Linear Programming Sensitivity Analysis s1

OSCM 230 Homework 04 Solution

H O M E W O R K # 4 L I N E A R P R O G R A M M I N G S E N S I T I V I T Y A N A L Y S I S

Problem 4-8.

In most cases, when the Allowable Increase or Allowable Decrease column for the objective function coefficient of a variable has a value of zero in the Adjustable Cells table, this indicates the presence of alternate optimal solutions.

Problem 4-15.

We use Sensitivity Report given in Screenshot 4-8 [the sensitivity report given in Program 4.8 (page 147, textbook)] to answer the following questions. a. If the daily allowance of protein is reduced to 2.9 units, the total cost will decrease by 0.1 x $0.038 = $0.0038. b. Revised cost of grain A = $0.33/1.05 = $0.314. Revised cost of grain B = $0.47/0.9 = $0.522. First, we check the 100% rule. (0.016/Infinity) + (0.052/Infinity) < 1. Therefore, the current optimal solution remains optimal. The new total 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 x $0) - (0.1 x $0.038), or a decrease of $0.0038.

Problem 4-16.

See file P4-16.XLS for the Excel solution and Solver sensitivity report. a. Four products (circuit boards, floppy drives, hard drives, and memory boards) are not included in the optimal production plan. The reduced costs indicate the minimum amounts by which the profit contributions of these products must increase before they would be included in the production mix. For example, the profit contribution of a circuit board should increase by at least $138.64 (from the current $135.50 to at least $274.14) before it becomes a viable product. b. The current production plan required only 62.07 hours of the available 100 hours on test device 3. Therefore, if the sister concern takes 35 hours of time on this test device, Quitmeyer’s optimal solution will not be affected. c. Each additional hour of time (up to 47.83 hours) on test device 1 increases Quitmeyer’s profit by $1,284.52. This shadow price value, however, assumes that time on test device 1 costs only $15 per hour. The 20 additional hours at a cost of

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$25 per hour will therefore increase Quitmeyer’s profit by ($1,284.52 x 20), less the premium of ($10 x 20), or by $25,490.4. The new profit will be $220,995.23. The deal is worthwhile. d. First, we check the 100% rule. (20/48) + (40/80) = 0.916 < 1. The current shadow prices are valid. By making this trade, Quitmeyer’s new profit will be $195,504.83 - 20 x $1,284.52 + 40 x $344.69 = $183,602.03. The deal is not worthwhile.

Problem 4-18.

See file P4-18.XLS for the Excel solution and Solver sensitivity report. a. The decrease of $0.01 per pound is beyond the allowable decrease limit of $0.005 per pound. Therefore, the optimal solution will change. b. The constraints prescribing the minimum daily requirements for ingredient A and ingredient D are binding. For each additional unit of ingredient A required in the mix, the cost will increase by $0.003. For each additional unit of ingredient D required in the mix, the cost will increase by $0.083. c. A 20% decrease in the cost of mineral implies that the cost is now 0.8 x $0.17 = $0.136, a decrease of $0.034. Since this is within the allowable decrease limit, the current solution remains optimal. The revised total cost is $0.57. d. The price of oats can fluctuate between $0.085 and $0.093 per pound for the current solution to remain optimal.

Problem 4-19.

See file P4-19.XLS for the Excel solution and Solver sensitivity report. a. Each additional gram of carbohydrates allowed in the meal will reduce the meal cost by $0.007. Each additional mg of iron required in the diet will cause the meal cost to increase by $0.074. b. Each pound of milk used in the diet will cause the meal cost to increase by $0.148 (reduced cost). c. Beans currently cost $0.58 per pound. Looking at the reduced cost, the price of beans would have to decrease by at least $0.261 (to $0.319) before Kathy can consider including it in the meal.

of 4 OSCM 230 Homework 04 Solution d. None of the allowable increase and allowable decrease values for the objective function coefficients is zero. Further, all items that are currently not in the meal (milk, fish, and beans) have non-zero reduced costs. The current solution is, therefore, a unique optimal solution.

Problem 4-21.

See file P4-21.XLS for the Excel solution and Solver sensitivity report. a. As we saw in Problem 3-45, there are approximately 2,791 medical patients and 2,105 surgical patients per year in the optimal solution. This translates to 61 medical beds and 29 surgical beds in the 90-bed addition. b. There are no empty beds with this optimal solution. Each additional patient day (over the current 32,850) will permit Mt. Sinai to increase revenue by $276.82. That is, by acquiring another bed (or 365 patient days), the revenue can be increased by $101,039. c. Labs have an unused capacity of 876.364. Acquiring more lab space is therefore not worthwhile. d. X-ray capacity is being utilized to its fullest extent. Each additional 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.

Problem 4-22.

[We use the information given in Programs 4.9A and 4.9B to answer the following questions.] a. The optimal production plan is to produce 540 Standard suitcases and 252 Deluxe suitcases, for a total profit of $7,668. At this point, no Luxury suitcases are scheduled for production. b. Given the optimal production plan, we can determine whether or not the new polishing process will have sufficient capacity 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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Therefore, there is sufficient capacity in the proposed polishing 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 have the capacity to support the production plan. In order to determine the impact, it is necessary to go back to the original formulation and add the constraint.

Problem 4-24.

We use the information in Screenshots 4-11A and 4-11B to answer questions to answer the questions problems 4-24 to 4-27. a. The optimal production plan is to make 100 TiniTote, 35 TubbyTote, and 90 ToddleTote strollers. The resulting profit is $2,086.25. The following constraints are binding: fabrication time, minimum production level for the TubbyTote model, and ratio of the ToddleTote model to the total production. b. All 620 hours are being used fabrication. Only 415 of the 500 hours are being used in Sewing, while only 385 of the 480 hours are being used in Assembly. c. Each additional hour of fabrication time (up to 110.50 hours) 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 not be interested in obtaining any additional sewing time. d. None of the products are being produced at their maximum level (demand). However, the TubbyTote model is being produced at its minimum level.