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Chapter 3 Problem Summary Prob. # Concepts Covered Level of Notes Difficulty 3.1 Maximization production model 2 3.2 Maximization model with "", "=", "" 5 constraints 3.3 Maximization production model, percentage 4 constraints, interpretation of shadow prices and range of feasibility 3.4 Maximization financial mix model, percentage 5 constraint, determining current rate of return and future rate of return, shadow prices 3.5 Maximization model, calculation of profit 4 coefficients, sensitivity analyses 3.6 Minimization model, determining correct RHS 4 3.7 Maximization model, sensitivity analyses, 6 evaluation of quantities outside the ranges of optimality and feasibility, many constraints 3.8 Maximization production model, analysis of why 6 a high profit item is not produced, sensitivity analyses, evaluation of purchasing additional resources 3.9 Minimization model with 8 variables and 14 6 functional constraints 3.10 Minimization diet problem, alternate optimal 6 solutions, interpretation of shadow prices, the effects of deleting a constraint 3.11 Minimization model 3 3.12 Maximization model, re-evaluation of RHS 4 3.13 Financial maximization model, weighted 6 averages, sensitivity analyses 3.14 Multiperiod production model 9 3.15 Maximization agriculture model, sensitivity 7 analyses, evaluation of an alternative based on shadow prices 3.16 Supply chain model 7 3.17 Supply Chain model 9 3.18 Maximization finance model 4 3.19 Blending problem, alternate optimal solutions, 6 shadow prices, determining the acceptability of the solution 3.20 Maximization model with 9 functional 4 constraints 3.21 Minimization transportation model 5 3.22 Classic trim loss model, definition of variables 5 3.23 Maximization advertising model, checking the 5 linear programming assumptions, elimination of Chapter 3- 1 a nonbinding constraint 3.24 Maximization retailing model, evaluating 5 changes to objective function coefficients, the profitability of added resources, and the addition of a constraint 3.25 Large design model, "tricky" constraints 9 3.26 Maximization production problem, calculation of 7 objective function coefficients, calculating coefficients used in the model 3.27 Maximization advertising model, sensitivity 8 analyses, adding constraints, changing the RHS values 3.28 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, 6 Change Tolerance in Options Dialogue box k out of n constraints 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 7 variables 3.34 Binary model with constraints requiring binary 6 variables 3.35 Minimization integer model 4 3.36 Maximization/Minimization integer advertising 5 model, fixed charge 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, 5 evaluation of purchasing additional resources 3.42 Maximization production model, infeasibility, 5 sensitivity analyses, addition of constraints 3.43 Maximization financial model 4 3.44 Large workforce integer model 9 3.45 Integer model 2 3.46 Scheduling model, redundant constraints, 7 alternate optimal solutions, shadow prices, adding constraints 3.47 Fixed charge model with additional constraints 7 Change Tolerance in Options Dialogue box requiring binary variables 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 Case 3.1 Maximization production model with input in 8 Excel, many percentage constraints, analysis of output Case 3.2 Maximization integer model, definitional 10 Solved as a linear program -- takes just a few variables, percentage constraints, rounding the seconds linear model, "slightly" violated constraints, time Solved as an integer program -- takes many Chapter 3- 2 issues minutes Case 3.3 Binary models 7 Case 3.4 “On the job training” model with many “linking” 9 constraints Case 3.5 Maximization financial mix model, linking 10 results between two periods, percentage constraints, analysis of output Case 3.6 Fixed charge model 4 Chapter 3- 3 Problem Solutions 3.1 See file Ch3.1.xls X1 = Number of 20-inch girls bicycles produced this week X2 = Number of 20-inch boys bicycles produced this week X3 = Number of 26-inch girls bicycles produced this week X4 = Number of 26-inch boys bicycles produced this week MAX 27X1 + 32X2 + 38X3 + 51X4 S.T. X1 + X3 200 (Min girls models) X2 + X4 200 (Min boys models) 12X1 + 12X2 + 9X3 + 9X4 4800 (Production minutes) 6X1 + 9X2 + 12X3 + 18X4 4800 (Assembly minutes) 2X1 + 2X2 500 (20-inch tires) 2X3 + 2X4 800 (26-inch tires) All X's 0 150 20-inch girls, 100 20-inch boys, 100 26-inch girls, 100 26-inch boys; profit = $16,150 Chapter 3- 4 3.2 See file Ch3.2.xls a. X1 = Number of stoves produced weekly X2 = Number of washers produced weekly X3 = Number of electric dryers produced weekly X4 = Number of gas dryers produced weekly X5 = Number of refrigerators produced weekly MAX 110X1 + 90X2 + 75X3 + 80X4 + 130X5 S.T. 5.5X1 + 5.2X2 + 5.0X3 + 5.1X4 + 7.5X5 4800 (Molding/pressing) 4.5X1 1200 (Stove assembly) 4.5X2 + 4.0X3 + 3.0X4 2400 (Washer/dryer assembly) 9.0X5 1200 (Refrigerator assembly) 4.0X1 + 3.0X2 + 2.5X3 + 2.0X4 + 4.0X5 3000 (Packaging) All X's 0 266.6667 stoves, 448.7179 Washers, 133.33333 refrigerators; Profit = $87,051.28 Fractional values are work in progress from one week to the next. Chapter 3- 5 b. Add the following constraints: X2 - X3 - X4 = 0 (Washers = Dryers) X3 - X4 100 (E. Dryers G. Dryers + 100) -X3 + X4 100 (G. Dryers E. Dryers + 100) 266.6667 stoves, 227.1545 washers, 63.57724 electric dryers, 163.5772 gas dryers, 133.3333 refrigerators; profit = $84,965.04 Chapter 3- 6 3.3 See file Ch3.3.xls X1 = the number of standard Z345’s produced weekly X2 = the number of industrial Z345’s produced weekly X3 = the number of standard W250’s produced weekly X4 = the number of industrial W250’s produced weekly X5 = the total number of products produced weekly MAX 400X1 + 560X2 + 560X3 + 700X4 S.T. 25X1 + 46X2 + 16X3 + 34X4 2500 (zinc) 50X1 + 30X2 + 28X3 + 12X4 2800 (iron) X1 + X2 20 (Min Z345’s) X1 + X2 + X3 + X4 - X5 = 0 (X5 definition) X2 + X4 - .50X5 0 (Industrial min.) X1 + X2 - .75X5 0 (Max Z345’s) X3 + X4 - .75X5 0 (Max W250’s) X1, X2, X3, X4, X5 0 Chapter 3- 7 a. Produce 22.93578 standard Z345’s, 22.93578 standard W250’s, 45.87156 industrial W250’s (fractional production quantities are work in progress carried over from one week to the next). Weekly profit = $52,752.29 b. 75% (Slack is 0 on that constraint.) If this restriction is loosened or eliminated, the weekly profit will increase. c. The shadow price of zinc is $21.10091743, which is valid for an additional 492.1568627 pounds. i) 100 pounds is worth $2110.09 > $1500; yes purchase 100 additional pounds. ii) $2110.09 < $2600, no, 100 additional pounds should not be purchased. iii) Cannot tell without resolving since 800 additional pounds is outside the range of feasibility for the shadow price. Re-solving (not shown) with 3300 pounds of zinc gives a profit of $68,055.87. Since this a $15,303.87 increase, then, yes, the 800 additional pounds should be purchased. 3.4 See file Ch3.4.xls X1 = amount invested in EAL stock X2 = amount invested in BRU stock X3 = amount invested in TAT stock X4 = amount invested in long term bonds X5 = amount invested in short term bonds X6 = amount invested in the tax deferred annuity X7 = the total amount invested in stocks only MAX .15X1 + .12 X2 + .09X3 + .11X4 + .085X5 + .06X6 S.T. X1 + X2 + X3 + X4 + X5 + X6 = 50,000 (Total) X6 10,000 (TDA) X1 + X2 + X3 - X7 = 0 (Stocks) X3 -.25X7 0 (Min TAT) X4 + X5 - X7 0(Bond stock) X3 + X5 + X6 12,500 (Low %) All X's 0 a. Invest in (See following screen): EAL $ 7,500 TAT $ 2,500 Long Term Bonds $30,000 Tax Deferred Annuity $10,000 Total return: $5,250 Chapter 3- 8 b. Rate of return for this investment = ($5,250/$50,000) = 10.5% The rate of return for additional funds = shadow price for the total investment constraint (the first constraint above) = 11% which is valid to + (1E+30). c. The return on EAL cannot fall below 12% from 15%; the return on BRU cannot increase above 15% from 12%; the rate on long term bonds cannot increase above 13.5% from 11%. d. Shadow price for: Total investment = .11 -- each additional dollar invested will earn 11% Minimum invested in taxed deferred annuity = -0.15 -- $0.15 lost for each extra dollar required to be invested in tax deferred annuities. Stock Definition -- a meaningless shadow price; the right hand side will not change. Minimum invested in TAT = -0.16 -- a $0.16 decrease in return for each extra dollar required to be in the low risk stock above 25%. Bonds >= Stocks = 0 -- no change in return for requiring bond investment to exceed stock investment by at least $1. Maximum invested in low yield investments = 0.1 -- $0.10 additional return for each extra dollar allowed to be invested in investments with returns less than 10%. Chapter 3- 9 3.5 See file Ch3.5.xls Unit Profit X1 = Number of full comforters produced daily 19-3(.50)-55(.20) = 6.50 X2 = Number of queen comforters produced daily 26-4(.50)-75(.20) = 9.00 X3 = Number of king comforters produced daily 32-6(.50)-95(.20) = 10.00 MAX 6.50X1 + 9.00X2 + 10.00X3 S.T. 3X1 + 4X2 + 6X3 2,700 (Stuffing) 55X1 + 75X2 + 95X3 48,000 (Fabric) 3X1 + 5X2 + 6X3 3,000 (Cutting minutes) 5X1 + 6X2 + 8X3 12,000 (Sewing minutes) All X's 120 a. 240 full, 312 queen, 120 king comforters; daily profit = $5,568 2688 pounds of stuffing, 48,000 sq. ft. of fabric, 3000 cutting minutes, 4032 sewing minutes used. b. Allowable decrease = .13636 -- Minimum selling price = $26-.14 = $25.86 c. (i) $0 2688 - (ii) $0.11 46,800 - 48,200 (iii) $0.15 2880 - 3080 (iv) $0 4032 - d. (See worksheet No Minimum) --300 full, 420 queen, 0 king; profit = $5,730 Chapter 3- 10 3.6 See file Ch3.6.xls X1 = number of 8-oz. portions of steak in the diet X2 = number of ounces of cheese in the diet X3 = number of apples in the diet X4 = number of 8-oz. portion of milk in the diet MIN 51X1 + 9X2 + 1X3 + 8X4 S.T. 692X1 + 110X2 + 81X3 + 150X4 1410 (=1800-390 minimum calories) 692X1 + 110X2 + 81X3 + 150X4 1610 (=2000-390 maximum calories) 57X1 + 6X2 + 1X3 + 8X4 80 (=100-20 grams of protein) 1X2 + 22X3 + 12X4 30 (= 45-15 grams of carbs.) All X's 0 Steak = 8(1.49566474) = 11.96532 oz., Milk = 8(2.5) = 20 oz. Fat = 96.2789 + 29 = 125.2789 grams Chapter 3- 11 3.7 See file Ch3.7.xls X1 = Number of Student models produced each week X2 = Number of Plus models produced each week X3 = Number of Net models produced each week X4 = Number of Pro models produced each week MAX 70X1 + 80X2 + 130X3 + 150X4 S.T. X3 100 (Contract) .4X1 + .5X2 + .6X3 + .8X4 750 (Production Hours) X1 + + X3 700 (Celeron) X2 + X4 550 (Pentium) X1 + X2 + X3 + 800 (20gb Hard Drives) X4 950 (30gb Hard Drives) X1 + X2 + 2X3 + X4 1600 (Floppy Drives) X1 + X2 + X4 1000 (Zip Drives) X1 + X3 + X4 1600 (CD R/W's) X2 + X3 + X4 900 (DVD's) X1 + X2 850 (15-in. monitors) X3 + X4 800 (17-in. monitors) X2 + X3 1250 (Mini-tower cases) X1 + X4 750 (Tower cases) All X's 0 Given the output shown on the next page: a. 325 Student models, 100 Plus models, 375 Net models, 425 Pro models; weekly profit = $143,250 b. Since Allowable Decrease for the Plus model is 55, the optimal solution could change if its profit coefficient is reduced below $80 - $55 = $25. But this does not necessarily guarantee the production will now be 0, only that it could change. But if $24.99 is substituted for $80 and Solver is called again (not shown--see worksheet Plus Profit $24.99), there is no change to the optimal solution. Now the Allowable Decrease for the Plus model is $24.99, so the optimal solution could change if its profit coefficient is reduced below $24.99 - $24.99 = $0. Set the profit for the Plus model to -$0.01 and call Solver again. (See worksheet Plus Profit -$0.01). Now the optimal production quantity for the Plus model is 0. So, its profit can fall to $0, before it becomes unprofitable to produce the Plus model. c. Yes, the shadow price for 17-inch monitors is $25. d. No, production hours is not a binding constraint; DO NOT HIRE THE WORKER. Chapter 3- 12 Chapter 3- 13 3.8 See file Ch3.8.xls X1 = the number of Delta assemblies produced daily X2 = the number of Omega assemblies produced daily X3 = the number of Theta assemblies produced daily MAX 800X1 + 900X2 + 600X3 S.T. X1 + X2 + X3 7 (X70686 chips) 2X1 + X2 + X3 8 (Production hours) 80X1 + 160X2 + 80X3 480 (Quality minutes) All X's 0 From the output on the next page: a. Produce 2 Delta’s and 4 Theta’s; daily profit = $4,000. No Omega’s are produced because the use too much quality control time compared to its slightly higher unit profit. b. Reduced cost = -100; thus minimum profit for production = $1,000. Since costs are $900, this implies a minimum price of $1,900. c. 6; there is a slack of 1. d. (i) There is slack on X70686 chips, thus additional profit is $0. (ii) Production hours are sunk costs. Since the shadow price for production hours is $200 and 3 additional hours is within the range of feasibility, these 3 hours gross 3($200) = $600 additional profit. Thus the net additional profit = $600 - $525 = $75. (iii) Quality control hours are also sunk costs. Since 1 additional hour = 60 additional minutes is also within its range of feasibility, the $5 shadow price is valid for all 60 minutes grossing 60($5) = $300 additional profit. The net additional profit is $300 - $200 = $100. OPTION (iii) is of the most value. Chapter 3- 14 Chapter 3- 15 3.9 See file Ch3.9.xls X1 = the number in group I contacted by telephone X2 = the number in group II contacted by telephone X3 = the number in group III contacted by telephone X4 = the number in group IV contacted by telephone X5 = the number in group I contacted in person X6 = the number in group II contacted in person X7 = the number in group III contacted in person X8 = the number in group IV contacted in person MIN 15X1 + 12X2 + 20X3 + 18X4 + 35X5 + 30X6 + 50X7 + 40X8 S.T. X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 = 2000 (Total) X1 + X2 + X5 + X6 1000 (W&R) X5 + X6 + X7 + X8 500 (In person) -.5X1 + .5X5 0 (W&R,ip) X2 + X4 + X6 + X8 800(Small) - .25X2 - .25X4+ .75X6 + .75X8 0 (Small,ip) X1 + X5 200 (Min I) X2 + X6 200 (Min II) X3 + X7 200 (Min III) X4 + X8 200 (Min IV) X1 + X5 1000 (Max I) X2 + X6 1000 (Max II) X3 + X7 1000 (Max III) X4 + X8 1000 (Max IV) All X's 0 As shown on the next page: Minimum Cost = $39,800; conduct survey as follows: Telephone Personal I 500 500 II 600 0 III 200 0 IV 200 0 Chapter 3- 16 3.10 See file Ch3.10 X1 = the number of ounces of Multigrain Cheerios in the mixture X2 = the number of ounces of Grape Nuts in the mixture X3 = the number of ounces of Product 19 in the mixture X4 = the number of ounces of Frosted Bran in the mixture X5 = the total number of ounces in the mixture MIN 12X1 + 9X2 + 9X3 + 15X4 S.T. 30X1 + 30X2 + 20X3 + 20X4 50 (Vitamin A) 25X1 + 2X2 + 100X3 + 25X4 50 (Vitamin C) 25X1 + 25X2 + 25X3 + 25X4 50 (Vitamin D) 25X1 + 25X2 + 100X3 + 25X4 50 (Vitamin B6) 45X1 + 45X2 + 100X3 + 25X4 50 (Iron) X1 + X2 + X3 + X4 - X5 = 0 (Total) X1 - .1X5 0 ( 10% M/G Cheerios) X2 - .1X5 0 ( 10% Grape Nuts) X3 - .1X5 0 ( 10% Product 19) X4 - .1X5 0 ( 10% Frosted Bran) All X's 0 Chapter 3- 17 a. From output below: Total sugar intake = 19.8 grams with this combination: .2 oz. Multigrain Cheerios 1.2244898 oz. Grape Nuts .3755102 oz. Product 19 .2 oz. Frosted Bran Total 2.0 oz. The Allowable Increase for the sugar coefficient for Grape Nuts is 0, and the Allowable Decrease for the sugar coefficient for Product 19 is 0. This indicates alternate optimal solutions. By setting cell H6 to 19.8, then changing the target cell to MIN D4 (on worksheet Alternate (not shown)), another optimal solution is: Chapter 3- 18 .2 oz. Multigrain Cheerios .8 oz. Grape Nuts .8 oz. Product 19 .2 oz. Frosted Bran Total 2.0 oz. b. Total = 2 oz. of cereal and 2 (1/2) = 1 cup of skim milk c. Extra % above 50% required of vitamin D adds .396 grams of sugar Extra ounce above 10% required of Multigrain Cheerios adds 3 grams of sugar Extra ounce above 10% required of Frosted Bran adds 6 grams of sugar 4.6d. Product 19 has less sugar and gives percentages that are at least as large as those for Frosted Bran for every vitamin and iron requirement. Re-solving gives the following alternate optimal solutions with 18 grams of sugar. Mixture 1 (Shown) Mixture 2 (On Worksheet Alternate No 10%) 1.53 oz. Grape Nuts 1.0 oz. Grape Nuts .47 oz. Product 19 1.0 oz. Product 19 Total 2.00 oz. Total 2.0 oz. Chapter 3- 19 3.11 See file Ch3.11.xls X1 = Number of refrigerator/ovens produced X2 = Number of French fry makers produced X3 = Number of French toast makers produced MIN 140X1 + 50X2 + 36X3 S.T. 100X1 + 35X2 + 27X3 2,000,000 (Min Profit) X1 5,000 (Min Refrig/oven) X2 4,000 (Min French fry maker) X3 2,300 (Min French toast maker) X1 15,000 (Max Refrig/oven) X2 15,000 (Max French fry maker) X3 15,000 (Max French toast maker) Make 14,550 refrigerator ovens, 4000 French fry makers, 15,000 French toast makers Total Variable Cost = $2,777,000. Chapter 3- 20 3.12 See file Ch3.12.xls a. X1 = Number of plates made per day X2 = Number of mugs made per day X3 = Number of steins made per day X4 = Total daily production MAX 2.50X1 + 3.25X2 + 3.90X3 S.T. 2X1 + 3X2 + 6X3 1920 ((4)(8)(60) Molding min.) 8X1 + 12X2 + 14X3 3840 ((8)(8)(60) Finishing min.) X2 150 (Minimum mugs) -2X1 - 2X2 + X3 0 (Steins 2(Plates + Mugs) X1 + X2 + X3 - X4 = 0 (Total Definition) X1 - .3X4 0 (Plates 30% Total Produced) All X's 0 101.8033 plates, 150 mugs, 87.54098 steins; total daily profit = $1083.42 Chapter 3- 21 b. Combine the first two constraints into one: 10X1 + 15X2 + 20X3 5760 128 plates, 298.6667 mugs, 0 steins; total daily profit = $1290.67 This is an increase of ($1290.67 - $1083.42) = $207.25 Chapter 3- 22 3.13 See file Ch3.13.xls X1 = $ invested in first trust deeds X2 = $ invested in second trust deeds X3 = $ invested in third trust deeds X4 = $ invested in commercial trust deeds X5 = $ invested in a savings account X6 = Total $ invested in residential trust deeds X7 = Total $ invested in all trust deeds MAX .0775X1 +.1125X2 +.1425X3 +.9875X4 +.0445X5 S.T. X1 + X2 + X3 + X4 + X5 = 68,000,000 (Total) X5 5,000,000 (Save) X1 + X2 + X3 + - X6 = 0 (Res Tr.) X1 + X2 + X3 + X4 -X7 = 0 (Total Tr) X6 - .8X7 0 (80% Res.) X1 -.6X6 0 (60% First) 4X1 + 6X2 + 9X3 + 3X4 340,000,000 (*) All X's 0 *Average Risk Factor is found by: X1 X2 X3 X4 X5 4 6 9 3 0 68,000,000 68,000,000 68,000,000 68,000,000 68,000,000 This expression must be 5. Multiplying both sides by 68,000,000 gives the above constraint. a. As seen on the next page the optimal allocation of funds is: $30,240,000.00 First Trust Deeds $ 66,666.67 Second Trust Deeds $20,093,333.33 Third Trust Deeds $12,600,000.00 Commercial Trust Deeds $ 5,000,000.00 Savings Account $68,000,000.00 Return = $6,539,400; Rate of return = $6,539,400/$68,000,000 =9.61676% b. Allowable Increase = .015, so could increase to 9.25% This would add (.015)($30,240,000) = $453,600 to the return. Now, return = $6,539,400 + $453,600 = $6,993,000 New rate of return = $6,993,000/$68,000,000 = 10.28382% Chapter 3- 23 Output: Chapter 3- 24 3.14 See file Ch3.14.xls XJR = Motor home cabinets produced in regular time in July XJO = Motor home cabinets produced in overtime in July XAR = Motor home cabinets produced in regular time in August XAO = Motor home cabinets produced in overtime in August XJS = Motor home cabinets produced in regular time in September XJS = Motor home cabinets produced in overtime in September YJR = Mobile home cabinets produced in regular time in July YJO = Mobile home cabinets produced in overtime in July YAR = Mobile home cabinets produced in regular time in August YAO = Mobile home cabinets produced in overtime in August YJS = Mobile home cabinets produced in regular time in September YJS = Mobile home cabinets produced in overtime in September SJ = Motor home cabinets stored in July SA = Motor home cabinets stored in August SS = Motor home cabinets stored in September TJ = Mobile home cabinets stored in July TA = Mobile home cabinets stored in August TS = Mobile home cabinets stored in September The objective function coefficients for the X's and Y's = (Material Cost) + (#Hours)(Hourly Cost for the Month and Period) Example for mobile home cabinets in August in Overtime = $210 + 5(1.5)(16) = $330 The following tables summarize these coefficients. Motor Home Regular Time Overtime July $188 $209 August $194 $218 September $200 $227 Mobile Home Regular Time Overtime July $280 $315 August $290 $330 September $300 $345 Objective function coefficients for the S's and T's Motor Home (S) Mobile Home (T) July $6 $9 August $6 $9 September $6 $9 Chapter 3- 25 Objective Function MIN 188XJR + 209XJO + 194XAR + 218XAO + 200 XSR + 227XSO + 280YJR + 315YJO + 290YAR + 330YAO + 300 YSR + 345YSO + 6SJ + 6SA + 6SS + 9TJ + 9TA + 9Ts Constraints Storage Constraints For each product: Amount Stored at the end of the month = (Amount Stored at the Beginning of the month) + (Production) - (Demand) SJ = 25 + XJR + XJO - 250 (Motor Home - July) SA = SJ + XAR + XAO - 250 (Motor Home - August) SS = SA + XSR + XSO - 150 (Motor Home - September) TJ = 20 + YJR + YJO - 100 (Mobile Home - July) TA = TJ + YAR + YAO - 300 (Mobile Home - August) TS = TA + YSR + YSO - 400 (Mobile Home - September) Required For September: SS 10 (Motor Home) TS 25 (Mobile Home) Maximum Storage in any Month SJ + TJ 300 (July) SA + TA 300 (August) SS + TS 300 (September) Production Regular Time 3XJR + 5YJR 2100 (July) 3XAR + 5YAR 1500 (August) 3XSR + 5YSR 1200 (September) Overtime 3XJO + 5YJO 1050 (July) 3XAO + 5YAO 750 (August) 3XSO + 5YSO 600 (September) Non-negativity All X's, Y's, S's, and T's 0 Chapter 3- 26 The optimal solution is shown in the cells C6:D8, G6:H8, and K6:L8. Total Cost = $367,969 Regular Time Begin Storage + Production - End Storage = Demand Overtime September Storage -- Motor Home Cabinets September Storage -- Mobile Home Cabinets Maximum Monthly Storage Chapter 3- 27 3.15 See file Ch3.15.xls X1 = the number of acres of wheat planted X2 = the number of acres of corn planted X3 = the number of acres of oats planted X4 = the number of acres of soybeans planted Profit coefficients are 210($3.20) - $50 = $622, 300($2.55) - $75 = $690, 180($1.45) - $30 = $231, and 240($3.10) - $60 = $684 respectively. MAX 622X1 + 690X2 + 231X3 + 684X4 S.T. 4X1 + 5X2 + 3X3 + 10X4 1,800 (Labor hours) 50X1 + 75X2 + 30X3 + 60X4 25,000 (Expenses) 2X1 + 6X2 + X3 + 4X4 1,200 (Water) 210X1 30,000 (Min. Wheat) 300X2 30,000 (Min. Corn) 180X3 25,000 (Max Oats) X1 + X2 + X3 + X4 300 (Total acres) All X's 0 a. (See Worksheet on next page) Plant 142.8571 acres of wheat, 142.8571 acres of corn and 14.28571 acres of soybeans; profit = $197,200. b. The net profit must rise to $675; adding the $30/acre in expenses = $705. If selling price remains $1.45 per bushel, then yield must increase to $705 per acre/$1.45 per bushel = 486.21 bushels per acre. If the yield remains 180 bushels per acre, then its price must rise to $705 per acre/180 bushels per acre = $3.92 per bushel. c. (See work sheet NO CORN (not shown).) Yes, corn would still be planted; there is currently slack on the corn constraint. If corn were not grown, the problem must be re- solved. Plant 200 acres of wheat and 100 acres of soybeans for a profit of $192,800 -- a decrease of $4,400. d. The range of feasibility for acres is only valid up to 318.57 acres (18.57 additional acres.) Thus the problem must be re-solved. See worksheet PARCEL. Plant 181.82 acres of wheat, 101.82 acres of corn, and 56.36 acres of soybeans for a profit of $221,898.18 -- an increase of $24,698.18. Yes Bill should lease this property for $2,000. Chapter 3- 28 Chapter 3- 29 3.16 See file Ch3.16.xls PSA = Number of professional sets produced in Sarasota and shipped to Anaheim PSD = Number of professional sets produced in Sarasota and shipped to Dallas PST = Number of professional sets produced in Sarasota and shipped to Toledo PCA = Number of professional sets produced in Carson and shipped to Anaheim PCD = Number of professional sets produced in Carson and shipped to Dallas PCT = Number of professional sets produced in Carson and shipped to Toledo DSA = Number of deluxe sets produced in Sarasota and shipped to Anaheim DSD = Number of deluxe sets produced in Sarasota and shipped to Dallas DST = Number of deluxe sets produced in Sarasota and shipped to Toledo DLA = Number of deluxe sets produced in Louisville and shipped to Anaheim DLD = Number of deluxe sets produced in Louisville and shipped to Dallas DLT = Number of deluxe sets produced in Louisville and shipped to Toledo DCA = Number of deluxe sets produced in Carson and shipped to Anaheim DCD = Number of deluxe sets produced in Carson and shipped to Dallas DCT = Number of deluxe sets produced in Carson and shipped to Toledo WLA = Number of weekender sets produced in Louisville and shipped to Anaheim WLD = Number of weekender sets produced in Louisville and shipped to Dallas WLT = Number of weekender sets produced in Louisville and shipped to Toledo WCA = Number of weekender sets produced in Carson and shipped to Anaheim WCD = Number of weekender sets produced in Carson and shipped to Dallas WCT = Number of weekender sets produced in Carson and shipped to Toledo Objective function coefficients = (Gross Profit) - (Shipping Cost) MAX 205PSA + 218PSD + 220PST + 242PCA + 210PCD + 200PCT + 135DSA + 147DSD + 150DST + 141DLA + 157DLD + 165DLT + 169DCA + 140DCD + 135DCT + 170WLA + 185WLD + 191WLT + 195WCA + 170WCD + 164WCT S.T. Minimum Professional PSA + PCA 480 PSD + PCD 320 PST + PCT 160 Minimum Deluxe DSA + DLA + DCA 640 DSD + DLD + DCD 800 DST + DLT + DCT 880 Minimum Weekender WLA + WCA 640 WLD + WCD 1200 WLT + WCT 800 Chapter 3- 30 Maximum Professional PSA + PCA 600 PSD + PCD 400 PST + PCT 200 Maximum Deluxe DSA + DLA + DCA 800 DSD + DLD + DCD 1000 DST + DLT + DCT 1100 Maximum Weekender WLA + WCA 800 WLD + WCD 1500 WLT + WCT 1000 Steel Availability 3.2PSA + 3.2PSD + 3.2PST + 3.6DSA + 3.6DSD + 3.6DST 5000 3.6DLA + 3.6DLD + 3.6DLT + 2.8WLA + 2.8WLD + 2.8WLT 9000 3.2PCA + 3.2PCD + 3.2PCT + 3.6DCA + 3.6DCD + 3.6DCT +2.8WCA + 2.8WCD + 2.8WCT 14000 Aluminum Availability 5PSA + 5PSD + 5PST + 4DSA + 4DSD + 4DST 7000 4DLA + 4DLD + 4DLT + 4.5WLA + 4.5WLD + 4.5WLT 13000 5PCA + 5PCD + 5PCT + 4DCA + 4DCD + 4DCT +4.5WCA + 4.5WCD + 4.5WCT 18000 Wood Availability 5.2PSA + 5.2PSD + 5.2PST + 4.8DSA + 4.8DSD + 4.8DST 10000 4.8DLA + 4.8DLD + 4.8DLT + 4.4WLA + 4.4WLD + 4.4WLT 18000 5.2PCA +5.2PCD + 5.2PCT + 4.8DCA + 4.8DCD + 4.8DCT +4.4WCA + 4.4WCD + 4.4WCT 20000 All P's, D's W's 0 Output is too large to display. Monthly Profit = $1,353,088.89 Professional Sarasota Carson Anaheim Dallas Toledo Anaheim Dallas Toledo 0 400 200 600 0 0 Deluxe Sarasota Louisville Carson Anaheim Dallas Toledo Anaheim Dallas Toledo Anaheim Dallas Toledo 0 855.5556 0 0 0 1100 800 144.4444 0 Weekender Louisville Carson Anaheim Dallas Toledo Anaheim Dallas Toledo 0 800 1000 800 700 0 Chapter 3- 31 3.17 See file Ch3.17.xls PSA = Number of professional sets produced in Sarasota and shipped to Anaheim PSD = Number of professional sets produced in Sarasota and shipped to Dallas PST = Number of professional sets produced in Sarasota and shipped to Toledo PCA = Number of professional sets produced in Carson and shipped to Anaheim PCD = Number of professional sets produced in Carson and shipped to Dallas PCT = Number of professional sets produced in Carson and shipped to Toledo DSA = Number of deluxe sets produced in Sarasota and shipped to Anaheim DSD = Number of deluxe sets produced in Sarasota and shipped to Dallas DST = Number of deluxe sets produced in Sarasota and shipped to Toledo DLA = Number of deluxe sets produced in Louisville and shipped to Anaheim DLD = Number of deluxe sets produced in Louisville and shipped to Dallas DLT = Number of deluxe sets produced in Louisville and shipped to Toledo DCA = Number of deluxe sets produced in Carson and shipped to Anaheim DCD = Number of deluxe sets produced in Carson and shipped to Dallas DCT = Number of deluxe sets produced in Carson and shipped to Toledo WLA = Number of weekender sets produced in Louisville and shipped to Anaheim WLD = Number of weekender sets produced in Louisville and shipped to Dallas WLT = Number of weekender sets produced in Louisville and shipped to Toledo WCA = Number of weekender sets produced in Carson and shipped to Anaheim WCD = Number of weekender sets produced in Carson and shipped to Dallas WCT = Number of weekender sets produced in Carson and shipped to Toledo YS = 1 if the Sarasota plant is operational YL = 1 if the Louisville plant is operational YC = 1 if the Carson plant is operational a. MAX 205PSA + 218PSD + 220PST + 242PCA + 210PCD + 200PCT + 135DSA + 147DSD + 150DST + 141DLA + 157DLD + 165DLT + 169DCA + 140DCD + 135DCT + 170WLA + 185WLD + 191WLT + 195WCA + 170WCD + 164WCT -250000YS - 350000YL - 500000YC S.T. Minimum Professional PSA + PCA 480 PSD + PCD 320 PST + PCT 160 Minimum Deluxe DSA + DLA + DCA 640 DSD + DLD + DCD 800 DST + DLT + DCT 880 Minimum Weekender WLA + WCA 640 WLD + WCD 1200 WLT + WCT 800 Chapter 3- 32 Maximum Professional PSA + PCA 600 PSD + PCD 400 PST + PCT 200 Maximum Deluxe DSA + DLA + DCA 800 DSD + DLD + DCD 1000 DST + DLT + DCT 1100 Maximum Weekender WLA + WCA 800 WLD + WCD 1500 WLT + WCT 1000 Steel Availability 3.2PSA + 3.2PSD + 3.2PST + 3.6DSA + 3.6DSD + 3.6DST -5000YS 0 3.6DLA + 3.6DLD + 3.6DLT + 2.8WLA + 2.8WLD + 2.8WLT -9000YL 0 3.2PCA + 3.2PCD + 3.2PCT + 3.6DCA + 3.6DCD + 3.6DCT +2.8WCA + 2.8WCD + 2.8WCT- 14000YC0 Aluminum Availability 5PSA + 5PSD + 5PST + 4DSA + 4DSD + 4DST - 7000YS 0 4DLA + 4DLD + 4DLT + 4.5WLA + 4.5WLD + 4.5WLT -13000YL 0 5PCA + 5PCD + 5PCT + 4DCA + 4DCD + 4DCT +4.5WCA + 4.5WCD + 4.5WCT -18000YC 0 Wood Availability 5.2PSA + 5.2PSD + 5.2PST + 4.8DSA + 4.8DSD + 4.8DST -10000YS 0 4.8DLA + 4.8DLD + 4.8DLT + 4.4WLA + 4.4WLD + 4.4WLT -18000YL 0 5.2PCA + 5.2PCD + 5.2PCT + 4.8DCA + 4.8DCD + 4.8DCT +4.4WCA + 4.4WCD + 4.4WCT-20000YC0 All P's, D's W's 0 Y's binary Output is too large to display. (See worksheet Part a.) Monthly Profit = $437,250 Professional Sarasota Carson Anaheim Dallas Toledo Anaheim Dallas Toledo 0 0 0 600 400 200 Deluxe Sarasota Louisville Carson Anaheim Dallas Toledo Anaheim Dallas Toledo Anaheim Dallas Toledo 0 0 0 0 0 820 800 857.5 60 Weekender Louisville Carson Anaheim Dallas Toledo Anaheim Dallas Toledo 0 1160 1000 800 340 0 Do not operate Sarasota. Chapter 3- 33 b. Add decision variables: ZA = 1 if the Anaheim center is operational ZD = 1 if the Dallas center is operational ZT = 1 if the Toledo center is operational MAX 205PSA + 218PSD + 220PST + 242PCA + 210PCD + 200PCT + 135DSA + 147DSD + 150DST + 141DLA + 157DLD + 165DLT + 169DCA + 140DCD + 135DCT + 170WLA + 185WLD + 191WLT + 195WCA + 170WCD + 164WCT -250000YS - 350000YL - 500000YC - 50000ZA - 100000ZB - 90000ZB S.T. Minimum Professional PSA + PCA - 480ZA 0 PSD + PCD - 320ZD 0 PST + PCT -160ZT 0 Minimum Deluxe DSA + DLA + DCA - 640ZA 0 DSD + DLD + DCD - 800ZD 0 DST + DLT + DCT - 880ZT 0 Minimum Weekender WLA + WCA - 640ZA 0 WLD + WCD -1200ZD 0 WLT + WCT - 800ZT 0 Maximum Professional PSA + PCA - 600ZA 0 PSD + PCD - 400ZD 0 PST + PCT - 200ZT 0 Maximum Deluxe DSA + DLA + DCA - 800ZA 0 DSD + DLD + DCD -1000ZD 0 DST + DLT + DCT -1100ZT 0 Maximum Weekender WLA + WCA - 800ZA 0 WLD + WCD -1500ZD 0 WLT + WCT -1000ZT 0 Steel Availability 3.2PSA + 3.2PSD + 3.2PST + 3.6DSA + 3.6DSD + 3.6DST -5000YS 0 3.6DLA + 3.6DLD + 3.6DLT + 2.8WLA + 2.8WLD + 2.8WLT -9000YL 0 3.2PCA + 3.2PCD + 3.2PCT + 3.6DCA + 3.6DCD + 3.6DCT +2.8WCA + 2.8WCD + 2.8WCT- 14000YC0 Aluminum Availability 5PSA + 5PSD + 5PST + 4DSA + 4DSD + 4DST - 7000YS 0 4DLA + 4DLD + 4DLT + 4.5WLA + 4.5WLD + 4.5WLT -13000YL 0 Chapter 3- 34 5PCA + 5PCD + 5PCT + 4DCA + 4DCD + 4DCT +4.5WCA + 4.5WCD + 4.5WCT -18000YC 0 Wood Availability 5.2PSA + 5.2PSD + 5.2PST + 4.8DSA + 4.8DSD + 4.8DST -10000YS 0 4.8DLA + 4.8DLD + 4.8DLT + 4.4WLA + 4.4WLD + 4.4WLT -18000YL 0 5.2PCA + 5.2PCD + 5.2PCT + 4.8DCA + 4.8DCD + 4.8DCT +4.4WCA + 4.4WCD + 4.4WCT- 20000YC0 All P's, D's W's 0 Y's, Z's binary Output is too large to display. (See worksheet Part b.) Monthly Profit = $197,250 Professional Sarasota Carson Anaheim Dallas Toledo Anaheim Dallas Toledo 0 0 0 600 400 200 Deluxe Sarasota Louisville Carson Anaheim Dallas Toledo Anaheim Dallas Toledo Anaheim Dallas Toledo 0 0 0 0 0 820 800 857.5 60 Weekender Louisville Carson Anaheim Dallas Toledo Anaheim Dallas Toledo 0 1160 1000 800 340 0 Do not operate Sarasota. Chapter 3- 35 3.18 See file Ch3.18.xls X1 = amount invested in Bonanza Gold X2 = amount invested in Cascade Telephone X3 = amount invested in the money market account X4 = amount invested in two-year treasury bonds MAX .15X1 + .09X2 + .07X3 + .08X4 S.T. X1 + X2 50,000 (Max stocks) X1 + X2 + X3 60,000 (Min potential 9%) X4 30,000 (Max treasury bonds) -.5X1 + .03X2 + .06X3 + .08X4 4,000 (Min return) X1 + X2 + X3 + X4 = 100,000 (Total) All X's 0 Invest $2,075.47 in Bonanza Gold, $47,924.53 is Cascade Telephone, $20,000 in the money market account, $30,000 in two year treasury bonds. The expected return is $8,424.53 (8.42453%). Chapter 3- 36 3.19 See file Ch3.19.xls Xij = number of gallons of crude i blended into grade j i = p (Pacific), g (Gulf), m (Middle East) j = r (Regular), p (Premium) Xj = total amount of grade j produced Example of profit coefficient: Selling price of regular = $0.52, purchase cost of Pacific crude = ($14.28)/42 = $0.34; thus profit on a gallon of Pacific crude = $0.52 - $0.34 = $0.18 MAX .18Xpr +.16Xgr +.05Xmr +.26Xpp +.24Xgp +.13Xmp S.T. Xpr + Xpp 126,000 (Pacific) Xgr + Xgp 84,000 (Gulf) Xmr + Xmp 336,000 (M. East) Xpr + Xgr + Xmr - Xr = 0 (Regular) Xpp + Xgp + Xmp – Xp = 0 (Premium) Xr 200,000 (Min Reg) Xp 100,000 (Min Prem) Xr + Xp 400,000 (Capacity) 85Xpr + 87Xgr + 95Xmr - 87Xr 0 (Reg Oct.) 85Xpp +87Xgp + 95Xmp - 91Xp 0 (Prem Oct.) All X's 0 From the sensitivity report (not shown), there are alternate optimal solutions giving a profit of $61,620. The one represented on the screen on the next page is: Pacific - Regular 46,000 Gulf - Regular 84,000 Mid East - Regular 70,000 Pacific - Premium 80,000 Gulf - Premium 0 Mid East - Premium 120,000 b. Shadow price (from Total Total (not shown)) = $0.13 per gallon; Thus 50,000 gallons is worth 50,000($0.13) = $6,500 > $5,000 -- Yes Caloco should secure the extra refining capacity. c. The solution uses all 190,000 gallons of Mid East Oil @ $0.47/gallon = $89,300. 8000 barrels @ $16.80 per barrel = $134,400 -- the Middle East distributors receive more. To see if it would be profitable for Caloco, the problem must be re-solved. $16.80 per barrel = $0.40 per gallon. The new profit coefficients for Xmr and Xmp would be $0.12 and $0.20 respectively. Change the third constraint to Xmr + Xmp = 336,000 Re-solve (See worksheet Mid East (not shown).) Chapter 3- 37 The new solution gives Caloco a profit of $67,840 -- thus it would be profitable for Caloco to accept this offer. There is now no Gulf crude purchased and only slightly more than 1500 barrels of Pacific crude purchased. These distributors would have to cut their prices to stay competitive. Minimum Requirements Availability Capacity Octane Requirements Chapter 3- 38 3.20 See file Ch3.20.xls Average Score X1 = the weight given to teaching (90+75+90)/3 = 85 X2 = the weight given to research (60+60+75)/3 = 65 X3 = the weight given to professional activities (90+95+85)/3 = 90 X4 = the weight given to service (80+95+95)/3 = 90 MAX 85X1 + 65X2 + 90X3 + 90X4 S.T. - X1 + X2 0 (Teaching research) - X1 + X3 0 (Teaching professional) - X1 + X4 0 (Teaching service) X2 .25 (Min. research) X1 + X2 .75 (Teaching + research .75) X1 + X2 .90 (Teaching + research .90) X3 - X4 0 (Service professional) X3 .05 (Professional .05) X1 + X2 + X3 + X4 = 1 (Total weights) All X's 0 Teaching = .50, Research = .25, Professional = .125, Service = .125--Average = 81.25 < 85. Professor Anna Sung should not get tenure. Chapter 3- 39 3.21 See file Ch3.21.xls -- a more compact Excel spreadsheet is discussed in Chapter 4 Xij = the number of Matey 20 catamarans made in plant I and sold to dealership j i = 1 (San Diego), 2 (Santa Ana), 3 (San Jose) j = 1 (Newport Beach), 2 (Long Beach), 3 (Ventura), 4 (S L O), 5 (San Fran) MIN 1265X11 + 1285X12 + 1345X13 + 1390X14 + 1565X15 + 1130X21 + 1130X22 + 1285X23 + 1355X24 + 1405X25 + 1365X31 + 1340X32 + 1275X33 + 1225X34 + 1075X35 S.T. X11 + X12 + X13 + X14 + X15 38 (San Diego) X21 + X22 + X23 + X24 + X25 45 (Santa Ana) X31 + X32 + X33 + X34 + X35 58 (San Jose) X11 + X21 + X31 = 42 (Newport Beach) X12 + X22 + X32 = 33 (Long Beach) X13 + X23 + X33 = 14 (Ventura) X14 + X24 + X34 = 10 (San Luis Obispo) X15 + X25 + X35 = 22 (San Francisco) All Xij’s 0 Build 30 in San Diego -- Ship them all to Newport Beach Build 45 in Santa Ana -- Ship 12 to Newport Beach and 33 to Long Beach Build 46 in San Jose -- Ship 14 to Ventura, 10 to San Luis Obispo, and 22 to San Francisco Chapter 3- 40 Total building and shipping costs = $142,550 3.22 See file Ch3.22.xls X1 = the number cut into 4 30-in. pieces X2 = the number cut into 2 30-in. and 1 42-in. pieces X3 = the number cut into 1 30-in. and 2 42-in. pieces X4 = the number cut into 2 30-in. and 1 56-in. pieces X5 = the number cut into 1 42-in. and 1 56-in. pieces X6 = the number cut into 2 56-in. pieces MIN 18X2 + 6X3 + 4X4 + 22X5 + 8X6 S.T. 4X1 + 2X2 + X3 + 2X4 1,500 (30-in.) X2 + 2X3 + X5 500 (42-in.) X4 + X5 + 2X6 600 (56-in.) All X's 0 From the Sensitivity Report, there are alternate optimal solutions. The one shown is 312.5 (rounded to 313) cut into 4 30-in. pieces; 250 cut into 1 30-in. and 2 42-in. pieces and 300 cut into 2 56-inch pieces. Total waste = 3900 inches. Two 30-in. pieces will remain in inventory with rounded solution. Chapter 3- 41 Chapter 3- 42 3.23 See file Ch3.23.xls a. This could violate the “no interaction” assumption of linear programming. b. X1 = $ spent on television advertising X2 = $ spent on radio advertising X3 = $ spent on newspaper advertising X4 = $ spent on its circulars MAX 28X1 + 18X2 + 20X3 + 15X4 S.T. X1 + X2 + X3 + X4 700,000 (Budget) X1 + X2 350,000 (TV/Radio) 10X1 + 7X2 + 8X3 + 4X4 2,500,000 (Yuppies) 5X1 + 2X2 + 3X3 + X4 1,200,000 (College) 5X1 + 8X2 + 6X3 + 9X4 1,800,000 (Audiophiles) X1 300,000 (Max TV) X2 300,000 (Max Radio) X3 300,000 (Max Newspapers) X4 300,000 (Max Circulars) All X's 0 $300,000 TV, $100,000 radio, $300,000 newspapers Total Exposure = 16,200,000 c. It would not affect the optimal solution; it is a non-binding constraint (there is surplus). Chapter 3- 43 3.24 See file Ch3.24 a. X1 = number of polyester suits ordered for the season X2 = the number of wool suits ordered for the season X3 = the number of cotton suits ordered for the season X4 = the number of imported suits ordered for the season MAX 35X1 + 47X2 + 30X3 + 90X4 S.T. .4X1 + .5X2 + .3X3 + X4 1,800 (Salesperson hours) 2X1 + 4X2 + 3X3 + 9X4 15,000 (Advertising budget) X1 + 1.5X2 + 1.25X3 + 3X4 18,000 (Square footage) All X's 0 Order 1500 polyester suits, 4000 cotton suits --Profit = $172,500 Chapter 3- 44 b. New Objective Within the Change to Coefficient Range of Optimality? Solution Profit (i) $34 Yes None (alternate optima) $171,000 $33 No 3000 wool, 1000 cotton* $171,000 (ii) $36 Yes None $174,000 $37 Yes None $175,500 *(See worksheet Part b (i-33) (not shown)): c. (i) No, there is a slack on the space constraint. (ii) Yes, $400 will net 400(2.5) = $1000 additional profit. (iii) Yes, 260 salesperson hours will add 260($75) = $19,500 for a cost of $3,600. d. Add the constraint X1 + X2 + X3 + X4 5000. (See worksheet Part d (not shown).) Order 1000 polyester, 1000 wool and 3000 cotton suits giving a $172,000 profit. 3.25 See file Ch3.25.xls X1 = Number of acres for zoo habitat attractions X2 = Number of acres for show areas X3 = Number of acres for restaurants X4 = Number of acres for retail establishments X5 = Number of acres for maintenance areas X6 = Number of acres for green areas X7 = Number of acres for roads/walkways X8 = Number of restaurants X9 = Number of retail stores Max gross profit per hour: 1000X1 + 900X2 + 800X8 + 750X9 S.T. Ttoal acreage = 350: X1 + X2 + X3 + X4 + X5 + X6 + X7 = 350 Habitat 40%(350): X1 140 Greens that required by habitat, show, restaurant, store and walkway areas: X6 .03X1 + .40X2 + .25X7 + .15X8 + .10X9 Show areas 5%(350): X2 17.5 Habitat + Show 70%(350): X1 + X2 245 (Show + its green) .20((Show + its green) + (Habitat + its green) : 1.4X2 .2(1.03X1 + 1.4X2) Green 25%(area not for habitat and show) : X6 .25(350-X1-X2) Maintenance require maintenance: X5 .01X1 + .10X2 + .08X3 + .06X4 + .02X6 + .04X7 Restaurants 20: X8 20 Restaurants 30: X8 30 Chapter 3- 45 Restaurant acreage = .25(# restaurants) X3 = .25X8 Stores 15 X9 15 Stores 25 X9 25 Restaurants Stores X8 X9 Store acreage = .20(#stores) X4 = .20X9 At least 10 acres walkways: X7 10 At least 100 acres of non-required green: X6 -(.03X1 + .40X2 + .25X7 + .15X8 + .10X9) 100 All X's 0 Summarizing: MAX 1000X1 + 900X2 + 800X8 + 750X9 S.T. X1 + X2 + X3 + X4 + X5 + X 6 + X7 = 350 X1 140 -.03X1 - .40X2 + X6 -.25X7 - .15X8 - .10X9 0 X2 17.5 X1 + X2 245 -2.06X1 + 1.12X2 0 .25X1 + .25X2 + X6 87.5 -.01X1 - .10X2 -.08X3 - .06X4 + X5 - .02X6 - .04X7 0 X8 20 X8 30 X3 -.25X8 = 0 X9 15 X9 25 X8 - X9 0 X4 -.20X9 = 0 X7 10 -.03X1 - .40X2 + X6 -.25X7 - .15X8 - .10X9 100 All X's 0 As seen on the next page: Habitat Area: 180.78 acres Show Area: 17.50 acres Restaurants (30) 7.50 acres Stores (25) 5.00 acres Maintenance 7.30 acres Green Area: 121.92 acres Roads/Walkways 10.00 acres TOTAL 350.00 acres Gross Hourly Profit = $239,280 Gross Daily Profit = $2,392,800 Net Daily Profit = $392,800 Annual Net Profit = 365($392,800) = $143,372,000. Chapter 3- 46 Chapter 3- 47 3.26 See file Ch3.26.xls X1 = the number of 2-oz. Go bars produced daily X2 = the number of 2-oz. Power bars produced daily X3 = the number of 2-oz. Energy bars produced daily X4 = the number of 8-oz. Energy bars produced daily X5 = the total number of 2-oz. bars produced daily Derivation of profit coefficients The costs per 2oz. are: Protein concentrate $0.40; Sugar substitute $0.175; Carob $0.325. Thus the unit profits are: 2 oz. Go bars: .68 - .03 - .2(.40) - .6(.175) - .2(.325) = $0.40 2 oz. Power bars: .84 - .03 - .5(.40) - .3(.175) - .2(.325) = $0.4925 2 oz. Energy bars: .76 - .03 - .3(.40) - .4(.175) - .3(.325) = $0.4425 8 oz. Energy bars: 3.00 - .05 - .3(1.60) - .4(.70) - .3(1.30) = $1.80 MAX .40X1 + .4925X2 + .4425X3 +1.80X4 S.T. .4X1 + X2 + .6X3 + 2.4X4 9,600 (Oz. protein) 1.2X1 + .6X2 + .8X3 + 3.2X4 16,000 (Oz. sugar sub.) .4X1 + .4X2 + .6X3 + 2.4X4 12,800 (Oz. carob) X1 + X2 + X3 - X5 = 0 (Total 2-oz.) X5 25,000 (Max 2-oz.) X4 2,000 (Max 8-oz.) X1 2,500 (Min 2-oz. Go) X2 2,500 (Min 2-oz. Power) X3 2,500 (Min 2-oz. Energy) X1 -.5X5 0 (Max 2-oz. Go) X2 -.5X5 0 (Max 2-oz. Power) X3 -.5X5 0 (Max 2-oz. Energy) 2X3 + 4X4 - X5 0 (Max Energy*) All X's 0 *The Maximum Energy Bar constraint is formulated as follows: Total Weight Energy Bars: 2X3 + 8X4 Total Weight All Bars: 8X4 + 2X5 Thus, 2X3 + 8X4 .5(8X4 + 2X5) or 2X3 + 4X4 - X5 0. Daily production, giving daily profit of $7,273.375 (see output on next page): 2-oz. Go: 7,550 2-oz. Power: 2,500 2-oz. Energy: 5,050 8-oz. Energy 437.5 Chapter 3- 48 Chapter 3- 49 3.27 See file Ch3.27.xls a. X1 = $ spent on advertising ketchup only X2 = $ spent on advertising spaghetti sauce only X3 = $ spent on advertising taco sauce only X4 = $ spent on joint advertising MAX 1.20X1 + 1.12X2 + 1.10X3 + 1.05X4 S.T. X1 + X2 + X3 + X4 2,000,000 (Total advertising) X4 400,000 (Max joint adv.) X4 100,000 (Min joint adv.) X3 + X4 1,000,000 (Min total taco sauce) X1 250,000 (Min ketchup only) X2 250,000 (Min spag sauce only) X3 750,000 (Min taco sauce only) X1 + X4 700,000 (Min total ketchup) X2 + X4 700,000 (Min total spag sauce) 4X1 + 3.2X2 + 11X3 + 4.2X4 7,500,000 (Min bottles sold) All X's 0 Chapter 3- 50 Advertising budget: $550,000 ketchup only, $450,000 spaghetti sauce only, $750,000 taco sauce only, $250,000 joint advertising Total return = $2,251,500 or (2,251,500/2,000,000) = 112.575% b. Shadow price = $1.20 c. Shadow price = -.03. Since 700,000 is within its range of feasibility, the profit would increase by (-50,000)(-$0.03) = $1500. Chapter 3- 51 3.28 See file Ch3.28.xls Xj = the number of workers that have shift j MIN 15X1 + 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. Chapter 3- 52 3.29 See 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. MIN X1 + 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. Chapter 3- 53 Chapter 3- 54 3.30 See 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 Chapter 3- 55 Build 30 cars in Texas, 20 vans in Texas and 10 buses in Tennessee; total cost $1,365,000. Chapter 3- 56 3.31 See 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 Sea Br.) 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 Change to .5 Chapter 3- 57 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. Chapter 3- 58 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 Change to .5 CHECK Chapter 3- 59 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. Chapter 3- 60 3.33 See 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.) Chapter 3- 61 Chapter 3- 62 3.34 See 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. Chapter 3- 63 3.35 See 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: MIN 1200X1 + 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. Chapter 3- 64 3.36 See file Ch3.36.xls X1 = the number of TV exposures X2 = the number of radio exposures X3 = the number of newspaper exposures a. MAX 500,000X1 + 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. Chapter 3- 65 c. See worksheet Part c (not shown). MAX 500,000X1 + 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. Chapter 3- 66 3.37 See 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 Chapter 3- 67 3.38 See 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 Chapter 3- 68 3.39 See 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 MIN 2000X1 + 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. Chapter 3- 69 3.40 See 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 Chapter 3- 70 Chapter 3- 71 3.41 See 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 MAX 2.75X1 + 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 Chapter 3- 72 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. 3.42 See 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 MAX 2000X1 + 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 Chapter 3- 73 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)). Chapter 3- 74 3.43 See 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. Chapter 3- 75 3.44 See 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) Chapter 3- 76 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. Chapter 3- 77 3.45 See 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 MAX 2X1 + 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 Chapter 3- 78 3.46 See 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 MIN X1 + 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 a. 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. Chapter 3- 79 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. Chapter 3- 80 (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. Chapter 3- 81 3.47 See 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 -1000000Y1 0 X2 -1000000Y2 0 X3 -1000000Y3 0 X4 -1000000Y4 0 X5 - 1000000Y5 0 X6 -1000000Y6 0 All X’s 0, All Y’s binary Chapter 3- 82 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. Chapter 3- 83 3.48 See 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. Chapter 3- 84 3.49 See file Ch3.49.xls Data 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 Y2 = Relative output weight applied to the campus graduation rate Y3 = Relative output weight applied to the campus percent employment For Cody MAX 2.63Y1 + .43Y2 + .78Y3 S.T. 2.63Y1 + .43Y2 + .78Y3 920X1 + .068X2 + 20.3X3 2.57Y1 + .55Y2 + .79Y3 960X1 + .059X2 + 24.6X3 2.81Y1 + .54Y2 + .83Y3 1000X1 + .061X2 + 35.2X3 920X1 + .068X2 + 20.3X3 = 1 All X's and Y's 0 Cody efficiency rating = 1; it is efficient. For Casper, the objective function is MAX 2.57Y1 + .55Y2 + .79Y3 and the last constraint is 960X1 + .059X2 + 24.6X3 = 1. See Worksheet Casper. Casper is also efficient. For Cheyenne, the objective function is MAX 2.81Y1 + .54Y2 + .83Y3 and the last constraint is 1000X1 + .061X2 + 35.2X3 = 1. See Worksheet Cheyenne. Cheyenne is also efficient. Chapter 3- 85 3.50 See file Ch3.50.xls Data Envelopment Analysis X1 = Relative input weight applied to income X2 = Relative input weight applied to number of employees X3 = Relative input weight applied to size Y1 = Relative output weight applied to men's clothing sales Y2 = Relative output weight applied to women's clothing sales Y3 = Relative output weight applied to cosmetics sales Y4 = Relative output weight applied to jewelry sales For Discount Store 1 MAX 9.6Y1 + 20.3Y2 + 16.3Y3 + 4.1Y4 S.T. 9.6Y1 + 20.3Y2 + 16.3Y3 + 4.1Y4 32987X1 + 275X2 + 23876X3 10.3Y1 + 17.9Y2 + 15.5Y3 + 4.6Y4 32987X1 + 215X2 + 28755X3 14.5Y1 + 55.2Y2 + 27.4Y3 + 22.3Y4 54321X1 + 185X2 + 19000X3 15.2Y1 + 44.8Y2 + 26.8Y3 + 28.4Y4 54321X1 + 180X2 + 18750X3 12.5Y1 + 45.9Y2 + 19.9Y3 + 35.1Y4 99765X1 + 85X2 + 11000X3 32987X1 + 275X2 + 23876X3 = 1 All X's, Y's 0 Discount Store 1 is DEA efficient. Chapter 3- 86 For Discount Store 2 MAX 10.3Y1 + 17.9Y2 + 15.5Y3 + 4.6Y4 S.T. 9.6Y1 + 20.3Y2 + 16.3Y3 + 4.1Y4 32987X1 + 275X2 + 23876X3 10.3Y1 + 17.9Y2 + 15.5Y3 + 4.6Y4 32987X1 + 215X2 + 28755X3 14.5Y1 + 55.2Y2 + 27.4Y3 + 22.3Y4 54321X1 + 185X2 + 19000X3 15.2Y1 + 44.8Y2 + 26.8Y3 + 28.4Y4 54321X1 + 180X2 + 18750X3 12.5Y1 + 45.9Y2 + 19.9Y3 + 35.1Y4 99765X1 + 85X2 + 11000X3 32987X1 + 215X2 + 28755X3 = 1 All X's, Y's 0 From worksheet DISC 2 (not shown), discount store 2 is also efficient. For Department Store 1 MAX 14.5Y1 + 55.2Y2 + 27.4Y3 + 22.3Y4 S.T. 9.6Y1 + 20.3Y2 + 16.3Y3 + 4.1Y4 32987X1 + 275X2 + 23876X3 10.3Y1 + 17.9Y2 + 15.5Y3 + 4.6Y4 32987X1 + 215X2 + 28755X3 14.5Y1 + 55.2Y2 + 27.4Y3 + 22.3Y4 54321X1 + 185X2 + 19000X3 15.2Y1 + 44.8Y2 + 26.8Y3 + 28.4Y4 54321X1 + 180X2 + 18750X3 12.5Y1 + 45.9Y2 + 19.9Y3 + 35.1Y4 99765X1 + 85X2 + 11000X3 54321X1 + 185X2 + 19000X3 = 1 All X's, Y's 0 From worksheet DEPT 1 (not shown), department store 1 is also efficient. For Department Store 2 MAX 15.2Y1 + 44.8Y2 + 26.8Y3 + 28.4Y4 S.T. 9.6Y1 + 20.3Y2 + 16.3Y3 + 4.1Y4 32987X1 + 275X2 + 23876X3 10.3Y1 + 17.9Y2 + 15.5Y3 + 4.6Y4 32987X1 + 215X2 + 28755X3 14.5Y1 + 55.2Y2 + 27.4Y3 + 22.3Y4 54321X1 + 185X2 + 19000X3 15.2Y1 + 44.8Y2 + 26.8Y3 + 28.4Y4 54321X1 + 180X2 + 18750X3 12.5Y1 + 45.9Y2 + 19.9Y3 + 35.1Y4 99765X1 + 85X2 + 11000X3 54321X1 + 180X2 + 18750X3 = 1 All X's, Y's 0 From worksheet DEPT 2 (not shown), department store 2 is also efficient. For the Upscale Store MAX 12.5Y1 + 45.9Y2 + 19.9Y3 + 35.1Y4 S.T. 9.6Y1 + 20.3Y2 + 16.3Y3 + 4.1Y4 32987X1 + 275X2 + 23876X3 10.3Y1 + 17.9Y2 + 15.5Y3 + 4.6Y4 32987X1 + 215X2 + 28755X3 14.5Y1 + 55.2Y2 + 27.4Y3 + 22.3Y4 54321X1 + 185X2 + 19000X3 15.2Y1 + 44.8Y2 + 26.8Y3 + 28.4Y4 54321X1 + 180X2 + 18750X3 12.5Y1 + 45.9Y2 + 19.9Y3 + 35.1Y4 99765X1 + 85X2 + 11000X3 99765X1 + 85X2 + 11000X3 = 1 All X's, Y's 0 From worksheet Upscale (not shown), the upscale store 2 is also efficient. Chapter 3- 87

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