Ch03 - 1 MP CHAPTER 3 SOLUTIONS SECTION 3.1 1. max z = s.t....

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MP CHAPTER 3 SOLUTIONS SECTION 3.1 1. max z = 30x 1 + 100x 2 s.t. x 1 + x 2 7 (Land Constraint) 4x 1 + 10x 2 40(Labor Constraint) 10x 1 30(Govt. Constraint) x 1 0, x 2 0 2a. No, government constraint is violated. 2b. No; Labor constraint is not satisfied. 2c. No, x 2 0 is not satisfied. 2d. Yes, all constraints and sign restrictions are satisfied. 3. 1 bushel of corn uses 1/10 acre of land and 4/10 hours of labor while 1 bushel of wheat uses 1/25 acre of land and 10/25 hours of labor. This yields the following formulation: max z = 3x 1 + 4x 2 s.t. x 1 /10 + x 2 /25 7 (Land Constraint) 4x 1 /10 + 10x 2 /25 40 (Labor Const.) x 1 30 (Govt. Const.) x 1 0, x 2 0 4. x 1 = Number of Type 1 Trucks produced daily x 2 = Number of Type 2 Trucks produced daily Expressing profit in hundreds of dollars we obtain the following formulation: max z = 3x 1 + 5x 2 s.t. x 1 /800 + x 2 /700 1 (Paint Shop Const.) x 1 /1500 + x 2 /1200 1 (Engine Shop Const.) x 1 0, x 2 0 5. If there are > and or < constraints, then a problem may have no optimal solution. Consider the problem max z = x st x<1. Clearly, this problem has no optimal solution (there is no largest number smaller than 1!!) 1
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Section 3.2 1. EF is 4x 1 + 10x 2 = 40, CD is x 1 = 3, and AB is x 1 + x 2 = 7. The feasible region is bounded by ACGH. The dotted line in graph is isoprofit line 30x 1 + 100x 2 = 120. Point G is optimal. At G the constraints 10x 1 30 and 4x 1 + 10x 2 40 are binding. Thus optimal solution has x 1 = 3, x 2 = 2.8 and z = 30(3) + 100(2.8) = 370. 2
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2. AB is x 1 /800 + x 2 /700 = 1. CD is x 1 /1500 + x 2 /1200 = 1. Feasible region is bounded by ABE. Dotted line is z = 3x 1 + 5x 2 = 1500. Moving isoprofit line up and to right we find optimal solution to be where x 1 0 and Paint Shop constraint are binding. Thus x 1 = 0, x 2 = 700, z = 3500(in hundreds) is optimal solution. 3
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3. x 1 = Number of hours of Process 1 and x 2 = Number of hours of Process 2. Then the appropriate LP is min z = 4x 1 + x 2 s.t. 3x 1 + x 2 10 (A constraint) x 1 + x 2 5 (B constraint) x 1 3 (C constraint) x 1 x 2 0 AB is 3x 1 + x 2 = 10. CD is x 1 + x 2 = 5. EF is x 1 = 3. The feasible region is shaded. Dotted line is isocost line 4x 1 + x 2 = 24. Moving isocost line down to left we see that H (where B and C constraints intersect) is optimal. Thus optimal solution to LP is x 1 = 3, x 2 = 2, 4
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4a. We want to make x 1 larger and x 2 smaller so we move down and to the right. 4b. We want to make x 1 smaller and x 2 larger so we move up and to left. 4c. We want to make both x 1 and x 2 smaller so we move down and to left. 5. Let x 1 = desks produced, x 2 = chairs produced. LP formulation is max z = 40x 1 + 25x 2 s.t. -2x 1 + x 2 0 4x 1 + 3x 2 20 x 1 , x 2 0 Graphically we find the optimal solution to be x 1 = 2, x 2 = 4 and z = 180. 5
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This note was uploaded on 04/02/2008 for the course IEOR 160 taught by Professor Hochbaum during the Fall '07 term at University of California, Berkeley.

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Ch03 - 1 MP CHAPTER 3 SOLUTIONS SECTION 3.1 1. max z = s.t....

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