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Chapter 4 - Sensitivity Analysis & The Simplex Method : S-1—————————————————————————————————————————————

Chapter 4 Sensitivity Analysis & The Simplex Method

Excel Solutions1. Howie's numbers are correct. Factory overhead is not a variable cost -- it is a sunk cost that must be paid

regardless of which type of hot tub is produced.

2. By contradiction, suppose that a variable assumes an optimal value that is somewhere between its upper and lower limits and that the reduced cost for this variable is not zero. This implies that if this variable is increased or decreased by one unit, the absolute value of the change in the objective function will be equivalent to the absolute value of the reduced cost. Increasing the value of this variable by one unit will either increase or decrease the value of the objective function while decreasing the value of the variable will have the opposite effect. Thus, by increasing or decreasing the value of this variable the objective function can be improved. Therefore, the solution could not have been optimal.

3. See file: Prb4_3.xlsm

Microsoft Excel Sensitivity Report (Obtained Using Standard LP/Quadratic Engine)

Adjustable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$C$4 Value: X1 5 0 4 1E+30 2.8$D$4 Value: X2 0 -4.666666667 2 4.666666667 1E+30

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$E$8 Used: 10 0 20 1E+30 10$E$9 Used: 15 1.333333333 15 15 15

a. The objective function coefficient for X1 can decrease by 2.8 (to 1.2) or increase by any amount without changing the optimal solution.

b. The optimal solution is unique. None of the allowable increase or decrease values for the objective coefficients are zero.

c. It has to increase by at least 4.666667d. If X2 were forced to equal 1, the optimal objective function value would be approximately 20 - 4.67

= 15.33.e. An increase of 10 in the RHS value of the second constraint is within the allowable range of increase

for the shadow price of this constraint. Therefore, if the RHS for the second constraint increases from 15 to 25 the new objective function value would be approximately 20 + 1.33 ×10=33.33.

f. The new reduced cost for X2 would be 2 - (4 × 0 + 1 × 1.333) = 0.67. Therefore, it would be profitable to increase the value of X2 and the current solution would no longer be optimal – but we can’t say what the new solution would be without re-solving the problem.

http://mihaylofaculty.fullerton.edu/sites/zgoldstein/Solutions/MS/Ragsdale/C04.zip





Cell Name Value Cost Coefficient Increase Decrease$C$4 Value: X1 12 0 2 1E+30 0$D$4 Value: X2 0 0 4 0 1E+30


Cell Name Value Price R.H. Side Increase Decrease$E$10 Used: 12 0 2 10 1E+30$E$8 Used: -12 0 8 1E+30 20$E$9 Used: 12 2 12 1E+30 10

a. Constraint 2 is binding. (Note that this constraint is reflected by the third row in the Constraints section of the report.)

b. There is an alternate optimal solution. Variable X2 is at its lower bound (of zero) but also has a reduced cost of zero. This indicates that the value of X2 could be increased while having zero impact on the optimal objective function value.

c. There is no way to answer this question directly from the sensitivity report. We know that the optimal solution would change since the allowable decrease in the objective function value of X1 is zero, but we cannot tell what the new optimal solution would be. If we re-solve the revised model the solution is X1=2, X2=5.

d. It can decrease by any amount without changing the solution.e. Constraint 2 is the only binding constraint. Therefore, we would want to increase its RHS value

before any other.




Cell Name Value Cost Coefficient Increase Decrease$C$4 Value: X1 0 3 5 1E+30 3$D$4 Value: X2 0 1 3 1E+30 1$E$4 Value: X3 1 0 4 2 4


Cell Name Value Price R.H. Side Increase Decrease$F$8 Produced: 2 2 2 1E+30 1$F$9 Produced: 2 0 1 1 1E+30

a. The allowable decrease is 4, so the smallest value of the objective for X3 without changing the solution is 0.

b. The new objective would be unbounded.


c. Increasing the RHS of the first constraint reduces the feasible region. Therefore, the objective function would be increased (made worse) at a rate of 2 per unit increase in the RHS value. Thus, the new objective function value would be 4 + 2 × 5 = 14.

d. 4 - 2×1 = 2e. Yes, 1/3 + 1/4 = 7/12 < 100%

6. a. MIN: 260X13 + 220X14 + 290X15 + 230X23 + 240X24 + 310X25

S.T.:X13 + X14 + X15 20X23 + X24 + X25 20X13 + X23 10X14 + X24 15X15 + X25 10Xij 0

b. See file: Prb4_6.xlsmc. See below (Total cost = $8,600)



Cell Name Value Cost Coefficient Increase Decrease$C$10 Eustis Miami 0 50 260 1E+30 50$D$10 Eustis Orlando 10 0 220 20 0$E$10 Eustis Tallahassee 10 0 290 0 310$C$11 Clermont Miami 10 0 230 50 230$D$11 Clermont Orlando 5 0 240 0 20$E$11 Clermont Tallahassee 0 0 310 1E+30 0


Cell Name Value Price R.H. Side Increase Decrease$C$12 Shipped Miami 10 230 10 5 10$D$12 Shipped Orlando 15 240 15 5 5$E$12 Shipped Tallahassee 10 310 10 5 5$F$10 Eustis Used 20 -20 20 5 5$F$11 Clermont Used 15 0 20 1E+30 5

d. No.e. No, alternate optima exist.

CapacityMiami Orlando Tallahassee Used Available

Eustis 0 15 5 20 20Clermont 10 0 5 15 20Shipped 10 15 10Demand 10 15 10

Total Cost $8,600f. The solution would not change. The current solution uses only 15 of the 20 tons of capacity

available at Clermont.


g. Reducing the capacity in Eustis would increase costs by $20 per unit yielding an objective function value of $8,600+20×5 = $8,700

h. Every additional ton of concentrate shipped from Eustis to Miami would increase costs by $50.


Total Profit $32,500



Cell Name Value Cost Coefficient Increase Decrease$C$4 Generators 130 0 250 1E+30 150$D$4 Alternators 0 -225 150 225 1E+30


Cell Name Value Price R.H. Side Increase Decrease$E$8 Wiring: 260 125 260 20 260$E$9 Testing: 130 0 140 1E+30 10

a. There is 0 excess wiring capacity and 10 excess units of testing.b. $32,500 + $125×10 = $33,750c. $225d. Using the 100% rule: 50/150 + 75/225 = 2/3 < 100%. The solution would not change.e. Let X = the increase in the price of alternators. We can be sure the optimal solution will not change

as long as 25/150 + X/225 < 1 or X < 225(125/150) = 187.5. So the maximum profit on alternators would be $150+$187.5 = $337.5.

f. The new reduced cost on alternators would be 150 - 125×1.5 - 0×2 = -37.5. Thus, it would still be unprofitable to produce alternators and the solution would not change.


Microsoft Excel Sensitivity Report (Obtained Using Standard LP/Quadratic Engine)Adjustable Cells

Final Reduced Objective Allowable AllowableCell Name Value Cost Coefficient Increase Decrease

$C$5 Razors 240 0 70 10 40$D$5 Zoomers 420 0 40 53.33 5


Cell Name Value Price R.H. Side Increase Decrease$E$9 Polymer Used 900 32 900 200 300$E$10 Production Time Used 2400 2 2400 200 800$E$11 Total Production Used 660 0 700 1E+30 40$E$12 Product Mix Used -180 0 300 1E+30 480

a. No, it could decrease by $40 without changing the solution.b. It could. At a profit of $35 there could be alternate optimal solutions.


c. Each additional unit of polymer (up to 200 more) would increase profit by $32 per unit, each unit decrease of polymer (up to 300 less) would decrease profit by $32 per unit.

d. This is a non-binding constraint. At the optimal solution, this constraint is 40 units below its RHS value (700) so changing the constraint RHS value (within its allowable limits) would not change the solution or allow for additional profit.

e. The allowable increase on the shadow price for labor is 200 units. So we can’t say what profit would result from a 300 unit increase in labor without re-solving the problem.


Total Profit = $26,740



Cell Name Value Cost Coefficient Increase Decrease$B$19 Acres to Plant Watermelons 60 0 256 28.5 66.33333333$C$19 Acres to Plant Cantaloupes 40 0 284.5 99.5 28.5


Cell Name Value Price R.H. Side Increase Decrease$F$11 Water Used per hour) 6,000 1 6000 1500 1000$F$19 Acres Planted per hour) 100 199 100 20 20

a. The profit per acre for watermelons can drop by $66.33 (to $189.67 per acre). That would occur if the price of watermelons dropped to about $2.26 per unit.

b. The profit per acre of cantaloupes would have to increase by $99.50 ($384 per acre). That would occur if the price of cantaloupes dropped to about $1.33 per unit.

c. 60/66.33 + 50/99.5 = 1.407 > 100%. Therefore we cannot guarantee that the solution is still optimal.d. The farmer should lease all 20 acres and be willing to pay up to $199 per acre (assuming he can use

his own water on the additional 20 acres or otherwise water it as needed for free).





Cell Name Value Cost Coefficient Increase Decrease$C$5 Qty Doors 20 0 500 300 233.3333$D$5 Qty Windows 40 0 400 350 150


Cell Name Value Price R.H. Side Increase Decrease$E$9 Cutting Used 40 350 40 40 13.33333


$E$10 Sanding Used 40 300 40 6.666667 20$E$11 Finishing Used 50 0 60 1E+30 10

a. No, the objective coefficient on doors could increase by $300 (to $800) without changing the optimal solution.

b. Yes, the objective coefficient on windows can only decrease by $150 (to $250) without changing the optimal solution.

c. The shadow price (marginal value) is $0 because there is a surplus of this resource.d. $350*20=$7000.e. The shadow price on sanding is $300 / unit (within the allowable increase/decrease). The allowable

decrease is 20 units. So if another company was willing to pay $400 for 15 hours of capacity, Sanderson would come out ahead by $1,500 (i.e., (400-300)*15). It would be profitable to Sanderson to offer up to 20 hours of their sanding capacity for this purpose. If the other company wanted 25 hours of capacity, the LP model would need to be modified to see if this would be profitable. (And that analysis shows it would not be economicallybeneficial to do so as total profit would drop from $26,000 to $25,000.)


Total cost = $453,300



Cell Name Value Cost Coefficient Increase Decrease$B$6 - Make Model 1 3,000 0.00 50 4 57$C$6 - Make Model 2 550 0.00 83 14 8$D$6 - Make Model 3 900 0.00 130 8 137$B$7 - Buy Model 1 0 4.00 61 1E+30 4$C$7 - Buy Model 2 1,450 0.00 97 8 14$D$7 - Buy Model 3 0 8.00 145 1E+30 8


Cell Name Value Price R.H. Side Increase Decrease$B$13 # Available Model 1 3,000 57.00 3000 380 2900$C$13 # Available Model 2 2,000 97.00 2000 1E+30 1450$D$13 # Available Model 3 900 137.00 900 211.1111 900$E$17 - Wiring Used 9,525 0.00 10000 1E+30 475$E$18 - Harnessing Used 5,000 (7.00) 5000 633.3333 1100

a. No.b. The cost of making model 1 slip rings can increase by $4 without changing the solution.c. No. The allowable decrease on the objective coefficient for buying model 2 slip rings is $14.d. No. There is presently a surplus of 475 hours in the wiring department. Overtime would only add to

this surplus.e. Yes. Harnessing represents a binding constraint with a shadow price of -$7. Each additional unit of

this resource (up to 633.33) will reduce costs by $7. Since workers are paid an additional $6 per hour for overtime, the company could save $633.33.

f. See spider chart in the file. Costs are most sensitive to reductions in the harnessing requirements for model 1 slip rings.



Total Revenue = $220,290



Cell Name Value Cost Coefficient Increase Decrease$C$4 Units Produced Country 405.80 0.00 350 1E+30 50$D$4 Units Produced Contemporary 173.91 0.00 450 75 1266.666667


Cell Name Value Price R.H. Side Increase Decrease$C$4 Units Produced Country 405.80 0.00 0 289.8550725 1E+30$D$4 Units Produced Contemporary 173.91 -65.22 0 238.0952381 114.8148148$E$8 Router Used 956.52 0.00 1000 1E+30 43.47826087$E$9 Sander Used 2,000.00 110.14 2000 90.90909091 2000$E$10 Polisher Used 1,275.36 0.00 1500 1E+30 224.6376812

a. No. There is already a surplus of routing capacity.b. Yes. They should be will to pay $110.14 per unit (above and beyond what they already pay) for up to

~90 additional units.c. Nothing. There is already a surplus of polishing capacity.d. The price would have to decrease by more than $1266.67. Currently, the company only makes $450

on this item. So this price reduction would actually result in the company selling this item at a loss. The constraint requiring at least 30% of the production to consist of contemporary tables creates this anomaly.


Microsoft Excel Sensitivity Report (Obtained Using Standard LP/Quadratic Engine) Decision Variable Cells


$B$6 Tahoe 23.57142857 0.00 450 300.0000007 50.00000007$C$6 Pacific 15 0.00 1150 42.85714296 714.2857144$D$6 Savannah 40.71428571 0.00 800 100.0000001 16.66666682$E$6 Aspen 0 -7.142857143 400 7.142857143 1E+30


Cell Name Value Price R.H. Side Increase Decrease$F$10 Glue Used 6,000 0.428571 6000 1500 2590.909091$F$11 Pressing Used 7,500 7.142857 7500 3300 1500$F$12 Pine chips Used 30,000 0.142857 30000 27333.33333 8250$F$13 Oak chips Used 33,214 0.000000 62500 1E+30 29285.71429


a. No.b. Yes.c. No, the price on Tahoe panels can go down by $50 without changing the optimal solution.d. Yes, the allowable increase on the profit coefficient for pallets of Aspen panels is $7.14, so the

solution would change.e. This is a binding constraint with a shadow price of ~$7.14 and an allowable increase of 3,300. They

should be willing to pay ~$7143 for 1,000 additional units of pressing capacity.f. Pine chips are a binding constraint worth ~$0.14 per pound. Reducing the amount available by 5,000

pounds would reduce profit by about $714. If they could sell the same 5,000 pounds for $1,250 they should do so as they would be ahead by $1250 - $714.

g. Profit is most sensitive to changes in pressing capacity.

h. As pressing capacity increases from 5000 to 6000, we make fewer Tahoes and more Savannahs. Beyond 6000, we start making Pacifics and fewer Savannahs.

Pressing Tahoe Pacific Savannah Aspen Profit$G$11 $B$6 $C$6 $D$6 $E$6 $F$7

5,000 42.85714286 0 28.57142857 0 $42,1435,500 38.57142857 0 35.71428571 0 $45,9296,000 34.28571429 0 42.85714286 0 $49,7146,500 30.71428571 5 42.14285714 0 $53,286

7,000 27.14285714 10 41.42857143 0 $56,857



Decision Variable Cells

Cell NameFinalValue

ReducedCost

ObjectiveCoefficient

AllowableIncrease

AllowableDecrease

$C$11 Feed 1 15 -3.250 2 3.25 1E+30

$D$11 Feed 2 9.5 0.000 2.5 1E+30 1.500000033


$E$11 Feed 3 8.5 0.000 3 4.5000001 6.5000002

Constraints

Cell NameFinalValue

ShadowPrice

ConstraintR.H. Side

AllowableIncrease

AllowableDecrease

$F$6 Min A 98 0.00 64 34 1E+30

$F$7 Min B 80 2.50 80 5.5 9

$F$8 Min C 32 0.00 16 16 1E+30

$F$9 Min D 200 0.00 128 72 1E+30

$F$6 Max A 98 0.00 128 1E+30 30

$F$7 Max B 80 0.00 160 1E+30 80

$F$8 Max C 32 -2.25 32 6.333333333 3.666666667

$F$9 Max D 200 0.00 256 1E+30 56

a. No.b. Yes.c. Feed 1 is at its upper bound, increasing it further (if possible) would reduce the objective function;

hence its negative reduced cost. Feeds 2 and 3 are between the upper and lower bounds and thefore have reduced costs of zero. (If there reduced costs were not zero, that would indicate further imporvements could be made in the objective by changing their values.)

d. A cost increase of $3 for Feed 3 is within the range of its allowable increase; so the optimal solution would not change but the objective function would change (increase) by $3*8.5=$25.50.

e. If possible, they should reduce the minimum requirement for nutrient B. It is binding at its lower bound and loosening that requirement would improve the objective by $2.5 per unit loosened (up to 9 units).

f. If possible, they should increase the maximum requirement for nutrient C. It is binding at its upper bound and loosening that requirement would improve the objective by $2.25 per unit loosened (up to 6.333 units).

g. All the costs coefficients vary in a positive linear fashion with total cost with changes in the cost of Feed 1 having the greatest impact.




Objective Cell (Max)

Cell NameFinal Value

$F$6 Total Profit 6925


Final Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease

$B$5 Quantity Product 1 6.67 0.00 90 8.33333E-09 3.7500006

$C$5 Quantity Product 2 41.25 0.00 120 6.6666678 1.11111E-08

$D$5 Quantity Product 3 9.17 0.00 150 30.000005 30.000003

Constraints

Final Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease

$E$9 Machine 1 Hrs Used 105 0 400 1E+30 295

$E$10 Machine 2 Hrs Used 240 15 240 6.66666667 18.3333333

$E$11 Machine 3 Hrs Used 320 0 320 3.33333333 27.5

$E$12 Total Labor Hrs Used 665 5 665 50.4166667 5


a. No. b. The shadow price is $0 indicating that additional units of this resource will not help increase profit as this

is a non-binding constratint.c. The shadow price is $5, indicating that additional units of labor would help to increase profits provided

they can be obtained at no more than $5 per hour above the current labor cost.d. The allowable decrease on the objective function coefficient for product 2 is $0. So if that coefficient is

an estimate it has no “wiggle room” in the downward direction without affecting the optimal solution.



Objective Cell (Min)

Cell NameFinal Value

$H$7 Cost / lb Total 5.85


Final Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease

$C$6 Amount (lbs) Raisins 0.75 0 2.5 0.2000000 1E+30

$D$6 Amount (lbs) Grain 0.1 0 1.5 1E+30 0.500000

$E$6 Amount (lbs) Chocolate 0.1 0 2 1E+30 0.500000

$F$6 Amount (lbs) Peanuts 0.95 0 3.5 1E+30 0.333333

$G$6 Amount (lbs) Almonds 0.1 0 3 1E+30 2.500000

Constraints

Final Shadow Constraint Allowable Allowable


$H$10 Vitamins (grams) Total 47.5 0 40 7.5 1E+30

$H$11 Minerals (grams) Total 15 0.5 15 0.1 0.5

$H$12 Protien (grams) Total 12.9 0 10 2.9 1E+30

$H$13 Calories Total 738.5 0 600 138.5 1E+30

$H$6 Amount (lbs) Total 2 -0.825 2 0.068965 0.013245

$C$6 Amount (lbs) Raisins 0.75 0 0 0.65 1E+30

$D$6 Amount (lbs) Grain 0.1 0.5 0 0.033333 0.1

$E$6 Amount (lbs) Chocolate 0.1 0.5 0 0.05 0.125

$F$6 Amount (lbs) Peanuts 0.95 0 0 0.85 1E+30

$G$6 Amount (lbs) Almonds 0.1 2.5 0 0.025 0.0833333

$C$6 Amount (lbs) Raisins 0.75 0 0 1E+30 0.25

$D$6 Amount (lbs) Grain 0.1 0 0 1E+30 0.9


$E$6 Amount (lbs) Chocolate 0.1 0 0 1E+30 0.9

$F$6 Amount (lbs) Peanuts 0.95 0 0 1E+30 0.05

$G$6 Amount (lbs) Almonds 0.1 0 0 1E+30 0.9

a. No.b. Yes, it has a maximum allowable increase of $0.20 (or to $2.70).c. No, it has a maximum allowable decrease of $0.33 (or to $3.17).d. The constraint on the mineral content of the mix is binding and, therefore, would lead to a reduction

in the price of the mix if this constraint could be loosened.





Cell Name Value Cost Coefficient Increase Decrease$B$4 Quantity Made HyperLink 500 -43.66666667 53 43.66666667 1E+30$C$4 Quantity Made FastLink 1000 0 48 16 1E+30$D$4 Quantity Made SpeedLink 1500 0 33 15 1$E$4 Quantity Made MicroLink 2250 0 32 1 5.272727273$F$4 Quantity Made EtherLink 500 -9.666666667 38 9.666666667 1E+30


Cell Name Value Price R.H. Side Increase Decrease$G$13 PC Board (sq in) Used 60,500 0 80000 1E+30 19500$G$14 Resistors Used 100,000 0 100000 10500 6000$G$15 Memory chips Used 30,000 7.5 30000 2000 2333.333333$G$16 Assembly Hours Used 3,688 0 5000 1E+30 1312.5$G$17 Fast/Hyper>2 Used 0 16 0 500 875

a. The constraints on the number of resistors and memory chips are binding. Also the constraint requiring twice as many Fastlink cards as Hyperlink cards is binding.

b. Hyperlinks. The company would earn $43.67 for each Hyperlink it could avoid producing.c. Yes. Profits would increase by $7.5×1,000=$7,500.d. The solution would change if the objective coefficients on the Speedlink decreased by $1 or if the

objective coefficient on the Microlink increased by $1. Thus, the objective coefficients on these products would be of greatest concern.

e. See the file. Total profit appears to be most sensitive to changes in the unit profit for Microlinks.


Total Cost = $2,9750,000




Cell Name Value Cost Coefficient Increase Decrease$B$4 Number to Make Electric 30,000 0 55 7.000000003 1E+30$C$4 Number to Make Gas 10,000 0 85 10 14.00000001$B$5 Number to Buy Electric 0 7 67 1E+30 7.000000003$C$5 Number to Buy Gas 5,000 0 95 14.00000001 10


Cell Name Value Price R.H. Side Increase Decrease$B$10 Total Available Electric 30,000 60 30000 20000 10000$C$10 Total Available Gas 15,000 95 15000 1E+30 5000$D$14 Production Used 10,000 -25 10000 800 4000$D$15 Assembly Used 14,000 0 15000 1E+30 1000$D$16 Packaging Used 4,000 0 5000 1E+30 1000

a. The current solution remains optimal until the cost of producing electric trimmers increases by $7 to $62.

b. The solution would not change. The allowable increase is $10.c. The shadow price in the assembly area is zero. There is already a surplus of assembly capacity so

the company should not pay anything to acquire more of it.d. The company should be willing to pay $25 per production hour above and beyond what it is already

paying. Production capacity is a binding constraint.e. See the file. Total profit is most sensitive to the cost of making the electric trimmer.f. Every 100 unit increase in production capacity results in 250 more gas trimmer being made instead

of bought.


Microsoft Excel Sensitivity Report (Obtained Using Standard LP/Quadratic Engine)Objective Cell (Max)

Cell Name Final Value$C$13 Total profit Baskets 1526.5

Decision Variable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$C$6 Baskets 3 0 2.5 1.59502E-07 2.23607E-07$D$6 Juice 87 0 1.75 2.23607E-07 1.59502E-07$C$7 Baskets 0 0 2.5 0 1E+30$D$7 Juice 225 0 1.75 1E+30 2.11058E-07$C$8 Baskets 114 0 2.5 1.11803E-07 5.31672E-08$D$8 Juice 186 0 1.75 5.31672E-08 1.11803E-07$C$9 Baskets 0 7.08329E-16 2.5 0 1E+30$D$9 Juice 100 0 1.75 1E+30 7.97508E-08$C$10 Baskets 75 0 2.5 1E+30 2.23607E-07$D$10 Juice 0 -7.4476E-16 1.75 7.4476E-16 1E+30

Constraints


Final Shadow Constraint Allowable AllowableCell Name Value Price R.H. Side Increase Decrease

$C$17Actual Wghtd Grade Baskets 720 -0.6 0 15 310

$D$17Actual Wghtd Grade Juice 1495 -0.6 0 103.6363636 10

$E$6 Total 90 0.85 90 69.09090909 6.666666667$E$7 Total 225 1.45 225 207.2727273 20$E$8 Total 300 2.05 300 20 207.2727273$E$9 Total 100 2.65 100 6.666666667 69.09090909$E$10 Total 75 3.25 75 248 12

a. $1,526.5b. Yes. Some students might already have this solution. You tend to get a different (alternate) optimal

solution depending on whether you use the Standard LP engine or the Gurobi engine. Either way, to make sure the solver is using the most grade 5 oranges for fruit baskets, students should try to maximize that quantity while holding the optimal profit at $1526.5 as a constraint.

Pounds used forGrade Baskets Juice

1 0 902 61.5 163.53 0 3004 55.5 44.55 75 0

c. Yes. The shadow price on grade 4 oranges is $2.65 with an allowable increase in the resource of 6.6667 (in 1000s).

d. See file. Increases in the requirements for the fruit baskets have the least negative impact on profit.



Objective Cell (Max)Cell Name Final Value

$E$16Total Profit 5012.5



$E$5 Regular 150.00 0.00 3.75 1E+30 0.0000001$F$5 Supreme 0.00 0.00 7.75 0 1E+30$E$6 Regular 95.65 0.00 5.25 2.09091E-08 2.0000001$F$6 Supreme 254.35 0.00 9.25 1E+30 2.09091E-08$E$7 Regular 54.35 0.00 3.25 1.76923E-07 2.09091E-08


$F$7 Supreme 195.65 0.00 7.25 2.09091E-08 1.76923E-07



$E$11Actual Octane Level Regular 97.6667 0.0000 90 7.666666667 1E+30

$F$11Actual Octane Level Supreme 97.0000 0.0000 97 2.777777778 4.888888889

$E$8 Available Regular 300 3.25 300 50 20.90909091

$F$8Available Supreme 450 7.25 450 20 11.36363636

$G$5 Used 150 0.5 150 54.34782609 50$G$6 Used 350 2 350 54.34782609 50$G$7 Used 250 0 300 1E+30 50

a. Yes, as reflected by the allowable increases and decreases of zero on an objective coefficients.b. Note that the sensitivity information cannot be used to answer this (and the following) questions.

Add a constraint holding the objective at its optimal value and then maximize E11 and F11 separately. Regular octane rating = 97.67, supreme octane rating = 97.

c. Regular octane rating = 90.0, supreme octane rating = 102.11.d. The solution in part c should be implemented. The company would receive negative publicity if the

answer in part b were implemented and the public learned that the inexpensive regular gas had a higher octane rating than the more expensive supreme gas.

e. This cannot be answered from the given information. Based on the above sensitivity report, they should be willing to buy an additional ~54 barrels at this price, but beyond that the shadow price it not guaranteed to hold.


Total Cost = $730



Cell Name Value Cost Coefficient Increase Decrease$C$10 Location 1 Location 3 0 26 54 1E+30 26$D$10 Location 1 Location 4 10 0 17 5 1E+30$E$10 Location 1 Location 5 1 0 23 2 5$F$10 Location 1 Location 6 5 0 30 5 2$C$11 Location 2 Location 3 9 0 24 2 5$D$11 Location 2 Location 4 0 5 18 1E+30 5$E$11 Location 2 Location 5 9 0 19 5 2$F$11 Location 2 Location 6 0 5 31 1E+30 5


Cell Name Value Price R.H. Side Increase Decrease$C$12 Actual Location 3 9 0 10 1E+30 1$D$12 Actual Location 4 10 -11 10 1 1$E$12 Actual Location 5 10 -5 10 4 1


$F$12 Actual Location 6 5 0 10 1E+30 5$C$12 Location 3 Received 9 0 5 4 1E+30$D$12 Location 4 Received 10 0 5 5 1E+30$E$12 Location 5 Received 10 0 5 5 1E+30$F$12 Location 6 Received 5 2 5 1 1$G$10 Location 1 Shipped 16 28 16 1 1$G$11 Location 2 Shipped 18 24 18 1 4

a. Yes. None of the allowable increases or decreases on the objective function coefficients are zero.b. Location 6.c. $26d. The allowable increase for the RHS of the last constraint is 1 so the shadow price of this constraint

will not hold if the RHS is increased from 5 to 8. Thus, this question cannot be answered with the information available in this sensitivity report.


Total Cost = $702



Cell Name Value Cost Coefficient Increase Decrease$C$10 Location 1 Location 3 0 31 54 1E+30 31$D$10 Location 1 Location 4 10 0 17 5.0000001 0.0000001$E$10 Location 1 Location 5 0 5 23 1E+30 5$F$10 Location 1 Location 6 5 0 30 0.0000001 7.0000001$C$11 Location 2 Location 3 8 0 24 7.0000001 5.0000001$D$11 Location 2 Location 4 0 0 18 0.0000001 5.0000001$E$11 Location 2 Location 5 10 0 19 5.0000001 1E+30$F$11 Location 2 Location 6 0 0 31 1E+30 0


Cell Name Value Price R.H. Side Increase Decrease$C$12 Actual Location 3 8 0 10 1E+30 2$D$12 Actual Location 4 10 -6 10 3 0$E$12 Actual Location 5 10 -5 10 3 2$F$12 Actual Location 6 5 0 10 1E+30 5$C$12 Actual Location 3 8 0 5 3 1E+30$D$12 Actual Location 4 10 0 5 5 1E+30$E$12 Actual Location 5 10 0 5 5 1E+30$F$12 Actual Location 6 5 7 5 3 0$G$10 Location 1 Shipped 15 23 15 0 3$G$11 Location 2 Shipped 18 24 18 2 3

a. No. At least one of the allowable increases or decreases on the objective function coefficients is zero.

b. $31.c. $31.



Total Cost = $3,011,360



Cell Name Value Cost Coefficient Increase Decrease$B$15 Macon Tacoma 0 0.55 38 1E+30 0.55$C$15 Macon San Diego 0 1.8 38.25 1E+30 1.8$D$15 Macon Dallas 0 0.25 37.25 1E+30 0.25$E$15 Macon Denver 0 1.1 37.5 1E+30 1.1$F$15 Macon St. Louis 0 1 37.6 1E+30 1$G$15 Macon Tampa 12000 0 37.3 0.1 0.85$H$15 Macon Baltimore 6000 0 37.15 0.25 0.1$B$16 Louisville Tacoma 600 0 39.35 0.1 0.05$C$16 Louisville San Diego 0 1.05 39.4 1E+30 1.05$D$16 Louisville Dallas 0 0.1 39 1E+30 0.1$E$16 Louisville Denver 0 0.8 39.1 1E+30 0.8$F$16 Louisville St. Louis 14400 0 38.5 0.05 0.15$G$16 Louisville Tampa 0 0.2 39.4 1E+30 0.2$H$16 Louisville Baltimore 0 0.3 39.35 1E+30 0.3$B$17 Detroit Tacoma 400 0 41.3 0.05 0.05$C$17 Detroit San Diego 0 0.95 41.25 1E+30 0.95$D$17 Detroit Dallas 10800 0 40.85 0.1 0.55$E$17 Detroit Denver 12600 0 40.25 0.05 1E+30$F$17 Detroit St. Louis 0 0.05 40.5 1E+30 0.05$G$17 Detroit Tampa 0 0.1 41.25 1E+30 0.1$H$17 Detroit Baltimore 1200 0 41 0.1 0.25$B$18 Phoenix Tacoma 5800 0 38.15 0.05 0.15$C$18 Phoenix San Diego 14200 0 37.15 0.15 0.05$D$18 Phoenix Dallas 0 0.15 37.85 1E+30 0.15$E$18 Phoenix Denver 0 0.9 38 1E+30 0.9$F$18 Phoenix St. Louis 0 0.95 38.25 1E+30 0.95$G$18 Phoenix Tampa 0 0.75 38.75 1E+30 0.75$H$18 Phoenix Baltimore 0 1.05 38.9 1E+30 1.05


Cell Name Value Price R.H. Side Increase Decrease$B$19 Total Shipped Tacoma 6800 0 8500 1E+30 1700$C$19 Total Shipped San Diego 14200 0 14500 1E+30 300$D$19 Total Shipped Dallas 10800 0 13500 1E+30 2700$E$19 Total Shipped Denver 12600 -0.05 12600 400 300$F$19 Total Shipped St. Louis 14400 0 18000 1E+30 3600$G$19 Total Shipped Tampa 12000 0 15000 1E+30 3000$H$19 Total Shipped Baltimore 7200 0 9000 1E+30 1800$B$19 Total Shipped Tacoma 6800 1 6800 1700 300$C$19 Total Shipped San Diego 14200 0 11600 2600 1E+30$D$19 Total Shipped Dallas 10800 0.55 10800 400 300


$E$19 Total Shipped Denver 12600 0 10080 2520 1E+30$F$19 Total Shipped St. Louis 14400 0.15 14400 600 300$G$19 Total Shipped Tampa 12000 0.85 12000 400 300$H$19 Total Shipped Baltimore 7200 0.7 7200 400 300$I$15 Macon Produced 18000 36.45 18000 300 400$I$16 Louisville Produced 15000 38.35 15000 300 600$I$17 Detroit Produced 25000 40.3 25000 300 400$I$18 Phoenix Produced 20000 37.15 20000 300 2600

a. Yes.b. Macon. Each additional unit of capacity there increases costs by $36.45 (which is the cheapest way

to increase capacity).c. Yes. The allowable increase in the objective coefficient for this variable is $0.05. So an $0.08

increase in shipping cost (from $1.90 to $1.98) should (and does) result in a new optimal solution.d. Yes. Several of the “80% constraints” are binding. Relaxing these constraints would allow the

company to distribute the products it can make in a less costly way to meet more of the demand that currently is not being met because of the “80% constraints”. As long as the product sells for the same price in all markets, lowering/relaxing the “80% constraints” should result in more profit.

e. This would cause the total cost to increase by $1 per unit or $1700 in total. So the company would likely want to pass this cost on to the distributor.

24. See file: Prb4_24.xlsmTotal Cost = $44,067.73



Cell Name Value Cost Coefficient Increase Decrease$C$10 Newspaper Newsprint 588.24 0.00 21.5 0.51984127 24.64285714$D$10 Newspaper Packaging 11.76 0.00 26 3.142857143 0.51984127$E$10 Newspaper Print Stock 0.00 18.14 15 1E+30 18.14285714$C$11 Mixed Paper Newsprint 0.00 0.55 25.75 1E+30 0.550420168$D$11 Mixed Paper Packaging 71.43 0.00 28.25 0.550420168 1.357142857$E$11 Mixed Paper Print Stock 428.57 0.00 25.5 1.357142857 26.39285714$C$12 White Office Paper Newsprint 0.00 1.87 23.75 1E+30 1.871848739$D$12 White Office Paper Packaging 300.00 0.00 26.75 1.551020408 1E+30$E$12 White Office Paper Print Stock 0.00 1.55 27.5 1E+30 1.551020408$C$13 Cardboard Newsprint 0.00 1.31 24.5 1E+30 1.306722689$D$13 Cardboard Packaging 397.78 0.00 25.5 1.21484375 0.78125$E$13 Cardboard Print Stock 0.00 17.00 17 1E+30 17


Cell Name Value Price R.H. Side Increase Decrease$C$14 Pulp Available Newsprint 500.00 28.99 500 1.651785714 295.8482143$D$14 Pulp Available Packaging 600.00 36.43 600 1.554621849 278.4453782$E$14 Pulp Available Print Stock 300.00 37.70 300 1.360294118 243.6397059$F$10 Newspaper Used 600.00 -3.14 600 348.0567227 1.943277311$F$11 Mixed Paper Used 500.00 -0.89 500 348.0567227 1.943277311$F$12 White Office Paper Used 300.00 -4.21 300 327.5827978 1.828966881$F$13 Cardboard Used 397.78 0.00 400 1E+30 2.220888355

a. No.


b. Yes.c. $0. There is a surplus of cardboard already.d. Yes. The recycler currently pays $15 per ton. The shadow price is $3.14. So the recycler should be

willing to pay up to $18.14 to acquire up to 348 additional tons.e. Newspaper $28.99, Packaging Paper $36.43, Print Stock Paper $37.70.f. $1.87.g. The yield on newsprint per ton of cardboard would have to increase from 0.0 to 24.499999/28.9915 =

0.845072. (This is the reduced cost of using cardboard for newsprint divided by the marginal value (shadow price) of newsprint.)

First of all, information about changes to constraint coefficients is generally not directly available from the sensitivity report. Usually you'll want to use ad hoc sensitivity analysis (like a Solver table) to answer these types of questions. However, in this case, from the sensitivity report we know that the cost of converting 1 ton of cardboard into 0.8 tons of newsprint is $24.5. Also, from the sensitivity report we know that the marginal value of a ton of newsprint is ~$28.9915. So, if we spend $24.5 to get 0.8 tons of newsprint worth $28.9915 per ton, we'll have exchanged $24.5 for something that is worth 0.8*$28.9915 = ~$23.193 -- so we would have lost ~$1.307 in that transaction. That's why we aren't using cardboard to make newsprint. Also note that ~1.307 is the reduced cost in the sensitivity report for the variable that represents the amount of cardboard being converted to newsprint.

So, in this case, $24.50 - 0.8*$28.9915 turned out to be negative (that is, a bad deal where we lose value). Question g boils down to asking "What would the 0.8 in this equation have to be in order to get a non-negative value?" So let's change the 0.8 factor into a variable called Q. Then, mathematically, we want to find the value of Q where $24.50-Q*$28.9915=0.

Doing a little algebra on that gives us Q = 24.5/28.9915 = 0.845072.

So if we could convert a ton of cardboard into about 0.845 tons of newsprint (instead of 0.8 tons of newsprint), it would become economically feasible to do so.


Total Cost = $664,200



Cell Name Value Cost Coefficient Increase Decrease$C$6 Forward 300 0 69 0 0.0000001

$D$6Forward Commodity 2500 0 50 1E+30 0.0000001

$E$6 Forward 185.7142857 0 60 0.0000001 8.07692E-08$F$6 Forward 0 0 80 0 1E+30$C$7 Center 4500 0 69 1E+30 6.66667E-08$D$7 Center Commodity 0 0 50 0 1E+30$E$7 Center 0 0 60 0 1E+30$F$7 Center 0 0 80 0 1E+30$C$8 Rear 0 0 69 0 1E+30$D$8 Rear Commodity 0 0 50 0 1E+30$E$8 Rear 1014.285714 0 60 0 0.0000001$F$8 Rear 1700 0 80 1E+30 0.0000001



Cell Name Value Price R.H. Side Increase Decrease$C$9 Loaded 4800 69 4800 22.5 300$D$9 Loaded Commodity 2500 50 2500 22.5 2130$E$9 Loaded 1200 60 1200 22.5 354.5454545$F$9 Loaded 1700 80 1700 27.27272727 354.5454545$G$6 Forward weight 2985.714286 0 3000 1E+30 14.28571429$G$7 Center weight 4500 0 6000 1E+30 1500$G$8 Rear weight 2714.285714 0 4000 1E+30 1285.714286$I$6 Forward volume 85642.85714 0 145000 1E+30 59357.14286$I$7 Center volume 180000 0 180000 12000 900$I$8 Rear volume 154357.1429 0 155000 1E+30 642.8571429$D$19 Forward max < 0 0.00 0.00 0 30 22.5$D$20 Center max < 0 -3134.29 0.00 0 1E+30 3134.285714$C$19 Forward min > 0 542.86 0.00 0 542.8571429 1E+30$C$20 Center min > 0 420.00 0.00 0 420 1E+30

a. No.b. No.c. $0. (Note that the allowable decrease for three different adjustable cells need to be considered.) d. Total profit is most sensitive to changes in the marginal profit/price of commodity one.


Total cost = $16,625



Cell Name Value Cost Coefficient Increase Decrease$C$6 15 0 300 175 50$C$7 0 125 300 1E+30 125$C$8 10 0 300 150 50$C$9 0 225 300 1E+30 225$C$10 0 175 300 1E+30 175$D$6 0 50 525 1E+30 50$D$7 0 50 525 1E+30 50$D$8 0 150 525 1E+30 150$D$9 0 325 525 1E+30 325$E$6 Length of Lease (in months) 0 0 775 50 150$E$7 Length of Lease (in months) 0 225 775 1E+30 225$E$8 Length of Lease (in months) 0 275 775 1E+30 275$F$6 5 0 850 125 75$F$7 0 175 850 1E+30 175$G$6 5 0 975 175 125

Constraints


Final Shadow Constraint Allowable AllowableCell Name Value Price R.H. Side Increase Decrease

$H$6 Avail. 25 300 25 1E+30 15$H$7 x Avail. 10 175 10 10 0$H$8 x Avail. 20 300 20 1E+30 10$H$9 x Avail. 10 75 10 0 5$H$10 x Avail. 5 125 5 5 5

a. Yes.b. Because of the degeneracy, we cannot tell from the sensitivity report.c. $425d. The solutions in months 1 and 2 do not change within this region.

27. See file: Prb4_27.xlsma.

$H$21$H$16 1265.47 1273.9 1282.3 1290.8 1299.2 1307.6 1309.9 1309.9 1309.9 1309.9 1309.9

b. 360,000 cf, Total profit ~$1,309,903c. ~$42,140d. At least $105,980

28. a. MAX 4 X1 + 2 X2ST 2 X1 + 4 X2 + S1 = 20

3 X1 + 5 X2 + S2 = 15X1, X2 , S1, S2 0

b. {X1, X2}, {X1, S1}, {X1, S2}, {X2, S1}, {X2, S2}, {S1, S 2}


c.X1 X2 S1 S2 Objective Feasible?

1 -20 15 0 0 -50 no2 5 0 10 0 20 yes3 10 0 0 -15 40 no4 0 3 8 0 6 yes5 0 5 0 -10 10 no6 0 0 20 15 0 yes

e. See table in c.f. X1 = 5, X2 = 0g. The second constraint is binding; the first is redundant.

29. a. MAX 2 X1 + 4 X2ST -1 X1 + 2 X2 + S1 = 8

1 X1 + 2 X2 + S2 = 121 X1 + 2 X2 - S3 = 2X1, X2 , S1, S2, S3 0

b. {X1, X2, S1}, {X1, X2, S2}, {X1, X2, S3}, {X1, S1, S2}, {X1, S1, S3},{X1, S2, S3}, {X2, S1, S2}, {X2, S1, S3}, {X2, S2, S3}, {S1, S2, S3}

c.X1 X2 S1 S2 S3 Objective Feasible?

1 -8 10 -20 0 0 24 no2 -1.33 3.33 0 6.67 0 10.67 no3 2 5 0 0 5 24 yes4 2 0 10 10 0 4 yes5 12 0 20 0 10 24 yes6 -8 0 0 20 -10 -16 no7 0 2 4 8 0 8 yes8 0 6 -4 0 4 24 no9 0 4 0 4 2 16 yes10 0 0 8 12 -2 0 no

e. See table in c.f. There are alternate optimal solutions. One is given by X1 = 2, X2 = 5, S3 = 5.g. The first and second constraints are binding.

30. a. MIN 5 X1 + 3 X2 + 4 X3ST X1 + X2 + 2 X3 - S1 = 2

5 X1 + 3 X2 + 2 X3 - S2 = 1X1, X2 , X3, S1, S2 0

b. {X1, X2}, {X1, X3}, {X1, S1}, {X1, S2}, {X2, X3},{X2, S1}, {X2, S2}, {X3, S1}, {X3, S2}, {S1, S2}

c. X1 X2 X3 S1 S2 Objective Feasible?

1 -2.5 4.5 0 0 0 1 no2 -0.25 0 1.125 0 0 3.25 no


3 0.2 0 0 -1.8 0 1 no4 2 0 0 0 9 10 yes5 0 -0.5 1.25 0 0 3.5 no6 0 0.33 0 -1.67 0 1 no7 0 2 0 0 5 6 yes8 0 0 0.5 -1 0 2 no9 0 0 1 0 1 4 yes10 0 0 0 -2 -1 0 no

d. See table in c.e. The optimal solution is X3 = 1, S2 = 1.f. The first constraint is binding.

31. a. 2X1 + 4X2 16b. i. S=4

ii. S=0iii. S=2iv. S=4

32. a. 31 + 4X2 12b. i. S=3

ii. S=2iii. S=13iv. S=0

Case 4-1: A Nut CaseSee file: Case4_1.xlsm.

1. No. The fixed costs should not be included in the calculation of unit profit margins.2. X1 = pounds of Whole product to produce

X2 = pounds of Cluster product to produceX3 = pounds of Crunch product to produceX4 = pounds of Roasted product to produce

MAX 1.93 X1 + 1.04 X2 + 1.15 X3 + 1.33 X4 ST 1 X1 + 1 X2 + 1 X3 + 1 X4 < 3600

2 X1 + 1.5 X2 + 1 X3 + 1.75 X4 < 36001 X1 + 0.7 X2 + 0.2 X3 + 0.00 X4 < 36002.5 X1 + 1.6 X2 + 1.25 X3 + 1.0 X4 < 36000.6 X1 + 0.4 X2 + 0.2 X3 + 1 X4 < 11000.4 X1 + 0.6 X2 + 0.8 X3 + 0 X4 < 8001,000 < X1 < 99,999400 < X2 < 5000 < X3 < 1500 < X4 < 200

3. See file: Case4_1.xlsm4. X1 = 1029, X2 = 400, X3 = 150, X4=200, Total profit = $2,839.76


5.

6. No.7. Yes.8. The company could increase profit by making fewer pounds of the Cluster product (it has a negative

reduced cost).9. Each additional unit of the Roasted product adds about $0.55 to profit.10. Packaging is the only binding constraint (aside from the upper and lower variable bounds). Each

additional unit of packaging material is worth $0.772 (above and beyond its normal variable cost).11. $0. It is a non-binding resource.12. No, it could decrease by $0.305 without changing the optimal solution.13.

Spider Plot

268027002720274027602780280028202840286028802900

0.65 0.75 0.85 0.95 1.05 1.15 1.25 1.35

% of Original

Cell C

13

Cell C17Cell D17Cell E17Cell F17


The Whole product is the only one that would be effected. The solution appears to be rather insensitive to changes in required roasting times.

14.

Note that scenarios 1 & 2 for Chocolate are infeasible.

Scenario % of Base Nuts Chocolate1 70% $2,404 Error2 75% $2,631 Error3 80% $2,747 $2,6124 85% $2,787 $2,8055 90% $2,827 $2,8246 95% $2,840 $2,8367 100% $2,840 $2,840

Profit as a function of


Case 4-2: Parket Sisters(contributed by Jack Yurkiewicz, Pace University)

Let X1 = ballpoints produced, X2 = pencils produced, X3 = fountain pens produced

The LP formulation for this problem is:

MAX 3.0X1 + 3.0X2 + 5.0X3

ST 1.2X1 + 1.7X2 + 1.2X3 10000.8X1 + 0.0X2 + 2.3X3 12002.0X1 + 3.0X2 + 4.5X3 2000X1, X2, X3 0

See file: Case4_2.xlsm.

Total Profit = $2,766.67

The sensitivity report is shown below:



Cell Name Value Cost Coefficient Increase Decrease$B$2 Number to Produce: Ballpoints 700.00 0.0000 3 2.000 0.778$C$2 Number to Produce: Pencils 0.00 -1.3833 3 1.383 1.00E+30$D$2 Number to Produce: Fountain

pens133.33 0.0000 5 1.750 2.000


Cell Name Value Price R.H. Side Increase Decrease$E$6 Plastic Total Used: 1000.00 1.1667 1000 200.000 466.667$E$7 Chrome Total Used: 866.67 0.0000 1200 1.00E+30 333.333$E$8 Stainless Steel Total Used: 2000.00 0.8000 2000 555.556 333.333

1. The optimal weekly product mix is: 700 ballpoints, 0 pencils, and 133.33 fountain pens. We could round the fountain pens off to 133, but we shall not, in order to avoid rounding problems. The profit is $2,766.67.

2. No. None of the allowable increases or decreases on the constraints are zero.

3. Yes, the answer is unique. In the absence of degeneracy, we see that there are no variables which have a value of zero and simultaneously a zero reduced cost.

4. What is the marginal values for one more unit of each of the resources are: $1.17 for plastic, 0 for chrome, and $0.80 for stainless steel. These are the shadow prices for our three resources.

5. We can consider buying the 500 additional ounces of stainless steel because 500 is less than the maximum allowable increase of 555.56 ounces. Thus, the shadow price for stainless steel (and the other


resources as well) remains the same. Now as to whether or not the deal is worthwhile, because the 80 cent shadow price for an additional ounce of stainless steel is more than the 60 cents the distributor is charging Parket (above which they ordinarily charge when the model was formulated), it pays to buy the 500 ounces they are willing to sell. Parket will “be ahead” 20 cents for each ounce they buy.

6. Because we are buying the 500 ounces, each ounce bought will increase our profit by 20 cents. Thus the 500 ounces will increase our profit by 500×$0.20 = $100. Our new profit is then $2766.67 + $100 = $2866.67. Once we change the RHS of a constraint, we generally cannot tell what the new product mix will be. To get that, we must re-solve the problem.

7. We cannot tell without re-solving the problem. Five hundred additional ounces of plastic is above the allowable increase of 200 ounces. That is, we are guaranteed that the shadow prices given will remain the same so long as we have no more than 1200 (1000 + 200) ounces of plastic. Above that value, the shadow prices are different that those given. Thus, we cannot make any analysis at this point.

8. We can buy as many as 200 additional ounces of plastic before we “run out of information” from the computer output. Parket ordinarily pays $5.00 per ounce, and now the distributor is charging $6.00 per ounce, or $1.00 more than Parket paid. Still, the shadow price for plastic is $1.17 (or $1.1667 to be exact) which is more than the $1.00 the distributor is “overcharging.” Parket will still be ahead about 17 cents (or 16.6667 cents) for each ounce they buy. Buying the 200 ounces will increase their profit by (200)($0.17) = $34 (or $33.33 to be exact). The new profit will thus be $2766.67 + $33.33 = $2800.00. Again, we cannot tell what the new product mix will be without re-solving the problem.

9. We can consider selling the 300 ounces of plastic because the allowable decrease of plastic is 466.67. That is, Parket can have as little as 533.33 ounces of plastic (1000 - 466.67) and still be assured that the shadow price plastic (and the others) will remain the same. Selling an ounce of plastic reduces Parket’s profit by $1.17 (or $1.1667 to be exact), yet the company is willing to pay $6.50 per ounce. This represents a yield of $1.50 above what Parket ordinarily pays (they pay $5.00 per ounce, if you recall). Thus Parket will be ahead by ($1.50 - $1.1667)×300 = $100 if they sell the plastic. The new profit will be $2766.67 + $100 = $2866.67. Once again, we cannot tell what the new product mix will be unless we solve the problem.

10. Parket had ordered 1200 ounces of chrome, yet the optimal solution uses only 866.67 ounces. There are 333.33 ounces left over. That is why the shadow price for chrome is zero. If only 1000 ounces are delivered, Parket will still have 133.33 ounces of chrome left over. Because the shadow price is zero, and the 200 ounce shortage from the distributor (1200 - 1000) is less than the allowable decrease of 333.33 ounces, the shadow price is still zero. Thus the new profit will not change and remain at $2766.67. We do not have to solve the problem again here to get the new product mix. The product mix stays the same as: 700 ballpoints, zero pencils, and 133.33 fountain pens. The only difference is that the slack for chrome is now 133.33, instead of the original 333.33.

11. Price out the new design. It is Profit - Cost = $3 - [(1.1)($1.17) + (2)($0) + (2)($0.80)] = $0.12, approximately. Because this value is positive, Parket should give the go-ahead for this new design.

12. If the profit on ballpoints were to decrease to $2.50, down from the original assumed $3.00, this $0.50 decrease is still less than the allowable decrease of $0.78 for ballpoints. The optimal solution stays the same, but the total profit now drops by ($3.00 - $2.50)(700 ballpoints) = $350. The new profit is $2416.67.

13. Price out the new felt tip pen: Profit -Cost = Profit - [(1.8)($1.167) + (0.5)($0) + (1.3)($0.80)] = Profit -3.14. If we are to produce the felt tip then this amount should be nonnegative. Thus, Profit -3.14 0, or Profit $3.14. At exactly $3.14, we have alternative optima and would be indifferent between producing or not producing the felt tip pens.

14. It should be at least $4.38 ($3.00 plus the reduced cost of $1.38).


15. Currently it does not pay to make pencils; the profit of $3.00 is not high enough. Yet if Parket insists on making 20 pencils per week, their profit would be down by (20)($1.383) = $27.66. That is, they would have a total profit of $2766.67 - $27.66 = $2739.

16. At $6.75, or $1.75 more than the assumed $5.00, we are just at the limit of the allowable increase for the profit of fountain pens. We have alternative optima at this point. Still, the optimal solution stays the same. The new profit is $2766.67 + ($1.75)(133.33 fountain pens) = $3000.

Case 4-3: Kamm Industries

See file: Case4_3.xlsm

1. Total cost: $1,902,864Make

Carpet Dobbie Pantera Buy1 0 na 14,0002 52,000 na 03 44,000 na 04 20,000 na 05 0 77,500 06 0 0 109,5007 0 120,000 08 0 54,075 5,9259 7,500 0 0

10 0 69,500 011 22,041 0 46,45912 0 83,000 013 0 10,000 014 0 381,000 015 0 64,000 0

Microsoft Excel Sensitivity Report


Cell Name Value Cost Coefficient Increase Decrease$C$6 Dobbie 0 0 2.66 1E+30 0.0228248$C$7 Dobbie 52,000 0 2.55 0.0550959 1E+30$C$8 Dobbie 44,000 0 2.64 0.0805898 1E+30$C$9 Dobbie 20,000 0 2.56 0.0292481 1E+30$C$10 Dobbie 0 0 1.61 1E+30 0.0080232$C$11 Dobbie 0 0 1.62 1E+30 0.0173936$C$12 Dobbie 0 0 1.64 1E+30 0.0248079$C$13 Dobbie 0 0 1.48 1E+30 0.0040811$C$14 Dobbie 7,500 0 1.5 0.0071038 1E+30$C$15 Dobbie 0 0 1.44 1E+30 0.0162417$C$16 Dobbie 22,041 0 1.64 0.0022523 0.00991$C$17 Dobbie 0 0 1.57 1E+30 0.0043207


$C$18 Dobbie 0 0 1.49 1E+30 0.0044589$C$19 Dobbie 0 0 1.31 1E+30 0.0033658$C$20 Dobbie 0 0 1.51 1E+30 0.0019973$D$10 na 77,500 0 1.6 0.0080232 1E+30$D$11 na 0 0 1.61 1E+30 0.0133342$D$12 na 120,000 0 1.61 0.0010843 1E+30$D$13 na 54,075 0 1.47 0.0054369 0.0008738$D$14 na 0 0 1.5 1E+30 0.0071038$D$15 na 69,500 0 1.42 0.0162417 1E+30$D$16 na 0 0 1.64 1E+30 0.0075932$D$17 na 83,000 0 1.56 0.0043207 1E+30$D$18 na 10,000 0 1.48 0.0044589 1E+30$D$19 na 381,000 0 1.3 0.0033658 1E+30$D$20 na 64,000 0 1.5 0.0019973 1E+30$E$6 na Buy 14,000 0 2.77 0.0228248 1E+30$E$7 na Buy 0 0 2.73 1E+30 0.0550959$E$8 na Buy 0 0 2.85 1E+30 0.0805898$E$9 na Buy 0 0 2.73 1E+30 0.0292481$E$10 Buy 0 0 1.76 1E+30 0.0415917$E$11 Buy 109,500 0 1.76 0.0133342 1E+30$E$12 Buy 0 0 1.76 1E+30 0.0010843$E$13 Buy 5,925 0 1.59 0.0008738 0.0054369$E$14 Buy 0 0 1.71 1E+30 0.0953073$E$15 Buy 0 0 1.63 1E+30 0.0913951$E$16 Buy 46,459 0 1.8 0.0075932 0.0022523$E$17 Buy 0 0 1.78 1E+30 0.0702167$E$18 Buy 0 0 1.63 1E+30 0.0089908$E$19 Buy 0 0 1.44 1E+30 0.0199327$E$20 Buy 0 0 1.69 1E+30 0.0401119


Cell Name Value Price R.H. Side Increase Decrease$F$6 na Total 14,000 2.770 14000 1E+30 14000$F$7 na Total 52,000 2.675 52000 28234.695 52000$F$8 na Total 44,000 2.769 44000 27251.544 44000$F$9 na Total 20,000 2.701 20000 25055.643 20000$F$10 Total 77,500 1.718 77500 54802.416 6004.1565$F$11 Total 109,500 1.760 109500 1E+30 109500$F$12 Total 120,000 1.759 120000 43575.392 4774.123$F$13 Total 60,000 1.590 60000 1E+30 5924.514$F$14 Total 7,500 1.615 7500 30748.502 7500$F$15 Total 69,500 1.539 69500 54711.549 5994.2012$F$16 Total 68,500 1.800 68500 1E+30 46458.57$F$17 Total 83,000 1.710 83000 43322.985 4746.4693$F$18 Total 10,000 1.621 10000 46018.683 5041.8101$F$19 Total 381,000 1.420 381000 54045.197 5921.1956$F$20 Total 64,000 1.650 64000 43292.697 4743.1509

$I$21Used Hr/Yd 32370 -0.5990 32370 12408.806 5887.1342


$K$21 Used na 172640 -0.6427 172640 1106.1453 10096.245

2. No. 3. Yes.4. The shadow price for capacity on the Dobbie looms is -0.599 per hour. Losing one loom for the quarter

results in 13×(7×24-2)=2158 lost hours of capacity and, as a result, increase costs by about $1,293.5. By the same reasoning, this would reduce costs by about $1,293.6. The shadow price for capacity on the Pantera looms is -0.6427 per hour. Losing one loom for the quarter

results in 13×(7×24-2)=2158 lost hours of capacity and, as a result, increase costs by about $1,387.7. By the same reasoning, this would reduce costs by about $1,387.8. The shawdow prices for carpets 1, 6, 8 and 11 are the same as the marginal prices of buying these

products. The allowable increase of 1E+30 indicate we can buy as much as we want at these prices.9. The company is paying $2.55 per yard to fill the order for carpet 2 on the Doobie loom. However, the

demand for carpet 2 is $2.675 per yard. Thus, if the company did not have to fill the order for carpet 2 it could save not only the $2.55 per yard cost of carpet 2 but then be able to use the extra capacity on Dobbie loom to fill the remaining orders in a more efficient manner – saving an additional $0.125 on every yard of carpet 2 it doesn't have to make.

10. The shadow price of carpet 14 is the smallest at $1.42 per yard. So selling more of this carpet would produce the highest marginal profit for the company.

11. Yes. It's current cost is $2.77 with an allowable increase of $0.02282. So a price increase to $2.80 would cause the optimal solution to change.

12. No. It's current cost is $1.69 with an allowable decrease of $0.0401. So a price decrease to $1.65 would not cause the optimal solution to change.

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