Management Chapter 8 Homework Amalgamated’s blending problem will have eight variables and 11 constraints

b. The computer-generated results are:

Trained

Technicians

Trainees

Month

Available

Beginning

Aug.

350

13.7 (actually 14)

Sept.

Oct.

72.2 (actually 72)

Nov.

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

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

5A

3

60

1

80

3

90

4

2

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