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b. The computer-generated results are:
Trained
Technicians
Trainees
Month
Available
Beginning
Aug.
350
13.7 (actually 14)
Total salaries paid over the five-month period = $3,627,279.
8-15. a. Let Xij = acres of crop i planted on parcel j
where i = 1 for wheat, 2 for alfalfa, 3 for barley
j = 1 to 5 for SE, N, NW, W, and SW parcels
Irrigation limits:
1.6X11 + 2.9X21 + 3.5X31 3,200 acre-feet in SE
Sales limits:
X11 + X12 + X13 + X14 + X15 2,200 wheat in acres (= 110,000 bushels)
Acreage availability:
X11 + X21 + X31 2,000 acres in SE parcel
Objective function:
maximize profit
( ) ( ) ( )( )
5 5 5
1, 2, 3,
1 1 1
$2 50 bushels $40 1.5 tons $50 2.2 tons
j j j
j j j
X X X
= = =
= + +
b. The solution is to plant
X12 = 1,250 acres of wheat in N parcel
X13 = 500 acres of wheat in NW parcel
Profit will be $337,862.10. Multiple optimal solutions exist.
c. Yes, need only 500 more water-feet.
8-16. Amalgamated’s blending problem will have eight variables and 11 constraints. The eight
variables correspond to the eight materials available (three alloys, two irons, three carbides) that
can be selected for the blend. Six of the constraints deal with maximum and minimum quality
limits, one deals with the 2,000 pound total weight restriction, and four deal with the weight
availability limits for alloy 2 (300 lb), carbide 1 (50 lb), carbide 2 (200 lb), and carbide 3 (100
lb).
manganese quality:
1 0.70X1 + 0.55X2 + 0.12X3 + 0.01X4 + 0.05X5 42 (2.1% of 2,000)
2 0.70X1 + 0.55X2 + 0.12X3 + 0.01X4 + 0.05X5 46 (2.3% of 2,000)
silicon quality:
carbon quality:
5 0.03X1 + 0.01X2 + 0.03X4 + 0.18X6 + 0.20X7 + 0.25X8 101 (5.05% of 2,000)
6 0.03X1 + 0.01X2 + 0.03X4 + 0.18X6 + 0.20X7 + 0.25X8 107 (5.35% of 2,000)
Availability by weight:
7 X2 300
One-ton weight:
11 X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 = 2,000
The solution is infeasible.
8-17. This problem refers to Problem 8-16’s infeasibility. Some investigative work is needed to
track down the issues. The two issues are:
1. Requiring at least 5.05% carbon is not possible.
8-18. X1 = number of medical patients
X2 = number of surgical patients
Maximize revenue = $2,280X1 + $1,515X2
subject to
X1, X2 0
Problem 8-18 solved by computer:
X1 = 2,791 medical patients
To convert X1 and X2 to number of medical versus surgical beds, find the total number of hospital
days for each type of patient:
medical = (2,791 patients)(8 days/patient)
= 22,328 days
8-19. This problem, suggested by Professor C. Vertullo, is an excellent exercise in report
writing. Here is a chance for students to present management science results in a management
format. Basically, the following issues need to be addressed in any report:
(a) As seen in Problem 8-18, there should be 61 medical and 29 surgical beds, yielding
$9,551,659 per year.
8-20.
For the Low Knock Oil Company example it was originally assumed that a one to one ratio of
raw materials (crude oil) to finished goods (gasoline). In reality, that ratio is closer to 46%.
Hence, the example problem needs to be modified with 0.46 as the coefficient throughout the
first two constraints as follows:
The rounded solution is X1 = 32609; X2 = 57971; X3 = 21739; X4 = 11594; Cost = 3877391
8-21. Minimize time = 12XA1 + 11XA2 + 8XA3 + 9XA4 + 6XA5 + 6XA6 + 6XG1 + 12XG2 + 7XG3 +
7XG4 + 5XG5 + 8XG6 + 8XS1 + 9XS2 + 6XS3 + 6XS4 + 7XS5 + 9XS6
subject to
XA1 + XA2 + XA3 + XA4 + XA5 + XA6 = 200
Solution:
Source
Destination
Number of
(Station)
(Wing)
Trays
5A
5
60
5A
6
80
8-22. Let
Xi = proportion of investment invested in stock i for i = 1, 2, . . . , 5
Minimize beta = 1.2X1 + 0.85X2 + 0.55X3 + 1.40X4 + 1.25X5
subject to
X1 + X2 + X3 + X4 + X5 = 1 total of the proportions must add to 1
Xi 0 for i = 1, 2, . . . , 5
b. Solving this on the computer, we have
X1 = 0
X2 = 0.10625
8-23. Let
A = 1,000 gallons of fuel to purchase in Atlanta
L = 1,000 gallons of fuel to purchase in Los Angeles
H = 1,000 gallons of fuel to purchase in Houston
Minimize cost = 4.15A + 4.25L + 4.10H + 4.18N
subject to
A + FA 24 minimum amount of fuel on board when leaving Atlanta
A + FA 36 maximum amount of fuel on board when leaving Atlanta
FL = A + FA – (12 + 0.05(A + FA – 24))
This says that the fuel on board when the plane lands in Los Angeles will equal the amount on
board at take-off minus the fuel consumed on that flight. The fuel consumed is 12 (thousand
gallons) plus 5% of the excess above 24 (thousand gallons). This simplifies to:
The optimal solution is
A = 18 (1,000 gallons of fuel to purchase in Atlanta)
FA = 6 (1,000 gallons of fuel remaining when plane lands in Atlanta)
L = 3 (1,000 gallons of fuel to purchase in Los Angeles)
SOLUTIONS TO INTERNET HOMEWORK PROBLEMS
8-24. Let X1 = number of Chaunceys mixed
X2 = number of Sweet Italians mixed
Maximize total drinks = X1 + X2 + X3 + X4
subject to
1X1 + 4X3 52 oz (bourbon limit)
1X1 + 1X2 38 oz (brandy limit)
Problem 8-27 solved by computer:
Mix 25.99 (or 26) Chaunceys (X1)
Mix 5.00 (or 5) Sweet Italians (X2)
This is a total of 51.75 drinks (in five iterations).
8-25. Minimize 6X11 + 8X12 + 10X13 + 7X21 + 11X22 + 11X23 + 4X31 + 5X32 + 12X33
subject to
X11 + X12 + X13 150
X21 + X22 + X23 175
8-26. Let Xi = number of BR54 produced in month i, for i = 1, 2, 3.
Yi = number of BR49 produced in month i, for i = 1, 2, 3.
Minimize cost = 80(X1 + X2 + X3) + 95(Y1 + Y2 + Y3) + 0.8(IX1 + IX2 + IX3) + 0.95(IY1 + IY2 + IY3)
Subject to:
X1 + Y1 1,100 maximum production level in August
X2 + Y2 1,100 maximum production level in September
X3 + Y3 1,100 maximum production level in October
X1 + IX0 = 320 + IX1 BR54 requirements for August
All variables 0
A computer solution to this results in IX0 = 50, IX1 = 190, IX2 = 130, IX3 = 100, IY0 = 50, IY3 = 150,
X1 = 460, X2 = 680, X3 = 470, Y1 = 400, Y2 = 420, Y3 = 630. All other variables = 0. The total cost
= $267,028.50.
SOLUTION TO CHASE MANHATTAN BANK CASE
This very advanced and challenging scheduling problem can be solved most expeditiously using
linear programming, preferably integer programming. Let F denote the number of full-time
employees. Some number, F1, of them will work 1 hour of overtime between 5 P.M. and 6 P.M.
each day and some number, F2, of the full-time employees will work overtime between 6 P.M.
The workforce requirements for the first two hours, 9 A.M. and 10 A.M., are:
F + P1 14
F + P1 + P2 25
At 11 A.M. half of the full-time employees go to lunch; the remaining half go at noon. For those
hours:
Starting at 1 P.M., some of the part-time employees begin to leave. For the remainder of the
straight-time day:
F + P1 + P2 + P3 + P4 + P5 – Q4 55
For the two overtime hours:
F1 + P1 + P2 + P3 + P4 + P5 + P6 + P7 – Q4 – Q5 – Q6 – Q7 – Q8 14
This also leads to the objective function. The total daily labor cost which must be minimized
is
Total overtime for a full-time employee is restricted to 5 hours or less, an average of 1 hour or
less per day per employee. Thus the number of overtime hours worked per day cannot exceed the
number of full-time employees:
F1 + F2 F
Since part-time employees must work at least 4 hours per day,
Similarly, for the remainder of the day,
Q6 P1 + P2 + P3 – Q4 – Q5
To ensure that all part-timers who began at 9 A.M. do not work more than 7 hours:
Q4 + Q5 + Q6 + Q7 P1
Similarly,
Q4 + Q5 + Q6 + Q7 + Q8 P1 + P2
The resulting problem has 16 integer variables and 22 constraints. If integer programming
software is not available, the linear programming problem can be solved and the solution
rounded, making certain that none of the constraints have been violated. Note that the integer
programming solution might also need to be adjusted—if F is an odd integer, 0.5F will not be an
integer and the requirement that “half” of the full-time employees go to lunch at 11 A.M. and the
other half at noon will have to be altered by assigning the extra employee to the appropriate hour.
1. The least-cost solution requires 29 full-time employees, 9 of whom work two hours of
overtime per day. In actuality, 18 of the full-time employees would work overtime on two
different days and 9 would work overtime on one day. Fourteen of the full-time workers would
of 232 hours of straight time, 18 hours of overtime, and 126 hours of part-time work is $3,476.28
per day.
This solution is not unique—other work assignments can be found that result in this same
cost.
2. The same staffing would be used every day. In fact, one would expect different patterns to
present themselves on different days; for example, Fridays are usually much busier bank days
