three types of Elasticity:
A. Price Elasticity of Demand
B. Income Elasticity of Demand C. Cross Elasticity of Demand
60
Types of Elasticity (Cont…)
A. Price Elasticity of Demand A measure of the relationship between a change in the quantity demanded of a particular good and a change in its price. The formula for calculating price elasticity of demand is: Price Elasticity of Demand = % Change in Quantity Demanded / % Change in Price If the price elasticity of demand is equal to 0, demand is perfectly inelastic (i.e., demand does not change when price changes). 61
Types of Elasticity (Cont…) Values between zero and one indicate that demand is
inelastic (this occurs when the percent change in demand is less than the percent change in price). When price elasticity of demand equals one, demand is unit elastic (the percent change in demand is equal to the percent change in price). Finally, if the value is greater than one, demand is perfectly elastic (demand is affected to a greater degree by changes in price). For example, if the quantity demanded for a good increases 15% in response to a 10% decrease in price, the price elasticity of demand would be 15% / 10% = 1.5. 62
Types of Elasticity (Cont…)
B. Income Elasticity of Demand A measure of the relationship between a change in the quantity demanded for a particular good and a change in real income. The formula for calculating income elasticity of demand is: Income Elasticity of Demand = % change in quantity demanded / % change in income If the price elasticity of demand is equal to 0, demand is perfectly inelastic (i.e., demand does not change when price changes). 63
Types of Elasticity (Cont…)
• For example, if the quantity demanded for a good increases for 15% in response to a 10%increase in income, the income elasticity of demand would be 15% / 10% = 1.5. • The degree to which the quantity demanded for a good changes in response to a change in income depends on whether the good is a necessity or a luxury.
64
Types of Elasticity (Cont…)
C. Cross Elasticity of Demand An economic concept that measures the responsiveness in the quantity demand of one good when a change in price takes place in another good. The measure is calculated by taking the percentage change in the quantity demanded of one good, divided by the percentage change in price of the substitute good: The formula for calculating income elasticity of demand is: Cross Elasticity of Demand = % change in quantity demanded of good-1 / % change in the price of good-2
65
Types of Elasticity (Cont…)
• The cross elasticity of demand for substitute goods will always be positive, because the demand for one good will increase if the price for the other good increases. • For example, if the price of coffee increases (but everything else stays the same), the quantity demanded for tea (a substitute beverage) will increase as consumers switch to an alternative. 66
•
Introduction to Engineering Economy Engineer Means: Problem solver, Find new ways doing things economically, Engineering is the art of doing that well with one dollar which any bungler can do with two (Arthur M. Wellington, 1887).
The challenge: Every problems has multiple solutions, How to choose the best one?
As engineers not only model the stress on a column, they must also model the economic impact of their recommendations.
Introduction to Engineering Economy • What is engineering economy? Engineering project must be not only physically realizable but also economically affordable. Engineering economy involves the systematic evaluation of the economic merits of proposed solutions to engineering problems.
Engineering is more than a problem solving activity
The Principles of Engineering Economy Develop the alternatives Identify, Define, Creativity, Innovation
Focus on the differences Differences in the future outcomes
Use a consistent viewpoint The prospective outcomes of the alternatives (stakeholders)
Use a common unit of measure Monetary ( usually USD) presentation of the alternatives and their outcomes
The Principles of Engineering Economy Consider all relevant criteria
Decision making based on several criteria (organisational objectives, e.g. long term interest)
Make risk and uncertainty explicit Identify, define, allocate, and mitigate
Revisit your decisions The initial projected outcomes of the selected alternatives should be subsequently compared with actual results achieved.
Engineering Economy and the design process Engineering economic analysis procedure Steps: 1. Problem recognition, definition, and evaluation. 2. Development of the feasible alternatives. 3. Development of the outcomes and cash flows for each alternative. 4. Selection of a criterion 5. Analysis and comparison of the alternatives. 6. Selection of the preferred alternatives. 7. Performance monitoring and post monitoring results.
Engineering Design Process Activity: 1. 2.
3. 4. 5. 6.
Problem/need definition. Problem/need formulation and evaluation. Synthesis of possible solutions. Analysis, optimisation, and evaluation. Specification of preferred alternatives. Communication.
Engineering Economy and the design process Engineering economic analysis procedure Problem recognition, definition, and evaluation. Problem must be well understood and stated in an explicit form before the project team proceeds with the rest of the analysis.
Development of the feasible alternatives. Searching for potential alternatives (creativity and resourcefulness), screening them to select a smaller group of feasible alternatives for detailed analysis.
Development of the outcomes and cash flows for each alternative. Cash flow approach (revenue and payments), nonmonetary factors e.g. meeting or exceeding customer expectations, safety to employees, employees satisfaction, etc.
Engineering Economy and the design process Engineering economic analysis procedure Selection of a criterion. Long-term interest of the client and the organisation, environmental concerns, etc. Analysis and comparison of the alternatives. Based on cash flows , exchange rate, inflation, regulatory, etc.
Selection of the preferred alternatives. A result of the total effort of the above mentioned 5 steps. It is the technical-economic modelling. Performance monitoring and post monitoring results. Accomplished during and after the time that the result achieved. the aim of post evaluation is to learn how to do better the job.
Using spreadsheets in Engineering Economy Spreadsheets are useful tool for solving engineering economy problems. Because of the following reasons: They consist of structured, repetitive calculations that can be expressed as formulas that rely on a few functional relationships. The parameters of the problems subjected to change. The result must be documented. Graphical output is often required, as well as control over the format of the graphs.
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
SCOPE OF ENGINEERING ECONOMICS • 1. Engineering economics plays a very major role in all engineering decisions. • 2. It is concerned with the monetary consequences, financial analysis of the projects, products and processes that engineers design. • 3. Engineering economics helps an engineer to assess and compare the overall cost of available alternatives for engineering projects. • 4. According to the analysis an engineer can take decision from the alternative which is more economic. • 5. Engineering economics concepts are used in the fields for improving productivity, reducing human efforts, controlling and reducing cost. 1
• 6. Engineering economics helps to understand the market conditions general economic environment in which the firm is working. • 7. It helps in allocating the resources. • 8. Engineering economics helps to deal with the identification of economic choices, and is concerned with the decision making of engineering problems of economic nature.
2
Introduction • The word Economics is derived from the Greek word “OKIOS NEMEIN” meaning household management • Man is a bundle of desires. Goods and services satisfy these wants. But almost all the goods are scares • To produce goods factors of production are needed and these are all scarce
3
The Study of Economics • Economics is the study of how individuals and societies choose to use the scarce resources that nature and previous generations have provided. • It is the study of economic problems. Wants are motive for economic activity. Wants leads to efforts and which lead to satisfaction
4
Why Study Economics? • To learn a way of thinking • Three fundamental concepts: – Opportunity cost – Marginalism, and – Efficient markets
5
Contd… Opportunity Cost • the best alternative that we forgo, or give up, when we make a choice or a decision • arises because time and resources are scarce.
Marginalism • In weighing the costs and benefits of a decision, it is important to weigh only the costs and benefits that arise from the decision
Efficient Market • is one in which profit opportunities are eliminated almost instantaneously • Profit opportunities are rare because, at any one time, there are many people searching for them
6
Economic Definitions • • • •
Adam Smith gave the Wealth Definition Alfred Marshall gave the Welfare Definition Lionel Ribbons gave the Scarcity Definition Paul Samuelson gave the Growth Definition
7
Wealth Definition • Adam Smith (Father of Economics) in his book “Wealth of nations 1776” defined economics is the study of wealth • J.B. Jay, J.S. Mill, Walker, B Price all agreed with Adam Smith • In this definition wealth is given the first place and man is given the second place
8
Meaning of wealth • Around the industrial revolution, merchants were the most powerful class in Western Europe, and wealth for them meant money only. Since money at that time was in the shape of gold, merchants declared gold as the only wealth, • This definition rendered merchants as the only productive class, as they created it by trade, • This definition harmed the interests of newly emerging class of petty industrialists and their hard working workers, • Adam Smith as spokesman of the emerging class widened the definition to include all material goods, • Activities which did not result in material goods production were unproductive.
9
1
Concept of Efficiency • In general business decision making, and policies three different types of efficiency concepts are encountered. – Engineering efficiency – Technical Efficiency
– Economic efficiency.
4/22/2017
2
Engineering Efficiency: • physical amount of some single key input used in production and is measured by the ratio of that input to output. Engineering efficiency (%) = (Output/Key Input) x 100
• Engineering efficiency does not take financial considerations. It is purely about physical relationships. Ex. engineering efficiency of an engine: Let a steam engine is 40 percent efficient means that 40 percent of the energy in the fuel is converted into work done, while the other 60 percent is lost in friction, heat loss, and other unavoidable sources of waste
4/22/2017
NHU 501 Dr N R Kidwai, JIT Barabanki
3
Technical Efficiency: • is related to physical amount of all resources used in producing a product. Technical efficiency (%) = (Output produced / All Input resources ) x 100
• Thus technical efficiency is about getting the most output from any given set of inputs; or, equivalently, about producing a given level of output using the least amount of physical inputs. Ex. Technical efficiency of a diesel engine Technical efficiency (%) = (Heat equivalent of mechanical energy produced/ Heat equivalent of fuel used ) x 100 EX., Let a firm is using 100 units of labour and 50 units of capital to produce a level of output. If the firm could maintain its output level by using only 90 units of labour without using more capital, then it is being technically inefficient in current methods as it is “wasting” 10 labour units
4/22/2017
4
Economic Efficiency • is related to the value or cost (rather than the physical amounts) of all inputs used in producing a given output. Economic efficiency (%) = (Output/ Input) x 100 = (Worth/ Cost) × 100
• The production of a given output is economically efficient if there are no other ways of producing the output that use a smaller total value of inputs. •
Ex. a firm have three alternative production methods. First require a lot of labour but only a little capital, Second requires a lot of capital and only a little labour, while third production method may require a lot of land but relatively little of both labour and capital. In order to be economically efficient (maximize its profits) the firm should choose the production method that costs the least.
• Economic efficiency is also called ‘productivity’
4/22/2017
5
Relationship among various efficiencies • Engineering efficiency aims to maximizes the output with respect to key input (Engineering efficiency may increase the cost i.e reduce the economic efficiency)
• Technical Efficiency aims to maximize the output with least quantities of resources (A high end technological process may be technically efficient but may significantly increase fixed cost and cost of skilled labor and may not be economic efficient)
• Economic efficiency is maximizing the profit in shortest time
(bottom-line is the only yardstick) (In the context a product or service need only be complete enough to satisfy the customer, especially if it makes it possible to sell the same product repeatedly, or sell a newer, more complete version of the product later. A business that sells a product which completely satisfies its customers indefinitely will put itself out of business; Thus, a business may have a strong disincentive to produce design which are efficient in the engineering sense.)
4/22/2017
6
Relationship among different efficiencies • Technical efficiency is desirable as long as long as inputs have a positive cost to the firm for being economically efficient. Thus achieving technical efficiency is clearly a necessary condition for producing any output at the least cost. But achieving technical efficiency, however, is not a sufficient condition for producing at the lowest possible cost. Ex., the engineering efficiency of a gas turbine engine can be increased by using more and stronger steel in its construction. Raising the engineering efficiency of an engine saves on fuel, but at the cost of using more of other inputs. To know whether this is worth doing, the firm must compare the value of the fuel saved with the value of the other inputs used. A Automobile which is strongly technically efficient (a longer life) may affect the repeat sale/sale of newer versions and may not be economically efficient to the company.
4/22/2017
7
Ways to increase Economics Efficiency (Productivity) Increased output for the same input Decreased input for the same output By a proportionate increase in the output which is more than the proportionate increase in the input By a proportionate decrease in the input which is more than the proportionate decrease in the output Through simultaneous increase in the output with decrease in the input.
4/22/2017
8
Definition
• Economic Efficiency: When goods are produced in the least costly manner and distributed to those who value them most. • Requires: – Productive Efficiency – Allocative Efficiency
Productive Efficiency
• There is no way to re-direct production among firms to increase total output.
Perfect Comp and Productive Efficiency • In LR firms produce at lowest possible LRAC. – There is no way to cut costs by changing plant size. • Since all firms take the same price, all firms have same MC (why?) – There is no way to re-direct production to other firms and get lower marginal costs. • Productive efficiency holds.
Allocative Efficiency
• Goods are consumed by those who most value them. • There is no alternative comb. of goods that could be produced that would increase society’s well-being.
Measuring Allocative Efficiency • The sum of consumers’ surplus and producers’ surplus.
What is the Efficiency? Efficiency also can be stated as the act of being adequate in performance with a minimum of waste, effort, time. 14
Time material
,
are
money
,
raw
limited.
So it makes sense to try to conserve them. 15
Efficiency can only be delivered by
improving productivity procurement processes
working practices 16
EFFICIENCY TYPES
17
Economic Efficiency Economic efficiency refers to the use of resources so as to maximize the production of goods and services.
18
A situation can be called economically efficient if : • No one can be made better off without making someone else worse off. • No additional output can be obtained without increasing the amount of inputs. • Production proceeds at the lowest possible per-unit cost. 19
Productive Efficiency • Productive efficiency, also known as technical efficiency, occurs when the economy is utilizing all of its resources efficiently, producing most output from least input.
20
Production Possibility Frontier ( PPF ) An example PPF: • Points B, C and D are all productively efficient, but an economy at A would not be.
21
Financial Market Efficiency Financial market efficiency refers to in financial markets is the efficiency of allocating resources. This includes producing the right goods for the right people at the right price.
22
There are three levels of market efficiency.
There are four market efficiency types.
These are:
These are:
• Weak-form efficiency
• Information arbitrage efficiency • Fundamental valuation efficiency • Full insurance efficiency • Functional/Operational efficiency
• Semi-strong efficiency • Strong-form efficiency
23
Allocative efficiency Allocative efficiency is a measure of the benefit or utility derived from a proposed or actual choice in the distribution or apportionment of resources.
24
Distributive efficiency Distributive efficiency occurs when goods and services are received by those who have the greatest need for them.
25
Dynamic efficiency Dynamic efficiency refers to an economy that appropriately balances short run concerns with concerns in the long run. Dynamic efficiency also refers to the ability to adapt to changed economic conditions at low cost quickly. 26
Technical efficiency Technical efficiency is the effectiveness with which a given set of inputs is used to produce an output. • When technical-efficiency is not being achieved due to a lack of competetive pressure, X-inefficiency occurs. 27
CALCULATING EFFICIENCY The efficiency of a process is calculated by dividing the output by the input, and then multiplying the result by 100.
28
Efficiency is usually given as a percentage and can be computed with the following formula:
• Efficiency = Work Output / Work Input
(1)
• Efficiency rating x 100=Percentage efficiency rating (2) 29
WAYS TO IMPROVE EFFICIENCY IN BUSINESSES • Train the workforce • Improve motivation • More capital equipment • Use better quality raw materials 30
Train the workforce Training the workforce increases improvements in the workers productivity levels. Training should enable workers to work more quickly and more accurately and to produce better quality products. 31
Improve motivation A better-motivated workforce will work harder and take pride in their work.
There are many different financial (e.g. bonuses) and non-financial ways (e.g. empowerment) for businesses to motivate their workers. 32
More capital equipment Investment into new, higher technological machinery can have number of advantages: • Longer hours can be worked. • Machine can perform repetitive and complicated tasks more quickly. • Accuracy incerases and therefore wastage can be less. 33
Use better quality raw materials This can reduce the amount of time wasted on rejected or defective products.
A business should ensure they find the supplier who can supply the best quality resources, but at a competitive price and also with reliable delivery. 34
What is the relationship between Efficiency and Effectiveness? Efficiency and effectiveness were originally industrial engineering concepts that came of age in the early twentieth century. • Effectiveness is ‘doing the right thing’ . • Efficiency is ‘doing the thing right’. 35
• Efficiency alone will put your company on the fast track to bankruptcy. • Effectiveness (efficiacy) alone *may* allow your company to survive. However the company will not reach its maximum potantial if it is inefficient. Effectiveness and Efficiency together will almost guarantee success! 36
1
Elements of Costs? In order to interpret the term cost correctly and to ascertain the cost with respect to the cost centers, the cost attached with the manufacturing process may be subdivided, known as Elements of Costs. 2
3
Material costs – The cost of commodities supplied to an undertaking.
Labor costs – The cost of remuneration of laborers. Overheads – These are the other expenses incurred. Examples: Direct materials - Raw materials used in manufacturing a product. Indirect materials - Lubricants and cotton waste used in maintaining machinery. Direct Labor - Wages of those workers who are engaged in production. Indirect Labor - Wages to those who are aiding manufacturing activities by way of supervision, maintenance, tools setting etc. Direct Overheads - The cost of special pattern, dyes drawings tools etc. made for specific product. Indirect Overheads- Office salaries, rent, electricity, advertisement expenses etc. 4
MATERIAL COST
5
DIRECT MATERIAL COST IT IS A COST OF MATERIALS WHICH ENTER INTO & FORM PART OF PRODUCT in furniture – making - clay in brick making
e.g. - timber
6
Direct Materials Raw materials that become an integral part of the product and that can be conveniently traced directly to it.
Example: A radio installed in an automobile
7
Direct materials Direct materials are the raw materials that become part of the product.
The cost of materials – the cost of materials used entering into and becoming the elements of a product or service E.g. fabrics in garments 8
INDIRECT MATERIAL COST IT IS THE COST OF MATERIAL WHICH DO NOT FORM THE PART OF PRODUCT BUT WHICH HELP THE PRODUCTION e.g - LUBRICATING OIL , FUEL etc. Small items like thread, gum , nails etc
Though forms part of product but difficult to calculate cost per unit of that material ----so consider as direct material 9
Indirect materials • Such as stationery, consumable supplies, spare parts for machine that assist to the production of final products
10
LABOUR COST
11
Direct Labor The definition of direct labor is pretty easy. Direct labor represents the people who do the core work of the business. For example, if the business is a construction company, direct labor would be the people actually constructing the building. They would be the people with hammers and saws in their hands. In a retail store, direct labor would be the people helping on the sales floor doing the basic work that takes place serving the customers. In a grocery bag factory, direct labor are the people running the machines actually making the bags. I think of direct labor as the people who make or build the product.
12
Labour cost The labour cost is the cost of remuneration of the e plo ees of a u dertaki g. su h as Wages, salaries, commissions, bonus, etc. There are two types of labour cost are 1. Direct Labour cost 2. Indirect Labour cost 13
Direct Labor Those labor costs that can be easily traced to individual units of product.
Example: Wages paid to automobile assembly workers
14
Direct labour • The cost of remuneration for working time • E.g. asse l orkers’ ages i to assembly
15
Indirect labour • Such as salaries of factory supervision and office staff that do not directly involve in production of the final product • If a person's labor cannot be easily linked to a specific customer's order or the person does not directly build the product, the person's labor is called "indirect labor." 16
EXPENSE COST
17
Expenses cost The ost of ser i es pro ided to a undertaking and the notional cost of the use of o ed assets. Expenses are two types 1. Direct Expenses 2. Indirect Expenses
18
Direct expenses • Other costs which are incurred for a specific product or service • E.g. royalties
19
Indirect expenses • Such as rent, rates, depreciation, maintenance expenses that do not have instant relationships with the manufacturing processes
20
OVERHEADS
21
The definition for overhead is easy. Here it is......
If a cost is not direct labor or direct materials, the cost is overhead.
22
Overheads The aggregate of I dire t aterial ost, I dire t ages a d I dire t e pe ses
Three types such as 1. Indirect material cost 2. Indirect labour cost 3. Indirect expenses 23
24
Manufacturing Overhead Manufacturing costs that cannot be easily traced directly to specific units produced. Examples: Indirect materials and indirect labor
Materials used to support the production process.
Wages paid to employees who are not directly involved in production work.
Examples: lubricants and cleaning supplies used in the automobile assembly plant.
Examples: maintenance workers, janitors, and security guards.
25
Indirect cost (overhead) • Cost that cannot be identified specifically with or traced to a given cost object • They are identified with cost centres as overheads – Indirect materials – Indirect labour – Indirect expenses
26
Indirect expenses overheads 1. Factory overheads 2. Office & administration overheads 3. Selling & distribution overheads
27
Selling and distribution overhead
• Selling and distribution overhead is also known as marketing or selling overhead. Distribution expenses usually begin when the factory costs end. Such expenses are generally incurred when the product is in saleable condition. It covers the cost of making sales and delivering/dispatching products. These costs include advertising, salesmen salaries and commissions, packing, storage, transportation, and sales administrative costs. • Administrative overhead includes costs of planning and controlling the general policies and operations of business enterprises. Usually, all costs which cannot be
28
Direct Material Cost + Direct Labour Cost + Direct Expenses Cost
Overheads
Prime Cost
Indirect Material Cost + Indirect Labour Cost + Indirect Expenses Cost 29
Rights reserved by Prof. Bhagyashree Kulkarni
Fixed
Behaviour
in short run & long run
Variable Varies with volume and constant per unit
Semi-variable A cost could be variable for one level of activity whereas it could be fixed for another.
Not inherently fixed or variable Many costs are semi-variable in nature
30
According to behaviour or variability
Semi-variable cost
Fixed cost
Variable cost
31
According to behaviour or variability
• Fixed cost----------costs which remain same at every level of output.these cost are related to time and not to production. • Variable cost-------costs which increase in volume with increase in volume of production.the per unit cost remains constant at every level of output. • Semi – variable cost—these are fixed as well as variable. Upto a particular level these costs are fixed but once the level is crossed even slightly these costs shift to the next higher level and so on.
32
Cost center & profit center • Cost center-a smallest segment of activity or area or responsibility for which costs are accumulated. • Profit center-a center whose performance is measured both in terms of expenses incurred and revenues earned by it.
33
Marginal Costing
1
Marginal Cost
“Marginal cost is amount at any given volume of out put by which aggregate costs are changed….. if volume of output is increased or decreased by one unit”
2
Marginal Costing • The term cost can be viewed from two angles basically. – Direct Cost and Indirect Cost – Fixed Cost and Variable Cost
• If fixed cost is included in the total cost, the per-unit cost varies from one cost period to another with the fluctuations in level of activities in two cost periods. • Thus, per unit cost becomes incomparable between two periods. • To avoid this, it will be necessary to eliminate the fixed costs from the determination of total cost. • This has resulted into concept of Marginal Costing
3
Marginal Cost “Marginal cost is amount at any given
volume of out put by which aggregate costs are changed if volume of output is increased or decreased by one unit”
1 Marginal Cost 100 x150= 15000 Fixed Cost = 5000 total 20000
2 1 Manufacture 100 radio Variable costs Rs150 p u Fixed cost Rs 5000 2 If Manufacture 101 radios
Marginal cost 150 x101=15150 Fixed Cost = 5000 TOTAL 20150
additional Cost=Rs 150 4
Marginal Costing
“marginal
costing is ascertainment of marginal cost by differentiating between fixed and variable costs and of the effect of changes in volume or type of output”
5
Marginal Costing
What Could be effects of Changes
In volume or Type of output
1 lakh units To 2 lakh units
6
Marginal Costing
What Could be effects of Changes
In volume or Type of output
From One Model of Car to Another
From One Size of product to another
7
Basics of marginal costing • Marginal cost – cost of producing an additional unit or output or service • Marginal costing differentiates the fixed and variable costs
8
Features Of Marginal Costing • Semi-variable costs are included in comparison of cost • Only variable costs are considered • Fixed costs are written off • Prices are based on variable and marginal contribution
9
Marginal Cost • Marginal cost is defined as the amount at any given volume of output by which aggregate costs are changed if the volume of output is increased or decreased by one unit.
10
Marginal Costing ---Characteristics
Fixed & Variable Costs
Inventory Valuation
MC Costs as Products Costs
Contribution
Fixed Costs as Period Costs
Pricing
Marginal Costing & Profit
11
Marginal Costing ---Characteristics
Segregation Fixed & Variable Costs
Semi-variable costs are segregated into fixed & variable
12
Marginal Costing ---Characteristics
Marginal Costs as Products Costs
Only Variable costs are charged to products
13
Marginal Costing ---Characteristics
Fixed Costs as Period Costs
Fixed costs treated Period costs Charged to costing P & L Account
14
Marginal Costing ---Characteristics
Inventory Valuation
WIP & F goods are Valued at Marginal Cost
15
Marginal Costing ---Characteristics
S-V=C Contribution
Profitability judged on Contribution made
16
Marginal Costing ---Characteristics
Pricing
Pricing is based on Contribution & Marginal Costs
17
Marginal Costing ---Characteristics
A Sales Less VC Contribution
B -
-
C -
-
-
Total ----------
Fixed Cost
----
Profit
-----
Marginal Costing & Profit
18
Basic equation of Marginal Costing • • • •
Profit = Sales – Total cost Profit = Sales – (Variable cost + Fixed cost) Profit + Fixed cost = Sales – Variable cost Sales – Variable cost = Contribution = Fixed cost + Profit • Contribution – Fixed cost = Profit
19
Value Of Marginal Costing To Management • It integrates with other aspects of management accounting. • Management can easily assign the costs to products. • It emphasizes the significance of key factors. • The impact of fixed costs on profits is emphasized. • The profit for a period is not affected by changes in absorption of fixed expenses. • There is a close relationship between variable costs and controllable costs classification. • It assists in the provision of relevant costs for decisionmaking. 20
Limitations Of Marginal Costing
• To segregate the total cost into fixed and variable components is a difficult task • Under marginal costing, the fixed costs are eliminated for the valuation of inventory , in spite of the fact that they might have been actually incurred. • In the age of increased automation and technological development, the component of fixed costs in the overall cost structure may be sizeable. • Marginal costing technique does not provide any standard for the evaluation of performance. • Fixation of selling price on marginal cost basis may be useful for short term only. • Marginal costing can be used for assessment of profitability only in the short run.
21
Absorption Costing
“Absorption cost is a total cost technique Under which total cost ie fixed & variable is charged to production. Inventory is also valued at total cost.
22
Absorption-Marginal Costing--differences
Fixed & Variable Costs
Valuation Of stock
Measurement Of Profitability
23
Absorption-Marginal Costing--differences
Fixed & Variable Costs
Marginal Costing
Absorption Costing
Only variable cost
Both F & V Costs Are charged
FC charged to P/L
24
Absorption-Marginal Costing--differences
Valuation Of stock
WIP & FS at Marginal Cost
Total Cost
25
Marginal Revenue • Marginal revenue ( MR ) is the change in total revenue resulting from selling an extra unit of goods. • MR = TR/Q, where TR = change in TR due to change in Q, Q = change in Q
1
To find T R from the M R curve • For a certain known quantity transacted, the area under the MR and above the horizontal axis is the T R . (I.e. the sum of the Marginal Revenues of all units of goods.) • The slope of the TR curve is MR. Why? • And, MR is always smaller Price for single pricing arrangement (I.e. MR < P) Why? (Hint MR MUV , and also downward-sloping 10
• • • • •
•
Revision on different pricing arrangement Single Pricing Arrangement with Consumer Surplus = TUV - TEV MUV = DD = AR = P [>MR] {MUV} 11
REVISION Single Pricing Arrangement
All-or-nothing Pricing Arrangement
DEMAND AUV
P
AUV
= AR
MUV
= MR
P DEMAND
MUV
MR
Q TUV= TEV + CS
= AR
Q TUV= TEV
12
A Price -Searcher = Price-Searcher’s Market
Price
MR cuts the midpoint of the perpendicular line drawn from the AR to the vertical axis
AR MR
Quantity 13
Marginal Revenue Monopolist sells one more unit. Price decreases from P(Q) to P(Q+1). Slope of demand is calculated between the two red dots on the demand curve.
P(Q) P(Q+1)
The slope of demand Is [P(Q) – P(Q+1) ]/ 1 demand Q
Q+1
Price falls by the slope of demand.
14
Marginal Revenue By adding 1 unit a monopolist gains The area is 1 wide by P high = P P(Q)
The monopolist looses This area is the decrease in price, which is the slope of demand times Q.
P(Q+1)
Q
Q+1
So MR is the sum of the two areas MR= P + Q (slope demand) 15
MR with linear Demand • • • •
MR(Q) = Q (slope of P(Q)) + P(Q) P(Q) = a - b Q; slope = -b MR(Q) = - Q b + (a -bQ) = a- 2bQ MR has twice as negative a slope as demand and the same intercept. – special property of linear demand
16
MR = MC • MR is amount revenue goes up for a unit more output • MC is additional cost • So if MR > MC make more • MR < MC make less • MR = MC determines Q; output • P(Q) demand, determines price 17
SUNK COSTS
Sunk costs are all costs incurred or committed in the past that cannot be changed by any decision made now or in the future. Sunk costs should not be considered in decisions. Dr. Varadraj Bapat, IIT Mumbai
1
SUNK COSTS
E.g. cost incurred on research of a product will be irrelevant while making decision whether to undertake production or not.
Dr. Varadraj Bapat, IIT Mumbai
2
Sunk Cost Sunk costs have been incurred and cannot be reversed. Historical costs are sunk costs. They play no role in decision making in the current period.
Dr. Varadraj Bapat, IIT Mumbai
3
Sunk Cost do not affect future costs and cannot be changed by any current or future action, hence these costs are irrelevant in decision making.
Ex. Spending on advertising during product launching is sunk for taking a decision on continuance of product Dr. Varadraj Bapat, IIT Mumbai
4
SUNK COST • In economics and business decision-making, a sunk cost is a cost that has already been incurred and cannot be recovered. Sunk costs are sometimes contrasted with prospective costs, which are future costs that may be incurred or changed if an action is taken.
• Both retrospective and prospective costs may be either fixed (continuous for as long as the business is in operation and unaffected by output volume) or variable (dependent on volume) costs. However, many economists consider it a mistake to classify sunk costs as "fixed" or "variable." For example, if a firm sinks $1 million on an enterprise software installation, that cost is "sunk" because it was a one-time expense and cannot be recovered once spent.
• The sunk cost is distinct from economic loss. For example, when a new car is purchased, it can subsequently be resold; however, it will probably not be resold for the original purchase price.
• Economists argue that sunk costs are not taken into account when making rational decisions. In the case of a Yankees ticket that has already been purchased, the ticket-buyer can choose between the following two end results if he realizes that he doesn't like the game: • Having paid the price of the ticket and having suffered watching a game that he does not want to see, or; • Having paid the price of the ticket and having used the time to do something more fun. • In either case, the ticket-buyer has paid the price of the ticket so that part of the decision no longer affects the future.
Three sources of sunk cost Ex-policy price ARF COE Premium
9
Source of sunk costs: Expolicy price Value of ex-policy price declines as soon as the car is out on the road Sunk cost is therefore the difference between the amount paid and the amount available if re-sold the very next day
10
Sunk cost of ex-policy price
COE Premium
ARF
Ex-policy price
0
5
10
Car age in years
11
Source of sunk costs: ARF Owners can purchase a new car by paying ARF at a preferential rate (PARF) if they dispose the car within 10 years If disposed within the first 5 years, a new car can be purchased by paying 25% of ARF (current policy) From the 6th year onward, the preferential rate increases by 5% per year (current policy) Therefore, 25% of ARF is sunk cost
12
Sunk cost of ARF
COE Premium
ARF
Ex-policy price
0
5
10
Car age in years
13
Source of sunk costs: COE premium COE is valid for 10 years If vehicle is disposed within 2 years of purchase, only 80% is refundable After 2 years the COE premium is depreciated on a monthly basis until the end of the 10th year.
Therefore, 20% of COE premium is sunk cost
14
Sunk costs of COE Premium
COE Premium
ARF
Ex-policy price
0
2
5
10
Car age in years
15
Advertising expenses is an example of a sunk cost. Let’s say you invest $1 million promoting a new product but find out that few people are interested in actually buying it. You could continue to pump money into hawking the good — not wanting to admit failure — or you could move on and promote a different product. Either way, you won’t be able to recover that initial $1 million investment. It is a sunk cost.
OPPORTUNITY COST
Understanding Opportunity Cost We make choices every day. We have to, as we have limited resrouces but so many wants. We therefore have to decide which wants we will satisfy and those we will not. All choices involve giving something up is called Opportunity Cost.
What is Opportunity Cost? Opportunity Cost is the cost of a decision in terms of the best alternative given up to achieve it. It is the best alternative forgone.
Opportunity Cost and ... Consumers
workers
• Consumers are buyers and users of goods and services. We all are consumers. The vast majority of us cannot buy everything we like. We can sellect the one with wides and the most accurate informative coverage.
• Undertaking one job involves an opportunity cost. People employed as teachers might also be able to work as civil servants. They need to carefully consider their preference for the jobs available.
Opportunity Cost and ... Producers
government
• Producers have to decide what to make. In deciding what to produce, private sector firms will tend to choose the option which will give them the maximum point.
• Government has to carefully consider, its expenditure of tax revenue on various things. To pay higher taxes, people may have to give up the opportunity to buy certain products to save.
Economic Goods This mean that it takes resources to produce them and hence, their production involves an opportunity cost. They are limited in supply. Almost every good and service is economic good.
Opportunity Cost • The opportunity cost of any alternative is defined as the cost of not selecting the "next-best" alternative. • Example: Suppose that you own a building that is worth $100,000 today and is expected to be worth $100,000 one year from today. If the interest rate is 10%, what is the opportunity cost of using this building for one year?
Example II • The opportunity cost of college attendance includes: – the cost of tuition, books, and supplies, – foregone income (this is usually the largest cost associated with college attendance), and – psychic costs.
• What about room and board?
Example III: • Opportunity cost of attending a movie: – opportunity cost of tickets – opportunity cost of time
BREAK EVEN-ANALYSIS
INTRODUCTION The break-even point has its origins in the economic concept of the "point of indifference." From an economic perspective, this point indicates the quantity of some good at which the decision maker would be indifferent, i.e., would be satisfied, without reason to celebrate or to opine. At this quantity, the costs and benefits are precisely balanced. Similarly, the managerial concept of break-even analysis seeks to find the quantity of output that just covers all costs so that no loss is generated. Managers can determine the minimum quantity of sales at which the company would avoid a loss in the production of a given good. If a product cannot cover its own costs, it inherently reduces the profitability of the firm
DEFINITION • The break even point is the point where the gains equal the losses. The point defines when an investment will generate a positive return. The point where sales or revenues equal expenses. The point where total costs equal total revenues. There is no profit made or loss incurred at the break even point. It is the lower limit of profit when prices are set and margins are determined.
DEFINITION • At this point the income of the business exactly equals its expenditure. If production is enhanced beyond this level, profit shall accrue to the business and if it is decreased from this level, loss shall be suffered by the business.
FORMULA
•Break even point = (fixed cost) / (contribution per unit) Where, Contribution=selling cost-variable cost Fixed cost= Contribution- profit
CALCULATION OF THE BREAK EVEN POINT
• VARIABLE COST They are directly related to the volume of sales: that is these cost increase in proportion to the increase in sales and vice versa.
• FIXED COST Fixed costs continue regardless of how much you can sell or not sell, and can be made up of such expenses as rent, wages, telephone account and insurance. These cost can be estimated by using last years figure as a basis, because they typically do not change.
• At break even point, the desired profit is zero. In case the volume of output or sales is to be computed for a desired profit, the amount of desired profit should be added to fixed cost is the formula given above. • Units for a desired profit= Fixed cost+ desired profit Contribution per unit
Margin Of Safety Margin of safety represents the strength of the business. It enables a business to know what is the exact amount it has gained or lost and whether they are over or below the break even point. It helps the management to estimate that how much their estimated sales can be reduced to even achieve some kind of profit from production and sales or how much costs can increase to even then company at profit point and can survive loss position. margin of safety = (current output - breakeven output)
MARGIN OF SAFETY • Margin of safety represents the strength of the business. It enables a business to know what is the exact amount it has gained or lost and whether they are over or below the break even point. • margin of safety = (current output - breakeven output) OR • Margin o safety = actual sales – BEP sales • margin of safety% = (current output - breakeven output)/current output × 100
Break-Even Analysis Costs/Revenue
TR (p = Rs. 3)
TR (p = Rs. 2)
Margin of safety shows how far sales can fall before losses made. If Q1 = 1000 and Q2 = 1800, sales could fall by 800 units before a loss would be made
TC
VC
Margin of Safety
FC
Q3
Q1
Q2
Output/Sales
A higher price would lower the break even point and the margin of safety would widen
Application of break-even analysis in market conditions Fixed Cost Monthly Rental
$100
Insurance(600 per year, so 600/12 = 50 )
$50
TOTAL MONTHLY FIXED COST
$150
Variable Cost Materials
$3
Lobour
$4
TOTAL VARIABLE COST
$7
Selling Price
$10
BREAK-EVEN POINT CALCULATION Break-Even Point Break -Even Point
Fixed Cost / (Selling cost – Variable Cost) = $150 / ($10 - $7) = 50
To break-even the company must sell 50 units per month. If the Company just broke even, then its Profit and Loss Statement would look like the following: Monthly Profit and Loss Statement Sales Gross Sales Less Cost of Goods Sold Net Sales Expenses Rent Insurance Total Expense Net Profit
($10 per unit times 50 units) ($7 per unit times 50 units)
$500 $350 $150
$100 $50 $150 $0
S A L E S
UNITS SOLD
BREAK-EVEN ANALYSIS • It refers to the ascertainment of level of operations where total revenue equals to total costs. • Analytical tool to determine probable level of operation. • Method of studying the relationship among sales, revenue, variable cost, fixed cost to determine the level of operation at which all the costs are equal to the sales revenue and there is no profit and no loss situation. • Important techniques in profit planning and managerial decision making.
DEPENDENCE Break even analysis depends on the following variables: • The fixed production costs for a product. • The variable production costs for a product. • The product's unit price. • The product's expected unit sales [sometimes called projected sales.]
USES OF BREAK EVEVN POINT • Helpful in deciding the minimum quantity of sales • Helpful in the determination of tender price. • Helpful i e a i i g effe ts upo orga izatio ’s profita ilit . • Helpful in deciding about the substitution of new plants. • Helpful in sales price and quantity. • Helpful in determining marginal cost.
ADVANTAGES • It is cheap to carry out and it can show the profits/losses at varying levels of output. • It provides a simple picture of a business - a new business will often have to present a break-even analysis to its bank in order to get a loan.
LIMITATIONS • Break-even analysis is only a supply side (i.e. costs only) analysis, as it tells you nothing about what sales are actually likely to be for the product at various prices. • It assumes that fixed costs (FC) are constant • It assumes average variable costs are constant per unit of output, at least in the range of likely quantities of sales. (i.e. linearity) • It assumes that the quantity of goods produced is equal to the quantity of goods sold (i.e., there is no change in the quantity of goods held in inventory at the beginning of the period and the quantity of goods held in inventory at the end of the period). • In multi-product companies, it assumes that the relative proportions of each product sold and produced are constant (i.e., the sales mix is constant).
Problems
PV ratio • Profit-volume ratio indicates the relationship between contribution and sales and is usually expressed in percentage. • The ratio shows the amount of contribution per rupee of sales. Since, in the short-term, fixed cost does not change, the profit-volume ratio also measures the rate of change of profit due to change in the volume of sales. • It is influenced by sales and variable or marginal cost. If the sale price increases without a corresponding increase in marginal cost, the contribution increases—and the profit-volume ratio improves. Similarly, if the marginal cost is reduced with sale price remaining same— profitvolume ratio improves. 1
2
A high P/V ratio indicates high profitability so that a slight increase in volume, without increase in fixed cost, would result in high profits. A low P/V ratio, on the other hand, is a sign of low profitability so that efforts should be made to improve P/V ratio.
3
Uses of P/V Ratio: (i) It helps in the determination of Break-even-point [BEP = Fixed cost ÷ P/V ratio] (ii) It helps in the determination of profit at any volume of sales [Sales x P/V ratio = Contribution, Profit = Contribution – Fixed Cost] (iii) It helps in the determination of sales to earn a desired amount of profit (vi) It helps in determining margin of safety [Margin of safety = Profit ÷ P/V ratio]
4
5
PV ratio PV ratio = CMU/SP a % figure a rate of profitability
Uses of PV ratio: – – – – –
1- P/V ratio = Variable cost ratio Sales X P/V ratio = Gross contribution Determining the sales mix BEP = FC / PV Ratio [FC+ Target Profit ] / PV ratio gives the volume of output to be sold to earn a desired level of output
6
Improving PV ratio improvement in P/V ratio will mean more profit – reduce variable cost – increase selling price – product mix to change in favour of high P/V ratio products – Change in FC?
7
• The P/v Ratio ,which establishes the relationship between contribution and sales is of vital importance for studying the profitability of operations of a business .It reveals the effect on profit in the volume . • Higher the P/V Ratio, more will be the profit and lower the P/V Ratio lesser will be the profit. 8
The ratio can be increased by increasing the contribution ….. 1) Increasing the selling price 2) Reducing the variable or marginal cost 3) Changing the sales mixture and selling more profitable products for which the P/V Ratio is higher.
9
• The concept of P/V Ratio is also useful to calculate the break even point , the profit at a given volume of sales ,the sales volume required to earn a given profit and the volume of sales required to maintain the present profits if the selling price is reduced by a specific percentage.
10
Profit –Volume Ratio (PV Ratio)
(Expresses the relation of Contribution to sales)
Sales= Rs 10,000 P/V Ratio =Contribution Sales
= C/S =S-V/S
V Cost=Rs 8,000
C = S XP/V Ratio C S = -------P/V Ratio
P/V Ratio=c/s =S-V/S =10,000-8000/10,000 =20%
11
Profit –Volume Ratio (PV Ratio)
When PV Ratio is Given C= SXPV Ratio C= 10000X20% =Rs 20,000
12
Profit –Volume Ratio (PV Ratio)
Change in Contribution P/V Ratio = --------------------------------Change in Sales
=
Another Method
Change in profit ----------------------Change in Sales
1600-1000 =-------------------x 100 22000-20000
Year
sales
2005
20,000
2006
22,000
net profit 1000 1600
600 = -----------x100=30% 2,0000 13
What Could be the Uses of PV Ratio? Break Even Point Profit at Given Sales Vol required to earn given Profit
14
How Improvement in PV Ratio Could be Achieved?
Increasing Selling Price Reducing Variable Cost
Changing Sales Mix
15
MATERIAL SELECTION FOR A PRODUT
• Material : Materials are commodities which are used directly or indirectly in producing a product. • Material Selection: • (a) Material Properties : Expected level of performance from the material • (b) Material Cost : Material must be available at a cheaper price • (c) Material availability: Should be easily available (d) Processing : Should be easily machinable. • (e) Environment : The environmental factors should not affect the raw material
MATERIAL SELECTION PROCEDURE
• Translation : Express design requirements as constraints and objectives. Example : Tie rod Function : Support the tensile load • Objective : Minimize mass Constraints : Required length, load carrying capacity. • Screening : Eliminate materials that cannot do the job. Ranking : Find the materials that can do the job best. • Selection : Select and verify the supporting materials
• Four Basic Steps 1) Translation: express design requirements as constraints and objectives • 2) Screening: eliminate materials that cannot do the job • 3) Ranking: find materials that best do the job • 4) Supporting Info: handbooks, expert systems, web, etc. • Step 1) Translation Function: What does the component do? Objective: What essentials conditions must be met? Constraints: What is to be maximized or minimized? Free Variables: Identify which design variables are free? Example: Tie Rod Function: Support a tensile load • Objective: Minimize mass Constraints: Required length Load carrying capability w/o/ failure Free Variables: Cross-sectional area Material m = A*L*Density F/A < Yield Stress Eliminate free variable m >= (F)(L) (Density/Yield Stress) therefore minimize weight by maximizing Yield Stress / Density
• Step 2) Screening Methods to evaluate large range of materials Material Bar Charts Material Property Charts (density vs. Young’s Modulus) Screen on Constraints Rank on Objectives Step 3) Ranking What if multiple materials remain after screening? Rank on Objectives Objectives define performance metrics Step 4) Select, then verify with any supporting materials
Knowledge of material properties • application • design of components • material protection (from corrosion, damage, etc.)
Material properties 1. Physical properties 2. Mechanical properties 3. Chemical properties
Physical properties • colour –light wave length • specific heat – the heat required to raise the temperature of one gram of a substance by one degree centigrade (J/kg K)
Physical properties • density – mass per unit volume expressed in such units as kg/cm 3 • thermal conductivity –rate at which heat flows through a given material (W/m K)
Physical properties • melting point – a temperature at which a solid begins to liquify • electrical conductivity – a measure of how strongly a material opposes the flow of electric current (Ω⋅m)
• coefficient of thermal expansion – degree of expansion divided by the change in temperature (m/°C)
Mechanical properties • tensile strength – measures the force required to pull something such as rope,wire or a structural beam to the point where it breaks • ductility – a measure of how much strain a material can take before rupturing
• malleability – the property of a material that can be worked or hammered or shaped without breaking • brittleness –breaking or shattering of a material when subjected to stress (when force is applied to it)
• elasticity – the property of a material that returns to its original shape after stress (e.g. external forces) that made it deform or distort is removed • plasticity - the deformation of a material undergoing non-reversible changes of shape in response to applied forces
Mechanical properties • toughness – the ability of a material to absorb energy and plastically deform without fracturing • hardness – the property of being rigid and resistant to pressure; not easily scratched
Mechanical properties • machinability – the property of a material that can be shaped by hammering, pressing, rolling
Chemical properties • corrosion resistance - a material's ability to resist deterioration caused by exposure to an environment
Which properties do the following materials possess? Material aluminium rubber
ceramics steel copper
lead nylon cast iron
wood
Properties
Which properties do the following materials possess? Material
Properties
aluminium
lightness ; strength
rubber
elasticity ; insulation
ceramics
thermal resistivity
steel
strength
copper
conductivity ; corrosion resistance
lead
high density; ductility
nylon
strength ; toughness
cast iron
damping capacity
wood
insulation ; environmental friendliness
Find application for the following engineering materials: Material aluminium rubber ceramics
steel copper lead nylon
cast iron wood
Application
Find application for the following engineering materials: Material
Application
aluminium
foil; aircraft; window frame
rubber
tyres,; seal; gasket
ceramics
furnace; brick
steel
section; pipe
copper
pipe; cables
lead
storage battery; radiation protection ballast; bullets
nylon
rope; clothing
cast iron
engine block; valves
wood
furniture; deck
material with greatest density gold - 19300 kg/m3 uranium - 19100 kg/m3
lead - 11340 kg/m3 steel - 7800 kg/m3
the strongest material Material
Tensile Strength
UTS (Ultimate Tensile Strength)
carbon nanotubes
62000 MPa (theoretical300000 MPa)
48000 kNm/kg
carbon fibre
5650 MPa
3200 kNm/kg
glass fibre
4700 MPa
1340 kNm/kg
spider web
1000 MPa
900 kNm/kg
high-strength steel
1200 MPa
154 kNm/kg
the best conductor Material
silver
Conductivity
63 x 106 S/m (1/ohm)
copper
59.6 x 106 S/m (1/ohm) gold 45.2 x 106 S/m (1/ohm) aluminium
37.8 x 106 S/m (1/ohm)
the best insulator Material
Specific resistance
polyethylene terephthalate (PET)
1020 ohm
glass
1014 ohm
rubber
1013 ohm
Processes: Forging - a manufacturing process where metal is shaped by plastic deformation under great pressure into high strength parts. Casting – pouring or injecting molten metal into a mold containing a cavity with the desired shape
Pricing issues of material • Pricing of materials may change from time to time. • Materials are usually acquired by several deliveries at different prices. • Actual costs can then take on several different values. • Therefore, the materials pricing system adopted should be the simplest and the most effective one.
Methods of stock valuation • • • •
First-in-first-out(FIFO) Last-in-first-out(LIFO) Weight average cost (WAVCO) Specific identification/unit cost method
First-in-first-out • This method assumes that the first stock to be received is the first to be sold. • The cost of materials used is based on the oldest prices. • The closing stock is valued at the most recent prices.
Last-in-first-out (LIFO) • This method assumes that the last stock to be received is the first to be sold. • Therefore, the cost of materials used is based on the most recent prices. • The closing stock is valued at the oldest prices.
Weight average cost (WAVCO)
This method assumes that the cost of materials used and closing stock are valued at the weighted average cost.
Specific identification/unit cost method • This method assumes that each item of the stock has its own identity. • The costs of materials used and closing stock are determined by associating the units of stock with their specific unit cost.
Economic Order Quantity (EOQ) • EOQ is the order quantity that minimizes total inventory carrying costs and ordering costs. • Ordering costs are costs that are incurred on obtaining additional inventories. They include costs incurred on communicating the order, transportation cost, etc. • Carrying costs represent the costs incurred on holding inventory in hand. They include the opportunity cost of money held up in inventories, storage costs, spoilage costs, etc. EOQ =
2*O*Q C
Where EOQ = Economic Order Quantity O= order cost per order Q = Annual quantity required in units C =Carrying cost per unit per annum
PROCESS PLANNING • Process : It is defined as a group of actions instrumental to the achievement of the output of an operating system • Process Planning : It is the systematic determination of the methods by which a product is to be manufactured economically and competitively. Process Planning procedure: • 1. Analyze the part drawing to get an overall picture on what is required. • 2. Consult with product engineers on product design changes. • 3. List the basic operations required to produce the part to the drawing or specifications • 4. Determine the most economical manufacturing method and form or tooling required to complete the product.
1
2
3
4
5
6
7
8
9
10
11
12
MAKE-OR-BUY DECISION
INTRODUCTION •
The beginning of the 21st century is proving to be an interesting time for the logistics industry. There has been a great deal of transformation in logistics outsourcing caused by the continued drive of manufacturers and retailers to cut costs and by a continued focus on core competencies.
•
Logistics and supply chain management have often been among the first functions to be outsourced. This has moved beyond the warehousing and trucking functions and spread to ancillary services. Usually, companies decide to outsource some or all of their logistics functions in order to reduce costs, make more effective use of the working capital and focus their energies creating differentiation and promoting revenue growth. In some cases, companies claim they can respond faster and more effectively to change when using a logistics service provider (LSP). Indeed, there are significant benefits to outsourcing logistics.
•
To maximize these benefits a review of make-or-buy decision must be implemented. The make-or-buy decision is the act of making a strategic choice between producing a product internally (in-house) or buying it externally (from an outside provider). Making the right choice can be the key factor in sustaining a company competitive advantage and is one of the most important tasks of a successful management.
•
Although the strategic implications of the make-or-buy decision have been discussed for many years these decisions are often made purely on the basis of costs. This paper aims to address this gap by developing a decision-making process that can provide managers with a way of managing the makeor-buy decisions more effectively. The key objectives that arise from the defined purpose are first to describe the set of factors which affect the make-or-buy process, second, to understand better the challenges and barriers that companies face when deciding whether or not to outsource a component or process and third, to suggest some tools and methods for addressing the make-or-buy decision process.
• • • •
•
Determination of whether it is more advantageous to make a particular item in house or to buy it from a supplier. The choice involves both qualitative (such as quality control) and quantitative (such as the relative cost) factors. The buy side of the decision also is referred to as outsourcing. Make-or-buy decisions usually arise when a firm that has developed a product or part—or significantly modified a product or part—is having trouble with current suppliers, or has diminishing capacity or changing demand. Issues like government regulation, competing firms, and market trends all have a strategic impact on the make-or buy decision.
Criteria for Make • • • • • • • • • • • • •
Cost considerations (less expensive to make the part) Desire to integrate plant operations Productive use of excess capacity to help absorb fixed overhead. Need to exert direct control over production and/or better quality Design secrecy is required to protect proprietary technology Unreliable suppliers or No competent suppliers Desire to maintain a stable workforce (in periods of declining sales) Quantity too small to interest a supplier Control of lead time, transportation, and warehousing costs Greater assurance of continual supply Provision of a second source Political, social or environmental reasons (union pressure) Emotion (e.g., pride)
Elements of the "make" analysis • • • • • • • •
Incremental inventory-carrying costs Direct labor costs Incremental factory overhead costs Delivered purchased material costs Incremental managerial costs Any follow-on costs stemming from quality. Incremental purchasing costs Incremental capital costs
Criteria for buy • • • • • • • • • •
Lack of expertise Suppliers' research and specialized know-how exceeds that of the buyer cost considerations (less expensive to buy the item) Small-volume requirements Limited production facilities or insufficient capacity Desire to maintain a multiple-source policy Indirect managerial control considerations Procurement and inventory considerations Brand preference Item not essential to the firm's strategy
• • • • •
Cost considerations for the "buy" analysis Purchase price of the part Transportation costs Receiving and inspection costs Incremental purchasing costs Any follow-on costs related to quality or service
Approaches for make or Buy Decisions Types of analysis followed in make or buy decision are as follows: • Simple cost analysis • Economic cost analysis • Break even analysis
Simple Cost Analysis • •
Quantitative factors deal with cost. The quantitative effects of the make-or-buy decision are best seen through the Relevant Cost Approach. For example, assume a firm has prepared the following cost estimates for the manufacture of a subassembly component based on an annual production of 8000 units:
The supplier has offered the subassembly at a price of $16 each. Two-thirds of fixed factory overhead, which represents executive salaries, rent, depreciation, and taxes, continue regardless of the decision. Should the company buy or make?
Economic Analysis The following inventory models are considered to illustrate this concept. • Purchase model • Manufacturing model • A basic problem for businesses and manufacturers is, when ordering supplies, to determine what quantity of a given item to order. The formula for Purchase model (EOQ) and TC for each model are given as:
Economic Analysis •
An item has yearly demand of 2000 units. The different costs in respect of make and buy are as follows. Determine the best option.
Break Even Analysis
Break Even Analysis •
Break-even point (BEP) is the point at which cost or expenses and revenue are equal: there is no net loss or gain. The main objective of break-even analysis is to find the cut-off production volume from where a firm will make profit.
FOUR STAGES PROCESS MAKEOR-BUY-DECISION •
•
•
•
The decision of whether to make or to buy is a problem that is frequently encountered by higher managers who want to reveal and exploit every competency within the links of the supply chain. Make or buy is a decision not to be made only on the basis of economic considerations, since acquisition or loss of core competencies may also be involved. Decisions regarding outsourcing significant functions are among the most strategic that can be made by an organization. They address the basic organization al choice of the functions for which internal expertise is developed and nurtured and those for which such expertise is purchased. Even an individual make-or-buy decision can affect company’s production methods, working capital, cost of borrowing or competitive position. Outsourcing provides companies with the freedom to concentrate their energies on key activities that are critical to maintaining their competitive edge. This results in improvement of industrial relations and rising labour productivity. For activities eligible for outsourcing, the key strategic question is whether the firm can perform those service activities on a level that is comparable with the best organizations in the world. If a service activity meets several criteria, the next step is deciding whether the service is central to the firm’s core strategic activities. Moreover, to make the best make-or-buy decision, companies must determine how that decision will affect the final product quality and the company’s technology [4].
• The following are the suggested stages to successful make-or-buy decision: • building incentive for outsourcing, • exploring strategic implications, • analyzing costs/performance, • selecting providers.
Planning stage •
•
As with any significant new planning activities also by planning initiative team member selection, the leader selection and outside adviser come into play. The project team assesses the risk and the resources, information and management skills needed to mitigate those risks, while the outsourcing adviser levels the playing field with the outsourcing providers [6]. The feasibility of outsourcing is determined by a series of screenings that every outsourcing initiative should pass before further, detailed evaluation: 1. Senior management and employee announcement: Because outsourcing involves a number of strategic issues, as well as significant risks and rewards, senior management’s clear and unequivocal stated support is prerequisite. For larger outsourcing initiatives top management must play a role. For smaller initiatives, middle-level managers might do the heavy lifting with the support of senior management. The team usually needs a mix of managerial and technical talent and representatives from user areas whose services will be directly impacted by outsourcing. User perspectives and objectives are essential for setting the scope and assessing the risks. Before selecting the project team and advisers the organisation must inform also the employees about the outsourcing initiative, explain why outsourcing should be explored and indicate that this is only a test to determine if outsourcing is a viable tool. If they do not know what is happening, they may react very negatively to outsourcing.
Planning stage •
• • • • • • • • •
2. Choosing the project leader: Early in the outsourcing evaluation process, the customer must identify who will take the leadership responsibility, perform the analysis, and make the decisions. The role of the outsourcing project leader is complex and challenging. It runs from managing the project, the team, and the individual team members to gathering technical support, facilitating discussions, and drawing out consensus to drafting reports, making presentations and so on [6]. In selecting the individuals for this position, the following characteristics should be considered: (1) ability to embrace and champion change, (2) earned credibility across the organization, (3) a desire to manage, not to do, (4) the ability to build trust, (5) strong communication skills, (6) strong negotiation skills, (7) strategic planning skills, (8) project and team management skills, (9) marketing skills, (10) process expertise
Planning stage •
• • • • • • •
3. Appoint process implementation team: Choosing team members is a critical step in successful outsourcing. Characteristics and skills that the outsourcing team members should possess are the following: (1) motivation to participate, (2) a record of delivering on promises, (3) strong communication skills, (4) ability to think creatively and strategically, (5) solid performance evolutions, (6) relevant specialization within the organization, (7) wide experience from outside the organization. The team members should be objective. They should represent a cross-section of the organization’s functions and should have access to and full cooperation of the individuals in the areas to be considered for outsourcing [6]. The size of the customer’s team depends on the scope and size of the project, but smaller teams are generally more effective. The team can be quite small in the planning phase and expanded when the analysis begins. Teams with full-time members are often more focused and effective than teams composed of people who work part-time, although full-time allocation may only make sense for large outsourcing projects. It helps tremendously to have persons experienced in outsourcing on the team for the insight they bring to the issues and the realism they bring to cost and benefit estimates.
Planning stage •
4. Choosing an independent advisor: Independent outsourcing advisors can help avoid failures. They will have seen different situations first hand and can help the client avoid the same pitfalls. Lack of planning, lack of follow-up in execution, miscommunication, not understanding cultural differences, poor process, etc., are just a few of the areas that an outside advisor can help with. If a client has not established an offshore outsourcing relationship, having an independent advisor can help formulate clearly articulated objectives and bring their expertise to the table with regard to process, selection, contract negotiations, and getting the transition going in an accelerated manner. Another benefit of an outside advisor is to help the client understand if they are getting a good deal (compared with the industry and other clients doing similar work with a similar type of outsourcer). This removes one potential area of mistrust in the relationship. Also, advisors who have people with good understanding of the outsourcer’s national culture can explain the cultural differences. There are steps one can take to avoid these problems. Independent advisors can most certainly help in establishing, repairing, maintaining, and improving client/provider relationships.
Evaluation stage • •
•
If the outsourcing idea has passed the initial screenings, planning can proceed to a detailed outsourcing evaluation: 1. Outsourcing implication on organizational vision and structure: Early in the outsourcing evaluation process, the customer must find out and understand how outsourcing can fit within the organisation’s strategies (organizational structure and vision) and how its implementation will affect those strategies. 2. Determining core competences: Core competencies are the source of competitive advantage and arise from the integration of multiple technologies and the coordination of diverse production skills. To achieve long-term growth, companies need to develop, protect, and leverage their core competencies and consider outsourcing any activities that do not confer a competitive advantage. If the function or functions to be outsourced contribute in central ways to the organization’s competitive success (core competencies) then those core functions are probably not strong candidates for outsourcing. Identifying which resources and capabilities to preserve and which activities to outsource, requires careful consideration and planning.
•
•
3. Defining the motive, objectives and scope: Before serious analysis of cost and performance can take place, the company should know the objectives of outsourcing and know the scope of what is to be considered for outsourcing. If the company does not know what it is trying to accomplish, any alternative can look good or bad. Companies that rush into outsourcing without fully understanding what they hope to gain soon find themselves in a mire of contractual battle or not receiving improved services. Sensible reasons to consider outsourcing are both strategic and tactical. It is essential to know and clearly define the objectives of the company and to document what the company expects from outsourcing. Clear objectives help lead to a sound decision on what to outsource and what not to outsource. The objectives for outsourcing are often derived directly from the motives for outsourcing that can be grouped into a few summarised categories: (1) concentration on core business, (2) investment reduction, (3) restructuring of the supply chain, (4) cost reduction and service improvements. Specifying outsourcing objectives that are more specific than the motives provides a basis for developing selection criteria for the provider and guides the requirements and conditions in the RFP and provides a common understanding for the persons who write the RFP, evaluate the proposals, and recommend the selected supplier(s). Objectives also provide the basis for evaluating provider proposals. The last important thing to be defined is the scope. Outsourcing can be divided into two general categories: total and selective. Total outsourcing involves contracting out 80% or more of the function. Selective outsourcing involves outsourcing a few functions that total less than 80% of the whole. Methods for identifying functions that might be selectively outsourced include opportunistic, problem-focused approaches and more methodical planning approaches.
Internal costs and performance analysis •
•
• •
•
The goals of this phase are to develop detailed cost analysis of the target function. When considering outsourcing part or all of any business process the following associated costs need to be taken into account: salaries, benefits, training/education, specialised software, travel, phone charges, depreciation/amortisation, mail costs and postage, office supplies, equipment, management time, information costs, occupancy charges. It is essential that the project team conducts activity-based analyses in order to understand the current costs of the activities that might be outsourced and those that are staying. To this are added the costs of invested capital and the estimated costs of poor performance [6]. The biggest challenge is to develop both costs and service requirements in light of expected technological and business change (which may be interrelated) over the expected life of the outsourcing arrangement. Forecasting the future is always difficult and often inaccurate, but still necessary. The project team must estimate which costs do not disappear with outsourcing and what new costs will be incurred as result of outsourcing. Without a good idea of future needs and the costs of meeting these needs, it is difficult to outsource effectively and efficiently. There are also financial benefits resulting from outsourcing and they are other than cost-related. These are estimated and used in the make-or-buy decision [6]. Current performance should also be measured and analyzed, since performance improvement is often a reason for outsourcing. The project team estimates the financial impact of the internal poor performance and projects future performance. Understanding the existing performance is fundamental to setting provider performance standards and monitoring their performance [6]. Once a company has estimated all the costs, it determines its position on who will bear each cost – the organization or the outsourcing provider, as shown in Figures 5 and 6.
Selecting service provider •
•
• • • • •
This process of first defining organisations needs, then engaging the marketplace of providers, preparing and delivering request of proposal (RFP) and ultimately selecting the right provider requires a good interplay of internal assessment, coupled with an effective and practical interaction with the marketplace of potential providers [5]. Once a decision has been made on which areas to consider for outsourcing, the project team should begin researching the providers. Following are a number of ways how to proceed in identifying potential providers: ––open a dialogue with outside organizations the company is already doing business with, ––use the organization’s professional network, ––direct research, ––use consultants [5]. The project team lists the criteria for “qualified” providers based on the reasons for outsourcing. Potential providers are identified and further investigations are made to determine their qualification. Their qualifications are then compared to the criteria and a decision is made on whether they should be invited to submit proposals. Requests for proposal are prepared and delivered to the targeted providers list and should include the reasons for outsourcing, scope of outsourcing, qualifications of providers, pricing models, performance measures. When the proposals are returned, they are evaluated and compared to the other proposals and further discussions are held with the providers on the short list. From this short list the prime provider is selected [6].
Decision time •
This is the step where the oversight of the project team and top management comes into active play. The goal here is to review the information gathered till this phase and to consider the recommendation on whether the service should be made internally or outsourced. If the decision is to buy the part family from an outside source (or sources), then the four steps shown in Figure 7 should be undertaken.
Value engineering
•
Value is the lowest price you must pay to provide a reliable function or service (L. D. Miles)
•
“The ratio of Function to Cost” Function(Utility)
Worth =
Value = Cost
Cost
What is Value Engineering? •
Value Engineering (VE, or Value Analysis) is a management technique that seeks the best functional balance between cost , reliability and performance of a product, project, process or service.
How is VA different from VE ? • Traditionally Value Analysis (VA) is used to describe the application of the 'techniques to an existing product or services or after the fact. • Value Engineering (VE) has been used to refer to the design stage or before the fact. Value Engineering (VE) approach is used for new products, and applies the same principles and techniques to premanufacturing stages such as concept development, design and prototyping. • Value Analysis and Value Engineering (VE) is a powerful Change Management and Problem Solving' tool with over a century of worldwide application track record. • VE is used to create functional breakthroughs by targeting value mismatches during product, process, and project design.
4
How is VA different from VE ? • VA is also a vital tool to deal with post product release problems and process improvement innovation. • Value Analysis (VA) is considered to be a process, as opposed to a simple technique, because it is both an organized approach to improving the profitability of product applications and it utilizes many different techniques in order to achieve this objective. • The techniques that support VA activities include 'common' techniques used for all VA exercises and some that are appropriate for the product under consideration. • A few other names for VA / VE are - Value Management, Value Planning, etc.
5
How is VA different from VE ? • Value Analysis process attacks unnecessary costs and is thus one of the most effective ways to increase an organization's profitability. • However that is only doing half the job. • A truly effective value improvement program cannot only reduce costs, but also improve operations and product performance. • The VA approach can be effectively used to analyze existing products or services offered by manufacturing companies and service providers alike. • The VA / VE methodology involves function analysis and everything has a function. • Therefore the methodology has universal application. • Value Analysis / Value Engineering can be applied with equal success to any cost generating areas. 6
THE VALUE EQUATION • Value analysis is evaluates a product utility, esteem and market values, each of which are defined below : • Utility value – how useful / functional the product is seen to be. • Esteem value – the value that customer / user gives to product attributes, not directly contributing to utility but more relating to aesthetic and subjective value. Esteem issues and functionality should not be overlooked or compromised. • Market value – what market is prepared to pay for the product. • Market value = Utility value + Esteem value
7
THE VALUE ANALYSIS TERMINOLOGY • • • •
•
Need : These are users expectations, may be expressed explicitly, or may be latent. Value : Value is an imprecise word, its meaning depends both on the user and on the context. For example a typewriter ribbon or a word – processing package may have good value while the typewriter or computer may not have. In an engineering context the distinction can be important, as any cosmetic changes brought about by Value Analysis or by means of any other technique are waste of time if the total product is unacceptable to the market. Value is a quantity, which enhances customer satisfaction or slashes the expense attributable to the product
8
THE VALUE ANALYSIS TERMINOLOGY
• •
• • 1. 2. 3. 4.
In value method terms : Value = Worth / Cost OR Value of an item = Performance of its function / Cost OR Value = Σ (+) / Σ (-) = Σ (Benefits) / Σ (Costs) Value greater than 1.0, the item is perceived to be fair or having good value. Value is less than 1.0, the item is perceived to be having poor value. When an item has a perceived worth that far exceeds the life – cycle cost, we usually consider purchasing the item. An item that does its function better than another, has more value. Between two items that perform their function equally well, the one that costs less is more valuable. 9
THE VALUE ANALYSIS TERMINOLOGY • • • • •
Different customers will interpret the value of a product in different ways. The perfor a ce of its fu ctio s could i clude that it is beautiful (where needed) or it lends an image to the user / possessor (where desired ) Its common characteristic is a high level performance, capabilities, emotional appeal, style, etc. relative to its cost. This can also be expressed as maximizing the function of product relative to its cost : Value = (performance + capability / cost = Function / cost 10
The Job Plan Value engineering is often done by systematically following a multi-stage job plan. Larry Miles' original system was a six-step procedure which he called the "value analysis job plan."
THE VA PROCESS (JOB PLAN) Orientation
Functional identification Functional analysis
Creative alternatives
Analysis & evaluation
implementation 12
The Job Plan The modern version has the following eight steps: 1. 2. 3. 4. 5. 6. 7. 8.
Orientation Information Functional Creative Evaluation Development Presentation Implementation and Follow-up
The Job Plan 1. Orientation Phase Identify issues Prioritize Issues Drafts scopes and objective Establish evaluation factors Determine Study Team Collect Data Prepare for value study ...
The Job Plan 2. Information Phase
Further familiarization of the project by the team; all team members participate in determine the true needs of the project.
Areas of high cost or low worth are identified.
The Job Plan 3. Functional Phase
Functional analysis outlines the basic function of a product using a verb and a noun such as ‘boil water’ as in the case of our kettle.
What is the Function?
“ Boil Water ” Verb
Noun
The Job Plan 4. Creative Phase
This step requires a certain amount of creative thinking by the team. A technique that is useful for this type of analysis is brainstorming. This stage is concerned with developing alternative.
The Job Plan
5. Evaluation Phase
In this phase of the workshop, the VA team judges the ideas developed during the creative phase. The VA team ranks the ideas. Ideas found to be irrelevant or not worthy of additional study are disregarded. Those ideas that represent the greatest potential for cost savings and improvements are selected for development.
The Job Plan
6. Development Phase
The team develops the selected ideas into alternatives (or proposals) with a sufficient level of documentation to allow decision makers to determine
if
implemented.
the
alternative
should
be
The Job Plan
7. Presentation Phase
1.
The presentation phase is actually presenting the best alternative (or alternatives) to those who have the authority to implement the proposed solutions that are acceptable.
The Job Plan 8. Implementation And Follow Up
1.
Develop an implementation plan
2.
Execute the plan
3.
Monitor the plan to completion
Objective: During the implementation and follow-up phase, management must assure that approved recommendations are converted into actions.
Purpose 1.
Determine the best design alternatives
2.
Reduce cost
3.
Improve quality
4.
Increase reliability and availability
5.
Enhance customer satisfaction
6.
Improve organizational performance
7.
Identify problems
8.
Develop recommended solutions
Potential Saving From VE Early changes are naturally less expensive than later ones, as shown in the diagram below.
Principles Value Engineering principles:
1
Systematic method for evaluating product performance and value
2
The use of multi-functional teams
3
Focus on a simplified product
Case Study Focus Adjustment Knob for Slit Lamp
Case Study Introduce the Product • In this presentation we have considered a medical instrument manufacturing company, Aadarsh Instruments, located in Ambala. • This firm is producing different types of microscopes which they export to various countries around the globe. • One of their model SL250 have a component named Focus Adjustment Knob for Slit Lamp in microscope. This microscope has found application in the field of eye inspection.
Case Study The steps used for this purpose are as follows:
1. Product selection plan 2. Gather information of product 3. Functional analysis 4. Creativity Worksheet 5. Evaluation sheet 6. Cost analysis
7. Result
Case Study 1. Plan For Product Selection •
This Product is used to adjust the focus of lens for magnification purpose.
•
The present specifications of this part and its material used are costlier than the average industry cost.
•
Value of this product can be increased by maintaining its functions and reducing its cost or keeping the cost constant and increasing the functionality of the product.
Case Study 2. Obtain Product Information i. ii. iii. iv. v. vi. vii. viii. ix. x.
Material – Aluminum Bronze Alloy Diameter of base plate –30 mm Thickness of plate--3 mm Cost of the scrap is – 293 rupee/Kg Pieces Produced annually – 8000 Process used – C.N.C. indexing milling Cycle time—2.5 min Anodizing—2/min Material cost—65 gm Total Present cost – 29.99 rupee /piece
*{1$=56 rupee}*
Case Study 3. Functional Analysis of Present Functions
Case Study 4. Develop Alternate Design Or Methods During brainstorming these ideas were listed: i. ii. iii. iv. v. vi.
Change design Change material Use plastic Make it lighter Change the production process Use nylon indexing unit
Case Study 5. Evaluation Phase
For judging the ideas, the following designs were considered: A. B. C. D. E.
Function Cost Maintainability Quality Space
each of the above criteria was compared with others , and depending on their relative importance, three categories were formed, major, medium, and minor.
Case Study
Comparing this criteria according to relative importance :
Case Study From the paired comparison we get the following result:
The above ideas were discussed and the best feasible ideas were separated which were: a) Change the material to steel b) Use Nylon unit c) Use existing material
Case Study 6. Cost Analysis
Case Study 7. Result The total savings after the implementation of value engineering are given below: • Cost before analysis – 29.99 rupee • Total Cost of nylon knob – 18.40 rupee • Saving per product – 11.59 rupee • Percentage saving per product – 38.64 % • Annual Demand of the product – 8000 • Total Annual Saving – 92,720 rupee • Value Improvement - 62.98 %
Conclusion
Three goals that we're looking at value engineering:
1.
Identify additional functions that aren’t attractive to customers.
2.
Add attractive functions for customers.
3.
Saving because of redundant functions.
the
elimination
of
Function • • • • • • • • • •
Function is the purpose for which the product is made. Identification of the basic functions and determination of the cost currently being spent on them are the two major considerations of value analysis. Function identifies the characteristics which make the product/component/ part/item/device to work or sell. “Work functions” lend performance value while “sell functions” provide esteem value. Verbs like “support”, “hold”, “transmit”, “prevent”, “protect”, “exhibits”, “control”, etc., are used to describe work functions, while “attract”, enhance”, “improve”, “create”, etc., are used to describe “sell” functions. For example, in a “bus driver cabin”, the functional analysis of some of the parts are given in Table
Classification of the functions Rarely do all functions assume equal importance. Usually, some functions are more important than • others. Functions can be classified into the following three categories: • 1. Primary function • 2. Secondary function • 3.Tertiary function • 1. Primary functions are the basic functions for which the product is specially designed to achieve. Primary functions, therefore, are the most essential functions whose non-performance would make the product worthless, e.g. a photo frame exhibits photographs, a chair supports weight, a fluorescent tube gives light. • 2. Secondary functions are those which, if not in-built, would not prevent the device from performing its primary functions, e.g., arms of a chair provide support for hands. Secondary functions are usually related to convenience. The product can still work and fulfill its intended objective even if these functions are not inbuilt and yet they may be necessary to sell the product. • 3. Tertiary functions are usually related to esteem appearance. For example, Sun mica top of a table gives esteem appearance for the table.
Examples Let us consider a single example of painting a company bus to explain all the above three functions. Here, the primary function of painting is to avoid corrosion. The secondary function is to identify the company to which the bus belongs by the colour ofthe paint (e.g. blue colour for Ashok Leyland Ltd.). • The tertiary function is to impart a very good appearance to the bus by using brilliant colours. Aims The aims of value engineering are as follows: • Simplify the product. • Use (new) cheaper and better materials. • Modify and improve product design. • Use efficient processes. • Reduce the product cost. • Increase the utility of the product by economical means. • Save money or increase the profits.
Value Engineering Procedure • • • • • • • • • • •
The basic steps of value engineering are as follows: (a) Blast (i) Identify the product. (ii) Collect relevant information. (iii) Define different functions. (b) Create (iv) Different alternatives. (v) Critically evaluate the alternatives. (c) Refine (vi) Develop the best alternative. (vii) Implement the alternative. Step 1: Identify theproduct. First, identify the component for study. In future, any design change should add value and it should not make the product as obsolete one. Value engineering can be applied to a product as a whole or to sub-units.
• • • • • • • • • • • •
Step 2: Collect relevant information. Information relevant to the following must be collected: Technical specifications with drawings Production processes, machine layout and instruction sheet Time study details and manufacturing capacity Complete cost data andmarketing details Latest development in related products Step 3: Define different functions. Identify and define the primary, secondary and tertiary functions of the product or parts of interest. Also, specify the value content of each function and identify the high cost areas. Step 4: Different alternatives. Knowing the functions of each component part and its manufacturing details, generate the ideas and create different alternatives so as to increase the value of the product.Value engineering should be done after a brain storming session. All feasible or non-feasible suggestions are recorded without any criticism; rather, persons are encouraged to express their views freely.
Interest formulae and its applications
1
• Interest Is the money earned (profit) on a savings account or investment. Principal or present value is the amount of money invested, sometimes referred to as the initial amount.
2
Simple Interest Formulas • Simple interest is the interest that is computed on the original principal only. • If I denotes the interest on a principal P (in dollars) at an interest rate of r per year for t years, then we have I = Prt • The accumulated amount A, the sum of the principal and interest after t years is given by A = P + I = P + Prt = P(1 + rt) and is a linear function of t.
Example • A bank pays simple interest at the rate of 8% per year for certain deposits. • If a customer deposits $1000 and makes no withdrawals for 3 years, what is the total amount on deposit at the end of three years? • What is the interest earned in that period? Solution • Using the accumulated amount formula with P = 1000, r = 0.08, and t = 3, we see that the total amount on deposit at the end of 3 years is given by
or $1240.
A P(1 rt ) 1000[1 (0.08)(3)] 1240
Example • A bank pays simple interest at the rate of 8% per year for certain deposits. • If a customer deposits $1000 and makes no withdrawals for 3 years, what is the total amount on deposit at the end of three years? • What is the interest earned in that period? Solution • The interest earned over the three year period is given by
or $240.
I Prt 1000(0.08)(3) 240
Applied Example: Trust Funds • An amount of $2000 is invested in a 10-year trust fund that pays 6% annual simple interest. • What is the total amount of the trust fund at the end of 10 years? Solution • The total amount is given by
A P (1 rt ) 2000[1 (0.06)(10)] 3200 or $3200.
Compound Interest • Frequently, interest earned is periodically added to the principal and thereafter earns interest itself at the same rate. This is called compound interest. • Suppose $1000 (the principal) is deposited in a bank for a term of 3 years, earning interest at the rate of 8% per year compounded annually. • Using the simple interest formula we see that the accumulated amount after the first year is
or $1080.
A1 P(1 rt ) 1000[1 0.08(1)] 1000(1.08) 1080
Compound Interest
m
mt
r reff 1 1 m
r A P 1 m
(4)(3)
0.08 1000 1 4 1000(1.02)12 1268.24
1
0.08 1 1 1 1.08 1 0.08
Compound Interest • To find the accumulated amount A2 at the end of the second year, we use the simple interest formula again, this time with P = A1, obtaining: A2 P (1 rt ) A1 (1 rt ) 1000[1 0.08(1)][1 0.08(1)] 1000(1 0.08)2 1000(1.08)2 1166.40
or approximately $1166.40.
Compound Interest • We can use the simple interest formula yet again to find the accumulated amount A3 at the end of the third year: A3 P(1 rt ) A2 (1 rt ) 1000[1 0.08(1)]2 [1 0.08(1)] 1000(1 0.08)3 1000(1.08)3 1259.71
or approximately $1259.71.
Compound Interest • Note that the accumulated amounts at the end of each year have the following form: A1 1000(1.08) A2 1000(1.08) 2 A3 1000(1.08)3
A1 P(1 r ) or:
A2 P(1 r )2 A3 P(1 r )3
• These observations suggest the following general rule: – If P dollars are invested over a term of t years earning interest at the rate of r per year compounded annually, then the accumulated amount is A P(1 r )t
Compounding More Than Once a Year
• The formula
A P(1 r )t
was derived under the assumption that interest was compounded annually. • In practice, however, interest is usually compounded more than once a year. • The interval of time between successive interest calculations is called the conversion period.
Compounding More Than Once a Year
• If interest at a nominal rate of r per year is compounded m times a year on a principal of P dollars, then the simple interest rate per Annual interest rate conversion period is i r m
Periods per year
r 0.08 i 0.02 m 4
• For example, if the nominal interest rate is 8% per year, and interest is compounded quarterly, then
Compounding More Than Once a Year
• To find a general formula for the accumulated amount, we apply A P(1 r)t repeatedly with the interest rate i = r/m. • We see that the accumulated amount at the end of each period is as follows: A1 P (1 i ) 2 Second Period: A2 A1 (1 i ) [ P (1 i )](1 i ) P (1 i ) 2 3 Third Period: A3 A2 (1 i ) [ P (1 i ) ](1 i ) P (1 i ) nth Period: An An 1 (1 i ) [ P (1 i ) n 1 ](1 i ) P(1 i ) n First Period:
Compound Interest Formula There are n = mt periods in t years, so the accumulated
amount at the end of t years is given by
r A P 1 m
n
Where n = mt, and A = Accumulated amount at the end of t years P = Principal r = Nominal interest rate per year m = Number of conversion periods per year t = Term (number of years)
Example • Find the accumulated amount after 3 years if $1000 is invested at 8% per year compounded a. b. c. d. e.
Annually Semiannually Quarterly Monthly Daily
Example Solution a. Annually. Here, P = 1000, r = 0.08, and m = 1. Thus, i = r = 0.08 and n =n 3, so r A P 1 m
0.08 1000 1 1 1000(1.08)3 1259.71
3
Example Solution b. Semiannually. i 0.082 Here, P = 1000, r = 0.08, and m = 2. Thus, and n = (3)(2) n= 6, so r A P 1 m
0.08 1000 1 2 1000(1.04)6 1265.32
6
Example Solution c. Quarterly. i 0.084 Here, P = 1000, r = 0.08, and m = 4. Thus, and n = (3)(4)n = 12, so r A P 1 m
12
0.08 1000 1 4 1000(1.02)12 1268.24
Example Solution i 0.08 12 d. Monthly. Here, P = 1000, r = 0.08, and m = 12. Thus, and n = (3)(12) n = 36, so r A P 1 m
0.08 1000 1 12
36
0.08 1000 1 12 1270.24
36
Example Solution i 0.08 e. Daily. 365 Here, P = 1000, r = 0.08, and m = 365. Thus, and n = (3)(365) = 1095, so n r A P 1 m
1095
0.08 1000 1 365
1095
0.08 1000 1 365 1271.22
Continuous Compounding of Interest • One question arises on compound interest: – What happens to the accumulated amount over a fixed period of time if the interest is compounded more and more frequently? • We’ve seen that the more often interest is compounded, the larger the accumulated amount.
• But does the accumulated amount approach a limit when interest is computed more and more frequently?
Continuous Compounding of Interest
• Recall that in the compound interest mt formula r A P 1 m
the number of conversion periods is m. • So, we should let m get larger and larger (approach infinity) and see what happens to the accumulated amount A.
Continuous Compounding of Interest
• If we let u = m/r so that m = ru, then the above formula becomes
1 A P 1 u
urt
or
1 A P 1 u
u
rt
The table shows us that when u gets
larger and larger the expression u
1 1 u approaches 2.71828 (rounding to five decimal places). It can be shown that as u gets larger and larger, the value of the expression approaches the irrational number 2.71828… which we denote by e.
u
10
1 1 u 2.59374
100
2.70481
1000
2.71692
10,000
2.71815
100,000
2.71827
1,000,000
2.71828
u
Continuous Compounding of Interest •
Continuous Compound Interest Formula A = Pert where
P = Principal r = Annual interest rate compounded continuously t = Time in years A = Accumulated amount at the end of t years
Examples • Find the accumulated amount after 3 years if $1000 is invested at 8% per year compounded (a) daily, and (b) continuously. Solution a. Using the compound interest formula with P = 1000, r = 0.08, m = 365, and t = 3, we find mt
r 0.08 A P 1 1000 1 365 m
(365)(3)
1271.22
b. Using the continuous compound interest formula with P = 1000, r = 0.08, and t = 3, we find A = Pert = 1000e(0.08)(3) ≈ 1271.25 Note that the two solutions are very close to each other.
Effective Rate of Interest • The last example demonstrates that the interest actually earned on an investment depends on the frequency with which the interest is compounded. • For clarity when comparing interest rates, we can use what is called the effective rate (also called the annual percentage yield): – This is the simple interest rate that would produce the same accumulated amount in 1 year as the nominal rate compounded m times a year. • We want to derive a relation between the nominal compounded rate and the effective rate.
Effective Rate of Interest • The accumulated amount after 1 year at a simple interest rate R per year is A P(1 R )
• The accumulated amount after 1 year at a nominal interest rate r per year compounded m times a year is r A P 1 m
m
• Equating the two expressions gives r P(1 R ) P 1 m r 1 R 1 m
m
m
Effective Rate of Interest Formula
• Solving the last equation for R we obtain the formula for computing the effective rate of interest: m r reff 1 1 m
where reff = Effective rate of interest r = Nominal interest rate per year m = Number of conversion periods per year
Example • Find the effective rate of interest corresponding to a nominal rate of 8% per year compounded a. b. c. d. e.
Annually Semiannually Quarterly Monthly Daily
Example Solution a. Annually. Let r = 0.08 and m = 1. Then 1 0.08 reff 1 1 1 1.08 1 0.08
or 8%.
Example Solution b. Semiannually. Let r = 0.08 and m = 2. 2Then 0.08 reff 1 1 2 1.0816 1 0.0816
or 8.16%.
Example Solution c. Quarterly. Let r = 0.08 and m = 4. Then 4 0.08 reff 1 1 4 1.08243 1 0.08243
or 8.243%.
Example Solution d. Monthly. Let r = 0.08 and m = 12.12 Then 0.08 reff 1 1 12 1.08300 1 0.08300
or 8.300%.
Example Solution e. Daily. Let r = 0.08 and m = 365. Then 365 0.08 reff 1 1 365 1.08328 1 0.08328
or 8.328%.
Effective Rate Over Several Years • If the effective rate of interest reff is known, then the accumulated amount after t years on an investment of P dollars can be more readily computed t A P (1 reff ) by using the formula
Present Value • Consider the compound interest formula: r A P 1 m
mt
• The principal P is often referred to as the present value, and the accumulated value A is called the future value, since it is realized at a future date. • On occasion, investors may wish to determine how much money they should invest now, at a fixed rate of interest, so that they will realize a certain sum at some future date. • This problem may be solved by expressing P in terms of A.
Present Value • Present value formula for compound n interest P A 1 i r i m
Where
n mt
and
Examples • How much money should be deposited in a bank paying a yearly interest rate of 6% compounded monthly so that after 3 years the accumulated amount will be $20,000? Solution • Here, A = 20,000, r = 0.06, m = 12, and t = 3. • Using the present value formula we get r P A 1 m
mt
0.06 20, 000 1 12 16, 713
(12)( 3)
Examples • Find the present value of $49,158.60 due in 5 years at an interest rate of 10% per year compounded quarterly. Solution • Here, A = 49,158.60, r = 0.1, m = 4, and t = 5. • Using the present value formula we get mt r P A 1 m
0.1 49,158.60 1 4 30, 000
(4)(5)
Present Value with Continuously Compounded Interest • If we solve the continuous compound interest formula A = Pert for P, we get P = Ae–rt • This formula gives the present value in terms of the future (accumulated) value for the case of continuous compounding.
Applied Example: Real Estate Investment • Blakely Investment Company owns an office building located in the commercial district of a city. • As a result of the continued success of an urban renewal program, local business is enjoying a mini-boom. • The market value of Blakely’s property is
V (t ) 300,000e
t /2
where V(t) is measured in dollars and t is the time in years from the present. • If the expected rate of appreciation is 9% compounded continuously for the next 10 years, find an expression for the present value P(t) of the market price of the property that will be valid for the next 10 years. • Compute P(7), P(8), and P(9), and then interpret your results.
Applied Example: Real Estate Investment Solution • Using the present value formula for continuous compounding P = Ae–rt with A = V(t) and r = 0.09, we find that the present value of the market price of the property t years from now is P(t ) V (t )e 0.09t 300, 000e 0.09t
• Letting t = 7, we find that P(7) 300,000e0.09(7) or $599,837.
7 /2
t /2
599,837
(0 t 10)
Applied Example: Real Estate Investment Solution • Using the present value formula for continuous compounding P = Ae–rt with A = V(t) and r = 0.09, we find that the present value of the market price of the property t years from now is P(t ) V (t )e 0.09t 300, 000e 0.09t
• Letting t = 8, we find that P(8) 300,000e0.09(8) or $600,640.
8/2
t /2
600,640
(0 t 10)
Applied Example: Real Estate Investment Solution • Using the present value formula for continuous compounding P = Ae–rt with A = V(t) and r = 0.09, we find that the present value of the market price of the property t years from now is P(t ) V (t )e 0.09t 300, 000e 0.09t
• Letting t = 9, we find that P(9) 300,000e0.09(9) or $598,115.
9/2
t /2
598,115
(0 t 10)
Applied Example: Real Estate Investment
Solution • From these results, we see that the present value of the property’s market price seems to decrease after a certain period of growth. • This suggests that there is an optimal time for the owners to sell. • You can show that the highest present value of the property’s market value is $600,779, and that it occurs at time t ≈ 7.72 years, by sketching the graph of the function P.
Example: Using Logarithms in Financial Problems
• How long will it take $10,000 to grow to $15,000 if the investment earns an interest rate of 12% per year compounded quarterly? Solution • Using the compound interest formula n
r A P 1 m 4t 0.12 15,000 10,000 1 4
with A = 15,000, P = 10,000, r = 0.12, and m = 4, we obtain 15,000 4t (1.03)
10,000
1.5
Example: Using Logarithms in Financial Problems
• How long will it take $10,000 to grow to $15,000 if the investment earns an interest rate of 12% per year compounded quarterly? 4t (1.03) 1.5 Solution Taking logarithms 4t ln(1.03) ln1.5 • We’ve got on both sides logbmn = nlogbm 4t ln1.03 ln1.5 So, it will take about ln1.5 3.4 years for the 4t ln1.03 investment to grow from $10,000 to ln1.5 $15,000. t 3.43 4 ln1.03
Annuities
(1 i )n 1 (1 0.01)12 1 S R 100 1268.25 i 0.01 1 (1 i ) n 1 (1 0.01) 36 P R 400 12,043 i 0.01
Future Value of an Annuity • The future value S of an annuity of n payments of R dollars each, paid at the end of each investment period into an account that earns (1interest at the rate of i i )n 1 per period, isS R i
Example • Find the amount of an ordinary annuity consisting of 12 monthly payments of $100 that earn interest at 12% per year compounded monthly. Solution • Since i is the interest rate per period and since interest is compounded monthly in this case, we have
0.12 i 0.01 12 • Using the future value of an annuity formula, with = 100, n = 12, and i = 0.01, we have
(1 i )n 1 (1 0.01)12 1 S R 100 1268.25 i 0.01 or $1268.25.
R
Present Value of an Annuity • The present value P of an annuity consisting of n payments of R dollars each, paid at the end of each investment period into an account that earns interest at the rate of i per period, is 1 (1 i ) n P R i
Example • Find the present value of an ordinary annuity consisting of 24 monthly payments of $100 each and earning interest of 9% per year compounded monthly. Solution • Here, R = 100, i = r/m = 0.09/12 = 0.0075, and n = 24, so 1 (1 i ) n 1 (1 0.0075) 24 P R 100 2188.91 i 0.0075
or $2188.91.
Applied Example: Saving for a College Education
• As a savings program towards Alberto’s college education, his parents decide to deposit $100 at the end of every month into a bank account paying interest at the rate of 6% per year compounded monthly. • If the savings program began when Alberto was 6 years old, how much money would have accumulated by the time he turns 18?
Applied Example: Saving for a College Education Solution • By the time the child turns 18, the parents would have made (18 6) 12 144 deposits into the account, so n = 144. • Furthermore, we have R = 100, r = 0.06, and m = 12, 0.06 so i 0.005 12
(1 i )n 1 (1 0.005)144 1 S R 100 21,015 i 0.005
• Using the future value of an annuity formula, we get
Applied Example: Financing a Car
• After making a down payment of $4000 for an automobile, Murphy paid $400 per month for 36 months with interest charged at 12% per year compounded monthly on the unpaid balance. • What was the original cost of the car? • What portion of Murphy’s total car payments went toward interest charges?
Applied Example: Financing a Car Solution • The loan taken up by Murphy is given by the present value of the annuity formula 1 (1 i ) n 1 (1 0.01) 36 P R 400 12,043 i 0.01
or $12,043. • Therefore, the original cost of the automobile is $16,043 ($12,043 plus the $4000 down payment). • The interest charges paid by Murphy are given by (36)(400) – 12,043 = 2,357 or $2,357.
Time value of money
1
Definition: time value of money is the premise that an investor prefers to receive a payment of a fixed amount of money today, rather than an equal amount in the future, all else being equal.
• Which would you rather have -- $1,000 today or $1,000 in 5 years? • Money received sooner rather than later allows one to use the funds for investment or consumption purposes.
• All other factors being equal, it is better to have $1,000 today. • Simply put this is the concept of the time value of money.
2
Importance of Time Factor
Why is TIME such an important element in your decision?
TIME allows one the opportunity to postpone consumption and earn INTEREST.
3
4
5
6
7
8
9
10
11
12
13
14
15
Calculations based on the time value of money
• Present Value (PV) of an amount that will be received in the future. • Future Value (FV) of an amount invested (such as in a deposit account) now at a given rate of interest. • Present Value of an Annuity (PVA) • Future Value of an Annuity (FVA)
16
Notations • PV (Present Value) is the value at time = 0 • FV (Future Value) is the value at time = n
• ‘r’ is the rate at which the amount will be compounded each period • ‘n’ is the number of periods
17
Notations
• PV(A) the value of the annuity at time = 0 • FV(A) the value of the annuity at time = n • ‘A’ the value of the individual payments in each compounding period
18
Formulas
• Present value of a future sum / Future value of a present sum
19
Formulas
• Present value of an annuity
20
Time value of money
• Consider 2 situations – Option A: You receive Rs. 10,000 today. – Option B: You receive Rs. 10,000 in 3 years time – Assume no inflation – Assume interest rate 10% (Compound Interest) – Assume no change in any other financial situation
21
Future Value Calculation
• Consider Option A • Let’s calculate the future value of Rs. 10,000 received at the present time.
22
Future Value Calculation
23
Present Value Calculation
• Similarly using the equation as
the present value of Rs. 10,000 received in 3 years when the interest rate is 10% can be calculated as Rs. 7513.1
24
Time Value of Money
25
Time Value Calculations using Tables
26
Benefits of the knowledge of the time value of money
• For investment analysis – To decide the financial benefits of projects • To compare investment alternatives • To analyze how time impacts business activities such as loans, mortgages, leases, savings, and annuities.
27
SPPW, SPCA, Equal Payment series, Uniform gradient
1
1. Single Payment Compound Amount (SPCA) • This formula simply gives the future value (S) of an amount of money with present value (P) after n periods at an interest rate i. S= P* SPCA at i,n
Where SPCA= (1+i)^n
4/22/2017
2
3
4
1. Single Payment Compound Amount (SPCA)
4/22/2017
5
2. Single Payment Present Worth (SPPW) • This gives the present value P in terms of future value S after n periods at an interest rate i and is just opposite of SPCA.
Where SPPW equals
4/22/2017
6
7
8
2. Single Payment Present Worth (SPPW)
4/22/2017
Dept.of EEE, SDMIT, Ujire
9
10
11
12
13
14
15
16
17
18
19
20
Example • A person is planning for his retired life. He has 10 more years of service. He would like to deposit 20% of his salary, which is Rs. 4,000, at the end of the first year, and thereafter he wishes to deposit the amount with an annual increase of Rs. 500 for the next 9 years with an interest rate of 15%. Find the total amount at the end of the 10th year of the above series.
21
22
23
Example • A person is planning for his retired life. He has 10 more years of service. He would like to deposit Rs. 8,500 at the end of the first year and thereafter he wishes to deposit the amount with an annual decrease of Rs. 500 for the next 9 years with an interest rate of 15%. Find the total amount at the end of the 10th year of the above series.
24
3. Uniform Series Compound Amount (USCA) • This will determine the amount S that an equal annual payment R will total in n years at i interest
4/22/2017
25
3. Uniform Series Compound Amount (USCA)
4/22/2017
26
4.Sinking Fund Payment (SFP) • This will determine the equal annual amount R that must be invested for n years at i interest to accumulate a specified future amount. • This is opposite of Uniform Series Compound Amount (USCA)
4/22/2017
27
4.Sinking Fund Payment (SFP)
4/22/2017
28
5. Uniform Series Present Worth (USPW) • This will determine the present amount P that can be paid by equal payment of R at i interest over n years . • USPW= USCA*SPPW
P= R* USPW at i ,n
USCA- Uniform Series Compound Amount SPPW- Single Payment Present Worth
29
5. Uniform Series Present Worth (USPW)
30
6. Capital Recovery (CR) • This will determine an annual payment R required to pay off a present amount P at i interest for n years • Capital Recovery technique is also opposite of Uniform Series Present Worth (USPW)
Where
31
6. Capital Recovery (CR)
32
7.Gradient Present Worth (GPW) • This will determine the present amount P that can be paid by annual amounts R which escalates at e% at i interest for n years •
33
7.Gradient Present Worth (GPW)
Percentage at which an annual change in the price levels of the goods and services occurs or is expected to occur is called escalation rate. Eg. Inflation & deflation of money will influence on escalation rate.
34
Payback Analysis • Simple payback is defined as the investment divided by annual saving after taxes. • The duration require to recover the capital amount/investment is called payback period. Payback period= (Net investment or capital cost) (Net annual cash flow or net annual saving)
35
Payback Analysis • If the sum of difference between capital cash flow to overall cost is positive implies the project will accept.Else the project will be rejected.
36
Payback Analysis Advantages • Can be able to judge availability of financial resource. • Helps to achieve successful investments for a project. Disadvantages • Time value of money is not directly considered • Effect of cash flow after pay back is not considered.
37
Effective Interest Rate • Let i be the nominal interest rate compounded annually. But, in practice, the compounding may occur less than a year. For example, compounding may be monthly, quarterly, or semi-annually. Compounding monthly means that the interest is computed at the end of every month. There are 12 interest periods in a year if the interest is compounded monthly. Under such situations, the formula to compute the effective interest rate, which is compounded annually, is
38
39
40
41
42
43
44
45
46
47
Comparison of Alternatives
1
• •
•
•
• • •
•
For most of the engineering projects, equipments etc., there are more than one feasible alternative. It is the duty of the project management team (comprising of engineers, designers, project managers etc.) of the client organization to select the best alternative that involves less cost and results more revenue. For this purpose, the economic comparison of the alternatives is made. The different cost elements and other parameters to be considered while making the economic comparison of the alternatives are initial cost, annual operating and maintenance cost, annual income or receipts, expected salvage value, income tax benefit and the useful life. When only one, among the feasible alternatives is selected, the alternatives are said to be mutually exclusive. As already mentioned in module-1, the cost or expenses are generally known as cash outflows whereas revenue or incomes are generally considered as cash inflows. Thus in the economic comparison of alternatives, cost or expenses are considered as negative cash flows. On the other hand the income or revenues are considered as positive cash flows. From the view point of expenditure incurred and revenue generated, some projects involve initial capital investment i.e. cash outflow at the beginning and show increased income or revenue i.e. cash inflow in the subsequent years. The alternatives having this type of cash flow are known as investment alternatives. So while comparing the mutually exclusive investment alternatives, the alternative showing maximum positive cash flow is generally selected.
2
• •
In this case, the investment is made at the beginning to gain profit at the future period of time. Example for such type alternatives includes purchase of a dozer by a construction firm. The construction firm will have different feasible alternatives for the dozer with each alternative having its own initial investment, annual operating and maintenance cost, annual income depending upon the production capacity, useful life, salvage values etc.
•
Thus the alternative which will yield more economic benefit will be selected by the
•
• •
•
construction firm. There are some other projects which involve only costs or expenses throughout the useful life except the salvage value if any, at the end of the useful life. The alternatives having this type of cash flows are known as cost alternatives. Thus while comparing mutually exclusive cost alternatives, the alternative showing minimum negative cash flow is generally selected. Example for such type alternatives includes construction of a government funded national highway stretch between two regions. For this project there will be different feasible alternatives depending upon length of the stretch, type of pavement, related environmental, social and regulatory aspects etc. Each alternative will have its initial cost of construction, annual repair and maintenance cost and some major repair cost if any, at some future point of time. The alternative that will exhibit lowest cost will be selected for the construction of the highway stretch.
3
•
•
•
•
The differences in different parameters namely initial capital investment, annual operation cost, annually generated revenue, expected salvage value, useful life, magnitude of output and its quality, performance and operational characteristics etc. may exist among the mutually exclusive alternatives. Thus the economic analysis of the mutually exclusive alternatives is generally carried out on the similar or equivalent basis since each of the feasible alternatives will meet the desired requirements of the project, if selected. The economic comparison of mutually exclusive alternatives can be carried out by different equivalent worth methods namely present worth method, future worth method and annual worth method. In these methods all the cash flows i.e. cash outflows and cash inflows are converted into equivalent present worth, future worth or annual worth considering the time value of money at a given interest rate per interest period.
4
•
• •
•
•
Comparison of alternatives by present worth method In the present worth method for comparison of mutually exclusive alternatives, the future amounts i.e. expenditures and incomes occurring at future periods of time are converted into equivalent present worth values at a certain rate of interest per interest period and are added to present worth occurring at „0‟ time. The converted equivalent present worth values are always less than the respective future amounts since the rate of interest is normally greater than zero. The cash flow of the mutually exclusive alternatives may consist of future expenditures and incomes in different forms namely randomly placed single amounts, uniform amount series commencing from end of year 1, randomly placed uniform amount series i.e. commencing at time period other than end of year 1, positive and negative uniform gradient series starting either from end of year 1 or at different time periods and geometric gradient series etc. The different compound interest factors namely single payment present worth factor, uniform series present worth factor and present worth factors for arithmetic and geometric gradient series etc. will be used to convert the respective future amounts to the equivalent present worth values for different alternatives. The methodology for the comparison of mutually exclusive alternatives by the present worth method depends upon the magnitude of useful lives of the alternatives. There are two cases; a) the useful lives of alternatives are equal and b) the useful lives of alternatives are not equal.
5
•
•
•
The alternatives having equal useful lives are designated as equal life span alternatives whereas the alternatives having unequal life spans are referred as different life span alternatives. a) Equal life span alternatives The comparison of mutually exclusive alternatives having equal life spans by present worth method is comparatively simpler than those having different life spans. In case of equal life span mutually exclusive alternatives, the future amounts as already stated are converted into the equivalent present worth values and are added to the present worth occurring at time zero. Then the alternative that exhibits maximum positive equivalent present worth or minimum negative equivalent present worth is selected from the considered feasible alternatives.
6
•
• •
• •
•
•
a) Different life span alternatives In case of mutually exclusive alternatives, those have different life spans, the comparison is generally made over the same number of years i.e. a common study period. This is because; the comparison of the mutually exclusive alternatives over same period of time is required for unbiased economic evaluation of the alternatives. If the comparison of the alternatives is not made over the same life span, then the cost alternative having shorter life span will result in lower equivalent present worth i.e. lower cost than the cost alternative having longer life span. Because in this case, the cost of the short span alternative is considered only for a shorter period of time, even though this alternative may not be economical. In case of mutually exclusive investment alternatives, the alternative with longer life span will result in higher equivalent present worth i.e. higher positive equivalent worth, as the costs, revenues, savings through reduced costs is considered over a longer period of time than the alternative with shorter life span. Thus in order to minimize the effect of such kind of discrepancy on the selection of best alternative from the considered feasible alternatives, the comparison is made over the same life span. The two approaches used for economic comparison of different life span alternatives are as follows; i) Comparison of mutually exclusive alternatives over a time period that is equal to least common multiple (LCM) of the individual life spans ii) Comparison of mutually exclusive alternatives over a study period which is not necessarily equal to the life span of any of the alternatives.
7
•
In the first approach the comparison is made over a time period equal to the least common multiple of the life spans of mutually exclusive alternatives. The cash flow of the alternatives i.e. cash flow of the first cycle is repeated and the number of repetitions depends upon the value of least common multiple of life spans between the mutually exclusive alternatives. It may be noted here that the cash flow i.e. all the costs and revenues of the alternatives in the successive cycle will be exactly same as that in the first cycle. For example if there are two alternatives with useful lives of 4 years and 5 years.
8
COMPARING COST ALTERNATIVES • For cost alternatives that are compared using the PW method, the alternative that has the least negative PW is most economically desirable. • For cost alternatives that are compared using the AW method, the alternative that has the least negative AW is most economically desirable. • For cost alternatives that are compared using the FW method, the alternative that has the least negative FW is most economically desirable.
9
THREE GROUPS OF MAJOR INVESTMENT ALTERNATIVES
1. Mutually exclusive : At most one project out of the group can be chosen 2. Independent : The choice of a project is independent of the choice of any other project in the group, so that all or none of the projects may be selected or some number in between 3. Contingent : The choice of the project is conditional on the choice of one or more other projects
10
COTERMINATED ASSUMPTION • You cannot compare alternatives using PW method if life is not equal. • A finite and identical study period is used for all alternatives. Cash flows are repeated per the least common multiple of lives. • A planning horizon is proposed when AW method is used. Truncate the remaining useful life by substituting a salvage value. This planning horizon, combined with appropriate adjustments to the estimated cash flows, puts the alternatives on a common comparable basis
11
PLANNING HORIZON • The selected time period over which mutually exclusive alternatives are compared -- study period • May be influenced by factors including: – service period required – useful life of the shorter-lived alternative – useful life of the longer-lived alternative – company policy • It is key that the study period be appropriate for the decision situation under investigation
12
COTERMINATED ASSUMPTION Guidelines when useful life(s) different in length than study period • Useful life < study period a. Cost alternatives -- each cost alternative must provide same level of service as study period : 1) contract for service or lease equipment for remaining time; 2) repeat part of useful life of original alternative until study period ends b. investment alternatives -- assume all cash flows reinvested in other opportunities at MARR to end of study period
13
Comparison by present worth method
1
Comparison by present worth method •
• • •
Now some examples showing the use of present worth method for comparison of mutually exclusive alternatives are presented. First the comparison of equal life span mutually exclusive alternatives by present worth method will be illustrated followed by comparison of different life span alternatives. The following examples are formulated only to demonstrate the use of different methods for comparison of alternatives. The values of different cost and incomes mentioned in the examples are not the actual ones pertaining to a particular item. In addition it may also be noted here that the cash flow diagrams have been drawn not to the scale. These are merely graphical representations.
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Examples • • •
•
•
There are two alternatives for purchasing a concrete mixer. Both the alternatives have same useful life. The cash flow details of alternatives are as follows; Alternative-1: Initial purchase cost = Rs.3,00,000, Annual operating and maintenance cost = Rs.20,000, Expected salvage value = Rs.1,25,000, Useful life = 5 years. Alternative-2: Initial purchase cost = Rs.2,00,000, Annual operating and maintenance cost = Rs.35,000, Expected salvage value = Rs.70,000, Useful life = 5 years. Using present worth method, find out which alternative should be selected, if the rate of interest is 10% per year.
22
Solution: •
Since both alternatives have the same life span i.e. 5years, the present worth of the alternatives will be compared over a period of 5 years. The cash flow diagram of Alternative-1 is shown in Fig. 2.1. As already mentioned Module-1, the cash outflows i.e. costs or expenditures are represented by vertically downward arrows whereas the cash inflows i.e. revenue or income are represented by vertically upward arrows. The same convention is adopted here.
23
24
• • • • • •
The equivalent present worth of Alternative-1 i.e. PW1 is calculated as follows; The initial cost, P = Rs.3,00,000 (cash outflow), Annual operating and maintenance cost, A = Rs.20,000 (cash outflow), Salvage value, F = Rs.1,25,000 (cash inflow). PW1 = - 3,00,000 – 20,000(P/A, i, n) + 1,25,000(P/F, i, n) PW1 = - 3,00,000 – 20,000(P/A, 10%, 5) + 1,25,000(P/F, 10%, 5) Now putting the mathematical expressions of different compound interest factors (as mentioned in Module-1) in the above expression for PW1 (in Rs.) results in the following;
25
26
27
• • • • • •
Now the equivalent present worth of Alternative-2 i.e. PW2 (in Rs.) is calculated as follows; The initial cost, P = Rs.2,00,000 (cash outflow), Annual operating and maintenance cost, A = Rs.35,000 (cash outflow), Salvage value, F = Rs.70,000 (cash inflow). PW2 = - 2,00,000 – 35,000(P/A, i, n) + 70,000(P/F, i, n) PW2 = - 2,00,000 – 35,000(P/A, 10%, 5) + 70,000(P/F, 10%, 5)
28
29
•
Comparing the equivalent present worth of both the alternatives, it is observed that Alternative-2 will be selected as it shows lower negative equivalent present worth compared to Alternative-1 at the interest rate of 10% per year.
30
equivalent present worth •
•
The equivalent present worth of both the alternatives can also be calculated by using the values of compound interest factors from interest tables. The equivalent present worth of Alternative-1 i.e. PW1 is calculated as follows;
31
•
•
•
The values of compound interest factors i.e. (P/A, i, n) and (P/F, i, n) can be obtained from the interest tables (discrete compounding) available in texts cited in the list of references [1, 7, 14]. Now referring to the interest table for 10% interest rate, the values of compound interest factors i.e. (P/A, 10%, 5) and (P/F, 10%, 5) at interest rate („i’ ) of 10% and for interest period („n’) of 5 years are obtained at the intersection of these factors and interest period „n‟ equal to 5 i.e. the values are obtained from P/A column and P/F column at „n‟ equal to 5 from the interest table (discrete compounding) corresponding to 10% interest rate. The obtained values of (P/A, 10%, 5) and (P/F, 10%, 5) are 3.7908 and 0.6209 respectively (same as those obtained using mathematical expressions of these factors).
32
•
Now putting the values of compound interest factors in the above expression, the equivalent present worth of Alternative-1 i.e. PW1 is calculated as follows;
33
•
Now the calculation of equivalent present worth of Alternative-2 i.e. PW2 (in Rs.) is presented below.
•
Now putting the values of compound interest factors in the above expression (same as above) the equivalent present worth of Alternative-2 i.e. PW2 is calculated as follows;
•
It may be noted that in the above example only cost components and the salvage value of the alternatives were considered for comparison. In the next example, same problem as mentioned in Example-1 will be discussed by taking into account the annual revenues of the alternatives. 34
Example 2 • • • • • • • • • •
•
Alternative-1: Initial purchase cost = Rs.300000, Annual operating and maintenance cost = Rs.20000, Expected salvage value = Rs.125000, Useful life = 5 years. Alternative-2: Initial purchase cost = Rs.200000, Annual operating and maintenance cost = Rs.35000, Expected salvage value = Rs.70000, Useful life = 5 years. The annual revenue to be generated from production of concrete (by concrete mixer) from Alternative-1 and Alternative-2 are Rs.50000 and Rs.45000 respectively. Compute the equivalent present worth of the alternatives at the same rate of interest as in Example- 1 i.e. 10% per year and find out the economical alternative. 35
Solution: •
The cash flow diagram of Alternative-1 is shown in
36
37
38
39
•
• •
Comparing the equivalent present worth of the both the alternatives, it is observed that Alternative-1 will be selected as it shows lower cost compared to Alternative-2. The annual revenue to be generated by the alternatives made the difference as compared to the outcome obtained in Example-1. When there are more than two alternatives for the selection of the best economical alternative by present worth method, the same procedure as mentioned earlier for the case of two alternatives is followed and illustrated in the next example.
40
Example -3 •
• • •
• •
A construction contractor has three options to purchase a dump truck for transportation and dumping of soil at a construction site. All the alternatives have the same useful life. The cash flow details of all the alternatives are provided as follows; Option-1: Initial purchase price = Rs.2500000, Annual operating cost Rs.45000 at the end of 1st year and increasing by Rs.3000 in the subsequent years till the end of useful life, Annual income = Rs.120000, Salvage value = Rs.550000, Useful life = 10 years.
41
• • • • • • • • • • • •
Option-2: Initial purchase price = Rs.3000000, Annual operating cost = Rs.30000, Annual income Rs.150000 for first three years and increasing by Rs.5000 in the subsequent years till the end of useful life, Salvage value = Rs.800000, Useful life = 10 years. Option-3: Initial purchase price = Rs.2700000, Annual operating cost Rs.35000 for first 5 years and increasing by Rs.2000 in the successive years till the end of useful life, Annual income = Rs.140000, Expected salvage value = Rs.650000, Useful life = 10 years. Using present worth method, find out which alternative should be selected, if the rate of interest is 8% per year. 42
Solution:
• The cash flow diagram of Option-1 is shown in
43
•
For Option-1, the annual operating cost is in the form of a positive uniform gradient series with gradient starting from end of year „2‟. The operating cost at the end of different years can be split into the uniform base amount of Rs.45000 and the gradient amount in multiples of Rs.3000 as shown in Fig. 2.6.
44
•
The present worth of the uniform gradient series will be located at the beginning i.e. in year „0‟ i.e. 2 years before the commencement of the uniform gradient.
45
•
Now putting the values of different compound interest factors (the expressions in terms of ‘i’ and ‘n’ already stated in Module-1) in the above expression for PW1 results in the following;
46
47
•
For Option-2, the annual income is in the form of a positive uniform gradient series with gradient starting from end of year „4‟. The annual income can be split into the uniform base amount of Rs.150000 and the gradient amount in multiples of Rs.5000 starting from end of year „4‟ and is shown in Fig. 2.8.
48
•
•
The equivalent present worth of the gradient series (of the annual income) starting from end of year „4‟ will be located at the end of year „2‟ i.e. 2 years before the start of the gradient. Further the present worth of this amount at beginning i.e. at time „0‟ will be obtained by multiplying the equivalent present worth „Pg’ (shown in Fig. 2.8) at the end of year „2‟ (which is a future amount) with the single payment present worth factor (P/F, i, n). Now the equivalent present worth (in Rs.) of Option-2 is determined as follows;
49
50
51
• •
•
•
For Option-3, the annual operating cost is in the form of a positive uniform gradient series with gradient starting from end of year „6‟. The annual operating cost can thus be split into the uniform base amount of Rs.35000 and the gradient amount in multiples of Rs.2000 starting from end of year „6‟ (shown in Fig. 2.10). The equivalent present worth of the gradient series for the annual operating cost starting from end of year „6‟ will be located at the end of year „4‟. Further the present worth of this amount at time „0‟ will be determined by multiplying the equivalent present worth „Pg’ (shown in Fig. 2.10) at the end of year „4‟ with the single payment present worth factor (P/F, i, n).
52
53
54
•
From the comparison of equivalent present worth of all the three mutually exclusive alternatives, it is observed that Option-3 shows lowest negative equivalent present worth as compared to other options. Thus Option-3 will be selected for the purchase of the dump truck.
55
Comparison by present worth method:•
After the illustration of comparison of equal life span mutually exclusive alternatives, now some examples illustrating the use of present worth method for comparison of different life span mutually exclusive alternatives are presented.
56
Example •
• • •
• • •
•
A material testing laboratory has two alternatives for purchasing a compression testing machine which will be used for determining the compressive strength of different construction materials. The alternatives are from two different manufacturing companies. The cash flow details of the alternatives are as follows; Alternative-1: Initial purchase price = Rs.1000000, Annual operating cost = Rs.10000, Expected annual income to be generated from testing of different construction materials = Rs.175000, Expected salvage value = Rs.200000, Useful life = 10 years. Alternative-2: Initial purchase price = Rs.700000, Annual operating cost = Rs.15000, Expected annual income to be generated from testing of different construction materials = Rs.165000, Expected salvage value = Rs.250000, Useful life = 5 years. Using present worth method, find out the most economical alternative at the interest rate of 10% per year. 57
Solution •
• •
•
The alternatives have different life spans i.e. 10 years and 5 years. Thus the comparison will be made over a time period equal to the least common multiple of the life spans of the alternatives. In this case the least common multiple of the life spans is 10 years. Thus the cash flow of Alternative-1 will be analyzed for one cycle (duration of 10 years) whereas the cash flow of Alternative-2 will be analyzed for two cycles (duration of 5 years for each cycle). The cash flow of the Alternative-2 for the second cycle will be exactly same as that in the first cycle.
58
59
•
The cash flow diagram of Alternative-2 is shown in Fig. 2.12. As the least common multiple of the life spans of the alternatives is 10 years, the cash flow of Alternative-2 is shown for two cycles with each cycle of duration 5 years. 60
61
•
•
In the cash flow diagram of Alternative-2, the initial purchase price of Rs.700000 is again located at the end of year „5‟ i.e. at the end of first cycle or the beginning of the second cycle. In addition the annual operating cost and the annual income are also repeated in the second cycle from end of year „6‟ till end of year „10‟. Further the salvage value of Rs.250000 is also located at end of year „10‟ i.e. at the end of second cycle.
62
•
•
Thus from the comparison of equivalent present worth of the alternatives, it is evident that Alternative-1 will be selected for purchase of the compression testing machine as it shows the higher positive equivalent present worth. In the following example, the comparison of different life span mutually exclusive alternatives having expenditure or income in the form of gradient series by present worth method is illustrated.
63
Example •
• • • • •
A construction firm has decided to purchase a dozer to be employed at a construction site. Two different companies manufacture the dozer that will fulfill the functional requirement of the construction firm. The construction firm will purchase the most economical one from one of these companies. The alternatives have different useful lives. The cash flow details of both alternatives are presented as follows; Company-A Dozer: Initial purchase cost = Rs.3050000, Annual operating cost Rs.40000 at end of 1st year and increasing by Rs.2000 in the subsequent years till the end of useful life, Annual income = Rs.560000, Expected salvage value = Rs.1050000, Useful life = 6 years.
64
• • •
• • •
Company-B Dozer: Initial purchase cost = Rs.4000000, Annual operating cost = Rs.55000, Annual revenue to be generated Rs.600000 at the end of 1st year and increasing by Rs.5000 in the subsequent years till the end of useful life, Expected salvage value = Rs.1000000, Useful life = 12 years. Using present worth method, find out the most economical alternative at the interest rate of 7% per year.
65
Solution: •
•
•
Since the alternatives have different life spans i.e. 6 and 12 years, the comparison will be made over a time period equal to the least common multiple of the life spans of the alternatives i.e. 12 years. The cash flow of Company-A Dozer will be analyzed for two cycles i.e. duration of 6 years for each cycle. The cash flow of Company-B Dozer will be analyzed for one cycle i.e. duration of 12 years. The cash flow diagram of Company-A Dozer is shown in Fig. 2.13. Since the least common multiple of the life spans of the alternatives is 12 years, the cash flow is shown for two cycles.
66
67
•
• •
•
For Company-A Dozer, the annual operating cost is in the form of a positive uniform gradient series which can be split into the uniform base amount of Rs.40000 and the gradient amount in multiples of Rs.2000 starting from end of year „2‟ for first cycle as shown in Fig. 2.14. The equivalent present worth of this gradient for cycle one will be located at the beginning i.e. in year „0‟. However for second cycle, the equivalent present worth of the gradient for the annual operating cost starting from end of year „8‟ (shown in Fig. 2.14) will be located at the end of year „6‟. Further the present worth of this amount at time „0‟ will be determined by multiplying the equivalent present worth of the gradient at the end of year „6‟ with the single payment present worth factor (P/F, i, n).
68
69
70
71
•
For Company-B Dozer, the annual revenue is in the form of a positive uniform gradient series that can be split into the uniform base amount of Rs.600000 and gradient amount in multiples of Rs.5000 as shown in Fig. 2.16. The equivalent present worth of this gradient amount will be located at the beginning i.e. in year „0‟.
72
73
Future worth method
1
Future worth method • •
Future worth method is used particularly in an investment situation where we need to compute the equivalent worth of the project at the end of its investment period •For Eg : Building a nuclear power plant, where it is time consuming. In such situation it is more common to measure the worth of the investment at the time of commercialization
2
3
4
5
6
7
• In the future worth method for comparison of mutually exclusive alternatives, the equivalent future worth (i.e. value at the end of the useful lives of alternatives) of all the expenditures and incomes occurring at different periods of time are determined at the given interest rate per interest period. • As already mentioned, the cash flow of the mutually exclusive alternatives may consist of expenditures and incomes in different forms. • Therefore the equivalent future worth of these expenditures and incomes will be determined using different compound interest factors namely single payment compound amount factor, uniform series compound amount factor and future worth factors for arithmetic and geometric gradient series etc.
8
• The use of future worth method for comparison of mutually exclusive alternatives will be illustrated in the following examples. Similar to present worth method, first the comparison of equal life span alternatives by future worth method will be illustrated followed by comparison of different life span alternatives. • Some of the examples already worked out by the present worth method will be illustrated using the future worth method in addition to some other examples.
9
Example • There are two alternatives for purchasing a concrete mixer. Both the alternatives have same useful life. The cash flow details of alternatives are as follows; • Alternative-1: Initial purchase cost = Rs.300000, • Annual operating and maintenance cost = Rs.20000, • Expected salvage value = Rs.125000, • Useful life = 5 years. Alternative-2: Initial purchase cost = Rs.200000, • Annual operating and maintenance cost = Rs.35000, • Expected salvage value = Rs.70000, • Useful life = 5 years. • Using future worth method, find out which alternative should be selected, if the rate of interest is 10% per year.
10
• The future worth of the mutually exclusive alternatives will be compared over a period of 5 years. • The equivalent future worth of the alternatives can be obtained either by multiplying the equivalent present worth of each alternative already obtained by present worth method with the single payment compound amount factor or determining the future worth of expenditures and incomes individually and adding them to get the equivalent future worth of each alternative.
11
12
13
• Now it can be seen that the calculated future worth of Alternative-1 by both ways is same. The minor difference between the values is due to the effect of decimal points in the calculations.
14
15
• Thus the future worth of Alternative-2 obtained by both methods is same. In this case also the minor difference between the values is due to the effect of the decimal points in the calculations. • Comparing the equivalent future worth of the both the alternatives, it is observed that Alternative-2 will be selected as it shows lower negative equivalent future worth as compared to Alternative-1. • This outcome of the comparison of the alternatives by future worth method is same as that obtained from the present worth method (Example-1). This is due to the equivalency relationship between present worth and future worth through compound interest factors at the given rate of interest per interest period.
16
• There are two alternatives for a construction firm to purchase a road roller which will be used for the construction of a highway section. The cash flow details of the alternatives are as follows; • Alternative-1: Initial purchase cost = Rs.1500000, • Annual operating cost = Rs.35000 starting from the end of year „2‟ (negligible in the first year) till the end of useful life, • Annual revenue to be generated = Rs.340000 for first 4 years and then Rs.320000 afterwards till the end of useful life, • Expected salvage value = Rs.430000, • Useful life = 8 years. • Alternative-2: Initial purchase cost = Rs.1800000, • Annual operating cost = Rs.25000, • Annual revenue to be generated = Rs.365000, • Expected salvage value = Rs.550000, Useful life = 8 years. • Find out the most economical alternative on the basis of equivalent future worth at the interest rate of 9.5% per year. 17
Solution
18
• From Fig. 2.17, it is observed that there are two uniform amount series for the annual income i.e. first series with Rs.340000 from end of year „1‟ till end of year „4‟ and second one with Rs.320000 from end of year „5‟ till end of year „8‟. • For the first series, the equivalent present worth at time „0‟ will be calculated first and then it will be multiplied with single payment compound amount factor i.e. (F/P, i, n) to calculate its equivalent future worth. • For the second uniform series with Rs.320000, the future worth will be calculated by multiplying the uniform amount i.e. Rs.320000 with uniform series compound amount factor by taking the appropriate „n’ i.e. number of years. • The annual operating cost is in the form of a uniform amount series, which starts from end of year „2‟ till the end of useful life i.e. the uniform amount series is shifted by one year.
19
The equivalent future worth of the Alternative-1 i.e.FW1 is computed as follows;
20
21
• Comparing the equivalent future worth of the alternatives, it is observed that Alternative- 1 shows higher positive equivalent future worth as compared to Alternative-2. Thus Alternative-1 will be selected for purchase of the road roller.
22
Comparison by future worth method • In the following example, the comparison of three mutually exclusive alternatives by future worth method will be illustrated. The data presented in Example-3 will be used for comparison of the alternatives by the future worth method.
23
• A construction contractor has three options to purchase a dump truck for transportation and dumping of earth at a construction site. All the alternatives have the same useful life. The cash flow details of all the alternatives are presented as follows; • Option-1: Initial purchase price = Rs.2500000, • Annual operating cost Rs.45000 at the end of 1st year and increasing by Rs.3000 in the subsequent years till the end of useful life, • Annual income = Rs.120000, • Salvage value = Rs.550000, • Useful life = 10 years. • Option-2: Initial purchase price = Rs.3000000, • Annual operating cost = Rs.30000, • Annual income Rs.150000 for first three years and increasing by Rs.5000 in the subsequent years till the end of useful life, • Salvage value = Rs.800000, • Useful life = 10 years. 24
• Option-3: • Initial purchase price = Rs.2700000, • Annual operating cost Rs.35000 for first 5 years and increasing by Rs.2000 in the successive years till the end of useful life, • Annual income = Rs.140000, • Expected salvage value = Rs.650000, • Useful life = 10 years. • Using future worth method, find out which alternative should be selected, if the rate of interest is 8% per year
25
26
27
28
29
30
• For the annual operating cost, the equivalent present worth of the gradient series starting from end of year „6‟ will be located at the end of year „4‟. • The future worth of this amount at end of year „10‟ will be determined by multiplying the equivalent present worth „Pg’ (shown in Fig. 2.10) at the end of year „4‟ with the single payment compound amount factor (F/P, i, n).
31
32
• Comparing the equivalent future worth of all the three alternatives, it is evident that Option-3 shows lowest negative equivalent future worth as compared to other options. • Thus Option-3 will be selected for the purchase of the dump truck. This outcome obtained by future worth method is same as that obtained from the present worth method (Example-3) i.e. • Option-3 is the most economical alternative. After carrying out the comparison of equal life span mutually exclusive alternatives, now the illustration of future worth method for comparison of different life span mutually exclusive alternatives is presented.
33
• A material testing laboratory has two alternatives for purchasing a compression testing machine which will be used for determining the compressive strength of different construction materials. The alternatives are from two different manufacturing companies. The cash flow details of the alternatives are as follows; • Alternative-1: • Initial purchase price = Rs.1000000, • Annual operating cost = Rs.10000, • Expected annual income to be generated from testing of different construction materials = Rs.175000, • Expected salvage value = Rs.200000, Useful life = 10 years. • Alternative-2: Initial purchase price = Rs.700000, • Annual operating cost = Rs.15000, • Expected annual income to be generated from testing of different construction materials = Rs.165000, Expected salvage value = Rs.250000, Useful life = 5 years. Find out the most economical alternative at interest rate of 10% per year by future worth method. 34
• As the alternatives have different life spans i.e. 10 years and 5 years, the comparison will be made over a time period equal to the least common multiple of the life spans of the alternatives i.e. 10 years. Thus the cash flow of Alternative-1 is analyzed for one cycle (duration of 10 years) whereas that of cash flow of Alternative-2 is analyzed for two cycles of duration 5 years each (already mentioned in Example-4).
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Annual Equivalent method
1
Annual Equivalent method • •
In the annual equivalent method of comparison, first the annual equivalent cost or the revenue of each alternative will be computed. Then the alternative with the maximum annual equivalent revenue in the case of revenue-based comparison or with the minimum annual equivalent cost in the case of cost- based comparison will be selected as the best alternative.
2
3
4
5
6
7
8
9
10
11
12
Revenue-Dominated Cash Flow Diagram •
A generalized revenue-dominated cash flow diagram to demonstrate the annual equivalent method of comparison is presented in Fig.
13
• •
•
In Fig. P represents an initial investment, Rj the net revenue at the end of the j th year, and S the salvage value at the end of the nth year. The first step is to find the net present worth of the cash flow diagram using the following expression for a given interest rate, i:
In the above formula, the expenditure is assigned with a negative sign and the revenues are assigned with a positive sign.
14
Cost-Dominated Cash Flow Diagram •
A generalized cost-dominated cash flow diagram to demonstrate the annual equivalent method of comparison is illustrated in Fig.
15
•
•
In Fig, P represents an initial investment, Cj the net cost of operation and maintenance at the end of the jth year, and S the salvage value at the end of the nth year. The first step is to find the net present worth of the cash flow diagram using the following relation for a given interest rate, i.
16
17
18
19
20
Example •
A company provides a car to its chief executive. The owner of the company is concerned about the increasing cost of petrol. The cost per litre of petrol for the first year of operation is Rs. 21. He feels that the cost of petrol will be increasing by Re.1 every year. His experience with his company car indicates that it averages 9 km per litre of petrol. The executive expects to drive an average of 20,000 km each year for the next four years. What is the annual equivalent cost of fuel over this period of time?. If he is offered similar service with the same quality on rental basis at Rs. 60,000 per year, should the owner continue to provide company car for his executive or alternatively provide a rental car to his executive? Assume i = 18%. If the rental car is preferred, then the company car will find some other use within the company.
21
• • • •
Solution Average number of km run/year = 20,000 km Number of km/litre of petrol = 9 km Therefore,
22
23
•
The proposal of using the company car by spending for petrol by the company will cost an annual equivalent amount of Rs. 49,543.28 for four years. This amount is less than the annual rental value of Rs. 60,000. Therefore, the company should continue to provide its own car to its executive.
24
Comparison by annual worth method: •
Now the comparison of alternatives with cash flows involving gradient series and randomly placed single amount by annual worth method will be illustrated followed by the comparison of different life span alternatives.
25
•
• • • • •
•
Compare the following equipment on the basis of the equivalent uniform annual worth and find out the most economical one at the interest rate of 9.5% per year. Equipment-A Cash flow details: Initial purchase cost = Rs.5000000 Annual operating cost = Rs.60000 at the end of year „1‟ and increasing by Rs.3000 in the subsequent years till the end of useful life. Annual income = Rs.770000 Cost of one time major repair = Rs.200000 at the end of year „8‟ Expected salvage value = Rs.1400000 Useful life = 12 years
26
• • • • • •
Equipment-B Cash flow details: Initial purchase cost = Rs.4600000 Annual operating cost = Rs.75000 Annual income = Rs.710000 for the first 5 years and increasing by Rs.5000 in the subsequent years till the end of useful life. Cost of one time major repair = Rs.230000 at the end of year „6‟ Expected salvage value = Rs.1200000 Useful life = 12 years
27
28
•
The annual operating cost is in the form of a positive uniform gradient series. This can be split into the uniform base amount of Rs.60000 and gradient amount in multiples of Rs.3000 starting from end of year „2‟ till the end of useful life as shown in Fig. 2.22.
29
30
31
•
For Equipment-B, the annual income is in the form of a positive uniform gradient series with gradient starting from end of year „6‟. The annual income is split into the uniform base amount of Rs.710000 and the gradient amount in multiples of Rs.5000 starting from end of year „6‟ till the end of useful life as shown in Fig. 2.24.
32
•
• •
For the gradient series (of annual income), the equivalent present worth will be located at the end of year „4‟ i.e. 2 years before the start of the gradient. T hen the present worth of this amount at time „0‟ will be calculated and after that the equivalent annual worth of this amount will be determined. The equivalent uniform annual worth of Equipment-B is determined as follows;
33
•
Now putting the values of different compound interest factors in the above expression;
•
From the comparison by annual worth method, it is observed that Equipment-A exhibits higher positive equivalent uniform annual worth as compared to Equipment-B. Thus the Equipment-A is the most economical alternative.
34
• • • • • • • • • • • • •
Using the annual worth method, compare the following machines having different life spans at an interest rate of 11.5% per year. Machine-1 Cash flow details: Initial purchase price = Rs.1200000 Annual operating cost = Rs.38000 Annual revenue = Rs.210000 for first 6 years and then Rs.225000 afterwards till the end of useful life. Expected salvage value = Rs.320000 Useful life = 14 years Machine-2 Cash flow details: Initial purchase price = Rs.1400000 Annual operating cost = Rs.26000 Annual revenue = Rs.290000 Expected salvage value = Rs.450000 Useful life = 7 years
35
36
•
•
It may be noted here that the annual revenue is in the form of two uniform annual amount series i.e. one with annual amount of Rs.210000 from beginning till end of year „6‟ and the second one with annual amount of Rs.225000 from end of year „7‟ till the end of useful life (as shown in Fig. 2.25). The annual revenue in the cash flow diagram can also be represented as annual amount of Rs.210000 from beginning till the end of useful life and the annual amount of Rs.15000 from end of year „7‟ till the end of useful life as shown in Fig. 2.26.
37
•
•
For the annual revenue of Rs.15000 from end of year „7‟ till the end of useful life, first the equivalent future worth is calculated followed by the calculation of the equivalent annual worth. The equivalent uniform annual worth of Machine-1 is computed as follows;
38
39
•
As already stated, the equivalent uniform annual worth for more than one cycle of cash flow will be same as that for the first cycle provided the cash flow in the successive cycles is exactly same as that in the first cycle. For illustration of this note, the cash flow of Machine-2 is repeated for two cycles with life span of 7 years each (least common multiple of life spans of the alternatives is 14 years). The cash flow diagram of Machine-2 for two cycles is shown in Fig. 2.28
40
41
•
Now it can be seen that, the calculated equivalent uniform annual worth of Machine-2 for two cycles of cash flow is same as that with only one cycle i.e. first cycle. The minor difference among the values is due to the effect of decimal points in the above calculations.
42
43
44
45
46
47
48
49
50
51
52
53
54
55
Rate of Return
1
Rate of Return method • The rate of return of a cash flow pattern is the interest rate at which the present worth of that cash flow pattern reduces to zero. • In this method of comparison, the rate of return for each alternative is computed. Then the alternative which has the highest rate of return is selected as the best alternative. • A generalized cash flow diagram to demonstrate the rate of return method of comparison is presented in Fig
2
• In the above cash flow diagram, P represents an initial investment, Rj the net revenue at the end of the jth year, and S the salvage value at the end of the nth year. • The first step is to find the net present worth of the cash flow diagram using the following expression at a given interest rate, i.
3
• The rate of return technique is one of the methods used in selecting an alternative for a project. In this method, the interest rate per interest period is determined, which equates the equivalent worth (either present worth, future worth or annual worth) of cash outflows (i.e. costs or expenditures) to that of cash inflows (i.e. incomes or revenues) of an alternative. • The rate of return is also known by other names namely internal rate of return (IRR), profitability index etc. It is basically the interest rate on the unrecovered balance of an investment which becomes zero at the end of the useful life or the study period. • In the following lectures, the rate of return is denoted by “ir”. Using present worth, the equation for rate of return can be written as follows;
4
• As already stated in earlier lectures, cost or expenditures are considered as negative cash flows whereas income or revenues are considered as positive cash flows. Equation (2.1) can be rewritten as;
• In the above equation the net present worth is zero. Now putting the expressions for present worth of cash outflows and that of cash inflows in equation (2.1) results in the following expression;
• On left hand side of the above equation, Po is the initial cost at time zero and FC (single amount) and AC (uniform amount series) are the expenditures occurring at future period of time. Similarly on the right hand side of the equation, FI (single amount) and AI (uniform amount series) are the incomes or revenues occurring at future period of time. The value of interest period „n‟ will vary depending upon the occurrence of the future amounts (either expenditure or income). 5
• Equation (2.3) can be rewritten as follows;
• The value of rate of return „ir’ can be calculated by solving the above equation. The equation (2.4) can be solved either manually through trial and error process or using Microsoft Excel spreadsheet. • The first method i.e. trial and error process for determination of the rate of return consumes more time whereas the second method is faster. However the trial and error method gives a clear understanding of the analysis of calculation for the rate of return. • Similar to equivalent present worth, the rate of return can also be determined by finding out the interest rate at which the net future worth or net annual worth is zero.
6
• After determination of the rate of return for a given alternative, it is compared with minimum attractive rate of return (MARR) to find out the acceptability of this alternative for the project. • If the rate of return i.e. ir is greater than or equal to MARR, then the alternative will be selected or else it will not be selected. • The MARR is the minimum rate of return from the investment, which is acceptable. In other words it is the minimum rate of return below which the investment alternatives are economically not acceptable. • The minimum attractive rate of return (MARR) serves as an important criteria while selecting a single alternative or comparing mutually exclusive alternatives whenever the investments are made. • For an organization, it is governed by various parameters namely availability of financially viable projects, amount of fund available for investment along with the associated risk, and type of organization (i.e. government, public sector, private sector etc.).
7
• The difference between equivalent worth methods (present worth method/future worth method/annual worth method) and rate of return method is that; in case of former, the equivalent worth of the cash inflows and cash outflows are determined at MARR whereas in case of latter, a rate is determined which equates the equivalent worth of cash inflows to that of the cash outflows and the resulting rate is compared against MARR. The rate of return and MARR are expressed in terms of percentage per period i.e. mostly percentage per year. In the following example, the illustration of the procedure for determination of rate of return for an alternative is presented.
8
Example • A construction firm is planning to invest Rs.800000 for the purchase of a construction equipment which will generate a net profit of Rs.140000 per year after deducting the annual operating and maintenance cost. • The useful life of the equipment is 10 years and the expected salvage value of the equipment at the end of 10 years is Rs.200000. • Compute the rate of return using trial and error method based on present worth, if the construction firm‟s minimum attractive rate of return (MARR) is 10% per year. • Solution: • The cash flow diagram of the construction equipment is shown in Fig. 2.29.
9
10
• For determination of rate of return „ir‟ of the construction equipment, first the equation for net present worth of cash inflows and cash outflows is equated to zero. • Then using the trial and error method the value of „ir‟ is determined. • The net present worth of cash inflows and cash outflows of the construction equipment is given by the following expression. • For determining the value of „ir‟ the net present worth is equated to zero. • Now the above equation will be solved through trial and error process to find out the value of ir. Basically a positive value and a negative value of the net present worth will be determined at rate of return values close to the actual one and then by linear interpolation between these two values, the actual rate of return will be calculated.
11
• For finding out the rate of return values (close to the actual one), those will give a positive value and a negative value of net present worth, one has to carry out a number of trial calculations at various values of ir. Since MARR is 10%, first assume a value of ir equal to 8% and compute the net present worth. Now putting the values of different compound interest factors in the expression for net present worth at ir equal to 8% results in the following;
• The above calculated net present worth at ir equal to 8% is greater than zero, now assume a higher value of ir i.e. 12% for the next trial and compute the net present worth.
12
• As observed from this calculation, the net present worth is decreased at higher value of ir. Thus for getting a negative value of net present worth, assume further higher value of ir than the previous trial and take 14% for the next trial and determine the net present worth.
• Since a negative value of net present worth at ir equal to 14% is obtained (as above), the actual value of rate of return is less than 14%. • The actual rate of return is now obtained by doing linear interpolation either between 8% and 14% or between 12% and 14%. • However for obtaining a more accurate value of rate of return, the linear interpolation is carried out between 12% and 14% and is given as follows;
13
14
• On solving the above expression, the value of ir is found to be 13.55% per year which is greater than MARR (10%). Now using the using Microsoft Excel spreadsheet and entering year-wise cash inflows and cash out flows, the value of rate of return is found to be 13.53% (using the function „IRR‟). • However this minor difference in the value of ir obtained from both the methods can be minimized by finding out the net present worth at narrow range of interest rate values and carrying out linear interpolation between these values (trial and error method) to find out the more precise value close to the actual rate of return. • The net present worth of the construction equipment at MARR i.e. 10% is given by;
15
• The net present worth of the construction equipment at MARR is greater than zero and the rate of return is greater than MARR. • Thus the purchase of the construction equipment is economically justified. • It may be noted here that when the equivalent worth of an investment is greater than zero at interest rate equal to MARR, then the rate of return of the investment is greater than MARR. • The rate of return „ir‟ can also be determined by equating the net annual worth to zero. For the above construction equipment, the net equivalent annual worth at different values of ir are calculated as follows;
16
17
• • • • •
•
•
•
On solving the above expression, the value of ir is found to be 13.52% per year. The minor difference in the values of ir from present worth and annual worth methods is due to the effect of decimal points in the calculations. Similar to present worth and annual worth methods, the rate of return „ir‟ can also be determined by equating the net future worth to zero. From the above example, a unique value of rate of return was obtained for the construction equipment (on the basis of its cash inflow and cash outflow). This is due to the fact that, there was only one sign change in the cash flows i.e. minus sign at time zero for the cash outflow followed by plus sign for cash inflows during the remaining periods. However in some cases, depending upon the cash flow it is possible to get multiple values of rate of return, those satisfy the rate of return equation of the equivalent worth of cash inflows and cash out flows. This may happen due to more than one sign change in the cash flows e.g. cash outflow (negative) at beginning (time zero) followed cash inflows (positive) at end of year 1 and 2 and then cash outflow (negative) at end of year 3 etc. Thus while selecting an alternative that has multiple values of rate of return (depending on the cash flow), other method of economic evaluation may be adopted to find out the economical suitability of the alternative. 18
Incremental Rate of return • When the best alternative (economically suitable) is to be selected from two or more mutually exclusive alternatives on the basis of rate of return analysis, the incremental investment analysis is used. • In incremental rate of return method, the alternative with larger investment is selected, provided the incremental (extra) investment over the lower investment alternative produces a rate of return that is greater than or equal to MARR. • In other words if the additional benefits i.e. increased productivity, increased income, reduced operating expenditure etc. achieved at the expense of extra investment (associated with larger investment alternative) are more than that could have been obtained from the investment of same amount at MARR elsewhere by the organization, then this additional capital should be invested.
19
• In incremental rate of return method, the economically acceptable lower investment alternative is considered as the base alternative against which the higher investment alternative is compared. • The cash flow of higher investment alternative is considered equal to the cash flow of lower investment alternative plus the incremental cash flow i.e. difference in cash flow between the higher investment and lower investment alternatives. • When using rate of return method for comparing two or more mutually exclusive alternatives, the analysis must be done correctly, otherwise it may lead to incorrect ranking of the alternatives. • However this problem is avoided in incremental investment rate of return analysis. In this technique, the individual rate of return values on total cash flow of the mutually exclusive alternatives are not compared against each other rather the rate of return (or IRR) of the mutually exclusive alternatives or the rate of return of the incremental investment is compared against MARR.
20
• The procedures for comparison of mutually exclusive cost alternatives and that of mutually exclusive investment alternatives using incremental investment rate of return analysis are mentioned below. The details about cost and investment alternatives are already stated in Lecture-1 of Module 2.
21
Steps for comparison of cost alternatives: •
• •
•
•
i) First arrange the mutually exclusive cost alternatives on the basis of increasing initial capital investment. The lowest capital investment alternative is considered as the base alternative (B). ii) The incremental cash flow is calculated between the base alternative (B) and the next higher capital investment alternative (H) over the useful life. iii) Then the rate of return „ir(H-B)‟ of this incremental investment is calculated (procedure as stated earlier) by equating the net equivalent worth (present worth or annual worth or future worth) to zero. iv) If the calculated „ir(H-B)‟ is greater than or equal to MARR, then alternative „B‟ is removed from further analysis. Alternative „H‟ now becomes the new base alternative and is compared against the next higher capital investment alternative. If „ir(H-B)‟ is less than MARR, then alternative „H‟ is removed from further analysis and alternative „B‟ remains as the base alternative and is compared against the next higher investment alternative (alternative with investment higher than „H‟). v) Steps ii) to iv) are repeated till only one alternative is left i.e. the best alternative which justifies the incremental investment associated with.
22
Steps for comparison of investment alternatives: • •
• •
•
i) Arrange the mutually exclusive investment alternatives on the basis of increasing initial capital investment. ii) Then the rate of return (IRR) on total cash flow of the lowest investment alternative is determined (procedure already stated earlier) to find out its acceptability as the base alternative. If the calculated rate of return is greater than or equal to MARR, this is selected as the base alternative. If the calculated rate of return is less than MARR, then this alternative is not considered for further analysis and the acceptability of the next higher investment alternative as base alternative is found out by calculating the rate of return on its total cash flow and comparing against MARR. This process is continued till the base alternative „B‟ (acceptable alternative for which rate of return greater than or equal to MARR) is obtained. If no alternative is obtained in this manner i.e. rate of return less than MARR, then do-nothing alternative is selected.
23
• The do-nothing alternative indicates that all the investment alternatives are rejected. Similar to the comparison of cost alternatives, the incremental cash flow is now calculated between the base alternative (B) and the next higher investment alternative (H) over the useful life. Steps iii) to v) as mentioned above for the comparison of cost alternatives are then followed to select the best alternative.
24
Example • A person is planning a new business. The initial outlay and cash flow pattern for the new business are as listed below. The expected life of the business is five years. Find the rate of return for the new business.
25
Solution • • • •
Initial investment = Rs. 1,00,000 Annual equal revenue = Rs. 30,000 Life = 5 years The cash flow diagram for this situation is illustrated in Fig.
26
27
Example • A construction company is planning to invest for the purchase of a heavy construction equipment which will be used at a construction site. There are four feasible alternatives and the detailed cash flow of all the alternatives are presented in Table 2.4. • Each alternative has the useful life of 8 years. If the company's MARR is 12% per year, select the best alternative using the incremental investment rate of return analysis.
28
29
• After arranging the alternatives in increasing order of investment, the acceptability of the lowest investment alternative i.e. alternative B1 as base alternative is checked by finding out the rate of return on its total cash flow. The rate of return is found out by equating net present worth of alternative B1 to zero.
30
• As already stated, the value of rate of return „ir’ is calculated by solving the above equation either manually through trial and error process with linear interpolation or using Microsoft Excel spreadsheet. The value of rate of return is found to be 10.91% (using the function „IRR‟ in Excel spreadsheet), which is less than company‟s MARR i.e. 12%. Thus alternative B1 is eliminated from further calculation and the acceptability of next higher investment alternative i.e. B3 is checked in the same manner as above. The rate of return on total cash flow of alternative B3 is determined by equating the net present worth to zero.
31
• From this equation the value of rate of return is found to be 13.84% (using the function „IRR‟ in Excel spreadsheet), which is greater than company's MARR i.e. 12%. Thus Alternative B3 becomes the base alternative and it is compared with the next higher investment alternative B4. Now the rate of return of the incremental cash flow between the two alternatives B3 and B4 is calculated and compared with MARR to check suitability of the incremental investment associated with alternative B4. The entire calculation for selecting the best alternative is now presented in Table 2.6.
32
33
• In the above table the incremental values between the alternatives indicate the difference in their cash flows. In Table 2.6, the total cash flows of the individual alternatives and incremental cash flows (comparison between two alternatives) are written in different colour fonts for ease of understanding. • For alterative B1 and B3, total cash flows and for comparison between B4 – B3 and B2 – B4 the incremental cash flows are provided in the above table.
34
• The development authority of a city has to select a pumping unit from four feasible mutually exclusive alternatives for supply of water to a particular location of the city. The details of cash flow and the useful life of all the alternatives are presented in the following table. The minimum attractive rate of return (MARR) is 20% per year. • Select the best alternative using the incremental investment rate of return analysis. Solution: The cash flow and useful life of all the alternatives are presented in Table 2.1.
35
• As seen from the above table, these are cost alternatives involving all cash outflows (negative cash flows) except for the salvage value (positive cash flow) at the end of useful life. • The alternatives are not in the increasing order of capital investment as observed from Table 2.1. • The alternatives are now arranged in the increasing order of capital investment as shown in Table 2.2 and cash outflows and cash inflows are shown with negative and positive signs respectively.
36
• After arranging the alternatives in increasing order of capital investment, alternative-2 (A2) now becomes the base alternative (lowest capital investment with Rs.6600000) and it is compared with the next higher investment alternative i.e. alternative-4 (A4) with capital investment of Rs.7400000. The incremental cash flow between the two alternatives A2 and A4 is given as follows; • Incremental capital investment = -Rs.7400000 – (-Rs.6600000) = Rs.800000 at beginning i.e. at time zero. • Incremental annual operating and maintenance cost from end of year 1 till end of year 10 = -Rs.970000 – (-Rs.1185000) = Rs.215000 Incremental salvage value = Rs.1865000 – Rs.1780000 = Rs.85000 at end of year 10. • In order to find out the rate of return (IRR) of this incremental cash flow, the net present worth is equated to zero.
37
• The value of rate of return „ir’ is now calculated by solving the above equation either manually through trial and error process with linear interpolation or using Microsoft Excel spreadsheet (already mentioned earlier). • For faster calculation, the rate of return is calculated using Microsoft Excel spreadsheet after entering year-wise cash inflows and cash out flows. The value of rate of return is found to be 24.06% (using the function „IRR‟ in Excel spreadsheet). • As rate of return of the incremental cash flow is greater than MARR (20%), the incremental investment associated with alternative-4 (A4) is justified and alternative-2 (A2) is now removed from further analysis. • Alternative-4 now becomes the new base alternative and is compared with next higher capital investment alternative i.e. alternative-1 (A1) with investment of Rs.7800000. • The rate return of this incremental investment is calculated in same manner as above. The entire calculation is now presented in the Table 2.3. 38
39
• In Table 2.3 the incremental values between the alternatives indicate the difference in cash flows between them. • The outcomes of the incremental investment analysis for the comparison of cost alternatives as presented in Table 2.3 are briefly described below. • Comparison between alternatives A2 (base alternative) and A4 (next higher capital investment alternative). The obtained rate of return form the incremental investment analysis is 24.06% which is greater than MARR (20%). Alternative-2 (A2) is eliminated from further analysis and alternative-4 (A4) is the new base alternative. • Now comparison between alternatives A4 and A1 (next higher capital investment alternative). The obtained rate of return form the incremental investment is 28.65% and is greater than MARR. Thus alternative-4 (A4) is eliminated from further analysis and alternative-1 (A1) is the new base alternative.
40
•
•
•
Finally comparison between alternatives A1 and A3 (next higher capital investment alternative). The rate of return obtained from the incremental investment analysis is 14.09% which is less than MARR (20%). Thus the incremental investment associated with alternative-3 (A3) i.e. largest capital investment alternative is not justified and hence alternative-1 (A1) is selected as the best alternative, as no other alternative is left for comparison. In addition, the present worth of the incremental investment associated with alternative-1 (A1) over alternative-4 (A4) at MARR i.e. 20% is greater than zero i.e. Rs.132978 > 0. It can be seen here that, the largest capital investment alternative (A3) is not selected because the incremental investment associated with it results in a rate of return which is less than MARR. In addition the present worth of the incremental investment associated with alternative-3 (A3) over alternative-1 (A1) at MARR i.e. 20% is less than zero i.e. –Rs.66150 < 0. Now the values of equivalent present worth of the total cash flow of the cost alternatives at MARR (20%) are found to be -Rs.11280643, -Rs.11165528, Rs.11098700 and -Rs.11032550 for alternatives A2, A4, A3 and A1 respectively. Thus alternative A1 (the best alternative) exhibits lowest negative equivalent present worth as compared to other cost alternatives.
41
42
43
44
45
46
Internal rate of returns • It is the rate at which the sum of discounted cash inflow equals the sum of discounted cash outflow. In other words it is the rate which discounts the cash flow to zero. Accept/reject criterion • The acceptance and rejection is done on the basis of the IRR rate.
NPV
IRR
a) It take interest as a known a) It take interest as afactor unknown factor b) It calculates the exact amt. of investment
b) It calculates maximum rate of interest
Conflicts • The project require different cash outlay. • The project have unequal lives. • The project have different pattern of cash flows. Merits &demerits: • Consider the time value of money. • Take the amount of expenses &revenue. • Gives more value to the present money value.
• It is very difficult. • Reinvestment presumption.
• Consider the time value of money. • Take the amount of expenses &revenue. • Gives more value to the present money value. • It is very difficult. • Reinvestment presumption.
Calculating IRR - Example • PWB/PWC = 1 • 2000(P/A, i, 5)/8200 = 1 • (P/A, i, 5) = 8200/2000 = 4.1
• From Table, IRR = 7% From Compound Interest Tables
Interest rate
(P/A,i,5)
6%
4.212
7%
4.100
8%
3.993 50
Calculating ROR • Where two mutually exclusive alternatives will provide the same benefit, ROR is performed using an incremental rate of return (DROR) on the difference between the alternatives. • You cannot simply choose the higher IRR alternative. Two-alternative situation
Decision
DROR MARR
Choose higher-cost alternative
DROR < MARR
Choose lower-cost alternative 51
Accounting Rate of Return
Definition • Accounting rate of return (also known as simple rate of return) is the ratio of estimated accounting profit of a project to the average investment made in the project. ARR is used in investment appraisal. • ARR = Average Accounting Profit Average Investment • Accept the project only if its ARR is equal to or greater than the required accounting rate of return. In case of mutually exclusive projects, accept the one with highest ARR.
Payback • Payback Analysis: Components • This calculation must take into account Incomes, Expenses and Taxes: – The shorter the payback period, the better; – The longer the payback period, the longer funds are locked up and the riskier the project probably is. • Note: Depreciation should not be included in the calculation. • Payback Analysis: Calculation • Payback period = When cumulative net cash flow reaches breakeven • Payback period = (Last year that will show a negative cash flow) + (Absolute cumulative net cash flow for that year / Total net cash flow in the following year)
Discounted Payback Period • What is the 'Discounted Payback Period' • A capital budgeting procedure used to determine the profitability of a project. In contrast to an NPV analysis, which provides the overall value of an project, a discounted payback period gives the number of years it takes to break even from undertaking the initial expenditure. Futurecash flows are considered are discounted to time "zero." This procedure is similar to a payback period; however, the payback period only measure how long it take for the initial cash outflow to be paid back, ignoring the time value of money.
Replacement analysis
1
2
The Theory • The Replacement Theory in Operations Research is used in the decision making process of replacing a used equipment with a substitute; mostly a new equipment of better usage. The replacement might be necessary due to the deteriorating property or failure or breakdown of particular equipment.
Replacement Models • The ‘Replacement Theory’ is used in the cases like; existing items have out-lived, or it may not be economical anymore to continue with them, or the items might have been destroyed either by accident or otherwise. The above discussed situations can be solved mathematically and categorised on some basis which may be called as Replacement Models.
Replacement Decisions •
How should an asset be replaced?
•
If an asset can be overhauled and used a little longer, should those funds be invested in the overhaul or should we buy a new machine?
•
Over a given study period, when should I replace an asset?
5
Reasons for Replacement Inadequacy – Doesn’t meet needs, increased demand. May buy a duplicate or replace with a larger asset. Obsolescence – Technological changes, better equipment is available (e.g., computers)
Economic Life – An asset at the end of its economic life. Deterioration – there is just not enough duct tape in the world to keep it going longer. The question is: when to replace the asset?
6
Asset Life Terminology Economic Service Life – For on-going need. How long an asset is used to minimize equivalent annual cost. Physical Life – How long an asset can be kept alive (performing services). Ownership Life – How long an asset is actually kept before selling / disposing. Accounting Life – For tax purposes, mostly. Defined by law, GAAP. May not correspond to physical, ownership, or economic service life.
7
Asset Value Terminology Market Value – What can I sell the asset for at a point in time?
Salvage Costs – Expenses to prepare an asset for sale. Trade-In Value – What a vendor will offer for the old asset, if you buy their
product. This is frequently more than the market value. Book Value – For tax purposes, mostly. Book Value = Original Cost – Sum of Depreciation.
8
Basics of Replacement Analysis 1.
Relevant costs ONLY. Do not include costs that are common to all of the alternatives. They just clutter the analysis. •
2.
Example: Paper for a copier
Sunk costs are NOT included. It only matters what an asset is worth today. What you originally paid for it is immaterial (except for tax purposes).
9
Terminology •
Defender – The existing asset
•
Challenger(s) – Currently available replacement alternatives
We will create cash flow diagrams for both the Defender and the Challenger
10
Replacement Terminology • Defender: an old machine • Challenger: a new machine • Current market value: selling price of the defender in the market place
• Sunk cost: any past cost unaffected by any future decisions • Trade-in allowance: value offered by the vendor to reduce the price of a new equipment
11
Replacement Analysis Decision Map Defender
Available
Defender Marginal Cost Increasing? Yes Analysis Technique 1: Defender’s next year marginal cost Challenger’s EUAC
Identify Alternatives
Defender Marginal Cost Data? No Find lowest EUAC for Defender
Analysis Technique 2: Defender’s lowest EUAC Challenger’s EUAC at its minimum cost life
Best Challenger
Not Available
Find EUAC over given life Analysis Technique 3: Defender’s EUAC over its remaining life Challenger’s EUAC at its min. cost life
Outsider’s Perspective We take the perspective of an outsider.
In order to use the defender, the outsider would have to purchase it.
Thus, the best estimate of the current value of the existing equipment will be used as a “fair market value” for the defender.
13
Opportunity Cost Approach The fair market value of the defender will be treated as a cost to retain the defender (this is the opportunity cost that is foregone by retaining the defender)
DEFENDER DIAGRAM:
0
Salvage 1
2
3
Overhaul Cost
n= Lifetime Operating Cost
Fair Mkt. Value
Overhaul Cost
Maintenance Cost 14
Valuing the Defender Biggest problem in reality! Priority Order: (& information sources) 1. Market Value • Independent Audit • Want Ads in paper/trade magazine • Selling/Asking price at auction 2. Trade-In Value • Must subtract difference with MV from Challenger 1st cost if trade-in is excessive • Can shop around to different vendors • “Blue Book” price • Bank / Savings & Loan estimate 3. Book Value • Prefer Book depreciation to MACRS value
16
Companies Using Replacement theory: Manufacturing – Machine tools, Automotive.
Construction – Renovation. Electronic – Electric parts.
Service Industries – Retrenchment.
CONVENTIONAL REPLACEMENT PROBLEM • The replacement problems are concerned with the issues that arises when the performance of an item decreases, failure or breakdown occurs. The decrease in performance or breakdown may be gradual or sometimes sudden. The need for replacement of items is felt when, 1. The existing item or system has become inefficient or require more maintenance. 2. The existing equipment has failed due to accident or otherwise and does not work at all. 3. The existing equipment is expected to fail shortly. 4. The existing equipment has become obsolete due to the availability of equipment with latest technology and better design.
18
• The solution to replacement problem is nothing but arriving at the best policy that determines the time at which the replacement is most economical instead of continuing at an increased maintenance cost. • The Main objective of replacement policy is to direct the organization in many situations so that it can take right decision. For Example, few situations are: • (i) Waiting for complete failure of item or to replace earlier at the expense of higher cost of the item. • (ii) Whether to replace the under performing equipment with the similar kind of item or by different kind (latest model) of item. • The problem of replacement occurs in the case of both men and machines. Using probability it is possible to estimate the chance of death (failure) at various ages.
19
Types of Maintenance
1
Maintenance Maintenance is work that is carried out to preserve an asset (such as a roof or a heating boiler), in order to enable its continued use and function, above a minimum acceptable level of performance, over its design service life, without unforeseen renewal or major repair activities
Reasons for Maintenance Maintenance serves to protect the owners’ real estate investment in a number of ways • Physical Integrity To keep the assets in good working order so as to minimize disruptions and downtimes. • Risk Management To keep the assets in a state of good repair for the owners’ health and safety. • Aesthetic Preservation To keep the assets from deteriorating in appearance and becoming unsightly. • Responsible Stewardship To ensure that the assets achieve their full potential service life.
• Duty of Care To satisfy a legislated duty that is owed to owners, occupants and guest on the property. • Duty to Mitigate To prevent unnecessary damage to assets that may result in their premature failure.
Introduction: • • • • •
•
Plant maintenance is an important service function of an efficient production system. It helps in maintaining and increasing the operational efficiency of plant facilities. Maintenance is the procedure of finding the faults in any equipment/Machine and also removal of fault. It may be before the breakdown or after the breakdown. Plant maintenance usually refers to the methods, strategies, and practices used to keep an industrial factory running efficiently. This can include anything from regular checks of equipment to make sure they are functioning properly. The general aim of plant maintenance is to create a productive working environment that is also safe for workers.
Maintenance Objectives: • • • • • •
To increase functional reliability of production facilities. To maximize the useful life of the equipment. To maximize production capacity from the given equipment. To minimize the total production cost. To minimize the frequency of interruption in production by reducing breakdowns. To enhance the safety of the manpower.
Maintenance objective
Importance of maintenance • • • •
It helps in identify the cause of failure , e.g whether the failure is due to design defect, or a wear out failure. It also helps in deciding the type of maintenance and maintenance decision like replace and repair. It provides the necessary information regarding the life and reliability of the equipment. With the help of this tool spare parts management got initiated.
Different type of maintenance:
Maintenance
Preventive Maintenance
Planned Maintenance Corrective Maintenance Predictive Maintenance Unplanned maintenance
Emergence maintenance
Running maintenance Shutdown maintenance Scheduled maintenance Breakdown maintenance Shutdown maintenance
Planned Maintenance-
In planned maintenance the maintenance action is carried out with some fore thoughts, prior planning , record keeping and control action. These can be further classified as : Preventive maintenance Corrective maintenance Predictive maintenance
Preventive Maintenance It is based upon the principle that ‘prevention is better than cure’. It is a set of activities that are performed on plant equipment, machinery, and systems before the occurrence of a failure in order to protect them and to prevent or eliminate any degradation in their operating conditions. Or the maintenance carried out at predetermined intervals or according to prescribed criteria and intended to reduce the probability of failure or the degradation of the functioning and the effects limited. It has three types Running maintenance Scheduled maintenance Shut down maintenance
Running maintenance: • •
Running maintenance which includes those maintenance activities that are carried out while the machine or equipment is running. Example – lubrication, adjustment of nuts and screws, tightening of loose nut and bolts.
Scheduled Maintenance: • •
•
Scheduled Maintenance is any variety of scheduled maintenance to an object or item of equipment. Specifically, Planned Maintenance is a scheduled service visit carried out by a competent and suitable agent, to ensure that an item of equipment is operating correctly and to therefore avoid any unscheduled breakdown and downtime. Good example of PM program is car maintenance. After so many kilometers or miles oil should be changed, parts renewed.
Shut Down Maintenance: • • •
Which is a set of preventive maintenance activities that are carried out when the production line is in total stoppage situation. These are performed generally after three or six months. These involves the inspection of plant items which are known or suspected to occur.
Corrective maintenance: In this type, actions such as repair, replacement, or restore will be carried out after the occurrence of a failure in order to eliminate the source of this failure or reduce the frequency of its occurrence.
It also include the different types of actions like typical adjustment of redesign equipment. The difference between corrective maintenance and preventive maintenance is that for the corrective maintenance, the failure should occur before any corrective action is taken. It is of two types 1. Breakdown maintenance 2. Shutdown maintenance
Breakdown Maintenance: It is an emergency based policy in which the plant or equipment is operated until it fails and then it is brought back into running condition by repair.
The maintenance staff locate any mechanical, electrical or any other fault to correct it immediately. It is feasible for the small factories where 1. There are few types of equipment. 2. Machine and equipments are simple and does not require any specialist. 3. Where sudden failure does not cause any serious financial loss.
Predictive maintenance: •
•
As the names implies it involves the prediction of the failure before it occurs, identifying the root cause for those failures symptoms and eliminating those causes before they result in extensive damage of the equipment. Type of maintenance performed continuously or at intervals according to the requirements to diagnose and monitor a condition or system. Also called condition based maintenance.
Unplanned Maintenance: • • •
•
Maintenance action which is carried out without any fore thoughts or prior planning is called unplanned maintenance. Emergency maintenance is one of the example of unplanned maintenance. In this type of maintenance the maintenance action is executed with the help of all available maintenance resources in least possible time, without any major time lag. Examples are gas leakage in chemical plant, fire hazards, breakdown of boiler, turbine etc.
Nature of maintenance problem: Maintenance can be classifies as follow 1. General classification of maintenance problems a) Mechanical Failure b) Thermal Failure c) Chemical Failure
1.
Classification of maintenance problem based on time span a) Short Run Maintenance Problem b) Long Run Maintenance Problem
General classification Maintenance problem. Mechanical Failure • Worn out bushes and bearings and other moving parts. • Fatigue of machine members • Creep of material at high temp • Excessive forced vibration, misalignments etc. Thermal Failure • Overheating of the component • Lack of lubrication • Inadequate of cooling • Electrical insulation failure Chemical Failure • Highly corrosive fluids containing abrasive particles • Failure of protective linings like glass , rubber etc.
Classification maintenance problem based on time span: Short run production problem
Maintenance problem which are carried out in a sort period of time are known as short run production system. It may be hourly, daily ,weekly and monthly. Example:-
Hourly- inspection of correct lubricant, level of coolant, sharpness of cutting tool. Daily- cleaning of m/c, tightening of nuts, correct cooling, inspection of various indicators, minor adjustment of parts. Weekly- Major adjustment, lubrication, tightening of parts. Monthly- checking for insulation, corrosion, safety guards, checking of worn-out and distorted parts.
Procedure of Preventive maintenance Job identification by preparing of Facility engineer
Preparation of maintenance schedule
Preparation of History record
Preparation of Job specification
Preparation of maintenance programed
Preparation of monthly, weekly Maintenance programme
Preparation of inspection report
Preparation of maintenance request
Corrective and control action
1. Job identification by preparing of facility engineer •
The very first step of PMP is to prepare the facility register which defines that what to be maintained.
2. Preparation of maintenance schedule In this step we prepare a maintenance schedule. It is simply a
comprehensive list of all the incident and their time of incidence. It gives the useful information regarding the method, time and place of
maintenance work, besides it also provides secondary information about maintenance man power requirement etc. It gives the various details regarding maintenance like what, when,
how, where.
3. Preparation of history card •
It not only gives the useful information about the result of maintenance events but also furnishes the essential details regarding the uses of machines, free of failures and failure modes.
4. Preparation of Job specification In this step the job specification is prepared. It is simply a document
which provide the essential information regarding the maintenance work to be done. In general practice these job details are specifications compiled from
maintenance schedules. They are a means of communicating the engineer’s requirement to
guide the workers. They are prepared separately for each job.
5. Preparation of maintenance Programming It is a sequential list which allocates specific maintenance work to a specific period. In order to apply the job specification, the maintenance programme is generally prepared for long run when the machines/ equipment are to be inspected.
It is not a good practice because an industry can not prepare a long run production well in advance since too many factors arise and they will result into change in production and maintenance requirement. It presents a overall picture of present and future maintenance commitments.
6. Preparation of weekly/Monthly maintenance programmes •
The next step under plant maintenance procedure is to prepare the weekly/ Monthly maintenance programmes. The maintenance programmes include the following topics. 1. Reconditioning or replacing the lubricating oil 2. Repairing and replacing worn out parts and tools etc 3. Checking all the electrical connections of the system 4. Checking the control system 5. Checking the performance of each parts. 6. Cleaning the interior parts like spark plugs, filters radiators, crankcase, cylinders etc. 7. It will lie under the long run maintenance programmes.
7. Preparing of inspection report This step is followed after the maintenance programme is over. This is simply a document which furnishes the useful information about the maintenance inspections which were performed in the past.
8. Preparing of maintenance request The next step under plant maintenance procedure is to preparation of
maintenance request. It is simply a document or various maintenance suggestions and recommendation given by the inspection report. Suggestions are the useful feedback information that comes from users end workers.
9. Feedback mechanism •
In the last step the application of corrective and control actions are available on the basis of feedback mechanism. These corrective actions should be applied to respective plant facilities at the initial stage of maintenance planning or design.
Schedules of preventive maintenance: It is simply a comprehensive list of all the incident and their time of incident. It gives the following useful information regarding the maintenance work. i. What is to be done or maintained? ii. How is to be maintained? iii. When is to be maintained? Besides this it may also provide the secondary information regarding the place of maintenance and maintenance requirement.
Maintenance Strategy : •
•
•
A maintenance strategy defines the rules for the sequence of planned maintenance work. It contains general scheduling information, and can therefore be assigned to as many maintenance task lists (PM task lists) and maintenance plans as required. A maintenance strategy contains maintenance packages in which the following information is defined: The cycle in which the individual work should be performed (for example, every two months, every 3,106.86 miles, every 500 operating hours) Other data which affects scheduling
Structure: A maintenance strategy consists of: • Strategy header: Name of the strategy, • Scheduling parameters: Contain the scheduling data for the respective maintenance strategy, which influence the scheduling of maintenance plans. • Scheduling indicators • Maintenance packages
Scheduling Indicators: Within a maintenance strategy, you can use different scheduling indicators to specify the type of scheduling you require or to define a cycle set: • Time-based (for example, every 30 days) • Time-based by key date (for example, every 30 days on the 30th day of the month) • Time-based by factory calendar (for example, every 30 working days) • Performance-based (for example, every 50 operating hours)
Advantages And Disadvantages Break Down Maintenance Advantages
Disadvantages
Lower start up cost
Unpredictability
Limited personnel requirement
Equipment not maximised
Reduced maintenance costs
Indirect costs
Potentially increased margins
Preventive Maintenance Advantages
Disadvantages
Over all very cost effective
Catastrophic failure still a risk
Flexibility can allow for adjustment of schedule to accommodate other work
Labour Intensive
Increased equipment life
Performance of maintenance based on schedule not required
Saved energy cost resulting from equipment running from pick efficiency
Risk of damage when conducting unneeded maintenance
Reduced equipment or process failure
Saving not readily visible without a base line
Over all saving between 12% to 18%
Predictive Maintenance Advantages
Disadvantages
Increased component operational life/availability
Increased investment of diagnostic equipment
Allows for pre-emptive corrective action
Increased staff training for analysing data
Decreased part and labour cost
Saving not readily visible without a baseline/history
Improved safety and environment
Energy savings Over all saving between 8% to 12% over preventive maintenance
Condition Monitoring Advantages
Disadvantages
Extend bearing service life
Monitoring equipment costs
Maximise machine productivity
Operational costs (running the program)
Minimise unscheduled downtime
Skilled personnel needed
Safely extend overhaul intervals
Strong management commitment needed.
Improve repair time
A significant run-in time to collect machine histories and trends is usually needed.
Increased machine life
Improve product quality Reduce product cost
Enhance product safety
Corrective Maintenance Advantages
Disadvantages
Lower short-term costs
Increased long-term costs due to unplanned equipment downtime.
Requires less staff since less work is being done
Possible secondary equipment or process damage.
Prone to neglect of assets
Case Study Aircraft Maintenance • • •
•
Aircraft maintenance is the overhaul, repair, inspection or modification of an aircraft and aircraft components Airlines authorities casually refer to the detailed inspections as "checks“ Different Types of checks 1. A check 2. B check 3. C check 4. D check Among all the checks A and B checks are lighter checks, while C and D are considered heavier checks.
• A Check This is performed approximately every 500 - 800 flight hours or 200 - 400 cycles. It needs about 20 - 50 man-hours and is usually performed overnight at an airport gate or hangar. The actual occurrence of this check varies by aircraft type, the cycle count (take off and landing is considered an aircraft "cycle"), or the number of hours flown since the last check. • B Check This is performed approximately every 4–6 months. It needs about 150 man-hours and is usually performed within 1–3 days at an airport hangar. A similar occurrence schedule applies to the B check as to the A check. B checks may be incorporated into successive A checks
•
C Check
This is performed approximately every 20–24 months or a specific amount of actual flight hours (FH) as defined by the manufacturer. This maintenance
check is much more extensive than a B Check, requiring a large majority of the aircraft's components to be inspected. This check puts the aircraft out of service and until it is completed, the aircraft must not leave the maintenance
site. It also requires more space than A and B Checks—usually a hangar at a maintenance base. The time needed to complete such a check is generally 1–2 weeks and the effort involved can require up to 6000 man-hours
• D Check This is by far the most comprehensive and demanding check for an airplane. It is also known as a Heavy Maintenance Visit (HMV). This check occurs approximately every 5 years. It is a check that, more or less, takes the entire airplane apart for inspection and overhaul. Also, if required, the paint may need to be completely removed for further inspection on the fuselage metal skin. Such a check can usually demand up to 50,000 manhours and it can generally take up to 2 months to complete, depending on the aircraft and the number of technicians involved. It also requires the most space of all maintenance checks, and as such must be performed at a suitable maintenance base. Given the elevated requirements of this check and the tremendous effort involved in it, it is also by far the most expensive maintenance check of all, with total costs for a single visit ending up well within the million-dollar range.
41
NDT In Aircraft Maintenance • During aircraft maintenance 'NONDESTRUCTIVE TESTING' (NDT) is the most economical way of performing inspection and this is the only way of discovering defects. • In order to maintain the aircraft defects free and ensure a high degree of quality & reliability and as a part of inspection programme, usually following NDT methods are applied; 1) Liquid penetrant 2) Magnetic particle 3) Eddy current 4) Ultrasonic 5) Radiography (x-ray/gamma ray) 6) Visual/Optical 7) Sonic/Resonance 8) Infrared Thermography.
Maintenance Packages: •
Maintenance activities that must be performed at a particular date or point in time are combined into maintenance packages
Total Productive Maintenance (TPM) •
• •
• • •
• •
Total productive maintenance (TPM) originated in Japan in 1971 as a method for improved machine availability through better utilization of maintenance and production resources It can be considered as the medical science of machines. Total Productive Maintenance (TPM) is a maintenance program which involves a newly defined concept for maintaining plants and equipment. The goal of the TPM program is to markedly increase production while, at the same time, increasing employee morale and job satisfaction. TPM brings maintenance into focus as a necessary and vitally important part of the business. It is no longer regarded as a non-profit activity. Down time for maintenance is scheduled as a part of the manufacturing day and, in some cases, as an integral part of the manufacturing process. The goal is to hold emergency and unscheduled maintenance to a minimum.
Why TPM : • • • • • •
TPM was introduced to achieve the following objectives. The important ones are listed below. Avoid wastage in a quickly changing economic environment. Producing goods without reducing product quality. Reduce cost. Produce a low batch quantity at the earliest possible time. Goods send to the customers must be non defective.
Replacement problem
1
INTRODUCTION • The problem of replacement arises when any one of the components of productive resources, such as • machinery, building and men deteriorates due to time or usage. The examples are: • (a) A machine, which is purchased and installed in a production system, due to usage some of its components wear out and its efficiency is reduced. • (b) A building in which production activities are carried out, may leave cracks in walls, roof etc, and needs repair. • (c) A worker, when he is young, will work efficiently, as the time passes becomes old and his work efficiency falls down and after some time he will become unable to work.
Replacement due
• Thus the problem of replacement is experienced in systems where machines, individuals or capital assets are the main production or job performing units. • The characteristics of these units is that their level of performance or efficiency decreases with time or usage and one has to formulate some suitable replacement policy regarding these units to keep the system up to some desired level of performance. • We may have to take different type of decision such as: • (a) We may decide whether to wait for complete failure of the item (which may result in some losses due to deterioration or to replace earlier at the expense of higher cost of the item, • (b) The expensive item may be considered individually to decide whether we should replace now or, if not, when it should be reconsidered for replacement, • (c) Whether the item is to be replaced by similar type of item or by different type for example item with latest technology
O.R. Methodology of solving replacement problems: Identify the items to be replaced & their failure mechanism
Mechanism
Gradual Failure
Sudden Failure
On the basis of Data
Cost
Depreciation
Maintenance
Nature of failure
Gradual failure: • In this class as the life of the machine increases or due continuous usage, due to wear and tear of components of the facility, its efficiency deteriorates due to
which the management can experience: • (a) Progressive Increase in maintenance expenditure or operating costs, • (b) Decreased productivity of the equipment and • (c) decrease in the value of the equipment i.e. resale value of the equipment/facility decreases. • Examples of this category are: Automobiles, Machine tools, etc.
Sudden failure: • In this case, the items ultimately fail after a period of time. The life of the equipment cannot be predicted and is some sort of random variable.
The period between installation and failure is not constant for any particular type of equipment but will follow some frequency distribution, which may be:
Progressive failure: • In this case probability of failure increases with the increase in life of an item. The best example is electrical bulbs and computer components.
8
Retrogressive failure: • Some items will have higher probability of failure in the beginning of their life, and as the time passes chances of failure becomes less. • That is the ability of the item to survive in the initial period of life increases its expected life. The examples are newly installed machines in production systems, new vehicles, and infant baby (The probability of survival is very less in infant age, but once the baby get accustomed to nature, the probability of failure decreases). (c) Random failure: • In this class, constant probability of failure is associated with items that fail from random causes such as physical shocks, not related to age. Example is vacuum tubes.
Costs Associated with Maintenance • (a) Purchase cost or Capital cost: ( C ) • This cost is independent of the age of the machine or usage of the machine. This is incurred at the beginning of the life of the machine, i.e. at the time of purchasing the machine or equipment. But the interest on the invested money is an important factor to be considered. • (b) Salvage value / Scrap value / Resale value / Depreciation: (S) • As the age of the machine increases, the resale value decreases as its operating efficiency decreases and the maintenance costs increases. It depends on the operating conditions of the machine and life of the machine. • (c) Running costs including maintenance, Repair and Operating costs: • These costs are the functions of age of the machine and usage of the machine. As the usage increases or the age increases, due to wear and tear, many components fail to work and they are to be replaced. As the age increases, failures also increase and the maintenance costs goes on increasing. At some period the maintenance costs are so high, which will indicate that the replacement of the machine or equipment is essential.
TYPES OF REPLACEMET PROBLEMS • Replacement of Capital equipment, which looses its operating efficiency due to aging (passage of time), or due to continuous usage (due to wear and tear of components). • Examples are: Machine tools, Transport and other vehicles, etc., Here the system can maintain the level of performance by installing a new unit at the beginning of some unit of time (year, month or week) and decide to keep it up to some suitable period so as to minimize the operating and maintenance costs.
• Replacement of items that fail completely all in a sudden in a random nature. We use Group replacement or Preventive maintenance technique
for these items and these are expensive to replace individually. • Examples are: Electric bulbs, Transistors, Electronic components etc., Here replacement of items are done in anticipation of failure, which is known as
preventive maintenance.
• Replacement of human beings in organizations, known as Staffing problem, or known as Human resource planning or Staffing problem. • This problem requires the knowledge of life distribution for service of staff
in a system. • Miscellaneous problems such as replacement of existing units due to availability of more effective and new and advanced technology. In these problems replacement will become necessary due to research of new and advanced and more effective technology and old technology becomes out of date.
Replacement Models • Model-I: Aging of Machines i.e. Replacement of items that deteriorate Gradually : Replacement of items whose efficiency deteriorate with time e.g. machine tools, vehicles, equipment buildings .. • Model-II: Replacement of items that fails suddenly and completely like electric bulb/ tube Replacement Policy Individual Replacement Policy- Items becoming out-of-date due to new developments like manual accounting by tally, computers, cars Group Replacement Policy • Model- III: Replacement of Human being in an organization or staffing problem
• NotationsC= the capital cost of certain item S(t)- the selling or scrap value of item F(t) =operating cost of the item at time t n=optimal replacement period
REPLACEMENT OF ITEMS WHOSE EFFICIENCY REDUCES OR MAINTENCNCE COST INCREASES WITH TIME OR DUE TO AGE AND MONEY VALUE IS NOT CONSIDERED
Terminologies • Let C = Purchase cost or Capital cost of the item, • S = Scrap value or resale value of the item, it is assumed that this cost will remain constant over time. • Let u (t) be the maintenance or running cost at the time ‘t’. • M(y) is the cumulative maintenance cost during year y’’. • The total cost incurred on the item during period Y” • ‘y’ = Capital cost + total maintenance cost in the period ‘y’ - Scrap value. = C + M(y) – S • Hence average cost per unit of time incurred during the period ‘y’ on the item is given by: • G (y) = {C + M (y) – S}/y, to find the value of ‘y’ for which G (y) is minimum
Practical Problems
The cost of an equipment is Rs.62,000 and its scrap value is Rs. 2,000. The life of the equipment is 8 years. The maintenance costs for each year are given below.
Year
1
2
3
4
5
6
7
8
Maintainan ce Cost in Rs. 1000 2000 3500 5000 8000 11000 16000 24000
When the equipment should be replaced.
Determination of Economic Life of an Asset
1
Any asset will have the following cost components:
Capital recovery cost (average first cost), computed from the first cost (purchase price) of the machine.
Average operating and maintenance cost (O & M cost)
Total cost which is the sum of capital recovery cost (average first cost) and average maintenance cost.
2
3
EXAMPLE • A firm is considering replacement of an equipment, whose first cost is Rs. 4,000 and the scrap value is negligible at the end of any year. Based on experience, it was found that the maintenance cost is zero during the first year and it increases by Rs. 200 every year thereafter. • (a) When should the equipment be replaced if i = 0%? • (b) When should the equipment be replaced if i = 12%? • (a) When i = 0%. In this problem – (i) First cost = Rs. 4,000 – (ii) Maintenance cost is Rs. 0 during the first year and it increases by Rs. 200 every year thereafter. This is summarized in column B of Table
4
5
• Column C summarizes the summation of maintenance costs for each replacement period. The value corresponding to any end of year in this column represents the total maintenance cost of using the equipment till
the end of that particular year. • Average total cost = First cost (FC) + Summation of maintenance cost Replacement period
• =n+n Average first cost he given period for + Averagemaintenance cost for the given period • Column F = Column E + Column D
6
• The value corresponding to any end of year (n) in Column F represents the average total cost of using the equipment till the end of that particular year. For this problem, the average total cost decreases till the end of year 6 and then it increases. Therefore, the optimal replacement period is six years, i.e. economic life of the equipment is six years. (b) When interest rate, i = 12%. When the interest rate is more than 0%, the steps to be taken for getting the economic life are summarized with reference to Table
7
8
9
Replacement of an asset with a new asset
1
•
In this section, the concept of comparison of replacement of an existing asset with a new asset is presented. In this analysis, the annual equivalent cost of each alternative should be computed first. Then the alternative which has the least cost should be selected as the best alternative. Before discussing details, some preliminary concepts which are essential for this type of replacement analysis are presented
2
Example •
A diesel engine was installed 10 years ago at a cost of Rs. 50,000. It has a present realizable market value of Rs. 15,000. If kept, it can be expected to last five years more, with operating and maintenance cost of Rs. 14,000 per year and to have a salvage value of Rs. 8,000 at the end of the fifth year. This engine can be replaced with an improved version costing Rs. 65,000 which has an expected life of 20 years. This improved version will have an estimated annual operating and maintenance cost of Rs. 9,000 and ultimate salvage value of Rs. 13,000. Using an interest rate of 15%, make an annual equivalent cost analysis to determine whether to keep or replace the old engine.
3
Solution • • • • • • • •
Alternative 1— Old diesel engine Purchase price = Rs. 50,000 Present value (P) = Rs. 15,000 Salvage value (F) = Rs. 8,000 Annual operating and maintenance cost (A) = Rs. 14,000 Remaining life (n) = 5 years Interest rate = 15% The cash flow diagram of the old diesel engine is shown in Fig
4
• • • • • • • • • • •
The formula for the annual equivalent cost is AE(15%) = (P – F)(A/P, 15%, 5) + F i + A = (15,000 – 8,000)(0.2983) + 8,000 0.15 + 14,000 = Rs. 17,288.10 Alternative 2—New diesel engine Present value (P) = Rs. 65,000 Salvage value (F) = Rs. 13,000 Annual operating and maintenance cost (A) = Rs. 9,000 Life (n) = 20 years Interest rate = 15% The cash flow diagram of the new diesel engine is shown in Fig.
5
• • • • •
The formula for the annual equivalent cost is AE(15%) = (P – F)(A/P, 15%, 20) + F i + A = (65,000 – 13,000)(0.1598) + 13,000 0.15 + 9,000 = Rs. 19,259.60 For comparing the engines based on equal lives (20 years), the annual equivalent figures are given in Fig. 8.8. Equal lives are nothing but the least common multiple of the lives of the alternatives.
6
•
Since the annual equivalent cost of the old diesel engine is less than that of the new diesel engine, it is suggested to keep the old diesel engine. Here, an important assumption is that the old engine will be replaced four times during the 20 years period of comparison.
7
Example •
A diesel engine was installed 10 years ago at a cost of Rs. 50,000. It has a present realizable market value of Rs. 15,000. If kept, it can be expected to last five years more, with operating and maintenance cost of Rs. 14,000 per year and to have a salvage value of Rs. 8,000 at the end of the fifth year. This engine can be replaced with an improved version costing Rs. 65,000 which has an expected life of 20 years. This improved version will have an estimated annual operating and maintenance cost of Rs. 9,000 and ultimate salvage value of Rs. 13,000. Using an interest rate of 15%, make an annual equivalent cost analysis to determine whether to keep or replace the old engine.
8
Example •
A steel highway bridge must either be reinforced or replaced. Reinforcement would cost Rs. 6,60,000 and would make the bridge fit for an additional five years of service. If it is reinforced, it is estimated that its net salvage value would be Rs. 4,00,000 at the time it is retired from service. The new prestressed concrete bridge would cost Rs. 15,00,000 and would meet the foreseeable requirements of the next 40 years. Such a bridge would have no salvage value. It is estimated that the annual maintenance cost of the reinforced bridge would exceed that of the concrete bridge by Rs. 96,000. If the bridge is replaced by a new prestressed concrete bridge, the scrap value of the steel would exceed the demolition cost by Rs. 4,20,000. Assume that the money costs the state 10%. What would you recommend?
9
Solution • • • • • • • • • • •
There are two alternatives: 1. Reinforce the existing bridge. 2. Replace the existing bridge by a new prestressed concrete bridge. Alternative 1— Reinforce the existing bridge Cost of reinforcement (P) = Rs. 6,60,000 Salvage value after 5 years (F) = Rs. 4,00,000 The excess annual maintenance cost over prestressed concrete bridge (A) = Rs. 96,000 Life (n) = 5 years Interest rate (i) = 10% The cash flow diagram of alternative 1 is illustrated in Fig.
10
11
• • • • •
• • • • • • • • •
The annual equivalent cost of the alternative 1 is computed as AE(10%) = (P – F)(A/P, 10%, 5) + F i + A = (6,60,000 – 4,00,000)(0.2638) + 4,00,000 0.10 + 96,000 = Rs. 2,04,588 Alternative 2—Replace the existing bridge by a new prestressed concrete bridge Cost of prestressed concrete bridge (P) = Rs. 15,00,000 Excess scrap value of steel over the demolition cost of the current bridge (X) = Rs. 4,20,000 Life (n) = 40 years Interest rate (i) = 10% Note that the excess maintenance cost of the reinforced bridge over the prestressed concrete bridge is included in alternative 1. The cash flow diagram for alternative 2 is shown in Fig.
12
13
•
Three years back, a municipality purchased a 10 hp motor for pumping drinking water. Its useful life was estimated to be 10 years. Due to the fast development of that locality, the municipality is unable to meet the current demand for water with the existing motor. The municipality can cope with the situation either by augmenting an additional 5 hp motor or replacing the existing10 hp motor with a new 15 hp motor. The details of these motors are now tabulated.
14
Solution • • • • • • • • • • • • • •
There are two alternatives to cope with the situation: 1. Augmenting the present 10 hp motor with an additional 5 hp motor. 2. Replacing the present 10 hp motor with a new 15 hp motor. Alternative 1—Augmenting the present 10 hp motor with an additional 5 hp motor Total annual equivalent cost = Annual equivalent cost of 10 hp motor + Annual equivalent cost of 5 hp motor Calculation of annual equivalent cost of 10 hp Motor Present market value of the 10 hp motor (P) = Rs. 10,000 Remaining life (n) = 7 years Salvage value at the end of motor life (F) = Rs. 1,500 Annual operation and maintenance cost (A) = Rs. 1,600 Interest rate, i = 15% The cash flow diagram of this alternative is shown in Fig. 15
16
• • • • • • • • • • •
The annual equivalent cost of the 10 hp motor is calculated as AE(15%) = (P – F)(A/P, 15%, 7) + F i + A = (10,000 – 1,500)(0.2404) + 1,500 0.15 + 1,600 = Rs. 3,868.40 Calculation of annual equivalent cost of 5 hp motor Purchase value of the 5 hp motor (P) = Rs. 10,000 Life (n) = 7 years Salvage value at the end of motor life (F) = Rs. 800 Annual operation and maintenance cost (A) = Rs. 1,000 Interest rate, i = 15% The cash flow diagram of the 5 hp motor is illustrated in Fig.
17
• • • • • • • • • • • • • •
The annual equivalent cost of the 5 hp motor is computed as AE(15%) = (P – F)(A/P, 15%, 7) + F i + A = (10,000 – 800)(0.2404) + 800 0.15 + 1,000 = Rs. 3,331.68 Total annual equivalent cost of the alternative 1 = Rs. 3,868.40 + Rs. 3,331.68 = Rs. 7,200.08 Alternative 2—Replacing the present 10 hp motor with a new 15 hp motor Purchase value of the 15 hp motor (P) = Rs. 35,000 Life (n) = 7 years Salvage value at the end of motor life (F) = Rs. 4,000 Annual operation and maintenance cost (A) = Rs. 500 Interest rate, i = 15% The cash flow diagram of this alternative is shown in Fig. 18
19
Example •
• •
A machine was purchased two years ago for Rs. 10,000. Its annual maintenance cost is Rs. 750. Its life is six years and its salvage value at the end of its life is Rs. 1,000. Now, a company is offering a new machine at a cost of Rs. 10,000. Its life is four years and its salvage value at the end of its life is Rs. 4,000. The annual maintenance cost of the new machine is Rs. 500. The company which is supplying the new machine is willing to take the old machine for Rs. 8,000 if it is replaced by the new machine. Assume an interest rate of 12%, compounded annually. (a) Find the comparative use value of the old machine. (b) Is it advisable to replace the old machine?
20
Solution • • • • • •
Old machine Let the comparative use value of the old machine be X. Remaining life (n) = 4 years. Salvage value of the old machine (F) = Rs. 1,000 Annual maintenance cost (A) = Rs. 750 Interest rate, i = 12% The cash flow diagram of the old machine is depicted in Fig.
21
• • • • • • • • • •
The annual equivalent cost of the old machine is computed as AE(12%) = (X – F)(A/P, 12%, 4) + F i + A = (X – 1,000)(0.3292) + 1,000 0.12 + 750 New machine Cost of the new Machine (P) = Rs. 10,000 Life (n) = 4 years. Salvage value of the new machine (F) = Rs. 4,000 Annual Maintenance cost (A) = Rs. 500 Interest rate, i = 12% The cash flow diagram of the new machine is illustrated in Fig.
22
• • • • • • • • •
The annual equivalent cost of the new machine is illustrated as AE(12%) = (P – F) (A/P, 12%, 4) + F i + A = (10,000 – 4,000)(0.3292) + 4,000 0.12 + 500 = Rs. 2,955.20 Now, equate the annual equivalent costs of the two alternatives and solve for X. (X – 1,000)(0.3292) + 1,000 0.12 + 750 = 2,955.20 X = Rs. 7,334.14 The comparative use value of the old machine is Rs. 7,334.14, which is less than the price (Rs. 8,000) offered by the company which is supplying the new machine in the event of replacing the old machine by the new machine. Therefore, it is advisable to replace the old machine with the new one.
23
Capital recovery with return
1
2
3
4
5
6
7
8
9
10
• In most engineering situations, the capital cost has two components • I = the investment required at time 0, and • S = the salvage value received at time N. S 0 N I
• Capital Recovery (CR) cost is the annual equivalent of the capital cost. CR is often described from the bank’s point of view and is a function of the MARR, i%. • CR(i) = I(A/P, i, N) + S(A/F, i, N) • Because of the similarities between the formulas for (A/P, i, N) and • (A/F, i, N) , we can also calculate CR(i) using • CR(i) = (I – S)(A/P, i, N) + iS, or • CR(i) = (I – S)(A/F, i, N) + iI 11
Example • Your company is planning to manufacture a new product which requires a machine that the company does not now own. The cost of the machine is $10,000, its life is 4 years, and its salvage value (at the end of the 4th year) is $1000. With this machine, the new profit from each product is $12. What annual production makes the investment worthwhile? The MARR is 10%. • Solution • We must find the capital recovery costs. The data are • I = $10,000 S = $1,000 • MARR = 10%Useful life N = 4 • CR = I(A/P, MARR, N) – S(A/F, MARR, N) • = $10,000(A/P,10%,4) – $1000(A/F, 10%, 4) • = $10,000(0.3155) – $1000(0.2155) • = $3155 – $215.50 • = $2939.5 12
• • • •
Using the other approaches, we get CR = (I – S)(A/P,10%,4) + (0.1)S = $9,000 (0.3155) + (0.1)$1000 = $2839.5 + $100 = $2939.5
• • • •
CR = (I – S)(A/F,10%,4) + (0.1)I = $9000 (0.2155) + (0.1)$10,000 = $1939.5 + $1000 = $2939.5
• To be profitable, revenues must equal or exceed the capital recovery costs. Let X be the number of products produced per year. • Revenues – CR = $12X – $2939.5 0. Solving for X, we get X 245 units.
13
Concept of challenger and defender
1
•
In replacement analysis, the existing (i.e. currently owned) asset is referred as defender whereas the new alternatives are referred as challengers.
•
In this analysis the ‘outsider perspective' is taken to establish the first
cost of the defender. •
This initial cost of the defender in replacement analysis is nothing but the estimated market value from perspective of a neutral party.
•
In other words this cost is the investment amount which is assigned to
the currently owned asset (i.e. defender) in the replacement analysis. •
2
• • •
•
•
•
The current market value represents the opportunity cost of keeping the defender i.e. if the defender is selected to continue in the service. In other words, if the defender is selected, the opportunity to obtain its current market value is forgone. Sometimes the additional cost required to upgrade the defender to make it competitive for comparison with the new alternatives is added to its market value to establish the total investment for the defender. Along with the market value, there will be revised estimates for annual operating and maintenance cost, salvage value and remaining service life of the defender, which are expected to be different from the original values those were estimated at the time of acquiring the asset. The past estimates of initial cost, annual operating and maintenance cost, salvage value and useful life of defender are not relevant in the replacement analysis and are thus neglected. The past estimates also incorporate a sunk cost which is considered irrelevant in replacement analysis.
3
•
•
•
•
Sunk cost occurs when the book value (as determined using depreciation method) of an asset is greater than its current market value, when the asset (i.e. defender) is considered for replacement. In other words it represents the amount of past capital investment which can not be recovered for the existing asset under consideration for replacement. Sunk cost may occur due to incorrect estimates of different cost components and factors related productivity of the defender, those were made at the time of original estimates in the past with uncertain future conditions. Since sunk cost represents a loss in capital investment of the asset, the income tax calculations can be done accordingly by considering this capital loss. In replacement analysis the incorrect past estimates and decisions should not be considered and only the cash flows (both present and future) applicable to replacement analysis should be included in the economic analysis.
4
•
•
•
•
•
For replacement analysis, it is important know about different lives of an asset, as this will assist in making the appropriate replacement decision. The different lives are physical life, economic life and useful life. Physical life of an asset is defined as the time period that is elapsed between initial purchase (i.e. original acquisition) and final disposal or abandonment of the asset. Economic life is defined as the time period that minimizes the total cost (i.e. ownership cost plus operating cost) of an asset. It is the time period that results in minimum equivalent uniform annual worth of the total cost of the asset. Useful life is defined as the time period during which the asset is productively used to generate profit. In replacement analysis the defender and challenger is compared over a study period. Generally the remaining life of the defender is less than or equal to the estimated life of the challenger. When the estimated lives of the defender and challenger are not equal, the duration of the study period has to be appropriately selected for the replacement analysis. When the estimated lives of defender and challenger are equal, annual worth method or present worth method may be used for comparison between defender and the challengers (new alternatives). 5
•
If an existing equipment is considered for replacement with a new equipment, then the existing equipment is known as the defender and the new equipment is known as challenger.
•
Assume that an equipment has been purchased about three years back for Rs. 5,00,000 and it is considered for replacement with a new equipment.
•
The supplier of the new equipment will take the old one for some
money, say, Rs. 3,00,000. •
This should be treated as the present value of the existing equipment and it should be considered for all further economic analysis.
•
The purchase value of the existing equipment before three years is now
known as sunk cost, and it should not be considered for further analysis. 6
Example •
•
•
A construction company has purchased a piece of construction equipment 3 years ago at a cost of Rs.4000000. The estimated life and salvage value at the time of purchase were 12 years and Rs.850000 respectively. The annual operating and maintenance cost was Rs.150000. The construction company is now considering replacement of the existing equipment with a new model available in the market. Due to depreciation, the current book value of the existing equipment is Rs.3055000. The current market value of the existing equipment is Rs.2950000. The revised estimate of salvage value and remaining life are Rs.650000 and 8 years respectively. The annual operating and maintenance cost is same as earlier i.e. Rs.150000.
7
•
•
• •
The initial cost of the new model is Rs.3500000. The estimated life, salvage value and annual operating and maintenance cost are 8 years, Rs.900000 and Rs.125000 respectively. Company's MARR is 10% per year. Find out whether the construction company should retain the ownership of the existing equipment or replace it with the new model, if study period is taken as 8 years (considering equal life of both defender and challenger). Solution: For the replacement analysis, initial cost (Rs.4000000), initial estimate of salvage value (Rs.850000) and remaining life (12 – 3 = 9 years) and current book value (Rs.3055000) of the existing equipment (i.e. defender) are irrelevant. Similarly sunk cost of Rs.105000 (Rs.3055000 – Rs.2950000) is also not relevant for the replacement analysis. For the replacement analysis the current revised estimates of the existing equipment will be used.
8
• • • • • • •
For existing equipment (defender), Current market value (P) = Rs.2950000, Salvage value (F) = Rs.650000, Annual operating and maintenance cost (A) = Rs.150000, Study period (n) = 8 years. For new model (challenger), Initial cost (P) = Rs.3500000, Salvage value (F) = Rs.900000, Annual operating and maintenance cost (A) = Rs.125000, Study period (n) = 8 years. Now the equivalent uniform annual worth of both defender (i.e. the existing equipment) and challenger (i.e. the new model) at MARR of 10% (i.e. i = 10%) are calculated as follows;
9
10
•
From the above calculations, it is observed that equivalent uniform annual cost of the defender is less than that of the challenger. Thus the construction company should continue in retaining the ownership of the defender against the challenger with above details. Since the useful lives of defender and challenger are equal, the same conclusion will also be obtained by using present worth method for economic evaluation.
11
Example 2 •
•
• • •
Two years ago, a machine was purchased at a cost of Rs. 2,00,000 to be useful for eight years. Its salvage value at the end of its life is Rs. 25,000. The annual maintenance cost is Rs. 25,000. The market value of the present machine is Rs. 1,20,000. Now, a new machine to cater to the need of the present machine is available at Rs. 1,50,000 to be useful for six years. Its annual maintenance cost is Rs. 14,000. The salvage value of the new machine is Rs. 20,000. Using an interest rate of 12%, find whether it is worth replacing the present machine with the new machine.
12
Solution •
•
Alternative 1— Present machine Purchase price = Rs. 2,00,000 Present value (P) = Rs. 1,20,000 Salvage value (F) = Rs. 25,000 Annual maintenance cost (A) = Rs. 25,000 Remaining life = 6 years Interest rate = 12% The cash flow diagram of the present machine is illustrated in Fig.
13
• • • • •
•
Annual maintenance cost for the preceding periods are not shown in this figure. The annual equivalent cost is computed as AE(12%) = (P – F)(A/P, 12%, 6) + F i + A = (1,20,000 – 25,000)(0.2432) + 25,000 0.12 + 25,000 = Rs. 51,104 Alternative 2 New machine Purchase price (P) = Rs. 1,50,000 Salvage value (F) = Rs. 20,000 Annual maintenance cost (A) = Rs. 14,000 Life = 6 years Interest rate = 12% The cash flow diagram of the new machine is depicted in Fig.
14
• • • • •
The formula for the annual equivalent cost is AE(12%) = (P – F)(A/P, 12%, 6) + F i + A = (1,50,000 – 20,000)(0.2432) + 20,000 0.12 + 14,000 = Rs. 48,016 Since the annual equivalent cost of the new machine is less than that of the present machine, it is suggested that the present machine be replaced with the new machine
15
Simple Probabilistic Model For Items Which Fail Completely
1
Introduction • Electronic items like transistors, resistors, tube lights, bulbs, etc. could fail all of a sudden, instead of gradual deterioration. The failure of the item may result in complete breakdown of the system. The system may contain a collection of such items or just one item, say a tube light. • Therefore, we use some replacement policy for such items which would avoid the possibility of a complete breakdown. • The following are the replacement policies which are applicable for this situation. • (i) Individual replacement policy. Under this policy, an item is replaced immediately after its failure. • (ii) Group replacement policy. Under this policy, the following decision is made:
2
Cont.. •
At what equal intervals are all the items to be replaced simultaneously with a provision to replace the items individually which fail during a fixed group replacement period?
• • There is a trade-off between the individual replacement policy and the group replacement policy. Hence, for a given problem, each of the replacement policies is evaluated and the most economical policy is selected for implementation. This is explained with two numerical problems.
3
Individual Replacement Policy • Under this policy an item is immediately replaced after its failure. • To determine the probability of failure (or life span of any item), mortality tables are used. To discuss such type of replacement policy, we consider the problem of human population. • Assumptions: (i) All deaths are immediately replaced by births, and (ii) There are no other entries or exits. • However in reality it is impossible to have these conditions. But, the reason for assuming the above two is that the analysis will be easier by keeping the virtual human population in mind. Such models can be applied to industrial items, where death of a person is equivalent to the failure of an item or part and birth of a person is equivalent to replacement. Thus, organizations also face a fairly common situation. The following Mortality Theorem will make the conceptions clear.
4
Mortality Theorem • A large population is subject to a given mortality law for a very long
period of time. All deaths are immediately replaced by births and there are no other entries or exits. Then the age distribution ultimately
becomes stable and that the number of deaths per unit time becomes constant ( which is equal to the size of the total population divided by the mean age at death).
5
Group Replacement Policy: • Group Replacement Of Items That Fail Completely • There are certain items viz. Light bulbs that either work or fail completely. In some cases a system made up of a big number of similar low cost items that are increasingly prone to failure with age. While replacing such failed items, always a set-up cost will be there for replacement. The said set-up cost is independent of the number of items to be replaced and hence it may be advantageous to replace entire group of items at fixed intervals. Such a policy is referred as group replacement policy and found attractive when the value of any individual item is so less and the cost of keeping records for age of individual items is not justifiable. 6
Example
• The cost of replacing an individual failed transistor is Rs. 9. If all the transistors are replaced simultaneously, it would cost Rs. 3.00 per transistor. Any one of the following two options can be followed to replace the transistors: • (a) Replace the transistors individually when they fail (individual replacement policy). • (b) Replace all the transistors simultaneously at fixed intervals and replace the individual transistors as they fail in service during the fixed interval (group replacement policy). • Find out the optimal replacement policy, i.e. individual replacement policy or group replacement policy. If group replacement policy is optimal, then find at what equal intervals should all the transistors be replaced. 7
Solution Assume that there are 100 transistors in use. Let, • pi be the probability that a transistor which was new when placed in position for use, fails during the ith week of its life. Hence,
8
• Since the sum of pis is equal to 1 at the end of the 7th week, the transistors are sure to fail during the seventh week. • Assume that (a) transistors that fail during a week are replaced just before the end of the week, and (b) the actual percentage of failures during a week for a sub-group of transistors with the same age is same as the expected percentage of failures during the week for that sub-group of transistors.
9
10
11
Determination of group replacement cost • Cost of transistor when replaced simultaneously = Rs. 3 • Cost of transistor when replaced individually = Rs. 9 • The costs of group replacement policy for several replacement periods are summarized in Table. • Table Calculations of Cost for Preventive Maintenance
12
• From Table it is clear that the average cost/week is minimum for the fourth week. Hence, the group replacement period is four weeks. • Individual replacement cost/week = Rs. 207
• Minimum group replacement cost/week = Rs. 196.50 • Since the minimum group replacement cost/week is less than the individual replacement cost/week, the group replacement policy is the best, and hence all the transistors should be replaced once in four weeks and the transistors which fail during this four-week period are to be replaced individually.
13
DEPRECIATION
1
Introduction •
The word depreciation is the diminution in the value
of assets due
to use or lapse of time. It is measure of exhaustion of the effective life of the assets from any cause during the given period. •
Depreciation is a reduction in the value of the assets which is permanent, gradual and of continuing nature, depreciation takes
place gradually unless there is a quice physical deterioration or obsolescence due to technological developments.
2
Meaning •
“Depreciation means decline or fall in the value of an assets, due to use, passage of time or obsolescence”.
•
In other words, depreciation may be described as a permanent, continuing and gradual shrinkage in the book value of fixed
assets. It is based on the cost of assets consumed in a business and not on its market value.
3
Definition of depreciation • Applies only to fixed assets. • The whole cost of the fixed assets must be spread over its useful life. • The portion of the cost allocated to a particular accounting period is charged as an expense against revenue (Matching principle).
• This portion of the cost is called Depreciation.
4
Definitions •
“Depreciation may be defined as the permanent and continuing diminution in the quality, quantity or the value of an assets”.
•
•
–W. Pickles.
“Depreciation may be defined as the permanent decrease in the value of an assets through wear and tear in the use or the passage of time.
•
-B.G. Wickery
5
Depreciation:Gradual decrease in the value of an asset is known as depreciation. It has two types:1- Internal depreciation. 2- External depreciation.
6
ACCUMULATED DEPRECIATION VS. DEPRECIABLE COST • •
• •
Accumulated Depreciation = Original Cost – Book Value Depreciable Cost = Original Cost – Salvage Value – Depreciable Cost > Accumulated Depreciation Book Value > Salvage Value – Depreciable Cost = Accumulated Depreciation Book Value = Salvage Value – Book Value NEVER SMALLER than Salvage Value Depreciable Cost is how much an asset CAN be depreciated Accumulated Depreciation is how much an asset HAS been depreciated already
The program has been delivered under the Partnership Agreement with Kent Institute of Business & Technology, Sydney An Australian RTO
Formula of Depreciation Cost of Asset Depreciation = No of year
8
Internal Depreciation:Depreciation which occurs for certain inherent normal causes, is known as internal depreciation. Such as wear and tear and depletion. External Depreciation:Depreciation caused by some external reasons is called external depreciation. Such as obsolescence, efflux of time and accident.
9
Assets that May Be Depreciated •A useful life of more than one year. •A determinable useful life but not an unlimited life. Examples: vehicles, machinery, equipment, building, fences, purchased breeding livestock, wells.
10
Depreciation Terms •
Cost :- The price paid for the asset.
•
Useful life :- Number of years the asset is expected to be used in business.
11
Features •
It is decline in the value of Fixed Assets.
•
It is a continuing process.
•
It is charged on fixed assets and not on floating assets.
•
It include loss of value due to effluxion of time, usage, or obsolescence.
•
It is a process of spreading the burden of cost of an assets over its life
period. •
The diminution in value takes place gradually.
12
CAUSES • Wear and Tear • Depletion
• Obsolescence • Efflux-ion of time • Permanent fall in market value • Accidents
13
Physical deterioration
• Caused by physical wear and tear - rust, erosion, rot and decay
Examples - office furniture - printing machines
14
Obsolescence • Fixed assets become out-of-date
- when new model or more efficient tool come into existence.
Examples - cars Pentium I
- computers Pentium IV 15
Depletion of fixed assets • An asset that depletes over time as resources are extracted from it. Little Guilin
Examples - Gold mines - Quarries - Little Guilin is a depleted granite quarry now turned into a beautiful lake.
16
Passage of time • Some assets confer upon their holders the exclusive rights to enjoy certain privileges for a fixed period of time.
Examples - copyrights - patent rights - leases on land 17
Wear and Tear:The change in the shape of an asset due to use in the business is known as wear and tear. Obsolescence:The decrease in the value of an asset due to new inventories, change in habit and taste of people, improvement and change in technology and fashion is known as obsolescence.
18
NEEDS •
Matching of Costs and Revenue
•
Consideration of Tax
•
True and Fair Financial Position
•
Compliance with Law
•
To Ascertain True Value of Assets
19
OBJECTIVES •
To recover the cost incurred on fixed assets over its life.
•
To facilitate the purchase of new asset, when the old asset is disposed.
•
Find out the correct profit or loss for the particular period.
•
Find out the correct financial position through balance sheet.
•
To show the proper value of Assets.
•
Provision of funds for replacement of assets.
•
Ascertaining accurate cost of production.
20
Factor affecting •
Cost of Assets
•
Estimated Net Residual Value
•
Estimated Useful life
•
Legal Provisions
21
Straight Line Method
1
2
Depreciation Terminology Definition: Book (noncash) method to represent decrease in value of a tangible asset over time Two types: book depreciation and tax depreciation Book depreciation: used for internal accounting to track value of assets Tax depreciation: used to determine taxes due based on tax laws
In USA only, tax depreciation must be calculated using MACRS; book depreciation can be calculated using any method
16-3
Common Depreciation Terms First cost P or unadjusted basis B: Total installed cost of asset Book value BVt: Remaining undepreciated capital investment in year t Recovery period n: Depreciable life of asset in years Market value MV: Amount realizable if asset were sold on open market
Salvage value S: Estimated trade-in or MV at end of asset’s useful life Depreciation rate dt: Fraction of first cost or basis removed each year t Personal property: Possessions of company used to conduct business Real property: Real estate and all improvements (land is not depreciable) Half-year convention: Assumes assets are placed in service in midyear 16-4
Straight Line Depreciation Book value decreases linearly with time
Dt =
B-S n
BVt = B - tDt
Where: Dt = annual depreciation charge t = year B = first cost or unadjusted basis S = salvage value n = recovery period
Where: BVt = book value after t years
SL depreciation rate is constant for each year: d = dt = 1/n © 2012 by McGraw-Hill Reserved
All Rights
16-5
Example: SL Depreciation An argon gas processor has a first cost of $20,000 with a $5,000 salvage value after 5 years. Find (a) D3 and (b) BV3 for year three. (c) Plot book value vs. time.
Solution:
(a ) D3 = (B – S)/n = (20,000 – 5,000)/5 = $3,000
(b) BV3 = B – tDt = 20,000 – 3(3,000) = $11,000
(c) Plot BV vs. time BVt 20,000 11,000 5,000 0
3
5
Year, t
7
Straight Line Method It is a very popular method because its simplicity & consistency. A fixed amount of original cost is charged as a depreciation every year.
Annual depreciation = Cost of the assets – Residual value Estimated Economic life
8
Straight Line method i.
It is Also Known as fixed installment method.
ii.
Every year during the useful life of asset, depreciation is calculated as certain fixed percentage of original cost of asset
9
Formula of straight line method Depreciation= Original cost of an asset- scrap value Estimated life For percentage of depreciation:Depreciation × Depreciation × 100 Original asset
Market price of an assets:-
Scarp value of an asset:-
Working Life of an asset:The period during which an asset will help earning income of business.
STRAIGHT LINE METHOD (I) A machine X costs $20,000 is expected to last 4 years. At the end of the 4th year, it can be sold for $2,000 as scrap. ( Scrap value is the same as residual value.)
Original cost - Residual value Expected useful life
Depreciation per year =
=
20,000
- 2000 4
= $4,500 12
Important features:
• Fixed asset account shows original cost of asset.
• Provision for Depreciation account shows accumulated depreciation of fixed asset.
• Net book value of fixed asset (in Balance Sheet) is original cost less Provision for Depreciation.
13
14
15
16
17
18
Straight Line Method Advantages
Disadvantages
•
•
• •
It is simple to calculate & easy to understand. It can reduce the book value of the asset to zero. The valuation of the asset each year in the balance sheet is reasonably fair.
•
It ignores the fact that the service yielding ability of the asset fall while the repairs & maintenance cost increase with the passage of time. If an additional asset is acquired, the amount to be charged as depreciation needs to be calculated.
19
20
Declining balance method of depreciation
1
Reducing balance method i. It is also known as write down value method. ii. This method depreciation is charged at certain percentage on original cost in the first year and subsequently on opening written down value of an asset every year. iii. This amount of depreciation changes every year and it goes on reducing. 2
DEFINITION • • •
A common depreciation-calculation system that involves applying the depreciation rate against the non-depreciated balance. Produces a decreasing annual depreciation expense over asset’s useful life. With this method, companies compute annual depreciation expense by multiplying the book value at the beginning of the year by decliningbalance depreciation rate.
• Sometimes a company knows that an asset will be more efficient in its early years than in its later years. For example, computerized equipment may be used extensively when first acquired, then either quickly becomes obsolete or is used much less when more advanced equipment becomes available.
FORMULA Depreciation = Book Value x Depreciation Rate Book Value = Cost – Accumulated Depreciation
5
6
7
8
9
10
11
ADVANTAGES • Declining Balance Depreciation methods better match cost revenues to because it takes more depreciation in the early years of an assets’ useful life compare to the straight line depreciation method. • It reflects better the difference in usage of an asset from one period to the other compared to the straight line depreciation method
DISADVANTAGES • Might be harder to compute compared to the straight line depreciation method • They have declining amounts of depreciation expense which creates greater disparity between the costs • Decreasing depreciation expense and increasing maintenance of an asset might smooth the income. • Ignores salvage value in determining the amount to which the decliningbalance rate is applied.
SAMPLE PROBLEMS 1.
On January 2011, a Company purchased an equipment at a cost of P140,000, having a life span of 5 years. The depreciation rate is 20% Calculate the depreciation from 2011 to 2015 using Declining Balance Depreciation Method. Also calculate the salvage value of the equipment at the end of the year 2015.
Answer (Prob. #1) Year Book Value in the Beginning 201 1
140,000
Depreciation (140,000) (20 / 100) = 28,000
201 140,000 – 28,000 = 112,000 2
(112,000) (20 / 100) = 22,400
201 3
112,000 – 22,400 = 89,600
(89,600) (20 / 100) = 17,920
201 4
89,600 – 27,920 = 71, 680
(71,680) (20/100) = 14,336
201 5
71, 680 – 14,336 = 57,344
(57, 344) (20 /100) = 11,469
At the end of the year 2015, the SALVAGE VALUE = 57,334 – 11,469 = 45,875
SAMPLE PROBLEMS 2.
A machine had a cost of $24,000. Salvage Value of $2,000. Estimated life of five years. Compute depreciation. Yea r
Cost
1
$24,000
2
24,000
3
Accumulat Book Value at Rate Depreciati Book Value ed on the Beginning at the End Depreciati of the Year of the Year on $24,000
40%
$9,600
$14,400
9,600
14,400
40%
5,760
8,640
24,000
15,360
8,640
40%
3,456
5,184
4
24,000
18,816
5,184
40%
2,073.60
3,110.40
5
24,000
20,889.60
3,110.40
1,110.40
2,000
SAMPLE PROBLEMS 3.
An asset has a useful life of 3 years. Cost of the asset is $2,000. Residual Value is $500. Rate of Depreciation is 50%. Depreciation expense for three years will be as follows:
Answer (Prob. #3) NBV
RV
Rate
Depreciation
Accumulated Depreciation
Year 1
( 200 0
-
500 )
x
50%
=
750
750
Year 2
( 125 0
-
500 )
x
50%
=
375
1125
Year 3
( 875
-
500 )
x
50%
=
375*
1500
*Depreciation for the last year of the asset’s useful life is the difference between net book value at the start of the period and the estimated residual value. This is to ensure that depreciation is charged in full.
Declining Balance Depreciation Method
• Depreciation = Book value x Depreciation rate – Book value = Cost - Accumulated depreciation – For double declining balance method: Depreciation rate = Straight line depreciation rate x 200%
Sum-of-the-Years-Digits Method of Depreciation
1
Sum-of-the-Years-Digits Method of Depreciation • A third commonly used depreciation method is the sum of the yearsdigits method. Like double declining-balance depreciation, this method expenses more of the purchase price in the early years. • In this method of depreciation also, it is assumed that the book value of the asset decreases at a decreasing rate. • If the asset has a life of eight years, first the sum of the years is computed as Sum of the years = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 = n(n + 1)/2
• The rate of depreciation charge for the first year is assumed as the highest and then it decreases. The rates of depreciation for the years 1–8, respectively are as follows: 8/36, 7/36, 6/36, 5/36, 4/36, 3/36, 2/36, and 1/36. • For any year, the depreciation is calculated by multiplying the corresponding rate of depreciation with (P – F). Dt = Rate (P – F) Bt= Bt–1 – Dt
Sum-of-Years’ Digits (SOYD) Method • Principle Depreciation concept similar to DB but with decreasing depreciation rate. Charges a larger fraction of the cost as an expense early years than of the later years. • Formula •Annual Depreciation
of the
•Book Value
Dn ( I S )( N n 1) / SOYD
Bn I j 1 D j n
where SOYD=N(N+1)/2
(c) 2001 Contemporary Engineering Economics
4
Example Annual Depreciation $10,000 D1
$6,000
$4,000
Total depreciation at end of life
$8,000
Book Value
D2 D3
B1 B2
D 4
$2,000
I = $10,000 N = 5 years S = $2,000 SOYD = 15
n Dn Bn 1 (5/15)(8,000)=$2,667$7,333 2 (4/15)(8,000)=$2,133 5,200 3 (3/15)(8,000)=$1,600 3,600 4 (2/15)(8,000)=$1,067 2,533 5 (1/15)(8,000)=$533 2,000
D5
B3
0 0 1
2
3
B4 4
B5 5
n
(c) 2001 Contemporary Engineering Economics
5
6
• The formulae for Dt and Bt for a specific year t are as follows:
Example of Sum-of-the-Years-Digits Method of Depreciation
13
Merits • In this method, the quantum of depreciation is greater in the earlier years in comparison with the later years because the benefits received from the use of the asset are greater in the early years than in the later years. • In the earlier year repairs are light but depreciation is heavy but in the later year, as the asset gets older the repairs are heavy but depreciation is light. So depreciation plus repairs will more or less constant every year and the charge to Profit and Loss Account should be uniform. • If the asset is retired earlier than anticipated as result of unforeseen obsolescence, the loss upon retirement will be less than if straight line depreciation were used, because, asset are recovered at a higher rate in the earlier years where as only a small fraction remains left for recovering them in the later years.
14
Merits •
For tax accounting purposes, these methods have a clear advantage over the straight line method. The larger deductions in the early years mean that at the very least tax payments are postponed for a considerable period. On the other hand, this method gives a tax postponement with the greatest present value. • In this method nearly three-fourth of total depreciation is charged within half of its life. That means three-fourth of blocked investment recovered within short span. • This method is very simple to understand and simple to calculate.
15
Demerits • This method also ignored the cost of capital on invested fund. • In the earlier year greater depreciation is charged at a result less profit is available for declaring dividend in the earlier year. This may created serious problem for organization to attract new investor. As dividend is one of the motivating factors for investment. Again more depreciation in the earlier year may resultant the high cost of production in the competitive market.
16
Sinking fund method of depreciation
1
Sinking Fund Method of Depreciation • In this method of depreciation, the book value decreases at increasing rates with respect to the life of the asset. Let • P = first cost of the asset, • F = salvage value of the asset, • n = life of the asset, • i = rate of return compounded annually, • A = the annual equivalent amount, • Bt = the book value of the asset at the end of the period t, and • Dt = the depreciation amount at the end of the period t.
• The loss in value of the asset (P – F) is made available an the form of cumulative depreciation amount at the end of the life of the asset by setting up an equal depreciation amount (A) at the end of each period during the lifetime of the asset. A = (P – F) [A/F, i, n] • The fixed sum depreciated at the end of every time period earns an interest at the rate of i% compounded annually, and hence the actual depreciation amount will be in the increasing manner with respect to the time period. A generalized formula for Dt is Dt = (P – F) (A/F, i, n) (F/P, i, t – 1) • The formula to calculate the book value at the end of period t is Bt = P – (P – F) (A/F, i, n) (F/A, i, t) • The above two formulae are very useful if we have to calculate Dt and Bt for any specific period. If we calculate Dt and Bt for all the periods, then the tabular approach would be better.
Example
Merits • a) Under this method, at the end of the specified time, a definite sum is available in cash to replace the old asset. • b) Since the amount is invested outside the business there is no need to drawn money from the business for replacement purpose at the end of the life of the asset. This helps to avoid pressure on working capital.
10
Demerits • a) From the management view point the method is inefficient. Generally the internal rate of return of the firm is higher than the return on investment. • As a result this causes substantial loss to the firm. • b) There is always a risk factor about the loss on realisation of investments. That means if the market price of the investment in which depreciation is invested fluctuate in that case the amount realised may be less than cost of the asset. • c) The firm may face the difficulty of finding suitable investments which provide the desired rate of return per annum. • d) It is difficult to estimate the exact working life of the asset which will be replaced.
11
Service output method of depreciation
1
Service Output Method of Depreciation • In some situations, it may not be realistic to compute depreciation based on time period. • In such cases, the depreciation is computed based on service rendered by an asset. Let P = first cost of the asset F = salvage value of the asset X = maximum capacity of service of the asset during its lifetime x = quantity of service rendered in a period.
Then, the depreciation is defined per unit of service rendered:
Example of Service Output Method of Depreciation
Annuity Method of Depreciation • Under annuity method of depreciation the cost of asset is regarded as investment and interest at fixed rate is calculated thereon. • Had the proprietor invested outside the business, an amount equal to the cost of asset, he would have earned some interest. • So as a result of buying the asset the proprietor loses not only cost of asset by using it, but also the above mentioned interest. • Hence depreciation is calculated in such a way as will cover both the above mentioned losses.
• The principle underlying this method is that in calculating depreciation regard should be held, not only to the cost of an asset but also to the interest which the capital blocked in that asset would have earned had it been invested outside the business. • So under this method, a fixed installment of depreciation is charged against revenue for each year of the life of the asset in such a way that at a given rate of interest the present value of the sum of all those installments equals to the cost of the asset. • Stated otherwise, depreciation for each year is made to include an interest on unrecovered capital outlay but the interest which is credited annually to Profit and Loss Account, gradually diminishes. • On the other hand, depreciation excluding interest i.e. net depreciation goes on increasing. • It may also be noted that Annuity Method of depreciation recovers more than the original cost of the asset. The excess being interest on investment.
7
8
Merits • a) In this method depreciation includes not only the recovery of invested capital but also the interest on capital outlay. So this is more logical and realistic than other methods. • b) At the time of inflation, it is possible for the firm to replace the asset as total depreciation charged is more than cost of the asset. • c) This method is most appropriate for the intangible such as leasehold rights, patents etc. In this cases depreciation varies with the passing of time rather than with the intensity of use.
11
Demerits • The use of the asset or its service rendering capacity is not at all considered for measuring depreciation. The cost of waiting time rather than the cost of physical deterioration is given more importance in such a case. • In this method, it is assume that the rate of interest on unrecovered capital outlay is equal to its cost of capital but in real world it is hardly happen. • The scrap value of the asset is not considered which is not justified. • Annual repairs and maintenance cost are assumed to remain constant throughout the life of an asset, but it is not so happen.
12
Evaluation of public alternatives
1
Evaluation of public alternatives • • • •
In evaluating alternatives of private organizations, the criterion is to select the alternative with the maximum profit. The profit maximization is the main goal of private organizations while providing goods/services as per specifications to their customers. But the same criterion cannot be used while evaluating public alternatives. Examples of some public alternatives are constructing bridges, roads, dams, establishing public utilities, etc.
• • • • •
The main objective of any public alternative is to provide goods/services to the public at the minimum cost. In this process, one should see whether the benefits of the public activity are at least equal to its costs. If yes, then the public activity can be undertaken for implementation. Otherwise, it can be cancelled. This is nothing but taking a decision based on Benefit-Cost ratio (BC) given by
BC ratio = Equivalent benefits/Equivalent costs The benefits may occur at different time periods of the public activity
• • • • •
BP = present worth of the total benefits BF = future worth of the total benefits BA = annual equivalent of the total benefits P = initial investment PF = future worth of the initial investment
• • • •
PA = annual equivalent of the initial investment C = yearly cost of operation and maintenance CP = present worth of yearly cost of operation and maintenance CF = future worth of yearly cost of operation and maintenance
Examples of Evaluation of public alternatives
Example for Evaluation of public alternatives
1
Inflation
1
INTRODUCTION Inflation is defined as a sustained increase in the price level or a fall in the value of money. When the level of currency of a country exceeds the level of production, inflation occurs. Value of money depreciates with the occurrence of inflation. Inflation is a rise in the general level of prices of goods and services in an economy over a period of time. Inflation also reflects an erosion in the purchasing power of money Inflation's effects on an economy can be positive or negative. The rate of inflation is measured by the annual percentage change in the level of prices as measured by the consumer price index.
2
DEFINITION
According to C.CROWTHER, “Inflation is State in which the Value of Money is Falling and the Prices are rising.”
In Economics, the Word inflation Refers to General rise in Prices Measured against a Standard Level of Purchasing Power.
3
TYPES OF INFLATION
4
•
•
Open Inflation -: The rate where Costs rise Spending Products and Services.
due
to
Economic
trends
of
Suppressed Inflation -: Existing inflation disguised by government Price controls or other interferences in the economy such as subsidies. Such suppression, nevertheless, can only be temporary because no governmental measure can completely contain accelerating inflation in the long run. It is Also Called Repressed Inflation.
5
•
Galloping Inflation -: Very Rapid Inflation which is almost impossible to reduce.
•
Creeping Inflation -: Circumstance where the inflation of a nation increases gradually, but continually, over time. This tends to be a typically pattern for many nations. Although the increase is relatively small in the short-term, as it continues over time the effect will become greater and greater.
•
Hyper Inflation -: Hyperinflation is caused mainly by excessive deficit spending (financed by printing more money) by a government, some economists believe that social breakdown leads to hyperinflation (not vice versa), and that its roots lie in political rather than economic causes.
6
Causes of inflation FACTORS ON DEMAND SIDE: o o o o
Increase in money supply Increase in disposable income Deficit financing Foreign exchange reserves
7
Factors on supply side o o o o
Rise in administered prices Erratic agriculture growth Agricultural price policy Inadequate industrial growth
8
Effects of inflation
Production On income distribution Consumption and welfare Foreign trade Social and political effects Manufacturers
9
10
Effect of inflation • • • • •
They add inefficiencies in the market, and make it difficult for companies to budget or plan long-term. Uncertainty about the future purchasing power of money discourages investment and saving. There can also be negative impacts to trade from an increased instability in currency exchange prices caused by unpredictable inflation. Higher income tax rates. Inflation rate in the economy is higher than rates in other countries; this will increase imports and reduce exports, leading to a deficit in the balance of trade.
11
HOW IS INFLATION MEASURED? The 2 ways of Measuring Inflation are -: Consumer Price Index
12
Inflation in India
13
Negative effects of inflation
Hoarding Social unrest and revolts Hyperinflation Allocative efficiency Shoe leather cost Menu costs Business cycles
14
CONSEQUENCES OF INFLATION Adverse effect on production Adverse effect on distribution of income Obstacle to development Changes in relative prices Adverse effect on the B.O.P
15
MEASURES TO CONTROL INFLATION
Monetary policy Fixed exchange rates Gold standard Wage and price controls Cost-of-living allowance
16
Measures of inflation 1. Monetary policy • Credit Control • Demonetization of Currency • Issue of New Currency 2. Fiscal policy • Reduction in Unnecessary Expenditure • Increase in Taxes • Increase in Savings • Surplus Budgets • Public Debt 3. Other Measures • To Increase Production • Rational Wage Policy • Price Control 17
Inflation adjusted economic life of machine & comparison
1
Inflation Adjusted Economic life of Machine • The productivity of any organization is a function of many factors. • It is largely affected by efficient and effective use of machinery and equipment. • So, operations and maintenance of these equipment are very important to the organization. • A machine which is purchased today cannot be used forever. • It has a definite economic lifetime. After the economic life, the machine should be replaced with a substitute machine with similar operational capabilities. • This kind of analysis is called replacement analysis.
The elements of costs involved in the replacement analysis are as follows: 1. Purchase cost (initial cost) 2. Annual operation and maintenance cost 3. Salvage value at the end of every year, if it is significant
• From Fig, it is clear that the sum of operation and maintenance cost increases with the life of the machine. • But the capital recovery with return decreases with the life of the machine. • The total cost of the machine goes on decreasing initially but it starts increasing after some years. • The year with the minimum total cost is called as the economic life of the machine.
Example Economic Life Determination without Inflationary Effect The determination of economic life of a machine without considering the effect of inflation is demonstrated using the following example. EXAMPLE A machine costs Rs. 5,00,000. Its annual operation cost during the first year is Rs. 40,000 and it increases by Rs. 5,000 every year thereafter. The maintenance cost during the first year is Rs. 60,000 and it increases by Rs. 6,000 every year thereafter. The resale value of the machine is Rs. 4,00,000 at the end of the first year and it decreases by Rs. 50,000 every year thereafter. Assume an interest rate (discounting factor) of 20%. The method of finding the economic life of the machine with a discounting factor of 20% at zero inflation rate is summarized in Table 11.2. From the table it is clear that the total annual equivalent cost is minimum if the machine is used for 14 years. Hence, the economic life of the machine is 14 years.
Economic Life Determination with Inflationary Effect The illustration in above Section is reconsidered for analyzing the effect of inflation on the economic life of the machine. An average annual inflation rate of 6% is assumed for discussion. The corresponding steps are explained in Table From the Table , it is clear that the total annual equivalent cost is minimum if the machine is used for three years. Thus, the economic life of the machine is three years.
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
