7

1. ? 7 ??????????????/???? ???? 7.2 ????:????????(7.1?) 7.2�7.11 ?? BIP ?????:Tazer????(7.2?) 7.12�7.15 ?? BIP ???????????:??????(7.3?) 7.16�7.19 ?? BIP ???????:????????(7.4?) 7.20�7.24 ???? BIP ????????????????????? (7.5?) 7.25�7.30 ???? ??????(?????????) 7.31�7.46 ???????(?????????) 7.47�7.59

2. 7-2 ???????? ???????(binary variable)?? 0 ? 1???????,????????????/????(yes-or-no decisions)? ??: ?????????????? ???????????????? ??????????????????

3. 7-3 ????????? ?????????????????,???????????????,??????????????? ???????????????????????,????????? ????????????????,????????????????? ??:?????????????????????(?)???

4. 7-4 ?????????? Table 7.1 Data for the California Manufacturing Company problem.Table 7.1 Data for the California Manufacturing Company problem.

5. 7-5 ??????? Table 7.2 Binary decision variables for the California Manufacturing Co. problem.Table 7.2 Binary decision variables for the California Manufacturing Co. problem.

6. 7-6 ???

7. 7-7 ????? 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.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.

8. 7-8 ?????????????? 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.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.

9. 7-9 ??????? ??????????????? 1,000 ???? ?????,??????????????????,??????? ?????????,???????? 900 ???,??? 100 ?????????????? ???????? 900 ?????,?????????(????? 1,300 ????? 900 ???)? ??????? 100 ???(? 1,000 ????? 1,100 ???),???? 400 ????????(? 1,300 ???? 1,700 ???)?????????????? ????????????,???????????????????,??????????(??????????? 1,700 ???)?

10. 7-10 ?????? ???? ?????????????? ??:Turkish Petroleum Refineries (1990), South African National Defense Force (1997), Grantham, Mayo, Van Otterloo and Company (1999) ???? ???????????????????? ??:AT&T (1990) ????????? ????????????????????????????????????????????????????????????????????????????????? ??:Ault Foods (1994), Digital Equipment Corporation (1995)

11. 7-11 ??????(?) ???? ?????????????????????????????????????????????????????? ??:Quality Stores (1987), Air Products and Chemicals, Inc. (1983), Reynolds Metals Co. (1991), Sears, Roebuck and Company (1999) ???????? ???????????????? ??:Texas Stadium (1983), China (1995) ???????? ???????????????? ??:Homart Development (1987) ?????? ???????????????????????????????????????? ??:American Airlines (1989, 1991), Air New Zealand (2001)

12. 7-12 ????:Tazer ???? Tazer ???????,??????????????? ??????? R&D ??: ????:????????????,??????????????? ????:??????????? ????:???????????????? ????:????? HIV ?????? ????:????????????? ????????12 ???(??????????)? ??:???????????

13. 7-13 Tazer ????????????? Table 7.3 Data for the Tazer Project Selection ProblemTable 7.3 Data for the Tazer Project Selection Problem

14. 7-14 Tazer ???????????? ? xi = 1 ????? i ; 0 ?? ( i = 1, 2, 3, 4, 5) ??? P = 300x1 + 120x2 + 170x3 + 100x4 + 70x5 (????) ??? ????: 400x1 + 300x2 + 600x3 + 500x4 + 200x5 = 1,200 (????) ? xi ???? ( i = 1, 2, 3, 4, 5)

15. 7-15 Tazer ?????????????? 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.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.

16. 7-16 ???????????:?????? ?????????,???????????????? ???????????,????????? ?????????????????????? ??:?????????????????? ???: ???? = 10 ??

17. 7-17 ????????????????? Table 7.4 Response-time and cost data for the Caliente City problem.Table 7.4 Response-time and cost data for the Caliente City problem.

18. 7-18 ????????? ? xj = 1 ???? j ?????;??? 0 (j = 1, 2, � , 8) ??? C = 350x1 + 250x2 + 450x3 + 300x4 + 50x5 + 400x6 + 300x7 + 200x8 ??? ?? 1: x1 + x2 + x4 = 1?? 2: x1 + x2 + x3 = 1?? 3: x2 + x3 + x6 =�1?? 4: x1 + x4 + x7 =�1?? 5: x5 + x7 = 1?? 6: x3 + x6 + x8 =�1?? 7: x4 + x7 + x8 = 1?? 8: x6 + x7 + x8 = 1 ? xj ???? ( j = 1, 2, � , 8)

19. 7-19 ??????????? 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.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.

20. 7-20 ??????:???????? ?????????????,?????????????????? ???????:??????????(San Francisco,?? SFO)???????? 11 ??????? ??: ???????????????????,?? 11 ???????????

21. 7-21 ??????? Figure 7.5 The arrows show the 11 Southwestern Airways flights that need to be covered by the three crews based in San Francisco.Figure 7.5 The arrows show the 11 Southwestern Airways flights that need to be covered by the three crews based in San Francisco.

22. 7-22 ??????????? Table 7.5 Data for example 3 (the Southwestern Airways problem).Table 7.5 Data for example 3 (the Southwestern Airways problem).

23. 7-23 ??????????

24. 7-24 ???????????? 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.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.

25. 7-25 ????????????? ?????????????: 1. ??????????????(???)????, ???????????(setup cost)? 2. ???????,???????????????????? D ? W ??????????????,??????????,?????????????

26. 7-26 ?????????? This graph summarizes the application of the graphical method to the original Wyndor problem.This graph summarizes the application of the graphical method to the original Wyndor problem.

27. 7-27 ?????????????? Table 7.6 Net profit ($) for Variation 1 of the Wyndor Problem with Setup CostsTable 7.6 Net profit ($) for Variation 1 of the Wyndor Problem with Setup Costs

28. 7-28 ?????????????? 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.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.

29. 7-29 ??????????????

30. 7-30 ???????????????? 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).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).

31. 7-31 ???? ?????????????? ????????? ??:??????? ????? ??:???? ?????????? ?????????????? ??????? ??:? 114.286 ?? 114 ?????? ??????????????? ??????? ??:? 2.6 ?? 2 ? 3 ?????? ????? ??????? 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). 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).

32. 7-32 ????????????? ???????????????? ???????????????????? ???????????????? ??:??????30 ?????? LP ????????????????,???????????? 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 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.

33. 7-33 ????????? 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.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.

34. 7-34 ????????(?) 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 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.

35. 7-35 ???????? ???/??????? ??????? ??????? ??????? ???????????? ?????? ???????????????? ???? ??????????,???????????? ?????????,?????????

36. 7-36 ?? # 1(????) Norwood ?????????????????? ???????????????? ??????????????????????????????

37. 7-37 Norwood ????????????

38. 7-38 Norwood ?????????????

39. 7-39 ?????(????????) ?????? 1?2?3 ????? ?????? 3,???????? 2? ?? 3 ??? 4 ?????,??????? ?????????????? ??:??????????,?????????? 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 = 2At 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

40. 7-40 ?? # 2(??????) ??????????????????????? ???????????,?????????????????? ????????,?????????????????????????????? ??:????????????

41. 7-41 ??????

42. 7-42 ???

43. 7-43 ???? 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).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).

44. 7-44 ?? # 3(????) Woodridge??(??????????)????????????: ?????????? ??????????????????????????????????????,????: ????? $400??,???? $250 ??,????? $300 ??? ?????????????????????????????????

45. 7-45 ??? 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 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.

46. 7-46 ???? Other applications: Warehouse location: total shipped to warehouse = capacityj * yjOther applications: Warehouse location: total shipped to warehouse = capacityj * yj

47. 7-47 ???????? ??????????? ??????? ??????? ??????? ???????????? ???? ???????,??????????? ???????,????????? ?????? (Either-or constraints) ????? = 0 ? = 100 ????? (Subset of constraints) 4 ???????????? 3 ? 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).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).

48. 7-48 ????????????(?????) ??????????????????? ??????? 10 ????????? ?????????????????,?????????????? ??,???????????(contingency constraints): ????????1?2?3?????? ?? 4 ? ?? 5 ???????? ???? 6 ???,???? 7 ????? ??:??????????? For each potential project, a binary variable is used to determine whether the project should be undertaken. For each potential project, a binary variable is used to determine whether the project should be undertaken.

49. 7-49 ???????????

50. 7-50 ???? The cumulative cash flows are accounted for, so that any money not used in a given year is available in the next. The cumulative cash flows are accounted for, so that any money not used in a given year is available in the next.

51. 7-51 ???????(????) ????????????? ????,???????(start up),????????????(startup cost)? ??????????????? 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).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).

52. 7-52 ????

53. 7-53 ????(??????) ????????: ?????????????????????????(?????)????, ?????????1,600 ?????????????? 9,000 ??????????????? 3 ?????? 12 ??????????? 6 ?????? 38 ?????????????????? $8 ? $18? ?????????????? 200 ?(??,?? 0 ? ?? 200 ?)? ??:?????????????????? 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 = MThe 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

54. 7-54 ????

55. 7-55 ??????? ??????????????????,????????? 4 ?????? 3 ???? 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). 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).

56. 7-56 ???????(?) ? yi = 1 ?????? i ; 0 ?? ???: 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) ?? M ??????? 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). 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).

57. 7-57 ???? ???????,?? 5 ???? 3 ???,?? 4 ????????? ??????,??????????????????????,???????? ???????????????????????????????????? ???????????????????????????????? ??: ???????????? ???????????? ??????????????? ??????????????,???????????????????

58. 7-58 ??????????

59. 7-59 ????