OPMA Chapter 4 HW

# OPMA Chapter 4 HW - Objection Function Maximize Profit...

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2] A) X 1 = # of business courses, X 1 ≥ 23 X 2 = # of nonbusiness courses, X 2 ≥ 20 Objective Function: Minimize: 120X 1 + 200X 2 (1) Total Courses: X 1 + X 2 ≥ 65 (2) Money: 60X 1 + 24X 2 ≤ \$3000 (3) Nonbusiness: X 2 ≥ 20 (4) Business: X 1 ≥ 23 B) Graphic Analysis Feasible region is between points A B and C based of the different functions and whether they are ≥ or ≤ C) Optimal Solution Optimal Solution is most likely at Point C. Point A = (23, 67.5) 60(23) + 24X 2 ≤ \$3000; x 2 = 67.5 Point B = (23, 42) Point C = (40, 25) Since we want to isolate X 2 we will multiply each number by 24 24(40) + 24(25) = 1560 60X 1 + 24X 2 = \$3000 -24X 1 + 24X 2 = 1560.0 = 36X 1 = 1440 =X 1 = 40 Therefore 60(24) + 24(X 2 ) = 3000; X 2 = 25 Optimal Solution = (40, 25)

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D) Slack or Surplus Variables (1) Nonbusiness: X 2 ≥ 20 [25-20=5] Surplus of 5 (2) Business: X 1 ≥ 23 [40-23 = 17] Surplus of 17 5] A) Decision Variables: X 1 = Routing 1 [100’s of feet of pipe] X 2 = Routing 2 [100’s of feet of pipe]
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Unformatted text preview: Objection Function: Maximize Profit using: 60X 1 + 80X 2 = Z[Profit] Constraints Melting: X 1 + X 2 ≤ 45 Extruder A: 3X 1 ≤ 90 Extruder B: 1X 2 ≤ 160 Raw Material: 5X 1 + 4X 2 ≤ 200 B) Graph Below. Optimal Solution at point D. [0,45] C) 60(0) + 80(45) = \$3600 = MAX Profit 8) Considering Problem 7 A) Looking at the shadow price on the chart I got these answers: Cutting Time: If costs were less than \$4.75, then we’d be willing to pay for an extra hour of cutting Sewing Time: I would not be willing to pay anything for an extra hour of sewing because there is a lot of slack already Yard of Material: I would be willing to pay less than \$2.75 for an extra yard of material. It is because the shadow price of an extra yard is \$2.75. B) Range of feasibility for each: Cutting Range is [60, 132] Material is [33.33, 100] Shadow Price is valid for those ranges....
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OPMA Chapter 4 HW - Objection Function Maximize Profit...

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