integer programming, goal programming, and nonlinear programming

Chapter 11

Integer Programming,

Goal Programming,

and Nonlinear

Programming. Prepared By: Salah A. Skaik

In this Chapter:

• Integer Programming.

•Goal Programming.

•Nonlinear Programming.

Integer Programming

IP is the extension of LP that solves problems requiring Integer Solutions.

Ex. (Airline)

There are two ways to solve IP Problems:

1- Graphically.

2- The Branch & Bound Method.

Goal Programming

GP is the extension of LP that permits

Multiple Objectives to be stated.

Ex. (Max profit & Max market share).

Nonlinear Programming

NLP is the case in which Objectives

or Constraints are Nonlinear.

Ex. (Maximizing profit =

25X1 - 0.4X²1 + 30X2 – 0.5X²2 ).

Ex. (12X1 - 0.6X²1 ≥ 3,500 ).

Integer Programming

• There are three types of Integer

Programs:

1- Pure Integer Programs.

2- Mixed-Integer Programs.

3- Zero-One Integer Programming.

(special cases).

Example of Integer Programming

Harrison Electric Company

Chandeliers

2 hrs to wire

6 hrs to assembly

Ceiling Fans

3 hrs to wire

5 hrs to assembly

• 12 hrs of wiring

• 30 hrs of assembly Production Capability

• 7$ each chandelier

• 6$ each fan Net Profit

X1 = Number of chandeliers produced.

X2 = Number of ceiling fans produced.

Objective Function:

Maximize Profit = $7 X1 + $6 X2

Subject to:

2X1 + 3X2 ≤ 12 (wiring hours)

6X1 + 5X2 ≤ 30 (assembly hours)

0 – | | | | | | |

1 2 3 4 5 6 X1

+ + + +

6X1 + 5X2 ≤ 30

2X1 + 3X2 ≤ 12

+ = Possible Integer Solution

Optimal LP Solution

(X1 =3.75, X2 = 1.5,

Profit = $35.25)

In case the company doesn’t produce a

fraction of the product:

1- Rounding Off.

2- Enumeration Method.

Rounding Off has two problems:

1- Integer solution may not be in feasible

region. (X1=4, X2=2). (unpractical solution)

2- May not be the optimal feasible Integer

solution. (X1=4, X2=1).

Enumeration Method

Optimal solution to integer programming problem

Solution if rounding is used

CHANDELIERS (X1) CEILING FANS (X2) PROFIT ($7X1 + $6X2)

0 0 $0

2 0 14

3 0 21

4 0 28

5 0 35

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

An integer solution can never be better

than the LP solution and is usually a lesser

solution.

Six Steps in Solving IP

Maximization Problems by

Branch and Bound

Branch-and-Bound Method

Step(1):

Solve the original problem using LP. If the

answer satisfies the integer constraints, we

are done. If not, this value provides an

initial upper bound.

Step(2):

Find any feasible solution that meets the

integer constraints for use as a lower

bound. Usually, rounding down each

variable will accomplish this.

Step(3):

Branch on one variable from step 1 that does

not have an integer value. Split the

problem into two sub-problems based on

integer values that are immediately above

or below the non-Integer value.

Step(4):

Create nodes at the top of these new

branches by solving the new problem.

Step(5-a):

If a branch yields a solution to the LP

problem that is not feasible, terminate

the branch.

Step(5-b):

If a branch yields a solution to the LP

problem that is feasible, but not an integer

solution, go to step 6.

Step(5-c):

If the branch yields a feasible integer solution, examine the value of the objective function. If this value equals the upper bound, an optimal solution has been reached. If it not equal to the upper bound, but exceeds the lower bound, set it as the new lower bound and go to step 6. finally, if it’s less than the lower bound terminate this branch.

Step(6):

Examine both branches again and set the

upper bound equal to the maximum value

of the objective function at all final nodes. If

the upper bound equals the lower bound,

stop. If not, go back to step 3.

Minimization problems involved reversing the

roles of the upper and lower bounds.

Branch-and-Bound Method Harrison Electric Company Revisited

Recall that the Harrison Electric Company’s integer

programming formulation is

Maximize profit = $7X1 + $6X2

subject to 2X1 + 3X2 ≤ 12

6X1 + 5X2 ≤ 30

X1 = number of chandeliers produced

X2 = number of ceiling fans produced

And the optimal non-integer solution is

X1 = 3.75 chandeliers, X2 = 1.5 ceiling fans

profit = $35.25

Since X1 and X2 are not integers, this solution is not

valid.

The profit value of $35.25 will provide the initial upper

bound.

We can round down to X1 = 3, X2 = 1, profit = $27,

which provides a feasible lower bound.

The problem is now divided into two sub-problems.

6X1 + 5X2 ≤ 30

X1 ≥ 4

Subproblem A

6X1 + 5X2 ≤ 30

X1 ≤ 3

Subproblem B

If you solve both sub-problems graphically

[X1 = 4, X2 = 1.2, profit = $35.20] Sub-problem A’s optimal solution:

Sub-problem B’s optimal solution: [X1 = 3, X2 = 2, profit = $33.00]

We have completed steps 1 to 4 of the branch-and-bound method.

Harrison Electric’s first branching:

subproblems A and B

Subproblem A

Next Branch (C)

Next Branch (D)

Upper Bound = $35.25 Lower Bound = $27.00 (From

Rounding Down)

X1 = 4

X2 = 1.2

P = 35.20

X1 = 3

X2 = 2

P = 33.00

Stop This Branch Solution Is Integer, Feasible Provides New Lower Bound of $33.00

X1 = 3.75

X2 = 1.5

P = 35.25

Infeasible (Noninteger) Solution Upper Bound = $35.20 Lower Bound = $33.00

Subproblem B

6X1 + 5X2 ≤ 30

X1 ≥ 4

X2 ≥ 2

Subproblem C

6X1 + 5X2 ≤ 30

X1 ≥ 4

X2 ≤ 1

Subproblem D

Subproblem A has branched into two new subproblems, C

and D.

Subproblem C has no feasible solution because the all the

constraints can not be satisfied

We terminate this branch and do not consider this

solution

Subproblem D’s optimal solution is X1 = 4.17, X2 = 1,

profit = $35.16

This noninteger solution yields a new upper bound of

$35.16

6X1 + 5X2 ≤ 30

X1 ≥ 4

X1 ≤ 4

X2 ≤ 1

Subproblem E

6X1 + 5X2 ≤ 30

X1 ≥ 4

X1 ≥ 5

X2 ≤ 1

Subproblem D

Finally we create subproblems E and F

Optimal solution to E:

X1 = 4, X2 = 1, profit = $34

Optimal solution to F:

X1 = 5, X2 = 0, profit = $35

Subproblem F

X1 = 5

X2 = 0

P = 35.00

Subproblem E

X1 = 4

X2 = 1

P = 34.00

Harrison Electric’s full branch and bound solution

Feasible, Integer Solution

Optimal Solution

Upper Bound = $35.25

Lower Bound = $27.00

Subproblem C

No Feasible Solution Region

Subproblem D

X1 = 4.17

X2 = 1

P = 35.16

Subproblem A

X1 = 4

X2 = 1.2

P = 35.20

Subproblem B

X1 = 3

X2 = 2

P = 33.00

X1 = 3.75

X2 = 1.5

P = 35.25

Using Software To Solve Harrison Integer

Programming Problem

QM for Windows input screen with Harrison Electric data

Programming Problem

QM for Windows solution screen for Harrison Electric data

Programming Problem

QM for Windows iteration results screen for Harrison Electric data

Programming Problem

Using Excel’s Solver to formulate Harrison’s integer programming model

Programming Problem

Integer variables are specified with a drop-down menu in Solver

Programming Problem

Excel solution to the Harrison Electric integer programming model

Mixed-Integer Programming Problem

There are many situations in which some of

the variables are restricted to be integers

and some are not.

Bagwell wants to maximize profit

We let X = number of 50-pound bags of xyline

We let Y = number of pounds of hexall

This is a mixed-integer programming problem as Y is not

required to be an integer.

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

The model is

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 QM for Windows and Excel to solve Bagwell’s IP model

Excel formulation of Bagwell’s IP problem with Solver

Excel solution to the Bagwell Chemical problem

Modeling With 0-1 (Binary) Variables

We can 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

Capital Budgeting Example

A common capital budgeting problem is selecting 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 can not do them all

They want to maximize net present value of projects undertaken

The basic model is

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

Quemo Chemical Company information

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

The mathematical statement of the integer programming

problem becomes

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

The decision variables are

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

Solved with computer software, the optimal

solution is X1 = 1, X2 = 0, and X3 = 1 with an

objective function value of 57,000

This means that Quemo Chemical should fund

the catalytic converter and warehouse expansion

projects only

The net present value of these investments will

be $57,000

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

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 constrain 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

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 in

◦ Baytown, Texas

◦ Lake Charles, Louisiana

◦ Mobile, Alabama

Once the plant or plants are built, the company want to

have capacity to produce at least 38,000 units each year

Fixed and variable costs for Sitka Manufacturing

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

We can define the decision variables as

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

The integer programming formulation becomes

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

The optimal solution is

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

Objective function value = $1,757,000

Financial Investment Example

Numerous financial applications exist with 0-1 variables

Simkin, Simkin, and Steinberg specialize in recommending

oil stock portfolios for wealthy clients

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

Oil investment opportunities

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

Model 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)

All variables must be 0 or 1

Using Excel to Solve the Simkin Example

Solver input for Simkin’s 0-1 variables

Complete Solver input for Simkin’s 0-1 integer

programming problem

Excel solution to Simkin’s 0-1 integer programming

problem

Solved Problem 11-1

Consider the 0-1 integer programming problem that follows:

Maximize = 50X1 + 45X2 + 48X3

Subject to: 19X1 + 27X2 + 34X3 ≤ 80

22X1 + 13X2 + 12X3 ≤ 40

X1, X2, X3 must be either 0 or 1

Solved Problem 11-1

Additional constraints:

1- No more than two of the three variables can take on a

value equal to 1 in the solution.

2- Make sure that if X1 = 1, then X2 = 1 also.

Solved Problem 11-1

The model is

Maximize = 50X1 + 45X2 + 48X3

Subject to: 19X1 + 27X2 + 34X3 ≤ 80

22X1 + 13X2 + 12X3 ≤ 40

X1 + X2 + X3 ≤ 2

X1 - X2 ≤ 0

X1, X2, X3 must be either 0 or 1

Solved Problem 11-1

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