Application of Duality in Merton trucks

# Application of Duality in Merton trucks - Duality:Merton...

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Duality:Merton Trucks

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Primal and Dual Max Z=3000 X 1 + 5000 X 2 Dual Price P1: X 1 + 2X 2 <=4000 (Y 1 ) P2: 2X 1 +2X 2 <=6000 (Y 2 ) P3: 2X 1 <=5000 (Y 3 ) P4: 3X 2 <=4500 (Y 4 ) X 1 , X 2 >=0 Min Z’=4000 Y 1 + 6000 Y 2 + 5000 Y 3 +4500 Y 4 Dual Price D1: Y 1 + 2Y 2 + 2Y 3 <=3000 (X 1 ) D2: 2Y 1 +2Y 2 + 3Y 4 <=5000 (X 2 ) Y 1 , Y 2 ,Y 3 ,Y 4 >=0
Working Backwards Lets say the optimum is known : X 1 * =2000 and X 2 * =1000 How to compute Y 1 * , Y 2 * ,Y 3 * ,Y 4 * ? Step 1: Lets find the binding constraints in the Primal by plugging the values of X 1 * and X 2 * Constraints P1 and P2 are binding . Constraints P3 and P4 are non-binding. Step 2: Apply Complementary Slackness (S*Vi) (since at optimum) to

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Application of Duality in Merton trucks - Duality:Merton...

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