B.SC MATHEMATICS AND STATISTICS. STD404 OPERATIONS ...... Situation 3.2: In a machine shop, a batch of jobs is to be assigned to a group of machines.
B.SC MATHEMATICS AND STATISTICS
STD404 OPERATIONS RESEARCH Zimbabwe Open University
Zimbabwe Open University
Contents 1 Introduction to Operations Research
1
1.1
Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.3
What Is Operations Research? . . . . . . . . . . . . . . . . . . . . . . .
1
1.4
Some Operations Research Techniques . . . . . . . . . . . . . . . . . . .
2
1.5
Exercises
2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 Decision Analysis
3
2.1
Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
2.2
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
2.3
Structuring the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
2.3.1
Decision Alternatives . . . . . . . . . . . . . . . . . . . . . . . . .
4
2.3.2
States of Nature . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
2.3.3
Payoff . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
2.3.4
Payoff Table
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
Decision Making Under Uncertainty . . . . . . . . . . . . . . . . . . . .
5
2.4.1
Optimistic Approach . . . . . . . . . . . . . . . . . . . . . . . . .
5
2.4.2
Conservative Approach . . . . . . . . . . . . . . . . . . . . . . .
5
2.4.3
Minimax Regret Approach
. . . . . . . . . . . . . . . . . . . . .
6
Decision Making Under Risk . . . . . . . . . . . . . . . . . . . . . . . .
6
2.5.1
Expected Monetary Value (EMV) . . . . . . . . . . . . . . . . .
7
2.5.2
Expected Value of Perfect Information (EVPI) . . . . . . . . . .
7
Decision Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
2.6.1
9
2.4
2.5
2.6 2.7
Sequential Decision Trees . . . . . . . . . . . . . . . . . . . . . .
Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Network Models
11 15
3.1
Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
3.2
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
3.3
Examples of Network Problems . . . . . . . . . . . . . . . . . . . . . . .
15
3.4
Terms and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
3.5
Shortest Spanning Tree . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
3.5.1
Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
The Shortest Route Problem . . . . . . . . . . . . . . . . . . . . . . . .
19
3.6.1
20
3.6
Acyclic Directed Network . . . . . . . . . . . . . . . . . . . . . . i
3.7
Maximum Flow Problem . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
3.8
Exercises
28
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4 Project Management
31
4.1
Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
4.2
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
4.3
Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
4.4
Drawing a Network . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
4.5
Critical Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
4.5.1
Step 1: Forward Pass . . . . . . . . . . . . . . . . . . . . . . . .
37
4.5.2
Step 2: Backward Pass . . . . . . . . . . . . . . . . . . . . . . . .
37
4.5.3
Step 3: Calculate Slack . . . . . . . . . . . . . . . . . . . . . . .
38
4.6
Project Scheduling with Uncertain Activity Times . . . . . . . . . . . .
41
4.7
Exercises
46
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 Deterministic Inventory Models
49
5.1
Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
5.2
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
5.3
Elements of Inventory Control . . . . . . . . . . . . . . . . . . . . . . . .
50
5.3.1
Demand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
5.3.2
Lead Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
5.3.3
Inventory Systems . . . . . . . . . . . . . . . . . . . . . . . . . .
50
5.4
Notation and Types of Inventory Costs . . . . . . . . . . . . . . . . . . .
51
5.5
Single Item Inventory Models . . . . . . . . . . . . . . . . . . . . . . . .
51
5.5.1
The Basic Economic Order quantity (EOQ) Model . . . . . . . .
51
EOQ Model, Modification 1 Manufacturing Situation . . . . . . . . . . . . . . . . . . . . . . . . . . .
56
EOQ Model, Modification II Shortages Allowed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58
Exercises
62
5.6 5.7 5.8
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 Stochastic Inventory Models
63
6.1
Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63
6.2
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63
6.3
Statistical Data in Inventory Control . . . . . . . . . . . . . . . . . . . .
63
6.4
Single Order Random Demand for Perishable Goods . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65
6.4.1
65
Discrete Demand . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
6.4.2 6.5
6.6
Continuous Demand . . . . . . . . . . . . . . . . . . . . . . . . .
69
Random Demand Model with Shortage Costs . . . . . . . . . . . . . . .
71
6.5.1
Discrete Demand . . . . . . . . . . . . . . . . . . . . . . . . . . .
72
6.5.2
Continuous Demand . . . . . . . . . . . . . . . . . . . . . . . . .
74
Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7 Queuing Theory I: Single Server Models
75 77
7.1
Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
77
7.2
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
77
7.3
Queueing Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
78
7.4
Kendall-Lee Notation
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
80
7.5
Definitions and Notation . . . . . . . . . . . . . . . . . . . . . . . . . . .
81
7.6
Instantaneous Event Rate for the Exponential Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . .
81
7.7
Birth and Death Processes . . . . . . . . . . . . . . . . . . . . . . . . . .
85
7.8
The (M/M/1) : (FIFO/∞/∞) Queueing Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
85
Parameters of a Queueing Model . . . . . . . . . . . . . . . . . . . . . .
87
7.9
7.10 Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 Queuing Theory II: Multiple Server Models
89 91
8.1
Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
8.2
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
8.3
Queueing Models with m Parallel Servers . . . . . . . . . . . . . . . . .
91
8.3.1
Model 1:(M/M/m):(GD/∞/∞) . . . . . . . . . . . . . . . . . . .
91
8.3.2
Model 2: (M/M/m): (GD/N/∞), m ≤ N . . . . . . . . . . . . .
93
8.3.3
Model 3: (M/M/m): (GD/R/R), m ≤ R . . . . . . . . . . . . . .
93
8.3.4
Model 4: (M/M/∞): (GD/∞/∞). . . . . . . . . . . . . . . . . .
94
8.4
Numerical Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
95
8.5
Models with Non-Markovian Input and Output . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
99
The Pollaczek-Khintchine Formulae . . . . . . . . . . . . . . . . . . . . .
99
8.6
8.6.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
iii
List of Tables 2.1
Pay-off Table for Robotic Microcomputer . . . . . . . . . . . . . . . . .
4
2.2
Optimistic Approach Table . . . . . . . . . . . . . . . . . . . . . . . . .
5
2.3
Conservative Approach Table . . . . . . . . . . . . . . . . . . . . . . . .
6
2.4
Opportunity Loss Table . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
2.5
Tigamuchire Investors Payoff Table . . . . . . . . . . . . . . . . . . . . .
9
2.6
Conditional Probabilities for Survey Conducted . . . . . . . . . . . . . .
10
2.7
Defective Rates Table . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
2.8
Payoff Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
3.1
Distance Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
3.2
Pre-step-Permanent label the origin node . . . . . . . . . . . . . . . . .
24
3.3
First iteration with recent permanent node 1 . . . . . . . . . . . . . . .
24
3.4
Second iteration with recent permanent node 3 . . . . . . . . . . . . . .
25
3.5
Third iteration with recent permanent node 2 . . . . . . . . . . . . . . .
25
3.6
Fourth iteration with recent permanent node 4 . . . . . . . . . . . . . .
25
3.7
Fifth iteration with recent permanent node 5 . . . . . . . . . . . . . . .
25
3.8
Values of permanent labels . . . . . . . . . . . . . . . . . . . . . . . . .
25
3.9
Arc Capacities for Mat Oil
. . . . . . . . . . . . . . . . . . . . . . . . .
27
4.1
Activity Duration Table . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
4.2
Activity Dependency Table . . . . . . . . . . . . . . . . . . . . . . . . .
33
4.3
Earliest Event Start Times . . . . . . . . . . . . . . . . . . . . . . . . .
38
4.4
Latest Event Start Times . . . . . . . . . . . . . . . . . . . . . . . . . .
38
4.5
Total Float of an Activity . . . . . . . . . . . . . . . . . . . . . . . . . .
39
4.6
Free Float of an Activuty . . . . . . . . . . . . . . . . . . . . . . . . . .
40
4.7
Independent Float of an Activity . . . . . . . . . . . . . . . . . . . . . .
40
4.8
Activity Time Estimates (in weeks) . . . . . . . . . . . . . . . . . . . . .
43
4.9
Expected Duration of an Activity . . . . . . . . . . . . . . . . . . . . . .
43
4.10 Total Activity Floats . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
4.11 Activity Variance Table . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
4.12 Activity Duration Table . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
4.13 Activity Precedence . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
6.1
Data for Example 6.1
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
64
6.2
Frequency Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64
6.3
Frequency of Bread Sales . . . . . . . . . . . . . . . . . . . . . . . . . .
68
6.4
Prices Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
68
iv
6.5
Demand Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
68
6.6
Demand Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69
6.7
Frequency of Spare Parts Requirements . . . . . . . . . . . . . . . . . .
73
6.8
Demand Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
73
6.9
Demand Distribution for Newspapers . . . . . . . . . . . . . . . . . . . .
75
6.10 Frequency of Bread Demand . . . . . . . . . . . . . . . . . . . . . . . . .
76
7.1
Solution Table for Example 7.1 . . . . . . . . . . . . . . . . . . . . . . .
78
7.2
Probability Distribution for Waiting Time . . . . . . . . . . . . . . . . .
84
7.3
Transition Probability Rate Matrix . . . . . . . . . . . . . . . . . . . . .
85
v
List of Figures 2.1
Decision Tree for Tigamuchire Problem . . . . . . . . . . . . . . . . . .
10
2.2
Decision Tree for EMVPI for Q3. . . . . . . . . . . . . . . . . . . . . . .
11
2.3
Decision Tree for Example 2.7. Q4. . . . . . . . . . . . . . . . . . . . . .
12
3.1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
3.2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
3.3
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
3.4
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
3.5
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
3.6
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
3.7
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
3.8
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
3.9
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
3.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
3.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
3.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
4.1
Activity diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
4.2
Network Drawing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34
4.3
Network with Dummy . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
4.4
Network with a Loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
4.5
Numbering a Network . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
4.6
Network for the Sports-shop problem . . . . . . . . . . . . . . . . . . . .
37
4.7
Network for ZOU Computing Problem . . . . . . . . . . . . . . . . . . .
42
4.8
EST and LST Network . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
5.1
Variation in Level of Inventory for Basic EOQ Model. . . . . . . . . . .
52
5.2
Inventory of Deserted Cars. . . . . . . . . . . . . . . . . . . . . . . . . .
55
5.3
Inventory for a Manufacturing Situation. . . . . . . . . . . . . . . . . . .
57
5.4
EOQ Model with Shortages Allowed. . . . . . . . . . . . . . . . . . . . .
58
5.5
Manufacturing Model with Shortages Allowed. . . . . . . . . . . . . . .
60
7.1
A Queuing System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
78
7.2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
7.3
State Transition in Birth and Death Process . . . . . . . . . . . . . . . .
85
8.1
Queuing System for Incoming Calls to a Switch Board . . . . . . . . . .
96
vi
Preface
Operations Research, which is concerned with the allocation of scarce resources, is both an art and a science. The art lies in the ability to reflect the concepts efficient and scarce in a well-defined mathematical model of a given situation; the science consists of the derivation of computational methods for solving such models. This module is meant to take you through both aspects of the field. It is assumed that you have already covered the required basic concepts in Calculus 1 (MTD101), Probability Theory 1 (STD101) and Linear Mathematics 1 (MTD102). Although Graph Theory (MTD302) is not a prerequisite for this course, basic knowledge of graph theory would be an added advantage when it to comes unit 3 which deals with Network Models. The module consists of 8 units. In unit 1 we introduce you to the concept of Operations Research. Some of the techniques that are used in Operations Research are discussed in this unit. Decision Analysis is covered in unit 2. In this unit we take you through the structuring of a decision problem. Decision making under uncertainty and decision making under risk are discussed. The use of decision trees is also illustrated. Many real life situations can be modeled as networks made up of nodes and links. In unit 3 we discuss network modelling and the analysis techniques. In unit 4 we discuss the problem of Project Management. In unit 5 and 6 we discuss the problem of Inventory Control. Unit 5 discusses Deterministic Inventory Control problems and unit 6 discusses Stochastic Inventory Control models. Units 7 and 8 deal with the issue of Queuing Theory. Queuing problems arise when the demands of customers cannot be matched perfectly. Unit 7 discusses simple Single Server Queuing models and unit 8 discusses Multi-Server Queuing models and a few applications. In both these units only analytical procedures are discussed. Queuing problems can also be solved by simulation techniques which are not discussed in this module. Worked examples and exercises are given in each unit for self instruction and assessment. You are encouraged to go through each worked example carefully and do as many exercises as possible to help you understand the techniques covered. We hope this module will help you appreciate the quantitative approach to decision making problems that arise in commerce and industry.
V. Chikwasha C. Sigauke S. Kumar. vii
Unit 1
Introduction to Operations Research 1.1
Objectives
After completing this unit, you should be able to; i define operations research. ii give examples of situations where OR techniques can be applied.
1.2
Introduction
In this unit we give you an overview of what Operations Research is all about and the techniques that are used to solve problems encountered in real life situations. We hope this will motivate you to study the material covered in this module.
1.3
What Is Operations Research?
One fact of modern life is that the organizations with which we all interact and in which many of us earn our livelihood are very complex. These complex organizations generate difficult problems that require decisions. The field of knowledge that encompasses quantitative approaches to decision making is very broad and includes several areas from applied mathematics. This area of study is called operations research or management science. We shall refer to these terms frequently and interchangeably throughout this module. Operations Research is the application of quantitative methods to decision making in commerce, governmental and military organizations, with an objective of improving the quality of managerial decisions. The Operations Research approach to problem solving generally incorporates most, or all, of the following characteristics:
1. Applies the scientific method to develop the problem-solving methodology. 2. Adopts a systems approach to the modelling process. 3. Utilizes a team concept in the modelling process. 4. Makes extensive use of computers in obtaining solutions to problems. 1
2
Introduction to Operations Research
1.4
Some Operations Research Techniques
Some of the quantitative techniques used in Operations Research are:
1. Probability concepts and probability distributions: These are useful for problems involving uncertainty. 2. Decision theory: This is applicable to situations in which the decision maker has several alternative courses of action but is also confronted with an uncertain future set of possible events. 3. Network models: These are applicable to subway, highway and transportation systems, to information retrieval and processing and the planning and control of research and development projects. 4. PERT and CPM, that is, Program Evaluation Review Technique and Critical Path Method: This enable managers to more effectively manage the resources associated with large, complex projects. 5. Queuing or waiting line theory: This is concerned with arrival of customers at one or more service facilities where the demand and timing of the demand of the customers, the time duration of the service operations and the behaviour of the customers as they arrive for service and/or wait in the queue are characterized by uncertainty. The objective of queuing theory is the provision of an adequate but not excessive service facility. 6. Markov analysis: This is a stochastic procedure in which the decision maker tries to analyse and describe the current behaviour of some variable in order to predict the future behaviour of that same variable. 7. Heuristics and metaheuristics: These are problem specific methods designed to solve complex combinatorial optimisation problems. These are some of the quantitative techniques used in Operations Research. The list is inexhaustive. In this module we shall concentrate on some of these techniques.
1.5
Exercises
1. What do you understand by the term Operations Research? 2. List some of the techniques that are used in Operations Research. 3. Give an example of a problem which can be solved using Operations Research techniques and give the Operations Research technique(s) that can be applied to solve such a problem.
Unit 2
Decision Analysis 2.1
Objectives
After completing this unit, you should be able to; i draw a payoff-table for a given decision problem. ii apply the concept of optimistic approach, conservative approach, minimax regret approach to decision making problems. iii calculate the value of perfect information. iv apply the expected monetary value approach to identify the best decision alternative. v draw a decision tree diagram.
2.2
Introduction
In this unit, we introduce you to decision analysis. Decision analysis can be used in situations where the decision maker has several alternative courses of action but is also faced with an uncertain future set of possible events. A decision-making process can fall into one of the following categories: 1. Decision Making Under Certainty: A decision-making environment where the future states of nature are known. 2. Decision Making Under Risk: A decision-making environment where several states of nature may occur as a result of a decision or alternative and the probabilities of these states of nature can be estimated. 3. Decision Making Under Uncertainty: A decision-making environment where several states of nature may occur, but the probabilities of these states of nature are not known. 4. Decision Making Under Conflict: A decision-making environment in which there are two or more decision makers who have goals for which they are competing. In this unit, we shall concentrate on decision making under risk and decision making under uncertainty. 3
4
2.3
Decision Analysis
Structuring the Problem
To illustrate the decision analysis approach, let us consider Example 2.1. Example 2.1 The Robotic Microcomputer company manufactures microcomputers. The company is contemplating the expansion of its manufacturing facilities. Table 2.1 is a payoff table for this company. The payoffs are in dollars. Table 2.1 Pay-off Table for Robotic Microcomputer Decision Alternatives High Demand (s1 ) Moderate Demand (s2 ) Expand present facility (d1 ) 1, 000, 000.00 500, 000.00 Build new facility (d2 ) 2, 500, 00.000 1, 200, 000.00 License another manufacturer (d3 ) 1, 500, 000.00 800, 000.00 Probability P(si ) 0.4 0.4
2.3.1
Low Demand (s3 )
−100, 000.00 −500, 000.00
−50, 000.00 0.2
Decision Alternatives
These are the possible courses of action the decision maker can take. They are denoted by di in Table 2.1, they are d1 , d2 and d3 .
2.3.2
States of Nature
These are all possible future events that might occur. These are denoted by sj . From Table 2.1, they are s1 , s2 and s3 . Note: It is assumed that these states of nature are mutually exclusive and collectively exhaustive.
2.3.3
Payoff
This is the outcome for a decision alternative - state of nature combination that results from selecting a certain decision alternative and then having a particular state of nature occur. We shall denote it by v(di , sj ). For example, from Table 2.1, v(d2 , s1 ) = $2, 500, 000.00 is a payoff.
2.3.4
Payoff Table
This is a tabular representation of the payoff estimates in terms of the interaction of the decision alternatives and the states of nature. Table 2.1 is a payoff table.
Decision Analysis
2.4
5
Decision Making Under Uncertainty
In this section, we consider approaches to decision making that do not require knowledge of the probabilities of the states of nature. Because different approaches sometimes lead to different decision recommendations, it is important for the decision maker to understand the approaches available and then select the specific approach that, according to the decision maker’s judgement, is the most appropriate.
2.4.1
Optimistic Approach
The optimistic approach evaluates each decision alternative in terms of the best payoff that can occur. The decision alternative that is recommended is the one that provides the best possible payoff. For a problem in which maximum profit is desired, the optimistic approach would lead the decision maker to choose the alternative corresponding to the largest profit. For problems involving minimization, this approach leads to choosing the alternative with the smallest payoff. The following example, Example 2.2 uses the data of Example 2.1 to illustrate the Optimistic Approach. Example 2.2 Table 2.2 Optimistic Approach Table Decision alternative d1 d2 d3
Maximum payoff 1,000,000.00 2,500,000.00 1,500,000.00
Since $2, 500, 000.00 corresponding to d2 is the largest profit or payoff (see Table 2.2), the decision to build a new facility is the recommended decision alternative.
2.4.2
Conservative Approach
This approach evaluates each decision alternative in terms of the worst payoff that can occur. The decision alternative recommended is the one that provides the best of the worst possible payoffs. The following example, Example 2.3 uses the data of Example 2.1 to illustrate the Conservative Approach. Example 2.3 Since −$50, 000.00 corresponding to d3 is the maximum of the minimum payoffs (see Table 2.3), the decision to license another manufacturer is recommended.
6
Decision Analysis
Table 2.3 Conservative Approach Table Decision alternative Minimum payoff d1 -100,000.00 d2 -500,000.00 d3 -50,000.00
2.4.3
Minimax Regret Approach
This is a decision criterion in which the decision maker identifies the ”maximum” opportunity loss or ”maximum” regret for each decision alternative and then selects the decision alternative associated with the ”minimum” maximum regret value. The opportunity loss or regret is the difference between the optimal payoff for a particular state of nature and the actual payoff received for a particular state of nature-decision alternative combination. We calculate the opportunity loss using the following formula: R(di , sj ) = V ∗ (sj ) − V (dj , sj ) where R(dj , sj ) = opportunity loss or regret associated with decision alternative dj and state of nature sj , ∗ V (sj ) = best payoff under state of nature sj , and V (dj , sj ) = payoff associated with decision alternative dj and state of nature sj . The following example, Example 2.4 uses the data of Example 2.1 to illustrate the Minimax Regret Approach. Example 2.4 From Table 2.4, the minimum of the maximum regrets is 450,000.00. Therefore the decision to build a new facility is recommended. Table 2.4 Opportunity Loss Table s1 s2 d1 1 500 000 700 000 d2 0 0 d3 1 000 000 400 000
2.5
s3 50 000 450 000 0
Maximum regret 1 500 000 450 000 1 000 000
Decision Making Under Risk
In many decision-making situations, it is possible to obtain probability estimates for each of the states of nature. When such probabilities are available, we use the expected monetary value (EMV) approach to identify the best decision alternative. This approach evaluates each decision alternative in terms of its expected value. The recommended decision alternative is the one that provides the best expected value. We
Decision Analysis
7
now define the expected value of a decision alternative.
2.5.1
Expected Monetary Value (EMV) EMV(di ) =
N X
P (sj )V (dj , sj )
(2.1)
j=1
where sj = state of nature j P (sj ) = probability of occurrence of state of nature sj . V (dj , sj ) = the payoff associated with decision alternative dj and state of nature sj , and N = number of possible states of nature; j = 1, 2, ..., N. NOTE • P (sj ) ≥ 0 PN • j=1 P (sj ) = 1. Example 2.5 Using Example 2.1 expected values for the three decision alternatives can be calculated as follows: EMV(d1 ) = (0.4)($1000000) + (0.4)($500000) + (0.2)(−$100000)
(2.2)
= $580000
(2.3)
EMV(d2 ) = $1380000
(2.4)
EMV(d3 ) = $910000
(2.5)
Similarly
The largest expected monetary value results from selecting the second decision alternative, ”build new facility.”
2.5.2
Expected Value of Perfect Information (EVPI)
Perfect information is complete and accurate information. EVPI is the value of information that would allow the decision maker to be sure that the correct alternative was being chosen EVPI is defined as follows: EVPI =
N X j=1
P (sj )R(d∗j , sj )
(2.6)
8
Decision Analysis
where sj
= state of nature j
P (sj ) = the probability of occurrence of state of nature j R(d∗j , sj ) = opportunity loss or regret associated with the best decision alternative d∗j and state of nature sj . N
= the number of possible states of nature j
j = 1, 2, ..., N.
Example 2.6 Using example 2.5, we have EVPI = (0.4)($0) + (0.4)($0) + (0.2)($450, 000) = $90, 000
(2.7)
Alternatively, EV P I = expected monetary value with perf ect inf ormation (EMVPI ) − expected monetary value (EMV ). EVPI
= (2, 500, 000)(0.4) + (1, 200, 000)(0.4) + (−50, 000)(0.2) − (1, 380, 000) = (1, 470, 000) − (138, 000) = (90, 000).
2.6
Decision Trees
A decision tree provides a graphical representation of the decision making process. Managers often prefer this graphical approach because it allows them to see exactly what is happening at each stage of the decision making process. Thus a payoff table can also be represented diagramatically in the form of a decision tree. A decision tree shows the alternative actions that a decision maker can follow as well as the events (states of nature) that can arise, by means of branches. decision trees incoporate points where a decision must be made and points indicating the events that follow on the decisions. These points are shown by means of the following symbols, known as nodes. = decision node that shows that at those points a choice must be made between one or more alternatives.
= the event node where one or more uncertain events can occur.
To illustrate the decision tree approach let us consider the example in Section 2.6.1.
2.6.1
Sequential Decision Trees
Example 2.7
Decision Analysis
9
Tigamuchire Investors of Bulawayo recently purchased land and wants to determine the size of a holiday resort it should build. The sizes of holiday resorts being considered are: small, medium and large. At the same time, an uncertain economy makes it difficult to ascertain the demand for the new resort area. Tigamuchire’s management realizes that a large development followed by a low demand could be very costly to the company. However, if Tigamuchire makes a conservative small development decision and then finds a high demand, the firm’s profits will be lower than they might have been. With the three levels of demand (low, medium and high), Tigamuchire’s management has prepared the profit ($1000s) payoff matrix in Table 2.5. The probabilities of each of the states of nature are: Table 2.5 Tigamuchire Investors Payoff Table Decision DEMAND Alternatives Low(s1 ) Medium(s2 ) Small(d1 ) 400 400 Medium(d2 ) 100 600 Large(d3 ) -300 300
High(s3 ) 400 600 900
P (High demand) = 0.45 P (M edium demand) = 0.35 P (Low demand) = 0.20
1. Construct a decision tree for this problem. 2. What is the recommended decision using, (a) the decision tree approach? (b) the expected value approach? (c) comment on the above recommended decisions. 3. What is the expected value of perfect information (EVPI)? 4. The resort company is conducting a survey that will help to evaluate the demand for the development. The survey will result in three indicators of demand: weak (I1 ), average (I2 ) and strong (I3 ). The conditional probabilities are shown in Table 2.6: Table 2.6 Conditional Probabilities for Survey Conducted P (Ik /sk ) I1 I2 I3 s1 0.6 0.3 0.1 s2 0.4 0.4 0.2 s3 0.1 0.4 0.5
10
Decision Analysis
(a) What is the resort company’s optimal strategy? (b) What is the value of the survey information? (c) What is the efficiency of the survey information? Solution (1). Figure 2.1 gives the decision tree for Tigamuchire problem. 0.45
Small
400
0.35
D1
Medium
500
0.35
D2
D3
M
(400)
L
(400)
H
(600)
M
(600)
L
(100)
0.20
0.45 0.35 450 0.20
Large
(400)
0.20
0.45 500
H
H (900) M
(300)
L (-300) Decide not to build Figure 2.1 Decision Tree for Tigamuchire Problem 2(a). Using Figure 2.1, the recommended decision is: Build medium resort. (b). Build medium resort. (Use Figure 2.1). (c). From (a) and (b) above, the same decision is arrived at. 3. Refer to Figure 2.2. EV P I = EM V P I − EM V = $695 − $500 = $195 4. Refer to Figure 2.3. P (I1 ) = 0.2 × 0.6 + 0.35 × 0.4 + 0.45 × 0.1 = 0.305 P (I2 ) = 0.2 × 0.3 + 0.35 × 0.4 + 0.45 × 0.4 = 0.380 P (I3 ) = 0.1 × 0.2 + 0.35 × 0.2 + 0.45 × 0.5 = 0.315. Then P (S1 /I1 ) =
T P (S1 I1 ) 0.2 × 0.6 = = 0.393 P (I1 ) 0.305
Decision Analysis
11
S 400 L (0.2)
S M (0.35)
695
M L
400
600
M
600
L S
H (0.45) 900
M L
900
Figure 2.2 Decision Tree for EMVPI for Q3.
P (S2 /I1 ) = P (S3 /I1 ) = P (S1 /I2 ) = P (S2 /I2 ) = P (S3 /I2 ) = P (S1 /I3 ) = P (S2 /I3 ) = P (S3 /I3 ) =
0.35 × 0.4 = 0.459 0.305 0.45 × 0.1 = 0.148 0.305 0.2 × 0.3 = 0.158 0.380 0.35 × 0.4 = 0.368 0.38 0.45 × 0.4 = 0.474 0.38 0.10 × 0.2 = 0.063 0.315 0.35 × 0.2 = 0.222 0.315 0.45 × 0.5 = 0.714 0.315
(a) Have a survey carried out (b) Value of survey information = $538492 − $500000 = $38492 EV SI 38492 (c) Efficiency of sample information = EV P I × 100 = 195000 × 100 = 19.7%
2.7
Exercises
1. Describe what is involved in a decision theory approach. 2. Discuss the difference between decision making under certainty, under risk and under uncertainty. 3. What is the difference between prior and posterior probabilities?
12
Decision Analysis
S 403.5
M L
I1
0.305 538.5 #
521
M
"! 0.38
survey
S
I2 Have a
S1 (0.393) S2 (0.459) S3 (0.148)
L
0.315
690.30
No survey
I3
538.5
S M
L
500
S1 (0.158) S2 (0.368) S3 (0.474)
S1 (0.063) S2 (0.222) S3 (0.714)
Figure 2.3 Decision Tree for Example 2.7. Q4. 4. A firm produces a perishable food product at a cost of $1, 000 per case. The product sells for $1, 500 per case. For planning purposes, the company is considering possible demands of 100, 200 or 300 cases. If the demand is less than production, the excess production is lost. If demand is more than production, the firm, in an attempt to maintain a good service image, will satisfy the excess demand with a special production run at a cost of $1800 per case. The product, however, always sells at $1500 per case. (a) Set up the payoff table for this problem. (b) If P (100) = 0.2, P(200) = 0.2 and P (300) = 0.6 should the company produce 100, 200 or 300 cases? (c) What is the EVPI? 5. The research and development manager for COSD company is trying to decide whether or not the company should fund the development of a new lubricant. It is assumed that the project will be a major technical success, a minor success or a failure. The company has estimated that the value of a major success is $150, 000, since the lubricant can be used in a number of products the company is making. If the project is a minor success, its value is $10, 000, since COSD feels that the knowledge gained will benefit some other ongoing projects. If the project is a failure, it will cost the company $100, 000. Based on the opinion of the scientists involved and the manager’s own subjective assessment, the assigned
Decision Analysis
13
prior probabilities are as follows: P (majorsuccess) = 0.15 P (minorsuccess) = 0.45 P (f ailure) = 0.40. (a) Using the expected value approach, should the project be funded? (b) Suppose that a group of expert scientists from a research institute could be hired as consultants to study the project and make a recommendation. If this study will cost $30, 000, should COSD company consider hiring the consultants? 6. A quality control procedure involves 100% inspection of parts received from a supplier. Historical records indicate that the defective rates shown in Table ?? have been observed. The cost to inspect 100% of the parts received is $250 for Table 2.7 Defective Rates Table Percent defective 0 1 2 3
Probability 0.15 0.25 0.40 0.20
each shipment of 500 parts. If the shipment is not 100% inspected, defective parts will cause rework problems later in the production process. The rework cost is $25 for each defective part. (a) Complete the payoff Table 2.8, where the entries represent the total cost of inspection and reworking. Table 2.8 Payoff Table Inspection 100% Inspection No inspection
Percent Defective 0 $250
1 $250
2 $250
3 $250
(b) The plant manager is considering eliminating the inspection process to save the $250 inspection cost per shipment. Do you support this action? Use the expected value to justify your answer. (c) Draw the decision tree for this problem. (d) Suppose a sample of five parts is selected from the shipment and one defect is found. Let I = 1 defective part in a sample of 5. Use the binomial probability distribution to compute P (I/s1 ), P (I/s2 ), P (I/s3 ) and P (I/s4 ). The binomial probability function is as follows: f (x) =
n! px (1 − p)n−x x!(n − x)!
14
Decision Analysis
where n = the sample size, x = the number of defects, and p = the proportion defective In this problem, n = 5, x = 1 and p = 0, 0.01, 0.02 and 0.03. (e) If I occurs, what are the revised probabilities for the states of nature? (f) Should the entire shipment be 100% inspected whenever one defective part is found in a sample of size 5? (g) What is the cost saving associated with the sample information?
Unit 3
Network Models 3.1
Objectives
After completing this unit, you should be able to; i obtain a shortest connected graph of a given network. ii select a method which is appropriate in a given situation to find the shortest route between two given nodes in a network and determine the route and its length. iii find the maximum flow in a given network. iv find the minimal spanning tree in a given network.
3.2
Introduction
Many problems can be presented and analyzed as network problems. In this unit, we discuss a few problems which can be classified as routine problems. We formulate some of these problems and also solve them, in some cases using more than one method.
3.3
Examples of Network Problems
In this section, we state a few situations, which can be formulated and analyzed as network problems. At this stage you are expected to think about how they could be represented and how they might be solved. The purpose is to motivate you. Situation 3.1: A nation-wide chain store is interested in supplying a special product to its retail outlets from its various warehouses. The problem is to minimize the total cost of transportation. A similar but different situation is the supply of a perishable product to a set of retail outlets. Here, the objective is to minimize the time of the slowest delivery. Situation 3.2: In a machine shop, a batch of jobs is to be assigned to a group of machines. The problem is to find the pair which will maximize the total efficiency. Situation 3.3: Oil is to be distributed from several oil fields to several refineries which are connected by a network of pipes. If the capacities of the pipes are known, what is the maximum flow that can reach a given refinery? Situation 3.4: A firm has a table of distances between each pair of cities. This firm wants to find the shortest distance and route between each pair of these cities.
15
16
Network Models
Situation 3.5: A salesman wants to visit a number of cities and return to his home city in the shortest time or cost or distance. Situation 3.6: A project consists of a large number of activities which must be done in a given sequence in order to complete the project. How do you find the minimum time to complete the project? Situation 3.7: A set of computer terminals are to be connected in a network. How should the connection be made so that terminals can access every other terminal but minimizing the cost of materials for connection? In the following section, we now consider a few specific problems which have been formulated as network models.
3.4
Terms and Definitions
We define some terms and concepts which will be used in this Unit. They are explained with reference to Figure 3.1. An Isolated Terminal is a node which has no connection with any other node
6
11
3
5
9
7
2
4
1
10`
8`
Figure 3.1 in the network. In Figure 3.1, node 2 is an isolated terminal. A Fragment is a subset of nodes connected by direct links. For example, (5-6, and 8-10-9-11) are two of the fragments in Figure 3.1. An Isolated fragment is one which at a given stage of construction has no connections made to the external nodes. For example, in Figure 3.1 the fragment made by nodes (5-6) is an isolated fragment.
Network Models
17
Distance of a node from a fragment of which it is not an element is the minimum of its distances from individual nodes forming the fragment. For example, in Figure 3.2, the distance from the node 2 to the fragment(1-4) is 3. A Connected graph has an unbroken chain between every pair of nodes of a graph. For example, the graph in Figure 3.1 is not a connected graph. However, if connected (1-2),(5-7) and (7-9), the graph in Figure 3.1 will be a connected graph. A Complete graph: a graph is said to be complete if there is a link between
3
6
7
1
4
5
3
2
Figure 3.2 each pair of nodes. A cyclic network is a graph which has no loops or cycles and links are directed. Cyclic network includes cycles or loops and links may be directed or non-directed. A tree is a non-directed network without loops. See Figure 3.3. Connected tree is a graph which is connected and free of loops. Shortest connected graph or shortest spanning tree is a connected graph with the smallest possible total length.
3.5
Shortest Spanning Tree
The problem is to find a connected graph with the smallest possible total length. These kind of trees are useful in situations where cost of construction is high but the cost of operating that system is not proportional to the length. For example, the construction of a canal or telephone lines is expensive but operation is not dependent on the distance. Methods are given by Kruskal (1965), Prim(1957) and Garg and Kumar (1968). We describe the procedure by Prim.
18
Network Models
4
2
7
5
1
8
6
3
Figure 3.3
Construction Principle 1. Any terminal can be connected to a nearest neighbour. 2. Any isolation fragment can be connected to a nearest neighbour by shortest available link. Example 3.1 Find the shortest connected graph for the distance Table 3.1. The distance Table 3.1 makes the graph shown in Figure 3.4. Table 3.1 Distance Table Terminal 1 2 3 4 5
2 3.6
3 4.6 5.6
4 3.2 6.7 5.1
5 4.4 2.8 8.0 7.3
6 4.0 5.2 8.5 3.7 3.4
Solution 3.1 We apply step 1 and get the link(1-4). Now apply step 2 on the fragment(1-4). Nearest distance is 3.6 from node 2 to node 1. Hence the new fragment is (2-1-4). Again apply step 2 to the fragment (2-1-4) to find the new node which is nearest, but does not form a loop. This is node 5 when connected to node 2. Thus, forming the new fragment (5-2-1-4). In an n node network, we have to continue to find n − 1 links. The shortest connected graph is shown in Figure 3.5. The minimum length of this graph is 17.6.
Network Models
19
1
6.7
2
3.6
7.3
5.6
3.2
2.8
6
5.7 5.2
5.1
4.0
3
4.4
8.5
4.6
5
8.0
3.4
4
Figure 3.4
3.5.1
Extensions
1. The distance associated with an edge may be negative, the method is still applicable. 2. The method is applicable when finding the largest spanning tree. Simply replace each distance lij with edge (ij) in the given network by (−lij ) and find the shortest spanning tree. The largest tree is obtained when the total distance is once again multiplied by (-1). Example 3.2 Find the longest spanning tree for the graph given in Figure 3.6. The required tree is shown in Figure 3.7. Activity 3.1 Obtain the shortest and the longest tree for the graph in Figure 3.6 and show that total length of all other trees satisfy the condition 17.6 ≤ T L ≤ 35.1.
3.6
The Shortest Route Problem
The shortest route problem is concerned with the determination of the shortest route between two fixed nodes of a given network for a given set of non-negative numbers associated with various branches of the network. These fixed nodes are called origin and destination nodes. The non-negative number associated with a branch may represent distance, profit or cost associated with that branch. Several methods have been developed for this problem. We present two in this section.
20
Network Models
4
2
3.6
6`
3.2
2.8
1
3.4 4.6
5
3
Figure 3.5
3.6.1
Acyclic Directed Network
An acyclic directed network, as defined earlier, contains directed arcs and no loops. Because of this special property (i.e. all arcs are directed and the network is free of loops), one can number each node in such a way that each arc leads from a smaller number to a node with a larger number assigned to it. Consider for example the network shown in Figure 3.8.
Simple Algorithm This problem can be solved by a simple algorithm. The steps are as follows:
• Step 1. Associate with node i a number ni which represents the shortest distance from node 1 to node i. Set n1 = 0 (3.1) • Step 2. Set nj = mini
# C = (2,3) ~ -
1
B = (1,3)
C
A
A = (1,2)
B
3
"!
Figure 4.5 Numbering a Network 1. Networks proceed from left to right i.e. time flows from left to right. 2. Networks are not drawn to scale. 3. Nodes should be progressively numbered from left to right with head nodes having a higher number than tail nodes, Figure 4.5. 4. Activities are often identified by their start and finish node numbers. The general notation is (i, j), e.g. (1,2) where i is the start (tail) node and j is the finish(head) node. Drawing networks can be difficult. You will nearly always need to draw the network out a few times in rough before you finally get it right. Remember once you have drawn the network to do a final check to see if the logic follows that of your activity dependency table. Hence the completed network for the sports shop problem will be as in Figure 4.6.
4.5
Critical Path
The main objective of project management is to establish the overall completion time of projects by identifying the critical path. The critical path gives the shortest time in which the whole project can be completed. It is the chain of activities, which has the longest duration; hence it is ’critical’ to complete these activities on time if the overall project is to be completed by the earliest time possible. It is worth noting that there may be more than one critical path in a network and it is possible for the critical path to run through a dummy activity. In order to find the critical path, we need to insert the activity duration times on the network. We can then follow a procedure to analyze the start and finish times of all the activities in order to establish the ’critical’ activities and find the critical path. The procedure is as follows:
Project Management
37
2
Decorate Exterior
Sandblast Exterior
X(0)
Decorate -
1
B(6)
C(15) Buy
D(10)
A(10) ?
3
Arrange Window Display -
G(2) 6
Y(0)
H(5)
Stock
q
5
Stock shelves -
I(10) >
7
7
F(10)
Plan Store Layout
Advertisement Campaign
w
4
E(3) -
6
Prepare Advertisements
Figure 4.6 Network for the Sports-shop problem
4.5.1
Step 1: Forward Pass
We carry out a forward pass, starting at event node 1, through the network, calculating the earliest event start times (EST) for each node. The EST is the earliest time which all the activities leading to that event can be completed. To determine the EST of an event node, add the duration of each activity entering the event node to the EST of that activity tail event node and set the EST for the event node to be the maximum of all these. NOTE; the EST of node 1 will be zero. Hence the earliest event start times for the sports shop problem will be calculated as in Table 4.3. Thus, it will take 30 working days to complete the tasks, because 30 is the earliest time by which all the activities will have been completed. Note: The Latest Finish Time of an activity entering a particular node is equal to the smallest of all latest start times for all activities leaving the node.
4.5.2
Step 2: Backward Pass
We then carry out a backward pass, starting at the final event node, through the network, calculating the latest event start times (LST) for each event node. The LST is the least possible time at which each particular event can be completed without holding back the completion date of the project as a whole.
38
Project Management
Table 4.3 Earliest Event Start Times Event Node Start Times Notes 1 0 project starts at time 0 2 0 + 10 = 10 Actvity A 4 0 + 15 = 15 Activity C 3 10 + 0 = 10 Activity X(dummy) 0+6=6 Activity B 15 + 0 = 15 Activity Y(dummy) 5 15 + 2 = 17 Activity G 15 + 5 = 20 Activity H 6 15 + 3 = 18 Activity E 7 10 + 10 = 20 Activity D 20 + 10 = 30 Activity I 18 + 10 = 28 Activity F
EST 0 10 15 15
20 18 30
To determine the LST of each event node, subtract the duration of each activity leaving the event node from the LST of that activity’s lead event and set the LST for the event node to be the minimum of all these. As a check, the LST of event node 1 will always be zero, i.e. the same as its EST. Hence the latest start times for the sports shop problem will be calculated as: Latest Event Start T ime = Latest F inish T ime − Duration of Activity Table 4.4 Latest Event Start Times Event Node Start Times 7 30 6 30 − 10 = 20 5 30 − 10 = 20 3 20 − 2 = 18 4 20 − 5 = 15 18 − 3 = 15 18 − 0 = 18 2 30 − 10 = 20 18 − 0 = 18 1 18 − 10 = 8 18 − 6 = 12 15 − 15 = 0
Notes from forward pass Activity F Activity I Activity G Activity H Activity E Activity Y (dummy) Activity D Activity X (dummy) Activity A Activity B Activity C
LST 30 20 20 18 15
18 0
Often these calculations are performed on the network itself.
4.5.3
Step 3: Calculate Slack
Some activities, such as B, have some flexibility for their scheduling. The interior decoration could have started any time up to 9 days after the start of the project without
Project Management
39
delaying the overall time for opening the shop. This flexibility is called the slack (or total float) of an activity. For each activity the total float can calculated by:
T otal f loat of an activity
=
LST of head event − EST of tail event − Duration of the activity.
If an activity has zero slack, it means that any delay of the activity will delay the overall project, hence such activities are called ’critical’ and form the critical path. Hence, for the sports shop example the total float of each activity can be calculated as in Table 4.5. Hence the critical path for this problem is: C − H − I that is, buying the Table 4.5 Total Float of an Activity Activity LST (head) EST (tail) A 18 0 B 18 0 C 15 0 D 30 10 E 20 15 F 30 18 G 20 15 H 20 15 I 30 20 X (dummy) 18 10 Y (dummy) 18 15
Duration 10 6 15 10 3 10 2 5 10 0 0
Total float 8 12 0 10 2 2 3 0 0 8 3
stock, planning the store layout and stocking the shelves. These activities must not be delayed at all in order for the shop to open in 30 days. However, we can delay the exterior decoration by up to 10 days and still finish on time. Note that the total float belongs to a sub-path in the network rather than to a particular activity, hence the completion time will not be delayed if any one activity is delayed by its maximum total float time. However, if more than one activity on the same sub-path, e.g. A and D, are both delayed by their maximum total float times, then the total project completion will be delayed. We can calculate the total time by which an activity can be delayed without affecting any succeeding activities in the network. This is called free float and is calculated by:
F ree f loat of an activity = EST of head event − EST of tail event − Duration of activity.
40
Project Management
Hence, for the sports shop example the free float of each activity can be calculated as follows as in Table 4.6. Note how the free float assigns all of the slack of a sub path to Table 4.6 Free Float of an Activuty Activity EST(Head) EST(Tail) A 10 0 B 15 0 C 15 0 D 30 10 E 18 15 F 30 18 G 20 15 H 20 15 I 30 20 X(dummy) 15 10 Y(dummy) 15 15
Duration 10 6 15 10 3 10 2 5 10 0 0
Free Float 0 9 0 10 0 2 3 0 0 5 0
the final activity on that sub path. We can also calculate the total time by which an activity can be delayed without affecting any other activities in the network. This is called the independent float, and is calculated by: Independent F loat of an activity = EST of head event − LST of tail event − Duration of activity. Hence, for the sports shop example the independent float of each activity can be calculated as in Table 4.7. Table 4.7 Independent Float of an Activity Activity EST(Head) LST(Tail) Duration A 10 0 10 B 15 0 6 C 15 0 15 D 30 18 10 E 18 15 3 F 30 20 10 G 20 18 2 H 20 15 5 I 30 20 10 X(dummy) 15 18 0 Y(dummy) 15 15 0
Independent Float 0 9 0 2 0 0 0 0 0 -3 0
Note that it is only total float that has any meaning as a negative float and hence the −3 here does not cause concern.
Project Management
4.6
41
Project Scheduling with Uncertain Activity Times
When a project has not been attempted previously, the time for completing each activity in the project are liable to be uncertain. This does not mean that the technique of project management cannot be used. However, we do have to consider the uncertainty in the times, particularly since many large projects may have a ”fine” imposed on them for late completion. The network for a project where some, or all of its activity times are uncertain is drawn in the same way as has previously been described. The difference comes with the activity duration times that are placed on the network. Usually, when an activity duration is uncertain, three time estimates will be given (or sought), namely: • optimistic time o = the activity time under ideal conditions; • most probable time m = the most probable activity time under normal conditions and; • pessimistic time p = the activity time if significant delays are encountered. These estimates are then treated as random variables from a probability distribution (the beta distribution). The three values are combined to give an expected value for the activity duration, which is then used to calculate the expected project duration and the critical path. The variances of each of the expected activity times are also calculated so that the probability of exceeding any particular project duration can be calculated.
Example 4.2 The product development group at ZOU Computing has been working on a new computer software product that has the potential to capture a large market share. Through outside sources, ZOU Computing’s management has learnt that a competitor is working to bring a similar product to the market. As a result, ZOU Computing’s management has increased its pressure on the product development group. The group’s leader has turned to project scheduling techniques as an aid to scheduling the activities remaining before the new product can be brought to the market. The project network is shown in Figure 4.7 and the activity time estimates (in weeks) are given in Table 4.8.
To be able to determine the expected project duration for this project, we first have to determine the expected duration for each activity involved in the project. The expected duration is the duration by which there is a 50% chance that the activity will exceed the specified duration and a 50% chance that it will be completed quicker than the expected duration.
42
Project Management
D
2
-
6 6
4
I A
G #
C
1
^
7
6
"! >
H B
R
3
E
J R
F
- 5
Figure 4.7 Network for ZOU Computing Problem The underlying assumption that is made when calculating the expected duration is that the distribution of activity duration follows a beta distribution. Using this assumption, the expected duration for each activity can be calculated as follows: Expected duration of activityi, ti =
o + 4m + p 6
(4.1)
The expected durations for each of the above activities are therefore given in the Table 4.9. It is these expected activity times that are then used on the network to determine the critical path and expected project duration. The calculations of these are carried out in the same away as previously when the exact activity durations were known. The earliest and latest start times for this project would therefore be as shown in Figure 4.8. Calculating the total floats for each activity gives the results shown in Table 4.10. Therefore the critical path is made up of the activities: B - E - H - J and the minimum project duration using the expected activity completion times is 26 weeks. Since these are expected activity times, it must be noted that there is a 50% chance that the project will take longer than this time. Since this only gives the project duration for which there is a 50% chance of completing 26 weeks, we also need to consider the variability on the network to determine the probability of completing the project within other time scales. To do this, we first calculate the variance for each activity duration. Once again, the assumption used is that the activity duration times follow a beta distribution.
Project Management
43
Table 4.8 Activity Time Estimates (in weeks) Activity Optimistic Most Probable A 3.0 4.0 B 3.0 3.5 C 4.0 5.0 D 2.0 3.0 E 6.0 10.0 F 7.5 8.5 G 4.5 6.0 H 5.0 6.0 I 2.0 2.5 J 4.0 5.0
Pessimistic 5.0 7.0 6.0 4.0 14.0 12.5 7.5 13.0 6.0 6.0
Table 4.9 Expected Duration of an Activity Activity Expected Duration (o + 4m + p)/6 A 4 B 4 C 5 D 3 E 10 F 9 G 6 H 7 I 3 J 5
The variance of each activity is calculated by: σi2
(p − o)2 = 36
(4.2)
The activity variances for ZOU Computing would therefore be as shown in Table 4.11.
If we want to know the probability of completing the project within 25 weeks say, we need to consider the variability in the critical activities. NOTE: Ordinarily the variation in the non-critical activities will not affect the overall completion time of the project, hence we consider only those critical activities. This is not always the case though. If one of the non-critical activities has a high variance and a small total slack, it is possible for that activity to become critical. All activities on the project should be monitored closely when the activity times are uncertain. To determine the probability that a project is completed within a specified time, we
44
Project Management
[9,20]
[20,23]
D
2
-
6 6
(3)
4
I A
G (4)
1
(6)
[14,14] #
C (5)
[26,26] ^ 7
6
H (10) (4)
"! >
[0,0] B
(3)
R
(7) E F
3
(9)
[4,4]
J (5)
R 6 - 5
[21,21]
Figure 4.8 EST and LST Network assume that the total project completion time (which itself is uncertain due to the uncertainty in the durations of individual activities) follows a normal distribution. The parameters of this distribution will be:
µ = σ =
X
µcritical
qX
2 σcritical
These can be used, along with the assumption that the times are normally distributed, to calculate the probabilities of completing the project in any given time. For ZOU Computing, the expected project completion time has already been calculated as : µ = 4 + 10 + 7 + 5 = 26 weeks The variance of the critical path is given by: σ 2 =√0.44 + 1.78 + 1.78 + 0.11 = 4.11 weeks σ = 4.11 = 2.03 weeks The calculation of the z-value for the normal distribution is : X −µ z= σ
(4.3)
Project Management
45
Table 4.10 Total Activity Floats Activity A B C D E F G H I J Table 4.11 Activity Variance Table Activity A B C D E F G H I J
Total Float 16 0 11 11 0 8 3 0 3 0
Variance (p − o)2 /36 0.11 0.44 0.11 0.11 1.78 0.69 0.25 1.78 0.44 0.11
where X is the desired project completion time. This z-value can be looked up in the Normal Distribution tables to give the probability of completing the project in that time. If ZOU Computing wish to complete their project in 25 weeks, the z-value will be: z=
25 − 26 = −0.49. 2.03
From Normal Distribution tables this gives a probability of 0.3121(31.21%) of completing the project within 25 weeks. Note that this is the probability of completing the project on or before the 25 weeks. The probability of completing the ZOU Computing project within 30 weeks is calculated by: z=
30 − 26 = 1.97 2.03
From Normal Distribution tables, this gives a value of 0.9756. Therefore there is a 97.56% probability of completing the project within 30 weeks.
46
Project Management
It is also possible to determine how long you should quote for a project duration if you want to have a known probability of completing the project in time. This can be particularly important when fines are imposed for late completion. If ZOU Computing want to be 95% certain of completing the project on time, the project duration that should be quoted is calculated by: From Normal Distribution tables the value of z giving a probability of 0.95 is approximately 1.645 (interpolating between 1.64 and 1.65). Hence: X − 26 z= = 1.645 2.03 Solving for X gives: X = 1.645(2.03) + 26 = 29.34 weeks Therefore, to be 95% certain of completing the project on time, they should quote the project duration as 29.4 weeks. Rounding off to the nearest whole number gives 30 weeks.
4.7
Exercises
1. What is the basic difference between CPM and PERT? 2. What are ”dummy” activities and when are they used? 3. What assumptions are made in computing the expected activity times and variances in a PERT network? 4. Describe briefly how you determine the critical path in a network. 5. A building contractor has recently been awarded a contract to build a three storey office block. The method of construction is by pre-fabricated units and after the shell for a particular floor has been put in place by the fabrication gang (this takes 3 weeks for each floor), the glazing, plumbing, electrical and painting work may be carried out by the services gang, who take 4 weeks on each floor. Once the shell for any floor has been put together, the fabrication gang may start work on the next floor above. When the services gang has finished a floor, it can move onto the next floor above provided that the fabrication gang has finished that floor. When the fabrication gang has finished the third floor, work on the roof can begin (this will take 3 weeks). The building is considered finished when the roof has been completed and the services gang has completed the work on each floor. Before the fabrication gang can begin work on the first floor, the foundations have to be laid, which will take 10 weeks. (a) Draw the network. (b) Calculate the total time required to build the three storey office block, identifying the critical activities. (c) Find the total float and free float for each task. 6. The economic plan of a country includes a scheme to improve the agriculture of a particular region. This is to be achieved by building a dam to irrigate the
Project Management
47
region and to generate hydroelectric power, which will be used in the production of fertilizers in a plant to be built. The building of the dam, the irrigation system and the plant may all start together and be carried out simultaneously. The irrigation system takes 7 years to construct and the fertilizer plant requires 5 years. The dam needs 5 years to built and then it takes 3 years to fill. The construction of the power station, which takes 4 years can start only after at least 3 years work on the dam construction. Once the dam is full, irrigation can start immediately but the hydroelectric station requires a year to commission before electricity is produced. Then fertilizer production can start and it will require a year to build up stocks before any are available for use. The land must be irrigated for 4 years and the fertilizer applied for 3 years before any crops are harvested. Only after a further 7 years is the annual crop sufficient to meet the region’s needs. How long will this be after the start of the whole project and by how much can each individual activity be delayed if the region is to be self-sufficient in the least possible time? 7. Table 4.12 is a list of activities required for a new product launch. Table 4.12 Activity Duration Table Activity Duration(weeks) A 6 B 3 C 4 D 4 E 9 F 8 G 10 H 15 I 5
Must be preceded by C A A,B,D A,B,D C E,F
(a) Draw the network diagram for this project. (b) What is the critical path and the minimum time for completion of the project? (c) How would the minimum project completion time be affected if, i. Activity B required 6 rather than 3 weeks? ii. Activity A required 8 rather than 6 weeks?
48
Project Management
Table 4.13 Activity Precedence Activity Immediate Optimistic Predecessor Time(hrs) A 4 B 1 C A 3 D A 4 E A 0.5 F B,C 3 G B,C 1 H E,F 5 I E,F 2 J D,H 2.5 K G,I 3
Most Likely Time(hrs) 6 4.5 3 5 1 4 1.5 6 5 2.75 5
Pessimistic Time(hrs) 8 5 3 6 1.5 5 5 7 8 4.5 7
8. The project shown in Table 4.13 has been analyzed. (a) Draw the network for this project. (b) Identify the critical path and give the estimated project completion time. (c) What is the probability that the project will be completed within one day (24 hrs)? (d) If the company does not want more than a 10% chance of the project over running, what time should they quote for the project duration? Give your answer to the nearest hour.
Unit 5
Deterministic Inventory Models 5.1
Objectives
After completing this unit, you should be able to; i lists the elements of inventory control. ii calculate the Economic Order Quantity (EOQ) for a single item inventory model. iii calculate the reorder point for a given inventory model. iv calculate the cost of inventory for a given model.
5.2
Introduction
An inventory is a stock of physical goods in the form of raw material or finished goods for current or future use. Inventories are kept to avoid cost, time and nuisance of having to obtain an item at the moment when it is required. Inventory management entails two types of closely related problems: 1. planning for inventories 2. control of inventories. In planning for inventories, one is concerned with what to store or what toproduce as well as from where to purchase and what are the most economical arrangements for procurement. These problems can be studied in the context of a specific industry. In this unit, we shall not be concerned with this kindof inventory problems. The control aspect of inventory management is concerned with decisions such as when to order or when to produce and how much to order or how much to produce. The objective in these problems is to minimize the total cost of the inventory. The control of inventory management has a mathematical structure which we shall study in this unit and the next unit on stochastic inventory models. Inventory control is mainly a function of two types of events: 1. demands which diminish the existing stock 2. replenishments which increase the existing stock. It is obvious that if one wants to avoid frequent replenishments, one will be required to hold large inventories. If however, one does not want to carry any inventory, one will be involved in continual replenishments. On the other hand, keeping excessive inventory is 49
50
Deterministic Inventory Models
costly. In most practical situations, a trade-off between these two extremes is desirable. Inventory control deals with the above trade-off between the two extreme situations. Reducing the cost of inventories is an important problem, as the costs involved can be enormous. For example, the value of inventories in the department stores like Woolworths, Sales House, CW, etc., would be in the region of hundreds of millions of dollars. Inventory control has applications not only in industrial processes, but also in many other areas such as personnel planning, libraries, hospitals, tertiary institutions, etc. In this unit, we study deterministic inventory models. Probabilistic models will be covered in the unit 6.
5.3 5.3.1
Elements of Inventory Control Demand
The main concern in inventory control is to meet the future demand at minimum cost. Therefore, we must be able to forecast the demand. This requires the study of demand processes. Sometimes a production process is required to meet contractual obligations. In such cases, the demand is known with certainty. In many other situations, different demand levels can be predicted with associated probabilities. We then have probabilistic demand. Demand must also take into account factors such as seasonality, competition, etc.
5.3.2
Lead Time
The time between ordering and actually receiving of the items is known as the lead time. In the case of production processes, lead time is the time interval between the set up of, and actual output from, the machines. Lead time can either be deterministic or probabilistic.
5.3.3
Inventory Systems
An inventory system is a framework for integrating the information for a given inventory situation in order to minimize the total inventory cost. We have two main inventory systems. These are: • fixed-quantity systems (continuous review system) • fixed-time systems (periodic review system). In a fixed-quantity system, the size of the order is determined by an Economic Order Quantity (EOQ) model. This order size is kept fixed. The purchase order is issued as soon as the supply on hand reaches a predetermined level called the re-order point. In a fixed-time system, the time between orders remains constant. The size of the
Deterministic Inventory Models
51
order fluctuates. These two systems in determinstic models do not vary greatly. However, in the probabilistic models, one has to be careful about which system to adopt.
5.4
Notation and Types of Inventory Costs
We use the following notation to develop the models in inventory control: T.C d
= Total inventory cost per unit time = Demand per unit time (units per unit of time, for example, per year or per month) C1 = Cost of production per unit or purchase price per unit ($/unit) Q = Quantity per order (number of units) Q∗ = Optimum quantity h = Holding or carrying cost per unit ($).(It is convenient to represent the holding cost as a percentage of average dollar inventory.) n = Number of orders per unit time t = Time between orders or length of an inventory cycle C0 = k = Cost of processing an order or cost of set up in the production process L = Lead time Cs = Cost of shortage In general, inventory costs can fall into one of the four categories: (i) basic costs of inventory items (ii) inventory ordering costs (iii) inventory holding or carrying costs (iv) inventory shortage costs
5.5 5.5.1
(5.1)
Single Item Inventory Models The Basic Economic Order quantity (EOQ) Model
Consider a very simple situation where demand is costant and replenishment is instantaneous. It is assumed that bulk quantity discounts are not available, i.e. purchase price per unit is constant, and shortages are not allowed, so that demand must be met. Situations where the above assumptions are applicable are, for example: 1. clerical supplies such as paper, penciles etc., 2. routine industrial supplies such as nuts, bolts, washers etc. In this model, inventory costs (i) and (iv) mentioned in (??) are not applicable and shortages are not allowed. Thus, the inventory costs consist of (ii) and (iii) only. Owing to constant demand rate, the variation in the level of the inventory is as shown in Figure 5.1. From Figure 5.1 it is easy to see that:
52
Deterministic Inventory Models
Points in time at which the orders are placed
Inventory level Maximum Invetory level Q
6 ?
?
t = Q/d
?
?
Average Inventory Q/2
-
- Time 6
Figure 5.1 Variation in Level of Inventory for Basic EOQ Model.
holding cost per unit = (average inventory) × (holding cost per unit time) = Qh 2 and ordering cost per unit = (number of orders per unit time ) × (cost per order) = nk = dk Q Then Total cost for one cycle = h
QQ +k 2 d
(5.2)
and Total cost per unit time(T C) =
Qh dk + . 2 Q
(5.3)
To obtain the minimum total cost we differentiate (5.3) with respect to Q and set the derivatives equal to zero. Then h dk − 2 =0 2 Q and so the Q∗ of Q which minimizes the total cost is given by: r ∗
Q =
2dk h
(5.4)
The optimal value of Q which minimizes the total cost is Q∗ . This is known as the Economic Order Quantity (EOQ). This result has a variety of applications. We shall discuss a few numerical examples.
Deterministic Inventory Models
53
Activity 5.1 1. Explain why the minimization of the expression in (5.2) leads to a meaningless result. 2. Explain what happens if h, d and k are very small or very large.
Example 5.1 The demand for an item is 18 000 units per month. The holding cost per unit is $14.40 per year, and the cost of ordering is $400. No shortages are allowed, and the replacement rate is instantaneous. Determine: (i) the optimum order quantity (ii) the total cost per year if the cost of one unit is $1.00 (iii) the number of orders per year (iv) the time between orders.
Solution 5.1 This is a single item static situation. We have: k = $400.00(processing or set-up cost) d = 18 000(demand rate per month) h = 14.40 12 (holding cost per unit per month). (i) The optimum order quantity is found by (5.4). r 2dk Q∗ = = 3, 465 units. h
(5.5)
(ii) hQ Total cost = C1 d + kd Q + 2 = 1 × 18, 000 + 400 × 18,000 3,465 + 1.2 × = $22156 per month.
3,465 2
Hence, the total cost per year is $265 872. (iii) The number of orders per year is given by: n = 12
d 18, 000 = 12 × = 62.4 Q 3, 465
orders/per year.
(iv) The time between orders is given by: t=
Q = 0.1925 months. q
54
Deterministic Inventory Models
Activity 5.2 1. In example 5.1, suppose d and k remain unaltered. Find Q∗ where h is: (a) $1.00 per unit per month (b) $0.75 per unit per month (c) $0.50 per unit per month. 2. In example 5.1, find the cost when the order size is: (a) 4 000 units per order (b) 3 000 units per order. Verify that the total cost is within 1 % of the optimum total cost. (This exercise illustrates the fact that the total cost is not very sensitive to the order size.)
Example 5.2 Consider the publishing department of a university, which periodically replenishes its supply of paper. Assume that the paper is purchased in large rolls and is used at a rate of two rolls per week. The cost of processing an order, which includes the cost of bookkeeping, trucking and handling is estimated as $50 per order. This cost is independent of the size of the order. The cost of holding the paper, which includes the costs towards rent for space, insurance, interest on the capital tied up, etc, is $2.00 per roll per week. Find the economic order quantity.
Solution 5.2 Let a week represent the unit of time in this problem. In terms of our notation, we have: Demand per unit time, d = 2 rolls. Holding cost h = $2 per roll per week. Cost of processing an order, k = $50. Using (5.4), we have: r Q∗ =
2 × 2 × 50 = 10 rolls per order. 2
Time between orders t =
Q 10 = = 5weeks. d 2
Cost of operating the supply of paper = Qh + dk Q + cost of paper √2 = 2dkh + cost of the paper (using 5.4) = 2 × 2 × 50 × 2 + cost of paper = $20per week + cost of the paper.
Deterministic Inventory Models
Maximum number of deserted cars x
55
6
Average number of deserted cars x/2
6
t 6
- Time 6
6
Points in time when deserted cars are removed Figure 5.2 Inventory of Deserted Cars.
Example 5.3 The State Police Department is responsible for removing the deserted cars left on public parking sites. The Police Department has observed that the cars are deserted at a rate of r cars per day. The Police Department finds it cheaper to remove the cars in bulk because of economies of scale. The cost function expressed in terms of the number of cars removed is given by a + bx + cx2 , where a, b and c are constants and x represents the number of cars removed in one operation. The Police Department also associates a social disutility cost of k dollars per car per day while in an abandoned state in a bublic place.
1. Find the time t between removal operations that will minimize the removing cost per car.
2. Find the optimum value of t and x given that r = 7 cars per day, a = $1, 000, b = $50, c = $10 and k = $100.
Solution 5.3 (i) If t is the time between removal operations, it is clear that x = rt. In this case, the inventory is built up by the deserted cars, i.e. the inventory is not instantaneous. This type of inventory is shown in Figure 5.2. The social disutility cost
56
Deterministic Inventory Models
for removing cars once in t days is given by: Z t krt2 rτ dτ = k 2 0 kx2 = . 2r Thus, the cost of removing x cars in one operation per unit time is given by: {a + bx + cx2 +
kx2 }/(x/r). 2r
To minimize this, we differentiate with respect to x and set the derivative equal to zero. This yeilds k ar − 2 + cr + = 0. x 2 Hence r 2ar ∗ x = (5.6) 2cr + k Thus, the optimum value, t∗ , of t is given by: r 2ar 1 ∗ . t = r 2cr + k (ii)Using 5.6, we obtain ∗
x =
r
2 × 1000 × 7 = 7.637 2 × 10 × 7 + 100
and
x = 1.091. 7 This suggests that the abandoned cars must be removed every day. t∗ =
5.6
EOQ Model, Modification 1 Manufacturing Situation
Suppose the demand for an item in a company is d units per unit time, and the company can produce them at a rate of R units per unit time, where R > d. The cost of one set up is $k and the holding cost per unit item per unit time is $h. We wish to determine the optimum manufacturing quantity and the total cost per year, assuming the cost of one unit is $C1 . It is assumed that shortages are not allowed. This model is similar to the EOQ model, but the replenishment rate is not instantaneous. The replenishment rate is finite and greater than the demand rate. The model is shown schematically in Figure 5.3. Here, R is the manufacturing rate, and Im is the maximum inventory. Cost per period(t1 + t2 ) = C1 Q + k + h(t1 + t2 )
Im . 2
Deterministic Inventory Models
57
Inventory 6
6
Slope R − d Q 6
Im
Slope d
9
?
?
t1
t2
-
- Time
-
Figure 5.3 Inventory for a Manufacturing Situation. But t1 + t2 =
Q d
Im = t1 (R − d) and t1 =
Q R
Note that t1 is the number of production days and t2 is the duration when the inventory is reduced to zero level. Thus, we obtain: Im =
Q d (R − d) = Q(1 − ). R R
Then Cost per period = C1 Q + k + h
d 1 Q (1 − ) . d R 2
Hence h Total cost per year(T C) = C1 Q + k + = C1 d +
kd Q
+
i
hQ2 d 2d (1 − R ) h d 2 Q(1 − R ).
d Q
(5.7)
For the optimum kd h d dT C = − 2 + (1 − ) = 0. dQ Q 2 R This yields s ∗
Q =
2dk h(1 − d/R)
(5.8)
58
Deterministic Inventory Models
6
6
Q
Im ? 6
s
?
t1
t
- t2 -
?
Figure 5.4 EOQ Model with Shortages Allowed.
Example 5.4 A company manufactures an item which is also used in the company. The demand for this item is 18 000 units/year, and the production rate is 3 000 per month. The cost of one set-up is $500 and the holding cost of 1 unit per month is 15 cents. Determine the optimum manufacturing quantity and the total cost per year, assuming the cost of one unit is $2.00. Shortages are not allowed.
Solution 5.4 For the given problem k = 500, d = 18000, h = 0.15 × 12, R = 3000 × 12. Substituting these values into (5.8), we obtain: s Q∗ =
2 × 500 × 18000 = 4472 units. 0.15 × 12 [1 − 18000/(3000 × 12)]
Then Total cost = 2 × 18000 + 500
18000 0.15 × 12 18000 + × 4472(1 − ) = $40.025 4472 2 3000 × 12
Note: The total cost is not sensitive to the value of Q∗ . The student may set Q∗ = 4470 and find the new total cost.
5.7
EOQ Model, Modification II Shortages Allowed
Consider the EOQ model where shortages are allowed, all other assumptions remaining the same. This situation is illustrated schematically in Figure 5.4. Cost per period = C1 Q + k + ht1
Im s + Cs t2 2 2
Deterministic Inventory Models
59
where Cs is the cost of shortage per unit time and s is the number of units short per period. It can be shown that: Im = Q − s t1 =
Q−s t Q
and
ts . Q
t2 = But t = Q/d gives t1 =
(Q − s) Q sQ and t2 = . Q d Qd
Using the values, we obtain Cost per period = C1 Q + k + h
(Q − s) Q (Q − s) s Qs + Cs . Q d 2 Qd2
Hence, 2
2
s d + Cs 2d ]Q Total cost per unit time = [C1 Q + k + h (Q−s) 2d 2
2
d s = C1 d + k Q + h (Q−s) + Cs 2Q 2Q
Taking partial derivatives with respect to Q and s gives ∂T C kd h s2 =− 2 + − (h + Cs ) ∂Q Q 2 2Q2 and
∂T C hs Cs s = −h + + . ∂s Q Q
Equating these to zero gives us: r ∗
Q =
2kd h
r
h + Cs Cs
(5.9)
and r s=
2kd Cs
r
h h + Cs
(5.10)
Example 5.5 Suppose that in Example 5.1 shortages are allowed, and that the cost of shortages is $5.00 per unit per month. Find: (i) The optimum order size. (ii) The optimum number of shortages. (iii) The total cost.
60
Deterministic Inventory Models
Inventory 6 6
Im Q
?
t1
-
t3- t4 -
t2
Time 6
S
? 6
? 6
Figure 5.5 Manufacturing Model with Shortages Allowed.
Solution 5.5 Using the data in Example 5.1, Cs = 5/12 = $0.42. Hence by (5.9) and (5.10), we obtain: (i) r r 2 × 400 × 18, 000 1.2 + 0.42 ∗ Q = = 3465 × 1.96 = 6791 units. 1.2 0.42 (ii) Shortages are given by s=
h 1.2 Q= 6791 = 5030 units. h + Cs 1.2 + 0.42
(iii) The total cost is given by: 18000 + 400 18000 6791 +
1.2(6791−5030)2 2×3860
+
0.42(5030)2 2×6791
= $20325 per month = $243907 per year
Example 5.6 In the manufacturing model discussed in section 5.5, suppose that shortages are allowed. Show that: s r 2dk h + Cs ∗ Q = (5.11) h(1 − d/R) Cs r r r 2dk h d (5.12) s= 1− Cs h + Cs R 2 1 d h d Cs s2 1 T C = C1 d + k + Q(1 − ) − s + (5.13) Q 2Q R 1 − d/R 2Q 1 − d/R
Solution 5.6 The situation is illustrated in Figure 5.5 where
Deterministic Inventory Models
t1 t2 t3 t4
= time = time = time = time
61
during during during during
which which which which
inventory increases inventory decreases backlog demand increases backlog demand decreases.
Since Im is used in time t2 at the rate d, it follows that: Im = dt2 = t1 (R − d). From Figure 5.7 we see that Q = (t1 + t4 )R and (t1 + t4 ) =
Q . R
Similary Im + s = (t2 + t3 )d = (t1 + t4 )(R − d) = Hence Im =
Q (R − d). R
Q (R − d) − s. R
Again, from Figure 5.7 we obtain: Im Im + R−d d 1 1 Q = + (R − d) − s . R−d d R
t1 + t2 =
Similarly t3 + t4 =
s s + . d R−d
Then the cost per cycle is given by: h Cs Im (t1 + t2 ) + s(t3 + t4 ) = 2 2 2 h Q 1 1 Cs 2 1 1 (R − d) − s + s + C1 Q + k + + 2 R R−d d 2 d R−d
C1 Q + k +
(5.14)
To obtain the total cost per unit, time we multiply this by d/Q to obtain: T C = C1 d +
kd h 1 Cs s2 1 + [Q(1 − d/R) − s]2 + Q 2Q 1 − d/R 2 Q 1 − d/R
(5.15)
Thus, (5.13) is achieved. To obtain the optimum values, we take the partial derivatives of (5.15) with respect to Q and s. Setting α = 1 − d/R, 5.15 reduces to: T C = C1 d +
kd h 1 Cs s2 1 + (Qα − s)2 + Q 2Q α 2 Qα
62
Deterministic Inventory Models
Then
∂T C hα Cs s2 1 kd hs2 − =0 =− 2 + − ∂Q Q 2 2αQ2 2α Q2
and
(5.16)
∂T C hs Cs s = −h+ = 0. ∂s αQ αQ
It follows that s=
hαQ h + Cs
(5.17)
Substituting this into (5.16) and solving for Q, we obtain: Q2 =
2dk(h + Cs ) h + Cs
(5.18)
The required results then follow from (5.17) and (5.18).
5.8
Exercises
1. An appliance dealer sells electric heaters at a rate of 100 per month. Its ordering costs are $25.00 for each order, which is independent of the size of the order. The cost of holding this item is $2.00 per unit per month. Find: (a) the optimum order quantity, (b) the optimum average cost per month, and (c) the optimum time between orders. 2. A dealer sells a furniture item at the rate of 6 700 per year. The holding cost for this item is $2.00 per unit per year. The shortage cost is, in his estimation, $4.00 per item per year. The cost of orders is $65.00 per order. (a) Identify the appropriate model. (b) Determine the optimal order quantity. (c) Determine the optimal shortage. (d) Find the optimal annual average cost. 3. A radio dealer produces and sells portable television sets. Demand for his sets is steady at a rate of 130 units per day. This dealer does not like shortages, hence shortages are not permitted. The cost of setting a production run is $400.00 and the production rate is 550 sets per day. The storage and interest costs are $0.10 per set per day. Find: (a) the optimal production quantity (b) the average number of production runs per year, assuming a year has business days (c) the optimal average cost per year.
260
Unit 6
Stochastic Inventory Models 6.1
Objectives
After completing this unit, you should be able to; i find the probability distribution of a given demand pattern. ii calculate the expected demand per given time period. iii distinguish between discrete demand and continuous demand models. iv calculate the optimal order quantity for a given model.
6.2
Introduction
In the previous unit, we discussed a variety of inventory models under deterministic assumptions. This means we assumed that demand and lead time were perfectly known to us. We have already indicated that not many situations exist where deterministic demand can be justified. The models in the previous unit were useful in the sense that they did provide some insight. However, when demand or lead time are not deterministic, we require probabilistic considerations to model the situation. We also require data to develop a probability density function to describe the stochastic demand or lead time. Determination of an appropriate density function and various parameters describing the probability distribution is not easy. We do not intend to discuss these problems here but refer the student to the literature in statistics. In this unit, we assume that demand or lead time is a random variable for which the probability distribution is known. A simple example illustrating the use of statistical data in inventory control is given in the section 6.3.
6.3
Statistical Data in Inventory Control
In the case of stochastic demand, we are interested in examining the demand pattern of an item. All inventory systems have some sort of accounting procedure for calculating the demand and the inventory on hand. We are interested not in the nature of the accounting procedure, but in the data we can obtain from the available records. We explain this by a simple example.
Example 6.1 The inventory history for an item for the past 20 weeks is given in Table 6.1
63
64
Stochastic Inventory Models
Table 6.1 Data for Example 6.1 Week Opening Inventory Order Received 1 400 0 2 400 400 3 600 50 4 500 300 5 800 50 6 800 0 7 600 200 8 700 50 9 600 100 10 500 50 11 500 100 12 400 50 13 300 300 14 500 50 15 400 200 16 400 50 17 400 200 18 600 100 19 600 200 20 700 200
Demand 0 200 150 0 50 200 100 150 200 50 200 150 100 150 200 50 0 100 100 200
Closing Inventory 400 600 500 800 800 600 700 600 500 500 400 300 500 400 400 400 600 600 700 700
(i) Find the probability distribution of weekly demand. (ii) Find the expected demand per week. Solution 6.1 From Table 6.1, we can construct the frequency Table 6.2.
Table 6.2 Frequency Table Demand/Week 0 50 100 150 200
Frequency 3 3 4 4 6
Relative Frequency 3/20 = 0.15 3/10 = 0.30 4/20 = 0.20 4/20 = 0.20 6/20 = 0.30
Hence the expected demand per week is given by: 0(0.15) + 50(0.30) + 100(0.20) + 150(0.20) + 200(0.30) = 125 items per week.
Stochastic Inventory Models
6.4
65
Single Order Random Demand for Perishable Goods
Consider a perishable product for which there is only one chance to order. Due to the nature of the item, if not used, it cannot be stored. The product does not have to be perishable in the strict sense of the word. Consider, for example, items like newspapers, flowers for a special day, etc. This problem is also known as the ”newspaper-boy” problem. The demand may be described by a discrete or continuous random variable depending on the nature of the item. We consider both cases in the ensuing discussion.
6.4.1
Discrete Demand
In this model, we assume the following: 1. Demand is discrete. It is represented by the random variable X. The probability that X = x is given by p(x). 2. Odering cost = 0. 3. Initial inventory = 0. 4. Order quantity to be determined = Q. 5. Cost price per unit = C1 . 6. Selling price per unit = C2 , (C2 > C1 ). 7. Net salvage value at the end of the period = s, (s < C2 ). By net salvage value, we mean the reduced disposal price less the cost of disposal. Infact s can be less than zero. Using the above notation, it is clear that the sales will depend on the number of units stocked, that is: x if x ≤ Q Sales = Q if x > Q The profit P will be a function of x and Q and is given by: C2 x + s(Q − x) − C1 Q if x ≤ Q P = C2 Q − C1 Q if x > Q The expected profit as a function of Q is given by: E{P (Q)} =
Q X x=0
=
Q X x=0
∞ X
{C2 x + s(Q − x) − C1 Q}p(x) +
{C2 x + s(Q − x) − C1 Q}p(x) +
(C2 Q − C1 Q)p(x)
x=Q+1
∞ X x=Q+1
C2 p(x) − C1 Q
(6.1)
66
Stochastic Inventory Models
since ∞ X
p(x) = 1.
x=0
We may write (6.1) as follows: E{P (x)} =
∞ X
∞ X
{C2 x + s(Q − x)}p(x) −
x=0
{C2 x + s(Q − x)}p(x)
x=Q+1
∞ X
+
C2 Qp(x) − C1 Q
x=Q+1
=
∞ X x=0
{C2 − s}xp(x) +
∞ X
∞ X
sQp(x) +
x=0
{C2 Q − (C2 − s)x − sQ}p(x) − C1 Q
x=Q+1
= (C2 − s)E(X) + sQ +
∞ X
{C2 Q − (C2 − s)x − sQ}p(x) − C1 Q
x=Q+1
= (C2 − s)E(X) + (s − C1 )Q +
∞ X
(C2 − s)(Q − x)p(x)
x=Q+1
= (C2 − s)E(X) + (s − C1 )Q + (C2 − s)
∞ X
(Q − x)p(x)
x=Q+1
= (C2 − s)E(X) + (s − C1 )Q + (C2 − s)
P ∞
x=0 (Q − x)p(x) −
PQ
x=0 (Q − x)p(x)
= (C2 − s)E(X) + (s − C1 )Q + (C2 − s)Q − (C2 − s)E(X)
−(C2 − s)
Q X
(Q − x)p(x)
x=0
= (C2 − C1 )Q − (C2 − s)
Q X
(Q − x)p(x)
x=0
For the optimal value of the profit, we must have: E{P (Q)} ≥ E{P (Q − 1)}
(6.2)
Stochastic Inventory Models
67
and E{P (Q)} ≥ E{P (Q + 1)}
(6.3)
From (6.2) and (6.3) we obtain: (C2 − C1 )Q − (C2 − s)
Q X
(Q − x)p(x) ≥ (C2 − C1 )(Q − 1) − (C2 − s)
Q−1 X
(Q − 1 − x)p(x)
x=0
x=0
(6.4) The right-hand side of (6.4) may be arranged as
(C2 − C1 )Q − (C2 − s)
Q−1 X
(Q − x)p(x) − (C2 − C1 ) + (C2 − s)
x=0
Q−1 X
p(x)
(6.5)
x=0
Now, when x = Q, then Q − x = 0 and so the upper limit of the second term can be changed to Q. Thus, from (6.4) and (6.5) we obtain: (C2 − C1 )Q − (C2 − s)
Q X
(Q − x)p(x) ≥
x=0
(C2 − C1 )Q − (C2 − s)
Q X
(Q − x)p(x) − (C2 − C1 ) + (C2 − s)
x=0
Q−1 X
p(x).
x=0
Hence −(C2 − C1 ) + (C2 − s)
Q−1 X
p(x) ≤ 0
x=0
and so
Q−1 X
p(x) ≤
x=0
C2 − C1 C2 − s
(6.6)
Similarly, using the condition E{P (Q)} ≥ E{P (Q + 1)} and (6.2), we have: Q X x=0
p(x) ≥
C2 − C1 . C2 − s
(6.7)
It follows from (6.6) and (6.7) that for optimal Q, we must have: Q−1 X x=0
Example 6.2
Q
C2 − C1 X p(x) ≤ ≤ p(x) C2 − s x=0
(6.8)
68
Stochastic Inventory Models
Chikwasha’s Bakery has the problem of determining the number of loaves of bread to stock each day. Chikwasha supplies two types of bread - white and brown. From past data, Chikwasha collected the information shown in Table 6.3. It is the reponsibility of Chikwasha to collect the unsold loaves on the next day. Table 6.3 Frequency of Bread Sales Number of Sales Frequency (%) loaves sold White Brown White Brown 90 60 7 5 91 61 10 8 92 62 12 10 93 63 16 15 94 64 14 15 95 65 10 10 96 66 9 10 97 67 8 8 98 68 6 7 99 69 5 6 100 70 3 6 Chikwasha has an arrangement to salvage the unsold loaves to a nearby fast-food where the demand for the one day old bread is unlimited.Chikwasha’s accountant has supplied the information shown in Table 6.4. Table 6.4 Prices Table Bread Cost price per loaf ($) White 300 Brown 350
Selling price per loaf ($) 550 570
Salvage value ($) 250 260
Find the optimum number of loaves that should be ordered by Chikwasha. Solution 6.2 From Table 6.4, we can calculate the demand distribution as given in Table 6.5.
Table 6.5 Demand Distribution X p(x) PX x=0 p(x)
90 0.07 0.07
91 0.10 0.17
X p(x) PX x=60 p(x)
60 0.05 0.05
61 0.08 0.13
White Bread 93 94 95 96 97 98 99 100 0.16 0.14 0.10 0.09 0.08 0.06 0.05 0.03 0.45 0.59 0.69 0.78 0.86 0.92 0.97 1.00 Brown Bread 62 63 64 65 66 67 68 69 70 0.10 0.15 0.15 0.10 0.10 0.08 0.07 0.06 0.06 0.23 0.38 0.53 0.63 0.73 0.81 0.88 0.94 1.00
92 0.12 0.29
Stochastic Inventory Models
69
From this, we see that: C1(w) = 300, C2(w) = 550, sw = 250 and C1(B) = 350, C2(B) = 570, sB = 260 From (6.8) and Table 6.5 we obtain • For the white bread, C2 − C1 = 0.8333 C2 − sw and therefore the optimum number of white loaves is 97. • For brown bread, the ratio is 0.709 and the optimum number of brown loaves is 66. Activity 6.1 1. Find the expected profit when Q = 94, 95, 96 and 98 for the white loaf and hence establish that the profit is maximum when Q = 97. Hint: Use relation (6.1) for calculation of the E{P (Q)}. 2. Suppose for an item, the demand distribution is given as in Table 6.6. Table 6.6 Demand Distribution Demand (X) 10 p(x) 0.15
11 0.20
12 0.19
1 13 0.18
14 0.17
15 0.11
It is also given that: C1 = $12.50, C2 = $15.00 and s = $11.25. Show that: (a) E{P (12)} = 28.125. (b) the optimum value of Q is 13.
6.4.2
Continuous Demand
In this section, we assume that the demand can be represented as a continuous random variable x with a density function f (x). Using the notation introduced in the previous model, we can write the expected profit as a function of Q as follows:
70
Stochastic Inventory Models
Z
Q
Z
∞
{C2 x + s(Q − x)}f (x)dx +
E{P (Q)} = 0
Z
C2 Qf (x)dx − C1 Q Q Z ∞
Q
{C2 x + s(Q − x)}f (x)dx + C2 Qf (x)dx − 0 0 Z Q (Q − x)f (x)dx + (C2 − C1 )Q. = (s − C2 )
Z
Q
C2 Qf (x)dx − C1 Q
=
0
0
To obtain the optimal value of Q, we differentiate (6.9) with respect to Q. Recall that if a and b are differentiable and
Z
b(z)
G(z) =
g(x, z)dx a(z)
then
dG = dz
Z
b(z)
a(z)
∂g db da dx + g(b, z) − g(a, z) . ∂z dz dz
Hence, differentiating (6.9) with respect to Q and equating the derivative to zero, we obtain Z (C2 − C1 ) + (s − C2 )
Q
f (x)dx + 0 × 1 − Qf (x) × 0 = 0.
0
Hence, the optimal value of Q is given by:
Z
Q
f (x)dx = 0
C2 − C1 . C2 − s
(6.9)
The result is similar to (6.8) for discrete demand. Example 6.3 Suppose the demand in Example 6.2 is a continuous random variable with density functions ( 1 fW (x) = 10 0 and
, 90 ≤ x ≤ 100 , otherwise
Stochastic Inventory Models
71
( 1 fB (x) = 10 0
, 60 ≤ x ≤ 70 , otherwise,
for white and brown bread respectively and assume C1 , C2 and s have the values given in Example 6.2. Find the optimum supply quantity for the white and brown loaves. Solution 6.3 (a) For white bread, the value of the right-hand side of (6.10) is 0.833. We wish to find Q for which Z
Q
Z
Q
fW (x)dx = 90
90
1 dx = 0.833. 10
Then Q = 98.33. (b) For brown bread, the ratio is 0.709. The optimal value of Q is obtained from Z
Q
60
1 dx = 0.709. 10
Then Q = 67.09. Activity 6.2 Suppose that the demand for an item is normally distributed with a mean of 5 and standard deviation 2 and it is given that:
C1 = $12.50, C2 = $15.00 and s = $11.25. Find the optimal value of Q.
6.5
Random Demand Model with Shortage Costs
This is a variant of the model discussed in the previous section. Here we assume that the supply is deterministic, demand is random and the organization suffers a shortage cost when it is unable to meet the demand. Suppose we place an order at the start of a period. Then the inventory equation for the periods (t, t+1) and (t+1, t+2) is given by:
Wt+1 = Wt + Yt − Xt ,
(6.10)
72
Stochastic Inventory Models
where Xt Wt Yt Wt+1
= the = the = the = the
random demand during theperiod (t, t + 1) inventory at the beginning of the period (t, t + 1) supply at the beginning of the period (t, t + 1) inventory at the beginning of the period (t + 1, t + 2).
From (6.11), it is clear that the inventory at the start of the next period will be nonnegative if Wt + Yt ≥ Xt , and the system will suffer shortages if Wt + Yt < Xt . Once again let
h = holding cost per unit, Cs = shortage cost per unit, and Xt denote a discrete or continuous random variable.
6.5.1
Discrete Demand
Define P {Xt = x} = p(x), x = 0, 1, 2, .... From (6.11), the stock at the beginning of period (t + 1, t + 2) is (Wt + Yt − Xt ) and the average stock may be taken as (Wt + Yt − Xt )/2. The expected cost of holding the average inventory is then given by: h
WX t +Yt
p(x)
0
Wt + Yt − x , 2
(6.11)
when the random demand does not exceed the supply. However, when demand exceeds the supply, for a part of the period we would be holding items and for the remanining period we would be short of items. The expected cost in this case would be given by:
h
∞ X x=Wt +Yt +1
(Wt + Yt Wt + Yt )( )p(x) + Cs 2 x
∞ X
(
x=Wt +Yt +1
x − Wt − Yt )2 p(x). 2x
From (6.12) and (6.13), the total expected cost would be given by:
T (Wt + Yt ) = h
WX t +Yt x=0
+ Cs
p(x)(Wt + Yt − x) +h 2 ∞ X
x=Wt +Yt +1
∞ X Wt +Yt +1
)2
(x − Wt − Yt p(x) 2x
(Wt + Yt )2 p(x) + 2x
(6.12)
Stochastic Inventory Models
73
From (6.13), following the treatment in the earlier section, one can establish that for optimum stock Qo , (Qo = Wt + Yt ), we have: Qo−1
X x=0
Qo
X Cs ρ(x) ≤ < ρ(x). h + Cs
(6.13)
x=0
Example 6.4 A car manufacturer has decided to order a new machine from Jack Heavy Equipment Private Ltd(JHEPL) for its production plant. One of the essential parts of this machine is very complicated and expensive. JHEPL has indicated that it would be desirable to order this part along with the machine as they feel that it would be impractical to order it at a future date. Each part is uniquely built for this machine and it is not suitable for use on any other machine. JHEPL has supplied the information shown in Table 6.7 to the car manufacturer. The cost of this spare part, if odered with the machine, is $5 000 per part but if required at a future date would be $25 000. JHEPL have also provided the information given in Table 6.8. Table 6.7 Frequency of Spare Parts Requirements No. of spare parts required % of the machines requiring indicated number of spare parts 0 60 1 20 2 10 3 5 4 3 5 2 6 or more 0 The manufacturer wants to know the number of spare parts they should incorporate in their order with the machine. Solution 6.4 In this example Cs = $25000 and h = $5000. The probability distribution, is given in Table 6.8. Table 6.8 Demand Distribution No. of spare parts required = x P {X = x} P {X ≤ x} Now 25000 Cs = = 0.83 h + Cs 5000 + 25000
0 0.60 0.60
1 0.20 0.80
2 0.10 0.90
3 0.05 0.95
4 0.03 0.98
5 0.02 1.00
74
Stochastic Inventory Models
The optimum value of Qo is 2. This means that the car manufacturer must order two spare parts of this item, along with the order of the machine. Activity 6.3 For the data in Example 6.4, calculate T C for Q = 4.
6.5.2
Continuous Demand
In this case, all the assumptions are the same as in the previous section, except for the demand which is assumed to be a continuous random variable. Denote by f (x)dx the probability that demand Xt in a period lies between x and x + dx. Suppose we stock a quantity Q at the start of each period. If the demand is less than Q, we pay a holding cost on items not used and if the demand exceeds Q, we pay a shortage cost. The total expected cost as a function of Q is given by:
Z
Q
Z
∞
(Q − x)f (x)dx + Cs
T C(Q) = h 0
Z
(x − Q)f (x)dx Q Z ∞
Q
(Q − x)f (x)dx + Cs
= h 0
(x − Q)f (x)dx + Q
Z Q (x − Q)f (x)dx − Cs (x − Q)f (x)dx 0 0 Z Q Z ∞ = (h + Cs ) (Q − x)f (x)dx + Cs (x − Q)f (x)dx. Z
Q
+Cs
0
0
Differentiating this and setting the derivative equal to zero, we obtain: dT C = (h + Cs ) dQ
Z
Q
Z f (x)dx − Cs
0
∞
f (x)dx = 0. 0
Hence Z
Q
f (x)dx = 0
Cs . h + Cs
The result is similar to (6.15), as was to be expected.
Example 6.5 A fast-food chain has introduced a new strategy of selling pizza by weight. If the product is not sold on the day it is baked, it is kept in storage. The cost of holding a kilogram of pizza in stock for one day is 20 cents. The profit on one kilogram of pizza, if sold on the day it is baked, is $1.00. Thus, we take the cost of shortage as $ 1.00 per kilogram. The
Stochastic Inventory Models
75
probability density function for demand as a function of x (x is demand in kilograms) is given by: f (x)d(x) = 0.03 − 0.0003x. How many kilograms of pizza should this fast-food chain bake daily? Solution 6.5 In this case h = 0.20 and Cs = 1.00. Hence Cs 1.00 = = 0.833. h + Cs 1.20 For the optimum value of Q, we must have: Z
Q
(0.03 − 0.0003x)dx = 0.833. 0
0.0003Q2 = 0.833. 2 0.00015Q2 − 0.03Q + 0.833 = 0. Then
or
0.03Q −
This has solutions Q1 = 168 and Q2 = 32. The first solution is discarded, since for Q = 168 the probability distribution is not applicable. Hence the optimal value of Q is 32 kilograms.
6.6
Exercises
1. The owner of a newsstand wants to determine the number of newspapers that must be stocked at the start of each day.It costs $100 to buy a copy, and the owner sells it for $165. Newspapers left at the end of the day are recycled for an income of $50 a copy. How many copies should the owner stock every morning, assuming that the demand for the day can be described as: (a) A normal distribution with mean 300 copies and standard deviation of 20 copies? (b) A discrete pdf, f (D), defined in Table 6.9? Table 6.9 Demand Distribution for Newspapers D 200 220 300 320 340 f (D) 0.1 0.2 0.4 0.2 0.1
2. The U of A Bookstore offers a program of reproducing class notes for participating professors. Professor Gututu teaches second year students, where an enrolment of between 200 and 250 students, uniformly distributed, is expected. It costs the bookstore $1000 to produce each copy, which it then sells to the students for $2500 a copy. The students purchase their books at the start of the semester.
76
Stochastic Inventory Models
Any unsold copies of professor Gututu’s notes are shredded for recycling. In the meantime, once the bookstore runs out of copies, no additional copies are printed and the students are responsible for securing the notes from other sources. If the bookstore wants to maximize its revenues, how many copies should it print? 3. QuickStop provides its customers with coffee and doughnuts at 6 : 00 A.M. each day. The convenience store buys the doughnuts for $70 a piece and sells them for $250 a piece until 8 : 00 A.M. After 8 : 00 A.M., the doughnuts sell for $50 a piece. The number of customers buying doughnuts each day is uniformly distributed between 30 and 50. Each customer usually orders 3 doughnuts with coffee. Approximately how many dozen doughnuts should QuickStop stock every morning to maximize revenue? 4. For the single-period model, suppose that the demand occurs uniformly during the period (rather than instantaneously at the start of the period). Develop the associated cost model and find the optimal order quantity. 5. It is desired to maximize the expected profit in a single-period model. The demand occurs instantaneously at the end of the period. Let c, r and v be the unit purchasing cost, unit selling price and unit salvage value for the item. Assuming that the demand D is continuous and is described by the pdf f (D), develop an expression for the total expected profit and determine the optimal order quantity. 6. A grocery store owner sells a tray of bread for $2500, where it costs him $1500 to purchase it. An unsold tray left over at the end of the day is sold for half-price, i.e. $1250. Demand in trays is discrete. Demand over the past has shown the pattern in Table 6.10. Table 6.10 Frequency of Bread Demand No. of trays 16 17 % time demanded 15 20
18 19
19 18
20 17
21 11
(a) Find the optimal number of trays the grocery store should stock each day. (b) Find the expected profit, using the optimal solution. (c) Find the expected profit for stocking 21 trays and compare your solution with the optimal solution.
Unit 7
Queuing Theory I: Single Server Models 7.1
Objectives
After completing this unit, you should be able to; i Define what constitutes a queue. ii Define a single Server Model. iii List the input and output units of a queuing system. iv Identify the service discipline of a give service system. v Use the Kendall-Lee notation. vi Define the terms used in the kendall-Lee notation.
7.2
Introduction
After shopping in a supermarket, a shopper takes his selected goods to a cash register for checking out and payment. Shoppers may be served immediately if the assisant at the cash register is free, otherwise they may have to wait for customers ahead of them to be served. The customer would like to be served promptly. In order to offer this prompt service, the manager of this store would have to create a few extra cash register channels. These extra chanels would cost money and at times the channels would not be busy. In order to keep the cost of the system down, the manager would like his cashiers to be busy all the time. The requirements of the customers and management conflict with each other. Morever, customers do not arrive for checking out and payment at fixed and regular intervals. Some customers take longer to be served than others, simply because of the volume of their goods. A queueing situation arises because of the unpredictable nature of the input and output from the system. Queueing theory deals with the above type of situation. Queueing situations occur in a wide range of cases, for example, aircraft landing on, and taking off from, a runway, handling of telephone calls in an exchange, cars passing through an intersection, etc. The basic features of a queueing problem are that there is some sort of input requiring service and when this input has been served, it forms the output of the queueing system. This is shown diagramatically in Figure 7.1. The input, service and output change from one queueing situation to another. We explain this in Example 7.1.
77
78
Queuing Theory I: Single Server Models
—
-
queue
-
-
output
service channel
Figure 7.1 A Queuing System
Example 7.1 Identify the input units and the service in each of the following situations: 1. A secretarial typing pool 2. An ambulance station 3. The toll booth for Beitbridge 4. A repair workshop 5. A shipping dock 6. An aircraft runway 7. Issuing of supplies from a factory tool store.
Solution 7.1 The answers are as in Table 7.1. Table 7.1 Solution Table for Example 7.1 Serial Number Input Unit 1 Letter for typing 2 An emergency call 3 Arriving cars 4 Items for repair 5 Ships for loading and unloading 6 Aircraft 7 workers demanding the tools
7.3
Server Secretaries Ambulance Toll collection gate The repairman The dockyard or berth The runway The clerk
Queueing Processes
As explained earlier, a queue is formed by customers requiring service. These customers form an input which may be finite or infinite. They enter the queueing system to join the queue and wait for their turn. At a certain point in time, some member of
Queuing Theory I: Single Server Models
79
the queue is selected for service by some rule known as the service discipline. The service is then performed on the selected customer by a service mechanism. The customer leaves the queueing system after completion of his service, thus forming an output of the queueing system. For the analytical treatment of a queueing situation, an infinite queue length is assumed. When the actual queue size is relatively large, the calculations for the infinite case are relatively easy. In this unit, we also assume that the input is generated according to a Poisson process. This simply means that arrivals take place in a random fashion but at a certain average rate. An equivalent statement is that the distribution of the time between consecutive arrivals is an exponential distribution. The time between two consecutive arrivals is known as the inter-arrival time. This means that the statements, 1. ”input to a queueing system is Poisson distributed”, and 2. ”the inter-arrival time in a queueing system follows an exponential distribution”
are equivalent. The term service discipline refers to the procedure for selection of units from the queue for service. For example, a commonly used service discipline is ”first-in-firstout” (FIFO) also known as ”first-come-first-served” (FCFS). In this unit we are mostly concerned with the FIFO discipline. Other examples are: ”last-in-first-out” (LIFO), ”service-in-random-order” (SIRO) and various priority disciplines. The service itself may comprise one or more parallel channels called servers. In this unit, we consider only single server queues. The case of more than one server is known as the multi-channel queues. In some cases it is also possible that customers receive service from more than one server in a sequence. These are known as server channels in series. The time from the start of a service to its completion for a customer at a service channel is known as the service time or holding time. In the analytical study of queuing systems, we must be able to specify the probability distribution of the service time. Frequently the service time is assumed to be exponentially distributed. A more general case is to assume that the service time follows a Gamma (or Erlang) distribution. However, we shall consider only exponentially distributed service times. From the above discussion, it may be noted that the characteristics of a queuing situation are: 1. The input distribution or inter-arrival distribution 2. The service distribution or departure (output) distribution 3. The number of service channels
80
Queuing Theory I: Single Server Models
4. The service discipline 5. The capacity of the queueing system, i.e. the maximum number allowed in the system 6. The calling source or the number of units that can demand that service.
7.4
Kendall-Lee Notation
The Kendall-Lee notation which describes in symbolic form the six characteristics for a given system is set out as follows: (a/b/c) : (d/e/f) where a describes the arrivals distribution b describes the service time distribution c represents the number of parallel service channels d describes the service discipline e indicates the maximum number allowed in the system (in the queue and in service). f indicates the number of people who may require that service. For the symbols a and b, it is conventional to use the following codes: M
to describe a Poisson arrival or exponential interarrival distribution or to describe an exponential service time distribution. Ek is used to represent a Gamma or Erlangian distribution with parameter k for arrival or service time distribution. G describes the general distribution of departures. GI denotes the general independent arrival distribution. For the symbol d, commonly used codes are:
• FCFS or FIFO for first-come-first-served or first-in-first-out • LCFS for last-come-first-served • SIRO for service-in-random-order • GD for any general service discipline.
Example 7.2 Explain what you understand by a queueing system described in Kendall-Lee notation as: (M/M/c) : (LCFS/N/∞). Solution 7.2 The above queueing system has Poisson arrival, exponential departure, c parallel servers, the discipline is last-come-first-served, the capacity of the system including
Queuing Theory I: Single Server Models
81
the customers in service is N and the number of customers who may call for that service is infinite.
7.5
Definitions and Notation
The following statistical terms are commonly used in queueing theory: A queueing system is said to be in transient or time dependent state if its behaviour varies with time. In the long run, a system which becomes independent of time is said to have a steadystate behaviour. In this unit, we shall concentrate on the steady-state behaviour.
Notation A commonly used notation in queueing theory is as follows: n = the number of customers or units in the system. pn (t) = the time dependent probability of exactly n units being in the system at time t, given that the system started at time zero. pn = the time independent or steady-state probability of exactly n units being in the system. λ = the number of customers or units arriving per unit time, i.e. the mean arrival rate. µ = the number of customers or units served per unit time, i.e. the mean service rate per busy server. c = the number of parallel service channels providing identical service. ρ = µλ = traffic intensity. ρ n = the utilization factor for c parallel servers. (In the case of one server, the traffic represents the utilization factor). Ws = the mean waiting time per unit in the system. Wq = the mean waiting time per unit in the queue. Ls = the mean number of units in the system. Lq = the mean number of units in the queue.
7.6
Instantaneous Event Rate for the Exponential Distribution
The instantaneous event rate Z(t) for a random variable T is defined as Z(t) =
f (t) 1 − F (t)
(7.1)
82
Queuing Theory I: Single Server Models
where f(t) is the probability density function of the random variable T and F(t) is the cumulative density function for the same random variable. For the exponential distribution f (t) = λe−λt (7.2) where λ is the rate of event, and F (t) = 1 − e−λt .
(7.3)
From (7.1) Z(t) =
λe−λt = λ. 1 − (1 − e−λt )
(7.4)
If we are interested in the probability that the event takes place in the interval (t, t+∆t) given that the event has not happened up to time t, we write
P {t < T < t + ∆t | T > t} = =
P {t < T < t + ∆t} P [T > t]
(7.5)
[1 − e−λ(t+∆t) ] − [1 − e−λt ] e−λt
(7.6)
eλt − e−λ(t+∆t) e−λt
(7.7)
= 1 − e−λ∆t
(7.8)
=
(λ∆t)2 ) 2! = λ∆t + (∆t)
= 1 − (1 − λ∆t +
(7.9) (7.10)
where (∆t) denotes any quantity which tends to zero as ∆t → 0. Suppose we are given the rate per unit time as λ and let t be the total duration. Divide the time interval t into small sub-intervals of length ∆t. Then the events are said to be random if for each interval the probability that an event occurs in it is λ∆t. This statement is the equivalent of (7.10). It also suggests that: (i) We neglect the possibility of more than one event in a small interval. (ii) The probability of no event in a small time ∆t is (1 − λ∆t). (iii) The time to the next event is independent of the time since the last event. We study two important probabilities: (a) The probability that the time to the next event is less than t. (b) The probability of exactly r events in time t.
Queuing Theory I: Single Server Models
83
Probability that Time to the Next Event is less than t Let the time t be subdivided into n sub-intervals each of length ∆t as shown in Figure 7.2. 0
1
n
2
-
∆t
∆t
t
∆t
Figure 7.2
Thus n∆t = t.
(7.11)
Now P {time to next event > t} = P {no events in n number of intervals}
(7.12)
= (1 − λ∆t)n .
(7.13)
P {time to next event > t} = lim (1 − λ∆t)n
(7.14)
Letting ∆t → 0, we have ∆t→0
= lim (1 − ∆n→∞
λt n ) n
= e−λt .
(7.15) (7.16)
Note that this is independent of the time when the last event took place. Further, it follows from (7.16) that P {time to next event < t} = 1 − e−λt .
(7.17)
The right-hand side of (7.17) is the cumulative distribution function for the exponential distribution. Probability of Exactly r Events in Time t Using (7.10) we have: P {r events in the first r intervals and no events in the remaining (n − r) intervals} (7.18) r n−r = (λ∆t) (1 − λ∆t) (7.19) In order to obtain the probability of exactly r in the n intervals, we multiply (7.19) by the number of ways in which the r events can occur within the n intervals. Example 7.3
84
Queuing Theory I: Single Server Models
Suppose that buses from a city terminus to a university campus come randomly at 8 minute intervals. If the service were not random but at fixed intervals of 8 minutes, we expect the average waiting time per passenger to be 4 minutes. However, due to traffic jams, the buses arrive in a random fashion, one every 8 minutes. Prove that the average waiting time is 8 minutes per passenger (i.e. double the amount for the case when arrivals were not random). Solution 7.3 Consider a general arrival rate λ. The given time t is once again regarded as comprising n intervals each of length ∆t as shown in Figure 7.2. Using (7.10), it is easy to obtain the probabilities in Table 7.2. Table 7.2 Probability Distribution for Waiting Time Waiting Time 0 ∆t 2∆t ... n∆t Probability λ∆t (1 − λ∆t)λ∆t (1 − λ∆t)2 λ∆t ... (1 − λ∆t)n λ∆t From Table 7.2 we obtain: Average Waiting Time (AWT) = =
n X
X
(Waiting Time) × P robability
i∆t(1 − λ∆t)i λ∆t.
(7.20) (7.21)
i=0
Setting x = 1 − λ∆t, it follows that 2
AWT = (1 − x)x∆t[1 + 2x + 3x + ...] d = x(1 − x)∆t [1 + x + x2 + ...] dx d 1 = x(1 − x)∆t [ ] dx 1 − x 1 = x(1 − x)∆t (1 − x)2 x = ∆t 1−x (1 − λ∆t)∆t = λ∆t 1 = − ∆t. λ Hence, letting ∆t → 0, AWT = λ1 . In our example, λ= per minute, hence Average Waiting Time = 8 minutes.
1 8
(7.22) (7.23) (7.24) (7.25) (7.26) (7.27) (7.28) (7.29)
Queuing Theory I: Single Server Models
7.7
85
Birth and Death Processes
The queueing models which we intend to discuss in this Unit as well as the next assume that the inputs (arriving units) and the outputs (departing units) are according to the birth-and-death processes (an important area of study in stochastic processes). This has many applications in various areas. In the context of queueing theory, an arriving customer is referred to as a birth and a departing customer as a death. If arrivals to, and departures from, a queueing system are at a constant rate of λ and µ repectively, the state transition in the birth and death process can be represented as in Figure 7.3.
λ
λ
λ
λ
λ
λ
λ
-
-
-
-
-
-
-
0
1
µ
...
2
µ
µ
n
n-1
µ
...
n+1
µ
µ
µ
Figure 7.3 State Transition in Birth and Death Process The transition probabilities between time t and t + ∆t for the birth and death process are summarized in Table 7.3, where we have set (1 − λ∆t)(1 − µ∆t) = A Table 7.3 Transition Probability Rate Matrix 0 1 2 3 ... n − 1 n 0 1 − λ∆t λ∆t ... 1 µ∆t A λ∆t ... 2 µ∆t A λ∆t ... . . . n−1 ... A λ∆t n ... µ∆t A n+1 ... µ∆t
7.8
n+1 -
... ... ... ...
λ∆t A
... ... ...
The (M/M/1) : (FIFO/∞/∞) Queueing Model
This is a simple birth and death model. From the Kendall-Lee notation, this model assumes that the inputs and outputs are at a constant rate and there is only one server. The service discipline is first-in-first-out with queue length and calling source unlimited. For this model, the transition probability rate matrix will be as given in Table 7.3. From Table 7.3, we have the following probability equations:
86
Queuing Theory I: Single Server Models
po (t + ∆t) = po (t)(1 − λ∆t) + p1 (t)µ∆t
(7.30)
and pn (t + ∆t) = pn (t)(1 − λ∆t)(1 − µ∆t) + pn+1 (t)µ∆t + pn−1 (t)λ∆t, n > 0.
(7.31)
It follows from (7.30) and (7.31) that lim
∆t→0
po (t + ∆t) − po (t) = p0o (t) = −λpo (t) + µp1 (t) ∆t
(7.32)
and lim
∆t→0
pn (t + ∆t) − pn (t) = p0n (t) = −(λ + µ)pn (t) + µpn+1 (t) + λpn−1 (t), n > 0. (7.33) ∆t
The solution of equations (7.32) and (7.33) will give us the values of the transient probabilities. In queueing theory, we are interested in steady-state solutions (i.e. in the solution as t → ∞ ). When a system is in its steady-state, the rate of change of probability will be zero, i.e. p0n (t) = p00 (t) = 0. From (7.32) and (7.33) the steady state or time independent probability relations result in the ordinary difference equations −λp0 + µp1 = 0
(7.34)
−(λ + µ)pn + µpn+1 + λpn−1 = 0, n > 0
(7.35)
and
From (7.34) we obtain p1 =
λ p0 = ρp0 . µ
(7.36)
Setting n = 1 in (7.35) and using (7.36), it follows that p 2 = ρ2 p 0 .
(7.37)
By mathematical induction, it follows that pn = ρn p0 , n = 1, 2, ...
(7.38)
However, we know that ∞ X
pn = 1.
(7.39)
n=0
Hence by (7.38) and (7.39) we deduce that p0 = 1 − ρ if ρ < 1.
(7.40)
Queuing Theory I: Single Server Models
87
This condition is necessary for the existence of a steady-state solution. In general, the probability pn , is given by: pn = (1 − ρ)ρn , n = 0, 1, 2, ...
(7.41)
Having obtained the probabilities pn , we now study a few parameters of the system. The Expected Number in the System Ls = E{n} =
∞ X
(7.42)
npn
(7.43)
n(1 − ρ)ρn
(7.44)
n=0
=
∞ X n=0
λ ρ = . 1−ρ µ−λ The Expected Number in the Queue =
Lq =
∞ X
(n − 1)pn
(7.45)
(7.46)
n=1
=
∞ X
(n − 1)(1 − ρ)ρn
(7.47)
n=1
=
λ2 ρ2 = . 1−ρ µ(µ − λ
(7.48)
From (7.45) and (7.48) one can show that Ls = Lq +
7.9
λ . µ
(7.49)
Parameters of a Queueing Model
It has been proved that under general conditions of arrival, departure and service discipline, the following relationships hold: Ls λ Lq Wq = λ Ws =
(7.50) (7.51) 1 µ
(7.52)
Ls = Lq + ρ.
(7.53)
Ws = Wq +
For the(M/M/1) : (FIFO/∞ / ∞) queueing model, we have discussed in detail the derivation of the results for Ls , Lq and Ws .
88
Queuing Theory I: Single Server Models
Example 7.4 Customers arrive at a drive-in bottle shop in a Poisson fashion. The mean rate of arrival is 15 per hour and servIce time per customer is exponential with a mean of 3 minutes. There is only one server to attend the customers. The space of the driveway can accommodate 5 cars including the car under service. However, other people can wait on the driveway on the road. 1. What is the probability that an arriving car can be served immediately? 2. What is the probability that an arriving car can drive onto the driveway of the bottleshop? 3. What is the probability that an arriving car has to wait on the roadside? 4. How long does an arriving car have to wait before starting service? 5. What is the mean waiting time in the system? 6. The management of the bottle shop desires that 99% of the cars should be able to drive onto their driveway. How many spaces would be enough to fulfil this requirement? Solution 7.4 The given situation represents a (M/M/1) : (FIFO/ ∞/∞) queque, where λ = 15 per hour, µ = 60 3 = 20 per hour, and 3 ρ = 4. 1. The probability of serving a car immediately is given by: 1 p0 = 1 − ρ = . 4
(7.54)
2. The probability that an arriving car can drive onto the driveway is given by: p0 + p1 + p2 + p3 + p4 = 1 − ρ + ρ(1 − ρ) + ρ2 (1 − ρ) + ρ3 (1 − ρ) + ρ4 (1 − ρ) (7.55) = 1 − ρ5
(7.56) 5
= 1 − (3/4) .
(7.57)
3. The probability that an arriving car has to wait on the roadside is given by: p5 + p6 + ... = 1 − (p0 + p1 + p2 + p3 + p4 )
(7.58)
5
(7.59)
λ 15 3 = = hours = 9 minutes µ(µ − λ) 20 × 5 20
(7.60)
= (3/4) . 4. The mean waiting time in the queue is by: Wq =
Queuing Theory I: Single Server Models
89
5. The mean waiting time in the system is by: Ws =
1 hours = 12 minutes. 5
(7.61)
6. Let n be the number of places needed to fulfil the management’s requirement. We require p0 + p1 + ... + pn ≥ 0.99.
(7.62)
1 − ρn ≥ 0.99
(7.63)
0.01 ≥ (3/4)n ,
(7.64)
This is satisfied if
or
which means that we require n to be greater than 16.
7.10
Exercises
1. Explain the following queueing systems: (a) (M/M/1) : (FIFO/∞/∞) (b) (M/Ek /1) : (SIRO/N/M) (c) (M/M/c) : (FIFO/∞/∞) (d) (GI/G/c) : (GD/N/M). 2. In a Poisson process with arrival rate of λ = 3 per unit time, it is given that an arrival occurs at time t = 10. (a) What is the probability that an arrival will occur by time t = 1? (b) What is the probability that an arrival will occur in the interval (14,15)? 3. Jobs for repair arrive according to a Poisson distribution with a mean rate of 3 per hour. These jobs are processed on a particular machine which is subject to breakdowns. It then takes 1 12 hours to repair. Find the probability that the number of new jobs that arrive during the down time of the machine is: (a) zero (b) three (c) three or more. 4. A grocery has a single checkout register with a full-time assistant manning it. Customers arrive at the checkout randomly in a Poisson fashion at a mean rate of 20 per hour. When there is only one customer at the checkout register, that customer is processed by the assistant alone, with an expected service rate of 2 minutes. However, the stockboy has been given instructions that whenever
90
Queuing Theory I: Single Server Models
there are two customers at the register, he should help the assistant by packing the groceries. This reduces the service time to 1.5 minutes. The service time distribution is exponential in both cases. (a) Write down the probability equations for the above system. (b) Obtain the values of the state probabilities, pn . (c) Find the value of the expected number of customers at the checkout register. (d) Find the values of Lq , Ws and Wq . 5. In an (M/M/1):(FIFO/ ∞ /∞) system, assume that the arrival and departure rates are λn and µn , where n is the number of customers in the system at that time. (a) Prove that the steady-state probabilities are given by: pn = p 0
(7.65)
(b) Let λn = λ for all n and µn = na µ where λ, a and µ are constant. Prove that, in the steady-state: i. ρn pn = (n!)a R
n = 1, 2, ...
∞ X ρn λ where R = and ρ = . a (n!) mµ n=0 (7.66)
ii. p0 =
1 . R
(7.67)
6. A counselling service has a staff of one counsellor. The arrivals at the clinic come in Poisson fashion at a mean rate of 2 per hour. The counselling time follows an exponential distribution with a mean time of 20 minutes. (a) What is the expected waiting time for each arrival? (b) What is the expected number of people waiting for counselling at any given time? (c) What is the expected number of people in the system? (d) What is the probability that an arriving person will find the counsellor free? (e) What proportion of time is the counsellor idle? (f) What is the probability that there is exactly one person in the queue at any given time? (g) What is the probability that there are three or more people waiting in the line? (h) What is the probability that there are three or more people in the system? (i) Due to an increase in the population, the present arrival rate at the clinic is 6 per hour. What is expected to happen to the stability of the system? Why?
Unit 8
Queuing Theory II: Multiple Server Models 8.1
Objectives
After completing this unit, you should be able to; i distinguish between single server models and multi-server models. ii construct a diagram for a given queuing situation. iii write down the probability equations and obtain the steady-state equations for a given queuing situation. iv distinguish between Markovian and non-Markovian queuing models. v identify the appropriate queuing model for a given situation.
8.2
Introduction
In unit 7, we studied queuing models with a single server. In this unit, multi-server models and their generalizations are considered. Once again, with the exception of one model, the arrival time and departure time are assumed to be Markovian. Solutions for the above models are obtained under steady state conditions. Finally, some applications of queuing models are discussed.
8.3 8.3.1
Queueing Models with m Parallel Servers Model 1:(M/M/m):(GD/∞/∞)
In this model, it is assumed that m parallel servers provide the service simultaneously (where m > 1). All service channels are identical and the service time distribution is exponential with mean µ per unit time. A customer is attended to by only one server, even if some of the other service channels are free. A typical example is a bank where more than one teller provides the same service. Using the notation of unit 7,we have the following probability equations: p0 (t + ∆t) = p0 (t)(1 − λ∆t) + p1 µ∆t
(8.1)
pn (t + ∆t) = pn (t)(1 − λ∆t)(1 − nµ∆t) + pn−1 (t)λ∆t + pn+1 (t)(n + 1)µ∆t, n < m (8.2) and pn (t + ∆t) = pn (t)(1 − λ∆t)(1 − mµ∆t) + pn−1 (t)λ∆t + pn+1 (t)mµ∆t, n < m. (8.3) 91
92
Queuing Theory II: Multiple Server Models
Letting ∆t → 0 in (8.1), (8.2) and (8.3), we have p0n (t) = −λp0 (t) + µp1 (t),
(8.4)
p0n (t) = −(λ + nµ)pn (t) + λpn−1 (t) + (n + 1)µpn+1 (t), 0 < n < m
(8.5)
p0n (t) = −(λ + mµ)pn (t) + λpn−1 (t) + mµpn+1 (t), n ≥ m.
(8.6)
and
Letting t → ∞ in (8.4), (8.5) and (8.6), it follows that −λp0 + µp1 = 0,
(8.7)
−(λ + nµ)pn + λpn−1 + (n + 1)µpn+1 = 0, 0 < n < m
(8.8)
(λ + mµ)pn + λpn−1 + mµpn+1 = 0, n ≥ m.
(8.9)
and
We can now solve (8.7), (8.8) and (8.9) together with the equation ∞ X
pn = 1
n=1
to obtain the values of pn as a function of p0 .
Activity 8.1 1. Explain the terms (1 − nµ∆t) and (1 − mµ∆t) in (8.2) and (8.3) respectively. 2. From (8.7), (8.8) and (8.9) show that for an (M/M/m):(GD/∞/∞) queue pn =
ρ n! p0n, ρ p , mn−m m! 0
0≤n≤m n≥m
(8.10)
and p0 =
(m−1 X ρn n=0
ρm + n! m!(1 − ρ/m)
)−1 (8.11)
where ρ = λ/µ and ρ/m < 1 and pn represents the probability that there are n customers in the system (n = 0, 1, 2, ...). 3. Prove that for the (M/M/m):(GD/∞/∞) queuing model (a) ρm+1 ρm Lq = p0 = pc 2 (m − 1)!(m − ρ) (m − ρ)2 (b)Ls = Lq + ρ.
(8.12)
Queuing Theory II: Multiple Server Models
8.3.2
93
Model 2: (M/M/m): (GD/N/∞), m ≤ N
In this model, the system can accommodate a limited number N , of people. When the system attains the state N , it does not allow any arrivals, i.e. the possible states of the system are n = 0, 1, 2, ..., N . The probability equations for this system are exactly the same as (8.1) to (8.3), but in addition we require a separate equation for the case n = N , which is pN (t + ∆t) = pN (1 − mµ∆t) + pN +1 (t)λ∆t. Letting ∆t → 0 and t → ∞, this becomes −mµpN + λpN −1 = 0. Hence the required difference equations for the (M/M/m):(GD/N/∞) system are −λp0 + µp1 −(λ + nµ)pn + λpn−1 + (n + 1)µpn+1 −(λ + mµ)pn + λpn−1 + mµpn+1 −mµpn + λpN −1
= = = =
0 0 , 00
n =
This can be expressed conveniently by introducing a unit function 1 , n=0 δ(n) = 0 , n>0
(8.33)
(8.34)
Then (8.33) with the help of (8.34) becomes n0 = n − 1 + δ(n) + m, for all n
(8.35)
It may be recalled that for a given length of service time t, the number of arrivals m, follows a Poisson distribution with mean λt. Therefore Z ∞ E{m \ t}f (t)dt E{m} = 0 ∞
Z =
λtf (t)dt 0
(8.36)
∞
Z =λ
tf (t)dt 0
= and E{m2 }
λ =ρ µ
∞
Z
E{m2 \ t}f (t)dt
= 0 ∞
Z
{λt + (λt)2 }f (t)dt
= 0
Z
∞
tf (t)dt + λ2
=λ
∞
Z
0
t2 f (t)dt
0
= λE{t} + λ2 E{t2 }. Since V = E(x2 ) − µ2 for a random variable X. But E{t} = and V = E{t2 } − Hence E{m2 } =
λ µ
1 µ 2 1 . µ
n + λ2 V +
= ρ + λ 2 V + ρ2
1 µ2
o (8.37)
Queuing Theory II: Multiple Server Models
101
Taking expectation of both sides of (8.35), applying the steady-state condition En = En0 and using (8.36), we obtain E{δ(n)} = 1 − E{m} =1−ρ
(8.38)
Squaring both sides of (8.35), we obtain (n0 )2 = n2 + δ(n)2 + m2 + 1 + 2mδ(n) − 2m + 2mn − 2δ(n) + 2nδ(n) − 2n. Taking expectations of both sides and substituting in the steady-state condition E{n2 } = E{(n0 )2 }, we obtain E{n} =
1 + E{m2 } + 2E{m}E{δ(n)} − 2E{m} − E{δ(n)} 2(1 − E{m})
(8.39)
Using (8.36), (8.37) and (8.38) in (8.39), we obtain E{n} = ρ
λ 2 + ρ2 , 2(1 − ρ)
which is the required result. To obtain the other two results, we use Lq = Ls − ρ and Ws =
Ls . λ
Example 8.5 A workshop intends hiring one of two repairmen to attend to their machines. The first repairman will be paid at the rate of $5.00 per hour and he can repair machines at a rate of 5 per hour. The second repairman will be paid at the rate of $7.00 per hour and he can repair 8 machines per hour. It is estimated that the machine down time cost is $20 per hour. Assuming that the repair time is exponential and the machines break down in a Poisson fashion at a mean rate of 4 per hour, which repairman should be hired?
Solution 8.5 This is an (M/M/1) : (GD/∞/∞) situation. The cost per hour of hiring the first repairman = wages per hour + cost of down time per hour = 5 + 20Ls . But ρ 4/5 Ls = = = 4. 1−ρ 1 − 4/5
102
Queuing Theory II: Multiple Server Models
Hence, the cost per hour of hiring the first man is given by: 5 + 20 × 4 = $85. Similarly, the cost of hiring the second person is $27, and we decide to have the second repairman.
8.6.1
Exercises
1. Consider the queuing model (M/M/1) : (GD/R/R). (a) For the process above, define states of the system and obtain the probability equations of the queuing process. (b) Solve the steady-state equations and prove that R pn = n!ρn ρ0 , n = 1, 2, ..., R. n (c) Find the value of p0 . (d) Prove that Ls = R + ρ1 (1 − p0 ). 2. A self-service station has three petrol pumps to serve its customers. The car arrival follows a Poisson process with a mean rate of 10 per hour. When all pumps are busy, an arriving car waits in a single queue and is served by the first available pump. The service time has an exponential distribution with mean of 5 minutes. (a) Construct a diagram for this queuing system. (b) Write down the probability equations and obtain the steady-state solutions. (c) Find the expected waiting time per car in the queue. (d) Find the expected time spent by a car at the petrol station. 3. At a car wash, cars arrive in Poisson fashion at an average rate of 20 per hour. One attendant can clean cars at the rate of 15 per hour. Customers do not wait for a wash if the queue size is three or more cars. In that event a sale is lost. The attendant is paid $4.50 per hour. If the opportunity cost of a lost sale is $6.00 per car, how many attendants should be employed by the car wash company? 4. A fast-food service has a single server with Poisson arrival and exponential service times. The arrival rate is 15 per hour and the mean service time is 3 minutes per customer. This food service is a member of a fast-food chain who will be renewing the franchise next month. The fast-food chain evaluates a particular store on the basis of four criteria. The store is naturally concerned about the quality of service and has hired you as a consultant to evaluate the following for them: (a) the percentage of time the server will be idle
Queuing Theory II: Multiple Server Models
103
(b) the mean waiting time for a customer (c) the average time a customer spends in the system (d) the average number of customers in the system. 5. A shopping centre has a parking lot which can accommodate 25 cars. Cars arrive according to a Poisson process with mean arrival rate of 10 cars per hour. The shopping duration follows a negative exponential distribution with a mean duration of 10 minutes. It has been noticed that an arriving car does not wait for a parking space if the car park is full. (a) Identify the queuing model which will represent the above situation. Write down the probability equations and obtain their solution. (b) Find the values of Ls , Lq , Ws and Wq . (c) Find the percentage of customers who shop elsewhere due to difficulties in parking.
Bibliography [1] W.L. Winston, Operations Research: Applications and Algorithms,second edition, Duxbury Press, 1991. [2] David R. Anderson, Dennis J. Sweeney, Thomas A. Williams, Quantitative Methods For Business 1989. [3] Gary E. White House, Ben L. Wechsler, Applied Operations Research: A Survey, 1976 [4] Hans G. Daellenbach , John A. George, Introduction to Operations Research Techniques. [5] H.A. Taha, Operations Research, sixth edition, Prentice Hall, 1997. [6] Lawrence L.Lapin, Quantitative Methods for Business Decisions with cases. [7] R.Markland ,J.R.Sweigart, Quantitative Methods Applications To Managerial Decision Making. [8] F.S. Hillier and G.J. Lieberman, Introduction to Operations Research, fifth edition, McGraw-Hill, Inc. 1990. [9] W.L. Winston, Introduction to Mathematical Programming, second edition, Duxbury Press, 1995. [10] S.Kumar, Queuing Theory and Inventory Control, NUST 1997. [11] W.T. Knowles,Management Science, first edition, Irwin, 1989. [12] T. Lucey, Quantitative Techniques, fourth edition, ELBS, 1994.
104
