Solution to Midterm

# Solution to Midterm - TIi = trucks in inventory at the end...

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Solution to IEOR 162 Midterm Fall 2007 (30 points total for any three questions) 1. The optimal solution is found to be z = 12, x 1 = 0, x 2 = 6. Grading: Feasible region (5) Isoprofit line (3) Unique solution (2) 2. Let x 1 = Units of Process 1. x 2 = Units of Process 2. A = Ounces of Chemical A produced. B1 = Ounces of Chemical B that are sold. B2 = Ounces of Chemical B that must be destroyed. Then a correct formulation is max z = 16A+14B1-2B2 s.t. 2x 1 + 3x 2 50 (Use at most 60 labor hours) x 1 + 2x 2 30 (Use at most 40 units of raw material) A = 2x 1 + 3x 2 (Expresses total amount of A produced) B1 + B2 = x 1 + 2x 2 (Expresses total amount of B produced) B1 20 (Can sell at most 20 ounces of B) x 1 , x 2 , A, B1,B2 0 Grading: Decision variables: 3 points Objective: 3 Constraints: 4 3. Let Ti = trucks produced during month i Ci = cars produced during month i Si = steel bought during month i CIi = cars in inventory at the end of month i

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Unformatted text preview: TIi = trucks in inventory at the end of month i min z = 400S1 + 600S2 + 150(TI1 + CI1 + TI2 + CI2) st CI1 = 200 + C1 - 800 CI2 = CI1 + C2 - 300 TI1 = 100 + T1 - 400, C1 + T1 1000 TI2 = TI1 + T2 - 300, C2 + T2 1000 2T1 + C1 S1, S1 1500 2T2 + C2 S2, S2 1500 (20C1 + 10T1)/ (C1 + T1) 16 or 4C1 - 6T1 (20C2 + 10T2)/ (C2 + T2) 16 or 4C2 - 6T2 ALL VARIABLES 0 Grading: 10 points total Missing each constraint: -2 (until it hits 0) 4. Decision Variables X 1 = Daytime Calls, X 2 = Evening Calls Objective Minimize Z = 2 X 1 + 5 X 2 Constraints 0.30 X 1 + 0.30 X 2 150 0.10 X 1 + 0.30 X 2 120 0.10 X 1 + 0.15 X 2 100 0.10 X 1 + 0.20 X 2 110 1 X 1 1 X 2 1 X 1 , 1 X 2 0 Grading: Decision variables: 3 points Objective: 3 Constraints: 4...
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## This note was uploaded on 09/06/2009 for the course IEOR 162 taught by Professor Zhang during the Spring '07 term at University of California, Berkeley.

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Solution to Midterm - TIi = trucks in inventory at the end...

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