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Unformatted text preview: Table of Contents Chapter 7 (Using Binary Integer Programming) A Case Study: California Manufacturing (Section 7.1) 7.2–7.11 Using BIP for Project Selection: Tazer Corp. (Section 7.2) 7.12–7.15 Using BIP for the Selection of Sites: Caliente City (Section 7.3) 7.16–7.19 Using BIP for Crew Scheduling: Southwestern Airways (Section 7.4) 7.20–7.24 Using Mixed BIP to Deal with Setup Costs: Revised Wyndor (Section 7.5)7.25–7.30 Introduction to Integer Programming (UW Lecture) 7.31–7.46 These slides are based upon a lecture from the MBA core­course in Management Science at the University of Washington (as taught by one of the authors). Applications of Integer Programming (UW Lecture) These slides are based upon a lecture from the MBA elective “Modeling with Spreadsheets” at the University of Washington (as taught by one of the authors). 7.47–7.59 1 Applications of Binary Variables Since binary variables only provide two choices, they are ideally suited to be the decision variables when dealing with yes­or­no decisions. Examples: Should we undertake a particular fixed project? Should we make a particular fixed investment? Should we locate a facility in a particular site? 2 California Manufacturing Company The California Manufacturing Company is a diversified company with several factories and warehouses throughout California, but none yet in Los Angeles or San Francisco. A basic issue is whether to build a new factory in Los Angeles or San Francisco, or perhaps even both. Management is also considering building at most one new warehouse, but will restrict the choice to a city where a new factory is being built. Question: Should the California Manufacturing Company expand with factories and/or warehouses in Los Angeles and/or San Francisco? 3 Data for California Manufacturing Decision Number 1 2 3 4 Yes­or­No Question Build a factory in Los Angeles? Build a factory in San Francisco? Build a warehouse in Los Angeles? Build a warehouse in San Francisco? Decision Variable x1 x2 x3 x4 Net Present Value (Millions) $8 5 6 4 Capital Required (Millions) $6 3 5 2 Capital Available: $10 million 4 Binary Decision Variables Decision Number 1 2 3 4 Decision Variable x1 x2 x3 x4 Possible Value 0 or 1 0 or 1 0 or 1 0 or 1 Interpretation of a Value of 1 Build a factory in Los Angeles Build a factory in San Francisco Build a warehouse in Los Angeles Build a warehouse in San Francisco Interpretation of a Value of 0 Do not build this factory Do not build this factory Do not build this warehouse Do not build this warehouse 5 Algebraic Formulation Let x1 = 1 if build a factory in L.A.; 0 otherwise x2 = 1 if build a factory in S.F.; 0 otherwise x3 = 1 if build a warehouse in Los Angeles; 0 otherwise x4 = 1 if build a warehouse in San Francisco; 0 otherwise Maximize NPV = 8x1 + 5x2 + 6x3 + 4x4 ($millions) subject to Capital Spent: 6x1 + 3x2 + 5x3 + 2x4 ≤ 10 ($millions) Max 1 Warehouse: x 3 + x4 ≤ 1 Warehouse only if Factory: x3 ≤ x1 x4 ≤ x2 and x1, x2, x3, x4 are binary variables. 6 Spreadsheet Model 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 B NPV ($millions) Warehouse Factory Capital Required ($millions) Warehouse Factory C LA 6 8 D SF 4 5 E F G LA 5 6 SF 2 3 Capital Spent 9 Total Warehouses 0 <= Capital Available 10 Maximum Warehouses 1 Build? Warehouse Factory LA 0 <= 1 Total NPV ($millions) SF 0 <= 1 13 <= 7 Sensitivity Analysis with Solver Table 23 24 25 26 27 28 29 30 31 32 33 34 35 36 B Capital Available ($millions) 5 6 7 8 9 10 11 12 13 14 15 C Warehouse in LA? 0 0 0 0 0 0 0 0 0 0 1 1 D Warehouse in SF? 0 1 1 1 1 0 0 1 1 1 0 0 E Factory in LA? 1 0 0 0 0 1 1 1 1 1 1 1 F Factory in SF? 1 1 1 1 1 1 1 1 1 1 1 1 G Total NPV ($millions) 13 9 9 9 9 13 13 17 17 17 19 19 8 Management’s Conclusion Management’s initial tentative decision had been to make $10 million of capital available. With this much capital, the best plan would be to build a factory in both Los Angeles and San Francisco, but no warehouses. An advantage of this plan is that it only uses $9 million of this capital, which frees up $1 million for other projects. A heavy penalty (a reduction of $4 million in total net present value) would be paid if the capital made available were to be reduced below $9 million. Increasing the capital made available by $1 million (to $11 million) would enable a substantial ($4 million) increase in the total net present value. Management decides to do this. With this much capital available, the best plan is to build a factory in both cities and a warehouse in San Francisco. 9 Some Other Applications Investment Analysis Site Selection Should we make a certain fixed investment? Examples: Turkish Petroleum Refineries (1990), South African National Defense Force (1997), Grantham, Mayo, Van Otterloo and Company (1999) Should a certain site be selected for the location of a new facility? Example: AT&T (1990) Should a certain plant remain open? Should a certain site be selected for a new plant? Should a distribution center remain open? Should a certain site be selected for a new distribution center? Should a certain distribution center be assigned to serve a certain market area? Examples: Ault Foods (1994), Digital Equipment Corporation (1995) Designing a Production and Distribution Network 10 10 Some Other Applications Dispatching Shipments Scheduling Interrelated Activities Should a certain route be selected for a truck? Should a certain size truck be used? Should a certain time period for departure be used? Examples: Quality Stores (1987), Air Products and Chemicals, Inc. (1983), Reynolds Metals Co. (1991), Sears, Roebuck and Company (1999) Should a certain activity begin in a certain time period? Examples: Texas Stadium (1983), China (1995) Should a certain asset be sold in a certain time period? Example: Homart Development (1987) Should a certain type of airplane be assigned to a certain flight leg? Should a certain sequence of flight legs be assigned to a crew? Examples: American Airlines (1989, 1991), Air New Zealand (2001) Scheduling Asset Divestitures Airline Applications: 11 11 Project Selection at Tazer Corp. Tazer Corporation is searching for a new breakthrough drug. Five potential research and development projects: $1.2 billion available for investment (enough for 2 or 3 projects) Project Up: Develop a more effect antidepressant that doesn’t cause mood swings Project Stable: Develop a drug that addresses manic depression Project Choice: Develop a less intrusive birth control method for women Project Hope: Develop a vaccine to prevent HIV infection Project Release: Develop a more effective drug to lower blood pressure Question: Which projects should be selected to research and develop? 12 12 Data for the Tazer Project Selection Problem 1 Up R&D ($million) Success Rate Revenue if Successful ($million) Expected Profit ($million) 400 50% 1,400 2 Stable 300 35% 1,200 3 Choice 600 35% 2,200 4 Hope 500 20% 3,000 5 Release 200 45% 600 300 120 170 100 70 13 13 Algebraic Formulation of Tazer Project Selection Let xi = 1 if approve project i ; 0 otherwise (for i = 1, 2, 3, 4, and 5) Maximize P = 300x1 + 120x2 + 170x3 + 100x4 + 70x5 ($million) subject to R&D Budget: 400x1 + 300x2 + 600x3 + 500x4 + 200x5 ≤ 1,200 ($million) and xi are binary (for i = 1, 2, 3, 4, and 5). 14 14 Spreadsheet for Tazer Project Selection Problem A 1 2 3 4 5 6 7 8 9 10 B C D E F G H I J Tazer Corp. Project Selection Problem Up Stable 400 300 50% 35% 1400 1200 300 120 1 0 Choice Hope Release Total Budget 600 500 200 1200 <= 1200 35% 20% 45% 2200 3000 600 170 100 70 540 1 0 1 R&D Investment ($million) Success Rate Revenue if Successful ($million) Expected Profit ($million) Do Project? 15 15 Selection of Sites for Emergency Services: The Caliente City Problem Caliente City is growing rapidly and spreading well beyond its original borders They still have only one fire station, located in the congested center of town The result has been long delays in fire trucks reaching the outer part of the city Goal: Develop a plan for locating multiple fire stations throughout the city New Policy: Response Time ≤ 10 minutes 16 16 Response Time and Cost Data for Caliente City Fire Station in Tract 1 Response times (minutes) for a fire in tract 1 2 3 4 5 6 7 8 Cost of Station ($thousands) 2 9 17 10 21 25 14 30 350 2 8 3 8 13 12 15 22 24 250 3 18 10 4 19 16 7 18 15 450 4 9 12 20 2 13 21 7 14 300 5 23 16 21 18 5 15 13 17 50 6 22 14 8 21 11 3 15 9 400 7 16 21 22 6 9 14 2 8 300 8 28 25 17 12 12 8 9 3 200 17 17 Algebraic Formulation of Caliente City Problem Let xj = 1 if tract j is selected to receive a fire station; 0 otherwise (j = 1, 2, … , 8) Minimize C = 350x1 + 250x2 + 450x3 + 300x4 + 50x5 + 400x6 + 300x7 + 200x8 subject to Tract 1: x1 + x2 + x4 ≥ 1 Tract 2: x1 + x2 + x3 ≥ 1 Tract 3: x2 + x3 + x6 ≥ 1 Tract 4: x1 + x4 + x7 ≥ 1 Tract 5: x5 + x7 ≥ 1 Tract 6: x3 + x6 + x8 ≥ 1 Tract 7: x4 + x7 + x8 ≥ 1 Tract 8: x6 + x7 + x8 ≥ 1 and xj are binary (for j = 1, 2, … , 8). 18 18 Spreadsheet Model for Caliente City Problem A B C D E F G H I J K L M N 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 Caliente City Fire Station Location Problem 1 2 9 17 10 21 25 14 30 2 8 3 8 13 12 15 22 24 250 Fire Station in Tract 3 4 5 18 9 23 10 12 16 4 20 21 19 2 18 16 13 5 7 21 15 18 7 13 15 14 17 450 300 50 6 22 14 8 21 11 3 15 9 400 7 16 21 22 6 9 14 2 8 300 8 28 25 17 12 12 8 9 3 200 Number Covering 1 1 1 1 1 1 2 2 Response Times (Minutes) for a Fire in Tract 1 2 3 4 5 6 7 8 Cost of Station 350 ($thousands) 1 2 3 4 5 6 7 8 1 1 0 1 0 0 0 0 Response Time <= 10 Minutes? 1 1 1 0 0 0 0 0 0 1 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 1 0 1 1 0 0 0 0 0 1 1 1 >= >= >= >= >= >= >= >= 1 1 1 1 1 1 1 1 Total Cost ($thousands) 750 Station in Tract? 1 0 Fire Station in Tract 2 3 4 5 6 1 0 0 0 0 7 1 8 1 19 19 Southwestern Airways Crew Scheduling Southwestern Airways needs to assign crews to cover all its upcoming flights. We will focus on assigning 3 crews based in San Francisco (SFO) to 11 flights. Question: How should the 3 crews be assigned 3 sequences of flights so that every one of the 11 flights is covered? 20 20 Southwestern Airways Flights Se at tl e (SE A) Sa n Fr an c is co (SFO ) De nv er (DEN ) C h i ca g o OR D) Lo s A n ge l es (LAX ) 21 21 Data for the Southwestern Airways Problem Feasible Sequence of Flights Flights 1. SFO–LAX 2. SFO–DEN 3. SFO–SEA 4. LAX–ORD 5. LAX–SFO 6. ORD–DEN 7. ORD–SEA 8. DEN–SFO 9. DEN–ORD 2 4 4 2 2 2 3 3 3 3 5 2 22 22 1 1 2 1 3 4 1 5 1 6 7 1 8 1 9 10 1 11 1 12 1 2 1 2 3 1 3 2 5 4 3 3 5 1 3 4 Algebraic Formulation Let xj = 1 if flight sequence j is assigned to a crew; 0 otherwise. (j = 1, 2, … , 12). Minimize Cost = 2x1 + 3x2 + 4x3 + 6x4 + 7x5 + 5x6 + 7x7 + 8x8 + 9x9 + 9x10 + 8x11 + 9x12 (in $thousands) subject to Flight 1 covered: Flight 2 covered: : Flight 11 covered: Three Crews: and xj are binary (j = 1, 2, … , 12). x1 + x4 + x7 + x10 ≥ 1 x2 + x5 + x8 + x11 ≥ 1 : x6 + x9 + x10 + x11 + x12 ≥ 1 x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 ≤ 3 23 23 Spreadsheet Model 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 B C 1 2 D 2 3 E 3 4 F Cost ($thousands) Includes Segment? SFO-LAX SFO-DEN SFO-SEA LAX-ORD LAX-SFO ORD-DEN ORD-SEA DEN-SFO DEN-ORD SEA-SFO SEA-LAX GH I J Flight Sequence 4 5 6 7 8 6 7 5 7 8 K 9 9 L M N O P Q 10 11 12 9 8 9 Total 1 1 1 1 1 1 1 1 1 1 1 Total Sequences 3 At Least One 1 1 1 1 1 1 1 1 1 1 1 Number of Crews 3 18 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 0 1 0 1 1 0 0 0 1 1 0 1 0 1 0 0 1 1 0 0 1 1 0 1 0 0 0 1 0 1 0 0 1 0 1 0 1 0 1 0 0 1 1 0 0 1 0 0 1 1 >= >= >= >= >= >= >= >= >= >= >= Fly Sequence? 1 0 2 0 3 1 4 1 5 0 6 0 7 0 8 0 9 0 10 11 12 0 1 0 <= Total Cost ($thousands) 24 24 Wyndor with Setup Costs Suppose that two changes are made to the original Wyndor problem: 1. For each product, producing any units requires a substantial one­time setup cost for setting up the production facilities. 2. The production runs for these products will be ended after one week, so D and W in the original model now represent the total number of doors and windows produced, respectively, rather than production rates. Therefore, these two variables need to be restricted to integer values. 25 25 Graphical Solution to Original Wyndor Problem Production rate for windows W 8 Optimal solution 6 (2, 6) 4 Feasible Region P = 3,600 = 300 D+ 500 W 2 0 4 2 Production rate for doors 6 8 10 D 26 26 Net Profit for Wyndor Problem with Setup Costs Net Profit ($) Number of Units Produced 0 1 2 3 4 5 6 Doors 0(300) – 0 = 0 1(300) – 700 = –400 2(300) – 700 = –100 3(300) – 700 = 200 4(300) – 700 = 500 Not feasible Not feasible Windows 0 (500) – 0 = 0 1(500) – 1,300 = –800 2(500) – 1,300 = –300 3(500) – 1,300 = 200 4(500) – 1,300 = 700 5(500) – 1,300 = 1,200 6(500) – 1,300 = 1,700 27 27 Feasible Solutions for Wyndor with Setup Costs W 8 Production quantity for windows 6 (0, 6) gives P = 1700 (2, 6) gives P = -100 + 1700 = 1600 4 (4, 3) gives P = 500 + 200 = 700 2 (0, 0) gives P = 0 0 (4, 0) gives P = 500 6 2 4 Production quantity for doors 8D 28 28 Algebraic Formulation Let D = Number of doors to produce, W = Number of windows to produce, y1 = 1 if perform setup to produce doors; 0 otherwise, y2 = 1 if perform setup to produce windows; 0 otherwise . Maximize P = 300D + 500W – 700y1 – 1,300y2 subject to Original Constraints: Plant 1: D≤4 Plant 2: 2W ≤ 12 Plant 3: 3D + 2W ≤ 18 Produce only if Setup: Doors: D ≤ 99y1 Windows: W ≤ 99y2 and D ≥ 0, W ≥ 0, y1 and y2 are binary. 29 29 Spreadsheet Model 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 B Unit Profit Setup Cost C Doors $300 $700 D Windows $500 $1,300 E F G H Plant 1 Plant 2 Plant 3 Hours Used Per Unit Produced 1 0 0 2 3 2 Doors 0 <= 0 0 Windows 6 <= 99 1 Hours Used 0 12 12 <= <= <= Hours Available 4 12 18 Units Produced Only If Setup Setup? Production Profit $3,000 - Total Setup Cost $1,300 Total Profit $1,700 30 30 Integer Programming When are “non­integer” solutions okay? Solution is naturally divisible Solution represents a rate e.g., $, pounds, hours e.g., units per week When is rounding okay? Solution only for planning purposes When numbers are large When is rounding not okay? e.g., rounding 114.286 to 114 is probably okay. When numbers are small Binary variables e.g., rounding 2.6 to 2 or 3 may be a problem. yes­or­no decisions 31 31 The Challenges of Rounding Rounded Solution may not be feasible. Rounded solution may not be close to optimal. There can be many rounded solutions. x2 5 4 3 2 1 Example: Consider a problem with 30 variables that are non­integer in the LP­ solution. How many possible rounded solutions are there? 1 2 3 4 5 x1 32 32 How Integer Programs are Solved x2 5 4 3 2 1 1 2 3 4 5 x1 33 33 How Integer Programs are Solved x2 5 4 3 2 1 1 2 3 4 5 x1 34 34 Applications of Binary Variables Making “yes­or­no” type decisions Set­covering problems Build a factory? Manufacture a product? Do a project? Assign a person to a task? Fixed costs Make a set of assignments that “cover” a set of requirements. If a product is produced, must incur a fixed setup cost. If a warehouse is operated, must incur a fixed cost. 35 35 Example #1 (Capital Budgeting) Norwood Development is considering the potential of four different development projects. Each project would be completed in at most three years. The required cash outflow for each project is given in the table below, along with the net present value of each project to Norwood, and the cash that is available each year. Cash Outflow Required ($million) Project 1 Project 2 Project 3 Project 4 Cash Available ($million) 28 13 10 Year 1 Year 2 Year 3 NPV 9 6 6 30 7 4 0 16 6 3 4 22 11 0 0 14 Question: Which projects should be undertaken? 36 36 Algebraic Formulation Let yi = 1 if project i is undertaken; 0 otherwise (i = 1, 2, 3, 4). Maximize NPV = 30y1 + 16y2 + 22y3 + 14y4 subject to Year 1: Year 2 (cumulative): Year 3 (cumulative): and yi are binary (i = 1, 2, 3, 4). 9y1 + 7y2 + 6y3 + 11y4 ≤ 28 ($million) 15y1 + 11y2 + 9y3 + 11y4 ≤ 41 ($million) 21y1 + 11y2 + 13y3 + 11y4 ≤ 51 ($million) 37 37 Spreadsheet Solution A 1 2 3 4 5 6 7 8 9 10 11 12 13 B C D E F G H I Norwood Development Capital Budgeting NPV ($million) Year 1 Year 2 Year 3 Project 1 30 Project 2 16 Project 3 22 Project 4 14 Cumulative Outflow 22 35 45 <= <= <= Cumulative Available 28 41 51 Total NPV ($million) 68 Cumulative Outflow Required ($million) 9 7 6 11 15 11 9 11 21 11 13 11 Project 1 1 Project 2 1 Project 3 1 Project 4 0 Undertake? 38 38 Additional Considerations (Logic and Dependency Constraints) At least one of projects 1, 2, or 3 Project 2 can’t be done unless project 3 is done Either project 3 or project 4, but not both No more than two projects total Question: What constraints would need to be added for each of these additional considerations? 39 39 Example #2 (Set Covering Problem) The Washington State legislature is trying to decide on locations at which to base search­and­rescue teams. The teams are expensive, so they would like as few as possible. Response time is critical, so they would like every county to either have a team located in that county or in an adjacent county. Question: Where should search­and­rescue teams be located? 40 40 The Counties of Washington State 9 10 1 2 7 3 8 4 5 14 15 16 23 17 24 25 13 6 12 21 11 19 20 29 30 18 26 27 28 22 31 33 34 35 32 36 37 1 . Clallum 2 . Jef f er son 3 . Gray s Harbor 4 . Pacif ic 5 . Wahkiakum 6 . Kit sap 7 . Mason 8 . Thur st on 9 . What com 1 0 . Skagit 1 1 . Snohom ish 1 2 . King 1 3 . Pierce 1 4 . Lew is 1 5 . Cow lit z 1 6 . Clark 1 7 . Skam ania 1 8 . Okanogan Counties 1 9 . Chelan 2 0 . Douglas 2 1 . Kit t it as 2 2 . Grant 2 3 . Yakim a 2 4 . Klickit at 2 5 . Bent on 2 6 . Fer ry 2 7 . St ev ens 2 8 . Pend Or eille 2 9 . Lincoln 3 0 . Spokane 3 1 . A dam s 3 2 . Whit m an 3 3 . Franklin 3 4 . Walla Walla 3 5 . Colum bia 3 6 . Gar f ield 3 7 . A sot in 41 41 Algebraic Formulation Let yi = 1 if a team is located in county i; 0 otherwise (i = 1, 2, … , 37). Minimize Number of Teams = y1 + y2 + … + y37 subject to County 1 covered: County 2 covered: County 3 covered: y 1 + y2 ≥ 1 y 1 + y2 + y 3 + y6 + y 7 ≥ 1 y2 + y3 + y4 + y7 + y8 + y14 ≥ 1 : County 37 covered: and yi are binary (i = 1, 2, … , 37). 42 42 y32 + y36 + y37 ≥ 1 Spreadsheet Solution A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 B C D E F G H I J K L M N Search & Rescue Location County Clallam Jefferson Grays Harbor Pacific Wahkiakum Kitsap Mason Thurston Whatcom Skagit Snohomish King Pierce Lewis Cowlitz Clark Skamania Okanogan Total Teams: Team? 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 8 # Teams Nearby 1 1 2 1 1 1 1 1 1 1 1 1 2 2 2 1 2 1 County Chelan Douglas Kittitas Grant Yakima Klickitat Benton Ferry Stevens Pend Oreille Lincoln Spokane Adams Whitman Franklin Walla Walla Columbia Garfield Asotin Team? 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 # Teams Nearby 2 1 1 1 3 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 >= >= >= >= >= >= >= >= >= >= >= >= >= >= >= >= >= >= 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 >= >= >= >= >= >= >= >= >= >= >= >= >= >= >= >= >= >= >= 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 43 43 Example #3 (Fixed Costs) Woodridge Pewter Company is a manufacturer of three pewter products: platters, bowls, and pitchers. The manufacture of each product requires Woodridge to have the appropriate machinery and molds available. The machinery and molds for each product can be rented at the following rates: for the platters, $400/week; for the bowls, $250/week; for the pitcher, $300/week. Each product requires the amounts of labor and pewter given in the table below. The sales price and variable cost are also given in the table. Labor Hours Platter Bowl Pitcher Available 3 1 4 130 Pewter (pounds) 5 4 3 240 Sales Price $100 85 75 Variable Cost $60 50 40 Question: Which products should be produced, and in what quantity? 44 44 Algebraic Formulation Let x1 = Number of platters produced, x2 = Number of bowls produced, x3 = Number of pitchers produced, yi = 1 if lease machine and mold for product i; 0 otherwise (i = 1, 2, 3). Maximize Profit = ($100–$60)x1 + ($85–$50)x2 + ($75–$40)x3 – $400y1 – $250y2 – $300y3 subject to Labor: 3x1 + x2 + 4x3 ≤ 130 hours Pewter: 5x1 + 4x2 + 3x3 ≤ 240 pounds Allow production only if machines and molds are purchased: x1 ≤ 99y1 x2 ≤ 99y2 x3 ≤ 99y3 and xi ≥ 0, and yi are binary (i = 1, 2, 3). 45 45 Spreadsheet Solution A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 B C D E F G H Woodridge Pewter Company Sales Price Variable Cost Fixed Cost Constraint Labor (hrs.) Pewter (lbs.) Lease Equipment? Production Quantity Produce only if Lease Platters $100 $60 $400 Bowls $85 $50 $250 Pitchers $75 $40 $300 Total 60 240 <= <= Available 130 240 $5,100 $3,000 $250 $1,850 Usage (per unit produced) 3 1 4 5 4 3 0 0 <= 0 1 60 <= 99 0 0 <= 0 Revenue Variable Cost Fixed Cost Profit 46 46 Applications of Binary Variables Making “yes­or­no” type decisions Fixed costs Build a factory? Manufacture a product? Do a project? Assign a person to a task? Either­or constraints If a product is produced, must incur a fixed setup cost. If a warehouse is operated, must incur a fixed cost. Production must either be 0 or ≥ 100. meet 3 out of 4 constraints. Subset of constraints 47 47 Capital Budgeting with Contingency Constraints (Yes­or­No Decisions) A company is planning their capital budget over the next several years. There are 10 potential projects they are considering pursuing. They have calculated the expected net present value of each project, along with the cash outflow that would be required over the next five years. Also, suppose there are the following contingency constraints: at least one of project 1, 2 or 3 must be done, project 4 and project 5 cannot both be done, project 7 can only be done if project 6 is done. Question: Which projects should they pursue? 48 48 Data for Capital Budgeting Problem Cash Outflow Required ($million) Project 1 2 3 4 5 6 7 8 9 10 Cash Available ($million) Year 1 Year 2 Year 3 Year 4 Year 5 NPV 1 2 3 4 1 20 4 2 2 4 1 25 0 2 5 5 0 22 4 2 2 4 6 30 4 2 4 5 5 42 3 4 2 3 5 25 2 2 3 1 5 18 8 3 4 2 1 35 2 3 8 1 1 28 6 6 2 1 2 33 25 25 25 25 25 ($million) 49 49 Spreadsheet Solution A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 B C D E F G H I J K L M N O Capital Budgeting with Contingency Constraints Project 1 20 Project 2 25 Project 3 22 Project 4 30 4 6 8 12 18 Project 4 0 Project 5 42 4 6 10 15 20 Project 5 1 Project 6 25 3 6 8 11 16 Project 6 1 Project 7 18 2 4 7 8 13 Project 7 1 Project 8 35 8 11 15 17 18 Project 8 0 Project 9 28 2 5 13 14 15 Project 9 1 Project 10 33 6 12 14 15 17 Project 10 1 Cumulative Total Outflow 22 44 73 97 117 <= <= <= <= <= Cumulative Available 25 50 75 100 125 Total NPV ($million) 213 NPV ($million) Cumulative Cash Outflow Required ($million) Year 1 1 4 0 Year 2 3 6 2 Year 3 6 8 7 Year 4 10 12 12 Year 5 11 13 12 Project 1 1 Project 2 1 >= <= <= Project 3 1 1 1 1 Undertake? Contingency Constraints Project 1,2,3 3 Project 4,5 1 Project 7 1 Project 6 50 50 Electrical Generator Startup Planning (Fixed Costs) An electrical utility company owns five generators. To generate electricity, a generator must be started up, and associated with this is a fixed startup cost. All of the generators are shut off at the end of each day. Generator A Fixed Startup Cost Variable Cost (per MW) Capacity (MW) $2,450 $3 2,000 B $1,600 $4 2,800 C $1,000 $6 4,300 D $1,250 $5 2,100 E $2,200 $4 2,000 Question: Which generators should be started up to meet the total capacity needed for the day (6000 MW)? 51 51 Spreadsheet Solution A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 B C D E F G H I J Electrical Utility Generator Startup Planning Fixed Startup Cost Cost per Megawatt Max Capacity (MW) Startup? MW Generated Capacity Generator A $2,450 $3 2,000 1 2,100 <= 2,000 Generator B $1,600 $4 2,800 1 3,000 <= 2,800 Generator C $1,000 $6 4,300 0 0 <= 0 Generator D $1,250 $5 2,100 1 900 <= 2,100 Generator E $2,200 $4 2,000 0 0 <= 0 Fixed Cost Variable Cost Total Cost Total MW 6000 >= MW Needed 6,000 $5,300 $22,800 $28,100 52 52 Quality Furniture (Either­Or Constraints) Reconsider the Quality Furniture Problem: Now suppose that they would not produce any fewer than 200 units of either product (i.e., either produce 0 or at least 200). The Quality Furniture Corporation produces benches and picnic tables. The firm has a limited supply of two resources: labor and wood. 1,600 labor hours are available during the next production period. The firm also has a stock of 9,000 pounds of wood available. Each bench requires 3 labor hours and 12 pounds of wood. Each table requires 6 labor hours and 38 pounds of wood. The profit margin on each bench is $8 and on each table is $18. Question: What product mix will maximize their total profit? 53 53 Spreadsheet Solution A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 B C D E F G Quality Furniture (with either-or constraints) Profit Min Production (if any) Benches $8.00 200 Tables $18.00 200 Resources Used 1600 6400 Resources Available 1,600 9,000 Labor Wood Produce? Min Production Production Quantities Max Production Max Possible Use of Resources 3 6 12 38 1 200 <= 533.33 <= 533 533 0 0 <= 0 <= 0 237 <= <= Total Profit $4,266.67 54 54 Meeting a Subset of Constraints Consider a linear programming model with the following constraints, and suppose that meeting 3 out of 4 of these is good enough 12x1 + 24x2 + 18x3 ≥ 2,400 15x1 + 32x2 + 12x3 ≥ 1,800 20x1 + 15x2 + 20x3 ≤ 2,000 18x1 + 21x2 + 15x3 ≤ 1,600 55 55 Meeting a Subset of Constraints Let yi = 1 if constraint i is enforced; 0 otherwise. Constraints: y1 + y2 + y3 + y4 ≥ 3 12x1 + 24x2 + 18x3 ≥ 2,400y1 15x1 + 32x2 + 12x3 ≥ 1,800y2 20x1 + 15x2 + 20x3 ≤ 2,000 + M (1 – y3) where M is a large number. 18x1 + 21x2 + 15x3 ≤ 1,600 + M (1 – y4) 56 56 Facility Location Consider a company that operates 5 plants and 3 warehouses that serve customers in 4 different regions. To lower costs, they are considering streamlining by closing one or more plants and warehouses. Associated with each plant are fixed costs, shipping costs, and production costs. Each plant has a limited capacity. Associated with each warehouse are fixed costs and shipping costs. Each warehouse has a limited capacity. Questions: Which plants should they keep open? Which warehouses should they keep open? How should they divide production among the open plants? How much should be shipped from each plant to each warehouse, and from each warehouse to each customer? 57 57 Data for Facility Location Problem Fixed Cost (per month) (Shipping + Production) Cost (per unit) WH #1 $650 500 450 400 WH #2 $750 350 450 500 WH #3 $850 550 350 600 Capacity (units per month) Plant 1 Plant 2 Plant 3 Plant 4 Plant 5 $42,000 50,000 45,000 50,000 47,000 Fixed Cost (per month) 400 300 300 350 375 Capacity (per month) 550 450 350 Shipping Cost (per unit) to customer C 1 $25 50 60 250 C 2 $65 25 20 225 C 3 $70 40 40 200 C 4 $35 60 45 275 WH #1 WH #2 WH #3 Demand : $45,000 25,000 65,000 600 400 900 58 58 Spreadsheet Solution A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 B C D E F G H I J K L M Plant to Warehouse Shipping + Production Cost Warehouse 1 Plant 1 $650 Plant 2 $500 Plant 3 $450 Plant 4 $400 Plant 5 $550 Shipment Quantities Plant 1 Plant 2 Plant 3 Plant 4 Plant 5 Total Shipped Warehouse 1 0 0 0 0 0 0 Warehouse 2 $750 $350 $450 $500 $450 Warehouse 2 0 300 0 0 0 300 Warehouse 3 $850 $550 $350 $600 $350 Warehouse 3 0 0 275 0 375 650 Total Shipped 0 300 275 0 375 Fixed Cost $42,000 $50,000 $45,000 $50,000 $47,000 Actual Capacity 0 300 300 0 375 Capacity 400 300 300 350 375 Open? 0 1 1 0 1 Total Costs $332,500 $37,375 $142,000 $90,000 $601,875 <= <= <= <= <= Shipping Cost (P-->W) Shipping Cost (W-->C) Fixed Cost (P) Fixed Cost (W) Total Cost Warehouse to Customer Shipping Cost Warehouse 1 Warehouse 2 Warehouse 3 Shipment Quantities Warehouse 1 Warehouse 2 Warehouse 3 Total Shipped Needed Customer 1 $25 $50 $60 Customer 1 0 250 0 250 >= 250 Customer 2 $65 $25 $20 Customer 2 0 0 225 225 >= 225 Customer 3 $70 $40 $40 Customer 3 0 50 150 200 >= 200 Customer 4 $35 $60 $45 Customer 4 0 0 275 275 >= 275 Shipped Out 0 300 650 Fixed Cost $45,000 $25,000 $65,000 Shipped In 0 300 650 Capacity 600 400 900 Actual Capacity 0 400 900 Open? 0 1 1 <= <= <= <= <= <= 59 59 ...
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This note was uploaded on 04/21/2009 for the course BUS 338W-1 taught by Professor White during the Winter '09 term at University of Michigan-Dearborn.

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