103
T
EACHINGS
UGGESTIONSTeaching Suggestion 8.1:Importance of Formulating Large LP Problems.
Since computers are used to solve virtually all business LP prob-lems, the most important thing a student can do is to get experi-ence in formulating a wide variety of problems. This chapter pro-vides such a variety.
Teaching Suggestion 8.2:Note on Production Scheduling Problems.
The Greenberg Motor example in this chapter is the largest prob-lem in the book in terms of constraints, so it provides a good prac-tice environment. An interesting feature to point out is that LP constraints are capable of tying one production period to the next. Teaching Suggestion 8.3:Solving Assignment Problems by LP. The example of the law firm of Ivan and Ivan in this chapter can clearly be solved more quickly using QM for Windows’ assign-ment program than by the LP program. Students should be asked why anyone would choose to use the LP approach. There are two answers: (1) many commercial LP programs do not contain as-signment algorithms (which are more popular in academic soft-ware such as QM for Windows); and (2) the LP program can pro-vide more sensitivity analysis and economic interpretation than is available in the assignment module. The assignment problem is treated in Chapter 10.
Teaching Suggestion 8.4:Labor Planning Problem—Arlington Bank.
This example is a good practice tool and lead-in for the Chase Manhattan Bank case at the end of the chapter. Without this exam-ple, the case would probably overpower most students.
Teaching Suggestion 8.5:Ingredient Blending Applications. Three points can be made about the two blending examples in this chapter. First, both the diet and fuel blending problems presented here are tiny compared to huge real-world blending problems. But they do provide some sense of the issues to be faced.
Second, diet problems that are missing the constraints that force variety into the diet can be terribly embarrassing. It has been said that a hospital in New Orleans ended up with an LP solution to feed each patient only castor oil for dinner because analysts ne-glected to add constraints forcing a well-rounded diet.
A
LTERNATIVEE
XAMPLESAlternative Example 8.1: Natural Furniture Company manu-factures three outdoor products, chairs, benches, and tables. Each product must pass through the following departments before it is
shipped: sawing, sanding, assembly, and painting. The time re-quirements (in hours) are summarized in the tables below.
The production time available in each department each week and the minimum weekly production requirement to fulfill con-tracts are as follows:
The production manager has the responsibility of specifying pro-duction levels for each product for the coming week. Let
X1Number of chairs produced
X2Number of benches produced
X3Number of tables produced
The objective function is
Maximize profit 15X110X220X3
Constraints
1.5X11.5X22.0X3450 hours of sawing available
1.0X11.5X22.0X3400 hours of sanding available
2.0X12.0X22.5X3625 hours of assembly available
1.5X12.0X22.0X3550 hours of painting available
X12.0X22.0X3100 chairs X22.0X350 benches
X350 tables
X1, X2, X30
Alternative Example 8.2: A phosphate manufacturer produces three grades, A, B, and C, which cost the firm $40, $50, and $60 per kilogram, respectively. The products require the labor and ma-terials per batch that are shown on the following page.
8
C H A P T E R
Linear Programming Modeling Applications: With
Computer Analyses in Excel and QM for Windows
Minimum Capacity Production Department (In Hours) Product Level
Sawing 450 Chairs 100
Sanding 400 Benches 50
Assembly 625 Tables 50
Painting 550
Unit Product Sawing Sanding Assembly Painting Profit
Chairs 1.5 1.0 2.0 1.5 $15
Benches 1.5 1.5 2.0 2.0 $10
Tables 2.0 2.0 2.5 2.0 $20
What mix of products would yield minimum cost? Objective function
Minimize cost 40A 50B 60C Constraints
Labor: 4A 4B 5C 80
Raw material #1 200A 300B 300C 6,000 Raw material #2 600A 400B 500C 5,000
S
OLUTIONST
OP
ROBLEMS8-1. Since the decision centers about the production of the two different cabinet models, we let
X1number of French Provincial cabinets produced
each day
X2number of Danish Modern cabinets produced each
day
Objective: maximize revenue $28X1$25X2
subject to
3X12X2360 hours (carpentry department)
11–
2X11X2200 hours (painting department) 3
–
4X13–4X2125 hours (finishing department)
X1 60 units (contract requirement)
X260 units (contract requirement)
X1, X20
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 X1dollars invested in Los Angeles municipal bonds
X2dollars invested in Thompson Electronics
X3dollars invested in United Aerospace
X4dollars invested in Palmer Drugs
X5dollars invested in Happy Days Nursing Homes
Maximize return 0.053X10.068X20.049X30.084X4
0.118X5
subject to X1X2X3X4X5$250,000 (funds) X1.2 (X1X2X3X4X5) (bonds)
or
.8X1.2X2.2X3.2X4.2X50
X2X3X4.4 (X1X2X3X4X5) (combination
of electronics, aerospace, and drugs) or
0.4X10.6X20.6X30.6X40.4X50
(X50.5X1) rewritten as
0.5X1X50 (nursing home as percent of bonds)
X1, X2, X3, X4, X50
Problem 8-2 solved by computer:
$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 X1X2X3X4 X5X6
where
Xinumber of workers reporting for start of work at period i (with i1, 2, 3, 4, 5, or 6)
X1X212
X2X316
X3X4 9
X4X511
X5X6 4
X1X6 3
All variables 0
The computer solution 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.
8-4. Let X1number of pounds of oat product per horse each
day
X2number of pounds of enriched grain per horse
each day
X3number of pounds of mineral product per horse
each day
Minimize cost 0.09X10.14X20.17X3
subject to
2X13X21X36 (ingredient A) 1
–
2X11X21–2X32 (ingredient B)
3X15X26X39 (ingredient C)
1X1121–X22X38 (ingredient D) 1
–
2X11–2X2121–X35 (ingredient E)
X1X2X36 (maximum feed/day)
All variables 0 Solution: X111–3
X20 X331–3
cost0.687 Grade Grade Grade Available
A B C Resources
Labor hours 4 4 5 80 hr
Raw material #1 200 300 300 6,000 kg
8-5. Let Xij1 if pitcher iis scheduled to go against opponent j, 0 otherwise
where i1, 2, 3, 4 stands for Jones, Baker, Parker, and Wilson, respectively, and
j1, 2, 3, 4 stands for Des Moines, Davenport, Omaha, and Peoria, respectively.
Objective: maximize sum of ratings
0.6X11 0.8X120.5X130.4X14
0.7X210.4X220.8X230.3X24
0.9X310.8X320.7X330.8X34
0.5X410.3X420.4X430.2X44
subject to
X11X12X13X141 (“Dead-Arm” Jones)
X21X22X23X241 (“Spitball” Baker)
X31X32X33X341 (“Ace” Parker)
X41X42X43X441 (“Gutter” Wilson)
X11X21X31X411 (Des Moines)
X12X22X32X421 (Davenport)
X13X23X33X431 (Omaha)
X14X24X34X441 (Peoria)
Solution: X121, X231, X341, X411, Total P2.9
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
R 10 10 10 R6
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 (see Chapter 11) could be used.
8-7. Let:X1number of newspaper ads placed
X2number of TV spots purchased
Minimize cost $925X1$2,000X2
subject to 0.04X10.05X20.40 (city exposure)
0.03X10.03X20.60 (exposure in
northwest suburbs) X1, X20
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).
Problem 8-7 solved by computer: 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 Xijnumber of new leases in month ifor j-months, i1, . . . , 6; j3, 4, 5
Minimize cost 1260X131260X231260X331260X43
840X53420X631600X141600X24
1600X341200X44800X54400X64 1850X151850X251480X351110X45
740X55370X65
subject to: X13X14X15420390 X13X14X15X23X24
X25400270
X13X14X15X23X24X25X33
X34X35430130
X14X15X23X24X25X33X34
X35X43X44X45460
X15X24X25X33X34X35X43
X44X45X53X54X55470 X25X34X35X43X44X45X53
X54X55X63X64X65440
X15X25X35X45X55X65
0.50(X13X14X15X23X24X25 X33X34X35X43X44X45
X53X54X55X63X64X65) All variables0
Solving this on the computer results in the following solution: X1530 5-month leases in March
X25100 5-month leases in April
X35170 5-month leases in May
X45160 5-month leases in June
X5510 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(X13X23X33X43X53X63)
1600(X14X24X34X44X54X64)
1850(X15X25X35X45X55
X65)
Solving this on the computer results in the following solution: X1530 5-month leases in March
X25100 5-month leases in April
X3465 4-month leases in May
X43160 3-month leases in June
X5310 3-month leases in July
All other variables equal 0. Total cost$752,950.
8-10. Let Xijnumber of students bused from sector ito school j Objective: minimize total travel miles
5XAB8XAC6XAE 0XBB4XBC12XBE 4XCB0XCC7XCE 7XDB2XDC5XDE 12XEB7XEC0XEE subject to
XABXACXAE700 (number of students in sector A) XBBXBCXBE500 (number of students in sector B) XCBXCCXCE100 (number of students in sector C) XDBXDCXDE800 (number of students in sector D)
XEBXECXEE400 (number students in sector E) XABXBBXCBXDBXEB900 (school Bcapacity) XACXBCXCCXDCXEC900 (school Ccapacity) XAEXBEXCEXDEXEE900 (school Ecapacity) All variables 0
Solution: XAB400 XAE300 XBB500 XCC100 XDC800 XEE400
Distance 5,400 “student miles”
8-11. Maximize number of rolls of Supertrex sold 20X16.8X212X365,000X4
where X1dollars spent on advertising
X2dollars spent on store displays
X3dollars in inventory
X4percent markup
subject to
X1X2X3$17,000 (budgeted)
X1 $3,000 (advertising constraint)
X2 0.05X3(or X20.05X30)
(ratio of displays to inventory)
(markup ranges)
X1, X2, X3, X40
Problem 8-11 solved by computer: Spend $17,000 on advertising (X1).
Spend nothing on in-store displays or on-hand inventory (X2and X3).
Take a 20% markup.
The store will sell 327,000 rolls of Supertrex.
X X
4
4 0 20
0 45 .
.
⎫ ⎬ ⎪ ⎭⎪
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 $0.60X12.35X21.15X3
2.25X40.58X51.17X60.33X7
subject to
295X11,216X2394X3358X4128X5
118X6279X71,500
295X11,216X2394X3358X4128X5
118X6279X7900
.2X1121.2X2.4.3X33.2X43.2X5
14.1X62.2X74
16X11,296X2.4.9X30.5X40.8X5
1.4X60.5X750
16X1 81X274X383X47X5
14X6 8X726
22X128X5 19X663X7 50
All Xi0
Problem 8-12 solved by computer: The meal plan for the evening is
No milk (X10)
0.499 pound of ground meat (X2)
0.173 pound of chicken (X3)
No fish (X40)
No beans (X50)
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 veg-etable, 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 potatoes0.133 lb If chicken and meat are omitted;
8-13. a. Let X1no. of units of internal modems produced per
week
X2no. of units of external modems produced
per week
X3no. of units of circuit boards produced per
week
X4no. of units of floppy disk drives produced
per week
X5no. of units of hard drives produced per
week
X6no. 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 is
maximize profit revenue material cost test cost 200X1120X2180X3130X4430X5
260X6
35X125X240X345X4170X560X6
This can be rewritten as
maximize profit $161.35X192.95X2135.50X3
82.50X4249.80X5191.75X6
subject to
All variables 0 b. The solution is
X1496.55 internal modems
X21,241.38 external modems
X3through X60
profit $195,504.80
5 1 3 2 9 2
60 100
1 2 3 4 5 6
X X X X X X
hours
2 5 3 2 15 17
60 120
1 2 3 4 5 6
X X X X X X
hours
7 3 12 6 18 17
60 120
1 2 3 4 5 6
X X X X X X
hours
185 1 3 2 9 2
60
1 2 3 4 5 6
X X X X X X
122 5 3 2 15 17
60
1 2 3 4 5 6
X X X X X X
157 3 12 6 18 17
60
1 2 3 4 5 6
X X X X X X
5 1 3 2 9 2
60
1 2 3 4 5 6
X X X X X X
2 5 3 2 15 17
60
1 2 3 4 5 6
X X X X X X
=7 3 12 6 18 17
60
1 2 3 4 5 6
X X X X X X
c. The shadow prices, as explained in Chapters 7 and 9, for additional time on the three test devices are $21.41, $5.75, and $0, respectively, per minute.
8-14. a. Let Xino. of trained technicians available at start of month i
Yino. of trainees beginning in month i Minimize total salaries paid $2,000X1
2,000X22,000X32,000X42,000X5 900Y1900Y2900Y3900Y4900Y5 subject to
130X190Y140,000 (Aug. need, hours)
130X290Y245,000 (Sept. need)
130X390Y335,000 (Oct. need)
130X490Y450,000 (Nov. need)
130X590Y545,000 (Dec. need)
X1350 (starting staff on Aug. 1)
X2X1Y10.05X1(staff on Sept. 1)
X3X2Y20.05X2(staff on Oct. 1)
X4X3Y30.05X3(staff on Nov. 1)
X5X4Y40.05X4(staff on Dec. 1)
All Xi,Yi0
b. The computer-generated results are:
Total salaries paid over the five-month period $3,627,279. 8-15. a. Let Xijacres of crop iplanted 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:
1.6X112.9X213.5X313,200 acre-feet in SE
1.6X122.9X223.5X323,400 acre-feet in N
1.6X132.9X233.5X33800 acre-feet in NW
1.6X142.9X243.5X34500 acre-feet in W
1.6X152.9X253.5X35600 acre-feet in SW
water acre-feet total Sales limits:
X11X12X13X14X15 2,200 wheat in acres
(110,000 bushels) X21X22X23X24X251,200 alfalfa in acres
(1,800 tons)
1 6 1 2 9 2 3 5 7 400
1 5
1 5
3 1 5
. X j . X ,j . X, , j
j
j j
∑
∑
∑
Trained
Technicians Trainees
Month Available Beginning
Aug. 350 13.7 (actually 14)
Sept. 346.2 0
Oct. 328.8 72.2 (actually 72)
Nov. 384.6 0
X31X32X33X34X35 1,000 barley in acres
(2,200 tons) Acreage availability:
X11X21X312,000 acres in SE parcel
X12X22X322,300 acres in N parcel
X13X23X33600 acres in NW parcel
X14X24X341,100 acres in W parcel
X15X25X35500 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 3121–2acres of wheat in W parcel
X15 1371–2acres 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 materi-als available (three alloys, two irons, three carbides) that can be se-lected 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 X1through X8represent pounds of alloy 1 through pounds
of carbide 3 to be used in the blend.
Minimize cost 0.12X10.13X20.15X30.09X4
0.07X50.10X60.12X70.09X8
subject to
manganese quality:
햲 0.70X10.55X20.12X30.01X40.05X542
(2.1% of 2,000)
햳 0.70X10.55X20.12X30.01X40.05X546
(2.3% of 2,000) silicon quality:
햴 0.15X10.30X20.26X30.10X40.025X5
0.24X60.25X70.23X886 (4.3% of 2,000) 햵 0.15X10.30X20.26X30.10X40.025X5
0.24X60.25X70.23X892 (4.6% of 2,000)
carbon quality:
햶 0.03X10.01X20.03X40.18X60.20X7
0.25X8101 (5.05% of 2,000)
햷 0.03X10.01X20.03X40.18X60.20X7
0.25X8107 (5.35% of 2,000)
$40(1.5 tons)X2,j ($500 2 2 3
1 5
1 5
)( . tons)X,j j
j
∑
∑
$ (2 50 bushels)X1,j
1 5
j
∑
Availability by weight:
햸 X2300
햹 X650
햺 X7200 햻 X8100
One-ton weight:
햽 X1X2X3X4X5X6X7X82,000
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. From a final simplex tableau, we find that constraints 5 and 11 still have artifi-cial variables in the final solution. 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 (alloy 2), X650 lb (carbide 1), X7$83.6
(car-bide 2), and X8100 lb (carbide 3). Cost $34.91.
Each student may take a different approach and other recom-mendations may result.
8-18. X1number of medical patients
X2number of surgical patients
Maximize revenue $2,280X1$1,515X2
subject to
8X12.5X2 32,850 (patient-days available
365 days 90 new beds) 3.1X12.6X2 15,000 (lab tests)
1X12.2X2 7,000 (x-rays)
X2 2,800 (operations/surgeries)
X1, X2 0
Problem 8-18 solved by computer: X12,791 medical patients
X22,105 surgical patients
revenue$9,551,659 per year
To convert X1and X2to 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
total32,853 days
This represents 68% medical days and 32% surgical days, which yields 61 medical beds and 29 surgical beds. (Note that an alterna-tive approach would be to formulate with X1, X2as number of beds.)
See the printout on the next page for the solution and sensi-tivity analysis.
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:
Printout for Problems 8-18 and 8-19
(b) Referring to the QM for Windows printout, there are no empty beds.
(c) There are 876 lab tests of unused capacity.
(d) The x-ray is used to its maximum and has a $65.45 shadow price.
(e) The operating room still has 695 operations available. 8-20. 8-20. Let
Si1 if Smith is assigned to Job ifor i1, 2, 3, 4 0 otherwise
Ji1 if Jones is assigned to Job ifor i1, 2, 3, 4 0 otherwise
Di1 if Davis is assigned to Job ifor i1, 2, 3, 4 0 otherwise
Ni1 if Nguyen is assigned to Job ifor i1, 2, 3, 4 0 otherwise
Minimize days 4S110S28S39S45J114J2
8J310J44D113D29D312D45N1
11N27N311N4 Subject to
S1S2S3S41
J1J2J3J41
D1D2D3D41 N1N2N3N41
S1J1D1N11
S2J2D2N21
S3J3D3N31
S4J4D4N41
All variables 0, 1
There are multiple optimal solutions. All of these require a total of 31 days. One solution is to assign Smith to Job 2, Jones to Job 4, Davis to Job 1, and Nguyen to Job 3.
8-21. a. Let
A1tons of ore from mine A to plant 1
A2tons of ore from mine A to plant 2
B1tons of ore from mine B to plant 1
B2tons of ore from mine B to plant 2
X1tons shipped to Builder’s Home from plant 1
X2tons shipped to Builder’s Home from plant 2
Y1tons shipped to Homeowners’ Headquarters from
plant 1
Y2tons shipped to Homeowners’ Headquarters from
plant 2
Z1tons shipped to Hardware City from plant 1
Z2tons shipped to Hardware City from plant 2
Minimize cost6A18A27B110B213X119X2
17Y122Y220Z121Z2 subject to
A1A2320 supply at A
B1B2450 supply at B
A1B1500 capacity at plant 1
A2B2500 capacity at plant 2
X1X2200 demand at Builder’s Home
Y1Y2240 demand at Homeowners’ Headquarters
Z1Z2330 demand at Hardware City
A1B1X1Y1Z1 units shipped into plant 1 must
equal units shipped out of plant 1
A2B2X2Y2Z2 units shipped into plant 2 must
equal units shipped out of plant 2
All variables0
b. Solving this on the computer, we find the following solution:
A150 tons of ore from mine Ato plant 1
A2270 tons of ore from mine Ato plant 2
B1450 tons of ore from mine Bto plant 1
X1200 tons shipped to Builder’s Home from plant 1
Y1240 tons shipped to Homeowners’ Headquarters
from plant 1
Z160 tons shipped to Hardware City from plant 1
Z2270 tons shipped to Hardware City from plant 2
All other variables equal 0. Minimum total cost$19,160
8-22. a. The formulation is the same as the formulation in prob-lem 8-21 except for a change in the objective function. We add the processing cost in the objective function, and the new objective function is:
Minimize cost28A130A225B128B213X1
19X217Y122Y220Z121Z2
All the constraints are the same as in the previous problem. b. The solution is the same as problem 8-21 except the value of the objective function is $34,300.
8-23. Minimize time 12XA1 11XA2 8XA3 9XA4
6XA56XA66XG112XG27XG37XG45XG58XG6
8XS19XS26XS36XS47XS59XS6
subject to
XA1XA2XA3XA4XA5XA6200
XG1XG2XG3XG4XG5XG6225
XS1 XS2 XS3 XS4 XS5XS6 275
XA1XG1XS1 80
XA2XG2XS2 120
XA3XG3XS3 150
XA4XG4XS4 210
XA5XG5XS5 60
XA6XG6XS6 80
AFA36 maximum amount of fuel board when leaving Atlanta
LFL15 minimum amount of fuel board when leaving Los Angeles
LFL23 maximum amount of fuel board when leaving Los Angeles
HFH9 minimum amount of fuel board when leaving Houston
HFH17 maximum amount of fuel board when leaving Houston
NFN11 minimum amount of fuel board when leaving New Orleans
NFN20 maximum amount of fuel board when leaving New Orleans
FLAFA(120.05(AFA24))
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:
0.95A0.95 FAFL10.8 Similarly,
FHLFL(70.05(LFL15)) becomes 0.95L0.95FLFH6.25
FNHFH(30.05(HFH9)) becomes 0.95H0.95FHFN2.55
FANFN(50.05(NFN11)) becomes 0.95N0.95FNFA4.45
All variables0 b. The optimal solution is
A 18 (1,000 gallons of fuel to purchase in Atlanta) FA6 (1,000 gallons of fuel remaining when plane lands in
Atlanta)
L 3 (1,000 gallons of fuel to purchase in Los Angeles) FL12 (1,000 gallons of fuel remaining when plane lands
in Los Angeles)
H 1 (1,000 gallons of fuel to purchase in Houston) FH8 (1,000 gallons of fuel remaining when plane lands in
Houston)
N 5 (1,000 gallons of fuel to purchase in New Orleans) FN6 (1,000 gallons of fuel remaining when plane lands in
New Orleans) Total cost112.45 (1,000)
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OLUTIONS TOI
NTERNETH
OMEWORKP
ROBLEMS8-26. To formulate this problem, we first add an activity G to represent the end of the project:
Objective minimize XG subject to: XA 2
XB 3 XC 1 XDXA 4 XFXB 1 8-24. Let
Xiproportion of investment invested in stock ifor i1, 2, . . . , 5
Minimize beta1.2X10.85X20.55X31.40X4
1.25X5
subject to
X1X2X3X4X51 total of the proportions
must add to 1 0.11X10.09X20.065X30.15X40.13X50.11
return should be at least 11% X10.35 no more than 35% in any single stock
X20.35
X30.35
X40.35
X50.35
Xi0 for i1, 2, . . . , 5
b. Solving this on the computer, we have X10
X20.10625
X30.35
X40.35
X50.19375
Minimum beta1.015
Return0.11(0)0.09(0.10625)0.065(0.35) 0.15(0.35)0.13(0.19375)0.11 8-25. Let
A1,000 gallons of fuel to purchase in Atlanta L1,000 gallons of fuel to purchase in Los Angeles H1,000 gallons of fuel to purchase in Houston N1,000 gallons of fuel to purchase in New Orleans FAfuel remaining when plane lands in Atlanta FLfuel remaining when plane lands in Los Angeles FHfuel remaining when plane lands in Houston FNfuel remaining when plane lands in New Orleans Minimize cost4.15A4.25L4.10H4.18N subject to
AFA24 minimum amount of fuel board when leaving Atlanta
Source Destination Number of (Station) (Wing) Trays
5A 5 60
5A 6 80
5A 3 60
3G 1 80
3G 3 90
3G 4 55
1S 4 155
1S 2 120
Optimal cost 4,825 minutes. Multiple optimal solutions exist.
XEXC5 XEXD 5 XGXE 0 XGXF 0 All variables 0 Solution with QM for Windows:
XA 2 XB 10 XC 6 XD 6 XE 11 XF 11 XG 11 Z 11
8-27. Let X1number of Chaunceys mixed
X2number of Sweet Italians mixed
X3number of bourbon on the rocks mixed
X4number of Russian martinis mixed
Maximize total drinks X1X2X3X4
subject to
1X14X352 oz (bourbon limit)
1X11X238 oz (brandy limit)
1X122–3X464 oz (vodka limit)
1X211–3X424 oz (dry vermouth limit)
1X12X236 oz (sweet vermouth limit)
All variables 0
Because a Chauncey (X1) is 1–4sweet vermouth, it requires 1 oz of
that resource (each drink totals 4 oz). Problem 8-27 solved by computer:
Mix 25.99 (or 26) Chaunceys (X1)
Mix 5.00 (or 5) Sweet Italians (X2)
Mix 6.50 (or 61–
2) bourbon on the rocks (X3)
Mix 14.25 (or 141–
4) Russian martinis (X4)
This is a total of 51.75 drinks (in five iterations).
8-28. Minimize 6X118X1210X137X2111X2211X23
4X315X3212X33
subject to
X11X12X13150
X21X22X23175
X31X32X33275
X11X21X31200
X12X22X32100
X13X23X33300
All variables 0 The solution is:
X1125, X13125, X23175, X31175, X32100
Cost $4,525.
8-29. Let Si1 if Smith is assigned to Job i, 0 otherwise, for i 1, 2, 3, 4
Ji1 if Jones is assigned to Job i, 0 otherwise, for i1, 2, 3, 4 Di1 if Davis is assigned to Job i, 0 otherwise, for i1, 2, 3, 4 Ni1 if Nguyen is assigned to Job i, 0 otherwise,
for i1, 2, 3, 4
Minimize days4S15J14D15N110S214J213D2
11N28S38J39D37N39S410J412D411N4
subject to
S1J1D1N11
S2J2D2N21
S3J3D3N31
S4J4D4N41
S1S2S3S41
J1J2J3J41
D1D2D3D41
N1N2N3N41
All variables0
Solving this with QM for Windows, we have S21, J41, D1
1, and N31. So, Smith does Job 2, Jones does Job 4, Davis does
Job 1, and Nguyen does Job 3. The total time is 31 days.
8-30. Let Xinumber of BR54 produced in month i, for i1, 2, 3.
Yi number of BR49 produced in month i, for i1, 2, 3. IXinumber of BR54 units in inventory at end of month i,
for i0, 1, 2, 3.
IYinumber of BR49 units in inventory at end of month i, for i0, 1, 2, 3.
Minimize cost 80(X1X2X3) 95(Y1Y2Y3)
0.8(IX1IX2IX3) 0.95(IY1IY2IY3)
Subject to:
IX050 initial inventory of BR54
IY050 initial inventory of BR49
IX3100 ending inventory of BR54
IY3 150 ending inventory of BR49
X1Y11,100 maximum production level in August
X2Y21,100 maximum production level in September
X3Y31,100 maximum production level in October
X1IX0320 IX1 BR54 requirements for August
X2IX1740 IX2 BR54 requirements for September
X3IX2500 IX3 BR54 requirements for October
Y1IY0450 IY1 BR49 requirements for August
Y2IY1420 IY2 BR49 requirements for September
Y3IY2480 IY3 BR49 requirements for October
All variables 0
A computer solution to this results in IX050, IX1190, IX2
130, IX3100, IY050, IY3150, X1460, X2680, X3
470, Y1400, Y2420, Y3630. All other variables 0. The
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OLUTION TOR
EDB
RANDC
ANNERSC
ASE1. The main issue in this case is how to allocate 3 million pounds of tomatoes. The overall objective is to maximize total sales less variable costs. These costs include production and selling ex-penses. Twenty percent of the crop was grade A and the rest was grade B. In setting up the constraints, the amount of grade A toma-toes cannot exceed 20% of 3 million pounds. Thus not more than 600,000 pounds of grade A tomatoes can be used. Similarly, not more than 2,400,000 pounds of grade B tomatoes can be used. Furthermore, the demand for 50,000 cases of tomato juice and 80,000 cases of tomato paste should be met. The demand for whole tomatoes is not a constraint in this problem. Finally, mini-mum quality requirements should be met. This includes an aver-age of 8 points per pound for whole tomatoes and 6 points per pound for tomato juice. There is no constraint for tomato paste.
Another issue is whether or not to buy 80,000 additional pounds of grade A tomatoes. This would increase the amount of available grade A tomatoes from 600,000 pounds to 680,000 pounds. To answer this question, a new formulation can be made using the new 680,000-pound constraint and a price of 8.5 cents per pound for the 80,000 additional pounds of grade A tomatoes in the objective function. A faster way to resolve this issue is to use postoptimality analysis, or shadow prices. Using this approach, you compare the value of the 80,000 additional tomatoes with the cost, which is 8.5 cents per pound.
2. The problem can be formulated using LP as follows: X1pounds of whole A tomatoes
X2pounds of whole B tomatoes
X3pounds of juice A tomatoes
X4pounds of juice B tomatoes
X5pounds of paste A tomatoes
X6pounds of paste B tomatoes
Maximize: 0.0822X1 0.0822X2 0.066X3 0.066X4
0.074X50.074X6
subject to
1X11X2 14,400,000
1X31X4 1,000,000
1X5 1X62,000,000
1X1 1X3 1X5 600,000
1X2 1X4 1X62,400,000
1X1 3X2 0
3X3 1X4 0
All variables 0
The first constraint refers to the 14 million pounds of whole tomatoes—800,000 cases at 18 pounds per case—that constitutes maximum demand. Similarly, the maximum demand for tomato juice is 50,000 cases at 20 pounds per case or 1 million pounds, and the maximum demand for tomato paste is 80,000 cases at 25 pounds per case or 2 million pounds, and these are constraints 2 and 3. Constraints 4 and 5 reflect the availability of grade A and grade B tomatoes, respectively, and the last two constraints are the quality constraints. The requirements that canned tomatoes must average at least 8 points means that at least three-fourths of the tomatoes must be grade A:
X10.75(X1X2) X13X20
Similarly, the requirements that tomato juice must average at least 6 points means that at least one-fourth of the tomato juice must be grade A, and that is the last constraint.
The coefficients in the objective function are the unit profits. A case of whole tomatoes (grade A and grade B) sells for $4. The vari-able cost (less the tomatoes) is $2.52. Since the tomatoes are already on hand (and no salvage appears to be possible), they represent a sunk cost and are not part of the decision process. Since there are 18 pounds per case, the unit profit is (4.00 2.52)/18 0.0822. Simi-lar analyses hold for the other terms in the objective function.
The solution of the linear programming problem is X1525,000 X2175,000
X375,000 X4225,000
X50 X62,000,000
The maximum profit is $225,340.
All of the grade A tomatoes are used. The shadow price for the slack variable in constraint 4 is 0.0903. Each additional pound of grade A tomatoes costing 8.5 cents will increase profits by 0.093 0.0850 0.0053. A sensitivity analysis indicates that up to an additional 600,000 pounds of grade A tomatoes could be purchased without affecting the solution basis.
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OLUTION TOC
HASEM
ANHATTANB
ANKC
ASEThis very advanced and challenging scheduling problem can be solved most expeditiously using linear programming, preferably integer programming. Let Fdenote the number of full-time em-ployees. 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. and 7 P.M. There will be seven sets of part-time employees; Pjwill be the number of part-time employees who begin their workday at hour j, j1, 2, . . . , 7, with P1 being the number of workers beginning at 9 A.M., P2 at 10 A.M., . . . , P7 at 3 P.M. Note that because part-time employees must work a minimum of 4 hours, none can start after 3P.M. since the entire operation ends at 7 P.M. Similarly, some number of part-time employees, Q j,leave at the end of hour j, j 4, 5, . . . , 9.
The workforce requirements for the first two hours, 9 A.M. and 10 A.M., are:
FP1 14
FP1 P2 25
At 11 A.M. half of the full-time employees go to lunch; the remain-ing half go at noon. For those hours:
0.5FP1 P2 P3 26 0.5FP1 P2 P3 P4 38
Starting at 1 P.M., some of the part-time employees begin to leave. For the remainder of the straight-time day:
FP1 P2 P3 P4 P5 Q4 55 FP1 P2 P3 P4
FP1 P2P5 P6 Q4 Q5 60 FP1 P2 P3 P4 P5
For the two overtime hours:
F1 P1 P2 P3 P4 P5 P6
F1 P1 P2 P7 Q4 Q5 Q6 Q7 Q8 14 F2 P1 P2 P3 P4 P5 P6 P7
F1 P1 P2 Q4 Q5 Q6 Q7 Q8 Q9 9 If the left-hand sides of these 10 constraints are added, one finds that 7Fhours of full-time labor are used in straight time (although 8F are paid for), F1 F2 full-time labor hours are used and paid for at overtime rates, and the total number of part-time hours is
10P1 9P2 8P3 7P4 6P5 5P6 4P7 6Q4 5Q5 4Q6 3Q7 2Q8 Q9 128.4 which is 40% of the day’s total requirement of 321 person-hours.
This also leads to the objective function. The total daily labor cost which must be minimized is
Z8(10.11)F8.08(F1 F2) 7.82(10P1 9P2 8P3 7P4 6P5 5P6 4P7 6Q4 5Q5 4Q6 3Q7 2Q8 Q9)
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 num-ber of full-time employees:
F1 F2 F
Since part-time employees must work at least 4 hours per day, Q4 P1
for those leaving at the end of the fourth hour. At the end of the fifth hour, those leaving must be drawn from the P1 Q4 remain-ing plus the P2 that arrived at the start of the second hour:
Q5 P1 P2 Q4
Similarly, for the remainder of the day, Q6 P1 P2 P3 Q4 Q5
Q7 P1 P2 P3 P4 Q4 Q5 Q6
Q8 P1 P2 P3 P4 P5 Q4 Q5 Q6 Q7
Q9 P1 P2 P3 P4 P5 P6 Q4 Q5 Q6 Q7 Q8
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 Q6Q7 Q8 P1 P2
Q4 Q5 QQ7 Q8 Q9 P1 P2 P3
Finally, to ensure that all part-time employees leave at some time: P1 P2 P3 P4 P5 P6 P7
Q4 Q5 Q6 Q7 Q8 Q9
The resulting problem has 16 integer variables and 22 con-straints. 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 ad-justed—if Fis an odd integer, 0.5Fwill not be an integer and the re-quirement 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 take lunch at 11 A.M. and the other 15 would take it
at noon. Eleven part-timers would begin at 11 A.M., with 9 of them leaving at 3 P.M. and the other 2 at 4 P.M. Fifteen part-time
em-ployees would work from noon until 4 P.M., and 5 would work from 2 P.M. until 6 P.M. The resulting cost 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 than the others. In addition, the person-hours required for each hour of the day are assumed to be deterministic. In a real situation, wide fluctuations will be experienced in a stochastic manner.
