Assignment10Solution

# Of the ampl book that the dual value or shadow price

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of the AMPL book that the dual value, or shadow price, corresponds to the profit gained by having one additional unit on the right-hand side of a constraint. Below is the result if we used [option display precision 0] : 2

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Fall 2010 Optimization I (ORIE 3300/5300) ampl: option display_precision 0; ampl: model prod.mod; data plonk.dat; solve; display X; MINOS 5.5: optimal solution found. 3 iterations, objective 2447.5 X [*] := ADWhite 200 BRed 300 lred 82.50000000000006 ; ampl: display Total_Profit + Limit["BRed"]; Total_Profit + Limit[’BRed’] = 2448.125 ampl: let u["BRed"] := 301; ampl: solve; MINOS 5.5: optimal solution found. 1 iterations, objective 2448.125 Grading scheme: Problem 3 (a) The dual problem: min 7 y 1 + y 2 s.t. 6 y 1 - 6 y 2 4 2 y 1 - y 2 5 - y 1 + 3 y 2 - 5 3 y 1 - 2 y 2 8 y 1 , y 2 free (b) Use complementary slackness to check whether or not each of the following x is an optimal solution to the primal problem: (i) To see if x = [1 , 2 , 3 , 0] T is optimal for the primal problem, we need to (1) check that this is a feasible solution for the primal (plugging in the values to the primal constraints, we easily verify that x is indeed feasible), and (2) find a vector y which satisfies the complementary slackness condition: A T j y = c j whenever x j > 0 . (That is, if x j is positive, then the j th dual constraint is satisfied with equality.) Since x 1 , x 2 , x 3 are positive, then the first, second, and third dual constraint must be satisfied with equality: 6 y 1 - 6 y 2 = 4 2 y 1 - y 2 = 5 - y 1 + 3 y 2 = - 5 . 3
Fall 2010 Optimization I (ORIE 3300/5300) Solving for y 1 , y 2 using the first two of the equality constraints above, we obtain y 1 = 4 1 3 , y 2 = 3 2 3 . However, this solution does not satisfy the third constraint with equality. So, there does not exists ( y 1 , y 2 ) that is feasible for the dual and satisfies the comple- mentary slackness condition with x . Therefore, we can conclude that the given x is not optimal for the primal problem.
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