Linear Programming

20

SOLUTIONS TO DISCUSSION QUESTIONS

AND PROBLEMS

4-1. In most real world situations that are modeled using LP,conditions are dynamic and changing. Hence, input data such asresource availabilities, prices, and costs used in the LP model areestimated, rather than known with certainty. In such environments,sensitivity analysis can be used to identify the ranges of values ofthese input data for which the current LP solution remains opti-mal. This is done without solving the problem again each time weneed to examine a change in an input data’s value.

When all model values are deterministic, that is, known withcertainty, sensitivity analysis may not be needed from the perspec-tive of evaluating data accuracy. This may be the case in a portfo-lio selection model in which we select from among a series ofbonds whose returns and cash-in values are set for long periods of time.

4-2. Sensitivity analysis is important in all decision modelingtechniques. For example, it is important in breakeven analysis totest the model’s sensitivity to selling price, fixed cost, and variablecost. Likewise, it is important in inventory models in which wetests the result’s sensitivity to changes in demand, lead time, costs,and so on.

4-3. A change in a resource’s availability (right-hand-side)changes the size of a feasible region. An increase means moreunits of that resource are available, causing the feasible region toincrease in size. A decrease means fewer units are available. Obvi-ously, if more units of a binding resource are available, it may bepossible for the optimal objective value to improve. In contrast, ifmore units of a non-binding resource are available, the additionalunits would just contribute to more slack and there would be noimprovement in the optimal objective value.

4-4. A change in an objective function coefficient changes theslope of the objective function, with respect to that variable. Thechange in the slope may be sufficient to make a different cornerpoint become the new optimal solution to the LP model.

4-5. Simultaneous changes in input data values are extremelylogical in many contexts. For example, it may be possible to tradeone type of resource for another, causing a decrease in the avail-ability of one resource and an increase in the other’s availability.Likewise, market conditions may cause us to simultaneously re-duce the price of all our products, not just in a single product.

4-6. We use the 100% rule to verify if the shadow prices in thecurrent sensitivity report are still valid to analyze the impact of a

proposed simultaneous change in input data values. To analyze asimultaneous change, we compute the ratio of each proposedchange in a parameter’s value to the maximum allowable changein its value, as given in the sensitivity report. The sum of these ra-tios must not exceed 1 (or 100%) in order for the informationgiven in the current sensitivity report to be valid. If the sum of theratios does exceed 1, the current information may still be valid; wejust cannot guarantee its validity. However, if the ratio does notexceed 1, the information is definitely valid.

4-7. The pricing-out procedure allows a firm to analyze if a newproduct will be profitable enough to be included in the optimal solution, without having to reformulate and solve the model afterincluding this additional variable. Alternatively, we can use thepricing-out procedure to determine the minimum profit contribu-tion required from a new product in order for it to be included inthe optimal production mix.

4-8. In most cases, when the Allowable Increase or AllowableDecrease column for the objective function coefficient of a vari-able has a value of zero in the Adjustable Cells table, this indicatesthe presence of alternate optimal solutions.

4-9. The shadow price of a resource indicates the marginal valueof each additional unit of that resource to the firm. In an LP modelwith several resources, this information helps the firm to prioritizeits resources in terms of their marginal value.

4-10. The Current Solution from Problem 2-14 is:

4C H A P T E R

Linear Programming Sensitivity Analysis

X

Y

0 25 50 75 1000

25

50

75

100

a

b

c

Isoprofit Line forX +Y = $66.67

optimalsolution

X = 33 , Y = 33 /31 /31 )(

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CHAPTER 4 LINEAR PROGRAMMING SENSIT IV ITY ANALYS IS 21

a. If the profit equation is $3X � $1Y, the optimal solu-tion switches to point c.

b. If X’s profit coefficient was overestimated, but shouldonly have been $1.25, it is easy to see graphically thatthe solution at point b remains optimal.

c.

The optimal solution is at point b, but profit has decreased from$66 to $57 , and the solution has changed considerably.

4-11. Using the isoprofit line or corner point method, we saw inProblem 2-15 that point b (where X � 37.5 and Y � 75) is optimalif the profit � $3X � $2Y.

17

23

If the profit changes to $4.50 per unit of X, the optimal solutionshifts to point c.

If the objective function becomes $3X � $3Y, the corner point bremains optimal.

4-12. The optimal solution of $26 profit lies at the point X � 2,Y � 3, in Problem 2-16.

a. If the first constraint is altered to 1X � 3Y � 8, the feasible re-gion and optimal solution shift considerably, as shown in thenext figure.

X

Y

0 50 100 1500

50

100

150

Isoprofit Line for3X + 1Y = $150

X + 2Y ≤ 100

2X + Y ≤ 100

Optimal Solution isNow Here

Y

X

0 25 50 75 1000

25

50

75

100

b

Optimal SolutionRemains at

Point b

X = 42 , Y = 14 ; Profit = $57

2X +Y ≤ 100

X + 4Y ≤ 100

/7 6 /72 /71( )

Y

X

Profit Line for 3X + 3Y

Profit Line for 4.50X + 2Y

Profit Line for3X + 2Y

0 50 100 1500

50

100

150

a

b

c

X

Y

0 2 4 6 80

2

4

6

8

Profit = 4X + 6Y = $26

6X + 4Y ≤ 24

X + 2Y ≤ 8

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22 CHAPTER 4 LINEAR PROGRAMMING SENSIT IV ITY ANALYS IS

b.

Using the corner point method, we determine that the optimal so-lution mix under the new constraint yields a $29 profit, or an in-crease of $3 over the $26 profit calculated.

4-13. We use the sensitivity report given in Program 4.6 to an-swer the following questions.

a. Each additional $ in radio advertising (up to $1,575) willincrease the audience by 2.03. Hence, if management ap-proves an increase of $200, audience coverage will increaseby 406 to 67,646.b. No. Since we are already placing 6.21 radio spots, thiscontractual agreement is not a binding constraint.c. The audience reached for each afternoon radio spot wouldhave to increase to at least 3,144.83 (� 2,800 � 344.83) inorder for these spots to become attractive. Hence, the pro-posed strategy will not change the current optimal solution.d. TV spots currently reach 5,000 contacts. Interestingly,the optimal number of TV spots to use does not change as

long the number of contacts is between 0 and 6,620.69. Obvi-ously, if the objective function coefficient is 0, it is not worth-while to use any TV spots. The reasons why TV spots arepart of the optimal solution even if the exposure is as low as0.001 are the limits on the number of newspaper ads and thetotal radio budget. See if you can recognize this fact by alter-ing the LP model in file 3-1.XLS.


a. Persons 30 or younger who live in a border state cur-rently cost $7.50 each. This cost must decrease to $6.90 orless (� $7.50 � $0.60) before it is worthwhile to includethese individuals in the survey.b. Based on the shadow price for the total households con-straint, each additional person who needs to be included inthe survey will cause the cost to increase by $5.98. Hence, ifthe sample size is increased to 3,000, the total cost will in-crease to $19,352 (� $15,166 � 700 x $5.98).c. Based on the reduced cost, each person 31–50 not livingin a border state who is included in the survey will increasethe total cost of the survey by $0.45.d. First, we check the 100% rule. (100/1000) � (50/700) �1. Hence, the shadow prices are valid. The revised total costis $15,166 � 100 x $0.92 � 50 x $0.82 � $15,115. Note thatthe reduction in the 30 and younger requirement will causethe total cost to go down, while the increase in the 31–50 re-quirement will cause the total cost to go up.


a. If the daily allowance of protein is reduced to 2.9 units, thetotal cost will decrease by 0.1 x $0.038 � $0.0038.b. Revised cost of grain A � $0.33/1.05 � $0.314. Revisedcost of grain B � $0.47/0.9 � $0.522. First, we check the100% rule. (0.016/Infinity) � (0.052/Infinity) � 1. There-fore, the current optimal solution remains optimal. The newtotal cost is $0.0529.c. Check the 100% rule. (0.1/0.25) � (0.2/0.35) � 0.97 �1. The revised total cost changes by (0.2 � $0) � (0.1 �$0.038), or a decrease of $0.0038.

4-16. See file P4-16.XLS for the Excel solution and Solver sen-sitivity report.

a. The decrease of $0.01 per pound is beyond the allowabledecrease limit of $0.005 per pound. Therefore, the optimalsolution will change.b. The constraints prescribing the minimum daily require-ments for ingredient A and ingredient D are binding. Foreach additional unit of ingredient A required in the mix, thecost will increase by $0.003. For each additional unit of in-gredient D required in the mix, the cost will increase by$0.083.c. A 20% decrease in the cost of mineral implies that thecost is now 0.8 x $0.17 � $0.136, a decrease of $0.034. Sincethis is within the allowable decrease limit, the current solu-tion remains optimal. The revised total cost is $0.57.d. The price of oats can fluctuate between $0.085 and $0.093per pound for the current solution to remain optimal.


X

Y

0 2 4 6 80

2

4

6

8

Profit = 4X + 6Y = $21.71

1X + 3Y ≤ 8

6X + 4Y ≤ 24

Optimal solution at

X = 2 , Y = 1/76 /75

Y

X

0 2 4 6 8 10 120

2

4

6

8

10

12

X = 5, Y = 1 ; Profit = $29

6X + 4Y ≤ 36

1X + 2Y ≤ 8

/21 )(

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a. Each additional gram of carbohydrates allowed in themeal will reduce the meal cost by $0.007. Each additional mgof iron required in the diet will cause the meal cost to in-crease by $0.074.b. Each pound of milk used in the diet will cause the mealcost to increase by $0.148 (reduced cost).c. Beans currently cost $0.58 per pound. Looking at the re-duced cost, the price of beans would have to decrease by atleast $0.261 (to $0.319) before Kathy can consider includingit in the meal.d. None of the allowable increase and allowable decreasevalues for the objective function coefficients is zero. Further,all items that are currently not in the meal (milk, fish, andbeans) have non-zero reduced costs. The current solution is,therefore, a unique optimal solution.


a. Four products (circuit boards, floppy drives, hard drives,and memory boards) are not included in the optimal produc-tion plan. The reduced costs indicate the minimum amountsby which the profit contributions of these products must in-crease before they would be included in the production mix.For example, the profit contribution of a circuit board shouldincrease by at least $138.64 (from the current $135.50 to atleast $274.14) before it becomes a viable product.b. The current production plan required only 62.07 hours ofthe available 100 hours on test device 3. Therefore, if the sis-ter concern takes 35 hours of time on this test device, Quit-meyer’s optimal solution will not be affected.c. Each additional hour of time (up to 47.83 hours) on testdevice 1 increases Quitmeyer’s profit by $1,284.52. Thisshadow price value, however, assumes that time on test de-vice 1 costs only $15 per hour. The 20 additional hours at acost of $25 per hour will therefore increase Quitmeyer’sprofit by ($1,284.52 x 20), less the premium of ($10 x 20), orby $25,490.4. The new profit will be $220,995.23. The deal isworthwhile.d. First, we check the 100% rule. (20/48) � (40/80) �0.916 � 1. The current shadow prices are valid. By makingthis trade, Quitmeyer’s new profit will be $195,504.83 � 20� $1,284.52 � 40 x $344.69 � $183,602.03. The deal is notworthwhile.


a. Several of the allowable increase and allowable decreasevalues for the objective function coefficients are zero. For ex-ample, see these values for the number of acres of wheat inthe SE parcel. Also, several variables (crops) that are cur-rently not in the crop plan (for example, wheat in the NWparcel) have zero reduced costs. The current solution is,therefore, not a unique optimal solution and there are alter-nate optimal solutions.b. There are several ways in which we can get Solver toidentify an alternate optimal solution. Perhaps the easiest ap-proach is to rearrange the order in which the variables and/orconstraints are presented in the model. For example, see sheetnamed Alternate in file P4-19.XLS, in which the columnscorresponding to Alfalfa SE and Alfalfa N have been movedfrom columns G and H to columns J and K, respectively.

When we solve the problem now using Solver, we get a solu-tion with the same profit ($337,862.07), but with a differentcrop plan. Other optimal crop plans are also possible.c. Increasing Barley sales by 10% would make the newlimit 2,420 tons, or 1,100 acres. Based on the shadow pricefor this constraint, each additional acre of Barley (up to108.57 acres) will increase Margaret’s profit by $37.59. The100 additional acres will therefore increase profit by$3,758.60, to $341,620.67.d. Each additional acre-feet of water will allow Margaret toincrease total profit by $20.69.

4-20. We use the answer and sensitivity reports given in Table4.3 to answer the following questions.

a. Each additional pound of material will increase profit by$0.5. The 2 pounds will therefore cause profit to increase to$29.b. Each additional hour of labor will increase profit by $1.The 1.5 hours will therefore cause profit to increase to $29.5.c. First, we check the 100% rule. (1/2)�(1.5/4) � 0.875 �1. The shadow prices are valid. New profit � $28 � 1.5 x$0.5 � 1 x $1 � $27.75. The deal is not worthwhile.d. First, we check the 100% rule. (0.75/1)�(0.25/1) � 1.The solution remains optimal. New profit � 2 x $4.75 � 4 x$4.75 � $28.50.e. Decrease in current profit if 1 unit of the new product isproduced � 1 x $1 � 1 x $0.5 � 2 x $0 � $1.5. Profit contri-bution of new product � $2. Hence, the net profit will in-crease by $0.50 for each unit produced of the new product.


a. As we saw in Problem 3-14, there are approximately2,791 medical patients and 2,105 surgical patients per year inthe optimal solution. This translates to 61 medical beds and29 surgical beds in the 90-bed addition.b. There are no empty beds with this optimal solution. Eachadditional patient day (over the current 32,850) will permitMt. Sinai to increase revenue by $276.82. That is, by acquir-ing another bed (or 365 patient days), the revenue can be in-creased by $101,039.c. Labs have an unused capacity of 876.364. Acquiringmore lab space is therefore not worthwhile.d. X-ray capacity is being utilized to its fullest extent. Eachadditional x-ray that can be handled will increase revenue by$65.45.e. The operating room has an unused capacity of 695.45.Acquiring more operating room is therefore not worthwhile.

4-22. We use the information given in Programs 4.9A and 4.9Bto answer the following questions.

a. The optimal production plan is to produce 540 Standardsuitcases and 252 Deluxe suitcases, for a total profit of$7,668. At this point, no Luxury suitcases are scheduled forproduction.b. Given the optimal production plan, we can determinewhether or not the new polishing process will have sufficientcapacity to support that plan. The constraint (in hours) is:

1/6 (Standard) � 1/4 (Deluxe) � 1/3 (Luxury) � 170

Now we substitute the previous optimal production decision:

1/6 (540) � 1/4 (252) � 1/3 (0) � 153 � 170

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24 CHAPTER 4 LINEAR PROGRAMMING SENSIT IV ITY ANALYS IS

Therefore, there is sufficient capacity in the proposed polish-ing operation to sustain the optimal production plan.c. The constraint for the water proofing process (in hours)would be

1 (Standard) � 1.5 (Deluxe) � 1.75 (Luxury) � 900

Substitution of the optimal production plan yields:

1 (540) � 1.5 (252) � 1.75 (0) � 918 � 900

Therefore, the proposed water proofing process does not havethe capacity to support the production plan. In order to deter-mine the impact, it is necessary to go back to the original for-mulation and add the constraint.

4-23. The profit contributions per unit for the two new productsare:

Compact: $30 � $5 � 0.5($10) � 0.75($6) � 0.75($9)� 0.2($8) � $7.15

Kiddo: $37.50 � $4.50 � 1.2($10) � 0.75($6) � 0.5($9) � 0.2($8) � $10.40

However, we need to remember that a positive profit contributionis not a sufficient condition for making the new product. That isbecause resources allocated to make the new product will have tobe reallocated from existing products. We need to convert thevalue of these resources using the shadow prices, as follows:

Compact: 0.5($4.38) � 0.75($0) � 0.75($6.94) �0.2($0) � $7.395

Kiddo: 1.2($4.38) � 0.75($0) � 0.5($6.94) � 0.2($0) �$8.726

Since the profit contribution of $7.15 is smaller than the $7.395value of the resources required, the Compact model is not attrac-tive to make. However, the Kiddo model will more efficiently con-vert resources to revenue (since the profit contribution of $10.40exceeds the $8.726 value of the resources). We must include theKiddo product in the formulation and solve the problem againusing Solver to determine the new production plan.

4-24. We use the information given in Programs 4.10A and4.10B to answer the questions problems 4-24 to 4-27.

a. The optimal production plan is to make 100 TiniTote, 35TubbyTote, and 90 ToddleTote strollers. The resulting profit is$2,086.25. The following constraints are binding: fabricationtime, minimum production level for the TubbyTote model, andratio of the ToddleTote model to the total production.b. All 620 hours are being used fabrication. Only 415 of the500 hours are being used in Sewing, while only 385 of the480 hours are being used in Assembly.c. Each additional hour of fabrication time (up to 110.50hours) will allow Strollers-to-Go to increase profit by $3.60.Hence, the firm would be willing to pay a premium of up to$3.60 for each additional hour of fabrication time. In contrast,since sewing is a non-binding constraint, the firm would notbe interested in obtaining any additional sewing time.d. None of the products are being produced at their maxi-mum level (demand). However, the TubbyTote model isbeing produced at its minimum level.

4-25a. From the sensitivity report, the profit contribution ofTiniTote can vary between $5.92 (� $9.25 � $3.33) and$14.25 (� $9.25 � $5) without affecting the current optimal

production mix. Assuming labor costs do not change, this im-plies that material costs can vary between �$1 (� $4�$5,rounded to $0) and $7.33 (� $4 � $3.33) without affectingthe current optimal production mix. Note that a decrease inmaterial cost translates to an increase in the profit contribu-tion, and vice versa.b. Given the optimal production plan, we can determinewhether or not the new polishing process will have sufficientcapacity to support that plan. The constraint (in hours) is:

1/6 (TiniTote) � 1/4 (TubbyTote) � 1/5 (ToddleTote) � 48

Now we substitute the previous optimal production decision:

1/6 (100) � 1/4 (35) � 1/5 (90) � 43.42 � 48

Therefore, there is sufficient capacity in the proposed polishingoperation to sustain the optimal production plan.

4-26a. Each additional hour of fabrication time (up to 110.50hours) will allow Strollers-to-Go to increase profit by $3.60.Hence, the firm would be willing to pay a premium of up to$3.60 for each additional hour of fabrication time, or up to$11.85 (� $8.25 � $3.60) per hour. If the firm can get addi-tional time for $10.50 per hour, it should take it. Each hourobtained this way (up to 110.50 hours) will increase profit by$1.35 [� $3.60 � ($10.50�$8.25)].b. The first two bundles (80 hours total) should definitely bepurchased at $10.50 per hour since the profit would increaseby 80 x $1.35 � $108. For the third bundle, we know that thefirst 30.5 hours will cause profit to increase by 30.5 � $1.35� $41.175. However, the shadow price of the last 9.5 hoursof this bundle will be less than $3.60 (can you see why this isso?). In the worst case, if Strollers-to-Go has to pay $10.50per hour for these hours too, and they remain unused, pur-chasing the third bundle becomes an unattractive option. Thefirm should therefore purchase only 2 bundles of 40 hourseach.

4-27. The profit contribution per unit for the new products is:

$72 � $5.75 � 3.5($8.25) � 1.75($8.5) � 1.5($8.75) � $9.375

However, we need to remember that a positive profit contributionis not a sufficient condition for making the new product. That isbecause resources allocated to make the new product will have tobe reallocated from existing products. We need to convert thevalue of these resources using the shadow prices.

If we assume that TwinTotes are not included in the 40% over-all production limitation on ToddleTotes, the computation is asfollows:

3.5($3.6) � 1.75($0) � 1.5($0) � $12.60

Since the profit contribution of $9.375 is smaller than the$12.60 value of the resources required, the TwinTote model is notattractive to manufacture. Each TwinTote made will decreaseprofit by $3.225 (subject to round off).

In contrast, if we assume that TwinTotes are also inclued in the40% overall production limitation on ToddleTotes, the computa-tion is now as follows:

3.5($3.6) � 1.75($0) � 1.5 ($0) � 0.4($3.85) � $11.06

In this case also, since the profit contribution of $9.375 is smallerthan the $11.06 value of the resources required, the TwinTote model

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is not attractive to manufacture. However, now each TwinTotemade will decrease profit by only $1.685 (subject to round off).

SOLUTION TO RED BRAND CANNERS CASE

1. The main issue in this case is how to allocate 3 million poundsof tomatoes. The overall objective is to maximize total sales lessvariable costs. These costs include production and selling ex-penses. Twenty percent of the crop was grade A and the rest wasgrade B. In setting up the constraints, the amount of grade A toma-toes cannot exceed 20% of 3 million pounds. Thus not more than600,000 pounds of grade A tomatoes can be used. Similarly, notmore than 2,400,000 pounds of grade B tomatoes can be used.Furthermore, the demand for 50,000 cases of tomato juice and80,000 cases of tomato paste should be met. The demand forwhole tomatoes is not a constraint in this problem. Finally, mini-mum quality requirements should be met. This includes an aver-age of 8 points per pound for whole tomatoes and 6 points perpound for tomato juice. There is no constraint for tomato paste.

Another issue is whether or not to buy 80,000 additional poundsof grade A tomatoes. This would increase the amount of availablegrade A tomatoes from 600,000 pounds to 680,000 pounds.

2. The problem can be formulated using LP as follows.

WA � pounds of whole A tomatoes

WB � pounds of whole B tomatoes

JA � pounds of juice A tomatoes

JB � pounds of juice B tomatoes

PA � pounds of paste A tomatoes

PB � pounds of paste B tomatoes

Maximize: 0.0822WA � 0.0822WB � 0.066JA � 0.066JB �0.074PA � 0.074PB

subject to

1WA � 1WB � 14,400,000

1JA � 1JB � 1,000,000

1PA � 1PB � 2,000,000

1WA � 1JA � 1PA � 600,000

1WB � 1JB� 1PB � 2,400,000

1WA � 3WB � 0

3JA � 1JB � 0

All Variables � 0

The coefficients in the objective function are the unit profits. Acase of whole tomatoes (grade A and grade B) sells for $4. The

variable cost (less the tomatoes) is $2.52. Since the tomatoes arealready on hand (and no salvage appears to be possible), they rep-resent a sunk cost and are not part of the decision process. Sincethere are 18 pounds per case, the unit profit is (4.00 � 2.52)/18 �0.0822. Similar analyses hold for the other terms in the objectivefunction.

The first constraint refers to the 14.4 million pounds of wholetomatoes—800,000 cases at 18 pounds per case—that constitutesmaximum demand. Similarly, the maximum demand for tomatojuice is 50,000 cases at 20 pounds per case or 1 million pounds,and the maximum demand for tomato paste is 80,000 cases at 25pounds per case or 2 million pounds, and these are constraints 2and 3. Constraints 4 and 5 reflect the availability of grade A andgrade B tomatoes, respectively, and the last two constraints are thequality constraints. The requirements that canned tomatoes mustaverage at least 8 points means that

9WA � 5WB � 8(WA � WB) ⇒ 1WA � 3WB � 0

Similarly, the requirements that tomato juice must average at least6 points is the last constraint.

3. See file P4-Red Brand.XLS for the Excel implementation ofthis model and its solution. The optimal solutions is: WA �525,000, WB � 175,000, JA � 75,000, JB � 225,000, PA � 0, andPB � 2,000,000. The maximum profit is $225,355.60.

4. From the answer report (see file P4-Red Brand.XLS), we seethat the following constraints are binding: grade A tomato avail-ability, grade B tomato availability, paste demand, canned tomatoquality, and juice quality.

5. From the sensitivity report (see file P4-Red Brand.XLS), theshadow price of grade A tomatoes is $0.0903 per pound, valid upto 600,000 additional pounds. The 80,000 pounds being offered at$0.085 per pound will therefore increase profit by $0.0053 perpound, or by a total of $424.

6. The demands for whole tomatoes and juice are not bindingconstraints. Hence, their underestimation will not affect the opti-mal solution. However, each additional pound of paste (up to200,000 pounds) demanded will increase profit by $0.0161. If de-mand is 5% more (that is, 4,000 cases more at 25 pounds percase), the additional demand for 100,000 pounds will increaseprofit by $1,610.

7. First, we check the 100% rule. (100,000)/(466,666.67) �(200,000)/(466,666.67) � 0.64 � 1. The shadow prices are valid.The new profit is: $225,355.56 � (200,000)($0.0579) �(100, 000)($0.0903) � $227,905.60. Gordon should accept thistrade.

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