Integer Programming Pure, Mixed-Integer, Zero-One Models Strategic Resource Allocation and Planning MGMT E-5050 Applied Management Science for Decision

Integer ProgrammingInteger Programming

PurePure, , Mixed-IntegerMixed-Integer, , Zero-OneZero-One

ModelsModels StrategicResourceAllocation

andPlanning

MGMT E-5050

Applied Management Science for Decision Making, 2e © 2014 Pearson Learning Solutions Philip A. Vaccaro , PhD


One assumption of linear programming is that decision variables can take on fractional values such as X1 = 0.33 or X3 = 1.57 . Yet a large number of business problems can be solved only if variables have integer values.

When an airline decides how many planes to purchase, it cannot place an order for 5.38 aircraft ; it must order 4, 5, 6, or some other integer amount.

Applied Management Science for Decision Making, 2e © 2014 Pearson Learning Solutions


An integer programming model is one that has constraints andan objective function identical to that formulated by LP.

The only difference is that one or more of the decision variableshas to take on an integer value in the final solution.

There are three three types of integer programming problems:

1. PurePure IP problems are cases in which all variables are required to have integer values.

2. MixedMixed IP problems are cases in which some , but not all, of the decision variables are required to have integer values.

3. Zero-oneZero-one IP problems are special cases in which all the decision variables must have integer values of 0 or 1 .


Solving an IP problem is much more difficult .

The solution time may be excessive, even on a computer.

The most common algorithm here is the branch and bound method.

Harrison Electric CompanyHarrison Electric CompanyPURE INTEGER PROBLEMPURE INTEGER PROBLEM

The Harrison Electric Company produces two products: old-fashionedchandeliers and ceiling fans. Both products require a two-step processinvolving wiring and assembly.

It takes 2 hours to wire each chandelier and 3 hours to wire a ceilingfan.

Final assembly of the chandeliers and fans requires 6 and 5 hours, re-spectively.

The production capability is such that only 12 hours of wiring time and30 hours of assembly time are available.

If each chandelier produced nets the firm $7.00 and each fan $6.00, theProduction mix decision can be formulated using LP as follows:

Harrison Electric CompanyHarrison Electric Company

Maximize profit = $7.00 X1 + $6.00 X2

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

The Model


0 1 2 3 4 5 6

6

5

4

3

2

1

0

X2

X1

+ +

+ +

+

+ +

+

6X6X1 1 + 5X+ 5X22 =< 30 ( assembly hours ) =< 30 ( assembly hours )

2X2X11 + 3X + 3X22 =< 12 ( wiring hours ) =< 12 ( wiring hours )

Optimal LP SolutionOptimal LP Solution( X( X11 = 3.75 , X = 3.75 , X22 = 1.5 , Profit = $35.25 ) = 1.5 , Profit = $35.25 )

+ = Possible Integer Solution


The optimal solution is X1 = 3.75 chandeliers and X2 = 1.5 ceiling fans.

Rounding to X1 = 4 and X2 = 2 makes the solution unfeasible.

Rounding to X1 = 4 and X2 = 2 is probably not the optimal feasible integer solution either .

There are 18 feasible integer solutions to this problem.

The optimal integer solution is X1 = 5 and X2 = 0 , with a total profit of $35.00 .

The integer restriction reduced profit from $35.25 to $35.00

An integer solution can never produce a greater profit than the LP solution to the same problem.

DISCUSSION

Harrison Electric CompanyHarrison Electric CompanyListing all feasible solutions and selecting the one with the best objective

function value is called the enumeration method. This can be virtuallyimpossible for large problems where the number of feasible solutions is

extremely large !

Chandeliers ( X1 ) Ceiling Fans ( X2 ) Profit ( Z )

0 0 $0.00

1 0 7

2 0 14

3 0 21

4 0 28

5 0 35

0 1 6

1 1 13

2 1 20

Integeroptimalsolution

Harrison Electric CompanyHarrison Electric CompanyListing all feasible solutions and selecting the one with the best objective

function value is called the enumeration method. This can be virtuallyimpossible for large problems where the number of feasible solutions is

extremely large !

Chandeliers ( X1 ) Ceiling Fans ( X2 ) Profit ( Z )

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

Roundingoptimalsolution

Branch-and-Bound Branch-and-Bound MethodMethod

1. Solve the original problem using linear programming. If the answer satisfies the integer constraints, we are done. If not, this value provides an initial upper bound for the objective function.

2. Find any feasible solution that meets the integer constraints for use as a lower bound. Usually, rounding down each variable will accomplish this.


3. Branch on one variable from step 1 that does not have an integer value. Split the problem into two subproblems based on integer values that are above and below the noninteger value.

For example, if X2 = 3.75 was in the optimal linear programming solution, introduce constraint X2 => 4 in the first subproblem, and X2 =< 3 in the second subproblem.


4. Create nodes at the top of these new branches by solving the new problems.


5. a If a branch yields a solution that is not feasible, terminate the branch.

5. b If a branch yields a solution that is feasible, but not an integer solution, go to step 6.

5. c If the branch yields a feasible integer solution, look at the objective function. If its value equals the upper bound, an optimal solution has been reached. If it is 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 is less than the lower bound, terminate this branch.


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 .


Maximize profit = $7.00 X1 + $6.00 X2

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

The Model

REVISITEDREVISITED


0 1 2 3 4 5 6

6

5

4

3

2

1

0

X2

X1

+ +

+ +

+

+ +

+

6X6X1 1 + 5X+ 5X22 =< 30 ( assembly hours ) =< 30 ( assembly hours )

2X2X11 + 3X + 3X22 =< 12 ( wiring hours ) =< 12 ( wiring hours )

Optimal Optimal NON-INTEGER NON-INTEGER LP SolutionLP Solution( X( X11 = 3.75 , X = 3.75 , X22 = 1.5 , Profit = $35.25 ) = 1.5 , Profit = $35.25 )

+ = Possible Integer Solution

REVISITEDREVISITED

Branch-and-Bound MethodBranch-and-Bound Method

Since X1 and X2 are not integers, the solution is not valid.

The profit of $35.25 will be the initial upper bound.

Rounding down gives X1 = 3, X2 = 1, profit = $27.00 , which is feasible and can be used as a lower bound.

X1=3.75X2=1.5

P=35.25

Upper Bound = $35.25

Lower Bound = $27.00 (rounding down)

OriginalNon-Integer

Solution


We divide the problem into two subproblems, A and B

We can branch on either the non-integer X1 or X2

We choose X1 this time

X1=3.75X2=1.5

P=35.25



OriginalNon-Integer

Solution


Subproblem A

Max Z = $7X1 + $6X2

s.t. 2X1 + 3X2 =< 12 6X1 + 5X2 =< 30 X1 => 4

X1=3.75X2=1.5

P=35.25



OriginalNon-Integer

Solution

Subproblem B

Max Z = $7X1 + $6X2

s.t. 2X1 + 3X2 =< 12 6X1 + 5X2 =< 30 X1 =< 3


X1=3.75X2=1.5

P=35.25



X1=4X2=1.2

P=35.20

Subproblem A

X1=3X2=2

P=33.00

Subproblem B

Noninteger Solution


Lower Bound = $33.00

This BranchSolution Is Integer

New Lower Bound $33.00


Subproblem A is now branched into two new subproblems, C and D

Subproblem C has the additional constraint of X2 => 2

Subproblem D has the additional constraint of X2 =< 1

The logic here is that since A’s optimal solution of X1 = 1.2 is not feasible, the integer feasible answer must lie at X2 => 2 or X2 =< 1


Subproblem C

Max Z = $7X1 + $6X2

s.t. 2X1 + 3X2 =< 12 6X1 + 5X2 =< 30 X1 => 4 X2 => 2

Subproblem D

Max Z = $7X1 + $6X2

s.t. 2X1 + 3X2 =< 12 6X1 + 5X2 =< 30 X1 => 4 X2 =< 1

Subproblem C has no feasible solution whatsoever because the firsttwo constraints are violated if X1 => 4 and X2 => 2 constraints areobserved. We terminate this branch and do not consider its solution.

Subproblem D’s solution is X1 = 4.17, X2 = 1, profit = $35.16. This non-integer solution yields a new upper bound of $35.16.


X1=3.75X2=1.5

P=35.25

X1=4X2=1.2

P=35.20

Subproblem A

X1=3X2=2

P=33.00

Subproblem B


Lower Bound = $33.00

NoFeasibleSolution

X1=4.17X2=1

P=35.16

Subproblem C

Subproblem DX1 => 4

X1 =< 3

X2 => 2

X2 =< 1


Subproblem E

Max Z = $7X1 + $6X2

s.t. 2X1 + 3X2 =< 12 6X1 + 5X2 =< 30 X1 => 4 X1 =< 4 X2 =< 1

Subproblem F

Max Z = $7X1 + $6X2

s.t. 2X1 + 3X2 =< 12 6X1 + 5X2 =< 30 X1 => 4 X1 => 5 X2 =< 1

Finally, we create subproblems E and F and solve for X1 and X2 with the additional constraints X1 =< 4 and X1 => 5 .

Full Branch-and-Bound SolutionFull Branch-and-Bound Solution

X1=3.75X2=1.5

P=35.25

X1=4X2=1.2

P=35.20

Subproblem A

X1=3X2=2

P=33.00

Subproblem B

NoFeasibleSolution

X1=4.17X2=1

P=35.16

Subproblem C

Subproblem D

X1 => 4

X1 =< 3

X2 => 2

X2 =< 1

X1=4X2=1

P=34.00

X1=5X2=0

P=35.00

Subproblem E

Subproblem F

X1 =< 4

X1 => 5

FeasibleInteger

Solution

FeasibleIntegerOptimalSolution

The stopping rule for the branching process is that we continue until the new upper bound is less than or equal to the lower bound or no further branching is possible. The latter is the case here since both branches yielded feasible integer solutions.


Integer Programming with Integer Programming with QM for WINDOWSQM for WINDOWS

Harrison Electric Company

TheHarrisonElectric

Company

Integer Programming UsingInteger Programming Using

Mixed Integer ProgrammingMixed Integer Programming

The Bagwell Chemical Company produces two industrial chemicals.

The first product, xyline, must be produced in 50-pound bags.

The second, hexall, is sold by the pound in dry bulk and hence can be produced in any quantity.

Both xyline and hexall are composed of three ingredients: A, B, and C as follows:

The Bagwell Chemical Co.The Bagwell Chemical Co.

Amount per

50-Pound Bag

of Xyline (lb)

Amount per

Pound

of Hexall (lb)

Amount of

Ingredients

Available

30 0.5 2,000 lb - A

18 0.4 800 lb - B

2 0.1 200 lb - C

Bagwell sells 50-pound bags of xyline for $85.00 and hexall in any weight for $1.50 per pound.Bagwell sells 50-pound bags of xyline for $85.00 and hexall in any weight for $1.50 per pound.

By formulating the usage coefficients for each 50-pound bag of Xyline, rather than for eachBy formulating the usage coefficients for each 50-pound bag of Xyline, rather than for eachpound of Xyline, the program will automatically compute the purchase quantity for Xyline inpound of Xyline, the program will automatically compute the purchase quantity for Xyline in

50-pound bag units, if an integer output is required.50-pound bag units, if an integer output is required.

The Bagwell Chemical Co.The Bagwell Chemical Co.

X = number of 50-pound bags of xyline produced ; Y = number of pounds of hexall (in dry bulk) mixed

Maximize Z = $85.00 X + $1.50 Y

s.t.

30 X + 0.5 Y <= 2,000

18 X + 0.4 Y <= 800

2 X + 0.1 Y <= 200

X,Y => 0 and X integer

The Model

The Bagwell Chemical Co.The Bagwell Chemical Co.The SolutionThe Solution

X = 44 bags of xyline

Y = 20 pounds of hexall

Z = $3,770.00

X = 44.444 bags of xyline

Y = 0 pounds of hexall

Z = $3,777.78

The Integer Solution The Optimal Solution The Integer Solution The Optimal Solution


Mixed Integer Programfor

Bagwell Chemical Company

Mixed Integer Programming UsingMixed Integer Programming Using

TheBagwell

ChemicalCompany

Modeling with Binary VariablesModeling with Binary Variables

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

Another name for a ‘0-1’ variable is a binary variable.

A common problem of this type is the assignment problem which involves deciding which individuals are assigned to a set of jobs. A value of 1 indicates a person is assigned to a specific job. A value of 0 indicates the assignment was not made.

Capital Budgeting ExampleCapital Budgeting Example

A common capital budgeting decision involves selecting from a set of possible projects when budget limitations make it impossible to select all of these.

A separate ‘0-1’ variable can be defined for each project.

XYZ Chemical CompanyXYZ Chemical Company

XYZ is considering three possible improvement projectsfor its plant: a new catalytic converter, a new softwareprogram for controlling operations, and expanding thewarehouse used for storage.

Capital requirements and budget limitations in the nexttwo years prevent the firm from underatking all of theseat this time.

The net present value of each of the projects , the capitalrequirements, and the available funds for the next twoyears are given on the next slide.

NPV is thefuture valueof a projectdiscountedback to the

present time


Net Net

Present ValuePresent ValueYear 1Year 1Capital Capital

RequirementRequirement

Year 2Year 2CapitalCapital

RequirementRequirement

Catalytic Converter $25,000. $8,000. $7,000.

Software $18,000. $6,000. $4,000.

Warehouse Expansion $32,000. $12,000. $8,000.

Budget Limitations $20,000. $16,000.

PROJECTPROJECT


Maximize NPV of projects undertaken

subject to:

Total funds used in year 1 =< $20,000.

Total funds used in year 2 =< $16,000.

We define the decision variables as:

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


Maximize NPV = 25,000 X1 + 18,000 X2 + 32,000 X3

subject to:

8,000 X1 + 6,000 X2 + 12,000 X3 =< 20,000

7,000 X1 + 4,000 X2 + 8,000 X3 =< 16,000

X1, X2, X3 = 0 or 1

The Model


X1 = 1 , X2 = 0 , X3 = 1 , Z = 57,000

The firm should fund the catalytic converter project, and the warehouse expansion project, but not the new software project.

The net present value of these investments will be $57,000.00

The Solution


Binary Variable Modelingfor the

XYZ Chemical Company

XYZChemicalCompany


Limiting the Number of Alternatives SelectedLimiting the Number of Alternatives Selected

Suppose that XYZ Chemical Company is required to select no more than two of the three projects regardless of the funds available.

This could be modeled by adding the following constraint to the problem:

X1 + X2 + X3 =< 2

If we wished to force the selection of exactly two of the three projects for funding, the following constraint should be used:

X1 + X2 + X3 = 2

This forces exactly two of the variables to have values of 1, whereas the other variable must have a value of 0.


Dependent SelectionsDependent Selections

At times the selection of one project depends in some way upon the selection of another project. This situation can be modeled with the use of 0-1 variables.

Suppose that the new catalytic converter could be purchased only if the software was purchased also. The following constraint would force this to occur:

X1 <= X2 or X1 - X2 <= 0

Thus, if the software is not purchased, the value of X2 is 0, and the value of X1 must be 0 also because of this constraint.

However, if the software is purchased ( X2 = 1 ), then it is possible that the catalytic converter could be purchased ( X1 = 1) also, although this is not required.


Dependent SelectionsDependent Selections

If we wished for the catalytic converter and the software projects to either both be selected or both not be selected, we should use the following constraint:

X1 = X2 or X1 - X2 = 0

Thus, if either of these variables is equal to 0, the other must be 0 also. If either of these variables is equal to 1, the other must be 1 also.

Fixed-Charge ProblemFixed-Charge Problem

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

Building a new factory or entering into a long- term lease on an existing facility would involve a fixed cost that might vary depending upon the size of the facility and the location.

Once a factory is built, the variable production costs will be affected by the labor cost in the particular city where it is located.

Acme Manufacturing, Inc.Acme Manufacturing, Inc.

Acme Manufacturing is planning to build at least one new plant, and three cities are being considered: Baytown, TX; Lake Charles, LA; and Mobile, AL.

Once the plant or plants have been constructed, the firm wishes to have sufficient capacity to produce at least 38,000 units each year.

The costs associated with the possible locations are shown on the following slide.

FIXED-CHARGE PROBLEM EXAMPLEFIXED-CHARGE PROBLEM EXAMPLE


SITE

ANNUAL

FIXED COST

VARIABLE COST PER UNIT

ANNUAL CAPACITY

Baytown $340,000. $32. 21,000

Lake Charles $270,000. $33. 20,000

Mobile $290,000. $30. 19,000


The objective function is to minimize the total of the fixed cost and the variable cost.

Total production capacity is at least 38,000 units.

The number of units produced at Baytown is ‘0’ if the plant is not built, and no more than 21,000 if the plant is built.

The number of units produced at Lake Charles is ‘0’ if the plant is not built, and no more than 20,000 if the plant is built.

The number of units produced at Mobile is ‘0’ if the plant is not built, and no more than 19,000 if the plant is built.


The decision variables:

X1 = ‘1’ if the factory is built in Baytown ; ‘0’ otherwise.

X2 = ‘1’ if the factory is built in Lake Charles ; ‘0’ otherwise.

X3 = ‘1’ if the 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.

TexasLouisiana


Minimize Cost = 340,000 X1 + $270,000 X2 + $290,000 X3 + 32 X4 + 33 X5 + 30 X6

subject to:

X4 + X5 + X6 => 38,000 X4 =< 21,000 X1 X5 =< 20,000 X2 X6 =< 19,000 X3

X1, X2, X3 = 0 or 1 ; X4, X5, X6 => 0 and integer

The Model


If X1 = 0 , meaning Baytown plant is not built, then X4 , number of units produced at Baytown plant, must equal 0 also, due to the second constraint.

If X1 = 1 , then X4 may be any integer value less than or equal to the limit of 21,000 units.

The third and fourth constraints are similarly used to guarantee that no units are produced at the other locations if the plants are not built.

Proposed Site

Acme Manufacturing, Inc.Acme Manufacturing, Inc.X1 = 0X2 = 1X3 = 1X4 = 0X5 = 19,000 X6 = 19,000 Z = 1,757,000.00

Optimal Solution

Factories will be built atLake Charles and Mobile.

Each of these wil produce19,000 units each year.

Total annual cost will be$1,757,000.00

Financial Investment ExampleFinancial Investment Example

Numerous financial applications exist with binary variables.

A very common type of problem involves selecting from a group of investment op- portunities.

Hawthorne InvestmentsHawthorne Investments

Hawthorne Investments specializes in recommending oilstock portfolios for wealthy clients. One such client hasmade the following specifications:

1. At least two Texas oil firms must be in the portfolio.2. No more than one investment can be made in foreign oil companies.3. One of the two California oil stocks must be purchased.4. The client has up to $3 million available for investments and insists on purchasing large blocks of shares of each company he invests in.5. The objective is to maximize annual return on investment subject to the constraints.


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

Stocks Being ConsideredStocks Being Considered

The firm is being forced by the clientThe firm is being forced by the client to buy shares in “blocks” so the to buy shares in “blocks” so the cost per “block” is shown below.cost per “block” is shown below.


To formulate this as a 0-1 integer programming problem, let Xi be a ‘0-1’ integer variable, where Xi = 1 if the stock is purchased and Xi = 0 if the stock is not purchased.


Maximize return = 50 X1 + 80 X2 + 90 X3 + 120 X4

+ 110 X5 + 40 X6 + 75 X7

subject to:

X1 + X4 + X5 => 2 ( Texas constraint ) X2 + X3 =< 1 ( foreign oil constraint ) X6 + X7 = 1 ( California constraint )

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

All variables must be 0 or 1 in value.The Model


X3 = 1X4 = 1X5 = 1X6 = 1X1 = 0X2 = 0X7 = 0

Z = 360,000

Hawthorne should invest inHawthorne should invest in Dutch ShellDutch Shell

Houston DrillingHouston DrillingTexas PetroleumTexas Petroleum

San Diego OilSan Diego Oiland not inand not in

Trans-Texas OilTrans-Texas OilBritish PetroleumBritish PetroleumCalifornia Petro California Petro

The expected return isThe expected return is$360,000.00$360,000.00

The Optimal Solution

0-1 Integer Programming Using0-1 Integer Programming Using

HawthorneInvestments

Problem

0-1 Integer Programming with 0-1 Integer Programming with QM for WINDOWSQM for WINDOWS

Hawthorne InvestmentsProblem

Applied Management Science for Decision Making, 2e © 2014 Pearson Learning Solutions

Documents

Integer Programming Pure, Mixed-Integer, Zero-One Models Strategic Resource Allocation and Planning MGMT E-5050 Applied Management Science for Decision