1. 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 core-course in Management Science at the University of Washington (as taught by one of the authors).
Applications of Integer Programming (UW Lecture) 7.47–7.59
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).
© The McGraw-Hill Companies, Inc., 2008

2. Applications of Binary Variables
Since binary variables only provide two choices, they are ideally suited to be the decision variables when dealing with yes-or-no 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?
© The McGraw-Hill Companies, Inc., 2008

3. 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?
© The McGraw-Hill Companies, Inc., 2008

4. Data for California Manufacturing
Decision Number
Yes-or-No Question
Decision Variable
Net Present Value (Millions)
Capital Required (Millions) 1
Build a factory in Los Angeles? x1 $8 $6 2
Build a factory in San Francisco? x2 5 3
Build a warehouse in Los Angeles? x3 6 4
Build a warehouse in San Francisco? x4
Capital Available: $10 million
Table 7.1 Data for the California Manufacturing Company problem.
© The McGraw-Hill Companies, Inc., 2008

5. Binary Decision Variables
Decision Number
Decision Variable
Possible Value
Interpretation of a Value of 1
Interpretation of a Value of 0 1 x1
0 or 1
Build a factory in Los Angeles
Do not build this factory 2 x2
Build a factory in San Francisco 3 x3
Build a warehouse in Los Angeles
Do not build this warehouse 4 x4
Build a warehouse in San Francisco
Table 7.2 Binary decision variables for the California Manufacturing Co. problem.
© The McGraw-Hill Companies, Inc., 2008

6. 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: x3 + x4 ≤ 1 Warehouse only if Factory: x3 ≤ x1 x4 ≤ x2 and x1, x2, x3, x4 are binary variables.
© The McGraw-Hill Companies, Inc., 2008

7. Spreadsheet Model
Figure 7.1 A spreadsheet formulation of the BIP model for the California Manufacturing case study where the changing cells, Build Factory? (C18:D18) and Build Warehouse? (C16:D16) give the optimal solution obtained by using the Excel Solver.
© The McGraw-Hill Companies, Inc., 2008

8. Sensitivity Analysis with Solver Table
Figure 7.2 An application of the Solver Table that shows the effect on the optimal solution and the resulting total net present value of systematically varying the amount of capital being made available for these investments.
© The McGraw-Hill Companies, Inc., 2008

9. 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.
© The McGraw-Hill Companies, Inc., 2008

10. Some Other Applications
Investment Analysis
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)
Site Selection
Should a certain site be selected for the location of a new facility?
Example: AT&T (1990)
Designing a Production and Distribution Network
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)
© The McGraw-Hill Companies, Inc., 2008

11. Some Other Applications
Dispatching Shipments
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)
Scheduling Interrelated Activities
Should a certain activity begin in a certain time period?
Examples: Texas Stadium (1983), China (1995)
Scheduling Asset Divestitures
Should a certain asset be sold in a certain time period?
Example: Homart Development (1987)
Airline Applications:
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)
© The McGraw-Hill Companies, Inc., 2008

12. Project Selection at Tazer Corp.
Tazer Corporation is searching for a new breakthrough drug.
Five potential research and development 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
$1.2 billion available for investment (enough for 2 or 3 projects)
Question: Which projects should be selected to research and develop?
© The McGraw-Hill Companies, Inc., 2008

13. Data for the Tazer Project Selection Problem
1 Up
2 Stable
3 Choice
4 Hope
5 Release
R&D ($million)
400
300
600
500
200
Success Rate
50%
35%
20%
45%
Revenue if Successful ($million)
1,400
1,200
2,200
3,000
Expected Profit ($million)
120
170
100 70
Table 7.3 Data for the Tazer Project Selection Problem
© The McGraw-Hill Companies, Inc., 2008

14. 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 = 300x x x x4 + 70x5 ($million)
subject to
R&D Budget: 400x x x x x5 ≤ 1,200 ($million)
and xi are binary (for i = 1, 2, 3, 4, and 5).
© The McGraw-Hill Companies, Inc., 2008

15. Spreadsheet for Tazer Project Selection Problem
Figure 7.3 A spreadsheet formulation of the BIP model for the Tazer Corp. project selection problem were the changing cells DoProject? (C10:G10) give the optimal solution obtained by the Excel Solver.
© The McGraw-Hill Companies, Inc., 2008

16. 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
© The McGraw-Hill Companies, Inc., 2008

17. Response Time and Cost Data for Caliente City
Fire Station in Tract 1 2 3 4 5 6 7 8
Response times (minutes) for a fire in tract 18 9 23 22 16 28 10 12 14 21 25 17 20 13 19 11 15 30 24
Cost of Station ($thousands)
350
250
450
300 50
400
200
Table 7.4 Response-time and cost data for the Caliente City problem.
© The McGraw-Hill Companies, Inc., 2008

18. 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 = 350x x x x4 + 50x x x x8
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).
© The McGraw-Hill Companies, Inc., 2008

19. Spreadsheet Model for Caliente City Problem
Figure 7.4 A spreadsheet formulation of the BIP model for the Caliente City site selection problem where the changing cells StationInTract? (D29:K29) show the optimal solution obtained by the Excel Solver.
© The McGraw-Hill Companies, Inc., 2008

20. 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?
© The McGraw-Hill Companies, Inc., 2008

21. Southwestern Airways Flights
Figure 7.5 The arrows show the 11 Southwestern Airways flights that need to be covered by the three crews based in San Francisco.
© The McGraw-Hill Companies, Inc., 2008

22. Data for the Southwestern Airways Problem
Feasible Sequence of Flights
Flights 1 2 3 4 5 6 7 8 9 10 11 12
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
10. SEA–SFO
11. SEA–LAX
Cost, $1,000s
Table 7.5 Data for example 3 (the Southwestern Airways problem).
© The McGraw-Hill Companies, Inc., 2008

23. 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: x1 + x4 + x7 + x10 ≥ 1 Flight 2 covered: x2 + x5 + x8 + x11 ≥ 1 : : Flight 11 covered: x6 + x9 + x10 + x11 + x12 ≥ 1 Three Crews: x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 ≤ 3 and xj are binary (j = 1, 2, … , 12).
© The McGraw-Hill Companies, Inc., 2008

24. Spreadsheet Model
Figure 7.6 A spreadsheet formulation of the BIP model for the Southwestern Airways crew scheduling problem, where Fly Sequence? (C22:N22) shows the optimal solution obtained by the Excel Solver.
© The McGraw-Hill Companies, Inc., 2008

25. Wyndor with Setup Costs
Suppose that two changes are made to the original Wyndor problem:
For each product, producing any units requires a substantial one-time setup cost for setting up the production facilities.
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.
© The McGraw-Hill Companies, Inc., 2008

26. Graphical Solution to Original Wyndor Problem
This graph summarizes the application of the graphical method to the original Wyndor problem.
© The McGraw-Hill Companies, Inc., 2008

27. Net Profit for Wyndor Problem with Setup Costs
Number of Units Produced
Doors
Windows
0(300) – 0 = 0
0 (500) – 0 = 0 1
1(300) – 700 = –400
1(500) – 1,300 = –800 2
2(300) – 700 = –100
2(500) – 1,300 = –300 3
3(300) – 700 = 200
3(500) – 1,300 = 200 4
4(300) – 700 = 500
4(500) – 1,300 = 700 5
Not feasible
5(500) – 1,300 = 1,200 6
6(500) – 1,300 = 1,700
Table 7.6 Net profit ($) for Variation 1 of the Wyndor Problem with Setup Costs
© The McGraw-Hill Companies, Inc., 2008

28. Feasible Solutions for Wyndor with Setup Costs
Figure 7.7 The dots are the feasible solutions for the revised Wyndor problem. Also shown is the calculation of the total net profit P (in dollars) for each corner point from the net profits given in Table 7.6.
© The McGraw-Hill Companies, Inc., 2008

29. 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.
© The McGraw-Hill Companies, Inc., 2008

30. Spreadsheet Model
Figure 7.8 A spreadsheet model for variation 1 of the Wyndor problem, where the Excel Solver gives the optimal solution shown in the changing cells, Units Produced (C14:D14) and Setup? (C17:D17).
© The McGraw-Hill Companies, Inc., 2008

31. Integer Programming When are “non-integer” solutions okay?
Solution is naturally divisible
e.g., $, pounds, hours
Solution represents a rate
e.g., units per week
Solution only for planning purposes
When is rounding okay?
When numbers are large
e.g., rounding to 114 is probably okay.
When is rounding not okay?
When numbers are small
e.g., rounding 2.6 to 2 or 3 may be a problem.
Binary variables
yes-or-no decisions
Slides 7.31–7.46 are based upon a lecture from the MBA core-course in Management Science at the University of Washington (as taught by one of the authors).
© The McGraw-Hill Companies, Inc., 2008

32. The Challenges of Rounding
Rounded Solution may not be feasible.
Rounded solution may not be close to optimal.
There can be many rounded solutions.
Example: Consider a problem with 30 variables that are non-integer in the LP-solution. How many possible rounded solutions are there?
The solution to the LP-relaxation shown on the graph is approximately (3.8, 4.9).
None of the possible rounded solutions, (3, 4), (4, 4), (3, 5), or (4, 5), are even feasible.
The optimal solution at (1, 3) is not even close to the LP-relaxation solution.
There are 230, or approximately 1 billion rounded solutions to a problem with 30 variables that are non-integer.
© The McGraw-Hill Companies, Inc., 2008

33. How Integer Programs are Solved
This slide and the next can be used to intuitively explain the branch-and-bound procedure for solving integer programs.
The first step is to solve the LP relaxation. For this problem (as can be seen graphically), the optimal solution to the LP-relaxation is approximately (3.6, 4.3).
Since neither variable is integer, the next step is to branch and bound.
Two subproblems are created. In the first subproblem, the constraint x1 ≤ 3 is added. In the second subproblem, the constraint x1 ≥ 4 is added. In a feasible solution, x1 must either be ≤3 or ≥4, so the optimal solution must lie in one of these subproblems (we have not eliminated any feasible solutions).
The next slide shows the two subproblems.
© The McGraw-Hill Companies, Inc., 2008

34. How Integer Programs are Solved
The left subproblem (call it subproblem #1, with x1 ≤ 3) has an optimal solution of (3, 4.3). This solution is not integer either, so it is split into two subproblems (#1a and #1b), one with the added constraint x2≤4, and one with the added constraint x2≥5. Subproblem #1a would have a solution of (3,4) which is feasible and integer. Subproblem #1b would have no feasible solutions. (3,4) becomes the “incumbent” solution (the best feasible solution found so far).
The right subproblem (subproblem #2) has a solution of (4, 3.2). This solution is not integer either, so it is split into two subproblems (#2a and #2b), one with the added constraint x2≤3, and one with the added constraint x2≥4. Subproblem #2a would have a solution of (4.1,3) which is not integer. However, it has a lower objective function value than the “incumbent” found in Subproblem #1a (3, 4), so we can eliminate it from further consideration. Subproblem #2b would have no feasible solutions.
Therefore, since all other subproblems have been eliminated, the “incumbent” (3,4) is the optimal solution.
The main point: We (or the Solver) have to solve SEVEN LP’s for this simple two-variable problem. With more variables, the number of potential subproblems can explode. This helps explain why integer programs are so difficult to solve.
© The McGraw-Hill Companies, Inc., 2008

35. Applications of Binary Variables
Making “yes-or-no” type decisions
Build a factory?
Manufacture a product?
Do a project?
Assign a person to a task?
Set-covering problems
Make a set of assignments that “cover” a set of requirements.
Fixed costs
If a product is produced, must incur a fixed setup cost.
If a warehouse is operated, must incur a fixed cost.
© The McGraw-Hill Companies, Inc., 2008

36. 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 ($million)
Cash Available ($million)
Project 1
Project 2
Project 3
Project 4
Year 1 9 7 6 11 28
Year 2 4 3 13
Year 3 10
NPV 30 16 22 14
Question: Which projects should be undertaken?
© The McGraw-Hill Companies, Inc., 2008

37. 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: 9y1 + 7y2 + 6y3 + 11y4 ≤ 28 ($million) Year 2 (cumulative): 15y1 + 11y2 + 9y3 + 11y4 ≤ 41 ($million) Year 3 (cumulative): 21y1 + 11y2 + 13y3 + 11y4 ≤ 51 ($million) and
yi are binary (i = 1, 2, 3, 4).
© The McGraw-Hill Companies, Inc., 2008

38. Spreadsheet Solution
© The McGraw-Hill Companies, Inc., 2008

39. 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?
At least one of projects 1, 2, or 3
y1 + y2 + y3 ≥ 1
Project 2 can’t be done unless project 3 is done
y2 ≤ y3
Either project 3 or project 4, but not both
y3 + y4 ≤ 1 (or = 1 if you must do one or the other)
No more than two projects total
y1 + y2 + y3 + y4 ≤ 2
© The McGraw-Hill Companies, Inc., 2008

40. Example #2 (Set Covering Problem)
The Washington State legislature is trying to decide on locations at which to base search-and-rescue 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 search-and-rescue teams be located?
© The McGraw-Hill Companies, Inc., 2008

41. The Counties of Washington State
© The McGraw-Hill Companies, Inc., 2008

42. 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: y1 + y2 ≥ 1 County 2 covered: y1 + y2 + y3 + y6 + y7 ≥ 1
County 3 covered: y2 + y3 + y4 + y7 + y8 + y14 ≥ 1 : County 37 covered: y32 + y36 + y37 ≥ 1 and
yi are binary (i = 1, 2, … , 37).
© The McGraw-Hill Companies, Inc., 2008

43. Spreadsheet Solution
There are multiple optima, with 8 teams, for this model.
Other things to consider:
Most populous counties (King, Pierce, and Snohomish: 12, 13, and 11) do not have a team. Might want to add a constraint to ensure that these counties have a team, or add an incentive based on the population of the counties.
Some counties (e.g. 23) are large while some are small. Might be better to use a different way to break up the state, rather than counties.
Other applications:
locating fire stations, police stations, etc.
assigning people to project teams (e.g., for project 1 we need someone with skills in finance, engineering, accounting, and management science. Fred has skills in finance and accounting, Ann has skills in finance and management science, Sarah has skills in engineering, etc. -- how should people be assigned to the various projects so that each project has all of its required skills covered).
© The McGraw-Hill Companies, Inc., 2008

44. 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
Pewter (pounds)
Sales Price
Variable Cost
Platter 3 5
$100
$60
Bowl 1 4 85 50
Pitcher 75 40
Available
130
240
Question: Which products should be produced, and in what quantity?
© The McGraw-Hill Companies, Inc., 2008

45. 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).
The 99 in the constraints needs to be big enough that it will not constrain the corresponding x variables so long as y is equal to 1.
The labor and pewter constraints are such that 99 is big enough.
Rather than 99, an upper bound could be computed based upon the labor and pewter constraints. For example, we know that no more than 60 bowls can be produced because of the pewter constraint. Thus, 60 would be large enough.
© The McGraw-Hill Companies, Inc., 2008

46. Spreadsheet Solution Other applications:
Warehouse location: total shipped to warehouse ≤ capacityj * yj
© The McGraw-Hill Companies, Inc., 2008

47. Applications of Binary Variables
Making “yes-or-no” type decisions
Build a factory?
Manufacture a product?
Do a project?
Assign a person to a task?
Fixed costs
If a product is produced, must incur a fixed setup cost.
If a warehouse is operated, must incur a fixed cost.
Either-or constraints
Production must either be 0 or ≥ 100.
Subset of constraints
meet 3 out of 4 constraints.
Slides 7.47–7.59 are based upon a lecture from the MBA elective “Modeling with Spreadsheets” at the University of Washington (as taught by one of the authors).
© The McGraw-Hill Companies, Inc., 2008

48. Capital Budgeting with Contingency Constraints (Yes-or-No 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?
For each potential project, a binary variable is used to determine whether the project should be undertaken.
© The McGraw-Hill Companies, Inc., 2008

49. Data for Capital Budgeting Problem
Cash Outflow Required ($million)
Cash Available ($million)
Project 1 2 3 4 5 6 7 8 9 10
Year 1 25
Year 2
Year 3
Year 4
Year 5
NPV 20 22 30 42 18 35 28 33
($million)
© The McGraw-Hill Companies, Inc., 2008

50. Spreadsheet Solution
The cumulative cash flows are accounted for, so that any money not used in a given year is available in the next.
© The McGraw-Hill Companies, Inc., 2008

51. 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 B C D E
Fixed Startup Cost
$2,450
$1,600
$1,000
$1,250
$2,200
Variable Cost (per MW) $3 $4 $6 $5
Capacity (MW)
2,000
2,800
4,300
2,100
For each: a continuous variable to determine how many MW to generate (e.g., MWA).
For each: a binary variable to determine whether or not to start up the generator (e.g., yA).
For each, a capacity constraint combined with enforcing that the binary variable equals 1 if any electricity is generated (e.g., 0 ≤ MWA ≤ 2,000yA).
Question: Which generators should be started up to meet the total capacity needed for the day (6000 MW)?
© The McGraw-Hill Companies, Inc., 2008

52. Spreadsheet Solution
© The McGraw-Hill Companies, Inc., 2008

53. Quality Furniture (Either-Or Constraints)
Reconsider the Quality Furniture Problem:
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.
Now suppose that they would not produce any fewer than 200 units of either product (i.e., either produce 0 or at least 200).
Question: What product mix will maximize their total profit?
The Quality Furniture problem was first introduced in the Chapter 4 powerpoint slides under the UW lecture.
Introduce a binary variable for each product (yB, yT): 1 if produce; 0 otherwise.
Add constraint (e.g., for benches B): 120yB ≤ B ≤ MyB, where M is a big number (must be bigger than the maximum possible).
When yB = 0, this results in 0 ≤ B ≤ 0.
When yB = 1, this results in 200 ≤ B ≤ M
© The McGraw-Hill Companies, Inc., 2008

54. Spreadsheet Solution
© The McGraw-Hill Companies, Inc., 2008

55. 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
Add binary variable for each constraint: yi = 1 if enforce; 0 otherwise.
Add constraint sum of yi ≥ 3 (meet at least 3 of the constraints).
© The McGraw-Hill Companies, Inc., 2008

56. 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)
18x1 + 21x2 + 15x3 ≤ 1,600 + M (1 – y4)
where M is a large number.
Add binary variable for each constraint: yi = 1 if enforce; 0 otherwise.
Add constraint sum of yi ≥ 3 (meet at least 3 of the constraints).
© The McGraw-Hill Companies, Inc., 2008

57. 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?
© The McGraw-Hill Companies, Inc., 2008

58. Data for Facility Location Problem
Fixed Cost (per month)
(Shipping + Production) Cost (per unit)
Capacity (units per month)
WH #1
WH #2
WH #3
Plant 1
$42,000
$650
$750
$850
400
Plant 2
50,000
500
350
550
300
Plant 3
45,000
450
Plant 4
600
Plant 5
47,000
375
Fixed Cost (per month)
Shipping Cost (per unit)
Capacity (per month)
Cust. 1
Cust. 2
Cust. 3
Cust. 4
WH #1
$45,000
$25
$65
$70
$35
600
WH #2
25,000 50 25 40 60
400
WH #3
65,000 20 45
900
Demand:
250
225
200
275
© The McGraw-Hill Companies, Inc., 2008

59. Spreadsheet Solution
© The McGraw-Hill Companies, Inc., 2008