Chapter 3
Problem Summary
Prob. # / Concepts Covered / Level of Difficulty / Notes3.1 / Maximization production model / 2
3.2 / Maximization model with "", "=", "" constraints / 5
3.3 / Maximization production model, percentage constraints, interpretation of shadow prices and range of feasibility / 4
3.4 / Maximization financial mix model, percentage constraint, determining current rate of return and future rate of return, shadow prices / 5
3.5 / Maximization model, calculation of profit coefficients, sensitivity analyses / 4
3.6 / Minimization model, determining correct RHS / 4
3.7 / Maximization model, sensitivity analyses, evaluation of quantities outside the ranges of optimality and feasibility, many constraints / 6
3.8 / Maximization production model, analysis of why a high profit item is not produced, sensitivity analyses, evaluation of purchasing additional resources / 6
3.9 / Minimization model with 8 variables and 14 functional constraints / 6
3.10 / Minimization diet problem, alternate optimal solutions, interpretation of shadow prices, the effects of deleting a constraint / 6
3.11 / Minimization model / 3
3.12 / Maximization model, re-evaluation of RHS / 4
3.13 / Financial maximization model, weighted averages, sensitivity analyses / 6
3.14 / Multiperiod production model / 9
3.15 / Maximization agriculture model, sensitivity analyses, evaluation of an alternative based on shadow prices / 7
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, shadow prices, determining the acceptability of the solution / 6
3.20 / Maximization model with 9 functional constraints / 4
3.21 / Minimization transportation model / 5
3.22 / Classic trim loss model, definition of variables / 5
3.23 / Maximization advertising model, checking the linear programming assumptions, elimination of a nonbinding constraint / 5
3.24 / Maximization retailing model, evaluating changes to objective function coefficients, the profitability of added resources, and the addition of a constraint / 5
3.25 / Large design model, "tricky" constraints / 9
3.26 / Maximization production problem, calculation of objective function coefficients, calculating coefficients used in the model / 7
3.27 / Maximization advertising model, sensitivity analyses, adding constraints, changing the RHS values / 8
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, 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
Case 3.1 / Maximization production model with input in Excel, many percentage constraints, analysis of output / 8
Case 3.2 / Maximization integer model, definitional variables, percentage constraints, rounding the linear model, "slightly" violated constraints, time issues / 10 / Solved as a linear program -- takes just a few seconds
Solved as an integer program -- takes many minutes
Case 3.3 / Binary models / 7
Case 3.4 / “On the job training” model with many “linking” constraints / 9
Case 3.5 / Maximization financial mix model, linking results between two periods, percentage constraints, analysis of output / 10
Case 3.6 / Fixed charge model / 4
Problem Solutions
3.1See 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
MAX27X1 + 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
3.2See 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
MAX110X1 + 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.
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
3.3See 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
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.
3.4See 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
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%.
3.5See 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 daily26-4(.50)-75(.20) = 9.00
X3 = Number of king comforters produced daily32-6(.50)-95(.20) = 10.00
MAX6.50X1 + 9.00X2 + 10.00X3
S.T.
3X1 + 4X2 + 6X3 2,700(Stuffing)
55X1 + 75X2 + 95X348,000(Fabric)
3X1 + 5X2 + 6X3 3,000(Cutting minutes)
5X1 + 6X2 + 8X312,000(Sewing minutes)
All X's 120
3.6See 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 85 (=100-15 grams of protein)
1X2 + 22X3 + 12X4 25 (= 45-20 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
3.7See 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.
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
MAX800X1 + 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.
3.9See 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
MIN15X1 + 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:
TelephonePersonal
I 500 500
II 600 0
III 200 0
IV 200 0
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
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:
.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 Nuts1.0 oz. Grape Nuts
.47 oz. Product 191.0 oz. Product 19
Total 2.00 oz. Total 2.0 oz.
3.11See file Ch3.11.xls
X1 = Number of refrigerator/ovens produced
X2 = Number of French fry makers produced
X3 = Number of French toast makers produced
MIN140X1 + 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.
3.12See 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
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
3.13See 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:
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:
$30,240,000.00First Trust Deeds
$ 66,666.67Second Trust Deeds
$20,093,333.33Third Trust Deeds
$12,600,000.00Commercial Trust Deeds
$ 5,000,000.00Savings 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%
Output:
3.14See 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
Objective Function
MIN188XJR + 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
The optimal solution is shown in the cells C6:D8, G6:H8, and K6:L8. Total Cost = $367,969
3.15See 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.
3.16See file Ch3.16.xls
PSA = Number of professional sets produced in Sarasota and shipped to Anaheim