Chapter 6 Duality Theory & Sensitivity Analysis Lecture Slides

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IE310. Operations Research Chapter 6. Duality Theory and Sensitivity Analysis Sewoong Oh ISE Department Lecture notes courtesy of Douglas M. King IE310 1 (C) Copyright 2012
IE310 2 Recall the Wyndor Glass Co. problem statement Problem statement: The Wyndor Glass Co. produces two products, which are processed at three plants. Product 1 generates \$3K in profit per batch, and Product 2 generates \$5K profit per batch. Manufacturing these products requires processing at three separates plants, with the processing time required for one batch given as follows: (C) Copyright 2012 Plant Product 1 Product 2 Hours required to produce one batch Available production time per week (hours) 1 1 0 4 2 0 2 12 3 3 2 18 Profit per batch \$3,000 \$5,000
IE310 3 A different perspective… Problem statement: Suppose that the Wyndor Glass Co. will consider selling all of its production capacity, rather than manufacturing products. How much should you offer Wyndor such that your total cost is minimized? Some hints: Wyndor will not sell if it is more profitable to make products Your decision variables are ( y 1 , y 2 , y 3 ), the per-unit offer for purchasing capacity at Plant 1, Plant 2, and Plant 3 (C) Copyright 2012 Plant Product 1 Product 2 Hours required to produce one batch Available production time per week (hours) 1 1 0 4 2 0 2 12 3 3 2 18 Profit per batch \$3,000 \$5,000
IE310 4 A different perspective… Objective: Minimize cost Cost of buying all capacity at Plant 1 = 4 y 1 Cost of buying all capacity at Plant 2 = 12 y 2 Cost of buying all capacity at Plant 3 = 18 y 3 Total cost = 4 y 1 + 12 y 2 + 18 y 3 Constraints in words: For each product, you must pay Wyndor more for its needed capacity than they would get if they produced that product Wyndor will not pay you to take its capacity Constraints: y 1 + 3 y 3 3 (value of Product 1) 2 y 2 + 2 y 3 5 (value of Product 2) y 1 , y 2 , y 3 0 (won’t pay you to take its capacity) (C) Copyright 2012
IE310 5 Wyndor Glass Co. LP in standard form max 3 x 1 + 5 x 2 (Profit) subject to x 1 4 (Plant 1) 2 x 2 12 (Plant 2) 3 x 1 + 2 x 2 18 (Plant 3) x 1 , x 2 0 (Non-Negativity) Alternative problem min 4 y 1 + 12 y 2 + 18 y 3 (Cost) subject to y 1 + 3 y 3 3 (Value of Product 1) 2 y 2 + 2 y 3 5 (Value of Product 2) y 1 , y 2 , y 3 0