1 applications of linear and integer programming models - 2 chapter 3

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1 Applications of Linear and Integer Programming Models - 2 Chapter 3

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1

Applications of Linear and Integer Programming

Models - 2

Applications of Linear and Integer Programming

Models - 2

Chapter 3

2

3.5 Applications of Integer Linear Programming Models

Many real life problems call for at least one integer decision variable.

There are three types of Integer models: Pure integer (AILP) Mixed integer (MILP) Binary (BILP)

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The use of binary variables in constraints

X 10 If a new health care plan is adopted If it is not

X 1 If a new police station is built downtown0 If it is not

X 1 If a particular constraint must hold0 If it is not

• AAny decision situation that can be modeled by “yes”/“no”, “good”/“bad” etc., falls into the binary category.

To illustrate

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Example A decision is to be made whether each of three plants should be built

(Yi = 1) or not built (Yi = 0)Requirement Binary Representation

At least 2 plants must be built Y1 + Y2 + Y3 2If plant 1 is built, plant 2 must not be built Y1 + Y2 1If plant 1 is built, plant 2 must be built Y1 – Y2 One, but not both plants must be built Y1+ Y2 = 1Both or neither plants must be built Y1 – Y2 =0Plant construction cannot exceed $17 million

given the costs to build plants are $5, $8, $10 million 5Y1+8Y2+10Y3 17

The use of binary variables in constraints

5

Example - continued Two products can be produced at a plant.

• Product 1 requires 6 pounds of steel and product 2 requires 9 pounds. • If a plant is built, it should have 2000 pounds of steel available.

The production of each product should satisfy the steel availability if the plant is opened, or equal to zero if the plant is not opened.

6X1 + 9X2 2000Y1

The use of binary variables in constraints

If the plant is built Y1 = 1.The constraint becomes6x1 + 9X2 2000

If the plant is not built Y1 = 0.The constraint becomes 6x1 + 9X2 0, and thus,X1 = 0 and X2 = 0

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3.5.1 Personnel Scheduling Models

Assignments of personnel to jobs under minimum required coverage is a typical integer problems.

When resources are available over more than one period, linking constraint link the resources available in period t to the resources available in a period t+1.

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The City of Sunset Beach staffs lifeguards 7 days a week.

Regulations require that city employees work five days. Insurance requirements mandate 1 lifeguard per 8000

average daily attendance on any given day.

The city wants to employ as few lifeguards as possible.

Sunset Beach Lifeguard Assignments

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Problem Summary Schedule lifeguard over 5 consecutive days. Minimize the total number of lifeguards. Meet the minimum daily lifeguard requirements

Sun. Mon. Tue. Wed. Thr. Fri. Sat. 8 6 5 4 6 7 9

Sunset Beach Lifeguard Assignments

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Decision VariablesXi = the number of lifeguards scheduled to begin on day “ i ” for i=1, 2, …,7 (i=1 is Sunday)

Objective FunctionMinimize the total number of lifeguard scheduled

ConstraintsEnsure that enough lifeguards are scheduled each day.

Sunset Beach Lifeguard Assignments

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To ensure that enough lifeguards are scheduled for each day, identify which workers are on duty. For example: …

Sunset Beach Lifeguard Assignments

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X1X1

Sunset Beach Lifeguard Assignments

X6X6 X5X5

X4X4 X3X3

Tue. Wed. Thu. Fri. Sat Sun.

Who works on Saturday ?Who works on Friday ? X2

Mon

X3X4X5X6

Repeat this procedure for each day of the week, and build the constraints accordingly.

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Sunset Beach Lifeguard Assignments

Min X1 + X2 + X3 + X4 + X5 + X6 + X7

S.T. X1 + X4 + X5 + X6 + X7 8

X1 + X2 + X5 + X6 + X7 6X1 + X2 + X3 + X6 + X7 5X1 + X2 + X3 + X4 + X7 4

X1 + X2 + X3 + X4 + X5 6

X2 + X3 + X4 + X5 + X6 7

X3 + X4 + X5 + X6 + X7 9

All the variables are non negative integers

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Sunset Beach Lifeguard Assignments

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Sunset Beach Lifeguard Assignments

17TOTAL LIFEGUARDS

OPTIMAL ASSIGNMENTSLIFEGUARDS

DAY PRESENT REQUIRED BEGIN SHIFT

SUNDAY 9 8 1MONDAY 8 6 0TUESDAY 6 5 1WEDNESDAY 5 4 1THURSDAY 6 6 3FRIDAY 7 7 2SATURDAY 9 9 2

10Note: An alternate optimal solution exists.

Sunset Beach Lifeguard Assignments

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These models involve a “go/no-go” situations, that can be modeled using binary variables.

Typical elements in such models are: Budget Space Priority conditions

3.5.2 Project selection Models

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The Salem City Council needs to decide how to allocate funds to nine projects such that public support is maximized.

Data reflect costs, resource availabilities, concerns and priorities the city council has.

Salem City Council – Project Selection

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Project Cost (1000) Jobs PointsHire seven new police officers 400.00$ 7 4176Modernize police headquarters 350.00$ 0 1774Buy two new police cars 50.00$ 1 2513Give bonuses to foot patrol officers 100.00$ 0 1928Buy new fire truck/support equipment 500.00$ 2 3607Hire assistant fire chief 90.00$ 1 962Restore cuts to sport programs 220.00$ 8 2829Restore cuts to school music 150.00$ 3 1708Buy new computers for high school 140.00$ 2 3003

Survey results

X1

X2

X3

X4

X5

X6

X7

X8

X9

Salem City Council – Project Selection

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Decision Variables: Xj- a set of binary variables indicating if a project j is selected (Xj=1) or not (Xj=0) for j=1,2,..,9.

Objective function:

Maximize the overall point score of the funded projects Constraints:

See the mathematical model.

Salem City Council – Project Selection

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Either police car or fire truck be purchased

Sports funds and music funds must be restored before computer equipment is purchased

Sports funds and music funds must be restored / not restored together

The maximum amounts of funds to be allocated is $900,000

The number of police-related activities selected is at most 3 (out of 4)

The number of new jobs created must be at least 10

Salem City Council – Project Selection The Mathematical Model

Max 4176X1+1774X2+ 2513X3+1928X4+3607X5+962X6+2829X7+1708X8+3003X9

S.T. 400X1+ 350X2+ 50X3+ 100X4+ 500X5+ 90X6+ 220X7+ 150X8+ 140X9 900

7X1+ X3+ 2X5+ X6+ 8X7+ 3X8+ 2x9 10

X1+ X2+ X3+ X4 3

X3+ X5 = 1

X7 - X8 = 0

X7 - X9 0

x8 - x9

(Xi = 0,1 for i=1, 2…, 9)

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Salem City Council – Project selection

=SUMPRODUCT(B4:B12,E4:E12)

=SUMPRODUCT(B4:B12,C4:C12)=SUMPRODUCT(B4:B12,D4:D12)=SUM(B4:B7)=B6+B8=B10-B11=B10-B12=B11-B12

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3.5.3 Supply Chain Management

Supply chain management models integrate the manufacturing process and the distribution of goods to customers.

The overall objective of these models is to minimize total system costs

The requirements concern (among others) Appropriate production levels Maintaining a transportation system to satisfy demand in timely

manner.

25

Globe Electronics, Inc. manufactures two styles of remote control cable boxes, G50 and H90.

Globe runs four production facilities and three distribution centers.

Each plant operates under unique conditions, thus has a different fixed operating cost, production costs, production rate, and production time available.

Globe Electronics, Inc.

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Demand has decreased, and management is contemplating closing one or more of its facilities.

Management wishes to:

– Develop an optimal distribution policy.– Determine which plant to close (if any).

Globe Electronics, Inc.

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Data

Fixed Cost Production Cost / unit Production Time (hr/unit) Available hrPlant per Month G50 H90 G50 H90 per Month

Philadelphia 40 10 14 0.06 0.06 640St. Louis 35 12 12 0.07 0.08 960New Orleans 20 8 10 0.09 0.07 480Denver 30 13 15 0.05 0.09 640

Fixed Cost Production Cost / unit Production Time (hr/unit) Available hrPlant per Month G50 H90 G50 H90 per Month

Philadelphia 40 10 14 0.06 0.06 640St. Louis 35 12 12 0.07 0.08 960New Orleans 20 8 10 0.09 0.07 480Denver 30 13 15 0.05 0.09 640

Production costs, Times, Availability

Monthly Demand ProjectionDemand

Cincinnati Kansas CitySan FranciscoG50 2000 3000 5000G90 5000 6000 7000

DemandCincinnati Kansas CitySan Francisco

G50 2000 3000 5000G90 5000 6000 7000

Globe Electronics, Inc.

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Transportation Costs per 100 units

At least 70% of the demand in each distribution center must be satisfied.

Unit selling price• G50 = $22; H90 = $28.

City FranciscoCincinnati Kansas San

Philadelphia $200 300 500

St.Louis 100 100 400New Orleans 200 200 300Denver 300 100 100

Globe Electronics, Inc.

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Ordering raw materialScheduling personnel

Production1. Production level for

each productin each plant.

2. Distribution plan.

Distribution centers 1. Storage 2. Sale and Dissemination to retail establishments

The Globe problem

Globe Electronics, Inc.

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3

2

11

2

3

4

Philadelphia

St. Louis

New Orleans

Denver

Cincinnati

Kansas City

San Francisco

G11, H11G12 , H

12

G22, H22

G 31, H 31

G13 , H

13

Transportation variables

Globe Electronics, Inc. - Variables

G 41, H 41

31+ +

3

2

11

2

3

4

Philadelphia

St. Louis

New Orleans

Denver

Cincinnati

Kansas City

San Francisco

G11, H11G12 , H

12

G22, H22

G 31, H 31

G13 , H

13

Total production of G50 in Philadelphia = GP =

G11

G11

G11

G11

G11G11G11G11G11G11

G11

G12

G12

G12

G12

G12G12G12G12

G11G13

G13

G13

G13

G13G13G13

G12

G13

G12

G13G12

Production variables in each plant

G13

G13

G12

G11

Globe Electronics, Inc. - Variables

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3

2

1

Philadelphia

St. Louis

New Orleans

Denver

Cincinnati

Kansas City

San Francisco

G11, H11G12 , H

12

G22, H22

G 31, H 31

G13 , H

13

Shipment variables to each distribution center

1

2

3

4

Total shipment of H90 to Cincinnati = HC = H11 + H21 + H31 +H41

Globe Electronics, Inc. - Variables

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Globe Electronics Model No. 1: All The Plants Remain Operational

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Objective functionMax Gross Profit = 22(Total G50)+28(Total H90) – Total

Production Cost – Total transportation Cost = Max 22G + 28H G = total number of

G50 producedH = total number of

H90 produced

Globe Electronics – all plants opened

35

Objective functionMax Gross Profit = 22(Total G50)+28(Total H90) – Total

Production Cost – Total transportation Cost = Max 22G + 28H

– 2H11 – 3H12 – 5H13

– 1H21 – 1H22 – 4H23

– 2H31 – 2H32 – 3H33

– 3H41 – 1H42 – 1H43

– 2G11 – 3G12 – 5G13

– 1G21 – 1G22 – 4G23

– 2G31 – 2G32 – 3G33

– 3G41 – 1G42 – 1G43

Transportation costs

– 10GP – 12GSL – 8GNO – 13GD

– 14HP – 12HSL – 10HNO – 15HDProduction costs

Globe Electronics – all plants opened

36

Constraints: Ensure that the amount shipped from a plant equals the amount

produced in a plant (summation constraints).For G50

G11 + G12 + G13 = GP

G21 + G22 + G23 = GSL

G31 + G32 + G33 = GNO

G41 + G42 + G43 = GD

For H90H11 + H12 + H13 = HP

H21 + H22 + H23 = HSL

H31 + H32 + H33 = HNO

H41 + H42 + H43 = HD

The amount received by a distribution center is equal to all the shipments made to this center (summation constraints).

For G50G11 + G21 + G31 + G41 = GC

G12 + G22 + G32 + G42 = GKC

G13 + G23 + G33 + G43 = GSF

For H90H11 + H21 + H31 + H41 = HC

H12 + H22 + H32 + H42 = HKC

H13 + H23 + H33 + H43 = HSF

Globe Electronics – all plants opened

37

Constraints The amount shipped to each distribution center is at least 70% of its

projected demand. The amount shipped to each distribution center does not exceed its

demand.• Cincinnati: GC 1400 GC 2000

HC 3500 HC 5000

• Kansas City GKC 2100 GKC 3000HKC 4200 HKC 6000

• San Francisco GSF 3500 GSF 5000HSF 4900 HSF 7000

Globe Electronics – all plants opened

38

Constraints: Production time used at each plant cannot exceed the time

available:

.06GP + .0 6HP 640

.07GSL+ .08HSL 960

.09GNO + .07HNO 480

.05GD + .09HD 640

All the variables are non negative

Globe Electronics – all plants opened

39

Globe Electronics – all plants opened spreadsheet

=F10*F9+F19*F18-SUMPRODUCT(G23:G26,F5:F8) SUMPRODUCT(H23:H26,F14:F17)-SUMPRODUCT(C5:E8,C23:E26)-SUMPRODUCT(C14:E17,C23:E26)-SUM(F23:F26)

=$I23*$F5+$J23*$F14Drag to L24:L26

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Globe Electronics 1 - Summary

The optimal value of the objective function is $356,571.43 Note that the fixed cost of operating the plants was not

included in the objective function because all the plants remain operational.

Subtracting the fixed cost of $125,000 results in a net monthly profit of $231,571.43

Rounding down several non-integer solution values results in an integral solution with total profit of $231,550.

This solution may not be optimal, but it is very close to it.

41

Globe Electronics Model No. 2:The number of plants that remain operational in each city is adecision variable.

42

High set up costs raise the question:Is it optimal to leave all the plants operational?

Using binary variables the optimal solution provides suggestions for: Production levels for each product in each plant, Transportation pattern from each plant to distribution

center, Which plant remains operational.

Globe Electronics – which plant remains opened?

43

• Binary Decision Variables

Yi = a binary variable that describes the number of operational plants in city i.

• Objective functionSubtract the following conditional set up costs from the previous objective function:

40,000YP + 35,000YSL + 20,000YND + 30,000YD

• ConstraintsChange the production constraints

.06GP + .0 6HP 640YP .07GSL+ .08HSL 960YSL

.09GNO + .07HNO 480YNO .05GD + .09HD 640YD

Globe Electronics – which plant remains opened?

44

=F10*F9+F19*F18-SUMPRODUCT(G23:G26,F5:F8) - SUMPRODUCT(H23:H26,F14:F17)-SUMPRODUCT(C5:E8,C23:E26)-SUMPRODUCT(C14:E17,C23:E26)-SUMPRODUCT(F23:F26,A5:A8)

Globe Electronics – which plant remains opened?

45

Globe Electronics 2 - Summary

The Philadelphia plant should be closed, while the other plants work at capacity.

Schedule monthly production according to the quantities shown in the Excel output.

The net monthly profit will be $266,083 (after rounding down the non-integer variable values), which is $34,544 per month greater than the optimal monthly profit obtained when all four plants are operational.

46

Appendix 3.4 (CD): Advertising Models

47

Many marketing situations can be modeled by linear programming models.

Typically, such models consist of: Budget constraints, Deadlines constraints, Choice of media, Exposure to target population.

The objective is to achieve the most effective advertising plan.

Appendix 3.4 (CD): Advertising Models

48

Vertex Software, Inc.

Vertex Software has developed a new software product, LUMBER 2000.

A marketing plan for this product is to be developed for the next quarter. The product will be promoted using black and white and colored full

page ads. Three publications are considered:

• Building Today• Lumber Weekly• Timber World

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Requirements A maximum of one ad should be placed in any one issue of any of

the publication during the quarter. At least 50 full-page ads should appear during the quarter. at least 8 color ads should appear during the quarter. One ad should appear in each issue of Timber World. At least 4 weeks of advertising should be placed in each of the

Building Today and Lumber Weekly publications. No more than $ 40000 should be spent on advertising in any one of

the trade publications.

Vertex Software, Inc.

50

Circulation and advertising costsPublication Frequency Circulat. Cost/AdBuilding Today 5 day/week 400,000 Full pg.: $800

Half pg.: $500Only B&W

Lumber Weekly Weekly 250,000 B&W pg.: $1500Color pg.: $4000

Timber WorldMonthly 200,000 B&W pg.: $2000Color pg.: $6000

Key reader attitudes Percentage of Readership Attribute Rating Bldng. Lumbr Timber

Computer data-base user .50 60% 80 90Large Firm (>2M sales) .25 40 80 80Location (city / suburb) .15 60 60 80

Age of firm (>5 years) .10 20 40 50

Vertex Software, Inc.

51

SolutionThe requirements are:

• Stay within a $90,000 budget for print advertising.• Place no more than 65 ads(=5 x 13 weeks) and no less than 20 ads

(=5 X 4 weeks) in Building Today.• Place no more than 13 and no less than 4 ads in Lumber Weekly.• Place exactly 3 ads in Timber World.• Place at least 50 full-page ads.• Place at least 8 color ads.• Spend no more than $40,000 on advertisement in any one of the trade

publications.

Vertex Software, Inc.

52

Variables X1 = number of full page B&W ads placed in

Building Today X2 = number of half page B&W ads placed in

Building Today X3 = number of full page B&W ads placed in

Lumber Weekly X4 = number of full page color ads placed in

Lumber Weekly X5 = number of full page B&W ads placed in

Timber World X6 = number of full page color ads placed in

Timber World

Vertex Software, Inc.

53

The Objective Function The objective function measures the effectiveness of the

promotion operation (to be maximized).

It depends on the number of ads in each publication, as

well as on the relative effectiveness per ad. A special technique (external to this problem) is applied to

evaluate this relative effectiveness.

Vertex Software, Inc.

54

Vertex Software, Inc.=SUMPRODUCT($B$6:$B$9,C6:C9)Drag to cells D11 and E11

=C$11*C$13*$B17Drag across to D17:E17 then down to C19:E19. Then delete formulas in cells C17,D19, and E19

55

Budget

• The Mathematical ModelMax 102000X1+40800X2+91250X3+182500X4+82000X5+164000X6

S.T.800X1 + 500X2+ 1500X3+ 4000X4+ 2000X5 + 6000X6 90000

X1 + X2 65

X1 + X2 20X3 + X4 13X3 + X4 4

X5 + X6 = 3X1 + X3 + X4 + X5 + X6 50X4 + X6 ³ 8

800X1 + 500X2 400001500X3 + 4000X4 40000

2000X5 + 6000X6 40000All variables non-negative

Vertex Software, Inc.

# of Building Today ads# of Lumber Weekly ads

Timber World adsFull PageColored

Maximum spent In each magazine

56

Publication Page Size Style Cost Per Ad Exposure Units Ads Cost Expsoure Units

Building Today Full B&W $800 102000 50 $40,000 5100000Half B&W $500 40800 0 $0 0

50 $40,000 5100000

Lumber Weekly Full B&W $1,500 91250 5 $7,500 456250Full Color $4,000 182500 8 $32,000 1460000

13 $39,500 1916250

Timber World Full B&W $2,000 82000 2 $4,000 164000Full Color $6,000 164000 1 $6,000 164000

3 $10,000 328000

Budget $90,000 TOTALS 66 $89,500 7344250Max Build Today 65Min Build Today 20 Size Totals Full Page 66 $89,500 7344250Max Lum Week 13 Half Page 0 $0 0Min Lum Week 4# Timber World 3 Style Totals B&W 57 $51,500 5720250Min Full Page 50 Color 9 $38,000 1624000

Min Color 8Max Any Pub $40,000

VERTEX SOFTWARE, INC.

Totals for Building Today

Totals for Lumber Weekly

LIMITS

Totals

Totals for Timber World

Vertex Software, Inc.

57

Copyright 2002 John Wiley & Sons, Inc. All rights reserved. Reproduction or translation of this work beyond that named in Section 117 of the United States Copyright Act without the express written consent of the copyright owner is unlawful. Requests for further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. Adopters of the textbook are granted permission to make back-up copies for their own use only, to make copies for distribution to students of the course the textbook is used in, and to modify this material to best suit their instructional needs. Under no circumstances can copies be made for resale. The Publisher assumes no responsibility for errors, omissions, or damages, caused by the use of these programs or from the use of the information contained herein.