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Quantitative business analysis text and cases

25/11/2021 Client: muhammad11 Deadline: 2 Day

Linear Programming Models: Graphical and Computer Methods

To accompany Quantitative Analysis for Management, Twelfth Edition,

by Render, Stair, Hanna and Hale

Power Point slides created by Jeff Heyl

After completing this chapter, students will be able to:

LEARNING OBJECTIVES

7 – 2

Understand the basic assumptions and properties of linear programming (LP).

Graphically solve any LP problem that has only two variables by both the corner point and isoprofit line methods.

Understand special issues in LP such as infeasibility, unboundedness, redundancy, and alternative optimal solutions.

Understand the role of sensitivity analysis.

Use Excel spreadsheets to solve LP problems.

7 – 3

7.1 Introduction

7.2 Requirements of a Linear Programming Problem

7.3 Formulating LP Problems

7.4 Graphical Solution to an LP Problem

7.5 Solving Flair Furniture’s LP Problem using QM for Windows, Excel 2013, and Excel QM

7.6 Solving Minimization Problems

7.7 Four Special Cases in LP

7.8 Sensitivity Analysis

CHAPTER OUTLINE

Introduction

Many management decisions involve making the most effective use of limited resources

Linear programming (LP)

Widely used mathematical modeling technique

Planning and decision making relative to resource allocation

Broader field of mathematical programming

Here programming refers to modeling and solving a problem mathematically

7 – 4

Requirements of a Linear Programming Problem

Four properties in common

Seek to maximize or minimize some quantity (the objective function)

Restrictions or constraints are present

Alternative courses of action are available

Linear equations or inequalities

7 – 5

LP Properties and Assumptions

PROPERTIES OF LINEAR PROGRAMS
1. One objective function
2. One or more constraints
3. Alternative courses of action
4. Objective function and constraints are linear – proportionality and divisibility
5. Certainty
6. Divisibility
7. Nonnegative variables
TABLE 7.1

7 – 6

Formulating LP Problems

Developing a mathematical model to represent the managerial problem

Steps in formulating a LP problem

Completely understand the managerial problem being faced

Identify the objective and the constraints

Define the decision variables

Use the decision variables to write mathematical expressions for the objective function and the constraints

7 – 7

Formulating LP Problems

Common LP application – product mix problem

Two or more products are produced using limited resources

Maximize profit based on the profit contribution per unit of each product

Determine how many units of each product to produce

7 – 8

Flair Furniture Company

Flair Furniture produces inexpensive tables and chairs

Processes are similar, both require carpentry work and painting and varnishing

Each table takes 4 hours of carpentry and 2 hours of painting and varnishing

Each chair requires 3 of carpentry and 1 hour of painting and varnishing

There are 240 hours of carpentry time available and 100 hours of painting and varnishing

Each table yields a profit of $70 and each chair a profit of $50

7 – 9

Flair Furniture Company

The company wants to determine the best combination of tables and chairs to produce to reach the maximum profit

HOURS REQUIRED TO PRODUCE 1 UNIT
DEPARTMENT (T) TABLES (C) CHAIRS AVAILABLE HOURS THIS WEEK
Carpentry 4 3 240
Painting and varnishing 2 1 100
Profit per unit $70 $50
TABLE 7.2

7 – 10

Flair Furniture Company

The objective is

Maximize profit

The constraints are

The hours of carpentry time used cannot exceed 240 hours per week

The hours of painting and varnishing time used cannot exceed 100 hours per week

The decision variables are

T = number of tables to be produced per week

C = number of chairs to be produced per week

7 – 11

Flair Furniture Company

Create objective function in terms of T and C

Maximize profit = $70T + $50C

Develop mathematical relationships for the two constraints

For carpentry, total time used is

(4 hours per table)(Number of tables produced) + (3 hours per chair)(Number of chairs produced)

First constraint is

Carpentry time used ≤ Carpentry time available

4T + 3C ≤ 240 (hours of carpentry time)

7 – 12

Flair Furniture Company

Similarly

Painting and varnishing time used ≤ Painting and varnishing time available

2 T + 1C ≤ 100 (hours of painting and varnishing time)

This means that each table produced requires two hours of painting and varnishing time

Both of these constraints restrict production capacity and affect total profit

7 – 13

Flair Furniture Company

The values for T and C must be nonnegative

T ≥ 0 (number of tables produced is greater than or equal to 0)

C ≥ 0 (number of chairs produced is greater than or equal to 0)

The complete problem stated mathematically

Maximize profit = $70T + $50C

subject to

4T + 3C ≤ 240 (carpentry constraint)

2T + 1C ≤ 100 (painting and varnishing constraint)

T, C ≥ 0 (nonnegativity constraint)

7 – 14

Graphical Solution to an LP Problem

Easiest way to solve a small LP problems is graphically

Only works when there are just two decision variables

Not possible to plot a solution for more than two variables

Provides valuable insight into how other approaches work

Nonnegativity constraints mean that we are always working in the first (or northeast) quadrant of a graph

7 – 15

Graphical Representation of Constraints

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100

Number of Chairs

Number of Tables

This Axis Represents the Constraint T ≥ 0

This Axis Represents the Constraint C ≥ 0

FIGURE 7.1 – Quadrant Containing All Positive Values

7 – 16

Graphical Representation of Constraints

The first step is to identify a set or region of feasible solutions

Plot each constraint equation on a graph

Graph the equality portion of the constraint equations

4T + 3C = 240

Solve for the axis intercepts and draw the line

7 – 17

Graphical Representation of Constraints

When Flair produces no tables, the carpentry constraint is:

4(0) + 3C = 240

3C = 240

C = 80

Similarly for no chairs:

4T + 3(0) = 240

4T = 240

T = 60

This line is shown on the following graph

7 – 18

Graphical Representation of Constraints

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100

Number of Chairs

Number of Tables

(T = 0, C = 80)

FIGURE 7.2 – Graph of Carpentry Constraint Equation

(T = 60, C = 0)

7 – 19

FIGURE 7.3 – Region that Satisfies the Carpentry Constraint

Graphical Representation of Constraints

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100

Number of Chairs

Number of Tables

Any point on or below the constraint plot will not violate the restriction

Any point above the plot will violate the restriction

(30, 40)

(30, 20)

(70, 40)

7 – 20

Graphical Representation of Constraints

The point (30, 40) lies on the line and exactly satisfies the constraint

4(30) + 3(40) = 240

The point (30, 20) lies below the line and satisfies the constraint

4(30) + 3(20) = 180

The point (70, 40) lies above the line and does not satisfy the constraint

4(70) + 3(40) = 400

7 – 21

Graphical Representation of Constraints

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100

Number of Chairs

Number of Tables

(T = 0, C = 100)

FIGURE 7.4 – Region that Satisfies the Painting and Varnishing Constraint

(T = 50, C = 0)

7 – 22

Graphical Representation of Constraints

To produce tables and chairs, both departments must be used

Find a solution that satisfies both constraints simultaneously

A new graph shows both constraint plots

The feasible region is where all constraints are satisfied

Any point inside this region is a feasible solution

Any point outside the region is an infeasible solution

7 – 23

Graphical Representation of Constraints

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100

Number of Chairs

Number of Tables

FIGURE 7.5 – Feasible Solution Region

Painting/Varnishing Constraint

Carpentry Constraint

Feasible Region

7 – 24

Graphical Representation of Constraints

For the point (30, 20)

Carpentry constraint 4T + 3C ≤ 240 hours available (4)(30) + (3)(20) = 180 hours used
Painting constraint 2T + 1C ≤ 100 hours available (2)(30) + (1)(20) = 80 hours used




Carpentry constraint 4T + 3C ≤ 240 hours available (4)(70) + (3)(40) = 400 hours used
Painting constraint 2T + 1C ≤ 100 hours available (2)(70) + (1)(40) = 180 hours used




For the point (70, 40)

7 – 25

Graphical Representation of Constraints

For the point (50, 5)

Carpentry constraint 4T + 3C ≤ 240 hours available (4)(50) + (3)(5) = 215 hours used
Painting constraint 2T + 1C ≤ 100 hours available (2)(50) + (1)(5) = 105 hours used




7 – 26

Isoprofit Line Solution Method

Find the optimal solution from the many possible solutions

Speediest method is to use the isoprofit line

Starting with a small possible profit value, graph the objective function

Move the objective function line in the direction of increasing profit while maintaining the slope

The last point it touches in the feasible region is the optimal solution

7 – 27

Isoprofit Line Solution Method

Choose a profit of $2,100

The objective function is

$2,100 = 70T + 50C

Solving for the axis intercepts, draw the graph

Obviously not the best possible solution

Further graphs can be created using larger profits

The further we move from the origin, the larger the profit

The highest profit ($4,100) will be generated when the isoprofit line passes through the point (30, 40)

7 – 28

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100

Number of Chairs

Number of Tables

$2,100 = $70T + $50C

Isoprofit Line Solution Method

(30, 0)

FIGURE 7.6 – Profit line of $2,100

7 – 29

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100

Number of Chairs

Number of Tables

FIGURE 7.7 – Four Isoprofit Lines

$2,100 = $70T + $50C

$2,800 = $70T + $50C

$3,500 = $70T + $50C

$4,200 = $70T + $50C

Isoprofit Line Solution Method

7 – 30

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100

Number of Chairs

Number of Tables

FIGURE 7.8 – Optimal Solution

$4,100 = $70T + $50C

Isoprofit Line Solution Method

Optimal Solution Point

(T = 30, C = 40)

Maximum Profit Line

7 – 31

Corner Point Solution Method

The corner point method for solving LP problems

Look at the profit at every corner point of the feasible region

Mathematical theory is that an optimal solution must lie at one of the corner points or extreme points

7 – 32

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100

Number of Chairs

Number of Tables

FIGURE 7.9 – Four Corner Points of the Feasible Region

Corner Point Solution Method

(0, 0)

(50, 0)

(0, 80)

(?)

7 – 33

Corner Point Solution Method

Solve for the intersection of the two constraint lines

Using the elimination method to solve simultaneous equations method, select a variable to be eliminated

Eliminate T by multiplying the second equation by –2 and add it to the first equation

– 2(2T + 1C = 100) = – 4T – 2C = –200

4T + 3C = 240 (carpentry)

– 4T – 2C = –200 (painting)

C = 40

7 – 34

Corner Point Solution Method

Substitute C = 40 into either equation to solve for T

4T + 3(40) = 240

4T + 120 = 240

4T = 120

T = 30

Thus the corner point is (30, 40)

NUMBER OF TABLES (T) NUMBER OF CHAIRS (C) PROFIT = $70T + $50C
0 0 $0
50 0 $3,500
0 80 $4,000
30 40 $4,100
TABLE 7.3 – Feasible Corner Points and Profits

7 – 35

Highest profit – Optimal Solution

Slack and Surplus

Slack is the amount of a resource that is not used

For a less-than-or-equal constraint

Slack = (Amount of resource available) – (Amount of resource used)

Flair decides to produce 20 tables and 25 chairs

4(20) + 3(25) = 155 (carpentry time used)

240 = (carpentry time available)

240 – 155 = 85 (Slack time in carpentry)

7 – 36

Slack is the amount of a resource that is not used

For a less-than-or-equal constraint

Slack = (Amount of resource available) – (Amount of resource used)

Slack and Surplus

Flair decides to produce 20 tables and 25 chairs

4(20) + 3(25) = 155 (carpentry time used)

240 = (carpentry time available)

240 – 155 = 85 (Slack time in carpentry)

At the optimal solution, slack is 0 as all 240 hours are used

7 – 37

Slack and Surplus

Surplus is used with a greater-than-or-equal-to constraint to indicate the amount by which the right-hand side of the constraint is exceeded

Surplus = (Actual amount) – (Minimum amount)

New constraint

T + C ≥ 42

If T = 20 and C = 25, then

20 + 25 = 45

Surplus = 45 – 42 = 3

7 – 38

Summaries of Graphical Solution Methods

ISOPROFIT METHOD
1. Graph all constraints and find the feasible region.
2. Select a specific profit (or cost) line and graph it to find the slope.
3. Move the objective function line in the direction of increasing profit (or decreasing cost) while maintaining the slope. The last point it touches in the feasible region is the optimal solution.
4. Find the values of the decision variables at this last point and compute the profit (or cost).
CORNER POINT METHOD
1. Graph all constraints and find the feasible region.
2. Find the corner points of the feasible reason.
3. Compute the profit (or cost) at each of the feasible corner points.
4. Select the corner point with the best value of the objective function found in Step 3. This is the optimal solution.
TABLE 7.4

7 – 39

Solving Flair Furniture’s LP Problem

Most organizations have access to software to solve big LP problems

There are differences between software implementations, the approach is basically the same

With experience with computerized LP algorithms, it is easy to adjust to minor changes

7 – 40

Using QM for Windows

Select the Linear Programming module

Specify the number of constraints (non-negativity is assumed)

Specify the number of decision variables

Specify whether the objective is to be maximized or minimized

For Flair Furniture there are two constraints, two decision variables, and the objective is to maximize profit

7 – 41

Using QM for Windows

PROGRAM 7.1A – QM for Windows Linear Programming Computer Input Screen

7 – 42

Using QM for Windows

PROGRAM 7.1B – QM for Windows Data Input

7 – 43

Using QM for Windows

PROGRAM 7.1C –

QM for Windows

Output and Graph

7 – 44

Using Excel’s Solver

The Solver tool in Excel can be used to find solutions to

LP problems

Integer programming problems

Noninteger programming problems

Solver is limited to 200 variables and, in some situations, 100 constraints

7 – 45

Using Solver

Recall the model for Flair Furniture is

Maximize profit = $70T + $50C

Subject to 4T + 3C ≤ 240

2T + 1C ≤ 100

To use Solver, it is necessary to enter data and formulas

7 – 46

Using Solver

Enter problem data

Variable names, coefficients for the objective function and constraints, RHS values for each constraint

Designate specific cells for the values of the decision variables

Write a formula to calculate the value of the objective function

Write a formula to compute the left-hand sides of each of the constraints

7 – 47

Using Solver

PROGRAM 7.2A – Excel Data Input

7 – 48

Using Solver

PROGRAM 7.2B – Formulas

7 – 49

Using Solver

PROGRAM 7.2C – Excel Spreadsheet

7 – 50

Using Solver

PROGRAM 7.2D – Starting Solver

7 – 51

Using Solver

PROGRAM 7.2E –

Solver Parameters

Dialog Box

7 – 52

Using Solver

PROGRAM 7.2F – Solver Add Constraint Dialog Box

7 – 53

Using Solver

PROGRAM 7.2G – Solver Results Dialog Box

7 – 54

Using Solver

PROGRAM 7.2H – Solution

7 – 55

Using Excel QM

PROGRAM 7.3A –

Excel QM in Excel 2013

7 – 56

Using Excel QM

PROGRAM 7.3B – Excel QM Input Data

7 – 57

Using Excel QM

PROGRAM 7.3C – Excel QM Output

7 – 58

Solving Minimization Problems

Many LP problems involve minimizing an objective such as cost

Minimization problems can be solved graphically

Set up the feasible solution region

Use either the corner point method or an isocost line approach

Find the values of the decision variables (e.g., X1 and X2) that yield the minimum cost

7 – 59

Holiday Meal Turkey Ranch

The Holiday Meal Turkey Ranch is considering buying two different brands of turkey feed and blending them to provide a good, low-cost diet for its turkeys

INGREDIENT COMPOSITION OF EACH POUND OF FEED (OZ.) MINIMUM MONTHLY REQUIREMENT PER TURKEY (OZ.)
BRAND 1 FEED BRAND 2 FEED
A 5 10 90
B 4 3 48
C 0.5 0 1.5
Cost per pound 2 cents 3 cents
TABLE 7.5 – Holiday Meal Turkey Ranch data

7 – 60

Minimize cost (in cents) = 2X1 + 3X2

subject to:

5X1 + 10X2 ≥ 90 ounces (ingredient A constraint)

4X1 + 3X2 ≥ 48 ounces (ingredient B constraint)

0.5X1 ≥ 1.5 ounces (ingredient C constraint)

X1 ≥ 0 (nonnegativity constraint)

X2 ≥ 0 (nonnegativity constraint)

Holiday Meal Turkey Ranch

X1 = number of pounds of brand 1 feed purchased

X2 = number of pounds of brand 2 feed purchased

Let

7 – 61

Holiday Meal Turkey Ranch

–

20 –

15 –

10 –

5 –

0 –

| | | | | |

5 10 15 20 25

Pounds of Brand 2

Pounds of Brand 1

Ingredient C Constraint

Ingredient B Constraint

Ingredient A Constraint

Feasible Region

FIGURE 7.10 – Feasible Region

7 – 62

Holiday Meal Turkey Ranch

Solve for the values of the three corner points

Point a is the intersection of ingredient constraints C and B

4X1 + 3X2 = 48

X1 = 3

Substituting 3 in the first equation, we find X2 = 12

Solving for point b we find X1 = 8.4 and X2 = 4.8

Solving for point c we find X1 = 18 and X2 = 0

7 – 63

Cost = 2X1 + 3X2

Cost at point a = 2(3) + 3(12) = 42

Cost at point b = 2(8.4) + 3(4.8) = 31.2

Cost at point c = 2(18) + 3(0) = 36

Holiday Meal Turkey Ranch

Substituting these values back into the objective function we find

The lowest cost solution is to purchase 8.4 pounds of brand 1 feed and 4.8 pounds of brand 2 feed for a total cost of 31.2 cents per turkey

7 – 64

Holiday Meal Turkey Ranch

Solving using an isocost line

Move the isocost line toward the lower left

The last point touched in the feasible region will be the optimal solution

7 – 65

Holiday Meal Turkey Ranch

–

20 –

15 –

10 –

5 –

0 –

| | | | | |

5 10 15 20 25

Pounds of Brand 2

Pounds of Brand 1

FIGURE 7.11 – Graphical Solution Using the Isocost Approach

Feasible Region

54¢ = 2X1 + 3X2 Isocost Line

Direction of Decreasing Cost

31.2¢ = 2X1 + 3X2

(X1 = 8.4, X2 = 4.8)

7 – 66

Holiday Meal Turkey Ranch

PROGRAM 7.4 – Solution in QM for Windows

7 – 67

Holiday Meal Turkey Ranch

PROGRAM 7.5A – Excel 2013 Solution

7 – 68

Holiday Meal Turkey Ranch

PROGRAM 7.5B – Excel 2013 Formulas

7 – 69

Four Special Cases in LP

Four special cases and difficulties arise at times when using the graphical approach

No feasible solution

Unboundedness

Redundancy

Alternate Optimal Solutions

7 – 70

No Feasible Solution

No solution to the problem that satisfies all the constraint equations

No feasible solution region exists

A common occurrence in the real world

Generally one or more constraints are relaxed until a solution is found

Consider the following three constraints

7 – 71

No Feasible Solution

8 –

–

6 –

–

4 –

–

2 –

–

0 –

| | | | | | | | | |

2 4 6 8

Region Satisfying First Two Constraints

FIGURE 7.12 – A problem with no feasible solution

Region Satisfying Third Constraint

7 – 72

Unboundedness

Sometimes a linear program will not have a finite solution

In a maximization problem

One or more solution variables, and the profit, can be made infinitely large without violating any constraints

In a graphical solution, the feasible region will be open ended

Usually means the problem has been formulated improperly

7 – 73

Unboundedness

15 –

10 –

5 –

0 –

| | | | |

5 10 15

Feasible Region

X1 ≥ 5

X2 ≤ 10

X1 + 2X2 ≥ 10

FIGURE 7.13 – A Feasible Region That Is Unbounded to the Right

7 – 74

Redundancy

A redundant constraint is one that does not affect the feasible solution region

One or more constraints may be binding

This is a very common occurrence in the real world

Causes no particular problems, but eliminating redundant constraints simplifies the model

7 – 75

Redundancy

Maximize profit = $1X1 + $2X2
subject to X1 + X2 ≤ 20
2X1 + X2 ≤ 30
X1 ≤ 25
X1, X2 ≥ 0
30 –

25 –

20 –

15 –

10 –

5 –

0 –

| | | | | |

5 10 15 20 25 30

FIGURE 7.14 – Problem with a Redundant Constraint

Redundant Constraint

Feasible Region

X1 ≤ 25

2X1 + X2 ≤ 30

X1 + X2 ≤ 20

7 – 76

Alternate Optimal Solutions

Occasionally two or more optimal solutions may exist

Graphically this occurs when the objective function’s isoprofit or isocost line runs perfectly parallel to one of the constraints

Allows management great flexibility in deciding which combination to select as the profit is the same at each alternate solution

7 – 77

Alternate Optimal Solutions

Maximize profit = $3X1 + $2X2
subject to 6X1 + 4X2 ≤ 24
X1 ≤ 3
X1, X2 ≥ 0
8 –

7 –

6 –

5 –

4 –

3 –

2 –

1 –

0 –

| | | | | | | |

1 2 3 4 5 6 7 8

FIGURE 7.15 – Example of Alternate Optimal Solutions

Feasible Region

Isoprofit Line for $8

Optimal Solution Consists of All Combinations of X1 and X2 Along the AB Segment

Isoprofit Line for $12 Overlays Line Segment AB

7 – 78

Sensitivity Analysis

Optimal solutions to LP problems thus far have been found under deterministic assumptions

We assume complete certainty in the data and relationships of a problem

Real world conditions are dynamic

Analyze how sensitive a deterministic solution is to changes in the assumptions of the model

This is called sensitivity analysis, postoptimality analysis, parametric programming, or optimality analysis

7 – 79

Sensitivity Analysis

Involves a series of what-if? questions concerning constraints, variable coefficients, and the objective function

Trial-and-error method

Values are changed and the entire model is resolved

Preferred way is to use an analytic postoptimality analysis

After a problem has been solved, we determine a range of changes in problem parameters that will not affect the optimal solution or change the variables in the solution

7 – 80

High Note Sound Company

The company manufactures quality speakers and stereo receivers

Products require a certain amount of skilled artisanship which is in limited supply

Product mix LP model

Maximize profit = $50X1 + $120X2
subject to 2X1 + 4X2 ≤ 80 (hours of electricians’ time available)
3X1 + 1X2 ≤ 60 (hours of audio technicians’ time available)
X1, X2 ≥ 0
Copyright ©2015 Pearson Education, Inc.

7 – 81

High Note Sound Company

b = (16, 12)

Optimal Solution at Point a

X1 = 0 Speakers

X2 = 20 Receivers

Profits = $2,400

a = (0, 20)

Isoprofit Line: $2,400 = 50X1 + 120X2

60 –

–

40 –

–

20 –

10 –

0 –

| | | | | |

10 20 30 40 50 60

(Receivers)

(Speakers)

c = (20, 0)

FIGURE 7.16 – The High Note Sound Company Graphical Solution

7 – 82

High Note Sound Company

Electrician hours used are

2X1 + 4X2 = 2(0) + 4(20) = 80

All hours are utilized so slack = 0

Additional units of a binding constraint will generally increase profits

Technician hours used are

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