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Goal programming using excel solver

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

Integer Programming, Goal Programming, and Nonlinear Programming

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

10 – 2

Understand the difference between LP and integer programming.

Understand and solve the three types of integer programming problems.

Formulate and solve goal programming problems using Excel and QM for Windows.

Formulate nonlinear programming problems and solve using Excel.

10.1 Introduction

10.2 Integer Programming

10.3 Modeling with 0-1 (Binary) Variables

10.4 Goal Programming

10.5 Nonlinear Programming

CHAPTER OUTLINE

10 – 3

Introduction

There are other mathematical programming models that can be used when the assumptions of LP are not met

A large number of business problems require variables have integer values

Many business problems have multiple objectives

Goal programming permits multiple objectives

Nonlinear programming allows objectives and constraints to be nonlinear

10 – 4

Integer Programming

An integer programming model is one where one or more of the decision variables has to take on an integer value in the final solution

Three types of integer programming problems

Pure integer programming – all variables have integer values

Mixed-integer programming – some but not all of the variables will have integer values

Zero-one integer programming – special cases in which all the decision variables must have integer solution values of 0 or 1

10 – 5

An integer programming model is one where one or more of the decision variables has to take on an integer value in the final solution

Three types of integer programming problems

Pure integer programming – all variables have integer values

Mixed-integer programming – some but not all of the variables will have integer values

Zero-one integer programming – special cases in which all the decision variables must have integer solution values of 0 or 1

Integer Programming

10 – 6

Solving an integer programming problem is much more difficult than solving an LP problem

Solution time required may be excessive

Harrison Electric Company Example of Integer Programming

Company produces two products, old-fashioned chandeliers and ceiling fans

Both require a two-step production process involving wiring and assembly

It takes about 2 hours to wire each chandelier and 3 hours to wire a ceiling fan

Final assembly of the chandeliers and fans requires 6 and 5 hours, respectively

Only 12 hours of wiring time and 30 hours of assembly time are available

Each chandelier produced nets the firm $7 and each fan $6

10 – 7

Harrison Electric Company Example of Integer Programming

Production mix LP formulation

10 – 8

Maximize profit = $7X1 + $6X2

subject to 2X1 + 3X2 ≤ 12 (wiring hours)

6X1 + 5X2 ≤ 30 (assembly hours)

X1, X2 ≥ 0

where

X1 = number of chandeliers produced

X2 = number of ceiling fans produced

Harrison Electric Company Example of Integer Programming

10 – 9

6 –

5 –

4 –

3 –

2 –

1 –

| | | | | |

0 1 2 3 4 5 6

+ = Possible Integer Solution

6X1 + 5X2 ≤ 30

Optimal LP Solution

(X1= 3.75, X2 = 1.5, Profit = $35.25

2X1 + 3X2 ≤ 12

FIGURE 10.1 – Harrison Electric Problem

Harrison Electric Company Example of Integer Programming

Production planner recognizes this is an integer problem

First attempt at solving it is to round the values to X1 = 4 and X2 = 2

However, this is not feasible

Rounding X2 down to 1 gives a feasible solution, but it may not be optimal

This could be solved using the enumeration method

Generally not possible for large problems

10 – 10

Harrison Electric Company Example of Integer Programming

10 – 11

CHANDELIERS (X1) CEILING FANS (X2) PROFIT ($7X1 + $6X2)
0 0 $0
1 0 7
2 0 14
3 0 21
4 0 28
5 0 35
0 1 6
1 1 13
2 1 20
3 1 27
4 1 34
0 2 12
1 2 19
2 2 26
3 2 33
0 3 18
1 3 25
0 4 24
Optimal solution to integer programming problem

Solution if rounding is used

TABLE 10.1 – Integer Solutions to the Harrison Electric Company Problem

Harrison Electric Company Example of Integer Programming

10 – 12

Solution if rounding is used

TABLE 10.1 – Integer Solutions to the Harrison Electric Company Problem

The optimal integer solution is less than the optimal LP solution of $35.25

An integer solution can never be better than the LP solution and is usually a lesser value

Using Software

10 – 13

PROGRAM 10.1A – QM for Windows Input Screen for Harrison Electric Problem

Using Software

10 – 14

PROGRAM 10.1B – QM for Windows Solution Screen for Harrison Electric Problem

Using Software

10 – 15

PROGRAM 10.2 – Excel 2013 Solver Solution for Harrison Electric Problem

Using Software

10 – 16

PROGRAM 10.2 – Excel 2013 Solver Solution for Harrison Electric Problem

Solver Parameter Inputs and Selections Key Formulas
Set Objective: D5 By Changing cells: B4:C4 To: Max Subject to the Constraints: D8:D9 >= F8:F9 B4:C4 = integer Solving Method: Simplex LP R Make Variables Non-Negative Copy D5 to D8:D9
Mixed-Integer Programming Problem Example

Many situations in which only some of the variables are restricted to integers

Bagwell Chemical Company produces two industrial chemicals

Xyline must be produced in 50-pound bags

Hexall is sold by the pound and can be produced in any quantity

Both xyline and hexall are composed of three ingredients – A, B, and C

Bagwell sells xyline for $85 a bag and hexall for $1.50 per pound

10 – 17

Mixed-Integer Programming Problem Example

10 – 18

AMOUNT PER 50-POUND BAG OF XYLINE (LB) AMOUNT PER POUND OF HEXALL (LB) AMOUNT OF INGREDIENTS AVAILABLE
30 0.5 2,000 lb–ingredient A
18 0.4 800 lb–ingredient B
2 0.1 200 lb–ingredient C
Objective is to maximize profit

Mixed-Integer Programming Problem Example

Let X = number of 50-pound bags of xyline

Let Y = number of pounds of hexall

A mixed-integer programming problem as Y is not required to be an integer

10 – 19

Maximize profit = $85X + $1.50Y
subject to 30X + 0.5Y ≤ 2,000
18X + 0.4Y ≤ 800
2X + 0.1Y ≤ 200
X, Y ≥ 0 and X integer
Using Software

10 – 20

PROGRAM 10.3 – QM for Windows Solution for Bagwell Chemical Problem

Using Software

10 – 21

PROGRAM 10.4 – Excel 2013 Solver Solution for Bagwell Chemical Problem

Using Software

10 – 22

Solver Parameter Inputs and Selections Key Formulas
Set Objective: D5 By Changing cells: B4:C4 To: Max Subject to the Constraints: D8:D10 <= F8:F10 B4 = integer Solving Method: Simplex LP R Make Variables Non-Negative Copy D5 to D8:D10
PROGRAM 10.4 – Excel 2013 Solver Solution for Bagwell Chemical Problem

Modeling With 0-1 (Binary) Variables

Demonstrate how 0-1 variables can be used to model several diverse situations

Typically a 0-1 variable is assigned a value of 0 if a certain condition is not met and a 1 if the condition is met

This is also called a binary variable

10 – 23

Capital Budgeting Example

Common capital budgeting problem – select from a set of possible projects when budget limitations make it impossible to select them all

A 0-1 variable is defined for each project

Quemo Chemical Company is considering three possible improvement projects for its plant

A new catalytic converter

A new software program for controlling operations

Expanding the storage warehouse

It cannot do them all

10 – 24

Capital Budgeting Example

Objective is to maximize net present value of projects undertaken

subject to Total funds used in year 1 ≤ $20,000

Total funds used in year 2 ≤ $16,000

10 – 25

PROJECT NET PRESENT VALUE YEAR 1 YEAR 2
Catalytic Converter $25,000 $8,000 $7,000
Software $18,000 $6,000 $4,000
Warehouse expansion $32,000 $12,000 $8,000
Available funds $20,000 $16,000
TABLE 10.2 – Quemo Chemical Company Information

Capital Budgeting Example

Decision variables

10 – 26

X1 =

1 if catalytic converter project is funded

0 otherwise

X2 =

1 if software project is funded

0 otherwise

X3 =

1 if warehouse expansion project is funded

0 otherwise

Formulation
Maximize NPV = 25,000X1 + 18,000X2 + 32,000X3
subject to 8,000X1 + 6,000X2 + 12,000X3 ≤ 20,000
7,000X1 + 4,000X2 + 8,000X3 ≤ 16,000
X1, X2, X3 = 0 or 1
Using Software

10 – 27

PROGRAM 10.5 – Excel 2013 Solver Solution for Quemo Chemical Problem

Using Software

10 – 28

PROGRAM 10.5 – Excel 2013 Solver Solution for Quemo Chemical Problem

Optimal Solution

X1 = 1, X2 = 0, X3 = 1

Fund the catalytic converter and warehouse projects but not the software project

NPV = $57,000

Using Software

10 – 29

Solver Parameter Inputs and Selections Key Formulas
Set Objective: E5 By Changing cells: B4:D4 To: Max Subject to the Constraints: E8:E9 <= G8:G9 B4:D4 = binary Solving Method: Simplex LP R Make Variables Non-Negative Copy E5 to E8:E9
PROGRAM 10.5 – Excel 2013 Solver Solution for Quemo Chemical Problem

Limiting the Number of Alternatives Selected

One common use of 0-1 variables involves limiting the number of projects or items that are selected from a group

Suppose Quemo Chemical is required to select no more than two of the three projects regardless of the funds available

This would require adding a constraint

X1 + X2 + X3 ≤ 2

If they had to fund exactly two projects the constraint would be

X1 + X2 + X3 = 2

10 – 30

Dependent Selections

At times the selection of one project depends on the selection of another project

Suppose Quemo’s catalytic converter could only be purchased if the software was purchased

The following constraint would force this to occur

X1 ≤ X2 or X1 – X2 ≤ 0

If we wished for the catalytic converter and software projects to either both be selected or both not be selected, the constraint would be

X1 = X2 or X1 – X2 = 0

10 – 31

Fixed-Charge Problem Example

Often businesses are faced with decisions involving a fixed charge that will affect the cost of future operations

Sitka Manufacturing is planning to build at least one new plant and three cities are being considered

Baytown, Texas

Lake Charles, Louisiana

Mobile, Alabama

10 – 32

Fixed-Charge Problem Example

Constraints

Total production capacity at least 38,000 units each year

Number of units produced at the Baytown plant is 0 if the plant is not built and no more than 21,000 if the plant is built

Number of units produced at the Lake Charles plant is 0 if the plant is not built and no more than 20,000 if the plant is built

Number of units produced at the Mobile plant is 0 if the plant is not built and no more than 19,000 if the plant is built

10 – 33

Fixed-Charge Problem Example

10 – 34

SITE ANNUAL FIXED COST VARIABLE COST PER UNIT ANNUAL CAPACITY
Baytown, TX $340,000 $32 21,000
Lake Charles, LA $270,000 $33 20,000
Mobile, AL $290,000 $30 19,000
TABLE 10.3 – Fixed and Variable Costs for Sitka Manufacturing

Fixed-Charge Problem Example

Decision variables

10 – 35

X1 =

1 if factory is built in Baytown

0 otherwise

X2 =

1 factory is built in Lake Charles

0 otherwise

X3 =

1 if factory is built in Mobile

0 otherwise

X4 = number of units produced at Baytown plant

X5 = number of units produced at Lake Charles plant

X6 = number of units produced at Mobile plant

Fixed-Charge Problem Example

Formulation

10 – 36

Minimize cost = 340,000X1 + 270,000X2 + 290,000X3

+ 32X4 + 33X5 + 30X6

subject to X4 + X5 + X6 ≥ 38,000

X4 ≤ 21,000X1

X5 ≤ 20,000X2

X6 ≤ 19,000X3

X1, X2, X3 = 0 or 1

X4, X5, X6 ≥ 0 and integer

Using Software

10 – 37

PROGRAM 10.6 – Excel 2013 Solver Solution for Sitka Manufacturing Problem

Using Software

10 – 38

PROGRAM 10.6 – Excel 2013 Solver Solution for Sitka Manufacturing Problem

Optimal solution

X1 = 0, X2 = 1, X3 = 1, X4 = 0, X5 = 19,000, X6 = 19,000

Objective function value = $1,757,000

Using Software

10 – 39

Solver Parameter Inputs and Selections Key Formulas
Set Objective: H5 By Changing cells: B4:G4 To: Min Subject to the Constraints: H8 >= J8 H9:H11 <= J9:J11 B4:D4 = binary E4:G4 = integer Solving Method: Simplex LP R Make Variables Non-Negative Copy H5 to H8:H11
PROGRAM 10.6 – Excel 2013 Solver Solution for Sitka Manufacturing Problem

Financial Investment Example

Simkin, Simkin, and Steinberg specialize in recommending oil stock portfolios

One client has the following specifications

At least two Texas firms must be in the portfolio

No more than one investment can be made in a foreign oil company

One of the two California oil stocks must be purchased

The client has $3 million to invest and wants to buy large blocks of shares

10 – 40

Financial Investment Example

10 – 41

STOCK COMPANY NAME EXPECTED ANNUAL RETURN ($1,000s) COST FOR BLOCK OF SHARES ($1,000s)
1 Trans-Texas Oil 50 480
2 British Petroleum 80 540
3 Dutch Shell 90 680
4 Houston Drilling 120 1,000
5 Texas Petroleum 110 700
6 San Diego Oil 40 510
7 California Petro 75 900
TABLE 10.4 – Oil Investment Opportunities

Financial Investment Example

Formulation

Maximize return = 50X1 + 80X2 + 90X3 + 120X4 + 110X5 + 40X6 + 75X7

subject to

X1 + X4 + X5 ≥ 2 (Texas constraint)

X2 + X3 ≤ 1 (foreign oil constraint)

X6 + X7 = 1 (California constraint)

480X1 + 540X2 + 680X3 + 1,000X4 + 700X5 + 510X6 + 900X7 ≤ 3,000 ($3 million limit)

Xi = 0 or 1 for all i

10 – 42

Using Software

10 – 43

PROGRAM 10.7 – Excel 2013 Solver Solution for Financial Investment Problem

Using Software

10 – 44

Solver Parameter Inputs and Selections Key Formulas
Set Objective: I5 By Changing cells: B4:H4 To: Max Subject to the Constraints: I7 >= K7 I8 <= K8 I9 = K9 I10 <= K10 B4:H4 = binary Solving Method: Simplex LP R Make Variables Non-Negative Copy I5 to I7:I10
PROGRAM 10.7 – Excel 2013 Solver Solution for Financial Investment Problem

Goal Programming

Firms often have more than one goal

In linear and integer programming methods the objective function is measured in one dimension only

It is not possible for LP to have multiple goals unless they are all measured in the same units

Highly unusual situation

Goal programming developed to supplement LP

10 – 45

Goal Programming

Typically goals set by management can be achieved only at the expense of other goals

Establish a hierarchy of importance so that higher-priority goals are satisfied before lower-priority goals

Not always possible to satisfy every goal

Goal programming attempts to reach a satisfactory level of multiple objectives

May not optimize but have to satisfice

10 – 46

Goal Programming

Main difference is in the objective function

Goal programming tries to minimize the deviations between goals and what can be achieved given the constraints

Objective is to minimize deviational variables

10 – 47

Harrison Electric Company Revisited

Production mix LP formulation

10 – 48

Maximize profit = $7X1 + $6X2

subject to 2X1 + 3X2 ≤ 12 (wiring hours)

6X1 + 5X2 ≤ 30 (assembly hours)

X1, X2 ≥ 0

where

X1 = number of chandeliers produced

X2 = number of ceiling fans produced

Harrison Electric Company Revisited

Moving to a new location and maximizing profit is not a realistic objective

A profit level of $30 would be satisfactory during this period

The goal programming problem is to find the production mix that achieves this goal as closely as possible given the production time constraints

Define two deviational variables

d1– = underachievement of the profit target

d1+ = overachievement of the profit target

10 – 49

Harrison Electric Company Revisited

Single-goal programming formulation

10 – 50

Minimize under or

overachievement = d1– + d1+

of profit target

subject to

$7X1 + $6X2 + d1– – d1+ = $30 (profit goal constraint)

2X1 + 3X2 ≤ 12 (wiring hours)

6X1 + 5X2 ≤ 30 (assembly hours)

X1, X2, d1–, d1+ ≥ 0

Minimize under or

overachievement = d1– + d1+

of profit target

subject to

$7X1 + $6X2 + d1– – d1+ = $30 (profit goal constraint)

2X1 + 3X2 ≤ 12 (wiring hours)

6X1 + 5X2 ≤ 30 (assembly hours)

X1, X2, d1–, d1+ ≥ 0

Single-goal programming formulation

Harrison Electric Company Revisited

10 – 51

Analyze each goal to see if underachievement or overachievement of that goal is acceptable

If overachievement is acceptable, eliminate the appropriate d+ variable from the objective function

If underachievement is okay, the d– variable should be dropped

If a goal must be attained exactly, both d– and d+ must appear in the objective function

Extension to Equally Important Multiple Goals

Achieve several goals that are equal in priority

Goal 1: to produce a profit of $30 if possible during the production period

Goal 2: to fully utilize the available wiring department hours

Goal 3: to avoid overtime in the assembly department

Goal 4: to meet a contract requirement to produce at least seven ceiling fans

10 – 52

Extension to Equally Important Multiple Goals

The deviational variables can be defined as

d1– = underachievement of the profit target

d1+ = overachievement of the profit target

d2– = idle time in the wiring department (underutilization)

d2+ = overtime in the wiring department (overutilization)

d3– = idle time in the assembly department (underutilization)

d3+ = overtime in the assembly department (overutilization)

d4– = underachievement of the ceiling fan goal

d4+ = overachievement of the ceiling fan goal

10 – 53

Extension to Equally Important Multiple Goals

Management is unconcerned about d1+, d2+, d3–, and d4+ so these may be omitted from the objective function

New objective function and constraints

10 – 54

Minimize total deviation = d1– + d2– + d3+ + d4–

subject to

$7X1 + $6X2 + d1– – d1+ = $30 (profit constraint)

2X1 + 3X2 + d2– – d2+ = 12 (wiring hours constraint)

6X1 + 5X2 + d3– – d3+ = 30 (assembly hours constraint)

X2 + d4– – d4+ = 7 (ceiling fan constraint)

All Xi, di variables ≥ 0

Ranking Goals with Priority Levels

In most goal programming problems, one goal will be more important than another

Lower-order goals considered only after higher-order goals are met

Priorities (Pis) are assigned to each deviational variable

P1 is the most important goal

P2 the next most important

P3 the third, and so on

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