chap11_2011

# chap11_2011 - 1 Chapter 11 Discrete Optimization Models...

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Unformatted text preview: 11/15/2011 1 Chapter 11 Discrete Optimization Models 11.1 Lumpy Linear Programs and Fixed Charges • Lumpy linear problems add ―either/or‖ constraints or objective functions to what is otherwise a linear programs. ILP Modeling of All-or-Nothing Requirements • All-or-nothing variable requirements of the form x j = 0 or u j Can be modeled by substituting x j = u j y j , with new discrete variable y j = 0 or 1. [11.1] • The new y j can be interpreted as the fraction of limit u j chosen. 11/15/2011 2 Swedish Steel Blending Example min 16 x 1 +10 x 2 +8 x 3 +9 x 4 +48 x 5 +60 x 6 +53 x 7 s.t. x 1 + x 2 + x 3 + x 4 + x 5 + x 6 + x 7 = 1000 0.0080 x 1 + 0.0070 x 2 + 0.0085 x 3 + 0.0040 x 4 6.5 0.0080 x 1 + 0.0070 x 2 + 0.0085 x 3 + 0.0040 x 4 7.5 0.180 x 1 + 0.032 x 2 + 1.0 x 5 30.0 0.180 x 1 + 0.032 x 2 + 1.0 x 5 30.5 0.120 x 1 + 0.011 x 2 + 1.0 x 6 10.0 0.120 x 1 + 0.011 x 2 + 1.0 x 6 12.0 0.001 x 2 + 1.0 x 7 11.0 0.001 x 2 + 1.0 x 7 13.0 x 1 75 x 2 250 x 1 …x 7 (11.1) Cost = 9953.67 x 1 * = 75, x 2 * = 90.91, x 3 * = 672.28, x 4 * = 137.31 x 5 * = 13.59, x 6 * = 0, x 7 * = 10.91 Swedish Steel Model with All-or-Nothing Constraints • Suppose that the first two ingredients (x 1 , x 2 ) had this lumpy character. We use either none or all 75 kg of ingredient 1 and none or all 250 kg of ingredient 2. • Let y j represents the discrete alternatives ? ? ≡ 1¡¡¡if¡scrap¡j¡is¡part¡of¡the¡blend 0¡¡¡otherwise¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ 11/15/2011 3 Swedish Steel Model with All-or-Nothing Constraints min 16 (75)y 1 +10 (250)y 2 +8 x 3 +9 x 4 +48 x 5 +60 x 6 +53 x 7 s.t. 75y 1 + 250y 2 + x 3 + x 4 + x 5 + x 6 + x 7 = 1000 0.0080 (75)y 1 + 0.0070 (250)y 2 +0.0085x 3 +0.0040x 4 6.5 0.0080 (75)y 1 + 0.0070 (250)y 2 +0.0085x 3 +0.0040x 4 7.5 0.180 (75)y 1 + 0.032 (250)y 2 + 1.0 x 5 30.0 0.180 (75)y 1 + 0.032 (250)y 2 + 1.0 x 5 30.5 0.120 (75)y 1 + 0.011 (250)y 2 + 1.0 x 6 10.0 0.120 (75)y 1 + 0.011 (250)y 2 + 1.0 x 6 12.0 0.001 (250)y 2 + 1.0 x 7 11.0 0.001 (250)y 2 + 1.0 x 7 13.0 x 1 75 x 2 250 x 3 …x 7 0 y 1 , y 2 = 0 or 1 (11.2) Cost = 9967.06 y 1 * = 1, y 2 * = 0, x 3 * = 736.44, x 4 * = 160.06 x 5 * = 16.50, x 6 * = 1.00, x 7 * = 11.00 ILP Modeling of Fixed Charges • Another common source of lumpy phenomena arises when the objective function involves fixed charges . (?) ≡ ? + ??¡¡¡¡¡ if x >0 ¡¡¡¡¡¡¡ 0¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ otherwise • Minimize objective functions with non-negative fixed charges for making variable x j >0 can be modeled by introducing new fixed charge variables ¢ £ ¤ 1¡¡¡¡¡ if x j >0 ¡¡¡¡¡¡ 0¡¡¡¡¡ otherwise The objective coefficient of y j is the fixed cost of x j , and the coefficient of x j is its variable cost. [11.2] 11/15/2011 4 ILP Modeling of Fixed Charges • Switching constraints model the requirement that continuous variable...
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## This note was uploaded on 12/03/2011 for the course ESI 6316 taught by Professor Staff during the Summer '11 term at FIU.

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chap11_2011 - 1 Chapter 11 Discrete Optimization Models...

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