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运筹学课后答案5

CHAPTER 5 WHAT-IF ANALYSIS FOR LINEAR PROGRAMMING

Review Questions

5.1-1 The parameters of a linear programming model are the constants (coefficients or right-hand sides)

in the functional constraints and the objective function.

5.1-2 Many of the parameters of a linear programming model are only estimates of quantities that cannot

be determined precisely and thus result in inaccuracies.

5.1-3 What-if analysis reveals how close each of these estimates needs to be to avoid obtaining an

erroneous optimal solution, and therefore pinpoints the sensitive parameters where extra care is

needed to refine their estimates.

5.1-4 No, if the optimal solution will remain the same over a wide range of values for a particular

coefficient, then it may be appropriate to make only a fairly rough estimate for a parameter of a

model.

5.1-5 Conditions that impact the parameters of a model, such as unit profit, may change over time and

render them inaccurate.

5.1-6 If conditions change, what-if analysis leaves signposts that indicate whether a resulting change in a

parameter of the model changes the optimal solution.

5.1-7 Sensitivity analysis is studying how changes in the parameters of a linear programming model

affect the optimal solution.

5.1-8 What-if analysis provides guidance about what the impact would be of altering policy decisions

that are represented by parameters of a model.

5.2-1 The estimates of the unit profits for the two products are most questionable.

5.2-2 The number of hours of production time that is being made available per week in the three plants

might change after analysis.

5.3-1 The allowable range for a coefficient in the objective function is the range of values over which the

optimal solution for the original model remains optimal.

5.3-2 If the true value for a coefficient in the objective function lies outside its allowable range then the

optimal solution would change and the problem would need to be resolved.

5.3-3 The Objective Coefficient column gives the current value of each coefficient. The Allowable

Increase column and the Allowable Decrease Column give the amount that each coefficient may

differ from these values to remain within the allowable range for which the optimal solution for the original model remains optimal.

5.4-1 The 100% rule considers the percentage of the allowable change (increase or decrease) for each

coefficient in the objective function.

5.4-2 If the sum of the percentage changes do not exceed 100% then the original optimal solution

definitely will still be optimal.

5.4-3 No, exceeding 100% may or may not change the optimal solution depending on the directions of

the changes in the coefficients.

5.5-1 The parameters in the constraints may only be estimates, or, especially for the right-hand-sides,

may well represent managerial policy decisions.

5.5-2 The right-hand sides of the functional constraints may well represent managerial policy decisions

rather than quantities that are largely outside the control of management.

5.5-3 The shadow price for a functional constraint is the rate at which the value of the objective function

can be increased by increasing the right-hand side of the constraint by a small amount.

5.5-4 The shadow price can be found with the spreadsheet by increasing the right-hand side by one,

and then re-solving to determine the increase in the objective function value. It can be found

similarly with a Solver Table by creating a table that shows the increase in profit for a unit increase in the right-hand side. The shadow price is given directly in the sensitivity report.

5.5-5 The shadow price for a functional constraint informs management about how much the total profit

will increase for each extra unit of a resource (right-hand-side of a constraint).

5.5-6 Yes. The shadow price also indicates how much the value of the objective function will decrease if

the right-hand side were to be decreased by 1.

5.5-7 A shadow price of 0 tells a manager that a small change in the right-hand side of the constraint will

not change the objective function value at all.

5.5-8 The allowable range for the right-hand side of a functional constraint is found in the Solver’s

sensitivity report by using the columns labeled ―Constraint R.H. Side‖, ―Allowable increase‖, and ―Allowable decrease‖.

5.5-9 The allowable ranges for the right-hand sides are of interest to managers because they tell them

how large changes in the right-hand sides can be before the shadow prices are no longer

applicable.

5.6-1 There may be uncertainty about the estimates for a number of the parameters in the functional

constraints. Also, the right-hand sides of the constraints often represent managerial policy

decisions. These decisions are frequently interrelated and so need to be considered simultaneously.

5.6-2 The spreadsheet can be used to directly determine the impact of several simultaneous changes.

Simply change the paremeters and re-solve.

5.6-3 Using Solver Table, trial values can be enumerated simultaneously for one or two data cells, with

the possibility of entering formulas for additional data cells in terms of these one or two data cells.

5.6-4 The right-hand sides of the constraints often represent managerial policy decisions. These

decisions are frequently interrelated and so need to be considered simultaneously.

5.6-5 The 100 percent rule basically says that we can safely use the shadow prices to predict the effect

of simultaneous changes in the right-hand sides if the sum of the percentages of the changes does not exceed 100 percent.

5.6-6 The data needed to apply the 100% rule for simultaneous changes in right-hand sides are given by

the Sensitivity Report (Constraint R.H. Side, Allowable Increase, and Allowable Decrease).

5.6-7 If the sum of the percentage changes does not exceed 100%, the shadow prices definitely will still

be valid.

5.6-8 If the sum of the percentages of allowable changes in the right-hand sides does exceed 100%,

then we cannot be sure if the shadow prices will still be valid.

Problems

5.1 a)

The estimate of the unit profit for toys can decrease by somewhere between $0 and $0.50

before the optimal solution will change. There is no change in the solution for an increase in the

unit profit for toys (at least for increase up to $1).

The estimate of the unit profit for subassemblies can decrease by somewhere between $0 and $0.50 before the optimal solution will change. There is no change in the solution for an

increase in the unit profit for subassemblies (at least for increases up to $1).

d) Solver Table for change in unit profit for toys (part b):

11 12 13 14 15 16 17 18 19 20 21 22

A B C D

Unit Profit

for Toys Toys Subassemblies Total Profit

2,0001,000$3,500 $2.0010000$2,000 $2.2510000$2,250 $2.5010000$2,500 $2.7520001000$3,000 $3.0020001000$3,500 $3.2520001000$4,000 $3.5020001000$4,500 $3.7520001000$5,000 $4.0020001000$5,500

Production

Solver Table for change in unit profit for subassemblies (part c):

11 12 13 14 15 16 17 18 19 20 21 22

A B C D

Unit Profit

for Subassemblies Toys Subassemblies Total Profit

2,0001,000$3,500 -$3.5010000$3,000

-$3.2510000$3,000

-$3.0010000$3,000

-$2.7520001000$3,250

-$2.5020001000$3,500

-$2.2520001000$3,750

-$2.0020001000$4,000

-$1.7520001000$4,250

-$1.5020001000$4,500

Production

e) The unit profit for toys can vary between $2.50 and $5.00 before the solution changes.

The unit profit for subassemblies can vary between (–$3.00) and (–$1.50) before the solution changes.

f) The allowable range for the unit profit for toys is $2.50 to $5.00.

The allowable range for the unit profit for subassemblies as (–$3.00) to (–$1.50).

Adjus table Cells

Final Reduced Obj ectiv e Allow able Allow able Cell Name Value Cost Coefficient

Increase

Decrease

$B$9Production Toys

2,0000320.5$C$9Production Subassemblies

1,000

-2.5

0.5

1112131415161718192021

C D E F G H I J K Total Profit

Unit Profit for Subass emblies $3,500-$3.50-$3.25-$3.00-$2.75-$2.50-$2.25-$2.00-$1.75-$1.50$2.00$2,000$2,000$2,000$2,000$2,000$2,000$2,000$2,250$2,500$2.25$2,250$2,250$2,250$2,250$2,250$2,250$2,500$2,750$3,000$2.50

$2,500$2,500$2,500$2,500$2,500$2,750$3,000$3,250$3,500Unit Profit $2.75$2,750$2,750$2,750$2,750$3,000$3,250$3,500$3,750$4,000for Toys $3.00

$3,000$3,000$3,000$3,250$3,500$3,750$4,000$4,250$4,500$3.25$3,250$3,250$3,500$3,750$4,000$4,250$4,500$4,750$5,000$3.50$3,500$3,750$4,000$4,250$4,500$4,750$5,000$5,250$5,500$3.75$4,000$4,250$4,500$4,750$5,000$5,250

$5,500$5,750$6,000$4.00

$4,500

$4,750

$5,000

$5,250$5,500$5,750$6,000$6,250$6,500

h) So long as the sum of the percentage change of the unit profit for the subassemblies does not

exceed 100% (where the allowable increase and decrease are given in part f), then the solution will not change. 5.2

a) Adjus table Cells

Final Reduced Obj ectiv e Allow able Allow able Cell Name

Value Cost Coefficient

Increase Decrease $B$9Solution Activity 16020.50.33333

$C$9Solution Activity 22

Constraints

Final Shadow Constraint Allow able Allow able Cell Name Value Price R.H. Side

Increase

Decrease

$D$5Resource 1 Used 1011022$D$6Resource 2 Used

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

A B C D Unit Profit for Total Activity 1Activity 1Activity 2Profit

62$22.00 $1.0004$20.00 $1.2004$20.00 $1.4004$20.00 $1.6004$20.00 $1.8062$20.80 $2.0062$22.00 $2.2062$23.20 $2.4062$24.40 $2.60100$26.00 $2.80100$28.00 $3.00100$30.00 Unit Profit for Total Activity 2Activity 1Activity 2Profit

62$22.00 $2.50100$20.00 $3.00100$20.00 $3.50100$20.00 $4.0062$20.00 $4.5062$21.00 $5.0062$22.00 $5.5062$23.00 $6.0004$24.00 $6.5004$26.00 $7.0004$28.00 $7.5004$30.00

Solution

The allowable range for the unit profit of activity 1 is approximately between $1.60 and $1.80 up to between $2.40 and $2.60.

The allowable range for the unit profit of activity 2 is between $3.50 and $4.00 up to between $5.50 and $6.00.

e) The allowable range for the unit profit of activity 1 is approximately between $1.67 and $2.50.

The allowable range for the unit profit of activity 2 is between $4 and $6.

f) The allowable range for the unit profit of activity 1 is approximately between $1.67 and $2.50.

The allowable range for the unit profit of activity 2 is between $4 and $6.

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

A B C D E F G H I J K L M Total Profit Unit Profit for Activity 2

$22$2.50$3.00$3.50$4.00$4.50$5.00$5.50$6.00$6.50$7.00$7.50

$1.00$11.00$12.00$14.00$16.00$18.00$20.00$22.00$24.00$26.00$28.00$30.00

$1.20$12.20$13.20$14.20$16.00$18.00$20.00$22.00$24.00$26.00$28.00$30.00

$1.40$14.00$14.40$15.40$16.40$18.00$20.00$22.00$24.00$26.00$28.00$30.00 Unit Profit$1.60$16.00$16.00$16.60$17.60$18.60$20.00$22.00$24.00$26.00$28.00$30.00 for$1.80$18.00$18.00$18.00$18.80$19.80$20.80$22.00$24.00$26.00$28.00$30.00 Activity 1$2.00$20.00$20.00$20.00$20.00$21.00$22.00$23.00$24.00$26.00$28.00$30.00 $2.20$22.00$22.00$22.00$22.00$22.20$23.20$24.20$25.20$26.20$28.00$30.00

$2.40$24.00$24.00$24.00$24.00$24.00$24.40$25.40$26.40$27.40$28.40$30.00

$2.60$26.00$26.00$26.00$26.00$26.00$26.00$26.60$27.60$28.60$29.60$30.60

$2.80$28.00$28.00$28.00$28.00$28.00$28.00$28.00$28.80$29.80$30.80$31.80

$3.00$30.00$30.00$30.00$30.00$30.00$30.00$30.00$30.00$31.00$32.00$33.00 Solution Unit Profit for Activity 2

(6,2)$2.50$3.00$3.50$4.00$4.50$5.00$5.50$6.00$6.50$7.00$7.50

$1.00(6,2)(0,4)(0,4)(0,4)(0,4)(0,4)(0,4)(0,4)(0,4)(0,4)(0,4)

$1.20(6,2)(6,2)(6,2)(0,4)(0,4)(0,4)(0,4)(0,4)(0,4)(0,4)(0,4)

$1.40(10,0)(6,2)(6,2)(6,2)(0,4)(0,4)(0,4)(0,4)(0,4)(0,4)(0,4) Unit Profit$1.60(10,0)(10,0)(6,2)(6,2)(6,2)(0,4)(0,4)(0,4)(0,4)(0,4)(0,4) for$1.80(10,0)(10,0)(10,0)(6,2)(6,2)(6,2)(0,4)(0,4)(0,4)(0,4)(0,4) Activity 1$2.00(10,0)(10,0)(10,0)(6,2)(6,2)(6,2)(6,2)(0,4)(0,4)(0,4)(0,4) $2.20(10,0)(10,0)(10,0)(10,0)(6,2)(6,2)(6,2)(6,2)(6,2)(0,4)(0,4)

$2.40(10,0)(10,0)(10,0)(10,0)(10,0)(6,2)(6,2)(6,2)(6,2)(6,2)(0,4)

$2.60(10,0)(10,0)(10,0)(10,0)(10,0)(10,0)(6,2)(6,2)(6,2)(6,2)(6,2)

$2.80(10,0)(10,0)(10,0)(10,0)(10,0)(10,0)(10,0)(6,2)(6,2)(6,2)(6,2)

$3.00(10,0)(10,0)(10,0)(10,0)(10,0)(10,0)(10,0)(6,2)(6,2)(6,2)(6,2)

5.3

Adjustable Cells

Final Reduced Obj ectiv e Allow able Allow able Cell Name Value Cost Coefficient Increase Decrease

$C$11Factory 1 Customer 11007001001E+30

$D$11Factory 1 Customer 220900100100

$E$11Factory 1 Customer 301008001E+30100

$C$12Factory 2 Customer 101008001E+30100

$D$12Factory 2 Customer 260900100100

$E$12Factory 2 Customer 3907001001E+30 Constraints

Final Shadow Constraint Allow able Allow able Cell Name Value Price R.H. Side Increase Decrease

$F$11Factory 1 Out1201201E+30

$F$12Factory 2 Out1501520

$C$13Total To Customer Customer 11070010010

$D$13Total To Customer Customer 28900802

$E$13Total To Customer Customer 39700902

a) All of the unit costs have a margin of error of 100 in at least one direction (increase or

decrease). Factory 1 to Customer 2 and Factory 2 to Customer 2 have the smallest margins

for error since it is 100 in both directions.

b) The allowable range for Factory 1 to Customer 1 is Unit Co st≤ $800.

The allowable range for Factory 1 to Customer 2 is $800 ≤ Unit Cost ≤ $1,000.

The allowable range for Factory 1 to Customer 3 is Unit Cost ≥ $700.

The allowable range for Factory 2 to Customer 1 is Unit Cost ≥ $700

The allowable range for Factory 2 to Customer 2 is $800 ≤ Unit Cost ≤ $900.

The allowable range for Factory 2 to Customer 3 is Unit Cost ≤ $800.

c) The allowable range for each unit shipping cost indicates how much that shipping cost can

change before you would want to change the shipping quantities used in the optimal solution.

d) Use the 100% rule for simultaneous changes in objective function coefficients. If the sum of

the percentage changes does not exceed 100%, the optimal solution definitely will still be

optimal. If the sum does exceed 100%, then we cannot be sure.

5.4 a) Optimal solution does not change.

d) The optimal solution does not change.

e) The optimal solution does not change.

Adjus table Cells

Final Reduced Obj ectiv e Allow able Allow able Cell Name Value Cost Coefficient Increase Decrease

$C$21Number Work ing Shift4801701E+3010

$D$21Number Work ing Shift31016010160

$E$21Number Work ing Shift3901755175

$F$21Number Work ing Shift4301801E+305

$G$21Number Work ing Shift1501951E+30195 Part a) Optimal solution does not change (within allowable increase of $10).

Part b) Optimal solution does change (outside of allowable decrease of $5).

Part c)

Percent of allowable increase for shift 2 is (165 – 160) / 10 = 50%

Percent of allowable decrease for shift 4 is (180 – 170) / 5 = 200%

Sum = 250%, so the optimal solution may or may not change.

Part d)

Percent of allowable decrease for shift 1 is (170 – 166) / 10 = 40%

Percent of allowable increase for shift 2 is (164 – 160) / 10 = 40%

Percent of allowable decrease for shift 3 is (175 – 171) / 175 = 2%

Percent of allowable increase for shift 4 is (184 –180) / ∞ = 0%

Percent of allowable increase fo shift 5 is (199 –195) / ∞ = 0%

The sum is 84%, so the optimal solution does not change.

Part e)

Percent of allowable increase for shift 1 is (173.40 –170) / ∞ = 0%

Percent of allowable increase for shift 2 is (163.20 – 160) / 10 = 32%

Percent of allowable increase for shift 3 is (178.50 – 175) / 5 = 70%

Percent of allowable increase for shift 4 is (183.60 –180) / ∞ = 0%

Percent of allowable increase for shift 5 is (198.90 –195) / ∞ = 0%

The sum is 102%, so the optimal solution may or may not change.

24 25 26 27 28 29 30 31 32 33 34 35 36 37

B C D E F G H Cost per Shift6am-2pm8am-4pm Noon-8pm4pm-midnight10pm-6am Total 6am-2pm Shift Shift Shift Shift Shift Cost 4831394315$30,610 $1555425394315$29,860 $1585425394315$30,022 $1614831394315$30,178 $1644831394315$30,322 $1674831394315$30,466 $1704831394315$30,610 $1734831394315$30,754 $1764831394315$30,898 $1794831394315$31,042 $1824831394315$31,186 $1854831394315$31,330

40 41 42 43 44 45 46 47 48 49 50 51 52 53

B C D E F G H Cost per Shift6am-2pm8am-4pm Noon-8pm4pm-midnight10pm-6am Total 8am-4pm Shift Shift Shift Shift Shift Cost 4831394315$30,610 $1454831394315$30,145 $1484831394315$30,238 $1514831394315$30,331 $1544831394315$30,424 $1574831394315$30,517 $1604831394315$30,610 $1634831394315$30,703 $1664831394315$30,796 $1694831394315$30,889 $1725425394315$30,970 $1755425394315$31,045

5657585960616263646566676869B

C D E

F G

Cost per Shift 6am-2pm 8am-4pm Noon-8pm 4pm-midnight 10pm-6am

Total Noon-8pm

Shift Shift Shift Shift Shift Cost 4831394315$30,610$1604831394315$30,025$1634831394315$30,142$1664831394315$30,259$1694831394315$30,376$1724831394315$30,493$1754831394315$30,610$1784831394315$30,727$1814831334915$30,838$1844831334915$30,937$1874831334915$31,036$1904831334915$31,135

7273747576777879808182838485B

C D E

F G

H Cost per Shift 6am-2pm 8am-4pm Noon-8pm 4pm-midnight 10pm-6am

Total 4pm-midnight

Shift Shift Shift Shift Shift Cost 4831394315$30,610$1654831334915$29,905$1684831334915$30,052$1714831334915$30,199$1744831334915$30,346$1774831394315$30,481$1804831394315$30,610$1834831394315$30,739$1864831394315$30,868$1894831394315$30,997$1924831394315$31,126$1954831394315$31,255

888990919293949596979899100101

C D E

F G

H Cost per Shift 6am-2pm 8am-4pm Noon-8pm 4pm-midnight 10pm-6am

Total 10pm-6am

Shift Shift Shift Shift Shift Cost 4831394315$30,610$1804831394315$30,385$1834831394315$30,430$1864831394315$30,475$1894831394315$30,520$1924831394315$30,565$1954831394315$30,610$1984831394315$30,655$2014831394315$30,700$2044831394315$30,745$2074831394315$30,790$210

4315

$30,835

5.5

b) The optimal solution does not change.

c) The optimal solution does not change.

d) The optimal solution does not change.

g) The optimal solution does not change.

Adjustable Cells

Final Reduced Obj ectiv e Allow able Allow able Cell Name Value Cost Coefficient Increase Decrease $C$16Participation Share Building0.00%-4.85%450.04851E+30 $D$16Participation Share Hotel16.50%0.00%700.45450.0543 $E$16Participation Share Center13.11%0.00%500.13890.3226 Constraints

Final Shadow Constraint Allow able Allow able Cell Name Value Price R.H. Side Increase Decrease $F$9Now Spent250.0097250.3049 4.3548 $F$10End of Year 1 Spent44.7570.0000451E+300.2427 $F$11End of Year 2 Spent60.5830.0000651E+30 4.4175 $F$12End of Year 3 Spent800.2233800.781218.8889 Part a) Optimal solution changes (not within allowable increase of $48,500).

Part b) Optimal solution does not change (within allowable increase of $454,500).

Part c) Optimal solution does not change (withi n allowable decrease of ∞).

Part d) Optimal solution does not change (within allowable decrease of $322,600).

Part e)

Percentage of allowable decrease for project 1 = (45 –40) / ∞ = 0%

Percentage of allowable increase for project 2 = (70.2 – 70) / 0.4545 = 44%

Percentage of allowable decrease for project 3 = (50 – 49.8) / 0.3226 = 62% Sum = 106%, so the solution may or may not change.

Part f)

Percentage of allowable increase for project 1 = (46 – 45) / 0.0485 = 2,062% Percentage of allowable decrease for project 2 = (70 – 69) / 0.0543 = 1,842% Percentage of allowable decrease for project 3 = (50 – 49) / 0.3226 = 310%

Sum = 4,214%, so the solution may or may not change.

Part g)

Percentage of allowable increase for project 1 = (54 – 45) / 0.0485 = 18,557% Percentage of allowable increase for project 2 = (84 – 70) / 0.4545 = 3,080% Percentage of allowable increase for project 3 = (60 – 50) / 0.1389 = 7,199% Sum = 28,836%, so the solution may or may not change.

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

B C D E F

Net Present Value Participation

Project 1 (Office)Office Shopping Total NPV ($millions)Building Hotel Center($millions)

0.00%16.50%13.11%18.11

400.00%16.50%13.11%18.11

410.00%16.50%13.11%18.11

420.00%16.50%13.11%18.11

430.00%16.50%13.11%18.11

440.00%16.50%13.11%18.11

450.00%16.50%13.11%18.11

4613.31% 6.12%15.65%18.23

4713.31% 6.12%15.65%18.36

4813.31% 6.12%15.65%18.49

4913.31% 6.12%15.65%18.63

5013.31% 6.12%15.65%18.76

36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

B C D E F

Net Present Value Participation

Project 2 (Hotel)Office Shopping Total NPV ($millions)Building Hotel Center($millions)

0.00%16.50%13.11%18.11

6513.31% 6.12%15.65%17.79

6613.31% 6.12%15.65%17.85

6713.31% 6.12%15.65%17.91

6813.31% 6.12%15.65%17.97

6913.31% 6.12%15.65%18.03

700.00%16.50%13.11%18.11

710.00%25.81%0.00%18.32

720.00%25.81%0.00%18.58

730.00%25.81%0.00%18.84

740.00%25.81%0.00%19.10

750.00%25.81%0.00%19.35

53 54 55 56 57 58 59 60 61 62 63 64 65 66 67

B C D E F

Net Present Value Participation

Project 3 (Shopping C.)Office Shopping Total NPV ($millions)Building Hotel Center($millions)

0.00%16.50%13.11%18.11

450.00%25.81%0.00%18.06

460.00%25.81%0.00%18.06

470.00%25.81%0.00%18.06

480.00%25.81%0.00%18.06

490.00%25.81%0.00%18.06

500.00%16.50%13.11%18.11

51 4.03%12.90%14.52%18.25

5213.31% 6.12%15.65%18.41

5313.31% 6.12%15.65%18.56

5413.31% 6.12%15.65%18.72

5513.31% 6.12%15.65%18.88

5.6 a) Optimal solution: produce no chocolate ice cream, 300 gallons of vanilla ice cream, and 75

gallons of banana ice cream. Total profit will be $341.25.

b) The optimal solution will change since $1.00 (an increase of $0.05) is outside the allowable

increase of $0.0214. The p rofit will go up, but how much can’t be determined without re-

solving.

c) The optimal solution will not change since $0.92 (a decrease of $0.03) is within the allowable

decrease ($0.05). Total profit will decrease by $2.25 ($0.03 x 75) to $339.

d) The optimal solution will change. Since the change is within the allowable range, we can

calculate the change in profit using the shadow price: ?Z = (Shadow Price)(?RHS) = ($1) x

(–3) = –$3. The new profit will be $338.25.

e) This increase is outside of the allowable increase so the total increase in profit with the extra

sugar can not be determined without re-solving. However, we know that the shadow price is

valid for the first increase of 10 pounds of sugar. For just this 10 pounds, the increase in profit

is ?Z = (Shadow Price)(?RHS) = ($1.875)(+10) = $18.75, so even just 10 pounds of sugar

would be worth the $15 price for 15 pounds.

f) The final value is 180 as shown in the E5 in the spreadsheet. The shadow price is 0 since we

are using less milk than we have available (there is slack in the constraint). The R.H.Side

value is 200 as given in cell G5. The allowable increase is infinity since the shadow price will

stay zero no matter how much we add to the right-hand side (since this would merely add to

the slack). The allowable decrease is 20 since the solution will change (and the shadow price

will change from zero) once the right-hand side drops below 180 (the amount currently being

used).

5.7 a) Let G = number of grandfather clocks produced

W = number of wall clocks produced

Maximize Profit = $300G + $200W

subject to 6G + 4W≤ 40 hours

8G + 4W≤ 40 hours

3G + 3W≤ 20 hours

and G≥ 0, W≥ 0.

b) 3.33 grandfather clocks and 3.33 wall clocks should be produced per week. If the unit profit

for grandfather clocks is changed from $300 to $375, the optimal solution does not change. If,

in addition, the estimated unit profit for wall clocks changes from $200 to $175, then the

optimal solution does change to 5 grandfather clocks and 0 wall clocks per week.

However, if the unit profit for wall clocks changes to $175 as well, then the optimal solution

15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

A B C D

Unit Profit

for Grandfather Grandfather Wall Total Clocks Clock Clock Profit

3.33 3.33$1,667

$1500 6.67$1,333 $1700 6.67$1,333 $1900 6.67$1,333 $210 3.33 3.33$1,367 $230 3.33 3.33$1,433 $250 3.33 3.33$1,500 $270 3.33 3.33$1,567 $290 3.33 3.33$1,633 $310 3.33 3.33$1,700 $330 3.33 3.33$1,767 $350 3.33 3.33$1,833 $370 3.33 3.33$1,900 $390 3.33 3.33$1,967 $41050$2,050 $43050$2,150 $45050$2,250

Production

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56

A B C D Unit Profit

for Wall Grandfather Wall Total Clocks Clock Clock Profit

3.33 3.33$1,667

$5050$1,500 $7050$1,500 $9050$1,500 $11050$1,500 $13050$1,500 $15050$1,500 $170 3.33 3.33$1,567 $190 3.33 3.33$1,633 $210 3.33 3.33$1,700 $230 3.33 3.33$1,767 $250 3.33 3.33$1,833 $270 3.33 3.33$1,900 $290 3.33 3.33$1,967 $3100 6.67$2,067 $3300 6.67$2,200 $3500 6.67$2,333

Production

1516171819202122232425

262728293031323334A B C

D E F G H Total Profit Unit Profit for Wall Clocks

$1,665$50$100

$150$200$250

$300$150$750$833$1,000$1,333$1,667$2,000$200$1,000$1,000$1,167$1,333$1,667$2,000Unit Profit $250$1,250$1,250$1,333$1,500$1,667$2,000for Grandfather $300$1,500$1,500$1,500$1,667$1,833$2,000Clocks $350$1,750$1,750$1,750$1,833$2,000$2,167$400$2,000$2,000$2,000$2,000$2,167$2,333$450$2,250$2,250

$2,250$2,250$2,333

$2,500

Production (Grandfather Clocks, Wall Clocks)Unit Profit for Wall Clocks (3.33,3.33)$50$100$150$200$250$300$150(5,0)(3.33,3.33)(3.33,3.33)(0,6.67)(0,6.67)(0,6.67)$200(5,0)(5,0)(3.33,3.33)(3.33,3.33)(0,6.67)(0,6.67)Unit Profit $250(5,0)(5,0)(3.33,3.33)(3.33,3.33)(0,6.67)(0,6.67)for Grandfather $300(5,0)(5,0)(5,0)(3.33,3.33)(3.33,3.33)(0,6.67)Clocks $350(5,0)

(5,0)(5,0)(3.33,3.33)(3.33,3.33)(3.33,3.33)$400(5,0)(5,0)(5,0)(5,0)(3.33,3.33)(3.33,3.33)$450(5,0)

(5,0)(5,0)(5,0)

(3.33,3.33)(3.33,3.33)

15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

A B C D Assembly Hours

Available Grandfather Wall

(David)Clock Clock Total Profit

3.33 3.33$1,667

35 3.33 3.33$1,667

37 3.33 3.33$1,667

39 3.33 3.33$1,667

41 3.33 3.33$1,667

43 3.33 3.33$1,667

45 3.33 3.33$1,667 Carving Hours

Available Grandfather Wall

(LaDeana)Clock Clock Total Profit

3.33 3.33$1,667

35 2.08 4.58$1,542

37 2.58 4.08$1,592

39 3.08 3.58$1,642

41 3.58 3.08$1,692

43 4.08 2.58$1,742

45 4.58 2.08$1,792 Shipping Hours

Available Grandfather Wall

(Lydia)Clock Clock Total Profit

3.33 3.33$1,667

15 5.000.00$1,500

17 4.33 1.33$1,567

19 3.67 2.67$1,633

21 3.00 4.00$1,700

23 2.33 5.33$1,767

25 1.67 6.67$1,833

《运筹学》课后习题答案

第一章线性规划1、由图可得：最优解为 2、用图解法求解线性规划： Min z=2x1+x2 ? ? ? ? ? ? ? ≥ ≤ ≤ ≥ + ≤ + - 10 5 8 24 4 2 1 2 1 2 1 x x x x x x 解：由图可得：最优解x=1.6,y=6.4

Max z=5x 1+6x 2 ? ?? ??≥≤+-≥-0 ,23222212 121x x x x x x 解：由图可得：最优解Max z=5x 1+6x 2, Max z= + ∞

Maxz = 2x 1 +x 2 ????? ? ?≥≤+≤+≤0,5242261552121211x x x x x x x 由图可得：最大值?????==+35121x x x ，所以?????==2 3 21x x max Z = 8.

12 12125.max 2328416412 0,1,2maxZ .j Z x x x x x x x j =+?+≤? ≤?? ≤??≥=?如图所示，在（4，2）这一点达到最大值为2 6将线性规划模型化成标准形式： Min z=x 1-2x 2+3x 3 ????? ??≥≥-=++-≥+-≤++无约束 321 321321321,0,05232 7x x x x x x x x x x x x 解：令Z ’=-Z,引进松弛变量x 4≥0，引入剩余变量x 5≥0，并令x 3=x 3’-x 3’’,其中x 3’≥ 0,x 3’’≥0 Max z ’=-x 1+2x 2-3x 3’+3x 3’’ ????? ? ?≥≥≥≥≥≥-=++-=--+-=+-++0 ,0,0'',0',0,05 232 '''7'''543321 3215332143321x x x x x x x x x x x x x x x x x x x

运筹学II习题解答

第七章决策论 1.某厂有一新产品，其面临的市场状况有三种情况，可供其选择的营销策略也是三种，每一钟策略在每一种状态下的损益值如下表所示，要求分别用非确定型（1）悲观法：根据“小中取大”原则，应选取的经营策略为s3；（2）乐观法：根据“大中取大”原则，应选取的经营策略为s1；（3）折中法（α=0.6）：计算折中收益值如下： S1折中收益值=0.6?50+0.4?(-5)=28 S2折中收益值=0.6?30+0.4?0=18 S3折中收益值=0.6?10+0.4?10=10 显然，应选取经营策略s1为决策方案。（4）平均法：计算平均收益如下： S1:x_1=（50+10-5）/3=55/3 S2:x_2=(30+25)/3=55/3 S3:x_3=(10+10)/3=10 故选择策略s1,s2为决策方案。（5）最小遗憾法：分三步第一，定各种自然状态下的最大收益值，如方括号中所示；第二，确定每一方案在不同状态下的最小遗憾值，并找出每一方案的最大遗憾值如圆括号中所示；第三，大中取小，进行决策。故选取S1作为决策方案。

2．如上题中三种状态的概率分别为: 0.3, 0.4, 0.3, 试用期望值方法和决策树方法决策。（1）用期望值方法决策：计算各经营策略下的期望收益值如下：故选取决策S2时目标收益最大。（2）用决策树方法，画决策树如下： 3. 某石油公司拟在某地钻井，可能的结果有三：无油(θ1)，贫油（θ2）和富油（θ3），估计可能的概率为：P (θ1) =0.5, P (θ2)=0.3，P (θ3)=0.2。已知钻井费为7万元，若贫油可收入12万元，若富油可收入27万元。为了科学决策拟先进行勘探，勘探的可能结果是：地质构造差(I1)、构造一般(I2)和构造好(I3)。根据过去的经验，地质构造与出油量间的关系如下表所示： P (I j|θi) 构造差(I1) 构造一般(I2) 构造好(I3) 无油(θ1) 0.6 0.3 0.1 贫油(θ2) 0.3 0.4 0.3 富油(θ3) 0.1 0.4 0.5 假定勘探费用为1万元, 试确定：

《管理运筹学》第二版课后习题参考答案

《管理运筹学》（第二版）课后习题参考答案第1章线性规划（复习思考题） 1．什么是线性规划线性规划的三要素是什么答：线性规划（Linear Programming ，LP ）是运筹学中最成熟的一个分支，并且是应用最广泛的一个运筹学分支。线性规划属于规划论中的静态规划，是一种重要的优化工具，能够解决有限资源的最佳分配问题。建立线性规划问题要具备三要素：决策变量、约束条件、目标函数。决策变量是决策问题待定的量值，取值一般为非负；约束条件是指决策变量取值时受到的各种资源条件的限制，保障决策方案的可行性；目标函数是决策者希望实现的目标，为决策变量的线性函数表达式，有的目标要实现极大值，有的则要求极小值。 2．求解线性规划问题时可能出现几种结果，哪种结果说明建模时有错误答：（1）唯一最优解：只有一个最优点；（2）多重最优解：无穷多个最优解；（3）无界解：可行域无界，目标值无限增大；（4）没有可行解：线性规划问题的可行域是空集。当无界解和没有可行解时，可能是建模时有错。 3．什么是线性规划的标准型松弛变量和剩余变量的管理含义是什么答：线性规划的标准型是：目标函数极大化，约束条件为等式，右端常数项0≥i b ，决策变量满足非负性。如果加入的这个非负变量取值为非零的话，则说明该约束限定没有约束力，对企业来说不是紧缺资源，所以称为松弛变量；剩余变量取值为非零的话，则说明“≥”型约束的左边取值大于右边规划值，出现剩余量。 4．试述线性规划问题的可行解、基础解、基可行解、最优解的概念及其相互关系。答：可行解：满足约束条件0≥=X b AX ，的解，称为可行解。基可行解：满足非负性约束的基解，称为基可行解。可行基：对应于基可行解的基，称为可行基。最优解：使目标函数最优的可行解，称为最优解。最优基：最优解对应的基矩阵，称为最优基。它们的相互关系如右图所示：

《管理运筹学》第四版课后习题解析(上)

《管理运筹学》第四版课后习题解析（上）第2章线性规划的图解法 1．解：（1）可行域为OABC 。（2）等值线为图中虚线部分。（3）由图2-1可知，最优解为B 点，最优解1x = 127，2157x =；最优目标函数值697 。图2-1 2．解：（1）如图2-2所示，由图解法可知有唯一解12 0.2 0.6x x =??=?，函数值为3.6。图2-2 （2）无可行解。（3）无界解。（4）无可行解。（5）无穷多解。

（6）有唯一解 12203 8 3x x ?=????=?? ，函数值为923。 3．解：（1）标准形式 12123max 32000f x x s s s =++++ 1211221231212392303213229,,,,0 x x s x x s x x s x x s s s ++=++=++=≥ （2）标准形式 1212min 4600f x x s s =+++ 12112212121236210764,,,0 x x s x x s x x x x s s --=++=-=≥ （3）标准形式 1 2212min 2200f x x x s s ''''=-+++ 12 211 2212221 2212355702555032230,,,,0x x x s x x x x x x s x x x s s '''-+-+=''''-+=''''+--=''''≥ 4．解：标准形式 1212max 10500z x x s s =+++ 1211221212349528,,,0 x x s x x s x x s s ++=++=≥ 松弛变量（0，0）最优解为 1x =1，x 2=3/2。 5．解：

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《运筹学》习题答案一、单选题 1.用动态规划求解工程线路问题时，什么样的网络问题可以转化为定步数问题求解（）B A.任意网络 B.无回路有向网络 C.混合网络 D.容量网络 2.通过什么方法或者技巧可以把工程线路问题转化为动态规划问题？（）B A.非线性问题的线性化技巧 B.静态问题的动态处理 C.引入虚拟产地或者销地 D.引入人工变量 3.静态问题的动态处理最常用的方法是？B A.非线性问题的线性化技巧 B.人为的引入时段 C.引入虚拟产地或者销地 D.网络建模 4.串联系统可靠性问题动态规划模型的特点是（）D A.状态变量的选取 B.决策变量的选取 C.有虚拟产地或者销地 D.目标函数取乘积形式 5.在网络计划技术中，进行时间与成本优化时，一般地说，随着施工周期的缩短，直接费用是( )。C A.降低的 B.不增不减的 C.增加的 D.难以估计的 6.最小枝权树算法是从已接接点出发，把( )的接点连接上C A.最远 B.较远 C.最近 D.较近 7.在箭线式网络固中，( )的说法是错误的。D A.结点不占用时间也不消耗资源 B.结点表示前接活动的完成和后续活动的开始 C.箭线代表活动 D.结点的最早出现时间和最迟出现时间是同一个时间 8.如图所示，在锅炉房与各车间之间铺设暖气管最小的管道总长度是( )。C A.1200 B.1400 C.1300 D.1700 9.在求最短路线问题中，已知起点到A，B，C三相邻结点的距离分别为15km，20km,25km，则（）。D A.最短路线—定通过A点 B.最短路线一定通过B点 C.最短路线一定通过C点 D.不能判断最短路线通过哪一点 10.在一棵树中，如果在某两点间加上条边，则图一定( )A A.存在一个圈 B.存在两个圈 C.存在三个圈 D.不含圈 11.网络图关键线路的长度( )工程完工期。C A.大于 B.小于 C.等于 D.不一定等于

管理学管理运筹学课后答案——谢家平

管理运筹学 ——管理科学方法谢家平第一章第一章 1. 建立线性规划问题要具备三要素：决策变量、约束条件、目标函数。决策变量（Decision Variable）是决策问题待定的量值，取值一般为非负；约束条件（Constraint Conditions）是指决策变量取值时受到的各种资源条件的限制，保障决策方案的可行性；目标函数（Objective Function）是决策者希望实现的目标，为决策变量的线性函数表达式，有的目标要实现极大值，有的则要求极小值。 2.（1）设立决策变量；（2）确定极值化的单一线性目标函数；（3）线性的约束条件：考虑到能力制约，保证能力需求量不能突破有效供给量；（4）非负约束。 3.（1）唯一最优解：只有一个最优点（2）多重最优解：无穷多个最优解（3）无界解：可行域无界，目标值无限增大（4）没有可行解：线性规划问题的可行域是空集无界解和没有可行解时，可能是建模时有错。 4. 线性规划的标准形式为：目标函数极大化，约束条件为等式，右端常数项bi≥0 , 决策变量满足非负性。如果加入的这个非负变量取值为非零的话，则说明该约束限定没有约束力，对企业来说不是紧缺资源，所以称为松弛变量；剩余变量取值为非零的话，则说明“≥”型约束的左边取值大于右边规划值，出现剩余量。 5. 可行解：满足约束条件AX =b,X≥0的解，称为可行解。基可行解：满足非负性约束的基解，称为基可行解。可行基：对应于基可行解的基，称为可行基。最优解：使目标函数最优的可行解，称为最优解。最优基：最优解对应的基矩阵，称为最优基。 6. 计算步骤：第一步，确定初始基可行解。第二步，最优性检验与解的判别。第三步，进行基变换。第四步，进行函数迭代。判断方式：唯一最优解：所有非基变量的检验数为负数，即σj< 0 无穷多最优解：若所有非基变量的检验数σj≤ 0 ，且存在某个非基变量xNk 的检验数σk= 0 ，让其进基，目标函数

最全的运筹学复习题及答案78213

四、把下列线性规划问题化成标准形式： 2、minZ=2x1-x2+2x3 五、按各题要求。建立线性规划数学模型 1、某工厂生产A、B、C三种产品，每种产品的原材料消耗量、机械台时消耗量以及这些资源的限量，单位产品的利润如下表所示：

根据客户订货，三种产品的最低月需要量分别为200，250和100件，最大月销售量分别为250，280和120件。月销售分别为 250，280和120件。问如何安排生产计划，使总利润最大。 2、某建筑工地有一批长度为10米的相同型号的钢筋，今要截成长度为3米的钢筋 90根，长度为4米的钢筋60根，问怎样下料，才能使所使用的原材料最省? 1．某运输公司在春运期间需要24小时昼夜加班工作，需要的人员数量如下表所示：起运时间服务员数 2—6 6—10 10一14 14—18 18—22 22—2 4 8 10 7 12 4 每个工作人员连续工作八小时，且在时段开始时上班，问如何安排，使得既满足以上要求，又使上班人数最少?

五、分别用图解法和单纯形法求解下列线性规划问题．并对照指出单纯形迭代的每一步相当于图解法可行域中的哪一个顶点。

六、用单纯形法求解下列线性规划问题：七、用大M法求解下列线性规划问题。并指出问题的解属于哪一类。

八、下表为用单纯形法计算时某一步的表格。已知该线性规划的目标函数为maxZ=5x1+3x2，约束形式为“≤”，X3，X4为松驰变量．表中解代入目标函数后得Z=10 X l X2X3X4 —10 b -1 f g X3 2 C O 1 1／5 X l a d e 0 1 (1)求表中a～g的值 (2)表中给出的解是否为最优解? （1）a=2 b=0 c=0 d=1 e=4/5 f=0 g=－5 （2）表中给出的解为最优解第四章线性规划的对偶理论五、写出下列线性规划问题的对偶问题 1．minZ=2x1+2x2+4x3

运筹学基础课后习题答案

运筹学基础课后习题答案 [2002年版新教材] 第一章导论 P5 1.、区别决策中的定性分析和定量分析，试举例。定性——经验或单凭个人的判断就可解决时，定性方法定量——对需要解决的问题没有经验时；或者是如此重要而复杂，以致需要全面分析（如果涉及到大量的金钱或复杂的变量组）时，或者发生的问题可能是重复的和简单的，用计量过程可以节约企业的领导时间时，对这类情况就要使用这种方法。举例：免了吧。。。 2、. 构成运筹学的科学方法论的六个步骤是哪些？ .观察待决策问题所处的环境； .分析和定义待决策的问题； .拟定模型； .选择输入资料； .提出解并验证它的合理性（注意敏感度试验）； .实施最优解； 3、．运筹学定义：利用计划方法和有关许多学科的要求，把复杂功能关系表示成数学模型，其目的是通过定量分析为决策和揭露新问题提供数量根据第二章作业预测P25 1、. 为了对商品的价格作出较正确的预测，为什么必须做到定量与定性预测的结合？即使在定量预测法诸如加权移动平均数法、指数平滑预测法中，关于权数以及平滑系数的确定，是否也带有定性的成分？答：（1）定量预测常常为决策提供了坚实的基础，使决策者能够做到心中有数。但单靠定量预测有时会导致偏差，因为市场千变万化，影响价格的因素很多，有些因素难以预料。调查研究也会有相对局限性，原始数据不一定充分，所用的模型也往往过于简化，所以还需要定性预测，在缺少数据或社会经济环境发生剧烈变化时，就只能用定性预测了。（2）加权移动平均数法中权数的确定有定性的成分；指数平滑预测中的平滑系数的确定有定性的成分。 2.、某地区积累了5 个年度的大米销售量的实际值（见下表），试用指数平滑法，取平滑系数α= 0.9，预测第6年度的大米销售量（第一个年度的预测值，根据专家估计为4181.9千公斤）年度 1 2 3 4 5 大米销售量实际值（千公斤）5202 5079 3937 4453 3979 。答： F6=a*x5+a(1-a)*x4+a(1-a)~2*x3+a(1-a)~3*x2+a(1-a)~4*F1 F6=0.9*3979+0.9*0.1*4453+0.9*0.01*3937+0.9*0.001*5079+0.9*0.0001*4181.9

运筹学(胡运权)第五版课后答案-运筹作业

47页1.1b 用图解法找不到满足所有约束条件的公共范围，所以该问题无可行解47页1.1d 无界解 1 2 3 4 5 4 3 2 1 - 1 -6 -5 -4 -3 -2 X2 X1 2x1- -2x1+3x 1 2 3 4 4 3 2 1 X1 2x1+x2=2 3x1+4x2= X

1.2（b）约束方程的系数矩阵A= 1 2 3 4 2 1 1 2 P1 P2 P3 P4 基基解是否可行解目标函数值X1 X2 X3 X4 P1 P2 -4 11/2 0 0 否 P1 P3 2/5 0 11/5 0 是43/5 P1 P4 -1/3 0 0 11/6 否 P2 P3 0 1/2 2 0 是 5 P2 P4 0 -1/2 0 2 否 P3 P4 0 0 1 1 是 5 最优解A=(0 1/2 2 0)T和(0 0 1 1)T 49页13题设Xij为第i月租j个月的面积 minz=2800x11+2800x21+2800x31+2800x41+4500x12+4500x22+4500x32+6000x1 3 +6000x23+7300x14 s.t. x11+x12+x13+x14≥15 x12+x13+x14+x21+x22+x23≥10 x13+x14+x22+x23+x31+x32≥20 x14+x23+x32+x41≥12 Xij≥0 用excel求解为： ( )

用LINDO求解： LP OPTIMUM FOUND AT STEP 3 OBJECTIVE FUNCTION V ALUE

运筹学习题答案

第一章习题 1.思考题（1）微分学求极值的方法为什么不适用于线性规划的求解？（2）线性规划的标准形有哪些限制？如何把一般的线性规划化为标准形式？（3）图解法主要步骤是什么？从中可以看出线性规划最优解有那些特点？（4）什么是线性规划的可行解，基本解，基可行解？引入基本解和基可行解有什么作用？（5）对于任意基可行解，为什么必须把目标函数用非基变量表示出来？什么是检验数？它有什么作用？如何计算检验数？（6）确定换出变量的法则是什么？违背这一法则，会发生什么问题？（7）如何进行换基迭代运算？（8）大M法与两阶段法的要点是什么？两者有什么共同点？有什么区别？（9）松弛变量与人工变量有什么区别？试从定义和处理方式两方面分析。（10）如何判定线性规划有唯一最优解，无穷多最优解和无最优解？为什么？ 2.建立下列问题的线性规划模型：（1）某厂生产A，B，C三种产品，每件产品消耗的原料和设备台时如表1-18所示：润最大的模型。（2）某公司打算利用具有下列成分（见表1-19）的合金配制一种新型合金100公斤，新合金含铅，锌，锡的比例为3：2：5。如何安排配方，使成本最低？（3）某医院每天各时间段至少需要配备护理人员数量见表1-20。

表1-20 假定每人上班后连续工作8小时，试建立使总人数最少的计划安排模型。能否利用初等数学的视察法，求出它的最优解？（4）某工地需要30套三角架，其结构尺寸如图1-6所示。仓库现有长6.5米的钢材。如何下料，使消耗的钢材最少？图1-6 3. 用图解法求下列线性规划的最优解： ?????? ?≥≤+-≥+≥++=0 ,425.134 1 2 64 min )1(21212 12121x x x x x x x x x x z ?????? ?≥≤+≥+-≤++=0 ,82 5 1032 44 max )2(21212 12121x x x x x x x x x x z ????? ????≥≤≤-≤+-≤++=0 ,6 054 4 22232 96 max )3(2122 1212121x x x x x x x x x x x z ??? ??≥≤+-≥+ +=0,1 12 34 3 max )4(2 12 12121x x x x x x x x z

《管理运筹学》课后习题答案

第2章线性规划的图解法 1．解： x ` A 1 (1) 可行域为OABC (2) 等值线为图中虚线部分 (3) 由图可知，最优解为B 点，最优解：1x = 712，7152=x 。最优目标函数值：769 2．解： x 2 1 0 1 (1) 由图解法可得有唯一解 6.02.021==x x ，函数值为3.6。 (2) 无可行解 (3) 无界解 (4) 无可行解 (5) 无穷多解

(6) 有唯一解 38320 21== x x ，函数值为392。 3．解： (1). 标准形式： 3212100023m ax s s s x x f ++++= 0,,,,9 2213 2330 2932121321221121≥=++=++=++s s s x x s x x s x x s x x (2). 标准形式： 21210064m in s s x x f +++= ,,,4 6710 26 3212121221121≥=-=++=--s s x x x x s x x s x x (3). 标准形式： 21''2'2'10022m in s s x x x f +++-= 0,,,,30 22350 55270 55321''2'2'12''2'2'1''2'2'11''2'21≥=--+=+-=+-+-s s x x x s x x x x x x s x x x 4．解：标准形式： 212100510m ax s s x x z +++= ,,,8259 432121221121≥=++=++s s x x s x x s x x 松弛变量（0，0）最优解为 1x =1，x 2=3/2.

管理运筹学第二版课后习题参考答案

管理运筹学第二版课后习题参考答案 Document number【980KGB-6898YT-769T8CB-246UT-18GG08】

如果加入的这个非负变量取值为非零的话，则说明该约束限定没有约束力，对企业来说不是紧缺资源，所以称为松弛变量；剩余变量取值为非零的话，则说明“≥”型约束的左边取值大于右边规划值，出现剩余量。 4．试述线性规划问题的可行解、基础解、基可行解、最优解的概念及其相互关系。答：可行解：满足约束条件0≥=X b AX ，的解，称为可行解。基可行解：满足非负性约束的基解，称为基可行解。可行基：对应于基可行解的基，称为可行基。最优解：使目标函数最优的可行解，称为最优解。最优基：最优解对应的基矩阵，称为最优基。它们的相互关系如右图所示： 5．用表格单纯形法求解如下线性规划。 . ??? ??≥≤++≤++0,,862383 21321321x x x x x x x x x 解：标准化 32124max x x x Z ++= . ?? ? ??≥=+++=+++0,,,,862385432153 214 321x x x x x x x x x x x x x 列出单纯形表

运筹学第五版课后答案,运筹作业

47页1.1b 用图解法找不到满足所有约束条件的公共范围，所以该问题无可行解47页1.1d 无界解

1.2（b）约束方程的系数矩阵 A= 1 2 3 4 ( ) 2 1 1 2 P1 P2 P3 P4 最优解A=(0 1/2 2 0)T和(0 0 1 1)T 49页13题设Xij为第i月租j个月的面积 minz=2800x11+2800x21+2800x31+2800x41+4500x12+4500x22+4500x32+6000x13 +6000x23+7300x14 s.t. x11+x12+x13+x14≥15 x12+x13+x14+x21+x22+x23≥10 x13+x14+x22+x23+x31+x32≥20 x14+x23+x32+x41≥12 Xij≥0 用excel求解为：

用LINDO求解： LP OPTIMUM FOUND AT STEP 3 OBJECTIVE FUNCTION VALUE 1) 118400.0 VARIABLE VALUE REDUCED COST Z 0.000000 1.000000 X11 3.000000 0.000000

X21 0.000000 2800.000000 X31 8.000000 0.000000 X41 0.000000 1100.000000 X12 0.000000 1700.000000 X22 0.000000 1700.000000 X32 0.000000 0.000000 X13 0.000000 400.000000 X23 0.000000 1500.000000 X14 12.000000 0.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 -2800.000000 3) 2.000000 0.000000 4) 0.000000 -2800.000000 5) 0.000000 -1700.000000 NO. ITERATIONS= 3 答若使所费租借费用最小，需第一个月租一个月租期300平方米，租四个月租期1200平方米，第三个月租一个月租期800平方米，

运筹学[胡运权]第五版课后答案,运筹作业

47页用图解法找不到满足所有约束条件的公共范围，所以该问题无可行解47页无界解

（b）约束方程的系数矩阵 A= 1 2 3 4 () 2 1 1 2 P1 P2 P3 P4 最优解A=(0 1/2 2 0)T和(0 0 1 1)T 49页13题设Xij为第i月租j个月的面积 minz=2800x11+2800x21+2800x31+2800x41+4500x12+4500x22+4500x32+6000x13 +6000x23+7300x14 . x11+x12+x13+x14≥15 x12+x13+x14+x21+x22+x23≥10 x13+x14+x22+x23+x31+x32≥20 x14+x23+x32+x41≥12 Xij≥0 用excel求解为：

用LINDO求解： LP OPTIMUM FOUND AT STEP 3 OBJECTIVE FUNCTION VALUE 1) VARIABLE VALUE REDUCED COST Z X11 X21 X31 X41 X12 X22 X32 X13 X23 X14 ROW SLACK OR SURPLUS DUAL PRICES 2) 3) 4) 5) NO. ITERATIONS= 3 答若使所费租借费用最小，需第一个月租一个月租期300平方米，租四个月租期1200平方米，第三个月租一个月租期800平方米， 50页14题设a1，a2，a3, a4, a5分别为在A1, A2, B1, B2, B3加工的Ⅰ产品数量，b1，b2，b3分别为在A1, A2, B1加工的Ⅱ产品数量，c1为在A2，B2上加工的Ⅲ产品数量。则目标函数为‘ maxz= a1+a2+a3)+( b3+( (a1+b1)- (a2+b2+c1)- (a3+b3)(a4+c1)-0.05a5 =0. 95a1+0. 97a2+0. 94a3++2.1c-0.11a-0.05a . 5a1+10b1≤6000 7a2+b2+12c1≤10000

运筹学[胡运权]第五版课后答案,运筹作业

运筹学[胡运权]第五版课后答案,运筹作业 -标准化文件发布号：（9456-EUATWK-MWUB-WUNN-INNUL-DDQTY-KII

47页1.1b 用图解法找不到满足所有约束条件的公共范围，所以该问题无可行解47页1.1d 无界解

用LINDO求解： LP OPTIMUM FOUND AT STEP 3 OBJECTIVE FUNCTION VALUE 1) 118400.0 VARIABLE VALUE REDUCED COST Z 0.000000 1.000000 X11 3.000000 0.000000

《运筹学》第五章习题及答案

《运筹学》第五章习题 1．思考题（1）试述动态规划的“最优化原理”及它同动态规划基本方程之间的关系。（2）动态规划的阶段如何划分？（3）试述用动态规划求解最短路问题的方法和步骤。（4）试解释状态、决策、策略、最优策略、状态转移方程、指标函数、最优值函数、边界函数等概念。（5）试述建立动态规划模型的基本方法。（6）试述动态规划方法的基本思想、动态规划的基本方程的结构及正确写出动态规划基本方程的关键步骤。 2．判断下列说法是否正确（1）动态规划分为线性动态规划和非线性动态规划。（2）动态规划只是用来解决和时间有关的问题。（3）对于一个动态规划问题，应用顺推法和逆推法可能会得到不同的最优解。（4）在用动态规划的解题时，定义状态时应保证各个阶段中所做的决策的相互独立性。（5）在动态规划模型中，问题的阶段等于问题的子问题的数目。（6）动态规划计算中的“维数障碍”，主要是由于问题中阶段数的急剧增加而引起的。 3．计算下图所示的从A 到E 的最短路问题 4．计算下图所示的从A 到E 的最短路问题 5．计算从A 到B、C、D 的最短路线。已知各线段的长度如下图所示。

6．设某油田要向一炼油厂用管道供应油料，管道铺设途中要经过八个城镇，各城镇间的路程如下图所示，选择怎样的路线铺设，才使总路程最短？ 7．用动态规划求解下列各题（1）．2 22211295m a x x x x x z -+-=； ?? ?≥≤+0,52 121x x x x ；（2）． 3 3 221m a x x x x z = ?? ?≥≤++0,,6321 321x x x x x x ； 8．某人外出旅游，需将3种物品装入背包，但背包重量有限制，总重量不超过 10千克。物品重量及其价值等数据见下表。试问每种物品装多少件，使整个背包的价值最大？ 913 千克。物品重量及其价值的关系如表所示。试问如何装这些物品，使整个背包价值最大？ 10 量和相应单位价值如下表所示，应如何装载可使总价值最大？ 30 30

运筹学(第五版) 习题答案

运筹学习题答案第一章（39页） 1.1用图解法求解下列线性规划问题，并指出问题是具有唯一最优解、无穷多最优解、无界解还是无可行解。（1）max 12z x x =+ 51x +102x ≤50 1x +2x ≥1 2x ≤4 1x ，2x ≥0 （2）min z=1x +1.52x 1x +32x ≥3 1x +2x ≥2 1x ，2x ≥0 （3）max z=21x +22x 1x -2x ≥-1 -0.51x +2x ≤2 1x ，2x ≥0 （4）max z=1x +2x 1x -2x ≥0 31x -2x ≤-3 1x ，2x ≥0 解：（1）（图略）有唯一可行解，max z=14 （2）（图略）有唯一可行解，min z=9/4 （3）（图略）无界解（4）（图略）无可行解 1.2将下列线性规划问题变换成标准型，并列出初始单纯形表。

（1）min z=-31x +42x -23x +54x 41x -2x +23x -4x =-2 1x +2x +33x -4x ≤14 -21x +32x -3x +24x ≥2 1x ，2x ，3x ≥0，4x 无约束（2）max k k z s p = 11 n m k ik ik i k z a x ===∑∑ 1 1(1,...,)m ik k x i n =-=-=∑ ik x ≥0 (i=1…n; k=1,…,m) （1）解：设z=-z ',4x =5x -6x , 5x ,6x ≥0 标准型： Max z '=31x -42x +23x -5(5x -6x )+07x +08x -M 9x -M 10x s. t . -41x +2x -23x +5x -6x +10x =2 1x +2x +33x -5x +6x +7x =14 -21x +32x -3x +25x -26x -8x +9x =2 1x ,2x ,3x ,5x ,6x ,7x ,8x ,9x ,10x ≥0

管理运筹学第三章习题答案

（1）解： , 5 3351042..715min 212 1 1 21 21≥≥+≥≥++=y y y y y y y t s y y ω （2）解：无限制 3213 21 3132 3213121,0,0 2 520474235323. .86max y y y y y y y y y y y y y y y t s y y ≤≥=++≤-=+≥+--≤++=ω 解：例3原问题 6 ,,1,0603020506070 ..min 166554433221654321Λ=≥≥+≥+≥+≥+≥+≥++++++=j x x x x x x x x x x x x x t s x x x x x x z j 对偶问题： 6 ,,1,0111111 ..603020506070max 655443322161654321Λ=≥≤+≤+≤+≤+≤+≤++++++=j y y y x y y y y y y y y y t s y y y y y y j ω

解：（1）由最优单纯形表可以知道原问题求max ，其初始基变量为54,x x ，最优基的逆阵为 ????? ? ??-=-316102 11 B 。由P32式（）（）（）可知b B b 1 -='，5,,1,,1Λ='-=='-j P C c P B P j B j j j j σ，其中b 和j P 都是初始数据。设???? ??=21b b b ，5,,1,21Λ=???? ??=j a a P j j j ，()321,,c c c C =，则 ?????? ??=???? ???????? ??-?='-2525316102 1 211 b b b B b ，即?????=+-=25316 12521211b b b ，解得???==10521b b ????? ? ??-=???? ???????? ??-?='-021******** 102 12322211312111 a a a a a a P B P j j ，即 ???????????????=+-=-=+-==+-=0 31 6 112121316121 211 316 1021 231313221212211111a a a a a a a a a ，解得???????????==-====12 1130231322 122111a a a a a a

第五版运筹学基础与应用-大题模拟试题及答案

计算题一一 1. 下列线性规划问题化为标准型。（10分） mi nZ x-|+5x 2-2x 3 min Z 4为 2x 2+3x 3 4x ,+5x 2 6X 3=7 8% 9x 2 10x 3 11 12% 13x 2 14 X 1 0,X 2 无约束，X 3 B1 B2 B3 B4 产量 A1 10 6 7 12 4 16 10 & 9 9 A3 5 4 10 10 4 销S 5 2 4 6 i （i 1,2,3）的投资额为x 时，其收益分别为 g 1（x 1） 4禺4区） g （x 3） 2x3 ，问应如何分配投资数额才能使总收益最大？（15分） 5.求图中所示网络中的最短路。（15分）计算题二 X 1 X 2 X 3 6 2x 1 X 2 3x 3 5 X 1 X 2 10 X 1 0,X 2 0,X 3符号不限满足〈 2. 写出下列问题的对偶问题（10分） 9x 2,

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(完整word版)运筹学课后习题解答_1.(DOC)

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单纯形法：原问题化成标准型为 121231241234 max z=10x 5x 349 ..528,,,0x x x s t x x x x x x x +++=?? ++=??≥? j c → 10 5 B C B X b 1x 2x 3x 4x 0 3x 9 3 4 1 0 0 4x 8 [5] 2 0 1 j j C Z - 10 5 0 0 0 3x 21/5 0 [14/5] 1 -3/5 10 1x 8/5 1 2/5 0 1/5 j j C Z - 1 0 - 2 5 2x 3/2 0 1 5/14 -3/14 10 1x 1 1 0 -1/7 2/7 j j C Z - -5/14 -25/14

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