4120_lecture15-F10

# 4120_lecture15-F10 - Computational Methods for Management...

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Computational Methods for Management and Economics Carla Gomes Lecture 15 Reading: 6.5- 6.7 of textbook (slides adapted from: M. Hillier’s, J. Orlin’s, and H. Sarper’s)

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Example of Sensitivity Analysis and Economic Interpretation of Duality Sensitivity Analysis: Computer Analysis
Glass Example • x 1 = # of cases of 6-oz juice glasses (in 100s) • x 2 = # of cases of 10-oz cocktail glasses (in 100s) • x 3 = # of cases of champagne glasses (in 100s) max 5 x 1 + 4.5 x 2 + 6 x 3 (\$100s) s.t 6 x 1 + 5 x 2 + 8 x 3 60 (prod. cap. in hrs) 10 x 1 + 20 x 2 + 10 x 3 150 (wareh. cap. in ft 2 ) x 1 8 (6-0z. glass dem.) x 1 0, x 2 0, x 3 0 (from AMP and slides from James Orlin)

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Z* = 51.4286 Decision Variables x 1 = 6.4286 (# of cases of 6-oz juice glasses (in 100s)) x 2 = 4.2857 (# of cases of 10-oz cocktail glasses (in 100s)) x 3 = 0 (# of cases of champagne glasses (in 100s)) Slack Variables s 1 * = 0 s 2 * = 0 s 3 * = 1.5714 Dual Variables y 1 * = 11/4 = 0.7857 y 2 * = 0.0286 y 3 * = 0 Complementary optimal slackness conditions
• Consider constraint 1. 6 x 1 + 5 x 2 + 8 x 3 60 (prod. cap. in hrs) Let’s look at the objective function if we change the production time from 60 and keep all other values the same. Production hours Optimal obj. value difference 60 51 3/7 61 52 3/14 11/14 62 53 11/14 63 53 11/14 11/14 The dual / shadow Price is 11/14.

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More changes in the RHS Production hours Optimal obj. value difference 64 54 4/7 11/14 65 55 5/14 11/14 66 56 1/11 * 67 56 17/22 15/22 The shadow Price is 11/14 until production = 65.5
The verbal description for the optimum basis for the glass problem: 1. Produce Juice Glasses and cocktail glasses only 2. Fully utilize production and warehouse capacity z = 5 x 1 + 4.5 x 2 6 x 1 + 5 x 2 = 60 10 x 1 + 20 x 2 = 150 x 1 = 6 3/7 (6.4286) x 2 = 4 2/7 (4.2857) z = 51 3/7 (51.4286)

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The verbal description for the optimum basis for the glass problem: 1. Produce Juice Glasses and cocktail glasses only 2. Fully utilize production and warehouse capacity z = 5 x 1 + 4.5x 2 6 x 1 + 5 x 2 = 60 + 10 x 1 + 20 x 2 = 150 x 1 = 6.4286 + 2 /7 x 2 = 4.2857 /7 z = 51.4286 + 11/14 For = 5.5, x 1 = 8, and the constraint x 1 8 becomes binding.
How do we interpret the intervals? If we change one coefficient in the RHS, say production capacity, by the “basis” remains optimal, that is, the same equations remain binding. So long as the basis remains optimal, the shadow prices are unchanged. The basic feasible solution varies linearly with . If is big enough or small enough the basis will change.

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Illustration with the glass example: max 5 x 1 + 4.5 x 2 + 6 x 3 (\$100s) s.t 6 x 1 + 5 x 2 + 8 x 3 60 (prod. cap. in hrs) 10 x 1 + 20 x 2 + 10 x 3 150 (wareh. cap. in ft 2 ) x 1 8 (6-0z. glass dem.) x 1 0, x 2 0, x 3 0 The shadow price is the “increase” in the optimal value per unit increase in the RHS. If an increase in RHS coefficient leads to an increase in optimal objective value, then the shadow price is positive.
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## This note was uploaded on 02/25/2011 for the course AEM 4120 taught by Professor Gomes,c. during the Fall '08 term at Cornell.

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4120_lecture15-F10 - Computational Methods for Management...

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