Slides by John Loucks St. Edward’s University

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Slides by

JohnLoucks

St. Edward’sUniversity

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Chapter 7Chapter 7Integer 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

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Types of Integer Programming ModelsTypes of Integer Programming Models

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

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

Binary variables are variables whose values are Binary variables are variables whose values are restricted to be 0 or 1. If all variables 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 0-1 or binary integer linear programor binary integer linear program. .

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Example: All-Integer LPExample: All-Integer LP

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

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LP RelaxationLP Relaxation

Solving the problem as a linear program ignoring the integer Solving the problem as a linear program ignoring the integer constraints, the optimal solution to the linear program gives constraints, the optimal solution to the linear program gives fractional values for both fractional values for both xx11 and and xx22. The optimal solution to the linear . The optimal solution to the linear program is:program is:

xx11 = 2.5, = 2.5, xx22 = 1.5, = 1.5,

Max 3Max 3xx11 + 2 + 2xx22 = 10.5 = 10.5

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 = 1.5) to the LP relaxation problem, we get relaxation problem, we get xx11 = 3 and = 3 and xx22 = 2. By checking the = 2. By checking the constraints, we see that this point lies outside the feasible region, constraints, we see that this point lies outside the feasible region, making this solution infeasible.making this solution infeasible.

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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 is an = 1, we see that this solution indeed is an integer solution within the feasible region, and integer solution within the feasible region, and substituting in the objective function, it gives 3substituting in the objective function, it gives 3xx11 + + 22xx22 = 8. = 8.

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

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

The answer is NO! The optimal solution is The answer is NO! The optimal solution is xx11 = = 3 and 3 and xx22 = 0 giving 3 = 0 giving 3xx11 + 2 + 2xx22 = 9, as evidenced in = 9, as evidenced in the next slide. the next slide.

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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 33xx11 + 2 + 2xx22

1.1. 0 0 0 0 0 0 2.2. 1 0 1 0 3 3 3.3. 2 0 2 0 6 6 4.4. 3 0 3 0 9 optimal 9 optimal

solutionsolution 5.5. 0 1 0 1 2 2 6.6. 1 1 1 1 5 5 7.7. 2 1 2 1 8 8

8.8. 1 2 1 2 7 7

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Example: Capital BudgetingExample: Capital Budgeting

The Ice-Cold Refrigerator Company is The Ice-Cold Refrigerator Company is considering investing in several projects that considering investing in several projects that have varying capital requirements over the next have varying capital requirements over the next four years. Faced with limited capital each four years. Faced with limited capital each year, management would like to select the most year, management would like to select the most profitable projects. The estimated net present profitable projects. The estimated net present value for each project, the capital requirements, value for each project, the capital requirements, and the available capital over the four-year and the available capital over the four-year period are shown on the next slide.period are shown on the next slide.

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Problem DataProblem Data

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Decision VariablesDecision Variables

The four 0-1 decision variables are as The four 0-1 decision variables are as follows:follows:

P P = 1 if the plant expansion project is accepted; = 1 if the plant expansion project is accepted;

0 if rejected0 if rejected WW = 1 if the warehouse expansion project is = 1 if the warehouse expansion project is

accepted;accepted; 0 if rejected0 if rejected MM = 1 if the new machinery project is accepted; = 1 if the new machinery project is accepted; 0 if rejected0 if rejected

RR = 1 if the new product research project is = 1 if the new product research project is accepted; accepted;

0 if rejected0 if rejected

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Problem FormulationProblem Formulation

Max 90Max 90PP + 40 + 40WW + 10 + 10MM + + 3737RR

s.t.s.t. 15 15P +P + 1 100WW + 10 + 10MM + 15 + 15RR << 40 (Yr. 1 40 (Yr. 1 capital avail.)capital avail.)

2020P +P + 15 15WW + 10 + 10RR << 50 (Yr. 2 50 (Yr. 2 capital avail.capital avail.

2020P +P + 20 20WW + 10 + 10RR << 40 (Yr. 3 40 (Yr. 3 capital avail.)capital avail.)

1515P +P + 55WW + 4 + 4MM + 10 + 10RR << 35 (Yr. 4 35 (Yr. 4 capital avail.)capital avail.)

PP, , WW, , MM, , RR = 0, 1 = 0, 1

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Optimal SolutionOptimal SolutionPP = 1, = 1, WW = 1, = 1, MM = 1, = 1, RR = 0. = 0.

Total estimated net present value = $140,000.Total estimated net present value = $140,000.

The company should fund the plant The company should fund the plant expansion, the warehouse expansion, and the expansion, the warehouse expansion, and the new machinery projects.new machinery projects.

The new product research project should The new product research project should be put on hold unless additional capital funds be put on hold unless additional capital funds become available.become available.

The company will have $5,000 remaining The company will have $5,000 remaining in year 1, $15,000 remaining in year 2, and in year 1, $15,000 remaining in year 2, and $11,000 remaining in year 4. Additional capital $11,000 remaining in year 4. Additional capital funds of $10,000 in year 1 and $10,000 in year 3 funds of $10,000 in year 1 and $10,000 in year 3 will be needed to fund the new product research will be needed to fund the new product research project. project.

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Example: Fixed CostExample: Fixed Cost

Three raw materials are used to produce 3 Three raw materials are used to produce 3 products: a fuel additive, a solvent base, and a products: a fuel additive, a solvent base, and a carpet cleaning fluid.carpet cleaning fluid. The profit contributions are The profit contributions are $40 per ton for the fuel additive, $30 per ton for $40 per ton for the fuel additive, $30 per ton for the solvent base, and $50 per ton for the carpet the solvent base, and $50 per ton for the carpet cleaning fluid. cleaning fluid.

Each ton of fuel additive is a blend of 0.4 Each ton of fuel additive is a blend of 0.4 tons of material 1 and 0.6 tons of material 3. tons of material 1 and 0.6 tons of material 3. Each ton of solvent base requires 0.5 tons of Each ton of solvent base requires 0.5 tons of material 1, 0.2 tons of material 2, and 0.3 tons of material 1, 0.2 tons of material 2, and 0.3 tons of material 3. Each ton of carpet cleaning fluid is a material 3. Each ton of carpet cleaning fluid is a blend of 0.6 tons of material 1, 0.1 tons of blend of 0.6 tons of material 1, 0.1 tons of material 2, and 0.3 tons of material 3. material 2, and 0.3 tons of material 3.

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RMC has 20 tons of material 1, 5 tons of RMC has 20 tons of material 1, 5 tons of material 2, and 21 tons of material 3, and is material 2, and 21 tons of material 3, and is interested in determining the optimal production interested in determining the optimal production quantities for the upcoming planning period.quantities for the upcoming planning period.

There is a fixed cost for production setup of There is a fixed cost for production setup of the products, as well as a maximum production the products, as well as a maximum production quantity for each of the three products.quantity for each of the three products.

Product Product Setup CostSetup Cost Maximum Maximum ProductionProduction

Fuel additiveFuel additive $200 $200 50 tons50 tons

Solvent baseSolvent base $ $ 5050 25 tons 25 tons

Cleaning fluidCleaning fluid $400 $400 40 tons 40 tons

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FF = tons of fuel additive produced = tons of fuel additive produced

SS = tons of solvent base produced = tons of solvent base produced

CC = tons of carpet cleaning fluid produced = tons of carpet cleaning fluid produced

SFSF = 1 if the fuel additive is produced; 0 = 1 if the fuel additive is produced; 0 if notif not

SSSS = 1 if the solvent base is produced; 0 = 1 if the solvent base is produced; 0 if notif not

SCSC = 1 if the cleaning fluid is produced; 0 = 1 if the cleaning fluid is produced; 0 if not if not

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Max 40Max 40FF + 30 + 30SS + 50 + 50CC – 200 – 200SFSF – 50 – 50SSSS – 400 – 400SCSC

s.t. 0.4s.t. 0.4F +F + 0.50.5SS + 0.6 + 0.6CC << 20 (Mat’l. 1)20 (Mat’l. 1)

0.20.2S S ++ 0.1 0.1CC << 5 (Mat’l. 2)5 (Mat’l. 2)

0.60.6F +F + 0.3 0.3SS + 0.3 + 0.3C C << 21 (Mat’l. 3)21 (Mat’l. 3)

F F - 50 - 50SFSF << 0 (Max.F)0 (Max.F)

S S - 25 - 25SSSS << 0 (Max. S)0 (Max. S)

C C - 50 - 50SFSF << 0 (Max. C)0 (Max. C)

FF, , SS, , CC >> 0; 0; SFSF, , SSSS, , SCSC = 0, 1 = 0, 1

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Optimal SolutionOptimal Solution

Produce 25 tons of fuel additive.Produce 25 tons of fuel additive.Produce 20 tons of solvent base.Produce 20 tons of solvent base.Produce 0 tons of cleaning fluid.Produce 0 tons of cleaning fluid.

The value of the objective function after The value of the objective function after deducting the setup cost is $1350. The setup deducting the setup cost is $1350. The setup cost for the fuel additive and the solvent base is cost for the fuel additive and the solvent base is $200 + $50 = $250. $200 + $50 = $250.

The optimal solution shows The optimal solution shows SCSC = 0, which = 0, which indicates that the more expensive $400 setup indicates that the more expensive $400 setup cost for the carpet cleaning fluid should be cost for the carpet cleaning fluid should be avoidedavoided.

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Example: Distribution System DesignExample: Distribution System Design

The Martin-Beck Company operates a plant The Martin-Beck Company operates a plant in St. Louis with an annual capacity of 30,000 in St. Louis with an annual capacity of 30,000 units. Product is shipped to regional distribution units. Product is shipped to regional distribution centers located in Boston, Atlanta, and Houston. centers located in Boston, Atlanta, and Houston. Because of an anticipated increase in demand, Because of an anticipated increase in demand, Martin-Beck plans to increase capacity by Martin-Beck plans to increase capacity by constructing a new plant in one or more of the constructing a new plant in one or more of the following cities: Detroit, Toledo, Denver, or following cities: Detroit, Toledo, Denver, or Kansas City. Kansas City.

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The estimated annual fixed cost and the The estimated annual fixed cost and the annual capacity for the four proposed plants are annual capacity for the four proposed plants are as follows:as follows:

Proposed PlantProposed Plant Annual Fixed CostAnnual Fixed Cost Annual Annual CapacityCapacity

DetroitDetroit $175,000 $175,000 10,00010,000

ToledoToledo $300,000 $300,000 20,00020,000

DenverDenver $375,000 $375,000 30,000 30,000

Kansas CityKansas City $500,000 $500,000 40,000 40,000

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The company’s long-range planning group The company’s long-range planning group developed forecasts of the anticipated annual developed forecasts of the anticipated annual demand at the distribution centers as follows: demand at the distribution centers as follows:

Distribution CenterDistribution Center Annual Annual DemandDemand

BostonBoston 30,000 30,000

AtlantaAtlanta 20,000 20,000

HoustonHouston 20,000 20,000

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The shipping cost per unit from each plant The shipping cost per unit from each plant to each distribution center is shown below.to each distribution center is shown below.

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yy11 = 1 if a plant is constructed in Detroit; 0 if = 1 if a plant is constructed in Detroit; 0 if

notnot

yy22 = 1 if a plant is constructed in Toledo; 0 if = 1 if a plant is constructed in Toledo; 0 if

notnot

yy33 = 1 if a plant is constructed in Denver; 0 if = 1 if a plant is constructed in Denver; 0 if

notnot

yy44 = 1 if a plant is constructed in Kansas City; 0 = 1 if a plant is constructed in Kansas City; 0

if not if not

xxijij = the units shipped (in 1000s) from plant = the units shipped (in 1000s) from plant ii to to

distribution center distribution center jj , with , with ii = 1, 2, 3, 4, 5 = 1, 2, 3, 4, 5 andand

jj = 1, 2, 3 = 1, 2, 3


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CConstruct a plant in Kansas City (onstruct a plant in Kansas City (yy44 = 1). = 1).

20,000 units will be shipped from Kansas 20,000 units will be shipped from Kansas City to Atlanta (City to Atlanta (xx4242 = 20), 20,000 units will be = 20), 20,000 units will be shipped from Kansas City to Houston (shipped from Kansas City to Houston (xx4343 = 20), = 20), and 30,000 units will be shipped from St. Louis and 30,000 units will be shipped from St. Louis to Boston (to Boston (xx5151 = 30). = 30).

The total cost of this solution including the The total cost of this solution including the fixed cost of $500,000 for the plant in Kansas fixed cost of $500,000 for the plant in Kansas City is $860,000.City is $860,000.


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Example: Bank LocationExample: Bank Location

The long-range planning department for The long-range planning department for the Ohio Trust Company is considering the Ohio Trust Company is considering expanding its operation into a 20-county region expanding its operation into a 20-county region in northeastern Ohio. Ohio Trust does not have, in northeastern Ohio. Ohio Trust does not have, at this time, a principal place of business in any at this time, a principal place of business in any of the 20 counties. of the 20 counties.

According to the banking laws in Ohio, if a According to the banking laws in Ohio, if a bank establishes a principal place of business bank establishes a principal place of business (PPB) in any county, branch banks can be (PPB) in any county, branch banks can be established in that county and in any adjacent established in that county and in any adjacent county. To establish a new PPB, Ohio Trust must county. To establish a new PPB, Ohio Trust must either obtain approval for a new bank from the either obtain approval for a new bank from the state’s superintendent of banks or purchase an state’s superintendent of banks or purchase an existing bank. existing bank.

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The 20 counties in the region and adjacent The 20 counties in the region and adjacent counties are listed on the next slide. For counties are listed on the next slide. For example, Ashtabula County is adjacent to Lake, example, Ashtabula County is adjacent to Lake, Geauga, and Trumbull counties; Lake County is Geauga, and Trumbull counties; Lake County is adjacent to Ashtabula, Cuyahoga, and Geauga adjacent to Ashtabula, Cuyahoga, and Geauga counties; and so on.counties; and so on.

As an initial step in its planning, Ohio Trust As an initial step in its planning, Ohio Trust would like to determine the minimum number of would like to determine the minimum number of PPBs necessary to do business throughout the PPBs necessary to do business throughout the 20-county region. A 0-1 integer programming 20-county region. A 0-1 integer programming model can be used to solve this model can be used to solve this location location problemproblem for Ohio Trust. for Ohio Trust.

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xxii = 1 if a PBB is established in county = 1 if a PBB is established in county ii; 0 ; 0

otherwiseotherwise



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For this 20-variable, 20-constraint problem:For this 20-variable, 20-constraint problem:

Establish PPBs in Ashland, Stark, and Geauga Establish PPBs in Ashland, Stark, and Geauga counties.counties.(With PPBs in these three counties, Ohio Trust (With PPBs in these three counties, Ohio Trust can place branch banks in all 20 counties.) can place branch banks in all 20 counties.)

All other decision variables have an optimal All other decision variables have an optimal value of zero, indicating that a PPB should not be value of zero, indicating that a PPB should not be placed in these counties. placed in these counties.


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Example: Product Design & Market ShareExample: Product Design & Market Share

Market Pulse Research has conducted a Market Pulse Research has conducted a study for Lucas Furniture on some designs for a study for Lucas Furniture on some designs for a new commercial office desk. Three attributes new commercial office desk. Three attributes were found to be most influential in determining were found to be most influential in determining which desk is most desirable: number of file which desk is most desirable: number of file drawers, the presence or absence of pullout drawers, the presence or absence of pullout writing boards, and simulated wood or solid color writing boards, and simulated wood or solid color finish. Listed on the next slide are the part-finish. Listed on the next slide are the part-worths for each level of each attribute provided worths for each level of each attribute provided by a sample of 7 potential Lucas customers.by a sample of 7 potential Lucas customers.

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File DrawerFile Drawer Pullout Writing Pullout Writing BoardsBoards

FinishFinish

ConsumConsumerer

00 11 22 PresentPresent AbsentAbsent Simulated Simulated WoodWood

Solid Solid ColorColor

11 55 2626 2020 1818 1111 1717 1010

22 1818 1111 55 1212 1616 1515 2626

33 44 1616 2222 77 1313 1111 1919

44 1212 88 44 1818 99 2222 1414

55 1919 99 33 44 1414 3030 1919

66 66 1515 2121 88 1717 2020 1111

77 99 66 33 1313 55 1616 2828


Part-WorthsPart-Worths

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Suppose the overall utility (sum of part-Suppose the overall utility (sum of part-worths) of the current favorite commercial office worths) of the current favorite commercial office desk is 50 for each customer. What is the desk is 50 for each customer. What is the product design that will maximize the share of product design that will maximize the share of choices for the seven sample participants? choices for the seven sample participants? Formulate and solve this 0 – 1 integer Formulate and solve this 0 – 1 integer programming problem.programming problem.

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There are 7 There are 7 llijij decision variables, one for each decision variables, one for each level of attribute.level of attribute.

lijlij = 1 if Lucas chooses level = 1 if Lucas chooses level ii for for attribute attribute jj; ;

0 otherwise.0 otherwise.

There are 7 There are 7 YYkk decision variables, one for each decision variables, one for each

consumer in the sample.consumer in the sample.

YYkk = 1 if consumer = 1 if consumer kk chooses the Lucas chooses the Lucas

brand;brand;

0 otherwise0 otherwise


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Objective FunctionObjective Function

Maximize the number of consumers preferring Maximize the number of consumers preferring the Lucas brand desk.the Lucas brand desk.

Max Max YY11 + + YY22 + + YY33 + + YY44 + + YY55 + + YY66 + + YY77

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ConstraintsConstraints

There is one constraint for each consumer in the There is one constraint for each consumer in the sample.sample.

55ll1111 + 26 + 26ll2121 + 20 + 20ll3131 + 18 + 18ll1212 + 11 + 11ll2222 + 17 + 17ll1313 + 10 + 10ll2323

– 50– 50YY11 >> 1 1 1818ll1111 + 11 + 11ll2121 + 5 + 5ll3131 + 12 + 12ll1212 + 16 + 16ll2222 + +

1515ll1313 + 26 + 26ll2323 – 50 – 50YY22 >> 1 1

44ll1111 + 16 + 16ll2121 + 22 + 22ll3131 + 7 + 7ll1212 + 13 + 13ll2222 + 11 + 11ll1313 + 19 + 19ll2323

– 50– 50YY33 >> 1 1 1212ll1111 + 8 + 8ll2121 + 4 + 4ll3131 + 18 + 18ll1212 + 9 + 9ll2222 + +

2222ll1313 + 14 + 14ll2323 – 50 – 50YY44 >> 1 1 1919ll1111 + 9 + 9ll2121 + 3 + 3ll3131 + +

44ll1212 + 14 + 14ll2222 + 30 + 30ll1313 + 19 + 19ll2323 – 50 – 50YY55 >> 1 1

66ll1111 + 15 + 15ll2121 + 21 + 21ll3131 + 8 + 8ll1212 + 17 + 17ll2222 + 20 + 20ll1313 + 11 + 11ll2323

– 50– 50YY66 >> 1 1

99ll1111 + 6 + 6ll2121 + 3 + 3ll3131 + 13 + 13ll1212 + 5 + 5ll2222 + 16 + 16ll1313 + 28 + 28ll2323

– 50– 50YY77 >> 1 1

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ConstraintsConstraints

There is one constraint for each attribute.There is one constraint for each attribute.

ll1111 + + ll2121 + + ll3131 = 1 = 1

ll1212 + + ll2222 = 1 = 1

ll1313 + + ll2323 = 1 = 1

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Lucas should choose these product features:Lucas should choose these product features:

1 file drawer (1 file drawer (ll2121 = 1) = 1)

No pullout writing boards (No pullout writing boards (ll2222 = 1) = 1)

Simulated wood finish (Simulated wood finish (ll1313 = 1) = 1)

Three sample participants would choose the Three sample participants would choose the Lucas design:Lucas design:

Participant 1 (Participant 1 (YY11 = 1) = 1)




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Modeling Flexibility Provided by 0-1 Modeling Flexibility Provided by 0-1 VariablesVariables

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 completed, the following special constraints may be formulated:be 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

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Example: Metropolitan MicrowavesExample: Metropolitan Microwaves

Metropolitan Microwaves, Inc. is planning toMetropolitan Microwaves, Inc. is planning to

expand its sales operation by offering other electronicexpand its sales operation by offering other electronic

appliances. The company has identified seven newappliances. The company has identified seven new

product lines it can carry. Relevant information aboutproduct lines it can carry. Relevant information about

each line follows on the next slide.each line follows on the next slide.

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

Product Line Product Line Invest. (Sq.Ft.) of ReturnInvest. (Sq.Ft.) of Return

1. TV/VCRs1. TV/VCRs $ 6,000 125$ 6,000 125 8.1% 8.1%2. TVs 2. TVs 12,000 150 12,000 150 9.0 9.0 3. Projection TVs 3. Projection TVs 20,000 200 20,000 200 11.0 11.0 4. VCRs4. VCRs 14,000 40 14,000 40 10.2 10.2 5. DVD Players5. DVD Players 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

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Metropolitan has decided that they should Metropolitan has decided that they should not stock projection TVs unless they stock either not stock projection TVs unless they stock either TV/VCRs or TVs. Also, they will not stock both VCRs TV/VCRs or TVs. Also, they will not stock both VCRs and DVD players, and they will stock video games and DVD players, and they will stock video games if they stock TVs. Finally, the company wishes to if they stock TVs. Finally, the company wishes to introduce at least three new product lines. introduce at least three new 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 return.maximize its overall expected return.

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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.

where:where:Product line 1 = TV/VCRsProduct line 1 = TV/VCRsProduct line 2 = TVsProduct line 2 = TVsProduct line 3 = Projection TVsProduct line 3 = Projection TVsProduct line 4 = VCRsProduct line 4 = VCRsProduct line 5 = DVD PlayersProduct line 5 = DVD PlayersProduct line 6 = Video GamesProduct line 6 = Video GamesProduct line 7 = Home ComputersProduct line 7 = Home Computers

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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 expected return:Maximize total 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

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Define the ConstraintsDefine the Constraints

1) Money: 1) Money:

66xx11 + 12 + 12xx22 + 20 + 20xx33 + 14 + 14xx44 + 15 + 15xx55 + 2 + 2xx66 + 32 + 32xx77 << 4545

2) Space: 2) Space:

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

3) Stock projection TVs only if stock TV/VCRs or TVs:3) Stock projection TVs only if stock TV/VCRs or TVs:

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

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

4) Do not stock both VCRs and DVD players: 4) Do not stock both VCRs and DVD players:

xx44 + + xx55 << 1 1

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

xx22 - - xx66 >> 0 0

6) Introduce at least 3 new lines:6) Introduce 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

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Introduce:Introduce: TV/VCRs, Projection TVs, and DVD PlayersTV/VCRs, Projection TVs, and DVD Players

Do Not IntroduceDo Not Introduce:: TVs, VCRs, Video Games, and Home ComputersTVs, VCRs, Video Games, and Home Computers

Total Expected ReturnTotal Expected Return:: $4,261$4,261

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Cautionary Note About Sensitivity Cautionary Note About Sensitivity AnalysisAnalysis

Sensitivity analysis often is more crucial for ILP Sensitivity analysis often is more crucial for ILP problems than for LP problems.problems than for LP problems.

A small change in a constraint coefficient can A small change in a constraint coefficient can cause a relatively large change in the optimal cause a relatively large change in the optimal solution.solution.

Recommendation: Resolve the ILP problem Recommendation: Resolve the ILP problem several times with slight variations in the several times with slight variations in the coefficients before choosing the “best” solution coefficients before choosing the “best” solution for implementation.for implementation.

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End of Chapter 7End of Chapter 7

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