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程序代写案例-000T
时间：2021-11-02
Solutions:
1. a. Let T = number of television spot advertisements
R = number of radio advertisements
N = number of newspaper advertisements
Max 100,000T + 18,000R + 40,000N
s.t.
2,000T + 300R + 600N 18,200 Budget
T 10 Max TV
R 20 Max Radio
N 10 Max News
-0.5T + 0.5R - 0.5N 0 Max 50% Radio
0.9T - 0.1R - 0.1N 0 Min 10% TV
T, R, N, 0
Budget $
Solution: T = 4 $8,000
R = 14 4,200
N = 10 6,000
$18,200 Audience = 1,052,000.
OPTIMAL SOLUTION
Optimal Objective Value
1052000.00000
Variable Value Reduced Cost
T 4.00000 0.00000
R 14.00000 0.00000
N 10.00000 11826.08695
Constraint Slack/Surplus Dual Value
1 0.00000 51.30430
2 6.00000 0.00000
3 6.00000 0.00000
4 0.00000 11826.08695
5 0.00000 5217.39130
6 1.20000 0.00000
Chapter 4
Objective Allowable Allowable
Coefficient Increase Decrease
100000.00000 20000.00000 118000.00000
18000.00000 Infinite 3000.00000
40000.00000 Infinite 11826.08695
RHS Allowable Allowable
Value Increase Decrease
18200.00000 13800.00000 3450.00000
10.00000 Infinite 6.00000
20.00000 Infinite 6.00000
10.00000 2.93600 10.00000
0.00000 2.93617 8.05000
0.00000 1.20000 Infinite
b. The dual value for the budget constraint is 51.30. Thus, a $100 increase in budget should provide an
increase in audience coverage of approximately 5,130. The right-hand-side range for the budget
constraint will show this interpretation is correct.
2. a. Let x1 = units of product 1 produced
x2 = units of product 2 produced
Max 30x1 + 15x2
s.t.
x1 + 0.35x2 100 Dept. A
0.30x1 + 0.20x2 36 Dept. B
0.20x1 + 0.50x2 50 Dept. C
x1, x2 0
Solution: x1 = 77.89, x2 = 63.16 Profit = 3284.21
b. The dual value for Dept. A is $15.79, for Dept. B it is $47.37, and for Dept. C it is $0.00. Therefore
we would attempt to schedule overtime in Departments A and B. Assuming the current labor
available is a sunk cost, we should be willing to pay up to $15.79 per hour in Department A and up
to $47.37 in Department B.
c. Let xA = hours of overtime in Dept. A
xB = hours of overtime in Dept. B
xC = hours of overtime in Dept. C
Max 30x1 + 15x2 - 18xA - 22.5xB - 12xC
s.t.
x1 + 0.35x2 - xA 100
0.30x1 + 0.20x2 - xB 36
0.20x1 + 0.50x2 - xC 50
xA 10
xB 6
xC 8
x1, x2, xA, xB, xC 0
x1 = 87.21
x2 = 65.12
Profit = $3341.34
Overtime
Dept. A 10 hrs.
Dept. B 3.186 hrs
Dept. C 0 hours
Increase in Profit from overtime = $3341.34 - 3284.21 = $57.13
3. x1 = $ automobile loans
x2 = $ furniture loans
x3 = $ other secured loans
x4 = $ signature loans
x5 = $ "risk free" securities
Max 0.08x1 + 0.10x2 + 0.11x3 + 0.12x4 + 0.09x5
s.t.
x5 600,000 [1]
x4 0.10(x1 + x2 + x3 + x4)
or -0.10x1 - 0.10x2 - 0.10x3 + 0.90x4 0 [2]
x2 + x3 x1
or - x1 + x2 + x3 0 [3]
x3 + x4 x5
or + x3 + x4 - x5 0 [4]
x1 + x2 + x3 + x4 + x5 = 2,000,000 [5]
x1, x2, x3, x4, x5 0
Solution:
Automobile Loans (x1) = $630,000
Furniture Loans (x2) = $170,000
Other Secured Loans (x3) = $460,000
Signature Loans (x4) = $140,000
Risk Free Loans (x5) = $600,000
Annual Return $188,800 (9.44%)
Chapter 4
4. a. x1 = pounds of bean 1
x2 = pounds of bean 2
x3 = pounds of bean 3
Min 0.50x1 + 0.70x2 + 0.45x3
s.t.
75 85 601 2 3
1 2 3
x x x
x x x
+ +
+ +
75
or 10x2 - 15x3 0 Aroma
86 88 751 2 3
1 2 3
x x x
x x x
+ +
+ +
80
or 6x1 + 8x2 - 5x3 0 Taste
x1 500 Bean 1
x2 600 Bean 2
x3 400 Bean 3
x1 + x2 + x3 = 1000 1000 pounds
x1, x2, x3 0
Optimal Solution: x1 = 500, x2 = 300, x3 = 200, Cost: $550
b. Cost per pound = $550/1000 = $0.55
c. Surplus for aroma: s1 = 0; thus aroma rating = 75
Surplus for taste: s2 = 4400; thus taste rating = 80 + 4400/1000 lbs. = 84.4
d. Dual value = $0.60. Extra coffee can be produced at a cost of $0.60 per pound.
5. a. Let W = # of servings of Wimpy to make
D = # of servings of Dial 911 to make
Max .45 W + .58 D
s.t.
.25 W + .25 D ≤ 20 (Beef)
.25 W + .4 D ≤ 15 (Onions)
5 W + 2 D ≤ 88 (Special Sauce)
5 D ≤ 60 (Hot Sauce)
W, D ≥ 0
b. Solution: W = 12.8, D = 12, Profit = $12.72.
c. Dual value for special sauce = $.09
d. Solution: W=13, D=12, Profit = $12.81. Note: $12.81-$12.72 = $.09. Dual value confirmed.
6. Let x1 = units of product 1
x2 = units of product 2
b1 = labor-hours Dept. A
b2 = labor-hours Dept. B
Max 25x1 + 20x2 + 0b1 + 0b2
s.t.
6x1 + 8x2 - 1b1 = 0
12x1 + 10x2 - 1b2 = 0
1b1 + 1b2 900
x1, x2, b1, b2 0
Solution: x1 = 50, x2 = 0, b1 = 300, b2 = 600 Profit: $1,250
7. a. Let F = total funds required to meet the six years of payments
G1 = units of government security 1
G2 = units of government security 2
Si = investment in savings at the beginning of year i
Note: All decision variables are expressed in thousands of dollars
MIN F
S.T.
1) F - 1.055G1 - 1.000G2 - S1 = 190
2) .0675G1 + .05125G2 +1.04S1 - S2 = 215
3) .0675G1 + .05125G2 + 1.04S2 - S3 = 240
4) 1.0675G1 + .05125G2 + 1.04S3 - S4 = 285
5) 1.05125G2 + 1.04S4 - S5 = 315
6) 1.04S5 = 460
OPTIMAL SOLUTION
Optimal Objective Value
1484.96660
Variable Value Reduced Cost
G1 232.39356 0.00000
G2 720.38782 0.00000
F 1484.96655 0.00000
S1 329.40353 0.00000
S2 180.18611 0.00000
S3 0.00000 0.02077
S4 0.00000 0.01942
S5 442.30769 0.00000
Constraint Slack/Surplus Dual Value
1 0.00000 1.00000
Chapter 4
2 0.00000 0.96154
3 0.00000 0.92456
4 0.00000 0.86903
5 0.00000 0.81693
6 0.00000 0.78551
Objective Allowable Allowable
Coefficient Increase Decrease
0.00000 0.02132 0.01973
0.00000 0.01963 Infinite
1.00000 Infinite 1.00000
0.00000 0.65430 0.01980
0.00000 1.28344 0.02026
0.00000 Infinite 0.02077
0.00000 Infinite 0.01942
0.00000 Infinite Infinite
RHS Allowable Allowable
Value Increase Decrease
0.00000 Infinite 1484.96655
0.00000 Infinite 342.57967
0.00000 Infinite 187.39356
0.00000 2762.03307 248.08012
0.00000 3824.23689 757.30769
0.00000 3977.20636 460.00000
The current investment required is $1,484,967. This calls for investing $232,394 in government
security 1 and $720,388 in government security 2. The amounts, placed in savings are $329,404,
$180,186 and $442,308 for years 1,2 and 5 respectively. No funds are placed in savings for years 3,
and 4.
b. The dual value for constraint 6 indicates that each $1 increase in the payment required at the
beginning of year 6 will increase the amount of money Hoxworth must pay the trustee by $0.78551.
A per unit decrease would therefore decrease the cost by $0.78551. The allowable decrease on the
right-hand-side range is $460,000 so a reduction in the payment at the beginning of year 6 by
$60,000 will save Hoxworth $60,000 (0.78551) = $47,131.
c. The dual value for constraint 1 shows that every dollar of reduction in the initial payment leads to a
drop in the objective function value of $1.00. So Hoxworth should be willing to pay anything less
than $40,000.
d. To reformulate this problem, one additional variable needs to be added, the right-hand sides for the
original constraints need to be shifted ahead by one, and the right-hand side of the first constraint
needs to be set equal to zero. The value of the optimal solution with this formulation is $1,417,739.
Hoxworth will save $67,228 by having the payments moved to the end of each year.
The revised formulation is shown below:
MIN F
S.T.
1) F - 1.055G1 - 1.000G2 - S1 = 0
2) .0675G1 + .05125G2 + 1.04S1 - S2 = 190
3) .0675G1 + .05125G2 + 1.04S2 - S3 = 215
4) 1.0675G1 + .05125G2 + 1.04S3 - S4 = 240
5) 1.05125G2 +1.04S4 - S5 = 285
6) 1.04S5 - S6 = 315
7) 1.04S6 - S7 = 460
8. Let x1 = the number of officers scheduled to begin at 8:00 a.m.
x2 = the number of officers scheduled to begin at noon
x3 = the number of officers scheduled to begin at 4:00 p.m.
x4 = the number of officers scheduled to begin at 8:00 p.m.
x5 = the number of officers scheduled to begin at midnight
x6 = the number of officers scheduled to begin at 4:00 a.m.
The objective function to minimize the number of officers required is as follows:
Min x1 + x2 + x3 + x4 + x5 + x6
The constraints require the total number of officers of duty each of the six four-hour periods to be at
least equal to the minimum officer requirements. The constraints for the six four-hour periods are as
follows:
Time of Day
8:00 a.m. - noon x1 + x6 5
noon to 4:00 p.m. x1 + x2 6
4:00 p.m. - 8:00 p.m. x2 + x3 10
8:00 p.m. - midnight x3 + x4 7
midnight - 4:00 a.m. x4 + x5 4
4:00 a.m. - 8:00 a.m. x5 + x6 6
x1, x2, x3, x4, x5, x6 0
Schedule 19 officers as follows:
x1 = 3 begin at 8:00 a.m.
x2 = 3 begin at noon
x3 = 7 begin at 4:00 p.m.
x4 = 0 begin at 8:00 p.m.
Chapter 4
x5 = 4 begin at midnight
x6 = 2 begin at 4:00 a.m.
9. Let Xi = the number of call center employees who start work on day i
(i=1 = Monday, i=2=Tuesday…)
Min X1 + X2 + X3 + X4 + X5 + X6 + X7
s.t.
X1 + X4 + X5 + X6 + X7 ≥ 75
X1 + X2 + X5 + X6 + X7 ≥ 50
X1 + X2 + X3 + X6 + X7 ≥ 45
X1 + X2 + X3 + X4 +X7 ≥ 60
X1 + X2 + X3 + X4 + X5 ≥ 90
X2 + X3 + X4 + X5 + X6 ≥ 75
X3 + X4 + X5 + X6 + X7 ≥ 45
X1 , X2 , X3 , X4 , X5 , X6 , X7 ≥ 0
Solution: X1 = 20 , X2 = 20 , X3 = 0 , X4 = 45, X5 = 5, X6 = 5, X7 = 0
Total Number of Employees = 95
Excess employees: Thursday = 25, Sunday = 10, all others = 0.
Note: There are alternative optima to this problem (Number of employees may differ from
above, but will have objective function value = 95).
10. a. Let S = the proportion of funds invested in stocks
B = the proportion of funds invested in bonds
M = the proportion of funds invested in mutual funds
C = the proportion of funds invested in cash
The linear program and optimal solution are as follows:
MAX 0.1S+0.03B+0.04M+0.01C
S.T.
1) 1S+1B+1M+1C=1
2) 0.8S+0.2B+0.3M<0.4
3) 1S<0.75
4) -1B+1M>0
5) 1C>0.1
6) 1C<0.3
OPTIMAL SOLUTION
Optimal Objective Value
0.05410
Variable Value Reduced Cost
S 0.40909 0.00000
B 0.14545 0.00000
M 0.14545 0.00000
C 0.30000 0.00455
Constraint Slack/Surplus Dual Value
1 0.00000 0.00000
2 0.00000 0.11818
3 0.34091 0.00000
4 0.00000 -0.00091
5 0.20000 0.00000
6 0.00000 0.00545
Objective Allowable Allowable
Coefficient Increase Decrease
0.10000 Infinite 0.01000
0.03000 0.00625 0.00200
0.04000 0.00167 Infinite
0.01000 Infinite 0.00455
RHS Allowable Allowable
Value Increase Decrease
1.00000 0.90000 0.20000
0.40000 0.16000 0.22500
0.75 Infinite 0.341
0.00000 0.32000 0.26667
0.10000 0.20000 Infinite
0.30000 0.20000 0.20000
The optimal allocation among the four investment alternatives is
Stocks 40.9%
Bonds 14.5%
Mutual Funds 14.5%
Cash 30.0%
The annual return associated with the optimal portfolio is 5.4%
The total risk = 0.409(0.8) + 0.145(0.2) + 0.145(0.3) + 0.300(0.0) = 0.4
b. Changing the right-hand-side value for constraint 2 to 0.18 and resolving we obtain the following
optimal solution:
Stocks 0.0%
Bonds 36.0%
Mutual Funds 36.0%
Cash 28.0%
The annual return associated with the optimal portfolio is 2.52%
Chapter 4
The total risk = 0.0(0.8) + 0.36(0.2) + 0.36(0.3) + 0.28(0.0) = 0.18
c. Changing the right-hand-side value for constraint 2 to 0.7 and resolving we obtain the following
optimal solution:
The optimal allocation among the four investment alternatives is
Stocks 75.0%
Bonds 0.0%
Mutual Funds 15.0%
Cash 10.0%
The annual return associated with the optimal portfolio is 8.2%
The total risk = 0.75(0.8) + 0.0(0.2) + 0.15(0.3) + 0.10(0.0) = 0.65
d. Note that a maximum risk of 0.7 was specified for this aggressive investor, but that the risk index for
the portfolio is only 0.65. Thus, this investor is willing to take more risk than the solution shown
above provides. There are only two ways the investor can become even more aggressive: increase
the proportion invested in stocks to more than 75% or reduce the cash requirement of at least 10% so
that additional cash could be put into stocks. For the data given here, the investor should ask the
investment advisor to relax either or both of these constraints.
e. Defining the decision variables as proportions means the investment advisor can use the linear
programming model for any investor, regardless of the amount of the investment. All the investor
advisor needs to do is to establish the maximum total risk for the investor and resolve the problem
using the new value for maximum total risk.
11. Let xij = units of component i purchased from supplier j
Min 12x11 + 13x12 + 14x13 + 10x21 + 11x22 + 10x23
s.t.
x11 + x12 + x13 = 1000
x21 + x22 + x23 = 800
x11 + x21 600
x12 + x22 1000
x13 + x23 800
x11, x12, x13, x21, x22, x23 0
Solution:
Supplier
1 2 3
Component 1
600
400
0
Component 2 0 0 800
Purchase Cost = $20,400
13. Let BR = pounds of Brazilian beans purchased to produce Regular
BD = pounds of Brazilian beans purchased to produce DeCaf
CR = pounds of Colombian beans purchased to produce Regular
CD = pounds of Colombian beans purchased to produce DeCaf
Type of Bean Cost per pound ($)
Brazilian 1.10(0.47) = 0.517
Colombian 1.10(0.62) = 0.682
Total revenue = 3.60(BR + CR) + 4.40(BD + CD)
Total cost of beans = 0.517(BR + BD) + 0.682(CR + CD)
Total production cost = 0.80(BR + CR) + 1.05(BD + CD)
Total packaging cost = 0.25(BR + CR) + 0.25(BD + CD)
Total contribution to profit = (total revenue) - (total cost of beans) - (total production cost)
Total contribution to profit = 2.033BR + 2.583BD + 1.868CR + 2.418CD
Regular % constraint
BR = 0.75(BR + CR)
0.25BR - 0.75CR = 0
DeCaf % constraint
BD = 0.40(BD + CD)
0.60BD - 0.40CD = 0
Pounds of Regular: BR + CR = 1000
Pounds of DeCaf: BD + CD = 500
The complete linear program is
Max 2.033BR + 2.583BD + 1.868CR + 2.418CD
s.t.
0.25BR - 0.75CR = 0
0.60BD - 0.40CD = 0
BR + CR = 1000
BD + CD = 500
BR, BD, CR, CD 0
The optimal solution is BR = 750, BD = 200, CR = 250, and CD = 300.
The value of the optimal solution is $3233.75
15. Let x11 = gallons of crude 1 used to produce regular
x12 = gallons of crude 1 used to produce high-octane
x21 = gallons of crude 2 used to produce regular
x22 = gallons of crude 2 used to produce high-octane
Chapter 4
Min 0.10x11 + 0.10x12 + 0.15x21 + 0.15x22
s.t.
Each gallon of regular must have at least 40% A.
x11 + x21 = amount of regular produced
0.4(x11 + x21) = amount of A required for regular
0.2x11 + 0.50x21 = amount of A in (x11 + x21) gallons of regular gas
0.2x11 + 0.50x21 0.4x11 + 0.40x21 [1]
-0.2x11 + 0.10x21 0
Each gallon of high octane can have at most 50% B.
x12 + x22 = amount high-octane
0.5(x12 + x22) = amount of B required for high octane
0.60x12 + 0.30x22 = amount of B in (x12 + x22) gallons of high octane.
0.60x12 + 0.30x22 0.5x12 + 0.5x22
0.1x12 - 0.2x22 0 [2]
x11 + x21 800,000 [3]
x12 + x22 500,000 [4]
x11, x12, x21, x22 0
Optimal Solution: x11 = 266,667, x12 = 333,333, x21 = 533,333, x22 = 166,667
Cost = $165,000
16. Let xi = number of 10-inch rolls of paper processed by cutting alternative i; i = 1,2...,7
Min x1 + x2 + x3 + x4 + x5 + x6 + x7
s.t.
6x1 + 2x3 + x5 + x6 + 4x7 1000 1 1/2" production
4x2 + x4 + 3x5 + 2x6 2000 2 1/2" production
2x3 + 2x4 + x6 + x7 4000 3 1/2" production
x1, x2, x3, x4, x5, x6, x7 0
x1 = 0
x2 = 125
x3 = 500 2125 Rolls
x4 = 1500
x5 = 0 Production:
x6 = 0 1 1/2" 1000
x7 = 0 2 1/2" 2000
3 1/2" 4000
Waste: Cut alternative #4 (1/2" per roll)
750 inches.
b. Only the objective function needs to be changed. An objective function minimizing waste
production and the new optimal solution are given.
Min x1 + 0x2 + 0x3 + 0.5x4 + x5 + 0x6 + 0.5x7
x1 = 0
x2 = 500
x3 = 2000 2500 Rolls
x4 = 0
x5 = 0 Production:
x6 = 0 1 1/2" 4000
x7 = 0 2 2/1" 2000
3 1/2" 4000
Waste is 0; however, we have over-produced the 1 1/2" size by 3000 units. Perhaps these can be
inventoried for future use.
c. Minimizing waste may cause you to over-produce. In this case, we used 375 more rolls to generate a
3000 surplus of the 1 1/2" product. Alternative b might be preferred on the basis that the 3000
surplus could be held in inventory for later demand. However, in some trim problems, excess
production cannot be used and must be scrapped. If this were the case, the 3000 unit 1 1/2" size
would result in 4500 inches of waste, and thus alternative a would be the preferred solution.
17. a. Let FM = number of frames manufactured
FP = number of frames purchased
SM = number of supports manufactured
SP = number of supports purchased
TM = number of straps manufactured
TP = number of straps purchased
Min 38FM + 51FP + 11.5SM + 15SP + 6.5TM + 7.5TP
s.t.
3.5FM + 1.3SM + 0.8TM 21,000
2.2FM + 1.7SM 25,200
3.1FM + 2.6SM + 1.7TM 40,800
FM + FP 5,000
SM + SP 10,000
TM + TP 5,000
FM, FP, SM, SP, TM, TP 0.
Solution:
Manufacture Purchase
Frames 5000 0
Chapter 4
Supports 2692 7308
Straps 0 5000
b. Total Cost = $368,076.91
c. Subtract values of slack variables from minutes available to determine minutes used. Divide by 60
to determine hours of production time used.
Constraint
1 Cutting: Slack = 0 350 hours used
2 Milling: (25200 - 9623) / 60 = 259.62 hours
3 Shaping: (40800 - 18300) / 60 = 375 hours
d. Nothing, there are already more hours available than are being used.
e. Yes. The current purchase price is $51.00 and the reduced cost of 3.577 indicates that for a purchase
price below $47.423 the solution may improve. Resolving with the coefficient of FP = 45 shows
that 2714 frames should be purchased.
The optimal solution is as follows:
OPTIMAL SOLUTION
Optimal Objective Value
361500.00000
Variable Value Reduced Cost
FM 2285.71429 0.00000
FP 2714.28571 3.57692
SM 10000.00000 0.00000
SP 0.00000 0.00000
TM 0.00000 1.15385
TP 5000.00000 0.00000
Constraint Slack/Surplus Dual Value
1 0.00000 -2.69231
2 3171.42857 0.00000
3 7714.28571 0.00000
4 0.00000 47.42308
5 0.00000 15.00000
6 0.00000 7.50000
19. a. Let x11 = amount of men's model in month 1
x21 = amount of women's model in month 1
x12 = amount of men's model in month 2
x22 = amount of women's model in month 2
s11 = inventory of men's model at end of month 1
s21 = inventory of women's model at end of month 1
s12 = inventory of men's model at end of month 2
s22 = inventory of women's model at end of month
The model formulation for part (a) is given.
Min 120x11 + 90x21 + 120x12 + 90x22 + 2.4s11 + 1.8s21 + 2.4s12 + 1.8s22
s.t.
20 + x11 - s11 = 150
or
x11 - s11 = 130 Satisfy Demand [1]
30 + x21 - s21 = 125
or
x21 - s21 = 95 Satisfy Demand [2]
s11 + x12 - s12 = 200 Satisfy Demand [3]
s21 + x22 - s22 = 150 Satisfy Demand [4]
s12 25 Ending Inventory [5]
s22 25 Ending Inventory [6]
Labor Hours: Men’s = 2.0 + 1.5 = 3.5
Women’s = 1.6 + 1.0 = 2.6
3.5 x11 + 2.6 x21 900 Labor Smoothing for [7]
3.5 x11 + 2.6 x21 1100 Month 1 [8]
3.5 x11 + 2.6 x21 - 3.5 x12 - 2.6 x22 100 Labor Smoothing for [9]
-3.5 x11 - 2.6 x21 + 3.5 x12 + 2.6 x22 100 Month 2 [10]
x11, x12, x21, x22, s11, s12, s21, s22 0
The optimal solution is to produce 193 of the men's model in month 1, 162 of the men's model in
month 2, 95 units of the women's model in month 1, and 175 of the women's model in month 2.
Total Cost = $67,156
Inventory Schedule
Month 1
63 Men's
0 Women's
Month 2 25 Men's 25 Women's
Labor Levels
Previous month
1000.00 hours
Month 1 922.25 hours
Month 2 1022.25 hours
b. To accommodate this new policy the right-hand sides of constraints [7] to [10] must be changed to
950, 1050, 50, and 50 respectively. The revised optimal solution is given.
Chapter 4
x11 = 201
x21 = 95
x12 = 154
x22 = 175 Total Cost = $67,175
We produce more men's models in the first month and carry a larger men's model inventory; the
added cost however is only $19. This seems to be a small expense to have less drastic labor force
fluctuations. The new labor levels are 1000, 950, and 994.5 hours each month. Since the added cost
is only $19, management might want to experiment with the labor force smoothing restrictions to
enforce even less fluctuations. You may want to experiment yourself to see what happens.
20. Let xm = number of units produced in month m
Im = increase in the total production level in month m
Dm = decrease in the total production level in month m
sm = inventory level at the end of month m
where
m = 1 refers to March
m = 2 refers to April
m = 3 refers to May
Min 1.25 I1 + 1.25 I2 + 1.25 I3 + 1.00 D1 + 1.00 D2 + 1.00 D3
s.t.
Change in production level in March
x1 - 10,000 = I1 - D1
or
x1 - I1 + D1 = 10,000
Change in production level in April
x2 - x1 = I2 - D2
or
x2 - x1 - I2 + D2 = 0
Change in production level in May
x3 - x2 = I3 - D3
or
x3 - x2 - I3 + D3 = 0
Demand in March
2500 + x1 - s1 = 12,000
or
x1 - s1 = 9,500
Demand in April
s1 + x2 - s2 = 8,000
Demand in May
s2 + x3 = 15,000
Inventory capacity in March
s1 3,000
Inventory capacity in April
s2 3,000
Optimal Solution:
Total cost of monthly production increases and decreases = $2,500
x1 = 10,250 I1 = 250 D1 = 0
x2 = 10,250 I2 = 0 D2 = 0
x3 = 12,000 I3 = 1750 D3 = 0
s1 = 750
s2 = 3000
学霸联盟
学霸联盟
1. a. Let T = number of television spot advertisements
R = number of radio advertisements
N = number of newspaper advertisements
Max 100,000T + 18,000R + 40,000N
s.t.
2,000T + 300R + 600N 18,200 Budget
T 10 Max TV
R 20 Max Radio
N 10 Max News
-0.5T + 0.5R - 0.5N 0 Max 50% Radio
0.9T - 0.1R - 0.1N 0 Min 10% TV
T, R, N, 0
Budget $
Solution: T = 4 $8,000
R = 14 4,200
N = 10 6,000
$18,200 Audience = 1,052,000.
OPTIMAL SOLUTION
Optimal Objective Value
1052000.00000
Variable Value Reduced Cost
T 4.00000 0.00000
R 14.00000 0.00000
N 10.00000 11826.08695
Constraint Slack/Surplus Dual Value
1 0.00000 51.30430
2 6.00000 0.00000
3 6.00000 0.00000
4 0.00000 11826.08695
5 0.00000 5217.39130
6 1.20000 0.00000
Chapter 4
Objective Allowable Allowable
Coefficient Increase Decrease
100000.00000 20000.00000 118000.00000
18000.00000 Infinite 3000.00000
40000.00000 Infinite 11826.08695
RHS Allowable Allowable
Value Increase Decrease
18200.00000 13800.00000 3450.00000
10.00000 Infinite 6.00000
20.00000 Infinite 6.00000
10.00000 2.93600 10.00000
0.00000 2.93617 8.05000
0.00000 1.20000 Infinite
b. The dual value for the budget constraint is 51.30. Thus, a $100 increase in budget should provide an
increase in audience coverage of approximately 5,130. The right-hand-side range for the budget
constraint will show this interpretation is correct.
2. a. Let x1 = units of product 1 produced
x2 = units of product 2 produced
Max 30x1 + 15x2
s.t.
x1 + 0.35x2 100 Dept. A
0.30x1 + 0.20x2 36 Dept. B
0.20x1 + 0.50x2 50 Dept. C
x1, x2 0
Solution: x1 = 77.89, x2 = 63.16 Profit = 3284.21
b. The dual value for Dept. A is $15.79, for Dept. B it is $47.37, and for Dept. C it is $0.00. Therefore
we would attempt to schedule overtime in Departments A and B. Assuming the current labor
available is a sunk cost, we should be willing to pay up to $15.79 per hour in Department A and up
to $47.37 in Department B.
c. Let xA = hours of overtime in Dept. A
xB = hours of overtime in Dept. B
xC = hours of overtime in Dept. C
Max 30x1 + 15x2 - 18xA - 22.5xB - 12xC
s.t.
x1 + 0.35x2 - xA 100
0.30x1 + 0.20x2 - xB 36
0.20x1 + 0.50x2 - xC 50
xA 10
xB 6
xC 8
x1, x2, xA, xB, xC 0
x1 = 87.21
x2 = 65.12
Profit = $3341.34
Overtime
Dept. A 10 hrs.
Dept. B 3.186 hrs
Dept. C 0 hours
Increase in Profit from overtime = $3341.34 - 3284.21 = $57.13
3. x1 = $ automobile loans
x2 = $ furniture loans
x3 = $ other secured loans
x4 = $ signature loans
x5 = $ "risk free" securities
Max 0.08x1 + 0.10x2 + 0.11x3 + 0.12x4 + 0.09x5
s.t.
x5 600,000 [1]
x4 0.10(x1 + x2 + x3 + x4)
or -0.10x1 - 0.10x2 - 0.10x3 + 0.90x4 0 [2]
x2 + x3 x1
or - x1 + x2 + x3 0 [3]
x3 + x4 x5
or + x3 + x4 - x5 0 [4]
x1 + x2 + x3 + x4 + x5 = 2,000,000 [5]
x1, x2, x3, x4, x5 0
Solution:
Automobile Loans (x1) = $630,000
Furniture Loans (x2) = $170,000
Other Secured Loans (x3) = $460,000
Signature Loans (x4) = $140,000
Risk Free Loans (x5) = $600,000
Annual Return $188,800 (9.44%)
Chapter 4
4. a. x1 = pounds of bean 1
x2 = pounds of bean 2
x3 = pounds of bean 3
Min 0.50x1 + 0.70x2 + 0.45x3
s.t.
75 85 601 2 3
1 2 3
x x x
x x x
+ +
+ +
75
or 10x2 - 15x3 0 Aroma
86 88 751 2 3
1 2 3
x x x
x x x
+ +
+ +
80
or 6x1 + 8x2 - 5x3 0 Taste
x1 500 Bean 1
x2 600 Bean 2
x3 400 Bean 3
x1 + x2 + x3 = 1000 1000 pounds
x1, x2, x3 0
Optimal Solution: x1 = 500, x2 = 300, x3 = 200, Cost: $550
b. Cost per pound = $550/1000 = $0.55
c. Surplus for aroma: s1 = 0; thus aroma rating = 75
Surplus for taste: s2 = 4400; thus taste rating = 80 + 4400/1000 lbs. = 84.4
d. Dual value = $0.60. Extra coffee can be produced at a cost of $0.60 per pound.
5. a. Let W = # of servings of Wimpy to make
D = # of servings of Dial 911 to make
Max .45 W + .58 D
s.t.
.25 W + .25 D ≤ 20 (Beef)
.25 W + .4 D ≤ 15 (Onions)
5 W + 2 D ≤ 88 (Special Sauce)
5 D ≤ 60 (Hot Sauce)
W, D ≥ 0
b. Solution: W = 12.8, D = 12, Profit = $12.72.
c. Dual value for special sauce = $.09
d. Solution: W=13, D=12, Profit = $12.81. Note: $12.81-$12.72 = $.09. Dual value confirmed.
6. Let x1 = units of product 1
x2 = units of product 2
b1 = labor-hours Dept. A
b2 = labor-hours Dept. B
Max 25x1 + 20x2 + 0b1 + 0b2
s.t.
6x1 + 8x2 - 1b1 = 0
12x1 + 10x2 - 1b2 = 0
1b1 + 1b2 900
x1, x2, b1, b2 0
Solution: x1 = 50, x2 = 0, b1 = 300, b2 = 600 Profit: $1,250
7. a. Let F = total funds required to meet the six years of payments
G1 = units of government security 1
G2 = units of government security 2
Si = investment in savings at the beginning of year i
Note: All decision variables are expressed in thousands of dollars
MIN F
S.T.
1) F - 1.055G1 - 1.000G2 - S1 = 190
2) .0675G1 + .05125G2 +1.04S1 - S2 = 215
3) .0675G1 + .05125G2 + 1.04S2 - S3 = 240
4) 1.0675G1 + .05125G2 + 1.04S3 - S4 = 285
5) 1.05125G2 + 1.04S4 - S5 = 315
6) 1.04S5 = 460
OPTIMAL SOLUTION
Optimal Objective Value
1484.96660
Variable Value Reduced Cost
G1 232.39356 0.00000
G2 720.38782 0.00000
F 1484.96655 0.00000
S1 329.40353 0.00000
S2 180.18611 0.00000
S3 0.00000 0.02077
S4 0.00000 0.01942
S5 442.30769 0.00000
Constraint Slack/Surplus Dual Value
1 0.00000 1.00000
Chapter 4
2 0.00000 0.96154
3 0.00000 0.92456
4 0.00000 0.86903
5 0.00000 0.81693
6 0.00000 0.78551
Objective Allowable Allowable
Coefficient Increase Decrease
0.00000 0.02132 0.01973
0.00000 0.01963 Infinite
1.00000 Infinite 1.00000
0.00000 0.65430 0.01980
0.00000 1.28344 0.02026
0.00000 Infinite 0.02077
0.00000 Infinite 0.01942
0.00000 Infinite Infinite
RHS Allowable Allowable
Value Increase Decrease
0.00000 Infinite 1484.96655
0.00000 Infinite 342.57967
0.00000 Infinite 187.39356
0.00000 2762.03307 248.08012
0.00000 3824.23689 757.30769
0.00000 3977.20636 460.00000
The current investment required is $1,484,967. This calls for investing $232,394 in government
security 1 and $720,388 in government security 2. The amounts, placed in savings are $329,404,
$180,186 and $442,308 for years 1,2 and 5 respectively. No funds are placed in savings for years 3,
and 4.
b. The dual value for constraint 6 indicates that each $1 increase in the payment required at the
beginning of year 6 will increase the amount of money Hoxworth must pay the trustee by $0.78551.
A per unit decrease would therefore decrease the cost by $0.78551. The allowable decrease on the
right-hand-side range is $460,000 so a reduction in the payment at the beginning of year 6 by
$60,000 will save Hoxworth $60,000 (0.78551) = $47,131.
c. The dual value for constraint 1 shows that every dollar of reduction in the initial payment leads to a
drop in the objective function value of $1.00. So Hoxworth should be willing to pay anything less
than $40,000.
d. To reformulate this problem, one additional variable needs to be added, the right-hand sides for the
original constraints need to be shifted ahead by one, and the right-hand side of the first constraint
needs to be set equal to zero. The value of the optimal solution with this formulation is $1,417,739.
Hoxworth will save $67,228 by having the payments moved to the end of each year.
The revised formulation is shown below:
MIN F
S.T.
1) F - 1.055G1 - 1.000G2 - S1 = 0
2) .0675G1 + .05125G2 + 1.04S1 - S2 = 190
3) .0675G1 + .05125G2 + 1.04S2 - S3 = 215
4) 1.0675G1 + .05125G2 + 1.04S3 - S4 = 240
5) 1.05125G2 +1.04S4 - S5 = 285
6) 1.04S5 - S6 = 315
7) 1.04S6 - S7 = 460
8. Let x1 = the number of officers scheduled to begin at 8:00 a.m.
x2 = the number of officers scheduled to begin at noon
x3 = the number of officers scheduled to begin at 4:00 p.m.
x4 = the number of officers scheduled to begin at 8:00 p.m.
x5 = the number of officers scheduled to begin at midnight
x6 = the number of officers scheduled to begin at 4:00 a.m.
The objective function to minimize the number of officers required is as follows:
Min x1 + x2 + x3 + x4 + x5 + x6
The constraints require the total number of officers of duty each of the six four-hour periods to be at
least equal to the minimum officer requirements. The constraints for the six four-hour periods are as
follows:
Time of Day
8:00 a.m. - noon x1 + x6 5
noon to 4:00 p.m. x1 + x2 6
4:00 p.m. - 8:00 p.m. x2 + x3 10
8:00 p.m. - midnight x3 + x4 7
midnight - 4:00 a.m. x4 + x5 4
4:00 a.m. - 8:00 a.m. x5 + x6 6
x1, x2, x3, x4, x5, x6 0
Schedule 19 officers as follows:
x1 = 3 begin at 8:00 a.m.
x2 = 3 begin at noon
x3 = 7 begin at 4:00 p.m.
x4 = 0 begin at 8:00 p.m.
Chapter 4
x5 = 4 begin at midnight
x6 = 2 begin at 4:00 a.m.
9. Let Xi = the number of call center employees who start work on day i
(i=1 = Monday, i=2=Tuesday…)
Min X1 + X2 + X3 + X4 + X5 + X6 + X7
s.t.
X1 + X4 + X5 + X6 + X7 ≥ 75
X1 + X2 + X5 + X6 + X7 ≥ 50
X1 + X2 + X3 + X6 + X7 ≥ 45
X1 + X2 + X3 + X4 +X7 ≥ 60
X1 + X2 + X3 + X4 + X5 ≥ 90
X2 + X3 + X4 + X5 + X6 ≥ 75
X3 + X4 + X5 + X6 + X7 ≥ 45
X1 , X2 , X3 , X4 , X5 , X6 , X7 ≥ 0
Solution: X1 = 20 , X2 = 20 , X3 = 0 , X4 = 45, X5 = 5, X6 = 5, X7 = 0
Total Number of Employees = 95
Excess employees: Thursday = 25, Sunday = 10, all others = 0.
Note: There are alternative optima to this problem (Number of employees may differ from
above, but will have objective function value = 95).
10. a. Let S = the proportion of funds invested in stocks
B = the proportion of funds invested in bonds
M = the proportion of funds invested in mutual funds
C = the proportion of funds invested in cash
The linear program and optimal solution are as follows:
MAX 0.1S+0.03B+0.04M+0.01C
S.T.
1) 1S+1B+1M+1C=1
2) 0.8S+0.2B+0.3M<0.4
3) 1S<0.75
4) -1B+1M>0
5) 1C>0.1
6) 1C<0.3
OPTIMAL SOLUTION
Optimal Objective Value
0.05410
Variable Value Reduced Cost
S 0.40909 0.00000
B 0.14545 0.00000
M 0.14545 0.00000
C 0.30000 0.00455
Constraint Slack/Surplus Dual Value
1 0.00000 0.00000
2 0.00000 0.11818
3 0.34091 0.00000
4 0.00000 -0.00091
5 0.20000 0.00000
6 0.00000 0.00545
Objective Allowable Allowable
Coefficient Increase Decrease
0.10000 Infinite 0.01000
0.03000 0.00625 0.00200
0.04000 0.00167 Infinite
0.01000 Infinite 0.00455
RHS Allowable Allowable
Value Increase Decrease
1.00000 0.90000 0.20000
0.40000 0.16000 0.22500
0.75 Infinite 0.341
0.00000 0.32000 0.26667
0.10000 0.20000 Infinite
0.30000 0.20000 0.20000
The optimal allocation among the four investment alternatives is
Stocks 40.9%
Bonds 14.5%
Mutual Funds 14.5%
Cash 30.0%
The annual return associated with the optimal portfolio is 5.4%
The total risk = 0.409(0.8) + 0.145(0.2) + 0.145(0.3) + 0.300(0.0) = 0.4
b. Changing the right-hand-side value for constraint 2 to 0.18 and resolving we obtain the following
optimal solution:
Stocks 0.0%
Bonds 36.0%
Mutual Funds 36.0%
Cash 28.0%
The annual return associated with the optimal portfolio is 2.52%
Chapter 4
The total risk = 0.0(0.8) + 0.36(0.2) + 0.36(0.3) + 0.28(0.0) = 0.18
c. Changing the right-hand-side value for constraint 2 to 0.7 and resolving we obtain the following
optimal solution:
The optimal allocation among the four investment alternatives is
Stocks 75.0%
Bonds 0.0%
Mutual Funds 15.0%
Cash 10.0%
The annual return associated with the optimal portfolio is 8.2%
The total risk = 0.75(0.8) + 0.0(0.2) + 0.15(0.3) + 0.10(0.0) = 0.65
d. Note that a maximum risk of 0.7 was specified for this aggressive investor, but that the risk index for
the portfolio is only 0.65. Thus, this investor is willing to take more risk than the solution shown
above provides. There are only two ways the investor can become even more aggressive: increase
the proportion invested in stocks to more than 75% or reduce the cash requirement of at least 10% so
that additional cash could be put into stocks. For the data given here, the investor should ask the
investment advisor to relax either or both of these constraints.
e. Defining the decision variables as proportions means the investment advisor can use the linear
programming model for any investor, regardless of the amount of the investment. All the investor
advisor needs to do is to establish the maximum total risk for the investor and resolve the problem
using the new value for maximum total risk.
11. Let xij = units of component i purchased from supplier j
Min 12x11 + 13x12 + 14x13 + 10x21 + 11x22 + 10x23
s.t.
x11 + x12 + x13 = 1000
x21 + x22 + x23 = 800
x11 + x21 600
x12 + x22 1000
x13 + x23 800
x11, x12, x13, x21, x22, x23 0
Solution:
Supplier
1 2 3
Component 1
600
400
0
Component 2 0 0 800
Purchase Cost = $20,400
13. Let BR = pounds of Brazilian beans purchased to produce Regular
BD = pounds of Brazilian beans purchased to produce DeCaf
CR = pounds of Colombian beans purchased to produce Regular
CD = pounds of Colombian beans purchased to produce DeCaf
Type of Bean Cost per pound ($)
Brazilian 1.10(0.47) = 0.517
Colombian 1.10(0.62) = 0.682
Total revenue = 3.60(BR + CR) + 4.40(BD + CD)
Total cost of beans = 0.517(BR + BD) + 0.682(CR + CD)
Total production cost = 0.80(BR + CR) + 1.05(BD + CD)
Total packaging cost = 0.25(BR + CR) + 0.25(BD + CD)
Total contribution to profit = (total revenue) - (total cost of beans) - (total production cost)
Total contribution to profit = 2.033BR + 2.583BD + 1.868CR + 2.418CD
Regular % constraint
BR = 0.75(BR + CR)
0.25BR - 0.75CR = 0
DeCaf % constraint
BD = 0.40(BD + CD)
0.60BD - 0.40CD = 0
Pounds of Regular: BR + CR = 1000
Pounds of DeCaf: BD + CD = 500
The complete linear program is
Max 2.033BR + 2.583BD + 1.868CR + 2.418CD
s.t.
0.25BR - 0.75CR = 0
0.60BD - 0.40CD = 0
BR + CR = 1000
BD + CD = 500
BR, BD, CR, CD 0
The optimal solution is BR = 750, BD = 200, CR = 250, and CD = 300.
The value of the optimal solution is $3233.75
15. Let x11 = gallons of crude 1 used to produce regular
x12 = gallons of crude 1 used to produce high-octane
x21 = gallons of crude 2 used to produce regular
x22 = gallons of crude 2 used to produce high-octane
Chapter 4
Min 0.10x11 + 0.10x12 + 0.15x21 + 0.15x22
s.t.
Each gallon of regular must have at least 40% A.
x11 + x21 = amount of regular produced
0.4(x11 + x21) = amount of A required for regular
0.2x11 + 0.50x21 = amount of A in (x11 + x21) gallons of regular gas
0.2x11 + 0.50x21 0.4x11 + 0.40x21 [1]
-0.2x11 + 0.10x21 0
Each gallon of high octane can have at most 50% B.
x12 + x22 = amount high-octane
0.5(x12 + x22) = amount of B required for high octane
0.60x12 + 0.30x22 = amount of B in (x12 + x22) gallons of high octane.
0.60x12 + 0.30x22 0.5x12 + 0.5x22
0.1x12 - 0.2x22 0 [2]
x11 + x21 800,000 [3]
x12 + x22 500,000 [4]
x11, x12, x21, x22 0
Optimal Solution: x11 = 266,667, x12 = 333,333, x21 = 533,333, x22 = 166,667
Cost = $165,000
16. Let xi = number of 10-inch rolls of paper processed by cutting alternative i; i = 1,2...,7
Min x1 + x2 + x3 + x4 + x5 + x6 + x7
s.t.
6x1 + 2x3 + x5 + x6 + 4x7 1000 1 1/2" production
4x2 + x4 + 3x5 + 2x6 2000 2 1/2" production
2x3 + 2x4 + x6 + x7 4000 3 1/2" production
x1, x2, x3, x4, x5, x6, x7 0
x1 = 0
x2 = 125
x3 = 500 2125 Rolls
x4 = 1500
x5 = 0 Production:
x6 = 0 1 1/2" 1000
x7 = 0 2 1/2" 2000
3 1/2" 4000
Waste: Cut alternative #4 (1/2" per roll)
750 inches.
b. Only the objective function needs to be changed. An objective function minimizing waste
production and the new optimal solution are given.
Min x1 + 0x2 + 0x3 + 0.5x4 + x5 + 0x6 + 0.5x7
x1 = 0
x2 = 500
x3 = 2000 2500 Rolls
x4 = 0
x5 = 0 Production:
x6 = 0 1 1/2" 4000
x7 = 0 2 2/1" 2000
3 1/2" 4000
Waste is 0; however, we have over-produced the 1 1/2" size by 3000 units. Perhaps these can be
inventoried for future use.
c. Minimizing waste may cause you to over-produce. In this case, we used 375 more rolls to generate a
3000 surplus of the 1 1/2" product. Alternative b might be preferred on the basis that the 3000
surplus could be held in inventory for later demand. However, in some trim problems, excess
production cannot be used and must be scrapped. If this were the case, the 3000 unit 1 1/2" size
would result in 4500 inches of waste, and thus alternative a would be the preferred solution.
17. a. Let FM = number of frames manufactured
FP = number of frames purchased
SM = number of supports manufactured
SP = number of supports purchased
TM = number of straps manufactured
TP = number of straps purchased
Min 38FM + 51FP + 11.5SM + 15SP + 6.5TM + 7.5TP
s.t.
3.5FM + 1.3SM + 0.8TM 21,000
2.2FM + 1.7SM 25,200
3.1FM + 2.6SM + 1.7TM 40,800
FM + FP 5,000
SM + SP 10,000
TM + TP 5,000
FM, FP, SM, SP, TM, TP 0.
Solution:
Manufacture Purchase
Frames 5000 0
Chapter 4
Supports 2692 7308
Straps 0 5000
b. Total Cost = $368,076.91
c. Subtract values of slack variables from minutes available to determine minutes used. Divide by 60
to determine hours of production time used.
Constraint
1 Cutting: Slack = 0 350 hours used
2 Milling: (25200 - 9623) / 60 = 259.62 hours
3 Shaping: (40800 - 18300) / 60 = 375 hours
d. Nothing, there are already more hours available than are being used.
e. Yes. The current purchase price is $51.00 and the reduced cost of 3.577 indicates that for a purchase
price below $47.423 the solution may improve. Resolving with the coefficient of FP = 45 shows
that 2714 frames should be purchased.
The optimal solution is as follows:
OPTIMAL SOLUTION
Optimal Objective Value
361500.00000
Variable Value Reduced Cost
FM 2285.71429 0.00000
FP 2714.28571 3.57692
SM 10000.00000 0.00000
SP 0.00000 0.00000
TM 0.00000 1.15385
TP 5000.00000 0.00000
Constraint Slack/Surplus Dual Value
1 0.00000 -2.69231
2 3171.42857 0.00000
3 7714.28571 0.00000
4 0.00000 47.42308
5 0.00000 15.00000
6 0.00000 7.50000
19. a. Let x11 = amount of men's model in month 1
x21 = amount of women's model in month 1
x12 = amount of men's model in month 2
x22 = amount of women's model in month 2
s11 = inventory of men's model at end of month 1
s21 = inventory of women's model at end of month 1
s12 = inventory of men's model at end of month 2
s22 = inventory of women's model at end of month
The model formulation for part (a) is given.
Min 120x11 + 90x21 + 120x12 + 90x22 + 2.4s11 + 1.8s21 + 2.4s12 + 1.8s22
s.t.
20 + x11 - s11 = 150
or
x11 - s11 = 130 Satisfy Demand [1]
30 + x21 - s21 = 125
or
x21 - s21 = 95 Satisfy Demand [2]
s11 + x12 - s12 = 200 Satisfy Demand [3]
s21 + x22 - s22 = 150 Satisfy Demand [4]
s12 25 Ending Inventory [5]
s22 25 Ending Inventory [6]
Labor Hours: Men’s = 2.0 + 1.5 = 3.5
Women’s = 1.6 + 1.0 = 2.6
3.5 x11 + 2.6 x21 900 Labor Smoothing for [7]
3.5 x11 + 2.6 x21 1100 Month 1 [8]
3.5 x11 + 2.6 x21 - 3.5 x12 - 2.6 x22 100 Labor Smoothing for [9]
-3.5 x11 - 2.6 x21 + 3.5 x12 + 2.6 x22 100 Month 2 [10]
x11, x12, x21, x22, s11, s12, s21, s22 0
The optimal solution is to produce 193 of the men's model in month 1, 162 of the men's model in
month 2, 95 units of the women's model in month 1, and 175 of the women's model in month 2.
Total Cost = $67,156
Inventory Schedule
Month 1
63 Men's
0 Women's
Month 2 25 Men's 25 Women's
Labor Levels
Previous month
1000.00 hours
Month 1 922.25 hours
Month 2 1022.25 hours
b. To accommodate this new policy the right-hand sides of constraints [7] to [10] must be changed to
950, 1050, 50, and 50 respectively. The revised optimal solution is given.
Chapter 4
x11 = 201
x21 = 95
x12 = 154
x22 = 175 Total Cost = $67,175
We produce more men's models in the first month and carry a larger men's model inventory; the
added cost however is only $19. This seems to be a small expense to have less drastic labor force
fluctuations. The new labor levels are 1000, 950, and 994.5 hours each month. Since the added cost
is only $19, management might want to experiment with the labor force smoothing restrictions to
enforce even less fluctuations. You may want to experiment yourself to see what happens.
20. Let xm = number of units produced in month m
Im = increase in the total production level in month m
Dm = decrease in the total production level in month m
sm = inventory level at the end of month m
where
m = 1 refers to March
m = 2 refers to April
m = 3 refers to May
Min 1.25 I1 + 1.25 I2 + 1.25 I3 + 1.00 D1 + 1.00 D2 + 1.00 D3
s.t.
Change in production level in March
x1 - 10,000 = I1 - D1
or
x1 - I1 + D1 = 10,000
Change in production level in April
x2 - x1 = I2 - D2
or
x2 - x1 - I2 + D2 = 0
Change in production level in May
x3 - x2 = I3 - D3
or
x3 - x2 - I3 + D3 = 0
Demand in March
2500 + x1 - s1 = 12,000
or
x1 - s1 = 9,500
Demand in April
s1 + x2 - s2 = 8,000
Demand in May
s2 + x3 = 15,000
Inventory capacity in March
s1 3,000
Inventory capacity in April
s2 3,000
Optimal Solution:
Total cost of monthly production increases and decreases = $2,500
x1 = 10,250 I1 = 250 D1 = 0
x2 = 10,250 I2 = 0 D2 = 0
x3 = 12,000 I3 = 1750 D3 = 0
s1 = 750
s2 = 3000
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