HW1_Solutions

# HW1_Solutions - OSCM 230 Homework 01 Solution LINEAR...

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OSCM 230 Homework 01 Solution H O M E W O R K   # 1 L I N E A R   P R O G R A M M I N G   ( L P )   F O R M U L A T I O N Problem 2-40   a. Let: X = number of pounds of stock X purchased per cow each month Y = number of pounds of stock Y purchased per cow each month Z = number of pounds of stock Z purchased per cow each month Four pounds of ingredient A per cow can be transformed to: 4 pounds × (16 oz/lb) = 64 oz per cow 5 pounds = 80 oz   Ingredient B 1 pound = 16 oz   Ingredient C 8 pounds = 128 oz   Ingredient D 3X + 2Y + 4Z ≥ 64 (ingredient A requirement) 2X + 3Y + 1Z ≥ 80 (ingredient B requirement) 1X + 0Y + 2Z ≥ 16 (ingredient C requirement) 6X + 8Y + 4Z ≥ 128 (ingredient D requirement) Z ≤ 5 (stock Z limitation) X, Y, Z ≥ 0 Minimize cost = \$3X + \$4Y + \$2.25Z   in 1 st  edition: [\$2X + \$4Y + \$2.50Z] b. Solution: (See file P2-40.XLS) Cost =\$112  [\$80] X = 16 lbs   [40 lbs]. of X Y = 16 lbs   [0 lbs]. of Y Z = 0 lbs.    [0 lbs] of Z

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## This note was uploaded on 09/21/2009 for the course B 362 taught by Professor Trus during the Spring '09 term at Washington University in St. Louis.

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HW1_Solutions - OSCM 230 Homework 01 Solution LINEAR...

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