Notes: Deterministic Models in Operations Research - Mathematics

Notes: Deterministic Models in Operations Research J.C. Chrispell Department of Mathematics Indiana University of Pennsylvania Indiana, PA, 15705, USA E-mail: [email protected] http://www.math.iup.edu/~jchrispe May 5, 2012

OR Notes Draft: May 5, 2012

ii

Preface These notes will serve as an introduction to the basics of solving deterministic models in operations research. Topics discussed will included optimization techniques and applications in linear programming. Specifically a discussion of sensitivity analysis, duality, and the simplex method will be given. Additional topics to be considered include non-linear and dynamic programming, transportation models, and network models. The majority of this course will follow the presentation given in the Operations Research: Applications and Algorithms text by Winston [8]. I will supplement the Winston text with additional material from other popular books on operations research. For further reading you may wish to consult: • Introduction to Operations Research by Hillier and Lieberman [2] • Operations Research: An Introduction by Taha [7] • Linear Programming and its Applications by Eiselt and Sandblom [1] • Linear and Nonlinear Programming by Luenberger and Ye [4] • Linear and Nonlinear Programming by Nash and Sofer [5] My Apologies in advance for any typographical errors or mistakes that are present in this document. That said, I will do my very best to update and correct the document if I am made aware of these inaccuracies. -John Chrispell

iii


iv

Contents

1 Introduction 1.1

1

Tabel and Chair Example . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 Basic Linear Algebra 2.0.1

7

The Gauss-Jordan Method for Solving Linear Systems . . . . . . . .

3 Linear Programming Basics 3.1

3.2

3.3

4.2

4.3

8 11

Parts of a Linear Program . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11

3.1.1

Linear Programming Assumptions . . . . . . . . . . . . . . . . . . . .

12

3.1.2

Feasible Region and Optimal Solution . . . . . . . . . . . . . . . . . .

12

Special Cases: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

15

3.2.1

Example: Multiple Optimal Solutions . . . . . . . . . . . . . . . . . .

15

3.2.2

Example: Infeasible Linear Program . . . . . . . . . . . . . . . . . .

16

3.2.3

Example: Unbounded Optimal Solution

. . . . . . . . . . . . . . . .

16

Setting up Linear Programs . . . . . . . . . . . . . . . . . . . . . . . . . . .

17

3.3.1

17

Work-Scheduling Problem: . . . . . . . . . . . . . . . . . . . . . . . .

4 Examples of Linear Programs 4.1

1

19

Diet Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

19

4.1.1

Solution Diet Problem . . . . . . . . . . . . . . . . . . . . . . . . . .

20

Scheduling Problem: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21

4.2.1

Solution Scheduling Problem: . . . . . . . . . . . . . . . . . . . . . .

21

A Budgeting Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

22

4.3.1

22

Solution Budgeting Problem . . . . . . . . . . . . . . . . . . . . . . . v

OR Notes Draft: May 5, 2012 5 The Simplex Algorithm 5.1

5.2

25

Standard Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

25

5.1.1

Example: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

28

Basic and Nonbasic Variables . . . . . . . . . . . . . . . . . . . . . . . . . .

28

5.2.1

30

Directions of Unboundedness

. . . . . . . . . . . . . . . . . . . . . .

6 Simplex Method Using Matrix-Vector Formulas

35

6.1

Problem Using Tableaus . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

35

6.2

Matrix Format of Linear Program . . . . . . . . . . . . . . . . . . . . . . . .

36

6.2.1

37

Initial Tableau (Matrix Form): . . . . . . . . . . . . . . . . . . . . . .

7 Duality 7.1

41

A Motivating Example: My Diet Problem: . . . . . . . . . . . . . . . . . . .

41

7.1.1

43

Cononical Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8 Basic Duality Theory

47

8.1

Relationship Between Primal and Dual . . . . . . . . . . . . . . . . . . . . .

47

8.2

Weak Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49

8.3

Strong Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

50

8.4

The Dual Simplex Method . . . . . . . . . . . . . . . . . . . . . . . . . . . .

57

9 Sensitivity Analysis 9.1

9.2

61

Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

61

9.1.1

Verify this graphically: . . . . . . . . . . . . . . . . . . . . . . . . . .

63

Sensitivity Analysis Using Matrices . . . . . . . . . . . . . . . . . . . . . . .

64

9.2.1

65

Illustrating Example . . . . . . . . . . . . . . . . . . . . . . . . . . .

10 Non Linear Programming

71

10.1 Data Fitting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.1 Linear Problems

71

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

72

10.1.2 Taylor series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75

10.1.3 Newton’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75

10.2 Non-linear Least Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

77

vi

OR Notes Draft: May 5, 2012 11 Network Flows

79

11.1 Dijkstra’s Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Appendices

79 83

12.1 Homework 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

83

12.2 Homework 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

88

12.3 Homework 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

93

12.4 Homework 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 12.5 Homework 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Bibliography

113

vii

Chapter 1 Introduction “If you choose not to decide you still have made a choice.” −Neil Peart

Operations Research, at its heart, uses modeling to solve applied mathematical problems in the hopes of making optimal decisions efficiently. The term Operations Research can trace its roots back to World War II when a scientific approach was adopted to distribute resources between various military operations (scientists doing research on military operations)[2, 8]. In order to do any scientific research on a problem a mathematical model is needed. The WInston text states: Mathematical Model: - is a mathematical representation of an actual situation that may be used to make a better decision or simply to understand the actual situation better. The goal with most models considered in this course will be to optimization some quantity (using an objective function based on decision variables) subject to problem constraints. Several examples where Operations Research techniques were implemented in order for different organizations to obtain optimal solutions and distribution of resources (CITGO Petroleum manufacturing, The San Francisco Police Department Scheduling, GE Capital credit card bill repayment).

1.1

Tabel and Chair Example

Consider the following modification of an example given in [6]: A furniture manufacturer produces two products: wooden tables and chairs. The unit profit for the tables is $6, and the unit profit for the chairs is $8. In order to simplify the problem assume that only resources used in the production of the tables and chairs are wood (in board feet, bf ) and labor (in hours, h). It takes 6 bf and 1 hour labor to make a table, and 4 bf and 2 hours labor to make a 1

OR Notes Draft: May 5, 2012 chair. There are 20 bf of wood available and 6 hours of labor available. If you wish to maximize profit what is the optimal distribution of these resources?

Formulate a Mathematical Model of the Problem The idea now becomes to solve the given problem using a linear model or linear programming. Thus, we need to translate the real world problem into a format with mathematical equations that represent: • An objective function: In this case maximize the profit. • Decision variables: Number of Tables and Chairs to produce. • Constraint set: We only have a limited number of resources available to use when manufacturing the tabels and chairs. For this example the objective function and the constraints will all be linear functions. We should also impose non-negativity conditions on the resources and decision variables in this instance. If we do not enforce these conditions it may be possible to achieve non-physical solutions to the given problem. The problem definition, and formulation are often the most difficult and import step in finding an optimal solution. Lets define the decision variables x1 and x2 as the number of tables and the number of chairs to produce respectively. It is sometimes useful to place all the information into a table: Resource Wood (bf ) Labor (hr) Unit Profit

Table x1 6 1 $6

Chair x2 4 2 $8

Available 20 6

Thus, the goal is to obtain the largest profit from the objective function by: Maximize 6x1 + 8x2 subject to: 6x1 + 4x2 ≤ 20 x1 + 2x2 ≤ 6 x1 ≥ 0 x2 ≥ 0 the problem constraints. 2


Solving the problem using a Graphical Approach Lets assume that the number of chairs to be produced, x1 is on the x-axis and the number of tables to be produced x2 is on the y−axis. First write the constraints as equalities, and then finding the intercepts of the feasible region. Thus, 3 6x1 + 4x2 ≤ 20 =⇒ x2 = − x1 + 5 2 1 x1 + 2x2 ≤ 6 =⇒ x2 = − x1 + 3 2 x1 ≥ 0 x1 = 0 x2 ≥ 0 x2 = 0

Figure 1.1.1: Plot of the constraints for the table and chair example. If all the wood is used to produce chairs (set x1 = 0 in 1.1.1): x2 = 5 this yields point C. If all the wood is used to produce tables (set x2 = 0 in 1.1.1): x1 =

10 this yields point E. 3

If all the labor is used to produce chairs (set x1 = 0 in 1.1.2): x2 = 3 this yields point B. 3

(1.1.1) (1.1.2) (1.1.3) (1.1.4) (1.1.5)

OR Notes Draft: May 5, 2012 If all the labor is used to produce tables (set x2 = 0 in 1.1.2): x1 = 6 this yields point D. Note we obtain a feasible region given by the polygon FEAB.

Plot instances of the objective function If we now pick two values for the objective function we can determine its slope and a direction of improvement. Let z be the value of the objective function: z = 6x1 + 8x2 Thus, if z = 0 then 3 6x1 + 8x2 = 0 =⇒ x2 = − x1 , 4 and if z = 8 we see

3 6x1 + 8x2 = 8 =⇒ x2 = − x1 + 1. 4

The direction of improvement is d = [4, 3]T , and is illustrated in Figure 1.1.2. The optimal solution will be obtained at the corner or edge of the feasible region that is farthest away from the origin on a line parallel to the objective function contours plotted for z = 0, and z = 8. Specifically from the figure the objective function is maximized at point A when 2 tables and 2 chairs are manufactured using the resources available, and yields a profit of $28.

Figure 1.1.2: Illustration of the direction of improvement. 4

OR Notes Draft: May 5, 2012 The optimal solution could have also been found by testing the objective function at each corner of the feasible region. Note if the same value of the objective function has been obtained at multiple corners the optimal solution of would have been any convex combination of the two corners. In this problem we are manufacturing discrete items (building a fraction of a table or chair doesn’t make since) so we are lucky that we obtained an integer solution to the mathematical model. Here we will solved this simple Linear Program graphically. In the future we will use more advanced algebraic techniques (such as the simplex method ) to obtain the optimal solution. For more complicated problems it will become necessary to use software packages such as LINGO. In the future we will also look to see how sensitive our optimal solution is to small changes.

5


6

Chapter 2 Basic Linear Algebra “Our weary eyes still stray to the horizon. Though down this road we’ve been so many times.” −Pink Floyd

Before looking into more complicated methods of solving linear programs we need to recall some of the basics of linear algebra. We will use the following notation. A matrix typical m × n matrix A written in the form:  a11 a12  a21 a22  A =  .. ..  . . am1 am2 The element of a matrix in the ith row and Thus, if the matrix  1  5 A= 34

is a rectangular array of numbers, with a . . . a1n . . . a2n .. .. . . . . . amn

    

(2.0.1)

j th column of matrix A will be denoted by aij .  1 2 3 8 13 21  55 89 144

then a23 = 13. We will think of a matrix with only one column as vectors (or column vectors). Similarly a matrix with only one row will be a row vector. The number of rows in a vector will define its dimension, and the number of columns in a row vector will define its dimension. An m-dimensional vector (row or column) where all elements are zero will be called a zerovector and denoted by 0. In two dimensions 0 0 = [0 0] and 0 = 0 are zero vectors. 7

OR Notes Draft: May 5, 2012 Vectors correspond to a directed line segment from the origin in the m-dimensional plane. Thus the vectors: 1 −3 u= , and w = 2 −4 are illustrated in Figure 2.0.1. Readers should reacquaint themselves with matrix multipli-

Figure 2.0.1: Illustration of two vectors in two-dimensional plane. cation, matrix transposition and inner products. The use of matrices and vectors will allow for mathematical models of linear systems to be developed and solved using linear algebra.

2.0.1

The Gauss-Jordan Method for Solving Linear Systems

The three basic types of elementary row operations (ERO) are used when solving systems of linear equation using the Gauss-Jordan Method. 1. Multiply any row of a matrix system by a nonzero scalar. 2. Multiply any row of a matrix system by a nonzero scalar and then add that row to a different row of the matrix. 3. Interchange any two rows of a matrix system. 8

OR Notes Draft: May 5, 2012 Consider the following example: x1 − 2x2 + 3x3 = 9 −x1 + 3x2 = −4 2x1 − 5x2 + 5x3 = 17

We may write this linear system in the augmented matrix system notation as   1 −2 3 9 A|b =  −1 3 0 −4  2 −5 5 17

9


10

Chapter 3 Linear Programming Basics “Remember to remember me Standing still in your past Floating fast like a hummingbird”

−Wilco

3.1

Parts of a Linear Program

We detailed earlier that the major components of a linear program: • Decision variables. • The objective function. • The problem constraints. It is also worthy to note that the coefficient of a variable in the objective function is referred to as an objective function coefficient, and less obviously the coefficients in the constraint functions are sometimes referred to as technological coefficients. It will often be useful to denote the coefficients on the right-hand-side of the constraints with rhs (representing the amount of a distinct resource that is available). We will separate non-negativity of decision variables as a separate type of constraint, and make note that decision variables may be unrestricted in sign. Definitions: • A function f (x1 , x2 , . . . , xn ) of x1 , x2 , . . . , xn is a linear function if and only if for some set of constants c1 , c2 , . . . , cn , f (x1 , x2 , . . . , xn ) = c1 x1 + c2 x2 + . . . + cn xn . 11

OR Notes Draft: May 5, 2012 • For any linear function f (x1 , x2 , . . . , xn ) and any number b, the inequalities f (x1 , x2 , . . . , xn ) ≤ b and f (x1 , x2 , . . . , xn ) ≥ b are linear inequalities. With these definitions we can define a linear programming problem or LP as an optimization problem for which we do the following: • Maximize or minimize a linear function of some decision variables (our objective function). • The decision variables are subject to satisfying a set of constraints each of which is a linear inequality. • There is also a set of sign restrictions that will be satisfied. For example decision variables may be non-negative or unrestricted in sign.

3.1.1

Linear Programming Assumptions

In all linear programs the objective function and constraints are effected proportionally by the decision variables (a constant coefficient multiplies the variable in each). Linear programming problems also have the benefit of each decision variable being independent of others in its contribution to the objective function. Similarly linear constraints have the benefit of independence. For example in the table and chair example we considered the number of hours of labor required to make a chair did not effect the number of hours labor required to manufacture a table. Thus, the left hand side of the labor constraint was the additive sum of constants multiples of the decision variables. In addition to the proportionality and additivity assumptions in order for LP’s to represent real situations, the divisibility assumption. That is the decision variables must be allowed to take on fractional values (we can not sell a fraction of a chair). In the future we may look at integer programming as a way of handling problems that require a discrete valued solution. We are also going to make a certainty assumption. That is that the model parameters objective function coefficients, the right hand side of constraints and technological coefficients are known with certainty.

3.1.2

Feasible Region and Optimal Solution

Assuming that a point is a set of distinct values for each decision variable in an LP. The feasible region is the collection of all possible points that satisfy the constraints and sign restrictions on the decision variables. Points not within the feasible for a linear program are considered infeasible. The optimal solution for a maximization problem then becomes the point in the LP’s feasible region that obtains the largest value of the objective function. Similarly for a minimization problem the point inside the LP’s feasible region that obtains the smallest value of the objective function is considered the optimal solution. 12


A second graphical example Consider the Linear Program: max z = x1 + x2 s.t. x1 + 5x2 ≤ 25 6x1 + 5x2 ≤ 40 x1 ≤ 5 with x1 , x2 ≥ 0. The graphical solution to the linear program is shown in Figure 3.1.1. Note the value of the objective function is is 7.4, and that only two of the constraints are binding at the optimal solution. We consider constraints binding if the left hand side of the constraint is equal to the right hand side when an optimal solution is achieved. The constraint x1 ≤ 5 is an example of a nonbinding constraint.

Figure 3.1.1: Graphical solution to LP with bounded Feasible Region.

The feasible region for this LP is a convex set. A set of points is aConvex set provided a line segment between any pair of points in the set is wholly contained in the set. 13

OR Notes Draft: May 5, 2012 For any convex set S, a point P is an extreme point if each line segment that lies completely in S and contains the point P has P as an endpoint of the line segment. Extreme points may also be called corner points. We make note that: Any LP that has an optimal solution has an extreme point that is optimal. See the Winston text for a proof of this result. It is an important result because it narrows our search for an optimal solution down from the whole feasible region to just the extreme points of the set.

A Minimization Problem Consider the following minimization linear programming problem: min z = x1 + x2 s.t. x1 + 4x2 ≥ 8 4x1 + x2 ≥ 6 with x1 , x2 ≥ 0. Here we again solve the LP graphically. Note 14that the optimal solution is found at point 16 26 , giving the objective function a value of 5 . The solution is illustrated in Figure 3.1.2. 15 15 Note that we could have used matrix notation to represent the previous problems. It will sometimes be useful to write our linear programs in this form.

Figure 3.1.2: Graphical solution to minimization LP with bounded Feasible Region. 14


3.2

Special Cases:

Linear programs are not always this nice! Sometimes the linear program has an infinite number of solutions. Sometimes the linear program has no feasible solution, and at other times the solution to the linear program is unbounded.

3.2.1

Example: Multiple Optimal Solutions

Consider the following minimization linear programming problem: max z = 3x1 + 2x2 s.t. 6x1 + 4x2 ≤ 32 −x1 + 2x2 ≤ 10 x1 ≤ 4 with x1 , x2 ≥ 0. When this LP is solved graphically as is done in Figure 3.2.3 it can be seen that there exist multiple optimal solutions. Thus, the solution to the LP becomes the convex combination of the two extreme points represented in the graph by C and D. We can write the optimal solution to the LP by letting α ∈ [0, 1], and stating it as a convex set: optimal set = (x, y) α (4, 2) + (1 − α) (1.5, 5.75) . (3.2.1)

We can note that any values in the optimal set will yield a value of 16 in the objective function. Use the following link to explore this problem using an interactive java applet: http://www.math.iup.edu/~jchrispe/MATH445_545/MultipleOptimalSolutionExample.html

Figure 3.2.3: Graphical solution to maximization LP with bounded Feasible Region and multiple optimal solutions. 15


3.2.2

Example: Infeasible Linear Program

Consider the following minimization linear programming problem: max z = 3x1 + 2x2 s.t. x1 + x2 ≥ 10 x2 ≤ 5.8 x1 ≤ 4 with x1 , x2 ≥ 0. Figure 3.2.4 shows the binding constraint contours and a plot of the objective functions

Figure 3.2.4: Plot of binding constraint contours and objective function contour when z = 10. contour for a value z = 10. Note that there is no region that will satisfy all of the linear programs constraints making this an infeasible LP.

3.2.3

Example: Unbounded Optimal Solution

Consider one last example (straight out of the Winston text). Solve again the following linear program graphically. max z = 2x1 − x2 s.t. x1 − x2 ≤ 1 2x1 + x2 ≥ 6 with x1 , x2 ≥ 0. In Figure 3.2.5 it can be seen that the feasible region is unbounded in the direction of increasing z contours. Thus the optimal solution is unbounded. 16


Figure 3.2.5: Plot of the feasible region for the maximization problem. Note that the feasible region is unbounded in the direction of increasing z contours.

3.3

Setting up Linear Programs

In many cases the difficult part of linear programming is translating a physical situation into a mathematical model. The following are a collection of word problems that may be formulated as a linear program.

3.3.1

Work-Scheduling Problem:

This example is from the Winston text: Consider a chain of computer stores. The number of skilled repair time the company requires during the next five months is given in the following table: Month Month Month Month Month

1 2 3 4 5

(January): 6,000 hours (February): 7,000 hours (March): 8,000 hours (April): 9,500 hours (May): 11,000 hours

At the beginning of January, 50 skilled technicians work for the company. Each skilled technician can work up to 160 hours per month. To meet future demands, new technicians must be trained. It takes one month to train a new technician. During the month of training the trainee must be supervised for 50 hours by an experienced technician. Each experienced technician is paid $2000 a month regardless of how much he works. During the month of training the trainee receives $1000 for the month. At the end of each month 5% of the companies experienced technicians quit to find other employment. Formulate an LP for the company that will minimize the labor costs incurred and meet the service demands for the next five months. 17


18

Chapter 4 Examples of Linear Programs “ The tall one wants white toast, dry, with nothin’ on it... And the short one wants four whole fried chickens, and a Coke.”

−Mrs. Murphy (The Blues Brothers)

In this section our goal will be to consider setting up some common types of linear programs. The list of examples given here is by no means exhaustive, and we should not for get about the work scheduling problem stated in the previous section.

4.1

Diet Problem

My diet may not be nearly as poor as that of the blues brothers; however, it is probably still pretty bad. Lets assume that my diet requires that all of my food come from my favourite food groups: pez candy, pop tarts, Cap’N crunch, and cookies. I have the following foods available for my consumption: red pez, strawberry pop-tarts, peanut butter crunch, and chocolate chip cookies. The red pez costs $0.20 per (per package), strawberry pop-tarts cost $0.50 (per package), peanut butter crunch costs $0.70 per bowl, and chocolate chip cookies cost $0.65 cents each. I must ingest at least 2500 calories a day in order to maintain my sugary lifestyle. I must also meet the following requirements I need 8 oz. of sugar, 10 oz. of fat, and 50 mg of yellow-5 food coloring. Each of my chosen foods has these required nutrients in the following quantities: Food red pez candy (per package) strawberry pop-tarts (per package) peanut butter Cap’N crunch (per bowl) chocolate chip cookies (1 cookie)

Calories 50 250 350 150

Sugar (oz.) 0 .1 0.5 1 0.3

How can I achieve my dietary constraints at a minimum cost? 19

Fat (oz.) 0.01 0.3 1 0.7

yellow-5 (mg) 5 0 4 1


4.1.1

Solution Diet Problem

The first step will be do define some decision variables. Here we will define x1 x2 x3 x4

as as as as

the the the the

number number number number

of pez packages to include in my diet per day strawberry pop-tart packages to consume per day of bowls of peanut butter crunch to consume per day of chocolate chip cookies to consume each day

Our objective function can then be defined as: min z = 0.20x1 + 0.50x2 + 0.70x3 + 0.65x4 which will minimize the cost of our diet. The next aim is to meet the special dietary needs. Here a constraint is set up for each of the “nutrients” (calories, sugar, fat, and yellow 5). 50x1 + 250x2 + 350x3 + 150x4 0.1x1 + 0.5x2 + x3 + 0.3x4 0.01x1 + 0.3x2 + x3 + 0.7x4 5x1 + 0x2 + 4x3 + x4

≥ ≥ ≥ ≥

2500 (calories required) 8 (sugar required) 10 (fat required) 50 (yellow-5 required)

It can also be noted that the decision variables will be non-negative in sign giving the xi ≥ 0 for all i ∈ 1, 2, 3, 4 sign restriction. This problem can be set up in vector notation using the above variable definitions and defining       x1 0.20 2500  x2       , c =  0.50  , b =  8  , and x=  x3   0.70   10  x4 0.65 50   50 250 350 150  0.1 0.5 1 0.3   A=  0.01 0.3 1 0.7  . 5 0 4 1 The optimal diet can now be found by considering: min z = cT x s.t. Ax ≥ b with x ≥ 0. The optimal solution can be quickly found using the LINGO software package and the old LINDO syntax. Open LINGO and type: 20

OR Notes Draft: May 5, 2012 min 0 . 2 x 1 + 0 . 5 x 2 + 0 . 7 x 3 + 0 . 6 5 x 4 s . t . 50 x 1 + 250 x 2 + 350 x 3 + 150 x 4 >= 2500 0 . 1 x 1 + 0 . 5 x 2 + x 3 + 0 . 3 x 4 >= 8 0 . 0 1 x 1 + 0 . 3 x 2 + x 3 + 0 . 7 x 4 >= 10 5 x 1 + 0 x 2 + 4 x 3 + x 4 >= 50 This yields the optimal objective function value of $7.39 when x1 = 2.016129, x2 = 0.000000, x3 = 9.979839, and x4 = 0.000000.

(4.1.1)

So the minimum cost diet has me eating nothing but approximately 2 packages of pez candy and 10 bowls peanut butter crunch to meet my dietary constraints.

4.2

Scheduling Problem:

Suppose that the math department wants to schedule tutors for a cram-day before final exams. The number of tutors needed for each four hour shift on this cram-day are: Shift Number 1 2 3 4 5 6

Time 12:00 am - 4:00 am 4:00 am - 8:00 am 8:00 am - 12:00 pm 12:00 pm - 4:00 pm 4:00 pm - 8:00 pm 8:00 pm - 12:00 am

Number of Tutors Needed 3 4 6 5 8 4

Each tutor will work two consequtive 4 hour shifts. Formulate a linear program that can be used to minimize the number of tutors needed to meet the cram-day demands.

4.2.1

Solution Scheduling Problem:

Its always best to think about the decision variable first! Here we can think about the starting shift for the tutors. Let xi be the number of tutors that start working on shift i. The objective of the linear program is to minimize the number of tutors needed to meet the demands, and we make note that summing the decision variables will count the total number of tutors used. We should also note that since we are only concerning ourselves with 8 hour shifts for a single day, we only need to start tutors for the first five shifts. This give the objective function: 5 X minimizez = xi i=1

21

OR Notes Draft: May 5, 2012 Then the constraints become: x1 x1 + x2 x2 + x3 x3 + x4 x4 + x5 x5

≥ ≥ ≥ ≥ ≥ ≥

3 4 6 5 8 4

where xi ≥ 0 for i ∈ {1, 2, . . . , 5}.

(4.2.2)

Note we also added the non-negativity constraint for all the decision variables. It may have also been nice to consider making the schedule for several cram-days in a row that again all had the same demands for the number of tutors needed during each shift.

4.3

A Budgeting Problem

Two investments with varying cash flows (in thousands of dollars) are available.

Investment 1 2

Cash Flow Year 0 -6 -8

Cash Flow Year 1 -5 -3

Cash Flow Year 2 7 9

Cash Flow Year 3 9 7

Assuming at time 0 that $10,000 is available to invest, and after one year there will be $7,000 available to invest. We will also assume an annual interest rate of 10% is available. We shall also assume that any fraction of an investment may be purchased. Lets find a linear program that will determine the maximum net present value obtained from the two investments.

4.3.1

Solution Budgeting Problem

We set up the linear program as follows. First the decision variables: Let xi be the amount of investment i to purchase (i = 1, 2). Our goal is to maximize the net present value of the investments purchased so we should find the net present value of each investment for use in our objective functions. Net present value: We need to discount each of the cash flows back to the present using a discount factor based on the interest rate. Note that $1.00 Today −→ $1.00 + 0.10($1.00) = $1.00(1.10) One Year From Now 22

OR Notes Draft: May 5, 2012 Thus, $1.00 one year from now −→ We can use

1 1+r

1 1.10

$1.00 ≈ $0.9090909 today.

n as a discount factor for rate r in year n.

Let N P V1 denote the net present value of investment 1. 1 2 3 1 1 1 N P V1 = −6000 + (−5000) + 7000 + 9000 ≈ 2001.503 1.1 1.1 1.1 Let N P V2 denote the net present value of investment 2. 1 2 3 1 1 1 N P V2 = −8000 + (−3000) + 9000 + 7000 ≈ 1969.947 1.1 1.1 1.1 The objective function for the linear program is then: maximize z = 2001.50x1 + 1969.94x2 . The constraints on the linear program are based on the amount of cash we have on hand for each of the first two years, and the additional constraint that we may only purchase upto 100% of each of the two investments. This gives 6000x1 + 8000x2 ≤ 10000 5000x1 + 3000x2 ≤ 7000 x1 ≤ 1 x2 ≤ 1

as our constraints, and we have the sign restriction that x1 and x2 are non-negative: x1 ≥ 0, and x2 ≥ 0. Note we can solve this linear program graphically and Figure 4.3.1 illustrates the feasible solution to the problem. We can see by looking for the optimal objective function contour that the optimal investment involves 100% of investment 1 and 50% of investment 2 giving the optimal net present value of the cash flow as: z = 2001.50(1) + 1969.94(0.5) = 2986.47

Note that there are an infinite number of different problems that may be formulated as linear programs. It is strongly recommended that the interested reader look over more examples presented in [8] and other texts. 23


Figure 4.3.1: Feasible region for the budgeting problem with an illustration of a z contour.

24

Chapter 5 The Simplex Algorithm “Don’t wanna wait ’til tomorrow Why put it off another day? One by one, little problems Build up and stand in our way, oh”

−Van Halen

The linear programs that we have been able to solve with out the assistance of a software package so far have all been in two variables. In general linear programs have many variables and the goals of this chapter will be: • Transform a linear programming problem into a standard form. • Look at basic and nonbasic variables. • Introduce the Simplex Algorithm for solving linear programming problems.

5.1

Standard Form

As we have seen there are several different ways of writing the objective function for a linear program (maximize profit, minimize cost). Additionally the constraints come in all different forms too (≤, ≥, =) so when discussing a systematic method for solving linear programs it will be advantages to having written or converted the linear program to a standard form. We will consider the following linear programs to be in standard form: minimize z = cT x, subject to Ax = b, x ≥ 0.

maximize z = −cT x, subject to Ax = b, x ≥ 0.

where b, c, and x represent vectors of length n, and A is an m × n constraint matrix, and b ≥ 0. 25

OR Notes Draft: May 5, 2012 Notice that: • The decision variables are constrained to be non-negative. • All of the constraints are set up to be equalities. • The components of the right hand side vector b are non-negative. The observation that a maximization problem could be converted to a minimization problem by multiplying every coefficient in the objective function by a ‘-1’ can be made. After the problem is solved the objective value would then need to be multiplied by ‘-1’ again in order to obtain the solution to the desired problem. The decision variable values obtained would be the same for both objective functions. Lets consider transforming the following linear program into a standard form minimization problem: maximize subject to

z =

3x1 +4x2 7x1 +x2 −2x1 +4x2 6x1 +3x2

−7x3 +4x3 +6x3 −4x3 x3

≤ 26 ≤ −2 ≥ 4 ≥ 5

with x1 ≥ 0 and x2 free. Objective function: The problem could be written using the equivalent objective function: minimize zˆ = −3x1 − 4x2 + 7x3 where z = −ˆ z once the optimal solution has been found. Non-negative right had side: In order to continue converting the given LP to standard form we next consider the constraints. We would like to have all the right hand side coefficients be non-negative. Thus, the third constraint in the given linear program should be multiplied by a ‘-1’ yielding: −2x1 + 4x2 + 6x3 ≤ −2 −→ 2x1 − 4x2 − 6x3 ≤ 2 Non-zero lower bounds: In the original statement of the problem we note that the 4th constraint has: x3 ≥ 5. To convert this to a standard form constraint we make a change of variables and define: xˆ3 = x3 − 5 This allows us to replace x3 with xˆ3 throughout and changes the constraint x3 ≥ 5 −→ xˆ3 + 5 ≥ 5 −→ xˆ3 ≥ 0. 26

OR Notes Draft: May 5, 2012 Note we need to modify the other constraints and objective function as well. For our example problem we may redefine zˆ = −z − 35 −→ z = −ˆ z − 35 for the modified objective function. For the moment we will leave upper bounds on variables right in the coefficient matrix. Free variables: In the given problem we make note that x2 is a free variable. Free variables may be converted to non-negative constraints by defining x2 = x02 − x002 with x02 , x002 ≥ 0. and we will have x02 to handle the positive values of x2 and the new variable x002 will take care of the negative pieces. There are other ways to handel free variables, and one is to use a free variable as a way of eliminating a constraint (think about this for a future homework assignment). After applying all of the discussed techniques to our linear programming problem we are left with the following LP: minimize subject to

zˆ = −3x1 −4x02 + 4x002 7x1 +x02 − x002 2x1 −4x02 + 4x002 6x1 +3x02 − 3x002

+7ˆ x3 +4ˆ x3 ≤ 6 −6ˆ x3 ≥ 32 −4ˆ x3 ≥ 24

with x1 , x02 , x002 , xˆ3 ≥ 0. Equality constraints: In order for the constraints we consider adding slack and excess variables to enforce the equality constraint. For the constraint: 7x1 + x02 − x002 + 4ˆ x3 ≤ 6 let s1 be a slack variable such that s1 ≥ 0 and 7x1 + x02 − x002 + 4ˆ x3 + s1 = 6. For the other two constraints we use excess variables e1 ande2 . Thus, x3 − e1 = 32 x3 ≥ 32 =⇒ 2x1 − 4x02 + 4x002 − 6ˆ 2x1 − 4x02 + 4x002 − 6ˆ and 6x1 + 3x02 − 3x002 − 4ˆ x3 ≥ 24 =⇒ 6x1 + 3x02 − 3x002 − 4ˆ x3 − e2 = 24 with e1 , e2 ≥ 0.

We can now restate the full linear program in standard form:

minimize subject to

zˆ = −3x1 −4x02 + 4x002 7x1 +x02 − x002 2x1 −4x02 + 4x002 6x1 +3x02 − 3x002

+7ˆ x3 +4ˆ x3 +s1 = 6 −6ˆ x3 −e1 = 32 −4ˆ x3 −e2 = 24

with x1 , x02 , x002 , xˆ3 , s1 , e1 , e2 ≥ 0. 27


5.1.1

Example:

Try writing the following linear programs in standard form: maximize subject to

z =

3x1 7x1 −2x1 5x1

+5x2 −2x2 +4x2 −3x2

−4x3 −3x3 ≥ 4 +8x3 = −3 −2x3 ≤ 9

with x1 ≥ 1, x2 ≤ 7, and x3 ≥ 0. After manipulation we obtain: maximize subject to

zˆ =

3ˆ x1 −7ˆ x1 2ˆ x1 5ˆ x1

+5x02 − 5x002 +2x02 − 2x002 −4x02 + 4x002 −3x02 + 3x002 x02 − x002

−4x3 +3x3 ≤ 3 −8x3 = 1 −2x3 ≤ 4 ≤ 7

with xˆ1 , x02 , x002 , x3 ≥ 0 and z = zˆ + 3. We need to take this one more step and include slack variables. Thus, maximize subject to

zˆ =

3ˆ x1 −7ˆ x1 2ˆ x1 5ˆ x1

+5x02 − 5x002 +2x02 − 2x002 −4x02 + 4x002 −3x02 + 3x002 x02 − x002

−4x3 +3x3 +s1 −8x3 −2x3 +s2 +s3

= = = =

3 1 4 7

with xˆ1 , x02 , x002 , x3 , s1 , s2 , s3 ≥ 0 and z = zˆ + 3.

5.2

Basic and Nonbasic Variables

Lets assume that we have a standard form linear program. minimize z = cT x, subject to Ax = b, x ≥ 0.

(5.2.1)

where b, c, and x represent vectors of length n, and A is an m × n constraint matrix (with n ≥ m), and b ≥ 0. The constraints to the linear program are given by: Ax = b. 28

OR Notes Draft: May 5, 2012 Definition 5.2.1 A basic solution to Ax = b is obtained by setting n − m variables equal to 0 and solving for the values of the remaining m variables. This assumes that the columns remaining m variables are linearly independent. To find a basic solution to Ax = b, we simply choose a set of n − m variables (and set these to zero) to be the nonbasic variables or NBV. The remaining variables will be our basic variables or BV that are used to satisfy the constraints. Consider the following example: x1 + x2 = 6 −x2 + x3 = 4 Here we get to pick one nonbasic variable as we have two equations and three unknowns. Thus, N BV = {x3 }, then BV = {x1 , x2 }. The values of the basic variables x1 , and x2 are found by solving the two equations with x3 set to zero. Thus, x2 = −4 and x1 = 10 If we choose x1 to be the nonbasic variable then the basic variables x2 = 6 and x3 = 10. We should note that not all sets of basic variables will yield a feasible solution to a set of constraints. Consider the constraints x1 + 2x2 + 4x3 = 4 x1 + 4x2 + 8x3 = 6.

If x1 is taken to be the nonbasic variable then the system becomes: 2x2 + 4x3 = 4 4x2 + 8x3 = 6 a system with no feasible solution and yielding no basic solution to BV = {x2 , x3 }. Definition 5.2.2 Any basic solution to (5.2.1) in which all variables are nonnegative is a basic feasible solution. When solving linear programs we are interested in sets of variables that will satisfy all of the constraints given by Ax = b as well as also satisfying the nonnegativity constraint for the linear program stated in standard form. We also know that 29

OR Notes Draft: May 5, 2012 Any LP that has an optimal solution has an extreme point that is optimal. So a good place to start looking for these optimal extreme points is at basic feasible solutions. It can be shown that: Theorem 5.2.1 A point in the feasible region of a linear program is an extreme point if and only if it is a basic feasible solution to the LP. A proof can be found in [3]. Example: The goal of this example is to show the relationship between extreme points and basic feasible solutions. Given maximize subject to

z = 4x1 + 3x2 x1 + x2 ≤ 4 2x1 + x2 ≤ 6

with x1 , x2 ≥ 0, write the LP in standard form by adding two slack variables s1 and s2 . Standard form: maximize subject to

z = 4x1 + 3x2 x1 + x2 + s 1 =4 2x1 + x2 + s2 = 6

with x1 , x2 , s1 , s2 ≥ 0. The feasible region for the linear program is shown if Figure 5.2.1. Note that all of the corner extreme points correspond to a basic feasible solution to the linear program. We note that the intersection at points E, and F are not basic feasible solutions because not all of the basic variables satisfy the nonnegativity constraint for the linear program in standard form.

5.2.1

Directions of Unboundedness

Assume that we have a standard form linear program. minimize z = cT x, subject to Ax = b, x ≥ 0. where b, c, and x represent vectors of length n, and A is an m × n constraint matrix (with n ≥ m), and b ≥ 0. The feasible solutions for this linear program will be denoted by S. 30


Figure 5.2.1: Definition 5.2.3 An n by 1 vector d is a direction of unboundedness if for all x in S, and c ≥ 0 then x + cd ∈ S. This means we can move as far as we desire in the direction of d and still have a feasible solution to the linear program. Consider the following example minimize z x1 + x2 , subject to 7x1 + 2x2 ≥ 28 2x1 + 12x2 ≥ 24 x1 , x2 ≥ 0. Written in standard form by adding excess variables we obtain, minimize z x1 + x2 , subject to 7x1 + 2x2 − e1 = 28 2x1 + 12x2 − e2 = 24 x1 , x2 , e1 , e2 ≥ 0. What are the basic feasible solutions for the LP. 31

OR Notes Draft: May 5, 2012 BV x1 , x2 x1 , e1 x1 , e2 x2 , e1 x2 , e2 e1 , e2

NBV e1 , e2 x2 , e2 x2 , e1 x1 , e2 x1 , e1 x1 , x2

bfs (Basic Feasible Solution) YES , x = (3.6, 1.4, 0, 0)T YES , x = (12, 0, 56, 0)T NO , x = (4, 0, 0, −16)T NO, x = (0, 2, −24, 0)T YES, x = (0, 14, 0, 144)T NO, x = (0, 0, −28, −24)T

Looking at the problem’s feasible region graphically we have the illustration in Figure 5.2.2. Specifically the feasible region for the linear program can be made up of a convex combination of the basic feasible solution points plus a positve constant multiple of a direction of unboundedness. Note that moving from the point basic feasible solution at point    3.6 4.6  1.4   2.4     0  and moving toward point  9 0 14

   

yields a direction of unboundedness given by: 

 1  1   d1 =   9 . 14

Note that the direction of unboundedness is not unique, as   2  4   d2 =   22  52 is a direction of unboundedness found by heading from the    12  0     toward interior point   56   0

basic feasible solution  14 4  . 0  52 P Specifically we could write the feasible set by letting α1 , α2 , α3 ∈ [0, 1] such that 3i=1 αi = 1, and c a nonnegative constant then         12 0 1 3.6  0   14   1   1.4         S = α1   0  + α2  56  + α3  0  + c  9  . 0 144 14 0 See 32


Figure 5.2.2: The feasible region is shown. Note that the region may be made up of a convex combination of basic feasible solutions plus a direction of unboundedness. http://www.math.iup.edu/~jchrispe/MATH445_545/DirectionOfUnboundedness.html for an interactive version of the solution.

33


34

Chapter 6 Simplex Method Using Matrix-Vector Formulas “Learn to be positive, it’s your only chance.” −The Kinks

Lets consider looking at the simplex algorithm using Matrices and compare that with the tableau format we have been using.

6.1

Problem Using Tableaus

Consider the following linear program: maximize subject to

z = 3x1 + 5x2 x1 ≤ 4 2x2 ≤ 12 3x1 + 2x2 ≤ 18 x1 , x2 ≥ 0.

Solution full Tableau Form: Adding slack variables in the constraints the initial tableau for the given linear program is written as:

Initial Tableau:

Row z 0 1 1 0 2 0 3 0

x1 -3 1 0 3

x2 -5 0 2 2

s1 0 1 0 0

s2 0 0 1 0

s3 0 0 0 1

RHS 0 4 12 18

Basis z s1 s2 s3

In the first pivot x2 enters the basis replacing s2 . Here we have not explicitly shown the ratio test. Thus, 35


Pivot 1 Tableau:

Row z 0 1 1 0 2 0 3 0

x1 -3 1 0 3

x2 0 0 1 0

s1 0 1 0 0

s2 5 2

0 1 2

-1

s3 0 0 0 1

RHS 30 4 6 6

Basis z s1 x2 s3

The next iteration of the simplex method shows that it is desirable to have x1 enter the basis and replace s3 . (Note again we are not showing the ratio text.) Doing this pivot yields:

Pivot 2 Tableau:

Row z 0 1 1 0 2 0 3 0

x1 0 0 0 1

x2 0 0 1 0

s1 0 1 0 0

s2 3 2 1 3 1 2 - 13

s3 1 - 13 0 1 3

RHS 36 2 6 2

Basis z s1 x2 x1

Here we have obtained an optimal tableau. We can see here that z is maximized at $36,000 when x1 = 2 and x2 = 6.

6.2

Matrix Format of Linear Program

A standard form linear program in matrix form is written as: maximize subject to

z = cT x Ax = b x ≥ 0.

Thus, for the example problem we make the following definitions:     x1 3    x2   5  4         12  x=  s1  , c =  0  , b =  s2   0  18 s3 0 and



 1 0 1 0 0 A =  0 2 0 1 0 . 3 2 0 0 1

In order to implement the simplex algorithm in matrix format we break the given coefficient matrix A into to parts: a basic part and a non-basic part. Thus, A = [N |B] and similarly we separate the vectors x, and c into their basic and non-basic parts:   s1 xN x1 x= , where xN = and xB =  s2  xB x2 s3 36

OR Notes Draft: May 5, 2012 

 0 cN 3 c= , where cN = and cB =  0  . cB 5 0 We can now make note of the correspondence between the simplex method as we know it in tableau format, and the defined matrices. Any current basis may be written as:

−cN

6.2.1

T

xN T xB T RHS Basis T −1 + cB B N 0 cB T B −1 b z −1 −1 B N I B b xB

Initial Tableau (Matrix Form):

For the starting basis  1  0 N= 3

xB = (s1 s2 s3 )T ,   0 1   2 , B= 0 2 0

and xN = (x1 x2 )T we have:    0 0 1 0 0 1 0  =⇒ B −1 =  0 1 0  . 0 1 0 0 1

with cN T =

3 5

cB T =

0 0 0

Evaluating each of the expressions in the tableau we have: −cN T + cB T B −1 N =

−3 0 cB T B −1 b =  4 B −1 b =  12 18  1 −1 B N =  0 3

−5

   0 2  2

Ratio Test: Note that both non-basic variables are attractive, and we choose x2 to enter the basis. Doing the ratio test using the second column of B −1 N and the current right hand side vector B −1 b: 12 18 min = 6, =9 2 2 and we choose s2 to leave the basis. Note we have ignored the comparison of 4 over 0 in the ratio test. Compare the above values with the tableau: 37


Initial Tableau:

Row z 0 1 1 0 2 0 3 0

x1 -3 1 0 3

x2 -5 0 2 2

s1 0 1 0 0

s2 0 0 1 0

s3 0 0 0 1

RHS 0 4 12 18

Basis z s1 s2 s3

Pivot 1 Tableau (Matrix Form): Note we have now replaced s2 in the basis with x2 . Thus, xB = (s1 x2 s3 )T , and xN = (x1 s2 )T we have: 

   1 0 1 0 0 N =  0 1 , B =  0 2 0  3 0 0 2 1



B −1

=⇒

 1 0 0 1 0 . = 0 2 0 −1 1

with cN T =

3 0

cB T =

0 5 0


−3

5 2

cB T B −1 b =

30   4 −1  6  B b = 6   1 0 1  B −1 N =  0 2 3 −1

Ratio Test: Note that the non-basic variable x1 should be picked to enter the basis. Doing the ratio test using the first column of B −1 N and the current right hand side vector B −1 b: 4 6 min = 4, = 2 1 3 and we choose s3 to leave the basis. Note we have ignored the comparison of 6 over 0 where x2 would potentially leave the basis in the ratio test. Compare values with the Piviot 1 Tableau: 38


Pivot 1 Tableau:

Row z 0 1 1 0 2 0 3 0

x1 -3 1 0 3

x2 0 0 1 0

s1 0 1 0 0

s2 5 2

0 1 2

-1

s3 0 0 0 1

RHS 30 4 6 6

Basis z s1 x2 s3

Pivot 2 Tableau (Matrix Form): Note we have now (s3 s2 )T we have:  0  0 N= 1

replaced s3 in the basis with x1 . Thus, xB = (s1 x2 x1 )T , and xN =    0 1 0 1 1 , B =  0 2 0  0 0 2 3

 1 − 13 1 3 1 0 . = 0 2 1 1 0 −3 3 

B −1

=⇒

with cN T =

0 0

cB T =

0 5 3

Evaluating each of the expressions in the tableau we have: 1 32 −cN T + cB T B −1 N = 36 cB T B −1 b =   2 −1  6  B b = 2   1 1 −3 3 1  B −1 N =  0 2 1 − 31 3 Note that the non-basic variables are nolonger attractive and we have reached an optimal solution to the problem with z = 36 when, x1 = 2 and x2 = 6. Compare with the Pivot 2 Tableau:

Pivot 2 Tableau:

Row z 0 1 1 0 2 0 3 0

x1 0 0 0 1

39

x2 0 0 1 0

s1 0 1 0 0

s2 3 2 1 3 1 2 - 13

s3 1 - 13 0 1 3

RHS 36 2 6 2

Basis z s1 x2 x1


40

Chapter 7 Duality “I’d like to change your mind by hitting it with a rock.” −They Might Be Giants

As is often the case you may look at problems form many different perspectives. This is the case with linear programs. Lets consider a problem form our past in two different perspectives.

7.1

A Motivating Example: My Diet Problem:

Recall from earlier this semester that I have a really bad diet that requires that all of my food come from my favourite food groups: pez candy, pop tarts, Cap’N crunch, and cookies. I have the following foods available for my consumption: red pez, strawberry pop-tarts, peanut butter crunch, and chocolate chip cookies. The red pez costs $0.20 per (per package), strawberry pop-tarts cost $0.50 (per package), peanut butter crunch costs $0.70 per bowl, and chocolate chip cookies cost $0.65 cents each. I must ingest at least 2500 calories a day in order to maintain my sugary lifestyle. I must also meet the following requirements I need 8 oz. of sugar, 10 oz. of fat, and 50 mg of yellow-5 food coloring. Each of my chosen foods has these required nutrients in the following quantities: Food Calories pez candy (per package) 50 pop-tarts (per package) 250 Cap’N crunch (per bowl) 350 cookies (1 cookie) 150

Sugar (oz.) 0 .1 0.5 1 0.3

Fat (oz.) 0.01 0.3 1 0.7

yellow-5 (mg) 5 0 4 1

How can I achieve my dietary constraints at a minimum cost?

Primal Solution To Diet Problem: The first step will be do define some decision variables. Here we will define 41

Price ($) 0.20 0.50 0.70 0.65

OR Notes Draft: May 5, 2012 x1 x2 x3 x4

as as as as

the the the the

number number number number

of pez packages to include in my diet per day strawberry pop-tart packages to consume per day of bowls of peanut butter crunch to consume per day of chocolate chip cookies to consume each day

Our objective function can then be defined as: min z = 0.20x1 + 0.50x2 + 0.70x3 + 0.65x4 which will minimize the cost of our diet. The next aim is to meet the special dietary needs. Here a constraint is set up for each of the “nutrients” (calories, sugar, fat, and yellow 5). 50x1 + 250x2 + 350x3 + 150x4 0.1x1 + 0.5x2 + x3 + 0.3x4 0.01x1 + 0.3x2 + x3 + 0.7x4 5x1 + 0x2 + 4x3 + x4

≥ ≥ ≥ ≥

2500 (calories required) 8 (sugar required) 10 (fat required) 50 (yellow-5 required)

It can also be noted that the decision variables will be non-negative in sign giving the xi ≥ 0 for all i ∈ {1, 2, 3, 4} sign restriction. This problem can be set up in vector notation using the above variable definitions and defining       x1 0.20 2500  x2       , c =  0.50  , b =  8  , and x=  x3   0.70   10  x4 0.65 50   50 250 350 150  0.1 0.5 1 0.3   A=  0.01 0.3 1 0.7  . 5 0 4 1 The optimal diet can now be found by considering: min z = cT x s.t. Ax ≥ b with x ≥ 0.

Dual of Diet Problem A second way to consider the problem. Suppose that I know a nutrient salesman (Willy Loman) who will sell me supplements that taste just as good as the items in my diet. Specifically Mr. Loman sells calories, sugar, Fat, and yellow-5 and wants to sell me these items in order to meet my daily needs at the maximum price I’m willing to pay. Then defining the decision variables: 42

OR Notes Draft: May 5, 2012 y1 is the price to charge me per calorie. y2 is the price to charge me per ounce of sugar. y3 is the price to charge me per ounce of fat. y4 is the price to charge me per mg of yellow-5. Loman’s objective function for my diet would look like: max w = 2500y1 + 8y2 + 10y3 + 50y4 The constraints for the salesman are found using the available foods. Mr. Loman being a great salesman needs to set his prices low enough that I will purchase his nutrients rather than my regular diet of pez and cookies. Thus, he is subject to the constraints:

50y1 + 0.1y2 + 0.01y3 + 5y4 250y1 + 0.5y2 + 0.3y3 + 0y4 350y1 + y2 + y3 + 4y4 150y1 + 0.3y2 + 0.7yy + y4

≤ ≤ ≤ ≤

0.20 ( 0.50 ( 0.70 ( 0.65 (

the the the the

price price price price

of of of of

pez) pop-tarts) Cap’N crunch) cookies)

where the sign restriction on the decision variables is: yi ≥ 0 for all i ∈ {1, 2, 3, 4} Writing this in matrix form in terms of the values defined for the primal problem we have: max w = bT y s.t. AT y ≤ c with y ≥ 0 where y = (y1 y2 y3 y4 )T . These two linear programs have the same optimal solution values.

7.1.1

Cononical Form

The cononical form of a primal and its corresponding dual linear program are given as follows: max w = bT y s.t. AT y ≤ c with y ≥ 0

min z = cT x s.t. Ax ≥ b with x ≥ 0. Primal

Dual 43


Taking Dual of General Linear Programs What if we don’t start in canonical form? Lets consider the constraints first: min z = cT x s.t. A1 x A2 x A3 x with x

≥ ≤ = ≥

b1 b2 b3 0.

Transform the above problem into standard canonical form (ignoring that the RHS may be negative for a moment). min z = cT x s.t. A1 x −A2 x A3 x −A3 x with x

≥ ≥ ≥ ≥ ≥

b1 −b2 b3 −b3 0.

Taking the dual we have: max w = bT1 y1 − bT2 y20 + bT3 y30 − bT3 y300 s.t. AT1 y1 − AT2 y20 + AT3 y30 − AT3 y300 ≤ c with y1 , y20 , y30 , y300 ≥ 0. Making the change of variables: y2 = −y20

and

y3 = y30 − y300

and we have: max w = bT1 y1 + bT2 y2 + bT3 y3 s.t. AT1 y1 + AT2 y2 + AT3 y3 ≤ c with y1 ≥ 0, y2 ≤ 0, y3 is free

What if the variables are not in standard form? min z = cT1 x1 + cT2 x2 + cT3 x3 s.t. A1 x1 + A2 x2 + A3 x3 ≥ b with x1 ≥ 0, x2 ≤ 0, x3 is free Here we make a change of variables and place the problem back into a canonical form. Let x2 = −x02

and 44

x3 = x3 0 − x003

OR Notes Draft: May 5, 2012 with x02 , x03 , and x003 ≥ 0. Then min z = cT1 x1 − cT2 x02 + cT3 x03 − cT3 x003 s.t. A1 x1 − A2 x02 + A3 x03 − A3 x003 ≥ b with x1 , x02 , x03 , x003 ≥ 0

We can now take the Dual: max z = bT y s.t. AT1 y −AT2 y AT3 y −AT3 y with y ≥ 0

≤ ≤ ≤ ≤

c1 −c2 c3 −c3

Adjusting the RHS to be non-negative: max z = bT y s.t. AT1 y ≤ c1 AT2 y ≥ c2 AT3 y = c3 with y ≥ 0

Relationship Between Primal and Dual The following table summarizes the relationship between the primal and the dual problem. Primal / Dual Constraint consistent with canonical form reversed from canonical form equality constraint

⇐⇒ ⇐⇒ ⇐⇒

Dual/Primal Variable variable ≥ 0 variable ≤ 0 variable is free

Examples: As an exercise find the dual of the following linear programs: 45

OR Notes Draft: May 5, 2012 • Problem 1: max z = x1 + x2 s.t. x1 − x2 ≤ 1 with x1 , x2 ≥ 0 We need a dual variable y for each constraint. Thus, min w = y1 s.t. y1 ≥ 1 −y1 ≥ 1 with y2 ≥ 0.

• Problem 2: min z = 4x1 − 9x2 + 10x3 s.t. 5x1 + 6x2 + 7x3 ≥ 3 3x1 + 2x2 + 1x3 ≤ 4 −1x1 + 8x2 + 2x3 ≤ 5 with x1 ≥ 0, x2 ≤ 0, x3 is free And the dual of the linear program is: max z = 3y1 + 4y2 + 5y3 s.t. 5y1 + 3y2 − y3 ≤ 4 6y1 + 2y2 + 8y3 ≥ −9 7y1 + y2 + 2y3 = 10 with y1 ≥ 0, y2 ≤ 0, y3 ≤ 0

46

Chapter 8 Basic Duality Theory “Read dozens of books about heroes and crooks, and I learned much from both of their styles.” − Jimmy Buffet

Here we will consider the major results that relate the primal and the dual linear programming problems.

8.1

Relationship Between Primal and Dual

As a warm up lets consider finding the dual of the linear program: max z = 2x1 + 3x2 + 4x3 s.t. 5x1 + 6x2 + 7x3 = 8 9x1 + 10x2 + 11x3 ≥ 12 13x1 + 14x2 + 15x3 ≤ 16 with x1 ≥ 0, x2 ≤ 0, x3 is free

Primal / Dual Constraint consistent with canonical form ⇐⇒ reversed from canonical form ⇐⇒ equality constraint ⇐⇒ min z = 8y1 + 12y2 + 16y3 s.t. 5y1 + 9y2 + 13y3 ≥ 2 6y1 + 10y2 + 14y3 ≤ 3 7y1 + 11y2 + 15y3 = 4 with y1 is free, y2 ≤ 0, y3 ≥ 0 47

Dual / Primal Variable variable ≥ 0 variable ≤ 0 variable is free

( as x1 ≥ 0 in primal problem) ( as x2 ≤ 0 in primal problem) ( as x3 is free in primal problem)

OR Notes Draft: May 5, 2012 What is the dual of a problem that is written in our typical standard form? min z = cT x s.t. Ax = b with x ≥ 0 Here we write the dual as: max w = bT y s.t. AT y ≤ c ( as x ≥ 0 in primal problem). with y is free

Graph the following Example: max z = 2x1 + x2 s.t. x1 ≤ 1 x2 ≤ 1 with x1 and x2 ≥ 0 Taking the dual: min w = y1 + y2 s.t. y1 ≥ 2 y2 ≥ 1 with y1 and y2 ≥ 0 Graphing the feasible region for each illustrates a nice relationship between the primal and the dual problem. The following plot illustrates the feasible region for each:

48

OR Notes Draft: May 5, 2012 Both linear programs are optimal when an objective function value of 3 is obtained. (z = 3, when x1 = 1, and x2 = 1) (w = 3, when y1 = 2, and y2 = 1)

8.2

Weak Duality

One of the major results of relating the two linear programs is “Weak Duality”. Here the primal objective values provide bounds for the dual objective values, and vice versa. Consider the primal problem to be the minimization problem. Then, Theorem 8.2.1 (Weak Duality) Let x be a feasible point for the primal problem in standard form, and let y be a feasible point for the dual problem. Then, z = cT x ≥ bT y = w. Proof: From the dual problem’s constraints and the primal problems sign restrictions we have cT ≥ yT A and x ≥ 0. Thus, z = cT x ≥ (yT A)x = yT b = bT y = w =⇒ z ≥ w Note this means that for a general primal dual min max pair that a feasible solution for the minimization problem will always have an objective function value that is greater than or equal to the objective function value for a feasible point in the dual maximization problem. Thus, • If the primal is unbounded then the dual is infeasible. If the dual is unbounded then the primal is infeasible. – In general if the primal is infeasible the dual may be infeasible or unbounded. • If x is a feasible solution to the primal problem, y is a feasible solution to the dual, and cT x = bT y then x and y are optimal for their respective problems. Example: Form last class we took the dual of the linear program: max z = x1 + x2 s.t. x1 − x2 ≤ 1 with x1 , x2 ≥ 0 49

OR Notes Draft: May 5, 2012 It can be seen that this linear program is unbounded. Taking the dual gave us: min w = y1 s.t. y1 ≥ 1 −y1 ≥ 1 with y1 ≥ 0. Note that this LP is infeasible.

Primal feasible region is unbounded, so the dual will be infeasible.

8.3

Strong Duality

Consider a pair of primal and dual linear programming problems. Theorem 8.3.1 (Strong Duality) If one of the problems has an optimal solution then so does the other, and the optimal objective values are equal. Proof: With out loss of generality we can make the following assumptions: • The primal problem has an optimal solution. • The primal problem is in standard form (min problem). • x∗ the solution to the primal problem is an optimal basic feasible solution. Here we write x∗ in terms of basic and non-basic variables: xB x∗ = xN 50

OR Notes Draft: May 5, 2012 and correspondingly we can write A = [B N ], and c =

cB cN

.

Recall any current basis may be written as: −cN T

xN T xB T RHS Basis T −1 T −1 + cB B N 0 cB B b z −1 −1 B N I B b xB

Then we can note that xB = B −1 b., and x∗ is optimal if −cN T + cB T B −1 N ≤ 0 =⇒ cB T B −1 N ≤ cN T the reduced costs for the non-basic variables are negative. For the dual problem we let y∗ = (B −1 )T cB . y∗ = (B −1 )T cB =⇒ y∗ T = cTB B −1 The goal now is to show that • y∗ is feasible for the dual • bT y∗ = cT x∗ (the objective functions have the same value). For feasibility we need to show AT y∗ ≤ c. Lets start with y∗ T A = cTB B −1 (B N ) = ≤

cTB cTB B −1 N cTB cTN

= cT Thus, AT y∗ ≤ c and we have a feasible solution for the dual. We now need to compare the value of the dual’s objective function with the value of the optimal primal objective function: z = cT x∗ = cTB xB = cTB B −1 b w = bT y∗ = y∗ T b = cTB B −1 b So y∗ is feasible in the dual and has the same objective value as the optimal primal problem objective value so y∗ is optimal for the dual. 51


More on Duality Theory Primal Problem Before starting into additional theory lets consider the following LP and for practice take its dual. min z = 20x1 + 15x2 + 54x3 s.t. x1 − 2x2 + 6x3 x2 + 2x3 2x1 − 3x3 x1 − x2 with x1 , x2 , x3 ≥ 0

≥ ≥ ≥ ≥

30 6 −5 18

The associated Dual problem is: max w = 30y1 + 6y2 − 5y3 + 18y4 s.t. y1 + 2y3 + y4 ≤ 20 −2y1 + y2 − y4 ≤ 15 6y1 + 2y2 − 3y3 ≤ 54 with y1 , y2 , y3 , y4 ≥ 0

Insight By Example Lets consider working through an example start to finish in both the primal and dual formulation and see how the two are related. Primal Problem: min z = 2x1 + 9x2 + 3x3 s.t. − 2x1 + 2x2 + x3 ≥ 1 x1 + 4x2 − x3 ≥ 6 with x1 , x2 , x3 ≥ 0 Solving the Primal Problem Adding excess variables and artificial variables to each of the primal constraints we may obtain the following initial simplex tableau. Note we will use the two-phase simplex algorithm 52

OR Notes Draft: May 5, 2012 as we are using artificial variables. The objective function for the first phase of the algorithm is: min zp1 = a1 + a2 where the index on each artificial variable denotes the constraint to which it corresponds. Row z 0 1 1 0 2 0

x1 0 -2 1

x2 0 2 4

x3 0 1 -1

e1 0 -1 0

e2 0 0 -1

a1 -1 1 0

a2 -1 0 1

RHS 0 1 6

Basis z a1 a2

Adjust the initial Phase one tableau so that a1 and a2 are truly basic. Row z 0 1 1 0 2 0

x1 -1 -2 1

x2 6 2 4

x3 0 1 -1

e1 -1 -1 0

e2 -1 0 -1

a1 0 1 0

a2 0 0 1

RHS 7 1 6

Basis z a1 a2

Here we see that x2 is attractive and should enter the basis by replacing a1 . Row z 0 1 1 0 2 0

x1 5 -1 5

x2 0 1 0

x3 -3 0.5 -3

e1 e2 2 -1 -0.5 0 2 -1

a1 a2 -3 0 0.5 0 -2 1

RHS 4 0.5 4

Basis z x2 a2

Next we see that x1 is attractive and should enter the basis in the place of a2 . Row z 0 1 1 0 2 0

x1 0 0 1

x2 0 1 0

x3 0 -0.1 -0.6

e1 0 -0.1 0.4

e2 0 -0.2 -0.2

a1 -1 0.1 -0.4

a2 -1 0.2 0.2

RHS 0 1.3 0.8

Basis z x2 x1

This completes the first phase of the two-phase simplex method. We can place back in the original objective function, and drop the columns that correspond to the artificial variables. min zp2 = 2x1 + 9x2 + 3x3 This gives Row z 0 1 1 0 2 0

x1 -2 0 1

x2 -9 1 0

x3 -3 -0.1 -0.6

e1 0 -0.1 0.4

e2 RHS 0 0 -0.2 1.3 -0.2 0.8

Basis z x2 x1

Update the tableau so that the basic variables are represented by identity columns. Gives: 53

OR Notes Draft: May 5, 2012 Row z 0 1 1 0 2 0

x1 0 0 1

x2 0 1 0

x3 -5.1 -0.1 -0.6

e1 -0.1 -0.1 0.4

e2 RHS -2.2 13.3 -0.2 1.3 -0.2 0.8

Basis z x2 x1

We have achieved the optimal solution to the primal linear programming problem. The minimum is achieved when z = 13.3 with x1 = 0.8 and x2 = 1.3 and x3 = 0. Solving the Dual Problem The dual for our example problem is given by max w = y1 + 6y2 s.t. − 2y1 + y2 ≤ 2 2y1 + 4y2 ≤ 9 y1 − y2 ≤ 3 with y1 , y2 ≥ 0 Here we need to add a slack variable to each of the constraints. This leads to the initial tableau: Row w 0 1 1 0 2 0 3 0

y1 -1 -2 2 1

y2 -6 1 4 -1

s1 0 1 0 0

s2 0 0 1 0

s3 0 0 0 1

RHS 0 2 9 3

Basis w s1 s2 s3

Note here we pick y2 to enter into the basis. It will replace s1 . Row w 0 1 1 0 2 0 3 0

y1 -13 -2 10 -1

y2 0 1 0 0

s1 6 1 -4 1

s2 0 0 1 0

s3 0 0 0 1

RHS 12 2 1 5

Basis w y2 s2 s3

Here we pick y1 to enter the basis in place of s2 . This gives, Row w 0 1 1 0 2 0 3 0

y1 0 0 1 0

y2 0 1 0 0

s1 s2 s3 0.8 1.3 0 0.2 0.2 0 -0.4 0.1 0 0.6 0.1 1

RHS 13.3 2.2 0.1 5.1

Basis w y2 y1 s3

The optimal solution has been obtained with w = 13.3 where y1 = 0.1 and y2 = 2.2. 54

OR Notes Draft: May 5, 2012 Observations Note we could have solved the Dual problem graphically:

Lets compare the optimal Tableau for each of the two problems: Row z 0 1 1 0 2 0

x1 0 0 1

Row w 0 1 1 0 2 0 3 0

x2 0 1 0 y1 0 0 1 0

x3 -5.1 -0.1 -0.6 y2 0 1 0 0

e1 -0.1 -0.1 0.4

e2 RHS -2.2 13.3 -0.2 1.3 -0.2 0.8

s1 s2 s3 0.8 1.3 0 0.2 0.2 0 -0.4 0.1 0 0.6 0.1 1

RHS 13.3 2.2 0.1 5.1

Basis z PRIMAL x2 x1 Basis w y2 DUAL y1 s3

• Note we can read the primal and dual solutions for the other problem using the reduced costs in row zero of the optimal tableau. – The optimal dual variable values are the same as the reduced costs of the slack (and excess variables with with signs reversed).

55


Complementary Slackness The optimal solutions are given by:    x1 0.8  x2   1.3    = 0 x PRIMAL =  3     e1   0 e2 0





    

  DUAL =   

and

y1 y2 s1 s2 s3





    =    

0.1 2.2 0 0 5.1

     

We can line up the constraints and sign restrictions for each problem and note which one is binding in each of the problems: PRIMAL PROBLEM −2x1 + 2x2 + x3 ≤ 1 x1 + 4x2 − x3 ≤ 6 x1 ≥ 0 x2 ≥ 0 x3 ≥ 0

BINDING DUAL PROBLEM PRIMAL y1 ≥ 0 PRIMAL y2 ≥ 0 DUAL −2y1 + y2 ≤ 2 DUAL 2y1 + 4y2 ≤ 9 PRIMAL y1 − y2 ≤ 3

This illustrates complementary slackness. Consider the primal-dual pair in standard form: PRIMAL: min z = cT x s.t. Ax = b with x ≥ 0 DUAL: max w = bT y s.t. AT y ≤ c with y is free There is an interdependence between the non-negativity constraints in the primal (x ≥ 0) and the constraints in the dual (AT y ≤ c). At the optimal solution it is not possible to have both: xj > 0

(AT y)j < cj .

and

At least one of the constraints must be binding giving that either • xj = 0 or • the j-th dual slack variable is zero (the corresponding dual constraint is binding). Note this gives that X

xj (c − AT y)j

j

56

OR Notes Draft: May 5, 2012 Complementary slackness is often written as: xT (c − AT y) = 0. as the primal and dual constraints ensure that the terms in the summation must be nonnegative. Thus, if the sum is zero then each term must be zero. Theorem 8.3.2 (Complementary Slackness) Consider a pair of primal and dual linear programs in standard form. If x is optimal for the primal and y is optimal for the dual then xT (c − AT y) = 0 Proof: If x and y are feasible for there respective problems then: z = cT x ≥ yT Ax = yT b = w. As x and y are also optimal we know that z = w so cT x = yT Ax =⇒ xT c = xT AT y =⇒ xT c − xT AT y = 0 =⇒ xT c − AT y = 0

8.4

The Dual Simplex Method

Consider the following linear program: min z = 5x1 + 4x2 s.t. 4x1 + 3x2 ≥ 10 3x1 − 5x2 ≥ 12 with x1 , x2 ≥ 0. If we add excess variables to the constraints and place the problem into a tableau we have: Row z 0 1 1 0 2 0

x1 -5 4 3

x2 -4 3 -5

e1 0 -1 0

e2 0 0 -1

RHS 0 10 12

BASIS z ? ?

Note that in the above Tableau we do not have any basic variables yet. However, the given tableau does show that the reduced costs for the problem are such that the optimality conditions are satisfied. Lets consider making e1 and e2 basic (multiply row 1 and row 2 by -1). This gives the following tableau. 57


x1 -5 -4 -3

x2 -4 -3 5

e1 0 1 0

e2 0 0 1

RHS 0 -10 -12

BASIS z e1 e2

Note that: • The optimality conditions are still satisfied. (Nothing is attractive to enter into the basis in the simplex method as we know it.) • The right hand side values are negative. This means that he current basis is infeasible in the primal problem. To move toward a feasible solution we could scan the RHS and see who is the most negative, and work to get them out of the basis. Here we need a candidate to come in in there place. Scan down the row corresponding to the mose negative RHS value and find negative coefficients. If there is more than one we would do a ‘Ratio’ test (reduced cost divided by the exiting row coefficient in the potential entering column) taking the minimum value to be our entering variable. Row z 0 1 1 0 2 0

x1 0 0 1

x2 - 37 3 - 29 3 - 53

e1 0 1 0

e2 - 53 - 43 - 13

RHS 20 6 4

BASIS z e1 x1

After doing the update we can now see that the optimality conditions and the primal feasibility conditions have now both been satisfied. Yielding the optimal solution of z = 20 when x1 = 4 and x2 = 0. This technique is called the dual simplex method . We can verify that this is indeed the solution to the problem by considering it graphically.

58


A Second Example Use the dual simplex method to solve the following linear program: min z = 5x1 + 2x2 + 8x3 s.t. 2x1 − 3x2 + 2x3 ≥ 3 −x1 + x2 + x3 ≥ 5 with x1 , x2 , x3 ≥ 0. Setting up the problem in tableau form using two excess variables: Row z 0 1 1 0 2 0

x1 -5 2 -1

x2 -2 -3 1

x3 -8 2 1

e1 0 -1 0

e2 0 0 -1

RHS 0 3 5

BASIS z ? ?

We negate the two constraints in order to make e1 and e2 basic. Row z 0 1 1 0 2 0

x1 -5 -2 1

x2 -2 3 -1

x3 -8 -2 -1

e1 0 1 0

e2 0 0 1

RHS 0 -3 -5

BASIS z e1 e2

Here we look to get e2 out of the basis. We see that x2 is the winner of the Ratio Test. Row z 0 1 1 0 2 0

x1 -7 1 -1

x2 0 0 1

x3 -6 -5 1

e1 0 1 0

e2 -2 3 -1

RHS 10 -18 5

BASIS z e1 x2

Note that after the pivot of the Dual Simplex algorithm there is no attractive candidate to enter into the basis using our normal “Primal Simplex” algorithm (the optimality conditions are still met). We now use Dual simplex to remove e1 from the basis. Note that x3 is the only choice to replace e1 in the basis. Row z 0 1 1 0 2 0

x 1 x2 -8.2 0 -0.2 0 -0.8 1

x3 0 1 0

e1 -1.2 -0.2 0.2

e2 RHS -5.6 31.6 -0.6 3.6 -0.4 1.4

BASIS z x3 x2

Note that the optimal solution of z = 31.6 is achieved when x1 = 0, x2 = 1.4 and x3 = 3.6. We make note here that the Dual Simplex method is especially useful when doing sensitivity analysis. If a change in the right hand side value of a constraint is made this method can be used to update the current simplex solution to one that is feasible if the change causes the primal problem to have negative right hand side values. 59


60

Chapter 9 Sensitivity Analysis “ I don’t know where the sun beams end and the star light begins it’s all a mystery...”

−The Flaming Lips

9.1

Sensitivity Analysis

Before looking at sensitivity analysis it is good to have a base problem to consider. We will base our initial pilgrimage into sensitivity analysis of linear programs on the following 2D linear program.

Base Problem Consider maximizing the profit from manufacturing two items on three different assembly lines. The profit from manufacturing item 1 is three thousand dollars, and from item 2 is five thousand dollars. Letting x1 and x2 be the number of each of these items to manufacture, and letting the following constraints denote the number of hours that it takes to manufacture each of the items at three different assembly lines with the given restrictions on the maximum number of hours available on each line. This gives the following linear program: maximize subject to

z = 3x1 + 5x2 x1 ≤ 4 2x2 ≤ 12 3x1 + 2x2 ≤ 18 x1 , x2 ≥ 0.

Assuming you are renting time on each assembly line what is the most you should pay for an additional hour of time on each of the assembly lines? 61

OR Notes Draft: May 5, 2012 Solution: Using the simplex algorithm to solve the given linear program we see that after adding two slack variables in the constraints the initial tableau for the given linear program is written as: Row z 0 1 1 0 2 0 3 0

x1 -3 1 0 3

x2 -5 0 2 2

s1 0 1 0 0

s2 0 0 1 0

s3 0 0 0 1

RHS 0 4 12 18

Basis z s1 s2 s3

Here it becomes desirable to pivot x2 into the basis replacing s2 . Thus, Row z 0 1 1 0 2 0 3 0

x1 -3 1 0 3

x2 0 0 1 0

s1 0 1 0 0

s2 5 2

0 1 2

-1

s3 0 0 0 1

RHS 30 4 6 6

Basis z s1 x2 s3

The next iteration of the simplex method shows that it is desirable to have x1 inter the basis and it will replace s3 . Doing this pivot yields: Row z 0 1 1 0 2 0 3 0

x1 0 0 0 1

x2 0 0 1 0

s1 0 1 0 0

s2 3 2 1 3 1 2 - 31

s3 1 - 31 0 1 3

RHS 36 2 6 2

Basis z s1 x2 x1

Here we have obtained an optimal tableau. We can see here that profit is maximized at $36,000 when x1 = 2 and x2 = 6 are manufactured between the different assembly lines.

Shadow prices: We are often interested in the marginal value of each of the given resources and their impact on the objective function. That is: Shadow Price : measures the marginal value of a given resource or the amount by which z will be increased by slightly increasing the amount of a given resource being made available. The shadow price for resource i is denoted by yi∗ and is the amount the objective function will be increased if the amount of resource bi is slightly increased. The shadow price for resource i is the zero row coefficient for the ith constraints slack variable in the optimal tableau. 62

OR Notes Draft: May 5, 2012 From our given example we can see that the shadow price for hours available for each assembly line (marginal increase in the objective function for slight increases in the amount the resource available) is: 3 (9.1.1) y1∗ = 0, y2∗ = y3∗ = 1. 2

9.1.1

Verify this graphically:

An interactive version of the solution is found at: http://www.math.iup.edu/~jchrispe/MATH445_545/ShadowPrices.html What happens if the constraint b2 goes from 12 to 13? y2∗ = ∆z = 37.5 − 36 = 1.5

(9.1.2)

How do we interpret this with regard to the linear program? An additional unit of labor on production line two will increase the profits by $1500. How far could we increase the value of b2 the amount of resource 2 and still stay optimal with the current basis? 63

OR Notes Draft: May 5, 2012 We could increase the units of resource b2 up to 18 and keep the same basic feasible solution. After which the constraint is no longer binding. Here the optimal objective function would be fixed at z =45. At this point a new basic feasible solution can be obtained, and a new set of shadow prices will come into play. Note that constraints that have zero shadow prices are not binding the objective function. In the given example we note that the first constraint is not binding and increasing the value of b1 will have no effect on the objective function’s optimal value. Resources for binding constraints are sometimes called scarce goods, and resources for non-binding constraints are called free goods. • Shadow prices can be also thought of as the maximum price you would want to pay for an additional unit of a given resource. • Shadow prices help identify the parameters in the model that need to be estimated carefully. If a shadow price is zero the model is not sensitive to (at least small) changes in the given resource. • You would want to monitor resources more closely if they have large shadow prices. How far can b2 be changed and keep the same optimal basic feasible solution? 6 ≤ b2 ≤ 18 keeps the same basic feasible solution, and an increase of 6 or a decrease of 6 is allowed. What information can we get out of LINDO/LINGO?

9.2

Sensitivity Analysis Using Matrices

We looked briefly as sensitivity of linear programs graphically. This is nice if the problem is small enough to handle with a software like Geogebra. However; it becomes more difficult to determine how the solution to a linear program is effected by perturbations of the objective function coefficients and constraints if the problem of interest is more than two dimensional. Lets recall that for a linear program in the form: (min or max) cT x s.t. Ax = b with x ≥ 0 then any current basis may be written using matrix form (using our previously defined notation) as: xN T xB T RHS Basis T T −1 T −1 −cN + cB B N 0 cB B b z −1 −1 B N I B b xB 64

OR Notes Draft: May 5, 2012 With this in mind any change in the initial linear program can be thought of in terms of these formulations. Specifically checking the feasibility, and optimality of any current basis will come down to looking at: B −1 b ≥ 0 for feasibility and −cN T + cB T B −1 N ≥ 0

for optimality in the maximization problem

−cN T + cB T B −1 N ≤ 0

for optimality in the minimization problem.

From these conditions we can now answer the following questions: • How much can the initial problem data be changed before the optimality conditions no longer hold? • If the current basis is no longer optimal we would apply primal simplex to restore it to optimality. • Changes in the data some times may affect the feasibility of the problem. How much can they change before the constraints are violated? We may use these formulations to do sensitivity analysis for a linear program allowing for the appropriate changes in the optimal solution to be made based on changes in the original problem parameters.

9.2.1

Illustrating Example

Consider the following linear program (A problem form the Nash and Sofer text): maximize subject to

z = 3x1 + 13x2 + 13x3 x1 + x2 ≤ 7 x1 + 3x2 + 2x3 ≤ 15 2x2 + 3x3 ≤ 9 x1 , x2 , x3 ≥ 0.

Thus, here the matrix form values are     1 1 0 1 0 0  A =  1 3 2 0 1 0 , c =    0 2 3 0 0 1  

65

3 13 13 0 0 0

     7   , and b =  15  .   9 

OR Notes Draft: May 5, 2012 If the optimal solution to the problem is found the optimal basis xB = (x1 x2 x3 )T , and xN = (s1 s2 s3 )T we have:       5 3 1 0 0 − 1 1 1 0 2 2 3 −1  . N =  0 1 0 , B =  1 3 2  =⇒ B −1 =  − 32 2 1 −1 1 0 0 1 0 2 3 with cN T =

0 0 0

cB T =

3 13 13

Lets answer the following questions: 1. What is the solution to the problem? Where we find the values of the decision variables as:      5 4 7 − 23 1 2 −1 3 3      3 . −1 15 −2 = Current RHS B b= 2 1 1 −1 1 9 

 4 3 13 13  3  = 1

cB T B −1 b =

Objective Value

64

Thus, x1 = 4, x2 = 3, and x3 = 1. 2. What are the optimal dual variables? Dual Variables

− cN T + cB T B −1 N 

= =

0 0 0 1 2 3

+

3 13 13



5 2 − 32

− 32

  1 0 0 1 3 −1   0 1 0  2 1 −1 1 0 0 1

Note these are the shadow prices, and if the problem had more context they would tell us the most we would wish to pay for an additional unit on the right hand side of each of the constraints. 3. What is the solution obtained by decreasing the right-hand side of the second constraint by 5? Now we add a vector of the form: ∆b = (0, −5, 0)T to the initial b vector:       7 0 7 bnew =  15  +  −5  =  10  . 9 0 9 66

OR Notes Draft: May 5, 2012 Where we find the values of the decision variables as:  5   − 32 1 7 2 3 −1   10  Current RHS B −1 b =  − 32 2 1 −1 1 9  23  2

=  − 92 . 6 Thus, x1 = 23 , x2 = − 92 , and x3 = 6. Here the RHS is no longer feasible for the 2 problem as x2 < 0. So we can implement an iteration of the dual simplex method to find a new basic feasible solution. We need to have a handle on the values of the non-basic columsn which are given by  5  − 23 1 2 3 −1  Non-Basic Columns B −1 N =  − 32 2 1 −1 1 as well as the new values for the dual variables: Dual Variables

− cN T + cB T B −1 N =

1 2 3

We do the Ratio Test for an iteration of the dual simplex method giving: 1 2 3 min col1 = −3 = , col3 = = 3 3 −1 2 So we pick x2 to leave the basis and we choose s1 to enter. This gives the new optimal solution to be: xB = (x1 s1 x3 )T , and xN = (x2 s2 s3 )T Where we now have: 

   1 1 0 1 0 0 B =  1 0 2  and N =  3 1 0  . 0 0 3 2 0 1

Dual Variables

− cN T + cB T B −1 N =

2 3

3

7 3



 4 B −1 b =  3 . 3

Current RHS

Objective Value

cB T B −1 bnew =

51

4. By how much can the right-hand side of the first constraint increase and decrease without changing the optimal basis? 67

OR Notes Draft: May 5, 2012 Here we are looking at the first component of B −1 (b + b∆ ) and want to see what range of values will keep the first component non-negative where:   ∆b b∆ =  0  0 Thus,   B

−1

(b + b∆ ) = 

5 2 − 32

− 32

   1 7 + ∆b  3   15  =  −1 2   1 −1 1 9

5 2

(7 + ∆b) − 23 (15) + (9) ≥ 0

  (7 + ∆b) + − (9) ≥ 0    (7 + ∆b) − (15) + (9) ≥ 0

−3 2

3 (15) 2

These constraints lead to the following conditions: ∆b ≥

−8 , ∆b ≤ 2 , and ∆b ≥ −1. 5

so the right hand side of the first constraint may increase by at most 2, and decrease by at most 1 with out changing the current basis. 5. What is the optimal solution to the linear program obtained by increasing the coefficient of x2 in the objective function by 15? Here the value of   3 cB =  28  13 For the different components of the problem we see that: cB T B −1 b =

Objective Value

109 ,

The values of the decision variables are: 

Current RHS

 4 B −1 b =  3 . 1

Thus, x1 = 4, x2 = 3, and x3 = 1, and the current dual variables Dual Variables

− cN T + cB T B −1 N =

− 43 2

49 2

−12

As we now have an attractive entering variable we can do a Primal simplex pivot with s1 entering the basis. The ratio test using:  5  − 23 1 2 3 −1  Current Non-Basic Columns B −1 N =  − 32 2 1 −1 1 68



OR Notes Draft: May 5, 2012 and the Current RHS yields that x3 should leave the basis. This gives xB = (x1 x2 s1 )T , and xN = (x3 s2 s3 )T where     0 0 0 1 1 1 N =  2 1 0  and B =  1 3 0  3 0 1 0 2 0 We can look at the updated dual variables and see that: cB T B −1 b =

Objective Value

261 2

,

The values of the decision variables are:  Current RHS

B −1 b = 

3 2 9 2

 .

1 Thus, x1 = 32 , x2 = 92 , and x3 = 0, and the current dual variables are Dual Variables

− cN T + cB T B −1 N =

43 2

3

19 2

Note that as the dual variable corresponding to s3 is zero it is also a potential candidate to enter the basis giving us multiple optimal solutions. 6. How much can the objective coefficient of x1 increase and decrease without changing the optimal basis? Here we consider the value of the dual variables that is: 1 2 3 Dual Variables − cN T + cB T B −1 N = Note that cN = 0 0 0 so we need only consider the following:  5   3 − 1 1 0 0 2 2 3 −1   0 1 0  ≥ 0 ((3 + ∆c) 13 13)  − 23 2 1 −1 1 0 0 1   5 − 32 1 2 3 −1  ≥ 0 =⇒ ((3 + ∆c) 13 13)  − 23 2 1 −1 1 =⇒

−3 5 (3 + ∆c) + (13)( ) + (13) ≥ 0, 2 2 −3 3 (3 + ∆c) + (13)( ) − (13) ≥ 0, 2 2 (3 + ∆c) − 13 + (13) ≥ 0

This gives that 2 4 x ≥ − , x ≤ , and x ≥ −3. 5 3 so the coefficient on x1 in the objective function can be increased by at most decreased by at most 25 if the same optimal basis is to be maintained. 69

4 3

and


70

Chapter 10 Non Linear Programming “ “Our opponent is an alien starship packed with atomic bombs,” I said. “We have a protractor.” ”

−Neal Stephenson, Anathem

There are many instances where the problem considered involves more that a linear relationship between decision variables and our objective. Here we will examine some of the preliminaries of non-linear programming starting with fitting functional models to sets of data.

10.1

Data Fitting

There are many instances where it would be advantageous to find a best fit model for a set of data. Here we have a set of points and we want to find in some sense the model or function that “best fits” the set of points 1 . For us best fit will be in the least squares sense. Thus given a set of n points (ti , yi ), we want to find a vector of m parameters x that will minimize the least square residual of: n minimize X (yi − f (ti , x))2 , x i=1

where f : Rm+1 → R. This may be a linear or non-linear problem depending on whether the function f is linear in its components of the parameter vector x. 1

Here we will follow the presentation in Heath’s Text.

71


10.1.1

Linear Problems

Consider for example where we are interested in finding a function: f (t, x) = x1 + x2 t + x3 t2 + . . . + xm tm−1 that best fits our set of n points we have to find the m coefficients for vector x that would minimize our residual in the least squares sense described above. This problem and others where the parameters x in f are linear combinations will result in a matrix system of the form Ax ∼ = b. Lets work a couple of examples. Example: Consider finding the best fitting linear and quadratic function for the following set of points: (1, 0), (2, 5), (3, 8), (4, 17), (5, 24) For the case of a linear function we want to find values of x = (x1 , x2 )T in f (t, x) = x1 + x2 t such that we minimize the sum of min

5 X

yi − f (ti , x)

i=1

Writing this out as a matrix system:  1  1  Ax =   1  1 1

t1 t2 t3 t4 t5



1 2 3 4 5





   x1  ∼   =  x2   

y1 y2 y3 y4 y5



0 5 8 17 24



   = b.  

Or with our given set of points:    Ax =   

1 1 1 1 1



   x1  ∼   =  x2   

So a residual vector could be found by looking at r = b − Ax 72

   = b.  

OR Notes Draft: May 5, 2012 if we are going look at the minimum of squared Euclidean norm of this residual can consider finding the minimum of φ(x) = krk22 = rT r = (b − Ax)T (b − Ax) = bT b − bT Ax − (Ax)T b + (Ax)T (Ax) = bT b − 2bT Ax + xT AT Ax If we use methods from multi variable calculus we know that the residual function will have a potential minimum provided (necessary condition): ∇φ(x) = 0 Thus, we will consider: 0 = ∇φ(x) = 0 − 2bT A + AT Ax + AT Ax = −2bT A + 2AT Ax

=⇒ AT Ax = AT b For this to indeed be a minimum note that we need the Hessian Matrix (matrix of second partial derivatives) given by AT A to be positive definite. (This can be proved by showing that AT A has rank n or that the columns are linearly independent). Note this system of equations is known as the Normal Equations. Looking back at our example we need to consider: 5 15 x1 54 = 15 55 x2 222 And solving the equation gives: x=

x1 x2

=

− 36 5 6

So the minimizing linear function is given by: f (t) = 6t −

36 5

and we make note that if we evaluate φ(x) gives: φ(x) = bT b − 2bT Ax + xT AT Ax = 10.8 73

OR Notes Draft: May 5, 2012 Now lets consider fitting a quadratic function to the data. Thus we want to find the best fit function of the form: f (t, x) = x1 + x2 t + x3 t2 The new system matrix is:    Ax =   

1 1 1 1 1

t21 t22 t23 t24 t25

t1 t2 t3 t4 t5



 



 x1      x2  ∼ =    x3 

After substituting our point into the matrix A we have:    1 1 1  1 2 4     x1  ∼   Ax =  1 3 9  =   1 4 16  x2  1 5 25

y1 y2 y3 y4 y5



0 5 8 17 24



   = b.  

   = b.  

  11  x1 −5 12     x = x2 = 7 5 x3 7 

Thus, the best fit quadratic function for the set of points is given by 12 11 5 f (t) = t2 + t − 7 7 5 Plotting our two found functions and the original data set gives:

74


Non-linear Problems In order to consider non linear problems we need to consider iterative methods that will help us find the optimal solution to our data fitting problem.

10.1.2

Taylor series

Lets recall in one dimension that we can expand a function about a close point (x + h) provided we know information about our function at point x. Thus a Taylor series expansion is: f (x) f 0 (x)h f 00 (x)h2 f (x + h) = + + + ··· 0! 1! 2! Truncate the expansion after the first term gives: f (x + h) ≈ f (x) + f 0 (x)h For a numerical scheme we could note that h is the distance between to points: h = x1 − x Thus, we can make a note that an approximation of the first derivative can be given by:

=⇒

10.1.3

f (x + x1 − x) ≈ f (x) + f 0 (x)(x1 − x) f (x1 ) − f (x) f 0 (x) ≈ (x1 − x)

Newton’s Method

Newton’s method is a way of iterating toward the root a function given that you have an initial guess that is reasonably close to the root. To find the root of f (x) = 0 we guess that x1 is our root. Then an iterative scheme for approaching a root would be given by rearranging the first derivative approximation that we just found (with f (x1 ) ≈ 0) to obtain: xn = xn−1 −

f (xn−1 ) f 0 (xn−1 )

where we iterate from some initial guess until some predetermined tolerance is achieved. In order to solve non-linear equations in multiple dimensions we can look at a generalization of Newton’s Method. Looking at a function f such that f : Rn → Rn , the truncated Taylor series expansion is given by: f (x + s) ≈ f (x) + Jf (x)s (10.1.1) 75

OR Notes Draft: May 5, 2012 where we are using Jf (x) to denote the Jacobian matrix of x. Specifically, Jf (x)i,j =

∂fi (x) . ∂xj

Noting that s is given as: s = xn − x =⇒ xn = s + x and that xn is an approximate zero to f , we can rewrite (10.1.1) as: xn ≈ x −

f (x) Jf (x)

An algorithm for solving a system of f (x) = 0 for some initial guess x0 could be written in the following manner: Algorithm for Newton’s Method (System of non-linear equations): While ksk2 > tolerance. • Solve Jf (xn−1 )sn = −f (xn−1 ) for sn . • Update xn = xn−1 + sn Lets now consider newtons method for a function f : Rm → R. Here a two term in our Taylor series expansion: 1 f (x + s) ≈ f (x) + ∇f (x)T s + sT Hf (x)s. 2 where Hf (x) is the Hessian matrix of second partial derivatives of f such that: Hf (x)i,j

∂ 2 f (x) = ∂xi ∂xj

It follows that the function f (x + s) will have a minimum when Hf (x)s = −∇f (x).

(10.1.2)

To see this you should look at this as if f is a quadratic function in s, and note that (10.1.2) is the vertex of the function. This allows us the oppertunity to think of Newton’s Method as: Algorithm for Newton’s Method : While ksk2 > tolerance. • Solve Hf (xn−1 )sn = −∇f (xn−1 ) for sn . • Update xn = xn−1 + sn 76


10.2

Non-linear Least Squares

Lets start by considering a biological system that involves fitting parameters to a mathematical model of a swimming lamprey. The work of Dr. Chia-Yu Hsu involves numerical simulations of these swimmers: http://www.ccs.tulane.edu/~chiayu/ Here the goal will be to fit a set of swimming data with a best fit model of the following form: f (x, tˆ) = (ax2 + bx + c) sin(kx + ω tˆ) where we need to find the values of a, b, c, k, and ω that will best fit our model. In order to solve this problem lets assume that we are given a set of m points of the form (ti , yi ) and we want to fit a nonlinear function with vector of parameters x ∈ Rk such that the function f (t, x), is the best fit to the given data set where f : Rk+1 → R. Thus, our goal is to minimize the residual r where r : Rk → Rm , and ri (x) = yi − f (ti , x), for i ∈ {1, 2, . . . , m}. where we are now hoping to minimize φ(x) =

1 1 krk22 = r(x)T r(x). 2 2

We have placed the 21 into the function φ for convenience. In order to apply Newton’s Method to this problem we need to find the following components: ∇φ(x), and Hφ (x). Finding the first of these expressions is not troubling. Denote the Jacobian matrix of r(x) by J(x) gives ∇φ(x) = JT (x)r(x). For the second term we calculate the Hessian Matrix of φ as T

Hφ (x) = J (x)J(x) +

m X

ri (x)Hri (x).

i=1

Calculating the Hri (x) is often difficult and computationally expensiveso it is often ignored or dropped fom computations. The result in what is known as a Gauss-Newton Method given by: Algorithm for Newton’s Method : While ksk2 > tolerance. 77

OR Notes Draft: May 5, 2012 • Solve JT (xn−1 )J(xn−1 )sn = −JT (xn−1 )r(xn−1 ) for sn . • Update xn = xn−1 + sn Note the solve step looks like the normal equations for a nonlinear least squares problem. The question now is what does J(x) look like for our given problem? This is the Jacobean matrix of the residual vector. For the swimmers we have: f (x, x, tˆ) = (ax2 + bx + c) sin(kx + ω tˆ) where we need to find the values of a, b, c, k, and ω that will best fit our model to a given set of data along a swimmer. Lets let φˆ = ω tˆ be a parameter all by itself and then back the value of ω out of the time dependent data fit (as we are doing this for a moving swimmer). Thus, ˆ and x ˆT ˆ = (a, b, c, k, φ) f (ˆ x, x) = (ax2 + bx + c) sin(kx + φ), So at any time we are looking at the residual vector for our m points where the components of r(x) are: ri (ˆ x) = yi − f (ˆ x, ti ) for i ∈ {1, 2, . . . , m}. and the components of the residuals Jacobean are J(ˆ x)i,1 =

∂ri (ˆ x) ˆ = −(t2i ) sin(kti + φ) ∂a

J(ˆ x)i,2 =

∂ri (ˆ x) ˆ = −ti sin(kti + φ) ∂b

J(ˆ x)i,3 =

∂ri (ˆ x) ˆ = − sin(kti + φ) ∂c

J(ˆ x)i,4 =

∂ri (ˆ x) ˆ i = −(at2i + bti + c) cos(kti + φ)t ∂k

J(ˆ x)i,5 =

∂ri (ˆ x) ˆ = −(at2i + bti + c) cos(kti + φ) ˆ ∂φ

At the end of the day we will have a five by five matrix system to iterate on and find our optimal parameters. Note as we are going to solve this over different time sets we will be able to plot the value computed for φˆ as a function of the different input times and back out a value for ω the speed of the swimmer.

78

Chapter 11 Network Flows “It’s hard to decide who’s truly brilliant; it’s easier to see who’s driven, which in the long run may be more important.”

−Michael Crichton

In this chapter a collection of Network Algorithms will be examined. The current treatment will only serve to illustrate some of the other areas that can be explored in the world of Operations Research.

11.1

Dijkstra’s Algorithm

Dijkstra’s Algorithm Can be used to find the shortest path spanning tree for a given network. Essentially the algorithm follows the following steps: 1. Set the distance to notes. (a) The starting (or current) node distance is set to zero, (b) All other nodes in the network are set to a distance of ∞. 2. Create a list of the unvisited nodes. Consisting of all nodes. 3. From the current node: Consider the distance to all the unvisited node neighbours in the network. If the distance from the current node to the neighbour is less than the listed value in you “distance list” update the distance and the predecessor array. 4. Mark the current node as visited by removing from “unvisited list”. 5. If the distance to all unvisited nodes is ∞ stop. Otherwise choose the closest unvisited node to be the current node, and go back to step 3. 79

OR Notes Draft: May 5, 2012 Lets consider the algorithm with the following example:

Example: Use Dijkstra’s algorithm to find a shortest path tree rooted at node 1 in the network below. Note that two of the arcs are bi-directional. Show at each iteration D(·), pred(·), and the selected node i∗ , as well as the final shortest path tree. 5

∞5

-

∞

-

2 3 2 3 Z 2 Z 2 > > 6 6 @ @ Z Z ∞ 6 6 6 6 Z Z @ @ ~ Z ~ Z @ @ 1 1 4 4 1 1 1 1 @ @ 8 3 8 3 Q Q > > @ @ Q Q 3 Q 3 Q @ @ R ? 2 @ R ? 2 @ s ? Q s ? Q - 5 - 5 6 6 5 5

∞

∞

Initialize node 1∗ 2 3 4 5 6 D( ) 0 ∞ ∞ ∞ ∞ ∞ pred( ) Note: The ∗ denotes the node selected and p will denote a nodes predecessor. 5

6 2

4 5

∞

-

11 -

3 2 3 Z 2 Z 2 > > 6 @ @ Z Z ∞ ∞ 6 6 6 6 Z Z @ @ ~ Z ~ Z @ @ 1 1 4 4 1 1 1 1 @ @ 8 3 8 3 Q Q > > @ @ Q Q 3 Q 3 Q @ @ R ? 2 @ R ? 2 @ s ? Q s ? Q - 5 - 5 6 6 5 5 6

∞

3

3

node 1p 2 3 4 5 6∗ D( ) 0 6 ∞ ∞ ∞ 3 pred( ) 0 1 1 4 5

8

node 1 2∗ 3 4 5 6p D( ) 0 4 11 ∞ 8 3 pred( ) 0 6 6 6 1 4 5

9

-

8

2 3 2 3 Z 2 > Z 2 6 @ Z > 6 9 @ Z 6 6 Z ∞ @ 6 6 ~ Z Z @ ~ Z @ 1 @ 4 1 1 1 4 @ 1 1 @ 8 3 Q > 8 @ Q > Q @3 ? ? 2 Q 3 Q @ R @ s Q ? ? 3 Q @ R @ 2 s Q - 5 6 - 5 6 5 5 3 7

3

-

7

80

OR Notes Draft: May 5, 2012 node 1 2 3∗ 4 5p 6 D( ) 0 4 8 9 7 3 pred( ) 0 6 5 5 2 1

node 1 2p 3 4 5∗ 6 D( ) 0 4 9 ∞ 7 3 pred( ) 0 6 2 2 1 4 5

4 5

8

-

8

2 3 2 3 Z 2 > Z 6 2 @ Z > 6 9 @ Z 6 6 Z @ 6 @ 6 ~ Z Z 9 ~ Z @ 1 @ 4 1 1 1 4 @ 1 1 @ 8 3 Q > 8 3 @ Q > Q @ Q 3 Q @ R ? 2 @ s ? Q ? ? 3 Q @ R @ 2 s Q 6 5 - 5 6 5 5 3 7

3

-

7

node 1 2 3 4∗ 5p 6 D( ) 0 4 8 9 7 3 pred( ) 0 6 5 5 2 1

node 1 2 3p 4 5 6 D( ) 0 4 8 9 7 3 pred( ) 0 6 5 5 2 1

The minimum distance spanning tree for the network is given by the final path shown in red.

81


82

Chapter 12 Appendices “Can you still have any famous last words if you’re somebody nobody knows?” − Ryan Adams

12.1

Homework 1

Answer the following questions. Please include the questions with the solution. I do not want to see scrap work. 1. Only three brands of beer (beer1, beer2, and beer3) are available for sale in Metropolis. From time to time, people try one or another of these brands. Suppose that at the beginning of each month, people change the beer they are drinking according to the following rules: • 30% of the people who prefer beer 1 switch to beer 2. • 20% of the people who prefer beer 1 switch to beer 3. • 30% of the people who prefer beer 2 switch to beer 3. • 30% of the people who prefer beer 3 switch to beer 2. • 10% of the people who prefer beer 3 switch to beer 1. For i = 1, 2, 3, let xi be the number who prefer beer i at the beginning of this month and let yi denote the number who prefer beer i at the beginning of next month. Use matrix notation to relate x = (x1 x2 x3 )T and y = (y1 y2 y3 )T . Note that the following equations will hold for the rules given above: 0.5x1 + 0.0x2 + 0.1x3 = y1 People switch to beer 1 0.3x1 + 0.7x2 + 0.3x3 = y2 People switch to beer 2. 0.2x1 + 0.3x2 + 0.6x + 3 = y3 People switch to beer 3. 83

where we needed to find the number of people that stayed loyal to any particular type over the course of the month. Let   0.5 0.0 0.1 A =  0.3 0.7 0.3  . 0.2 0.3 0.6 and note that Ax = y.

2. Show that for any two matrices A and B, (AB)T = B T AT . Assume that A is an m × n matrix and B is an n × q matrix so that there product is defined. Thus, AB is an m × q matrix, and its transpose is a q × m matrix. We can also note that B T AT is the product q × n matrix and a n × m matrix resulting in again a q × m. So each side of the equality is of the same size. We now only need to show that the arbitrary element in row i and column j of (AB)T is the same as the row i column j element in B T AT . Denoting the entry in row i column j in a matrix M as Mi,j we can consider the following algebra: (AB)Ti,j = (AB)j,i n X = Aj,k Bk,i k=1

= =

n X k=1 n X

T ATk,j Bi,k

T Bi,k ATk,j

k=1 T

= (B AT )i,j which concludes the proof. 3. Model each of the following decision making situations as a linear program(LP). For each LP clearly define your decision variables, objective function and constraints. (a) Suppose that the latest scientific studies indicate that cattle need certain amounts b1 , . . . , bm of nutrients N1 , . . . , Nm respectively. More over, these nutrients are currently found in n commercial feed materials F1 , . . . , F n as indicated by coefficients aij , i = 1, . . . , m, j = 1, . . . , n, that denote the number of units of nutrientNi per pound of feed material Fj . Each pound of Fj costs the rancher cj dollars. How can a rancher supply these minimal nutrient requirements to his prize bull while minimizing his feed bill? Linear Program: The goal of the problem is to minimize the cost of the bulls food while maintaining the bulls healthy diet.

OR Notes Draft: May 5, 2012 Let xj be the number of pounds that the farmer should purchase of commercial feed Fj where , j = 1, . . . , n. Thus the farmer wishes to minimize the following cost equation (where z denotes the farmers total cost): Minimize:

z = c1 x 1 + c2 x 2 + . . . + cn x n =

n X

cj x j

j=1

In order to meet the nutrient requirments the farmer needs to impose the following constrictions on his cost equation: a11 x1 a21 x1 a31 x1 .. .

+ a12 x2 + a22 x2 + a32 x2 .. .

+ a13 x3 + a23 x3 + a33 x3 .. .

am1 x1 + am2 x2 + am3 x3

+ . . . + a1n xn + . . . + a2n xn + . . . + a3n xn .. ... . + . . . + amn xn

≥ b1 ≥ b2 ≥ b3 .. . ≥ bm

where xj ≥ 0, j = 1, . . . , n Let x = [x1 , x2 , · · · , xn ]T , and c = [c1 , c2 , · · · , cn ]T . We can also denote [b1 , b2 , · · · , bm ]T by b, and coefficient matrix   a11 a12 a13 . . . a1n  a21 a22 a23 . . . a2n     a31 a32 a33 . . . a3n    by A.  .. .. .. . . .. ..    . . . am1 am2 am3 . . . amn Thus the farmer is minimizing z = cT x subject to, Ax ≥ b where xj ≥ 0, j = 1, . . . , n (b) Suppose that a small chemical company produces one chemical product which it stores in m storage facilities and sells in n retail outlets. The storage facilities are located in different communities. The shipping costs from storage facility Fi to retail outlet Oj is cij dollars per unit of the chemical product. On a given day, each outlet Oj requires exactly dj units of the product. Moreover, each facility Fi has si units of the product available for shipment. The problem is to determine how many units of the product should be shipped from each storage facility to each outlet in order to minimize the total daily shipping costs. Linear Program: The goal of the problem is to minimize the cost of product shipping while maintaining product invintory and not overdrawing from supply. Let xij be the number of units of chemical shipped from facility Fi to retail outlet Oj , where i ∈ {1, . . . , m}, and j ∈ {1, . . . , n}. Let z denote the total cost of 85

OR Notes Draft: May 5, 2012 shipping. Thus, the chemical company wishes to optimize the followin objective function, m X n X Minimize: z = cij xij i=1 j=1

In minimizing the above equation the company must meet the following constraints: P x11 + x21 + . . . + xm1 = Pm i=1 xi1 ≥ d1 m x12 + x22 + . . . + xm2 = i=1 xi2 ≥ d2 .. .. .. . . Pm. x1n + x2n + . . . + xmn = i=1 xin ≥ dn That is each outletj requires dj of chemical where j ∈ {1, . . . , n}. Each Facility i also only has a limited supply of the chemical it can ship si where i ∈ {1, . . . , m}. Thus, the problem has the constrictions: Pn x1j ≤ s1 x11 + x12 + . . . + x1n = Pj=1 n ≤ s2 x21 + x22 + . . . + x2n = j=1 x2j .. .. .. . . Pn . xm1 + xm2 + . . . + xmn = j=1 xmj ≤ sm where xij ≥ 0 , i ∈ {1, . . . , m} and j ∈ {1, . . . , n} The notation of the problem  x11  x21   ..  . xm1     

can be simplified if matrix notation is used:  x12 x13 · · · x1n x22 x23 · · · x2n   .. .. ..  = X .. . . . .  xm2 xm3 · · · xmn

d1 d2 .. .





     = d, and   

dn

s1 s2 .. .

    = s. 

sm

Denote a column of 1’s that is k rows long by 1k and he linear program can now take the following form: Minimize: z =

m X n X

cij xij

i=1 j=1

Subject to:

X1n ≤ s X T 1m ≥ d

where all elements of X are greator than or equal to zero.

86

OR Notes Draft: May 5, 2012 (c) An investor has decided to invest a total of $50,000 among three investment opportunities: savings certificates, municipal bonds, and stocks. The annual return in each investment is estimated to be 7%, 9%, and 14%, respectively. The investor does not intend to invest her annual interest returns (that is, she plans to use the interest to finance her desire to travel). She would like to invest a minimum of $10,000 in bonds. Also, the investment in stocks should not exceed the combined total investment in bonds and savings certificates. And, finally, she should invest between $5,000 and $15,000 in savings certificated. The problem is to determine the proper allocation of the investment capital among the three investment opportunities in order to maximize her yearly return. Linear Program: The goal of the problem is to maximize the return on investment subject to several constraints. Let xc be the amount to be invested in savings certificates. Let xb be the amount to be invested in bonds, and Let xs be the amount to be invested in stocks. The goal of the problem is to now maximize the following return on investment objective function given by z: Maximize: z = .07xc + .09xb + .14xs The problem also has the followning constraints, or is Subject to:

xc + xb + xs = 50, 000 xb ≥ 10, 000 xs ≤ xb + xc 5, 000 ≤ xc ≤ 15, 000 with xs ≥ 0.

This can be solved using LINDO and the following values are obtained (See Attached LINDO print out ): xc xb xs

= = =

$5,000 $20,000 $25,000

Table 12.1.1: Amount to Place in Each Investment (d) The owner of a small chicken farm must determine a laying and hatching program for 100 hens. There are currently 100 eggs in the hen house, and the hens can be used either to hatch existing eggs or to lay new ones. In each 10-day period, a hen can either hatch 4 eggs or lay 12 new eggs. Chicks that are hatched can be sold for 60 cents each, and every 30 days an egg dealer gives 10 cents each for the eggs accumulated to date. Eggs not being hatched in one period can be kept in a special incubator room for hatching in a later period. The problem is to determine how many hens should be hatching and how many should be laying in each of the next three periods so that total revenue is maximized. Linear 87

OR Notes Draft: May 5, 2012 Program: The goal of the problem is to maximize the total revenu over the next 30 day period subject to several constraints. Let Let Let Let

Hhi be the number of hens hatching eggs in period i, i ∈ {1, 2, 3}. Hli be the number of hens laying eggs in period i, i ∈ {1, 2, 3}. Es0 be the number of eggs initially in storage (100 eggs). Esi be the number of eggs in storage at the end of period i, i ∈ {1, 2, 3}.

The goal of the problem is to now maximize the profit made on the eggs and chicks produced by the chickens defined by the objective function z. That is: Maximize: z = .6 × 4 (Hh1 + Hh2 + Hh3 ) + .1Es3 Subject to:

Hhi + Hli = 100 4Hhi ≤ Es(i−1) Es(i−1) − 4Hhi + 12Hli = Esi

Hhi , Hli , Esi ≥ 0, where i ∈ {1, 2, 3} When the above program is optimized using LINDO(see the attached print out). After solving the following values are obtained: Hh1 Hh2 Hh3 Hl1 Hl2 H13 Es0 Es1 Es2 Es3

= = = = = = = = = =

25 100 100 75 0 0 100 900 500 100

This is sensible, because the return from a hatched chicken is 60 cents, and that of an egg is only 10 cents. Thus, a Hen laying 12 eggs a period can generate $1.20 and a Hen hatching 4 eggs can generate $2.40.

12.2

Homework 2

Answer the following questions. Please include the question with the solution (write or type them out doing this will help you digest the problem). I do not want to see scrap work. Note the problems are separated into two sections a set for all students and a bonus set for those taking the course at the 545 level. 88


All Students 1. Solve the following linear program graphically. maximize subject to

z = x1 + 2x2 2x1 + x2 x1 + x2 −x1 + 3x2 −2x1 + x2

≥ ≥ ≤ ≥

12 5 3 −12

(12.2.1)

with x1 , x2 ≥ 0. (a) Clearly label each of the constraints, and the feasible region. (b) List all the extreme corner points. (c) Write the feasible region as a convex convex set. (d) State the optimal solution to the linear program. (e) Comment about the solution to the liner program if the constraint −2x1 + x2 ≥ −12 is removed. Solution:

Figure 12.2.1 shows the feasible region obtained from the linear programs constraints, as well as a contour of the optimal objective function. See http://www.math.iup.edu/~jchrispe/MATH445_545/ProblemOneGraph.html for an interactive version of the solution. The extreme corner points of the solution set are: 33 18 39 18 , , , and C = A = (6, 0), B = 5 5 7 7

Figure 12.2.1: Graphical solution to the linear program described in problem 1. 89

OR Notes Draft: May 5, 2012 The feasible region can be written as a convex set by defining α1 , α2 , and α3 such that α1 , α2 , α3 ∈ [0, 1], and

3 X

αi = 1.

i=1

Then using the extreme corner points A, B, and C any point in the feasible region of the linear program is then given by: feasible region = (x, y) (x, y) = α1 A + α2 B + α3 C. , 18 with a maxiThe optimal solution to the linear program occurs at point B = 39 5 5 mum objective function value of z = 15. If the constrain −2x1 + x2 ≥ −12 is removed from the linear program then the solution becomes unbounded with a direction of unboundedness being given by the vector: 3 d= . 1

2. You have $2000 to invest over the next five years. At the beginning of each year, you can invest money in one or two year bank time deposits, yielding 8% (at the end of one year) and 17% (at the end of the two years) respectively. Also you can invest in threeyear certificates offered only at the start of the second and third years, yielding 26% (total) at the end of the third year. Assume that you invest all money available each year, and want to withdraw all cash at the end of year 5. Formulate the situation as a linear programming problem; carefully define your decision variables. Then solve the problem using lindo or lingo (as we have done in class), and attach the lindo output. Be sure to interpret your results. Solution: Let xi,j be the amount of money invested in an account at the start of year i for j years. Then for example x5,1 is the amount of money invested into an account for one year at the start of year five. There will then be 11 possible decision variables: decision variables = {x1,1 , x2,1 , x3,1 , x4,1 , x5,1 , x1,2 , x2,2 , x3,2 , x4,2 , x2,3 , and x3,3 } The objective is to maximize the money received during the fifth year. Thus, maximize z = 1.08x1,5 + 1.17x2,4 + 1.26x3,3 we are now subject to the following constraints: x1,1 + x1,2 = 2000 x2,1 + x2,2 + x2,3 = 1.08x1,1 x3,1 + x3,2 + x3,3 = 1.17x1,2 + 1.08x2,1 x4,1 + x4,2 = 1.17x2,2 + 1.08x3,1 x5,1 = 1.26x2,3 + 1.17x3,2 + 1.08x4,1

90

start start start start start

1st year 2nd year 3rd year 4th year 5th year

OR Notes Draft: May 5, 2012 with the set of decision variables all nonnegative. Solving the problem with lingo yields: x1,1 = 2000, x2,2 = 2160, x4,2 = 2527.20, and all other xi,j = 0. The optimal objective function is z = $2956.82. You should invest all your money the first year into a one year account. Then in the second year put all your money into a two year account, and in the fourth year put all your money again into a two year account. 3. Write the following linear program in the standard problem. maximize z = 6x1 − 3x2 subject to 2x1 + 5x2 3x1 + 2x2 x1 , x2

form, posed as a minimization

≥ 10 ≤ 40 ≤ 15.

Solution: Start by placing adding the x1 , x2 ≤ 15 variables into the standard constraints. We convert to looking at the minimization problem by multiplying the objective function by ‘-1’. minimize subject to

−z = −6x1 + 3x2 2x1 + 5x2 3x1 + 2x2 x1 x2 with x1 , x2 free.

≥ ≤ ≤ ≤

10 40 15 15

Now we deal with the variables x1 and x2 being free. Thus, we use the change of variables x1 = x01 − x001 and x2 = x02 − x002 with x01 , x001 , x02 , x002 ≥ 0. minimize subject to

−z = −6x01 + 6x001 + 3x02 − 3x002 2x01 − 2x001 + 5x02 − 5x002 3x01 − 3x001 + 2x02 − 2x002 x01 − x001 x02 − x002 with x01 , x001 , x02 , x002 ≥ 0.

≥ ≤ ≤ ≤

10 40 15 15

Last we add slack and excess variables to obtain the problem in standard form. minimize subject to

−z = −6x01 + 6x001 + 3x02 − 3x002 2x01 − 2x001 + 5x02 − 5x002 − e1 3x01 − 3x001 + 2x02 − 2x002 + s1 x01 − x001 + s2 0 00 x2 − x2 + s 3 0 00 0 00 with x1 , x1 , x2 , x2 , e1 , s1 , s2 , s3 ≥ 0.

91

= = = =

10 40 15 15

OR Notes Draft: May 5, 2012 4. Write the following linear program in the standard form, posed as a maximization problem. Do not use the substitution x2 = x02 − x002 , use the free variable instead to state the problem using one less constraint. maximize subject to

Solution: Thus,

z = 6x1 − 3x2 2x1 + 5x2 ≥ 10 3x1 + 2x2 ≤ 40 x1 ≥ 0 and x2 free .

Here we can start by adding a slack variable to the second constraint. maximize subject to

z = 6x1 − 3x2 2x1 + 5x2 ≥ 10 3x1 + 2x2 + s1 = 40 x1 , s1 ≥ 0 and x2 free .

We can now make note that the second constraint can be solved for x2 as x2 =

1 3 1 40 3 − x1 − s1 = 20 − x1 − s1 . 2 2 2 2 2

This expression for x2 can be placed into both the objective function and into the remaining constraints. This gives: maximize subject to

z = 6x1 − 3(20 − 23 x1 − 12 s1 ) 2x1 + 5(20 − 23 x1 − 12 s1 ) ≥ 10 x1 , s1 ≥ 0.

Simplify and add a slack varaible to get, maximize subject to

z + 60 = 21 x + 32 s1 2 1 11 x + 52 s1 + s2 = 90 2 1 x1 , s1 , s2 ≥ 0.

Note we could do a transform of variables on z to keep track of the additional 60 units for us.

545 Additional Homework 1. For a linear program written in standard form with constraints Ax = b and x ≥ 0 show that d is a direction of unboundedness if and only if Ad = 0 and d ≥ 0. Solution: First lets recall what a direction of unboundedness is. An n by 1 vector d is a direction of unboundedness if for all x in S (an LP’s feasible region), and c ≥ 0 then x + cd ∈ S. 92

OR Notes Draft: May 5, 2012 • (=⇒) Assume here that d is a direction of unboundedness, and that c is a positive constant (c = 0 is trivial). Then we know that for any x ∈ S that x + cd is in S. This gives A(x + cd) = b −→ −→ −→ −→

Ax + cAd = b b + cAd = b cAd = 0 Ad = 0.

We also know that x + cd ≥ 0 for all x ≥ 0 ∈ S. Thus, cd ≥ 0, and d ≥ 0. • (⇐=) Assume that Ad = 0, and d ≥ 0, and consider any point x in the feasible region of our linear program. As x is in the feasible region we know that Ax = b. Thus, for c > 0 consider the point x + cd A(x + cd) = Ax + cAd = Ax + 0 = Ax = b. Note we also need our new point “x + cd00 to be non-negative which is true as x ∈ S, and d ≥ 0. Thus, x + cd ∈ S, and d is a direction of unboundedness.

12.3

Homework 3

Answer the following questions. Please include the question with the solution (write or type them out doing this will help you digest the problem). I do not want to see scrap work. Note the problems are separated into two sections a set for all students and a bonus set for those taking the course at the 545 level.

All Students 1. Solve the following linear program. minimize subject to

z = −x1 − 3x2 x1 − 2x2 ≤ 4 −x1 + x2 ≤ 3 x1 , x2 ≥ 0.

Solution: After adding two slack variables in the constraints the initial tableau for the given linear program is written as: Row z 0 1 1 0 2 0

x1 1 1 -1

x2 3 -2 1 93

s1 0 1 0

s2 0 0 1

RHS 0 4 3

Basis z s1 s2

OR Notes Draft: May 5, 2012 As the given problem is a minimization problem we scan the non-basic variables in row zero of the tableau and find that x2 should be picked to enter the basis. The ratio test shows that s2 should leave the basis. After basic row operations we obtain the following tableau: Row z 0 1 1 0 2 0

x1 4 -1 -1

x2 0 0 1

s1 0 1 0

s2 -3 2 1

RHS -9 10 3

Basis z s1 x2

From the tableau it can be seen that it is favourable for x1 to enter into the basis; however, as both variables have negative coefficients in the x1 column for the constraints we can see there is a direction of unboundedness. Thus, the linear program is unbounded, and no minimum value of the objective function will be obtained. Specifically we can see that the constraints give: s1 = 10 + x1 x2 = 3 + x1 Note that if the value of s1 and x2 are both increased by one then increasing the value of x1 by one will keep these constraints satisfied. Thus, from the feasible point:   0  3   x=  10  0 There is a direction of unboundedness given by:   1  1   d=  1  0 and for all c ≥ 0 we note that x+cd is feasible for the linear program and the objective function is unbounded. 2. Solve the following linear program. maximize subject to

z = x1 + x2 x 1 + x2 + x3 ≤ 1 x1 + 2x3 ≤ 1 x1 , x2 , x3 ≥ 0.

Solution: After adding in the slack variables we obtain the following standard form initial tableau: 94


x1 -1 1 1

x2 -1 1 0

x3 0 1 2

s1 0 1 0

s2 0 0 1

RHS 0 1 1

Basis z s1 s2

Note that x1 and x2 are both attractive to the simplex method. Thus, using Bland’s rule we will pick x1 as the entering variable. The ratio test has a tie between s1 and s2 as leaving variables. Here we will pick s1 as the leaving variable. The updated tableau follows: Row z 0 1 1 0 2 0

x1 0 1 0

x2 0 1 -1

x3 1 1 1

s1 1 1 -1

s2 0 0 1

RHS 1 1 0

Basis z x1 s2

Here we see that we have obtained an optimal simplex tableau; however, we make note that there are multiple optimal solutions as x2 is a non-basic variable that has a zero coefficient in row zero of the tableau. Pivot x2 into the basis variables in place of x1 to obtain a second optimal solutions. Row z 0 1 1 0 2 0

x1 0 1 1

x2 0 1 0

x3 1 1 2

s1 1 1 0

s2 0 0 1

RHS 1 1 1

The optimal solutions to the linear program are:    1 0  0   1      X1 =   0  and X2 =  0  0   0 0 1

Basis z x2 s2

     

For α ∈ [0, 1] the optimal objective function value z = 1 is found with solutions of the form: αX1 + (1 − α)X2 .

3. Solve the following linear program. minimize subject to

z = 4x1 + 4x2 + x3 x1 + x2 + x3 ≤ 2 2x1 + x2 ≤ 3 2x1 + x2 + 3x3 ≥ 3 x1 , x2 , x3 ≥ 0. 95

(12.3.2)

OR Notes Draft: May 5, 2012 Solution: For this problem we need to add slack and excess variables. For use in the Big-M method for solving this problem we also need to add an artificial variable for the constraint with an excess variable that will allow us to have a starting basic feasible solution. The problem written in standard form then becomes: minimize subject to

z = 4x1 + 4x2 + x3 + M a1 x1 + x2 + x 3 + s 1 = 2 2x1 + x2 + s2 = 3 2x1 + x2 + 3x3 − e1 + a1 = 3 x1 , x2 , x3 , s1 , s2 , a1 ≥ 0.

where M denotes a large positive constant. Note we have penalized the objective function if a1 is a basic variable. The initial simplex tableau for the problem is: z 1 0 0 0

x1 -4 1 2 2

x2 -4 1 1 1

x3 -1 1 0 3

s1 0 1 0 0

s2 0 0 1 0

e1 0 0 0 -1

a1 RHS -M 0 0 2 0 3 1 3

Basic z s1 s2 a1

After adjusting the values of the top row using elementary column operations the following tableau is obtained. z 1 0 0 0

x1 2M-4 1 2 2

x2 M-4 1 1 1

x3 s1 3M-1 0 1 1 0 0 3 0

s2 0 0 1 0

e 1 a1 -M 0 0 0 0 0 -1 1

RHS 0 2 3 3

Basic z s1 s2 a1

The simplex method now scans the coefficients in the zero row looking for the most positive value (as this is a minimization problem). Here we see that x3 is the most attractive. The ratio test shows that a1 should leave the basis. After the pivot we obtain the tableau: z 1 0 0 0

x1 −10 3 1 3

x2 −11 3 2 3

2

1

2 3

1 3

x3 0 0 0 1

s1 0 1 0 0

s2 0 0 1 0

e1 −1 3 1 3

a1 −M + −1 3

0

0

−1 3

1 3

1 3

RHS 1 1 3 1

Basic z s1 s2 x3

This is an optimal solution as there is no longer any positive coefficients in the zero row of the tableau. The optimal solution to the given linear program is z = 1 when: x3 = 1, s1 = 1, s2 = 3, and all other variables are set to zero.

96

OR Notes Draft: May 5, 2012 4. A cargo plane has three compartments that can be used for storing cargo: front, center, and back. These compartments have the capacity limits on total weight and space available given by the table below: Compartment Front Center Back

Weight (tons) 12 18 10

Space (cu-ft) 7,000 9,000 5,000

Also the weight of the cargo actually placed in the compartments must be in the same proportion as the compartments’ weight capacity, in order to maintain the balance of the plane. The following four cargoes have been offered for shipment on an upcoming flight as space is available. Any portion of these cargos can be accepted. The objective is to maximize the total profit for the flight. Cargo 1 2 3 4

Weight (tons) 20 16 25 13

Volume (cu-ft) 8,000 8,000 14,000 13,000

Profit ($/ton) 280 360 320 310

(a) Formulate as a linear program and solve using LINDO/LINGO (attach output). Solution: Let xi,j be the amount of cargo i in tons to ship in compartment j where i ∈ {1, 2, 3, 4} and j ∈ {f, c, b} (here f, c, and b denote front, center, and back respectively. Thus, the objective function (maximize profit z) becomes:         X X X X maxz = 280  x1,j +360  x2,j +320  x3,j +310  x4,j  j∈{f,c,b}

j∈{f,c,b}

j∈{f,c,b}

j∈{f,c,b}

The problem is subject to the following constraints: x1,f + x2,f + x3,f + x4,f ≤ 12 x1,c + x2,c + x3,c + x4,c ≤ 18 x1,b + x2,b + x3,b + x4,b ≤ 10

x1,f x2,f x3,f x4,f

+ x1,c + x1,b + x2,c + x2,b + x3,c + x3,b + x4,c + x4,b

≤ ≤ ≤ ≤

20 16 25 13

Max Max Max Max

Weight limit front Weight limit center Weight limit back

amount amount amount amount

400x1,f + 500x2,f + 560x3,f + 1000x4,f ≤ 7000 400x1,c + 500x2,c + 560x3,c + 1000x4,c ≤ 9000 400x1,b + 500x2,b + 560x3,b + 1000x4,b ≤ 5000

97

of of of of

cargo cargo cargo cargo

1 2 3 4

Space limit front Space limit center Space limit back

OR Notes Draft: May 5, 2012 Last the weight of the cargo must be in the same proportion as the compartments’ weigh capacity. There are several ways to write this constraint, here we inforce the front to back and center to back ratio: x1,f + x2,f + x3,f + x4,f 12 = , x1,b + x2,b + x3,b + x4,b 10

and

x1,c + x2,c + x3,c + x4,c 18 = x1,b + x2,b + x3,b + x4,b 10

These constraints can be written in a more standard form as: 10x1,f + 10x2,f + 10x3,f + 10x4,f − 12x1,b − 12x2,b − 12x3,b − 12x4,b = 0 10x1,c + 10x2,c + 10x3,c + 10x4,c − 18x1,b − 18x2,b − 18x3,b − 18x4,b = 0 We additionally note that all the decision variables xi,j ≥ 0 for all i ∈ {1, 2, 3, 4} and j ∈ {f, c, b}. Solving the problem using LINDO gives the results on the suplmental pdf document. Note that the optimal solution was found in 9 iterations of the simplex method. Here profit obtains a maximum value of $13260.00 when x1,c = 4.5 x2,c = 6.0 x2,b = 10.0 x3,f = 12.0 x3,c = 7.5 and all other decision variables are set to zero. (b) Is the optimal solution unique? Why or why not? If not, determine the set of all optimal solutions. [The LINDO/LINGO report tableau may help here.] Solution: From the obtained optimal simplex tableau we can notice that x3,b is a non-basic variable with a zero coefficient in row zero of the tableau. Thus, if we pivot x3,b into the basis we will obtain an adjacent optimal basic feasible solution to the one found using LINDO. Note from the tableau that we have the following constraints: x1,b x3,c x2,c x2,b x1,c x3,f

= = = = = =

0 + 0.6x3,b 7.5 − x3,b 6.0 + 1.6x3,b 10.0 − 1.6x3,b 4.5 − 0.6x3,b 12

with again all other decision variables set to zero. From the ratio test we see that the maximum balue that x3,b can take on is 6.26. Thus, any solution written as above with x3,b ∈ [0, 6.25] will yield the stated optimal solution of $13,260.00.

98


545 Additional Homework 1. Consider the simplex method, implemented with the usual rules, applied to a linear program of the form: maximize z = cT x, subject to Ax ≤ b, x ≥ 0. • Can a variable xj that enters at one iteration leave at the very next iteration? Either provide a numerical example (in two variables) in which this happens or prove that it cannot occur. Solution: Yes a variable that enters the basis at one iteration can leave the basis on the next iteration of the simplex method. Consider the following example: maximize subject to

z = 2x1 + x2 x1 + 14 x2 ≤ 1 x1 − 3x2 ≤ 5 x1 − 2x2 ≤ 4 x1 , x2 ≥ 0.

After adding in the needed slack variables the initial simplex tableau is: Row z 0 1 1 0 2 0 3 0

x1 -2 1 1 1

x2 s 1 -1 0 0.25 1 -3 0 -2 0

s2 0 0 1 0

s3 0 0 0 1

RHS 0 1 5 4

Basis z s1 s2 s3

The first pivot of the simplex algorithm has x1 enter the basis in place of s1 . Row z 0 1 1 0 2 0 3 0

x1 0 1 0 0

x2 s1 -0.5 2 0.25 1 -3.25 -1 -2.25 -1

s2 0 0 1 0

s3 0 0 0 1

RHS 2 1 4 3

Basis z x1 s2 s3

On the next pivot of the algorithm we see that x1 is replaced in the basis by x2 . Row z 0 1 1 0 2 0 3 0

x1 2 4 13 9

x2 0 1 0 0

s1 s2 4 0 4 0 12 1 8 0

s3 0 0 0 1

RHS 4 4 17 12

Basis z x2 s2 s3

This is the optimal tableau, and the optimal solution to the stated maximization problem is z = 4 when x1 = 0, and x2 = 4. 99

OR Notes Draft: May 5, 2012 • Can a variable xi that leaves at one iteration enter at the very next iteration of the simplex method? Either provide a numerical example (in two variables) in which this happens or prove that it cannot occur. Solution: A variable xi that leaves at one iteration of the simplex method can not enter at the very next iteration of the simplex method. Before writing a proof lets recall the rules of the simplex method for a standard form maximization problem. (a) Scan the non-basic variables and find the one with the most negative coefficient in the current tableau (row zero of the tableau in our standard form). This will give the greatest improvement to the objective function value z upon entering the basis. (b) Use the ratio test to determine the maximum amount that the incoming basic variable may be with out causing other basic variables to take on negative values. (c) Use elementary row operations to make the incoming variable basic in the updated tableau. Proof: Assuming a standard form maximization problem, let xi be the variable leaving the basis and xj be the variable entering the basis at some pivot of the simplex method (Note Row: i out, Column j in). The relevant portion of a simplex tableau is: Row z 0 1 n 0

... ... ...

xi 0 1

. . . xj . . . cj . . . aj

... ... ...

RHS b0 bn

Basis z xi

From the stated rules of the simplex method note that the coefficient in row zero for xj will be negative in the simplex tableau (call this value cj , with cj < 0 in the tableau). If during the ratio test xi has been picked to be the leaving variable we note that its current value is given by the RHS value in row i call this bn , and we also know that the value of aj is positive. After completing the simplex pivot the updated relevant simplex tableau is given by: Row z 0 1 n 0

... ... ...

xi −cj aj 1 aj

... ... ...

xj 0 1

... RHS c b . . . b0 − jajn bn ... aj

Basis z xj

In the updated tableau we note that the zero row coefficient for xi (the variable that just became non-basic) is given by: −cj >0 aj As this value is positive in the current tableau, then xi will not enter the basis on the next pivot of the simplex algorithm. This completes the proof. 100


12.4

Homework 4

Answer the following questions. Please include the question with the solution (write or type them out doing this will help you digest the problem). I do not want to see scrap work. Note the problems are separated into two sections a set for all students and a bonus set for those taking the course at the 545 level.

All Students 1. Use the simplex method in matrix form to solve the following linear program: maximize subject to

z = 2x1 − x2 + x3 3x1 + x2 + x3 ≤ 60 x1 − x2 + 2x3 ≤ 10 x1 + x2 − x3 ≤ 20 x1 , x2 , x3 ≥ 0.

For a problem in the standard form given in class recall that any current tableau in the simplex algorithm can be represented in matrix form using the basis as follows: −cN T

xN T xB T RHS Basis T −1 T −1 + cB B N 0 cB B b z −1 −1 B N I B b xB

Solution: Lets first solve the problem using tableaus so we have something to use as comparison. Row z 0 1 1 0 2 0 3 0

x1 -2 3 1 1

x2 1 1 -1 1

x3 -1 1 2 -1

s1 0 1 0 0

s1 0 0 1 0

s3 0 0 0 1

RHS 0 60 10 20

Basis z s1 s2 s3

x3 3 -5 2 -3

s1 0 1 0 0

s1 2 -3 1 -1

s3 0 0 0 1

RHS 20 30 10 10

Basis z s1 x1 s3

x3 1.5 1 0.5 -1.5

s1 0 1 0 0

s1 s3 RHS 1.5 0.5 25 -1 -2 10 0.5 0.5 15 -0.5 0.5 5

Choose x1 to enter in place of s2 . Row z 0 1 1 0 2 0 3 0

x1 0 0 1 0

x2 -1 4 -1 2

Choose x2 to enter in place of s3 . Row z 0 1 1 0 2 0 3 0

x1 0 0 1 0

x2 0 0 0 1

101

Basis z s1 x1 s3

OR Notes Draft: May 5, 2012 Here we have achieved the optimal solution with x1 = 15, x2 = 5, and x3 = 0 the optimal objective is 25. For this problem we make the following definitions in order to solve the problem using the matrix simplex method:     x1 2  x2   −1        60  x3   1      10  x=  s1  , c =  0  , b =     20  s2   0  s3 0 and



 3 1 1 1 0 0 2 0 1 0 . A =  1 −1 1 1 −1 0 0 1

In order to implement the simplex algorithm in matrix format we break the given coefficient matrix A into to parts: a basic part and a non-basic part. Thus, A = [N |B] In matrix form we start by defining xB = (s1 s2 s3 )T , and xN = (x1 x2      3 1 1 1 0 0 1 −1      1 −1 2 , B= 0 1 0 0 N= =⇒ B = 1 1 −1 0 0 1 0

x3 )T we have:  0 0 1 0 . 0 1

with cN T = cB T =

2 −1 1 0 0 0


−2 1 −1 0 cB T B −1 b =   60 B −1 b =  10  20   3 1 1 2  B −1 N =  1 −1 1 1 −1

Note that the non-basic variable x1 should be picked to enter the basis. Doing the ratio test using the first column of B −1 N and the current right hand side vector B −1 b: 10 20 60 = 20, = 10, = 20 min 3 1 1 102

OR Notes Draft: May 5, 2012 and we choose s2 to leave the basis. This leads to the new values for xB = (s1 x1 s3 )T , and xN = (s2 x2 x3 )T we have: 

   0 1 1 1 3 0 2 , B =  0 1 0  N =  1 −1 0 1 −1 0 1 1



=⇒

B −1

 1 −3 0 1 0 . = 0 0 −1 1

with cN T = cB T =

0 −1 1 0 2 0


2 −1 3 20 cB T B −1 b =   30 B −1 b =  10  10   −3 4 −5 2  B −1 N =  1 −1 −1 2 −3

Note that the non-basic variable x2 should be picked to enter the basis. Doing the ratio test using the second column of B −1 N and the current right hand side vector B −1 b: min

10 30 = 7.5, =5 4 2

and we choose s3 to leave the basis. This leads to the new values for xB = (s1 x1 x2 )T , and xN = (s2 s3 x3 )T we have: 

   0 0 1 1 3 1 2  , B =  0 1 −1  N = 1 0 0 1 −1 0 1 1



=⇒

with cN T =

0 0 1

cB T =

0 2 −1

103

B −1

 1 −1 −2 1 1  = 0 . 2 2 1 0 − 12 2

OR Notes Draft: May 5, 2012 Evaluating each of the expressions in the tableau we have: 1 3 3 −cN T + cB T B −1 N = 2 2 2 25 cB T B −1 b =   10 B −1 b =  15  5   −1 −2 1 1 1  B −1 N =  12 2 2 1 − 12 − 32 2 We note here that none of the reduced costs given in −cN T + cB T B −1 N are negative, and we have reached the optimal solution z = 25 when s1 = 10, x1 = 15, and x2 = 5.

545 Additional Homework 1. The following tableau for a maximization problem was obtained when solving a linear program: Row z 0 1 1 0 2 0 3 0

x1 0 0 1 0

x2 a -2 g 0

x3 0 1 0 0

x4 b e -2 h

x5 c 0 0 1

x6 3 2 1 4

RHS d f 1 3

Basis z x3 x1 x5

Find conditions on the parameters a, b, . . . , h so that the following statements are true. (In the solution parameters that can take on any value will not be mentioned). (a) The current basis is optimal. Here the current basis is optimal if: • c := 0 as x5 is basic. • a and b need to be non-negative making no non-basic variable attractive. (b) The current basis is the unique optimal basis. • c := 0 as x5 is basic. • a and b must be positive to guarantee no alternate optimal solutions. (c) The current basis is optimal but alternative optimal bases exist. • c := 0 as x5 is basic. • Either a and b must be zero with the following sub-cases. 104

OR Notes Draft: May 5, 2012 – If a = 0 then for g > 0 would give an alternate optimal solution. When g ≤ 0 the ratio test would not be restrictive and there would be a direction of alternate optimal solutions. – If b = 0 then for when either e or h or both are positive we would have an alternate optimal solution. If both e and h are negative then ratio test would not be restrictive and there would be a direction of alternate optimal solutions. (d) The problem is unbounded. Assume again that c = 0 as x5 is basic. The problem will be unbounded in two cases: • a < 0 and g ≤ 0. • b < 0 and e, h ≤ 0. In both cases we would have an attractive variable that would enter into the basis, and the ratio test would not have to restrict the value of the decision variables to be non-negative. (e) The current solution will improve if x4 is increased. When x4 is entered into the basis, the change in the objective function is zero. The current solution will improve if x4 is increased means that b will be the most attractive. We will rig the other parameters such that after entering the objective function does not change. • c = 0 as x5 is basic. • b < 0 and a ≥ 0 making x4 attractive. • e > 0, f = 0 and h > 0. Will assure e wins the Ratio Test, and will have no effect on the value of d when Row 1 is added to Row 0.

12.5

Homework 5

The following question comes from Nash and Sofer’s text book. 1. The following questions apply to the linear program: minimize z = −101x1 + 87x2 + 23x3 subject to 6x1 − 13x2 − 3x3 ≤ 11 6x1 + 11x2 + 2x3 ≤ 45 x1 + 5x2 + x3 ≤ 12 with x1 , x2 , x3 ≥ 0. with the following optimal basic solution: Row z 0 1 1 0 2 0 3 0

x1 0 1 0 0

x2 0 0 1 0

x3 0 0 0 1

s1 s2 -12 -4 1 -2 -4 9 19 -43 105

s3 -5 7 -30 144

RHS -372 5 1 2

Basis z x1 x2 x3

OR Notes Draft: May 5, 2012 All of the following questions are independent. Before addressing each of the following questions we should note that the problem can be written in matrix form by defining xB = (x1 x2 x3 )T , and xN = (s1 s2 s3 )T , with the optimal basic solution given above represented as: −cN T where

Note that

xN T xB T RHS Basis T −1 T −1 + cB B N 0 cB B b z B −1 N I B −1 b xB



   6 −13 −3 1 0 0 11 2 ,N =  0 1 0 , B= 6 1 5 1 0 0 1       11 −101 0 87  , and cN =  0  . b =  45  , cB =  12 23 0 

 5 B −1 b =  1  , and cTB B −1 b = 2

−372

It now becomes easy to use sensitivity analysis to answer the following questions. • What is the solution of the linear program obtained by decreasing the right hand side of the second constraint by 15? Here we adjust the value of the original RHS b to the new value:   11 bnew =  30  12 We look to see if the RHS of the LP’s constraints are still feasible:      1 −2 7 11 35 9 −30   30  =  −134  B −1 bnew =  −4 19 −43 144 12 647 As the value for x2 = −134 it is no longer a feasible value. We make note that the optimality condition is still satisfied. We can do a pivot of the Dual simplex method in order to obtain a feasible solution. Here we scan the second row of B −1 N in order to find negative values on which we can pivot. We conduct the ratio test by looking at the minimum of the corresponding reduced cost divided by the possible pivot entry. This gives,      1 −2 7 1 0 0 1 −2 7 9 −30   0 1 0  =  −4 9 −30  , and B −1 N =  −4 19 −43 144 0 0 1 19 −43 144 106

OR Notes Draft: May 5, 2012 −cN T + cB T B −1 N = Ratio Test:

−12 −4 −5

−5 1 −12 = 3, min −30 = 6 , −4

this shows that s3 should gives:  6  6 B= 1

enter the basis in the place of x2 . Doing this update    0 −3 1 0 −13 0 2  , and N =  0 1 11  1 1 0 0 5

with  B −1 bnew = 

1 15 2 15 − 15

1 10 3 − 10 1 5

   0 11 1   30  =  0 12

56 15 67 15 19 5

  , and cB T B −1 bnew =

− 869 3

and the reduced costs are now: −cN T + cB T B −1 N =

− 34 − 11 − 16 3 2

Here we have found a feasible and optimal solution to the updated problem with and x3 = 19 yielding an objective function value the decision variables: x1 = 56 15 5 of z = −869 . 3 • By how much can the right hand side of the second constraint increase and decrease without changing the optimal basis? For this problem we need to consider keeping B −1 b∆b ≥ 0, where



b∆b

   11 0 =  45  +  ∆b  12 0

by finding the values of x that satisfy the given inequality. This specifically gives:        1 −2 7 11 11 − 2(45 + ∆b) + 7(12) 0  −4       9 −30 45 + ∆b −44 + 9(45 + ∆b) − 30(12) 0  = ≥ 19 −43 144 12 19(11) − 43(45 + ∆b) + 144(12) 0 Simplifying the three in equalities gives: 5 ∆b ≤ , 2

∆b ≥

−1 , 9

∆b ≤

2 43

This shows that the change in the right hand side value of the second constraint ∆b must be −1 2 −0.111111111 ≈ ≤ ∆b ≤ ≈ 0.0465116 9 43 in order for the given basis to remain optimal. 107

OR Notes Draft: May 5, 2012 • What is the solution of the linear program obtained by increasing the coefficient of x1 in the objective by 25? Increasing the coefficient may allow for some other value to be more attractive in the objective function. To see if this has occurred we need to recalculate the value of the nonbasic variable reduced costs. After the adjustment     −76 0 cB =  87  , and cN =  0  =⇒ −cN T + cB T B −1 N = 13 −54 170 23 0 This shows that s3 is now the most attractive to enter into the basis. A pivot of the standard simplex method is now done. Here 5 2 , min 7 144 gives that x3 should leave the basis so the new basic and nonbasic variable sets are xB = (x1 , x2 , s3 )T and xN = (s1 , s2 , x3 )T . Checking the optimality and feasibility of this new solution gives: 233 85 − − −cN T + cB T B −1 N = − 679 72 72 72 and

 B −1 b = 

11 144 1 − 24 19 144

13 144 1 24 43 − 144

   11 0 0   45  =  1 12

353 72 17 12 1 72

 

as the optimality and feasibility conditions respectively, and the optimal value of the objective function is:  11   13 11 0 144 144 1 1 0   45  = − 8977 cB T B −1 b =  − 24 24 36 43 19 − 144 1 12 144 with x1 =

353 72

≈ 4.9072, and x2 =

17 12

≈ 1.4167.

• By how much can the objective coefficient of x3 increase and decrease without changing the optimal basis? To solve this we keep N , B, and cN as given initially and consider finding values of ∆c that keep − cN T + cB T B −1 N ≤ 0T (12.5.3) when the vector



 −101 . 87 cB =  23 + ∆c

From (12.5.3) we have 

−

    1 −2 7 1 0 0 0      0 0 0 +(−101 87 (23+∆c)) −4 9 −30 0 1 0 0  ≤ 19 −43 144 0 0 1 0 108

OR Notes Draft: May 5, 2012 

   1 −2 7 0 9 −30  ≤  0  =⇒ (−101 87 (23 + ∆c))  −4 19 −43 144 0 This gives the following bounds on ∆c: −4 5 12 , ∆c ≤ ∆c ≤ 19 43 144 Thus the coefficient on x3 may vary as follows and the given basis will still be optimal: −4 5 0.09302 ≈ ≤ ∆c ≤ ≈ 0.04372. 43 144 ∆c ≤

• Would the current basis remain optimal if a new variable x4 were added to the model with objective coefficient c4 = 46 and constraint coefficients A4 = (12, −14, 15)T ? Here we add a new entry in with the non-basic variables, and check the optimality conditions. Thus, −cN T + cB T B −1 N = −209 −12 −4 −5 where we have use the following values     46 12 1 0 0  0    −14 0 1 0  cN =   0  , and N = 15 0 0 1 0 and we note that the current basis will remain optimal.

545 Additional Homework These questions are a continuation of the problem listed above (Note they are still independent questions): 1. Determine the solution of the linear program obtained by adding the constraint 5x1 + 7x2 + 9x3 ≤ 50. For problems where new constraints are added we can add the as new rows to the given optimal tableau. Update the tableau and do needed pivots in order to obtain optimality. Thus, Row z 0 1 1 0 2 0 3 0 new 0

x1 0 1 0 0 5

x2 0 0 1 0 7

x3 0 0 0 1 9

s1 s2 -12 -4 1 -2 -4 9 19 -43 0 0 109

s3 -5 7 -30 144 0

s4 0 0 0 0 1

RHS -372 5 1 2 50

Basis z x1 x2 x3 s4

OR Notes Draft: May 5, 2012 Updating the tableau to accommodate the new constraint gives: Row z 0 1 1 0 2 0 3 0 new 0

x1 0 1 0 0 0

x2 0 0 1 0 0

x3 0 0 0 1 0

s1 -12 1 -4 19 -148

s2 -4 -2 9 -43 334

s3 -5 7 -30 144 -1121

s4 0 0 0 0 1

RHS -372 5 1 2 0

Basis z x1 x2 x3 s4

Note that adding a new constraint allows us to add an additional variable to the constraints. Choosing xB = (x1 , x2 , x3 , s4 ), and xN = (s1 , s2 , s3 ) gives:     6 −13 −3 0 1 −2 7 0  6  11 2 0  9 −30 0   =⇒ B −1 =  −4  B=  1  5 1 0  19 −43 144 0  5 7 9 1 −148 334 −1121 1 and



 1 0 0  0 1 0   =⇒ −cN T + cB T B −1 N = −12 −4 −5 N =  0 0 1  0 0 0 It can be seen from either treatment that the objective function is not changed by the addition of the new constraint to the problem. 2. Determine the solution of the linear program obtained by adding the constraint x1 + x2 + x3 ≥ 10. For this problem we need to note that an excess variable is added to the constraint before it is placed in with the other constraints. We have two options to see how this constraint effects the problem. The easiest is to add the new constraint at the bottom of the given optimal tableau, and then update the tableau as needed to regain an optimal feasible solution. Row z 0 1 1 0 2 0 3 0 new 0

x1 0 1 0 0 1

x2 0 0 1 0 1

x3 0 0 0 1 1

s1 s2 -12 -4 1 -2 -4 9 19 -43 0 0

s3 -5 7 -30 144 0

e1 0 0 0 0 -1

RHS -372 5 1 2 10

Basis z x1 x2 x3 ?

Multiply the new constraint by a ‘-1’ in order to allow e1 to be basic. Row z 0 1 1 0 2 0 3 0 new 0

x1 0 1 0 0 -1

x2 0 0 1 0 -1

x3 0 0 0 1 -1

s1 s2 -12 -4 1 -2 -4 9 19 -43 0 0 110

s3 -5 7 -30 144 0

e1 0 0 0 0 1

RHS -372 5 1 2 -10

Basis z x1 x2 x3 e1

OR Notes Draft: May 5, 2012 Updating the basis columns. Row z 0 1 1 0 2 0 3 0 new 0

x1 0 1 0 0 0

x2 0 0 1 0 0

x3 0 0 0 1 0

s1 s2 s3 e1 -12 -4 -5 0 1 -2 7 0 -4 9 -30 0 19 -43 144 0 16 -36 121 1

RHS -372 5 1 2 -2

Basis z x1 x2 x3 e1

A pivot of the dual simplex method is needed. Thus, we pivot s2 into the basis in the place of e1 .

Row z 0 1 1 0 2 0 3 0 new 0

x1 0 1 0 0 0

x2 0 0 1 0 0

x3 0 0 0 1 0

s1 s2 -13.77777778 0 0.111111111 0 0 0 -0.111111111 0 -0.444444444 1

s3 e1 RHS -18.44444444 -0.111111111 -371.78 0.277777778 -0.055555556 5.111111111 0.25 0.25 0.5 -0.527777778 -1.194444444 4.388888889 -3.361111111 -0.027777778 0.055555556

Note here we have found an updated optimal solution to the linear program with values , x2 = 12 , x3 = 79 giving an objective value of z = −3346 . of x1 = 46 9 18 9 3. Determine the solution of the linear program obtained by adding the constraint x1 + x2 + x3 = 30. This problem is approached in a manner similar to the previous only we use a penalty of M in the objective function to insure that the added artificial variable doesn’t stay in the optimal basis. Considering the problem in matrix form and adding an artificial variable a1 into the base for the newly added constraint gives: xB T = (x1 , x2 , x3 , a1 )T and xN T = (s1 , s2 , s3 )T Thus, 

  6 −13 −3 0 1 −2 7  6   11 2 0  −4 9 −30 B= =⇒ B −1 =   1  19 −43 5 1 0  144 1 1 1 1 −16 36 −121 

1  0 N =  0 0

0 1 0 0

     0 −101 0  87  0   , cB =   , cN =  0   23  1  0 0 M 111

 0 0   0  1

Basis z x1 x2 x3 s2

OR Notes Draft: May 5, 2012 and using the matrix formulations given we have: −cN T + cB T B −1 N =

−16 M − 12 36 M − 4 −121 M − 5 T −1 22 M − 372 cB B b =   5  1   B −1 b =   2  22   1 −2 7  −4 9 −30   B −1 N =   19 −43 144  −16 36 −121

Note that it is now attractive to have s2 enter the basis. Thus, conducting a pivot of primal simplex gives 1 22 , min 9 36 for the ratio test, and we see that x2 will leave. xB T = (x1 , s2 , x3 , a1 )T and xN T = (s1 , x2 , s3 )T Thus, 

6 −13 −3  6 11 2 B=  1 5 1 1 1 1  1 −13 0  0 11 0 N =  0 5 1 0 1 0

  1  1 0 0 0 9 3 10  4  0   =⇒ B −1 =  − 19 1 − 32 0    0 0  −9 0 3 1 0 0 −1 1      −101 0    0   , cB =  , c =  87    23  N 0 M

and using the matrix formulations given we have: −cN T + cB T B −1 N =

−4 M + − 124 9 T −1 18 M − 3344 cB B b = 9  47  −1

 B b =  

9 1 9 61 9

4 9

  

18   B −1 N =  

1 9 − 49 − 19

2 9 1 9 43 9

0 −4

112

1 3 − 10 3 2 3

−1

   

−M −

55 3

OR Notes Draft: May 5, 2012 We note here that we have reached a point where no variable is attractive to enter into the basis; however, we still have a basic artificial variable. Thus, the solution to the problem is not feasible.

113


114

Bibliography [1] H. A. Eiselt and C.-L. Sandblom. Linear programming and its applications. Springer, Berlin, 2007. [2] Frederick S. Hillier and Gerald J. Lieberman. Introduction to operations research. HoldenDay Inc., Oakland, Calif., third edition, 1980. [3] D. Luenberger. Linear and nonlinear programming. Addison-Wesley, Reading, Mass., second edition, 1984. [4] David G. Luenberger and Yinyu Ye. Linear and nonlinear programming. International Series in Operations Research & Management Science, 116. Springer, New York, third edition, 2008. [5] S. Nash and A Sofer. Linear and nonlinear programming. McGraw-Hill, New York, 1996. [6] J. Reeb and S. Leavengood. Using the graphical method to solve linear programs. Oregon State University: Extension Service, Performance Excellence in the Wood Products Industry:1–24, 1998. [7] Hamdy A. Taha. Operations research. An introduction. The Macmillan Co., New York, 1971. [8] Wayne L. Winston. Operations research: applications and algorithms. Duxbury Press, Boston, MA, 1987.

115

Index complementary slackness, 56 dual simplex method, 58 elementary row operation, 8 sensitivity analysis, 61 simplex method, 5 strong duality, 50

116