Presentation on theme: "Applications of Linear and Integer Programming Models - 2"— Presentation transcript:
1 Applications of Linear and Integer Programming Models - 2 Chapter 3Applications of Linear and Integer Programming Models - 2
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)
3 The use of binary variables in constraints AAny decision situation that can be modeled by “yes”/“no”, “good”/“bad” etc., falls into the binary category.To illustrate
4 The use of binary variables in constraints ExampleA decision is to be made whether each of three plants should be built (Yi = 1) or not built (Yi = 0)Requirement Binary RepresentationAt least 2 plants must be built Y1 + Y2 + Y3 ³ 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 £ 0One, but not both plants must be built Y1+ Y2 = 1Both or neither plants must be built Y1 – Y2 =0 Plant construction cannot exceed $17 milliongiven the costs to build plants are $5, $8, $10 million 5Y1+8Y2+10Y3 £ 17
5 The use of binary variables in constraints Example - continuedTwo 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 £ 2000Y1If the plant is built Y1 = 1.The constraint becomes6x1 + 9X2 £ 2000If the plant is not built Y1 = 0. The constraint becomes 6x1 + 9X2 £ 0, and thus,X1 = 0 and X2 = 0
6 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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9 Sunset Beach Lifeguard Assignments 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.
10 Sunset Beach Lifeguard Assignments Problem SummarySchedule lifeguard over 5 consecutive days.Minimize the total number of lifeguards.Meet the minimum daily lifeguard requirementsSun. Mon. Tue. Wed. Thr. Fri. Sat.
11 Sunset Beach Lifeguard Assignments 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 scheduledConstraintsEnsure that enough lifeguards are scheduled each day.
12 Sunset Beach Lifeguard Assignments To ensure that enough lifeguards are scheduled for each day, identify which workers are on duty.For example: …
13 Sunset Beach Lifeguard Assignments Who works on Friday ?Who works on Saturday ?X2X3X3X4X5X6X4X5X6X1MonTue. Wed. Thu. Fri. Sat Sun.Repeat this procedure for each day of the week, and build the constraints accordingly.
18 3.5.2 Project selection Models These models involve a “go/no-go” situations, that can be modeled using binary variables.Typical elements in such models are:BudgetSpacePriority conditions
19 Salem City Council – Project Selection 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.
20 Salem City Council – Project Selection Survey resultsX1X2X3X4X5X6X7X8X9
21 Salem City Council – Project Selection 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 projectsConstraints:See the mathematical model.
22 Salem City Council – Project Selection The Mathematical ModelMax 4176X1+1774X X3+1928X4+3607X5+962X6+2829X7+1708X8+3003X9S.T. 400X1+ 350X X3+ 100X4+ 500X X6+ 220X7+ 150X8+ 140X9 £ 9007X1+ X X5+ X X7+ 3X x9 ³ 10X1+ X2+ X3+ X £ 3X3+ X = 1X7 - X8 = 0X X9 ³ 0x8 - x9 ³ 0(Xi = 0,1 for i=1, 2…, 9)The maximum amounts of funds to be allocated is $900,000The number of new jobs created must be at least 10The number of police-related activities selected is at most 3 (out of 4)Either police car or fire truck be purchasedSports funds and music funds must be restored / not restored togetherSports funds and music funds must berestored before computer equipmentis purchased
23 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
24 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 costsThe requirements concern (among others)Appropriate production levelsMaintaining a transportation system to satisfy demand in timely manner.
25 Globe Electronics, Inc.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.
26 Globe Electronics, Inc.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).
27 Globe Electronics, Inc. Data Production costs, Times, Availability Monthly Demand Projection
28 Globe Electronics, Inc. Transportation Costs per 100 units At least 70% of the demand in each distribution center must be satisfied.Unit selling priceG50 = $22; H90 = $28.CityFranciscoCincinnatiKansasSanPhiladelphia$200300500St.Louis100400New Orleans200Denver
29 Globe Electronics, Inc. The Globe problem Ordering raw material Scheduling personnelProductionProduction level for each product in each plant.Distribution plan.Distribution centers1. Storage2. Sale and Dissemination to retail establishments
31 Globe Electronics, Inc. - Variables Production variables in each plant1G11, H11G11PhiladelphiaG12G12, H12G11G121CincinnatiG13G13, H13G12G13G112St. LouisG22, H22G13G12G13G12G31, H31G112G13Kansas CityG12G113G13New OrleansG11G12G13G113San FranciscoG12G114G13DenverG11G12G11G13G11G12G11G13Total production of G50 in Philadelphia = GP =G12G11G12G13
32 Globe Electronics, Inc. - Variables Shipment variables to each distribution center1PhiladelphiaG11, H11G12, H121CincinnatiG13, H132St. LouisG22, H22G31, H312Kansas City3New Orleans3San Francisco4DenverTotal shipment of H90 to Cincinnati = HC = H11 + H21 + H31 +H41
33 Globe Electronics Model No. 1: All The Plants Remain Operational
34 Globe Electronics – all plants opened Objective functionMax Gross Profit = 22(Total G50)+28(Total H90) – Total Production Cost – Total transportation Cost =Max 22G + 28HG = total number of G50 producedH = total number of H90 produced
36 Globe Electronics – all plants opened Constraints:Ensure that the amount shipped from a plant equals the amount produced in a plant (summation constraints).For G50G11 + G12 + G13 = GPG21 + G22 + G23 = GSLG31 + G32 + G33 = GNOG41 + G42 + G43 = GDFor H90H11 + H12 + H13 = HPH21 + H22 + H23 = HSLH31 + H32 + H33 = HNOH41 + H42 + H43 = HDThe amount received by a distribution center is equal to all the shipments made to this center (summation constraints).For G50G11 + G21 + G31 + G41 = GCG12 + G22 + G32 + G42 = GKCG13 + G23 + G33 + G43 = GSFFor H90H11 + H21 + H31 + H41 = HCH12 + H22 + H32 + H42 = HKCH13 + H23 + H33 + H43 = HSF
37 Globe Electronics – all plants opened ConstraintsThe 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 ³ GC £ HC ³ HC £ 5000Kansas City GKC ³ GKC £ HKC ³ HKC £ 6000San Francisco GSF ³ GSF £ HSF ³ HSF £ 7000
38 Globe Electronics – all plants opened Constraints:Production time used at each plant cannot exceed the time available:.06GP HP £ GSL+ .08HSL £ GNO + .07HNO £ GD HD £ 640All the variables are non negative
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
40 Globe Electronics 1 - Summary The optimal value of the objective function is $356,571.43Note 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 $231,571.43Rounding 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 Globe Electronics Model No. 2: The number of plants that remain operational in each city is a decision variable.
42 Globe Electronics – which plant remains opened?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.
43 Globe Electronics – which plant remains opened?Binary Decision VariablesYi = a binary variable that describes the number of operational plants in city i.Objective function Subtract the following conditional set up costs from the previous objective function: 40,000YP + 35,000YSL + 20,000YND + 30,000YDConstraints Change the production constraints .06GP HP £ 640YP GSL+ .08HSL £ 960YSL.09GNO + .07HNO £ 480YNO .05GD HD £ 640YD
44 Globe Electronics – which plant remains opened?=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)
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.
47 Appendix 3.4 (CD): Advertising Models 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.
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 TodayLumber WeeklyTimber World
49 Vertex Software, Inc. 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 $ should be spent on advertising in any one of the trade publications.
50 Vertex Software, Inc. Circulation and advertising costs Publication Frequency Circulat. Cost/AdBuilding Today 5 day/week 400,000 Full pg.: $800Half pg.: $500Only B&WLumber Weekly Weekly 250,000 B&W pg.: $1500Color pg.: $4000Timber World Monthly ,000 B&W pg.: $2000Color pg.: $6000Key reader attitudes Percentage of ReadershipAttribute Rating Bldng. Lumbr TimberComputer data-base user %Large Firm (>2M sales)Location (city / suburb)Age of firm (>5 years)
51 Vertex Software, Inc. Solution The 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 $40,000 on advertisement in any one of the trade publications.
52 Vertex Software, Inc. Variables X1 = number of full page B&W ads placed in Building TodayX2 = number of half page B&W ads placed in Building TodayX3 = number of full page B&W ads placed in Lumber WeeklyX4 = number of full page color ads placed in Lumber WeeklyX5 = number of full page B&W ads placed in Timber WorldX6 = number of full page color ads placed in Timber World
53 Vertex Software, Inc. The Objective Function The objective function measures the effectiveness of the promotion operation (to be maximized).It depends on the number of ads in each publication, as well as on the relative effectiveness per ad.A special technique (external to this problem) is applied to evaluate this relative effectiveness.
54 Vertex Software, Inc. =SUMPRODUCT($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 Vertex Software, Inc. The Mathematical Model Max X X X X X X6 S.T. 800X X X X X X6 £ 90000X1 + X £ X1 + X ³ X3 + X £ 13X3 + X ³ 4X5 + X6 = 3X1 + X3 + X4 + X5 + X6 ³ 50X X6 ³ 8800X X £1500X X4 ³2000X X6 £All variables non-negativeBudget# of BuildingToday ads# of LumberWeekly adsTimber World adsFull PageColoredMaximum spentIn each magazine
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