Chapter 3
Problem Summary
Prob. # / Concepts Covered / Level of Difficulty / Notes3.28 / Integer crew scheduling model / 5
3.29 / Minimization sector assignment model / 7
3.30 / Fixed charge model / 5
3.31 / Maximization integer model, effects of rounding, k out of n constraints / 6 / Change Tolerance in Options Dialogue box to .5%; Excel may incorrectly print that part c is infeasible, but it gives an optimal solution.
3.32 / Integer Maximization model / 5 / Change Tolerance in Options Dialogue box to .5% and check Use Automatic Scaling
3.33 / Binary model with constraints requiring binary variables / 7
3.34 / Binary model with constraints requiring binary variables / 6
3.35 / Minimization integer model / 4
3.36 / Maximization/Minimization integer advertising model, fixed charge / 5
3.37 / Mixed integer model with binary variables / 4
3.38 / Mixed integer financial model / 3
3.39 / Maximization integer model / 4
3.40 / Binary model, k out of n constraints / 5
3.41 / Maximization problem, calculation of net profit, evaluation of purchasing additional resources / 5
3.42 / Maximization production model, infeasibility, sensitivity analyses, addition of constraints / 5
3.43 / Maximization financial model / 4
3.44 / Large workforce integer model / 9
3.45 / Integer model / 2
3.46 / Scheduling model, redundant constraints, alternate optimal solutions, shadow prices, adding constraints / 7
3.47 / Fixed charge model with additional constraints requiring binary variables / 7 / Change Tolerance in Options Dialogue box to .5% and check Use Automatic Scaling
3.48 / Binary model / 2
3.49 / Data envelopment analysis model / 6
3.50 / Data Envelopment Analysis model / 5
3.28See file Ch3.28.xls
Xj = the number of workers that have shift j
MIN15X1 + 25X2 + 52X3 + 22X4 + 54X5 + 24X6 + 55X7 + 23X8 + 16X9
S.T. X1 + X2 + X3 8
X2 + X3 10
X3 + X4 + X5 22
X3 + X4 + X5 15
X5 + X6 + X7 10
X5 + X6 + X7 20
X7 + X8 16
X7 + X8 + X9 8
X3 2
X7 2
.6X3 - .4X4 + .6X5 0
.6X5 - .4X6 + .6X7 0
All X’s 0, and integer
Shift 2--7, Shift 3--3, Shift 4--13, Shift 5--6, Shift 6--12, Shift 7--2, Shift 8--14
Total Cost = $1,661.
3.29See file Ch3.29.xls
Xj = the number of patrols in sector j
For each sector, there must be at least one patrol in that sector or one in an adjacent sector. Thus there are 15 constraints -- one for each sector.
MINX1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10 +X11 + X12 + X13 + X14 + X15
S.T.X1 + X2 + X9 + X10 + X11 1(Sector 1)
X1 + X2 + X3 + X9 1(Sector 2)
X2 + X3 + X4 + X8 + X9 1(Sector 3)
X3 + X4 + X5 + X6 + X8 1(Sector 4)
X4 + X5 + X6 + X7 1(Sector 5)
X4 + X5 + X6 + X7 + X8 1(Sector 6)
X5 + X6 + X7 + X8 + X13 + X14 + X15 1(Sector 7)
X3 + X4 + X6 + X7 + X8 + X9 + X13 1(Sector 8)
X1 + X2 + X3 + X8 + X9 + X10 + X13 1(Sector 9)
X1 + X9 + X10 + X11 + X12 + X13 1(Sector 10)
X1 + X10 + X11 + X12 1(Sector 11)
X10 + X11 + X12 + X13 + X14 1(Sector 12)
X7 + X8 + X9 + X10 + X12 + X13 + X14 1(Sector 13)
X7 + X12 + X13 + X14 + X15 1(Sector 14)
X7 + X14 + X15 1(Sector 15)
All X's binary
3 units -- Place a squad car (patrol unit) in sectors 3, 7, and 10.
3.30See file Ch3.30.xls
X1 = Number of cars produced in Michigan
X2 = Number of cars produced in Tennessee
X3 = Number of cars produced in Texas
X4 = Number of cars produced in California
X5 = Number of vans produced in Michigan
X6 = Number of vans produced in Tennessee
X7 = Number of vans produced in Texas
X8 = Number of vans produced in California
X9 = Number of buses produced in Michigan
X10 = Number of buses produced in Tennessee
X11 = Number of buses produced in Texas
X12 = Number of buses produced in California
Y1 = Number of Michigan plants producing vehicles
Y2 = Number of Tennessee plants producing vehicles
Y3 = Number of Texas plants producing vehicles
Y4 = Number of California plants producing vehicles
Note: In this formulation, since only 60 total vehicles need be produced, we use 100 as a large enough number so that if a plant is operational, there would not be a restriction on the number of vehicles produced at the plant.
MIN 15X1 + 15X2 + 10X3 + 14X4 + 20X5 + 28X6 + 24X7 + 15X8 + 40X9 + 29X10 + 50X11 + 25X12 + 150Y1 + 170Y2 + 125Y3 + 500Y4
S.T.X1 + X2 + X3 + X4 = 30(Cars)
X5 + X6 + X7 + X8 = 20(Vans)
X9 + X10 + X11 + X12 = 10(Buses)
X1 + X5 + X9 - 100Y1 0(Michigan)
X2 + X6 + X10- 100Y2 0(Tennessee)
X3 + X7 + X11- 100Y3 0(Texas)
X4 + X8 + X12 - 100Y4 0(California)
All X's 0 and integer
All Y's binary
Build 30 cars in Texas, 20 vans in Texas and 10 buses in Tennessee; total cost $1,365,000.
3.31See file Ch3.31.xls NOTE: Change Tolerance in Options Dialogue box to .5%.
X1 = the number of Tropic homes built
X2 = the number of Sea Breeze homes built
X3 = the number of Orleans homes built
X4 = the number of Grand Key homes built
MAX 40,000X1 + 50,000X2 + 60,000X3 + 80,000X4
S.T. .20X1 + .27X2 + .22X3 + .35X4 20 (Acres)
X1 + X2 40(One story)
X2 + X3 + X4 50(3+ BR)
X1 10 (Min Trop.)
X2 10 (Min SeaBr.)
X3 10(Min Orleans)
X4 10(Min Gr. Key)
All X's 0 and integer
a.Build 29 Tropic, 11 Sea Breeze, 35 Orleans, 10 Grand Key models; profit = $4,610,000
b.(See worksheet Atlantic Standard - Linear (not shown))
30 Tropic, 10 Sea Breeze, 35 Orleans, 10 Grand Key models; profit = $4,600,000. This solution satisfies all the constraints but is $10,000 less than the optimal solution.
c.Let Yi = 1 if the constraint holds and Yi = 0 if it does not
Add the following constraints:
X1 - 12Y1 0
X2 - 12Y2 0
X3 - 12Y3 0
X4 - 12Y4 0
Y1 + Y2 + Y3 + Y4 3
NOTE: Excel may incorrectly print that the problem is infeasible. But the solution below is feasible and optimal.
Build 30 Tropic, 10 Sea Breeze, 32 Orleans, 12 Grand Key models; profit = $4,580,000.
Note: There are alternate optimal solutions.
3.32 See file Ch3.32.xls In Options dialogue box, Change tolerance to .5% and check Use Automatic Scaling
X1 = the number of Nissan vans Logitech should purchase
X2 = the number of Toyota vans Logitech should purchase
X3 = the number of Plymouth vans Logitech should purchase
X4 = the number of Ford stretch vans Logitech should purchase
X5 = the number of Mitsubishi minibuses Logitech should purchase
X6 = the number of General Motors minibuses Logitech should purchase
X7 = the total number of vehicles Logitech should purchaase
MAX 7X1 + 8X2+ 9X3 + 11X4 + 20X5 + 24X6
S.T.26000X1 + 30000X2+ 24000X3 + 32000X4 + 50000X5 + 60000X6 250,000
5000X1 + 3500X2+ 6000X3 + 8000X4 + 7000X5 + 11000X6 50,000
X1 + X2+ X3 + X4 + X5 + X6 - X7 = 0
X7 8
X5 + X6 1
X1 + X2+ X3 + X4 3 X3 + X4 + X6 -.5X7 0
All X’s 0, and integer
a.Maximum Capacity = 97 using 2 Plymouth vans, 1 Ford van, 1 Mitsubishi minibus, 2 General Motors minibuses.
b.See worksheets b-253900, b-254000, b-249900, b-259900, b-260000 (not shown)
Capacity Nissan Toyota Plymouth Ford Mitsubishi General Motors
(i) 97 2 1 1 1
(ii) 98 3 1 3
(iii) 96 4 3
(iv) 100 4 2 1
(v) 100 4 2 1
Sensitivity of the right hand side gives non-smooth jumps to new optimal solutions.
c.The problem is infeasible. The minimum van and mini-bus constraints require a minimum budget of $122,000.
3.33See file Ch3.33.xls
Yj = the number of product line j eliminated
MIN 10Y1 + 8Y2 + 20Y3 + 12Y4 + 25Y5 + 4Y6 + 15Y7 + 5Y8 + 18Y9 + 6Y10 ($1000’s)
S.T. Y1 + Y2 + Y3 + Y4 + Y5 + Y6 + Y7 + Y8 + Y9 + Y10 4
( 4 eliminated)
50(1-Y1) + 60(1-Y2) + .... + 125(1-Y10) 600 (Floor space)
or, 50Y1 + 60Y2 + ... + 125Y10 225 (Floor space)
Y2 - Y3= 0(Compaq line)
Y4 - Y6= 0(P. Bell line)
Y5 - Y8 = 0(Apple line)
(1-Y1) + (1-Y2) + (1-Y3) + (1-Y4) + (1-Y5) 2(Computers)
or, Y1 + Y2 + Y3 + Y4 + Y5 3(Computers)
(1-Y6) + (1-Y7) 1(Monitors)
or, Y6 + Y7 1(Monitors)
(1-Y8) + (1-Y9) + (1-Y10) 1(Printers)
or, Y8 + Y9 + Y10 2 (Printers)
15000(1-Y1) + 12000(1-Y2) + .... + 10000(1-Y10) 75000 (Restock)
or,15000Y1 + 12000Y2 + ... + 10000Y10 93000 (Restock)
Y1 - Y10 0(Tosh/Epson)
All Yi’s are binary
Eliminate the Toshiba Notebook computers, Packard Bell PC’s and monitors, Apple Macs and printers, and Epson printers -- Cost = $62,000 (Note: There are alternate optimal solutions giving $62,000.)
3.34See file Ch3.34.xls
a.Xj = the number of software application j developed (j = 1, 2, 3, 4, 5, 6)
MAX 2X1 + 3.6X2 + 4X3 + 3X4 + 4.4X5 + 6.2X6(in $millions)
S.T. 6X1 + 18 X2 + 20X3 + 16X4 + 28X5 + 34X6 60 (Programmers)
.4X1 + 1.1X2 + .94X3 + .76X4 + 1.26X5 + 1.8X6 3.5 (Budget in $millions)
All Xj’s binary
Korvex should develop applications 1, 2, 3, and 4; net present worth = $12,600,000.
b. Add the following constraints to the formulation in part a.
X4 - X5= 0(Proj 4 = Proj 5)
- X1 + X2 0 (Proj 2 only if Proj 1)
X3 + X6 1(Not both Proj 3 and Proj 6)
X1 + X2 + X3 + X4 + X5 + X6 3(Max 3 applications)
See worksheet Part b (not shown).
Korvex should develop applications 1, 2, and 6; net present worth = $11,800,000.
3.35See file Ch3.35.xls
X1 = the number of CPA’s hired
X2 = the number of experienced accountants hired
X3 = the number of junior accountants hired
The total number of accounts that can be serviced is 6X1 + 6X2 + 4X3. This must be greater than or equal to 100 plus the number of corporate accounts serviced: 6X1 + 6X2 + 4X3 100 + (3X1 + X2); and the number of corporate accounts serviced must be at least 25: 3X1 + X2 25. Also the number of CPA’s and experienced accountants (X1 + X2) must be at least two-thirds of all employees hired (X1 + X2 + X3) or X1 + X2 2/3(X1 + X2 + X3). Rearranging terms gives the following:
MIN1200X1 + 900X2 + 600X3
S.T. 3X1 + 5X2 + 4X3 100(Service 100 personal accounts)
3X1 + X2 25 (Service 25 corporate accounts)
1/3X1 + 1/3X2 - 2/3X3 0( 2/3 are CPA’s or experienced)
All X's 0 and integer
Hire 2 CPA’s and 19 experienced accountants; total payroll = $19,500.
3.36See file Ch3.36.xls
X1 = the number of TV exposures
X2 = the number of radio exposures
X3 = the number of newspaper exposures
a.MAX500,000X1 + 50,000X2 + 200,000X3
S.T. X1 250 (Max TV)
X2 250 (Max radio)
X3 250 (Max newspapers)
4,000X1 + 500X2 + 1,000X3 500,000 (Budget)
All X's 0 and integer
Use 62 TV exposures, 4 radio exposures, and 250 newspaper exposures; total audience reached = 81,200,000.
b.Change the objective function to: MIN 4,000X1 + 500X2 + 1,000X3 and change the third constraint to: 500,000X1 + 50,000X2 + 200,000X3 30,000,000. See worksheet Century -- Min Cost (not shown). Use 150 newspaper exposures only; total cost $150,000.
c.See worksheet Part c (not shown).
MAX500,000X1 + 50,000X2 + 200,000X3
S.T.X1 - 250Y1 0
X2- 250Y2 0
X3- 250Y3 0
4,000X1 + 500X2 + 1,000X3 + 500,000Y1 + 50,000Y2 + 100,000Y3 1,000,000
All X's 0 and integer All Y's binary
Use 38 TV exposures, 0 radio exposures, and 248 newspaper exposures; total audience reached = 68,600,000.
3.37See file Ch3.37.xls
X1 = the number of casings produced in Springfield
X2 = the number of casings produced in Oak Ridge
X3 = the number of casings produced in Westchester
Y1 = the number of Springfield locations used
Y2 = the number of Oak Ridge locations used
Y3 = the number of Westchester locations used
MIN.224X1 + .280X2 + .245X3 + 1200Y1 + 1100Y2 + 1000Y3
S.T. X1 - 65000Y1 0 (Spring.)
X2 - 50000Y2 0 (Oak R.)
X3 - 55000Y3 0 (Westch.)
X1 + X2 + X3= 100,000
All X’s 0, All Y’s binary
Produce 65,000 in Springfield and 35,000 in Westchester. Total cost = $25,335
3.38See file Ch3.38.xls
X1 = Number of shares of TCS purchased
X2 = $ invested in MFI
X3 = Total $ invested
MIN X3
S.T.55X1 + X2 - X3= 0(Total invested)
13X1 +.09X2 250(Minimum expected return)
55X1 - .4X3 0(TCS 40% of total investment)
55X1 750(Max $ in TCS)
X1, X2 0; X1 integer
Buy 112 shares of TCS, invest 1044.44 in MFI for a total investment of $1704.44
3.39See file Ch3.39.xls
X1 = the number of Fords leased
X2 = the number of Chevrolets leased
X3 = the number of Dodges leased
X4 = the number of Macks leased
X5 = the number of Nissans leased
X6 = the number of Toyotas leased
MIN2000X1 + 1000X2 + 5000X3 + 9000X4 + 2000X5
S.T. X1 + X2 + X3 + X4 + X5 + X6 = 5,000 (Total)
X1 + X2 + X3 + X4 3,000 (U.S.)
500X1 + 600X2 + 300X3 + 900X4 + 200X5 + 400X6 2,750,000 (Budget)
X1 + X2 + .75X3 + 5X4 + .5X5 + .75X6 10,000(Payload)
All X’s 0 and integer
Lease 1581 Fords, 8 Chevrolets, 1411 Macks, 576 Nissans, 1424 Toyotas -- capital outlay = $17,021,000.
3.40See file Ch3.40.xls
X1 = 1 if 7 new police officers are hired
X2 = 1 if the police headquarters is modernized
X3 = 1 if two new police cars are bought
X4 = 1 if bonuses are given to foot patrolmen
X5 = 1 if a new fire truck and fire support equipment is purchased
X6 = 1 if an assistant fire chief is hired
X7 = 1 if cuts to the sports program are restored
X8 = 1 if cuts to the music program are restored
X9 = 1 if new computers are purchased for the high school
Y1 = 1 if the goal of spending only $650,000 is not met
Y2 = 1 if fewer than 3 police projects are funded
Y3 = 1 if 7 new police officers are not hired
Y4 = 1 if less than15 new jobs are created
Y5 = 1 if fewer than 3 education projects are funded
MAX 4176X1 + 1774X2 + 2513X3 + 1928X4 + 3607X5 +
962X6 + 2829X7 + 1708X8 + 3003X9
S.T.400X1 +350X2 + 50X3 + 100X4 + 500X5 + 90X6 + 220X7 + 150X8 + 140X9 900
7X1 + X3 + 2X5 + X6 + 8X7 + 3X8 + 2X9 10
X1 + X2 + X3 + X4 3
X3 + X5 = 1
X7 - X8 = 0
X7 - X9 0
X8 - X9 0
400X1 +350X2 + 50X3 + 100X4 + 500X5 + 90X6 + 220X7 + 150X8 + 140X9 - MY1 650
X1 + X2 + X3 + X4 + X5 + X6 -MY2 3
X1 +MY3 1
X1 -MY3 1
7X1 + X3 + 2X5 + X6 + 8X7 + 3X8 + 2X9 +MY4 15
X7 + X8 X9 +MY5 3
Y1 + Y2 + Y3 + Y4 + Y5 2
All X's and Y's binary
Fund the following projects: (Total Points = 12,943)
2 police cars
Bonuses for foot patrolmen
Hire an assistant fire chief
Restore sports funding
Restore music funding
Purchase new computers for the high school
3.41See file Ch3.41.xls
X1 = the number of Turkey De-Lite sandwiches made daily
X2 = the number of Beef Boy sandwiches made daily
X3 = the number of Hungry Ham sandwiches made daily
X4 = the number of Club sandwiches made daily
X5 = the number of All Meat sandwiches made daily
MAX2.75X1 + 3.5X2 + 3.25X3 + 4X4 + 4.25X5
S.T. 4X1 + 2X4 + 3X5 384 (Turkey)
4X2 + 2X4 + 3X5 576 (Beef)
4X3 + 2X4 + 3X5 480 (Ham)
X1 + X2 + 2X3 + 2X4 384 (Cheese)
X1 + X2 + X3 + X4 + X5 300 (Rolls)
All X's 0
a.Make 52 Turkey De-Lites, 100 Beef Boys, 76 Hungry Hams, 40 Clubs, 32 All Meat
Total Daily Revenue= $1,036
- Daily Supplies = $ 700
Net Daily Profit = $ 336 Net Annual Profit = 200($336) = $67,200
b.Shadow price for cheese =$0.3333; range of feasibility = 336 - 480
As long as the price of cheese is within its range of feasibility, its shadow price will not change.
c. Turkey: Additional 8 lbs. = 128 oz. is within its range of feasibility, so this will add:
128(.2292) = $29.33 to revenue - $20 cost = $9.33 net additional profit
Beef: Additional 12 lbs. = 192 oz. is within its range of feasibility, so this will add:
192(.4167) = $80 to revenue - $42 cost = $38 net additional profit
Ham: Additional 10 lbs. = 160 oz. is within its range of feasibility, so this will add:
160(.2708) = $43.33 to revenue - $30 cost = $13.33 net additional profit
Cheese: Additional 8 lbs. = 128 oz. is NOT within its range of feasibility. The problem must be re-solved.
For 8 additional pounds (128 oz.) of cheese, the new optimal revenue = $1068. (See worksheet 128 Extra Ounces of Cheese (not shown).) This is an increase of $1068- $1036 = $32 in additional revenue - $18 cost = $14 net additional profit.
Buy the beef.
3.42See file Ch3.42.xls
X1 = 100’s of men’s jackets produced in the week
X2 = 100’s of women’s jackets produced in the week
X3 = 100’s of men’s pants produced in the week
X4 = 100’s of women’s pants produced in the week
MAX2000X1 + 2800X2 + 1200X3 + 1500X4
S.T. 150X1 + 125X2 + 200X3 + 150X4 2500 (Denim)
3X1 + 4X2 + 2X3 + 2X4 36 (Cutting)
4X1 + 3X2 + 2X3 + 2.5X4 36 (Stitching)
.75X1 + .75X2 + .50X3 + .50X4 8 (Boxing)
All X's 0
Produce 450 women’s jackets, 900 women’s pants; weekly profit = $26,100
b. Add to the formulation X1 5, X2 5, X3 5, X4 5; the problem is now infeasible (See worksheet 500 (not shown).) 500 of each requires a minimum of 55 cutting hours and 57.5 stitching hours which exceeds the limit of 36 hours each.
c.Add to the formulation X1 3, X2 3, X3 3, X4 3. (See worksheet 300 (not shown).) Produce 300 men’s jackets, 350 women’s jackets, 300 men’s pants, and 300 women’s pants; weekly profit = $23,900.
d. Currently all items produced are women’s items. Thus adding a constraint requiring that at least 50% of the items produced be women’s items would be a nonbinding constraint and the solution would not change.
To add the constraint that at least 50% of the items produced be men’s items:
Define: X5 = total number of outfits produced weekly.
Add:
X1 + X2 + X3 + X4 - X5 = 0
X1 + X3 - .5X5 0.
Produce 514 men’s jackets and 514 women’s jackets (rounded) for a profit of $24,672
(see worksheet 50% Mens (not shown)).
3.43See file Ch3.43.xls
X1 = Amount invested in first trust deeds
X2 = Amount invested in second trust deeds
X3 = Amount invested in automobile loans
X4 = Amount invested in business loans
X5 = Amount invested in securities
X6 = Total amount invested in loans
MAX .09X1 + .105X2 + .1225X3 + .1175X4 + .0675X5
S.T. X5 3,333,333.33 (Max sec.)
X1 + X2 - X5 0(Trust sec)
X1 + X2 + X3 + X4 - X6 = 0 (Total Loans)
X4 -.49X6 0 (Max bus. loan)
-.50X1 - .50X2 + X3 0 (Max auto loan)
X1 + X2 + X3 + X4 + X5= 10,000,000 (Total)
All X's 0
Invest $2,537,313.43 in second trust deeds, $1,268,656.72 in auto loans, $3,656,716.42 in business loans, $2,537,313.43 in securities. Total return = $1,022,761.19.
3.44See file Ch3.44.xls
X1 = the number of operations managers kept
X2 = the number of department managers kept
X3 = the number of section heads kept
X4 = the number of engineers kept
X5 = the number of technicians kept
X6 = the number of business support personnel kept
X7 = the number of secretaries kept
X8 = the total number of workers kept
X9 = total overhead
X10 = total direct costs
MIN 1600X1 + 1200X2 + 1000X3 + 800X4 + 600X5 + 500X6 + 350X7
S.T.X1 + X2 + X3 + X4 + X5 + X6 + X7 - X8 = 0 (X8 definition of total workers)
X1 + X2 120 (Managers)
X3 = 0(Section heads)
X1 - .2X2 0(Operations/Department managers)
-20X1 - 20X2 + X4 + X5 (Technician/Management)
X7 - .05X8 0 (Min Clerical)
X7 - .1X8 0 (Max Clerical)
X6 - .01X8 0 (Min Administration)
X6 - .02X8 0 (Max Administration)
1280X1 + 840X2 + 350X6 + 245X7 -X9 = 0 (X9 definition of total overhead)
320X1 + 360X2 + 800X3 + 800X4 + 600X5 + 150X6 + 105X7 - X10 = 0
(X10 definition of total direct costs)
X9 - .05X10 0 (Min overhead)
X9 - .1X10 0 (Max overhead)
X10 = 4800000 (Fixed direct costs)
X1 6 (Min operations managers)
360X2 -80X3 - 60X5 0 (Department managers/ Technicians)
X4 - 4X5 0 (Engineers/Technicians)
X1 + X2 + X3 + X4 - .5104X8 0 (Min grade 100 personnel)
X1 + X2 + X3 + X4 - .7656X8 0 (Max grade 100 personnel)
X5 - .2145X8 0 (Min technicians)
X5 - .3217X8 0 (Max technicians)
X6 - .0107X8 0 (Min support)
X6 - .0161X8 0 (Max support)
X7 - .0643X8 0 (Min secretaries)
X7 - .0965X8 0 (Max secretaries)
X1 40(Current operations managers)
X2 200(Current department managers)
X3 900(Current section heads)
X4 6000(Current engineers)
X5 3000(Current technicians)
X6 150(Current business support personnel)
X7 900(Current secretaries)
All X’s 0, X1, X2, X3, X4, X5, X6 X7 integer
Keep 9 operations managers, 109 department managers, 0 section heads, 4764 engineers, 1480 technicians, 79 business support staff, and 446 secretaries;
total weekly salary = $5,040,000.
3.45See file Ch3.45.xls
X1 = the number of assistant professor positions recruited
X2 = the number of associate professor positions recruited
X3 = the number of full professor positions recruited
X4 = the total number of positions recruited
MAX2X1 + 7X2 + 14X3
S.T. X1 + X2 + X3 - X4 = 0(Definition of X4)
X4 20(Max recruiting)
X1 - .5X4 0(Min. assistant prof.)
X1 + X2 - .7X4 0(Min. below full prof.)
55003X1 + 69885X2 + 93471X3 1,275,000(Total salaries)
All X's 0 and integer
Hire 9 assistant professors, 4 associate professors, and 5 full professors
-- 116 total years of experience
3.46See file Ch3.46.xls
X1 = the number of guards whose shift begins at 12:00 midnight
X2 = the number of guards whose shift begins at 3:00AM
X3 = the number of guards whose shift begins at 6:00AM
X4 = the number of guards whose shift begins at 9:00AM
X5 = the number of guards whose shift begins at 12:00 noon
X6 = the number of guards whose shift begins at 3:00PM
X7 = the number of guards whose shift begins at 6:00PM
X8 = the number of guards whose shift begins at 9:00PM
MINX1 + X2 + X3 + X4 + X5 + X6 + X7 + X8
S.T.X1 + X7 + X8 5 (mid - 1AM) Redundant
X1 + X7 + X8 5 (1AM - 2AM)Redundant
X1 + X8 5 (2AM - 3AM)
X1 + X2 + X8 5 (3AM - 4AM)Redundant
X1 + X2 + X8 5 (4AM - 5AM)Redundant
X1 + X2 8 (5AM - 6AM)
X1 + X2 + X3 8 (6AM - 7AM)Redundant
X1 + X2 + X3 12 (7AM - 8AM)Redundant
X2 + X3 12 (8AM - 9AM)
X2 + X3 + X4 10 (9AM-10AM)Redundant
X2 + X3 + X4 10 (10AM-11AM)Redundant
X3 + X4 15 (11AM - noon)
X3 + X4 + X5 15 (noon - 1PM)Redundant
X3 + X4 + X5 15 (1PM - 2PM)Redundant
X4 + X5 9 (2PM - 3PM)
X4 + X5 + X6 9 (3PM - 4PM)Redundant
X4 + X5 + X6 12 (4PM - 5PM)Redundant
X5 + X6 12 (5PM - 6PM)
X5 + X6 + X7 12 (6PM - 7PM)Redundant
X5 + X6 + X7 7 (7PM - 8PM)Redundant
X6 + X7 7 (8PM - 9PM)
X6 + X7 +X8 7 (9PM -10PM)Redundant
X6 + X7 +X8 7 (10PM -11PM)Redundant
X7 +X8 7 (11PM - mid)
All X’s 0 and integer
- There are many solutions that give 42 total guards. This can be seen by the numerous adjustable cells with Allowable Increases or Decreases of 0. The screen on the next page shows one of them.
b.Shadow prices give an increase in the number of officers needed per time period given no additional changes in the constraints.
(i) Changing the number of guards required from midnight to 5AM to 7 changes the right hand side of the first non-redundant constraint from 5 to 7. But there was 3 slack and thus this does not require more guards.
(ii)Changing the number of officers required from 9AM to 11AM to 12 changes nothing; X3 + X4 + X5 12 would still be a redundant constraint.
(iii)Changing the number of officers required from 11AM to 2 PM changes the right hand side of the fourth non-redundant constraint to 17. Re-solving (see worksheet 11AM-2PM (not shown)) changes the total guards required to 44.
(c)If the objective function coefficients had been expressed in terms of dollars, each dollar coefficient would have been the same, say $K. From the range of optimality for the objective function coefficients for these two shifts, we see that their ranges of optimality would for at least one of them in each solution would be from $K to $K, i.e. any change will change the optimal solution. Thus a change of $5 will change the optimal solution.
(d)We do not need to do this since integer values are already obtained. If we do, we get the same answer. (See worksheet Guardsman Services-Integer (not shown).) The shadow prices and ranges of optimality do make sense because they are integers.
3.47See file Ch3.47.xls Change tolerance to .5% in Options dialogue box.
X1 = the number of Liltrykes produced per year
X2 = the number of Pinktrykes produced per year
X3 = the number of Herotrykes produced per year
X4 = the number of Robinhoods produced per year
X5 = the number of Jeeptrykes produced per year
X6 = the number of Monsters produced per year
Y1 = the number of setups of Liltrykes each year
Y2 = the number of setups of Pinktrykes each year
Y3 = the number of setups of Herotrykes each year
Y4 = the number of setups of Robinhoods each year
Y5 = the number of setups of Jeeptrykes each year
Y6 = the number of setups of Monsters each year
a.
MAX 1.50X1 + 2.00X2 + 2.25X3 + 2.75X4 + 3.00X5 + 3.50X6
-16,500Y1 - 18,000Y2 - 17,500Y3 - 18,000Y4 - 20,000Y5 - 17,000Y6
S.T.
3X1 + X2 + 2X3 + 2X4 + 2X5 120,000 (Small wheels/year)
2X2 + X3 + X4 + X5 + 3X6 96,000 (Big wheels/year)
.8X1+1.2X2 +1.5X3 +2.1X4 +1.8X5 + 3X6 108,000 (Plastic/year)
X1 -1000000Y10
X2 -1000000Y20
X3 -1000000Y30
X4 -1000000Y40
X5- 1000000Y50
X6 -1000000Y60
All X’s 0, All Y’s binary
a.Produce 60,000 Jeeptrykes only; yearly profit = $160,000
b.In addition to the variables already defined, define Zj = 0 if goal j is met, Zj = 1 if it is not.
Add the following.
Goal (1):Max $70,000 for new setups
Goal (2):If Herotryke is produced, Robinhood will not be produced
Goal (3):At least four new models produced
Goal (4):If Jeeptrykes are produced, Herotrykes will also be produced
Goal (5):At least 1500 lbs. per month of plastic should be left over (at most 7500 pounds per month or 90,000 pounds per year used)
16500Y1 + 18000Y2 + 17500Y3 + 18000Y4 + 20000Y5 + 17000Y6 -1000000Z1 75000 (1)
Y3 - Y4 -1000000Z2 1 (2)
Y1 + Y2 + Y3 + Y4 + Y5 + Y6-1000000Z3 4 (3)
Y5 - Y6-1000000Z4 0 (4)
.8X1 + 1.2X2 + 1.5X3 + 2.1X4 + 1.8X5 + 3X6-1000000Z5 90000 (5)
and, at least 4 out of 5 goals should be satisfied: (1-Z1)+(1- Z2)+(1- Z3)+(1-Z4)+(1-Z5) 4 or,
Z1 + Z2 + Z3 + Z4 + Z5 1
Z’s are binary
See worksheet Little Trykes -- Part b (not shown). The solution remains the same. All goals except goal 3 are satisfied.
3.48See file Ch3.48.xls
X1 = the number of Alpha car projects implemented
X2 = the number of Beta car projects implemented
X3 = the number of Delta car projects implemented
X4 = the number of Gamma car projects implemented
X5 = the number of Kappa car projects implemented
X6 = the number of Sigma car projects implemented
MAX 12X1 + 11X2 + 9X3 + 15X4 + 7X5 + 20X6
S.T. X1 + X2 + X3 + X4 + X5 + X6 3 Proj.
15000X1 + 18000X2 + 19000X3 + 20000X4 + 8000X5 + 22000X6 125000 eng-yr1
5000X1 + 5000X2 + 3000X3 + 4000X4 + 2000X5 + 5000X6 16000 staff-yr1
40000X1 + 25000X2 + 19000X3 + 25000X4 + 12000X5 + 27000X6 150000 eng-yr 2
5000X1 + 4000X2 + 3000X3 + 6000X4 + 2000X5 + 7000X6 24000 staff-yr 2
40000X1 + 30000X2 + 19000X3 + 30000X4 + 18000X5 + 32000X6 187500 eng-yr 3
8000X1 + 7000X2 + 3000X3 + 7500X4 + 3000X5 + 8000X6 40000 staff-yr 3
X1 + X2 1 (if alpha, no beta)
X1- X6 0 (if sigma, alpha)
All X's binary
Develop Alpha, Gamma, Kappa, Sigma; total present net worth = $54 million.
3.49See file Ch3.49.xlsData Envelopment Analysis
X1 = Relative input weight applied to the campus SAT score
X2 = Relative input weight applied to the campus faculty/student ratio
X3 = Relative input weight applied to the campus budget
Y1 = Relative output weight applied to the campus average GPA score