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hw2solution_2011

# hw2solution_2011 - MS&E 211 Linear and Nonlinear...

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Prof. Yinyu Ye Problem 1 (a) Let p 1 , p 2 , p 3 be the number of production cycles that Process 1, Process 2 and Process 3 finish, respectively. Then, the objective function is 38(4 p 1 + p 2 + 3 p 3 ) + 33(3 p 1 + p 2 + 4 p 3 ) - 51 p 1 - 11 p 2 - 40 p 3 = 200 p 1 + 60 p 2 + 206 p 3 and the linear program is formulated as follows: maximize 200 p 1 + 60 p 2 + 206 p 3 subject to 3 p 1 + p 2 + 5 p 3 8 × 10 6 5 p 1 + p 2 + 3 p 3 5 × 10 6 p 1 , p 2 , p 3 0 To use simplex method, rewrite the above LP as the standard form: minimize - 200 p 1 - 60 p 2 - 206 p 3 subject to 3 p 1 + p 2 + 5 p 3 + p 4 = 8 × 10 6 5 p 1 + p 2 + 3 p 3 + p 5 = 5 × 10 6 p 1 , p 2 , p 3 , p 4 , p 5 0 We use (8,5) to represent (8 × 10 6 , 5 × 10 6 ) temporally in iterations for simplicity, and choose the highlighted numbers as the pivot elements in iterations. Its initial simplex tableau is: x 1 x 2 x 3 x 4 x 5 B -200 -60 -206 0 0 0 x 4 3 1 5 1 0 8 x 5 5 1 3 0 1 5 Iteration 1: x 1 x 2 x 3 x 4 x 5 B 0 -20 -86 0 40 200 x 4 0 2/5 16/5 1 -3/5 5 x 1 1 1 / 5 3/5 0 1/5 1 MS&E 211 Linear and Nonlinear Optimization Homework Assignment 2 Solution

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Iteration 2: x 1 x 2 x 3 x 4 x 5 B 100 0 -26 0 60 300 x 4 -2 0 2 1 -1 3 x 2 5 1 3 0 1 5 Iteration 3: x 1 x 2 x 3 x 4 x 5 B 74 0 0 13 47 339 x 3 -1 0 1 1/2 -1/2 3/2 x 2 8 1 0 -3/2 5/2 1/2 The optimal solution to the original LP problem is ( p 1 , p 2 , p 3 ) = 10 6 × (0 , 0 . 5 , 1 . 5) and the optimal value is 339 × 10 6 = 3 . 39 × 10 8 . (b) Assume now we increase the selling price of gasoline to 38+ w , w > 0. Then the objective function becomes (38 + w )(4 p 1 + p 2 + 3 p 3 ) + 33(3 p 1 + p 2 + 4 p 3 ) - 51 p 1 - 11 p 2 - 40 p 3 = (200 + 4 w ) p 1 + (60 + w ) p 2 + (206 + 3 w ) p 3 Then our initial simplex tableau becomes (since we don’t need to worry about the objective value, it’ll be omitted in the tableau) x 1 x 2 x 3 x 4 x 5 B - (200 + 4 w ) - (60 + w ) - (206 + 3 w ) 0 0 * x 4 3 1 5 1 0 8 x 5 5 1 3 0 1 5 Iteration 1: x 1 x 2 x 3 x 4 x 5 B 0 - (20 + w/ 5) - (86 + 3 w/ 5) 0 40 + 4 w/ 5 * x 4 0 2/5 16/5 1 -3/5 5 x 1 1 1 / 5 3/5 0 1/5 1 Iteration 2:
x 1 x 2 x 3 x 4 x 5 B 100 + w 0 -26 0 60 + w * x 4 -2 0 2 1 -1 3 x 2 5 1 3 0 1 5 Iteration 3: x 1 x 2 x 3 x 4 x 5 B 74 + w 0 0 13 47 + w * x 3 -1 0 1 1/2 -1/2 3/2 x 2 8 1 0 -3/2 5/2 1/2 Therefore, the first row of the last tableau is nonnegative for all w > 0, yielding that the optimal solution will not change for all w > 0, i.e., no matter how high the selling price of gasoline goes, the optimal solution will remain the same. (c) We need to add the following constraint to the original linear program: 4 p 1 + 3 p 2 + 5 p 3 10 × 10 6 Adding the new constraint reduces the feasible region so the optimal value must decrease or remain the same. Note that the previous optimal solution doesn’t violate the new constraint so it’s also feasible for the new LP problem and the corresponding objective value is maximum over the new feasible region. Hence, the optimal solution remains the same. (d) The optimal solution won’t change. To see that, notice that maximizing r is equivalent to maximizing r. Thus, we can just solve the linear program.

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hw2solution_2011 - MS&E 211 Linear and Nonlinear...

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