Managerial finance answer

CHAPTER 8Linear Programming Applications

SOLUTIONS TO PROBLEMS

8-1. Since the decision centers about the production of the two different cabinet models, we let

X1 = number of French Provincial cabinets produced each day

X2 = number of Danish Modern cabinets produced each day

Objective: maximize revenue = $28X1 + $25X2

subject to

3X1 + 2X2 360 hours (carpentry department)

1 X1 + 1X2 200 hours (painting department)

X1 + X2 125 hours (finishing department)

X1 60 units (contract requirement)

X2 60 units (contract requirement)

X1, X2 0

Problem 8-1 solved by computer:

Produce 60 French Provincial cabinets (X1) per day

Produce 90 Danish Modern cabinets (X2) per day

Revenue = $3,930

8-2. Let X1 = dollars invested in Los Angeles municipal bonds

X2 = dollars invested in Thompson Electronics

X3 = dollars invested in United Aerospace

X4 = dollars invested in Palmer Drugs

X5 = dollars invested in Happy Days Nursing Homes

Maximize return = 0.053X1 + 0.068X2 + 0.049X3 + 0.084X4 + 0.118X5

subject to X1 + X2 + X3 + X4 + X5 $250,000 (funds)

X1 .2 (X1 + X2 + X3 + X4 + X5) (bonds)

or

.8X1 – .2X2 – .2X3 – .2X4 – .2X5 0

X2 + X3 + X4 .4 (X1 + X2 + X3 + X4 + X5) (combination of electronics, aerospace, and drugs)

Copyright ©2015 Pearson, Inc.8-1

or

–0.4X1 + 0.6X2 + 0.6X3 + 0.6X4 – 0.4X5 0

(X5 0.5X1) rewritten as

–0.5X1 + X5 0 (nursing home as percent of bonds)

X1, X2, X3, X4, X5 0


$50,000 invested in Los Angeles municipal bonds (X1)

$0 invested in Thompson Electronics (X2)

$0 invested in United Aerospace (X3)

$175,000 invested in Palmer Drugs (X4)

$25,000 invested in Happy Days (X5)

This produces an annual return on investment of $20,300.

8-3. Minimize staff size = X1 + X2 + X3 + X4 +X5 + X6

where

Xi = number of workers reporting for start of work at period i (with i = 1, 2, 3, 4, 5, or 6)

X1 + X2 12

X2 + X3 16

X3 + X4 9

X4 + X5 11

X5 + X6 4

X1 + X6 3

All variables 0

There are multiple optimal solutions for this problem. One solution found using computer software is to hire 30 workers:

16 begin at 7 A.M.

9 begin at 3 P.M.

2 begin at 7 P.M.

3 begin at 11 P.M.

An alternative optimum is

3 begin at 3 A.M.

9 begin at 7 A.M.

7 begin at 11 A.M.

2 begin at 3 P.M.

9 begin at 7 P.M.

0 begin at 11 P.M.

All of the solutions would have the same value for the objective function, so 30 workers would be needed.

8-4. Let X1 = number of pounds of oat product per horse each day


X2 = number of pounds of enriched grain per horse each day

X3 = number of pounds of mineral product per horse each day

Minimize cost = 0.09X1 + 0.14X2 + 0.17X3

subject to

2X1 + 3X2 + 1X3 6 (ingredient A)

X1 + 1X2 + X3 2 (ingredient B)

3X1 + 5X2 + 6X3 9 (ingredient C)

1X1 + 1X2 + 2X3 8 (ingredient D)

X1 + X2 + 1X3 5 (ingredient E)

X1 + X2 +X3 6 (maximum feed/day)

All variables 0

Solution: X1 = 1.33

X2 = 0

X3 = 3.33

cost = 0.687


8-5.

Let E1, E2, and E3 represent the ending inventory for the three months respectively. Let RT1, RT2, and RT3 represent the regular production for the three months and OT1, OT2, and OT3 represent the overtime production quantities during the three months respectively. Then the formulation is:

Minimize cost: 300RT1 + 300RT2 + 300RT3 + 325OT1 + 325OT2 + 325OT3 + 20E1 + 20E2 + 20E3

subject to

RT1 < 200 June regular production

RT2 < 200 July regular production

RT3 < 200 August regular production

OT1 < 15 June overtime production

OT2 < 15 July overtime production

OT3 < 15 August overtime production

E1 + RT1 + OT1 = 195 Ending inventory from first month

E2 + E1 + RT2 + OT2 = 215 Ending inventory from second month

E3 + E2 + RT3 + OT3 = 205 Ending inventory from third month

{All variables} ≥ 0 Non-negativity constraints

The optimal production schedule is to produce 200 each month during regular production and to use overtime to produce 10 units in July and 5 in August for a total cost of $184,975.

8-6. Let

T = number of TV ads

R = number of radio ads

B = number of billboard ads

N = number of newspaper ads

Maximize total audience = 30,000T + 22,000R + 24,000B + 8,000N

Subject to

800T + 400R + 500B + 100N 15,000

10

R10

10

10

+R 6

500B + 100N 800T

, R, , 0

Solution: T = 6.875; R = 10; B = 9; N = 10; Audience reached = 722,250.

If integer solutions are necessary, integer programming could be used.

8-7. Let: X1 = number of newspaper ads placed


X2 = number of TV spots purchased

Minimize cost = $925X1 + $2,000X2

subject to 0.04X1 + 0.05X2 0.40 (city exposure)

0.03X1 + 0.03X2 0.60 (exposure in northwest suburbs)

X1, X2 0

Note that the problem is not limited to unduplicated exposure (e.g., one person seeing the Sunday newspaper three weeks in a row counts for three exposures).


Buy 20 Sunday newspaper ads (X1)

Buy 0 TV ads (X2)

This has a cost of $18,500. Perhaps the paint store should consider a blend of TV and newspaper, not just the latter.

8-8. Let Xij = number of new leases in month i for j-months, i = 1,. . ., 6; j = 3, 4, 5

Minimize cost =1260X13 + 1260X23 + 1260X33 + 1260X43 + 840X53 + 420X63 + 1600X14 + 1600X24 + 1600X34 + 1200X44 + 800X54 + 400X64+ 1850X15 + 1850X25 + 1480X35 + 1110X45 + 740X55 + 370X65

subject to: X13 + X14 + X15 420 – 390

X13 + X14 + X15 + X23 + X24 + X25 400 – 270

X13 + X14 + X15 + X23 + X24 + X25 + X33 + X34 + X35 430 – 130

X14 + X15 + X23 + X24 + X25 + X33 + X34 + X35 + X43 + X44 + X45 460

X15 + X24 + X25 + X33 + X34 + X35 + X43 + X44 + X45 + X53 + X54 + X55 470

X25 + X34 + X35 + X43 + X44 + X45 + X53 + X54 + X55 + X63 + X64 + X65 440

X15 + X25 + X35 + X45 + X55 + X65 0.50(X13 + X14 + X15 + X23 + X24 + X25 + X33 + X34

+ X35 + X43 + X44 + X45 + X53 + X54 + X55 + X63 + X64 + X65)

All variables 0

Solving this on the computer results in the following solution:

X15 = 30 5-month leases in March

X25 = 100 5-month leases in April

X35 = 170 5-month leases in May

X45 = 160 5-month leases in June

X55 = 10 5-month leases in July

All other variables equal 0.

Total cost = $677,100.

As a result of this, there are 440 cars remaining at the end of August.

8-9. The linear program has the same constraints as in problem 8-8. The objective function changes and is now:


Minimize cost = 1260(X13 + X23 + X33 + X43 + X53 + X63) + 1600(X14 + X24 + X34 + X44 + X54 + X64) + 1850(X15 + X25 + X35 + X45 + X55 + X65)


X15 = 30 5-month leases in March

X25 = 100 5-month leases in April



X43 = 160 3-month leases in June

X53 = 10 3-month leases in July

All other variables equal 0.

Total cost = $752,950.

8-10. Let Xij = number of students bused from sector i to school j

Objective: minimize total travel miles =

5XAB + 8XAC + 6XAE

+ 0XBB + 4XBC + 12XBE

+ 4XCB + 0XCC + 7XCE

+ 7XDB + 2XDC + 5XDE

+ 12XEB + 7XEC + 0XEE

subject to

XAB + XAC + XAE = 700 (number of students in sector A)

XBB + XBC + XBE = 500 (number of students in sector B)

XCB + XCC + XCE = 100 (number of students in sector C)

XDB + XDC + XDE = 800 (number of students in sector D)

XEB + XEC + XEE = 400 (number students in sector E)

XAB + XBB + XCB + XDB + XEB 900 (school B capacity)

XAC + XBC + XCC + XDC + XEC 900 (school C capacity)

XAE + XBE + XCE + XDE + XEE 900 (school E capacity)

All variables 0

Solution: XAB = 400

XAE = 300

XBB = 500

XCC = 100

XDC = 800

XEE = 400

Distance = 5,400 “student miles”


8-11. Maximize number of rolls of Supertrex sold = 20X1 + 6.8X2 + 12X3 – 65,000X4

where X1 = dollars spent on advertising

X2 = dollars spent on store displays

X3 = dollars in inventory

X4 = percent markup

subject to

X1 + X2 + X3 $17,000 (budgeted)

X1 $3,000 (advertising constraint)

X2 0.05X3 (or X2 – 0.05X3 0) (ratio of displays to inventory)

(markup ranges)

X1, X2, X3, X4 0


Spend $17,000 on advertising (X1).

Spend nothing on in-store displays or on-hand inventory (X2 and X3).

Take a 20% markup.

The store will sell 327,000 rolls of Supertrex.

This solution implies that no on-hand inventory or displays are needed to sell the product, probably due to an oversight on Mr. Kruger’s part. Perhaps a constraint indicating that X3 $3,000 of inventory should be held might be needed.

8-12. Minimize total cost = $0.60X1 + 2.35X2 + 1.15X3 + 2.25X4 + 0.58X5 + 1.17X6 + 0.33X7

subject to

295X1 + 1,216X2 + 394X3 +358X4 + 128X5+ 118X6 + 279X7 1,500

295X1 + 1,216X2 + 394X3 +358X4 + 128X5+ 118X6 + 279X7 900

.2X1 + .2X2 + .4.3X3 + 3.2X4 + 3.2X5 + 14.1X6 + 2.2X7 4

16X1 + 96X2 + 9X3 + 0.5X4 + 0.8X5+ 1.4X6 + 0.5X7 50

16X1 +81X2 + 74X3 + 83X4 + 7X5+ 14X6 + 8X7 26

22X1 + 28X5 +19X6 + 63X7 50

All Xi 0


The meal plan for the evening is

No milk (X1 = 0)

0.499 pound of ground meat (X2)

0.173 pound of chicken (X3)

No fish (X4 = 0)

No beans (X5 = 0)


0.105 pound of spinach (X6)

0.762 pound of white potatoes (X7)

Each meal has a cost of $1.75.

The meal is fairly well-balanced (two meats, a green vegetable, and a potato). The weight of each item is realistic. This problem is very sensitive to changing food prices.

Sensitivity analysis when prices change:

Milk increases 10 cents/lb: no change in price or diet

Milk decreases 10 cents/lb: no change in price or diet

Milk decreases 30 cents/lb (to 30 cents): potatoes drop out and milk enters, price = $1.42/meal

Ground meat increases from $2.35 to $2.75: price = $1.93 and spinach leaves the optimal solution

Ground meat increases to $5.25/lb: price = $2.07 and meat leaves; milk, chicken, and potatoes in solution

Fish decreases from $2.25 to $2.00/lb: no change

Chicken increases to $3.00/lb: price = $1.91 and meat, fish, spinach, and potatoes in solution

If meat and fish are omitted from the problem, the solution is

chicken = 0.774 lb

milk = 1.891 lb

potatoes = 0.133 lb

If chicken and meat are omitted;

fish = 0.679 lb

spinach = 0.0988 lb

milk = 2.188 lb

8-13. a. Let X1 = no. of units of internal modems produced per weekX2 = no. of units of external modems produced per weekX3 = no. of units of circuit boards produced per weekX4 = no. of units of CD drives produced per weekX5 = no. of units of hard drives produced per weekX6 = no. of units of memory boards produced per week

Objective function analysis: First find the time used on each test device:hours on test device 1

hours on test device 2

hours on test device 3


Thus, the objective function ismaximize profit = (revenue) – (material cost) – )test cost)

= (200X1 + 120X2 + 180X3 + 130X4 + 430X5 + 260X6 – 35X1 – 25X2 – 40X3 – 45X4 – 170X5

– 60X6)

This can be rewritten asmaximize profit = $161.35X1 + 92.95X2 + 135.50X3 + 82.50X4 + 249.80X5 + 191.75X6

subject to7X1 + 3X2 + 12X3 + 6X4 + 18X5 + 17X6 < 120(60) Minutes on test device 12X1 + 5X2 + 3X3 + 2X4 + 15X5 + 17X6 < 120(60) Minutes on test device 25X1 + 1X2 + 3X3 + 2X4 + 9X5 + 2X6 < 100(60) Minutes on test device 3

All variables 0

b. The solution is

X1 = 496.55 internal modems

X2 = 1,241.38 external modems

X3 through X6 = 0

profit = $195,504.80

c. The shadow prices, as explained in Chapter 7 and Module 7, for additional time on the three test devices are $21.41, $5.75, and $0, respectively, per minute.

8-14. a. Let Xi = no. of trained technicians available at start of month i

Yi = no. of trainees beginning in month i

Minimize total salaries paid = $2,000X1

+ 2,000X2 + 2,000X3 + 2,000X4 + 2,000X5 + 900Y1 + 900Y2 + 900Y3 + 900Y4 + 900Y5

subject to

130X1 – 90Y1 40,000 (Aug. need, hours)

130X2 – 90Y2 45,000 (Sept. need)

130X3 – 90Y3 35,000 (Oct. need)

130X4 – 90Y4 50,000 (Nov. need)

130X5 – 90Y5 45,000 (Dec. need)

X1 = 350 (starting staff on Aug. 1)

X2 = X1 + Y1 – 0.05X1 (staff on Sept. 1)

X3 = X2 + Y2 – 0.05X2 (staff on Oct. 1)


X4 = X3 + Y3 – 0.05X3 (staff on Nov. 1)

X5 = X4 + Y4 – 0.05X4 (staff on Dec. 1)

All Xi, Yi 0

b. The computer-generated results are:

TrainedTechnicians Trainees

Month Available BeginningAug. 350 13.7 (actually 14)Sept. 346.2 0Oct. 328.8 72.2 (actually 72)Nov. 384.6 0Dec. 365.4 0

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

1.6X12 + 2.9X22 + 3.5X32 3,400 acre-feet in N

1.6X13 + 2.9X23 + 3.5X33 800 acre-feet in NW

1.6X14 + 2.9X24 + 3.5X34 500 acre-feet in W

1.6X15 + 2.9X25 + 3.5X35 600 acre-feet in SW

water acre-feet total

Sales limits:

X11 + X12 + X13 + X14 + X15 2,200 wheat in acres (= 110,000 bushels)

X21 + X22 + X23 + X24 + X25 1,200 alfalfa in acres (= 1,800 tons)

X31 + X32 + X33 + X34 + X35 1,000 barley in acres (= 2,200 tons)

Acreage availability:

X11 + X21 + X31 2,000 acres in SE parcel

X12 + X22 + X32 2,300 acres in N parcel

X13 + X23 + X33 600 acres in NW parcel

X14 + X24 + X34 1,100 acres in W parcel

X15 + X25 + X35 500 acres in SW parcel

Objective function:


maximize profit

b. The solution is to plant

X12 = 1,250 acres of wheat in N parcel

X13 = 500 acres of wheat in NW parcel

X14 = 312 acres of wheat in W parcel

X15 = 137 acres of wheat in SW parcel

X25 = 131 acres of alfalfa in SW parcel

X31 = 600 acres of barley in SE parcel

X32 = 400 acres of barley in N 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).

Let X1 through X8 represent pounds of alloy 1 through pounds of carbide 3 to be used in the blend.

Minimize cost = 0.12X1 + 0.13X2 + 0.15X3 + 0.09X4 + 0.07X5 + 0.10X6 + 0.12X7 + 0.09X8

subject to

manganese quality:

0.70X1 + 0.55X2 + 0.12X3 + 0.01X4 + 0.05X5 42 (2.1% of 2,000)

0.70X1 + 0.55X2 + 0.12X3 + 0.01X4 + 0.05X5 46 (2.3% of 2,000)

silicon quality:

0.15X1 + 0.30X2 + 0.26X3 + 0.10X4 + 0.025X5 + 0.24X6 + 0.25X7 + 0.23X8 86 (4.3% of 2,000)

0.15X1 + 0.30X2 + 0.26X3 + 0.10X4 + 0.025X5 + 0.24X6 + 0.25X7 + 0.23X8 92 (4.6% of 2,000)

carbon quality:

0.03X1 + 0.01X2 + 0.03X4 + 0.18X6 + 0.20X7 + 0.25X8 101 (5.05% of 2,000)

0.03X1 + 0.01X2 + 0.03X4 + 0.18X6 + 0.20X7 + 0.25X8 107 (5.35% of 2,000)

Availability by weight:

X2 300

X6 50


X7 200

X8 100

One-ton weight:

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.

2. Producing 1 ton from the materials is not possible.

If constraints 5 and 11 are relaxed (or removed), one solution is X2 = 83.6 lb (alloy 2), X6 = 50 lb (carbide 1), X7 = 83.6 lb (carbide 2), and X8 = 100 lb (carbide 3). Cost = $34.91.

Each student may take a different approach and other recommendations may result.

8-18. X1 = number of medical patients

X2 = number of surgical patients

Maximize revenue = $2,280X1 + $1,515X2

subject to

8X1 + 5X2 32,850 (patient-days available = 365 days 90 new beds)

3.1X1 + 2.6X2 15,000 (lab tests)

1X1 + 2X2 7,000 (x-rays)

X2 2,800 (operations/surgeries)

X1, X2 0

Problem 8-18 solved by computer results in the following solution (rounded):

X1 = 2,791 medical patients

X2 = 2,105 surgical patients

revenue= $9,551,659 per year

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

surgical = (2,105 patients)(5 days/patient)

= 10,525 days

total = 32,853 days (The rounding causes this to be slightly higher than the limit.)

This represents 68% medical days and 32% surgical days, which yields 61 medical beds and 29 surgical beds. (Note that an alternative approach would be to formulate with X1, X2 as number of


beds.)


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.

(b) There are no empty beds because the slack for constraint 1 has a value of 0.

(c) There are 876 (the slack for constraint 2) lab tests of unused capacity.

(d) The x-ray is used to its maximum (slack for constraint 3 is 0) and has a $65.45 dual price. The revenue would increase by this amount for each additional x-ray.

(e) The operating room still has 695 operations available (the slack for constraint 4).

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:

Minimize 30X1 + 30X2 + 34.80X3 + 34.80X4

subject to:

0.46X1 + 0.46X3 25000

0.46X2 + 0.46X4 32000

-0.10 X1 + 0.15X3 0

0.05X2 – 0.25X4 0

The rounded solution is X1 = 32609; X2 = 57971; X3 = 21739; X4 = 11594; Cost = 3877391

8-21. The objective is to minimize the total number of 10-inch rolls used. There are three constraints, indicating that the number of each size roll (2.5, 3, and 3.5 inches) generated by the cuts must be at least the number needed to fill the orders. Define the decision variables as

Xi = number of 10-inch rolls cut with cutting pattern i for i = 1, 2, 3, 4, 5.

Minimize number of rolls = X1 + X2 + X3 + X4 + X5

subject to4X1 + X4 + X5 > 2000 number of 2.5-inch rolls needed to fill orders 3X2 + X3 + X4 > 4000 number of 3-inch rolls needed to fill orders2X3 + X4 + 2X5 > 5000 number of 3.5-inch rolls needed to fill orders

Xi 0 for i = 1, 2,. . ., 5


X1 = 500; X2 = 500; X3 = 2500; X4 = 0; X5 = 0 with a total number of rolls used equal to 3,500.


If the company wanted to minimize waste instead of minimizing the number of rolls used, waste generated from each cutting patterned is considered. Patterns 1 and 3 have no waste, patterns 2 and 4 each have 1 inch of waste, and pattern 5 has 0.5 inches of waste. Thus, the objective func-tion would becomeMinimize total waste = X2 + X4 + 0.5X5

The constraints would not change. The optimal solution is:X1 = 500; X2 = 0; X3 = 4,000; X4 = 0; X5 = 0 with total waste equal to 0 inches. However, with this solution, there are 8,000 extra 3.5” rolls.

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

0.11X1 + 0.09X2 + 0.065X3 + 0.15X4 + 0.13X5 0.11 return should be at least 11%

X1 0.35 no more than 35% in any single stock

X2 0.35

X3 0.35

X4 0.35

X5 0.35

Xi 0 for i = 1, 2,. . ., 5

b. Solving this on the computer, we have

X1 = 0

X2 = 0.10625

X3 = 0.35

X4 = 0.35

X5 = 0.19375

Minimum beta = 1.015

Return = 0.11(0) + 0.09(0.10625) + 0.065(0.35) + 0.15(0.35) + 0.13(0.19375) = 0.11

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

N = 1,000 gallons of fuel to purchase in New Orleans

FA = fuel remaining when plane lands in Atlanta

FL = fuel remaining when plane lands in Los Angeles

FH = fuel remaining when plane lands in Houston

FN = fuel remaining when plane lands in New Orleans

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

L + FL 15 minimum amount of fuel on board when leaving Los Angeles

L + FL 23 maximum amount of fuel on board when leaving Los Angeles

H + FH 9 minimum amount of fuel on board when leaving Houston

H + FH 17 maximum amount of fuel on board when leaving Houston

N + FN 11 minimum amount of fuel on board when leaving New Orleans

N + FN 20 maximum amount of fuel on board when leaving New Orleans

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:

0.95A + 0.95 FA – FL = 10.8

Similarly,

FH = L + FL – (7 + 0.05(L + FL – 15)) becomes 0.95L + 0.95FL – FH = 6.25

FN = H + FH – (3 + 0.05(H + FH – 9)) becomes 0.95H + 0.95FH – FN = 2.55

FA = N + FN – (5 + 0.05(N + FN – 11)) becomes 0.95N + 0.95FN – FA = 4.45

All variables 0

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)

FL = 12 (1,000 gallons of fuel remaining when plane lands in Los Angeles)

H = 1 (1,000 gallons of fuel to purchase in Houston)

FH = 8 (1,000 gallons of fuel remaining when plane lands in Houston)

N = 5 (1,000 gallons of fuel to purchase in New Orleans)

FN = 6 (1,000 gallons of fuel remaining when plane lands in New Orleans)

Total cost = 112.45 ( 1,000)


SOLUTIONS TO INTERNET HOMEWORK PROBLEMS

8-24. Let X1 = number of Chaunceys mixed

X2 = number of Sweet Italians mixed

X3 = number of bourbon on the rocks mixed

X4 = number of Russian martinis mixed

Maximize total drinks = X1 + X2 + X3 + X4

subject to

1X1 + 4X3 52 oz (bourbon limit)

1X1 + 1X2 38 oz (brandy limit)

1X1 + 2.6667X4 64 oz (vodka limit)

1X2 + 1.3333X4 24 oz (dry vermouth limit)

1X1 + 2X2 36 oz (sweet vermouth limit)

All variables 0

Because a Chauncey (X1) is 25% sweet vermouth, it requires 1 oz of that resource (each drink totals 4 oz).


Mix 25.99 (or 26) Chaunceys (X1)

Mix 5.00 (or 5) Sweet Italians (X2)

Mix 6.50 (or 6) bourbon on the rocks (X3)

Mix 14.25 (or 14) Russian martinis (X4)

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

X31 + X32 + X33 275

X11 + X21 + X31= 200

X12 + X22 + X32= 100

X13 + X23 + X33= 300

All variables 0

The solution is:

X11 = 25, X13 = 125, X23 = 175, X31 = 175, X32 = 100

Cost = $4,525.

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.

IXi = number of BR54 units in inventory at end of month i, for i = 0, 1, 2, 3.

IYi = number of BR49 units in inventory at end of month i, for i = 0, 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:

IX0 = 50 initial inventory of BR54

IY0 = 50 initial inventory of BR49

IX3 = 100 ending inventory of BR54

IY3 = 150 ending inventory of BR49

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

X2 + IX1 = 740 + IX2 BR54 requirements for September

X3 + IX2 = 500 + IX3 BR54 requirements for October

Y1 + IY0 = 450 + IY1 BR49 requirements for August

Y2 + IY1 = 420 + IY2 BR49 requirements for September

Y3 + IY2 = 480 + IY3 BR49 requirements for October

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.


Managerial finance answer

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