1 1 © 2003 thomson /south-western slide slides prepared by john s. loucks st. edward’s...

1 1© 2003 Thomson© 2003 Thomson/South-Western/South-Western Slide Slide

Slides Prepared bySlides Prepared by

JOHN S. LOUCKSJOHN S. LOUCKS

St. Edward’s UniversitySt. Edward’s University


Chapter 8Chapter 8Integer Linear ProgrammingInteger Linear Programming

Types of Integer Linear Programming ModelsTypes of Integer Linear Programming Models Graphical and Computer Solutions for an All-Graphical and Computer Solutions for an All-

Integer Linear ProgramInteger Linear Program Applications Involving 0-1 VariablesApplications Involving 0-1 Variables Modeling Flexibility Provided by 0-1 VariablesModeling Flexibility Provided by 0-1 Variables


Types of Integer Programming ModelsTypes of Integer Programming Models

An LP in which all the variables are restricted An LP in which all the variables are restricted to be integers is called an to be integers is called an all-integer linear all-integer linear program program (ILP).(ILP).

The LP that results from dropping the integer The LP that results from dropping the integer requirements is called the requirements is called the LP RelaxationLP Relaxation of the of the ILP.ILP.

If only a subset of the variables are restricted If only a subset of the variables are restricted to be integers, the problem is called a to be integers, the problem is called a mixed-mixed-integer linear programinteger linear program (MILP). (MILP).

Binary variables are variables whose values Binary variables are variables whose values are restricted to be 0 or 1. If all variables are are restricted to be 0 or 1. If all variables are restricted to be 0 or 1, the problem is called a restricted to be 0 or 1, the problem is called a 0-1 or binary integer linear program0-1 or binary integer linear program. .


Example: All-Integer LPExample: All-Integer LP

Consider the following all-integer linear Consider the following all-integer linear program:program:

Max 3Max 3xx11 + 2 + 2xx22

s.t. 3s.t. 3xx11 + + xx22 << 9 9

xx11 + 3 + 3xx22 << 7 7

--xx11 + + xx22 << 1 1

xx11, , xx22 >> 0 and integer 0 and integer



LP RelaxationLP Relaxation

Solving the problem as a linear program Solving the problem as a linear program ignoring the integer constraints, the optimal ignoring the integer constraints, the optimal solution to the linear program gives fractional solution to the linear program gives fractional values for both values for both xx11 and and xx22. From the graph on . From the graph on the next slide, we see that the optimal solution the next slide, we see that the optimal solution to the linear program is:to the linear program is:

xx11 = 2.5, = 2.5, xx22 = 1.5, = 1.5, zz = 10.5 = 10.5



LP RelaxationLP Relaxation

LP Optimal (2.5, 1.5)LP Optimal (2.5, 1.5)

Max 3Max 3xx11 + 2 + 2xx22

--xx11 + + xx22 << 1 1

xx22

xx11

33xx11 + + xx22 << 9 9

11

33

22

55

44

1 2 3 4 5 6 71 2 3 4 5 6 7

xx11 + 3 + 3xx22 << 7 7



Rounding UpRounding Up

If we round up the fractional solution (If we round up the fractional solution (xx11 = 2.5, = 2.5, xx22 = 1.5) to the LP relaxation problem, = 1.5) to the LP relaxation problem, we get we get xx11 = 3 and = 3 and xx22 = 2. From the graph on = 2. From the graph on the next slide, we see that this point lies the next slide, we see that this point lies outside the feasible region, making this outside the feasible region, making this solution infeasible.solution infeasible.



Rounded Up SolutionRounded Up Solution

LP Optimal (2.5, 1.5)LP Optimal (2.5, 1.5)

Max 3Max 3xx11 + 2 + 2xx22

--xx11 + + xx22 << 1 1

xx22

xx11

33xx11 + + xx22 << 9 9

11

33

22

55

44

1 2 3 4 5 6 71 2 3 4 5 6 7

ILP Infeasible (3, 2)ILP Infeasible (3, 2)

xx11 + 3 + 3xx22 << 7 7



Rounding DownRounding Down

By rounding the optimal solution down to By rounding the optimal solution down to xx11 = 2, = 2, xx22 = 1, we see that this solution indeed = 1, we see that this solution indeed is an integer solution within the feasible is an integer solution within the feasible region, and substituting in the objective region, and substituting in the objective function, it gives function, it gives zz = 8. = 8.

We have found a feasible all-integer We have found a feasible all-integer solution, but have we found the OPTIMAL all-solution, but have we found the OPTIMAL all-integer solution?integer solution?

------------------------------------------

The answer is NO! The optimal solution The answer is NO! The optimal solution is is xx11 = 3 and = 3 and xx22 = 0 giving = 0 giving zz = 9, as evidenced = 9, as evidenced in the next two slides. in the next two slides.



Complete Enumeration of Feasible ILP SolutionsComplete Enumeration of Feasible ILP Solutions

There are eight feasible integer solutions There are eight feasible integer solutions to this problem:to this problem:

xx11 xx22 zz

1. 0 0 01. 0 0 0 2. 1 0 32. 1 0 3 3. 2 0 63. 2 0 6 4. 3 0 9 optimal 4. 3 0 9 optimal

solutionsolution 5. 0 1 25. 0 1 2 6. 1 1 56. 1 1 5 7. 2 1 87. 2 1 8

8. 1 2 78. 1 2 7



ILP Optimal (3, 0)ILP Optimal (3, 0)

Max 3Max 3xx11 + 2 + 2xx22

--xx11 + + xx22 << 1 1

xx22

xx11

33xx11 + + xx22 << 9 9

11

33

22

55

44

1 2 3 4 5 6 71 2 3 4 5 6 7

xx11 + 3 + 3xx22 << 7 7



Partial Spreadsheet Showing Problem DataPartial Spreadsheet Showing Problem Data

A B C D12 Constraint X1 X2 RHS

3 #1 3 1 9

4 #2 1 3 7

5 #3 -1 1 1

6 Obj.Coefficients 3 2

LHS Coefficients



Partial Spreadsheet Showing FormulasPartial Spreadsheet Showing Formulas

A B C D

8 X1 X2

9

10 =B6*C9+C6*D9

11

12 Constraints LHS RHS

13 #1 =B3*$C$9+C3*$D$9 <= 9

14 #2 =B4*$C$9+C4*$D$9 <= 7

15 #3 =B5*$C$9+C5*$D$9 <= 1

Maximized Objective Function

Optimal Decision Variable Values



Partial Spreadsheet Showing Optimal SolutionPartial Spreadsheet Showing Optimal Solution

A B C D

8 X1 X2

9 3.000 2.9419E-12

10 9.000

11

12 Constraints LHS RHS

13 #1 9 <= 9

14 #2 3 <= 7

15 #3 -3 <= 1

Maximized Objective Function

Optimal Decision Variable Values


Special 0-1 ConstraintsSpecial 0-1 Constraints

When When xxii and and xxjj represent binary variables represent binary variables designating whether projects designating whether projects ii and and jj have been have been completed, the following special constraints may be completed, the following special constraints may be formulated:formulated:

•At most At most kk out of out of nn projects will be completed: projects will be completed: xxjj << kk jj

•Project Project jj is is conditionalconditional on project on project ii: :

xxjj - - xxii << 0 0

•Project Project ii is a is a corequisitecorequisite for project for project jj: :

xxjj - - xxii = 0 = 0

•Projects Projects ii and and jj are are mutually exclusivemutually exclusive: :

xxii + + xxjj << 1 1


Example: Metropolitan MicrowavesExample: Metropolitan Microwaves

Metropolitan Microwaves, Inc. is planning Metropolitan Microwaves, Inc. is planning to expand its operations into other electronic to expand its operations into other electronic appliances. The company has identified seven appliances. The company has identified seven new product lines it can carry. Relevant new product lines it can carry. Relevant information about each line follows on the next information about each line follows on the next slide. slide.



Initial Floor Space Exp. Rate Initial Floor Space Exp. Rate

Product Line Product Line Invest. (Sq.Ft.) of Invest. (Sq.Ft.) of ReturnReturn

1. Black & White TVs1. Black & White TVs $ 6,000 125$ 6,000 125 8.1% 8.1%2. Color TVs 2. Color TVs 12,000 150 12,000 150

9.0 9.0 3. Large Screen TVs 3. Large Screen TVs 20,000 200 20,000 200 11.0 11.0 4. VHS VCRs 4. VHS VCRs 14,000 40 14,000 40 10.2 10.2 5. Beta VCRs 5. Beta VCRs 15,000 40 15,000 40

10.5 10.5 6. Video Games 6. Video Games 2,000 20 2,000 20

14.1 14.1 7. Home Computers 7. Home Computers 32,000 100 32,000 100 13.2 13.2



Metropolitan has decided that they should Metropolitan has decided that they should not stock large screen TVs unless they stock not stock large screen TVs unless they stock either B&W or color TVs. Also, they will not stock either B&W or color TVs. Also, they will not stock both types of VCRs, and they will stock video both types of VCRs, and they will stock video games if they stock color TVs. Finally, the games if they stock color TVs. Finally, the company wishes to introduce at least three new company wishes to introduce at least three new product lines. product lines.

If the company has $45,000 to invest and If the company has $45,000 to invest and 420 sq. ft. of floor space available, formulate an 420 sq. ft. of floor space available, formulate an integer linear program for Metropolitan to integer linear program for Metropolitan to maximize its overall expected rate of return.maximize its overall expected rate of return.



Define the Decision VariablesDefine the Decision Variables

xxjj = 1 if product line = 1 if product line jj is introduced; is introduced;

= 0 otherwise.= 0 otherwise.

Define the Objective FunctionDefine the Objective Function

Maximize total overall expected return:Maximize total overall expected return:

Max .081(6000)Max .081(6000)xx11 + .09(12000) + .09(12000)xx22 + .11(20000)+ .11(20000)xx33

+ .102(14000)+ .102(14000)xx4 4 + .105(15000)+ .105(15000)xx55 + .141(2000)+ .141(2000)xx66

+ .132(32000)+ .132(32000)xx77



Define the ConstraintsDefine the Constraints

1) Money: 1) Money:

66xx11 + 12 + 12xx22 + 20 + 20xx33 + 14 + 14xx44 + 15 + 15xx55 + 2 + 2xx66 + + 3232xx77 << 45 45

2) Space: 2) Space:

125125xx11 +150 +150xx22 +200 +200xx33 +40 +40xx44 +40 +40xx55 +20 +20xx66 +100+100xx77 << 420 420

3) Stock large screen TVs only if stock B&W or 3) Stock large screen TVs only if stock B&W or color:color:

xx11 + + xx22 > > xx33 or or xx11 + + xx22 - - xx33 >> 0 0



Define the Constraints (continued)Define the Constraints (continued)

4) Do not stock both types of VCRs: 4) Do not stock both types of VCRs:

xx44 + + xx55 << 1 1

5) Stock video games if they stock color TV's: 5) Stock video games if they stock color TV's:

xx22 - - xx66 >> 0 0

6) At least 3 new lines:6) At least 3 new lines:

xx11 + + xx22 + + xx33 + + xx44 + + xx55 + + xx66 + + xx77 >> 3 3

7) Variables are 0 or 1: 7) Variables are 0 or 1:

xxjj = 0 or 1 for = 0 or 1 for jj = 1, , , 7 = 1, , , 7



Partial Spreadsheet Showing Problem DataPartial Spreadsheet Showing Problem Data

A B C D E F G H I

1

2 Constraints X1 X2 X3 X4 X5 X6 X7 RHS

3 #1 6 12 20 14 15 2 32 45

4 #2 125 150 200 40 40 20 100 420

5 #3 1 1 -1 0

6 #4 1 1 1

7 #5 1 -1 0

8 #6 1 1 1 1 1 1 1 3

9 Obj.Func.Coeff. 486 1080 2200 1428 1575 282 4224

LHS Coefficients



Partial Spreadsheet Showing Example FormulasPartial Spreadsheet Showing Example Formulas

A B C D E F G H

12 X1 X2 X3 X4 X5 X6 X7

13 Dec.Values 0 0 0 0 0 0 0

14 Maximized Total Expected Return #

15

16 LHS RHS

17 0 <= 45

18 0 <= 420

19 0 >= 0

20 0 <= 1

21 ## >= 0

22 0 >= 3

Video

Lines

VCRs

TVs

Constraints

Money

Space



Solver Parameters Dialog BoxSolver Parameters Dialog Box

Solver ParametersSolver Parameters

Set Target Cell:Set Target Cell:

Equal To:Equal To:

By Changing Cells:By Changing Cells:

Reset AllReset All

X?

HelpHelp

CloseClose

SolveSolve

OptionsOptions

MMaxax MiMinn VValue of:alue of:

$B$13:$H$13$B$13:$H$13

Subject to the Constraints:Subject to the Constraints:

DeleteDelete

ChangeChange

GuessGuess

AddAdd$B$13:$H$13 = binary$D$17:$D$18 <= $H$17:$H$18$D$19 >= $H$19$D$20 <= $H$20

$B$13:$H$13 = binary$D$17:$D$18 <= $H$17:$H$18$D$19 >= $H$19$D$20 <= $H$20



Solver Options Dialog BoxSolver Options Dialog Box

•Select Select Assume Linear ModelAssume Linear Model

•Select Select Assume Non-negativeAssume Non-negative

•Set Set ToleranceTolerance equal to 0 equal to 0


Example: Tom’s TailoringExample: Tom’s Tailoring

Tom's Tailoring has five idle tailors and four Tom's Tailoring has five idle tailors and four custom garments to make. The estimated time (in custom garments to make. The estimated time (in hours) it would take each tailor to make each hours) it would take each tailor to make each garment is shown in the next slide. (An 'X' in the garment is shown in the next slide. (An 'X' in the table indicates an unacceptable tailor-garment table indicates an unacceptable tailor-garment assignment.)assignment.)

TailorTailor

GarmentGarment 11 22 33 44 55

Wedding gown 19 23 20 21 18Wedding gown 19 23 20 21 18 Clown costume 11 14 X 12 10Clown costume 11 14 X 12 10 Admiral's uniform 12 8 11 X 9Admiral's uniform 12 8 11 X 9 Bullfighter's outfit X 20 20 18 21Bullfighter's outfit X 20 20 18 21



Formulate an integer program for Formulate an integer program for determining the tailor-garment assignments that determining the tailor-garment assignments that minimize the total estimated time spent making minimize the total estimated time spent making the four garments. No tailor is to be assigned the four garments. No tailor is to be assigned more than one garment and each garment is to more than one garment and each garment is to be worked on by only one tailor.be worked on by only one tailor.

----------------------------------------

This problem can be formulated as a 0-1 This problem can be formulated as a 0-1 integer program. The LP solution to this problem integer program. The LP solution to this problem will automatically be integer (0-1).will automatically be integer (0-1).



Define the decision variablesDefine the decision variables

xxijij = 1 if garment i is assigned to tailor = 1 if garment i is assigned to tailor jj

= 0 otherwise.= 0 otherwise.

Number of decision variables = Number of decision variables =

[(number of garments)(number of tailors)] [(number of garments)(number of tailors)]

- (number of unacceptable assignments) - (number of unacceptable assignments)

= [4(5)] - 3 = 17= [4(5)] - 3 = 17



Define the objective functionDefine the objective function

Minimize total time spent making garments:Minimize total time spent making garments:

Min 19Min 19xx1111 + 23 + 23xx1212 + 20 + 20xx1313 + 21 + 21xx1414 + 18 + 18xx1515 + + 1111xx2121

+ 14+ 14xx2222 + 12 + 12xx2424 + 10 + 10xx2525 + 12 + 12xx3131 + 8 + 8xx3232 + + 1111xx3333

+ 9x+ 9x3535 + 20 + 20xx4242 + 20 + 20xx4343 + 18 + 18xx4444 + 21 + 21xx4545



Define the ConstraintsDefine the Constraints

Exactly one tailor per garment:Exactly one tailor per garment:

1) 1) xx1111 + + xx1212 + + xx1313 + + xx1414 + + xx1515 = 1 = 1

2) 2) xx2121 + + xx2222 + + xx2424 + + xx2525 = 1 = 1

3) 3) xx3131 + + xx3232 + + xx3333 + + xx3535 = 1 = 1

4) 4) xx4242 + + xx4343 + + xx4444 + + xx4545 = 1 = 1



Define the Constraints (continued)Define the Constraints (continued)

No more than one garment per tailor:No more than one garment per tailor:

5) 5) xx1111 + + xx2121 + + xx3131 << 1 1

6) 6) xx1212 + + xx2222 + + xx3232 + + xx4242 << 1 1

7) 7) xx1313 + + xx3333 + + xx4343 << 1 1

8) 8) xx1414 + + xx2424 + + xx4444 << 1 1

9) 9) xx1515 + + xx2525 + + xx3535 + + xx4545 << 1 1

Nonnegativity: Nonnegativity: xxijij >> 0 for 0 for ii = 1, . . ,4 and = 1, . . ,4 and jj = = 1, . . ,5 1, . . ,5


End of Chapter 8End of Chapter 8

1 1 © 2003 thomson /south-western slide slides prepared by john s. loucks st. edward’s...

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