Recitation4Solution - Fall 2010 Optimization I (ORIE...

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Fall 2010 Optimization I (ORIE 3300/5300) Recitation 4 Solution Winery Problem First, we solve the winery problem again using the model and data files provided ( prod.mod , plonk.dat ): ampl: model prod.mod; data plonk.dat; solve; MINOS 5.5: optimal solution found. 3 iterations, objective 2447.5 ampl: display X; X [*] := ADWhite 200 BRed 300 lred 82.5 ; Following the instruction, ampl: display Limit.body, Limit; : Limit.body Limit := ADWhite 200 0.5 BRed 300 0.625 lred 82.5 0 ; As we can read on page 35, Limit.body displays the “body” in the Limit constraints. In this case, it displays x[j] for each j in P . Meanwhile, as we can read on page 17, Limit displays the “dual values” or the marginal values associated with the Limit constraints. These dual values describe how the objective function (in our case, total profit) would change if the right hand side of the corresponding constraint is increased, given the current optimal solution. For instance, Limit[’ADWhite’] = 0.5 means that if the demand for AD White wine had increased (so we can sell more each day), then we would gain an additional profit of $0.50 for each additional unit up to some level. Limit[’BRed’] = 0.625 means that if the demand for Big Red wine had increased (so we can sell more each day), then we would gain an additional profit of $0.625 for each additional
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Recitation4Solution - Fall 2010 Optimization I (ORIE...

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