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Unformatted text preview: Homework 6 Additional Problem: Solution 1. My model file can be found in hw6apsoln.mod . The optimal solution is: black 1 carnitas 0.909091 cheese 0.0909091 chicken 0.0909091 lettuce 1 med_tom 1 mild 1 (All other variables are 0.) The optimal objective value is 153.18. 2. The optimal burrito contains black beans, cheese, chicken, hot salsa, lettuce, and mild salsa. Its objective value is 120. The solution tree looks like this: All 1 1 1 Ingredient: carnitas barbacoa 144.09 153.18 infeas F(3) infeas F(3) 133.5 med_corn 133.5 Z*=120 F(1) 1 infeas 120 F(2) hot F(3) The first incumbent solution is found on the x = 1 branch for med_corn . At the next step, the same solution is found, and it can be fathomed for reason #2. In total, 9 branchandbound nodes were required. 3. (153 . 18 120) / 120 = 27 . 7% 4. Optimal solution and objective value are the same as given in part 1. 5. The new constraint for If_Mild_Then_Spicy is given by subject to If_Mild_Then_Spicy {m in MEATS: is_spicy[m] = 0}:...
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 Spring '07
 Storer
 Optimization, LP, objective value, 20.8%, 27.7%, hw6apsoln.mod

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