Tutorial_4 - 4 1 = X 2 2 = X complementary slackness...

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IELM202’05 Tutorial 4 1. Consider the linear program (P) Max Z= 2 1 5 3 X X + Subject to 4 1 X 12 2 2 X 18 2 3 2 1 + X X 0 , 2 1 X X Write down the dual of this LP According to the following table Primal model (MAX) Dual model (MIN) Constraint j is ≤ Variable yj ≥ 0 Constraint j is = Variable yj is unrestricted Constraint j is ≥ Variable yj ≤ 0 Variable xi ≥ 0 Constraint i is ≥ Variable xi is unrestricted Constraint i is = Variable xi ≤ 0 Constraint i is ≤ The Dual is Min W= 3 2 1 18 12 4 y y y + + St. 0 , 0 , 0 5 2 2 3 3 3 2 1 3 2 3 1 + + y y y y y y y 2. Max Z= 2 1 2 3 X X + s.t. 10 2 2 1 + X X 6 2 3 2 1 + - X X 6 2 1 + X X 0 , 0 2 1 X X (a) Construct a dual for this primal problem. (b) Show that , 4 1 = X 2 2 = X is an optimal to (P) by solving its dual using the Complementary Slackness Theorem.
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IELM202’05 Tutorial 4 (a) the dual is Min W= 3 2 1 6 6 10 y y y + + s.t. 3 3 2 3 2 1 + - y y y 2 2 3 2 1 + + y y y 0 1 y 2 y 0, 3 y 0 (b) Check primal feasibility: 2*4+2=10<=10 -3*4+2*2=-8<=6 4+2=6<=6. Hence the give primal solution is feasible. Obtain dual solution using complementary slackness implies. Since
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Unformatted text preview: , 4 1 = X 2 2 = X , complementary slackness implies S 1 =10-10=0 S 2 =6-(-8)=14 S 3 =6-6=0 Since S i *y i =0 then we got the conclusion 2 = y (1) 3 3 2 3 2 1 1-+-= y y y e 2 2 3 2 1 2-+ + = y y y e * 1 1 = e x * 2 2 = e x Since X 1 =4 and X 2 =2 so 1 = e and 2 = e We got 3 3 2 3 2 1 =-+-y y y (2) And 2 2 3 2 1 =-+ + y y y (3) Solving equation (1), (2), and (3), we have 1 1 = y = 2 y 0, = 3 y 1 ( ) , , * 3 * 2 * 1 y y y =(1,0,1) W ( ) , , * 3 * 2 * 1 y y y = 3 2 1 6 6 10 y y y + + =16 Z( ) , * 2 * 1 x x =3*4+2*2=16 Z*=W* Based on Strong Duality Theorem: In a primal-dual LP pair, if either the Primal or the Dual has an optimal feasible bounded solution, then the two optimal objective values are equal...
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Tutorial_4 - 4 1 = X 2 2 = X complementary slackness...

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