Chapter 2 solutions

Chapter 2 solutions - Chapter 2 An Introduction to Linear...

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2 - 1 Chapter 2 An Introduction to Linear Programming Solutions: 1. a, b, and e, are acceptable linear programming relationships. c is not acceptable because of 2 2 B d is not acceptable because of 3 A f is not acceptable because of 1AB c, d, and f could not be found in a linear programming model because they have the above nonlinear terms. 2. a. 8 4 4 8 0 B A b. B A 8 4 4 8 0 c.
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Chapter 2 2 - 2 B A 8 4 4 8 0 Points on line are only feasible points 6. 7 A + 10 B = 420 is labeled (a) 6 A + 4 B = 420 is labeled (b) -4 A + 7 B = 420 is labeled (c) 80 40 40 80 B A 20 60 100 20 60 0 100 -20 -40 -60 -80 -100 (c) (a) (b) 7.
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An Introduction to Linear Programming 2 - 3 0 B A 150 50 100 200 250 50 100 10. 0 1 2 3 4 5 B 1 2 3 4 5 6 Optimal Solution A = 12/7, B = 15/7 Value of Objective Function = 2(12/7) + 3(15/7) = 69/7 A
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Chapter 2 2 - 4 A + 2 B = 6 (1) 5 A + 3 B = 15 (2) (1) × 5 5 A + 10 B = 30 (3) (2) - (3) - 7 B = -15 B = 15/7 From (1), A = 6 - 2(15/7) = 6 - 30/7 = 12/7 12. a. (0,0) 1 2 3 4 5 B 1 2 3 4 5 6 Optimal Solution A = 3, B = 1.5 Value of Objective Function = 13.5 6 (4,0) (3,1.5) A
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An Introduction to Linear Programming 2 - 5 b. (0,0) 1 2 3 B 1 2 3 4 5 6 Optimal Solution A = 0, B = 3 Value of Objective Function = 18 A 7 8 9 10 c. There are four extreme points: (0,0), (4,0), (3,1,5), and (0,3). 13. a. 0 2 4 6 8 B 2 4 6 8 Feasible Region consists of this line segment only A b. The extreme points are (5, 1) and (2, 4).
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Chapter 2 2 - 6 c. 0 2 4 6 8 B 2 4 6 8 Optimal Solution A A = 2, B = 4 14. a. Let F = number of tons of fuel additive S = number of tons of solvent base Max 40 F + 30 S s.t. 2/5 F + 1 / 2 S 200 Material 1 1 / 5 S 5 Material 2 3 / 5 F + 3 / 10 S 21 Material 3 F , S 0
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An Introduction to Linear Programming 2 - 7 b. c. Material 2: 4 tons are used, 1 ton is unused. d. No redundant constraints. 16. a. A variety of objective functions with a slope greater than -4/10 (slope of I & P line) will make extreme point (0, 540) the optimal solution. For example, one possibility is 3 S + 9 D . b. Optimal Solution is S = 0 and D = 540. c. Department Hours Used Max. Available Slack Cutting and Dyeing 1(540) = 540 630 90 Sewing 5 / 6 (540) = 450 600 150 Finishing 2 / 3 (540) = 360 708 348 Inspection and Packaging 1 / 4 (540) = 135 135 0 S F
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Chapter 2 2 - 8 17. Max 5 A + 2 B + 0 S 1 + 0 S 2 + 0 S 3 s.t. 1 A - 2 B + 1 S 1 = 420 2 A + 3 B + 1 S 2 = 610 6 A - 1 B + 1 S 3 = 125 A , B , S 1 , S 2 , S 3 0 18. a. Max 4 A + 1 B + 0 S 1 + 0 S 2 + 0 S 3 s.t. 10 A + 2 B + 1 S 1 = 30 3 A + 2 B + 1 S 2 = 12 2 A + 2 B + 1 S 3 = 10 A , B , S 1 , S 2 , S 3 0 b.
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An Introduction to Linear Programming 2 - 9 0 2 4 6 8 B 2 4 6 8 Optimal Solution A A = 18/7, B = 15/7, Value = 87/7 10 12 14 10 c. S 1 = 0, S 2 = 0, S 3 = 4/7 20. a.
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This note was uploaded on 02/24/2010 for the course MGT 621 taught by Professor Smith during the Spring '10 term at Stevens.

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Chapter 2 solutions - Chapter 2 An Introduction to Linear...

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