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hw1solutions

# hw1solutions - w 4 Thus we pivot with x 1 as the entering...

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ORF 307 Homework 1 Solutions Exercise 1. 2.1 Initialization: Write down the initial dictionary 0 +6 x 1 +8 x 2 +5 x 3 +9 x 4 w 1 = 5 - 2 x 1 - x 2 - x 3 - 3 x 4 w 2 = 3 - x 1 - 3 x 2 - x 3 - 2 x 4 The largest positive coeﬃcient in the objective is 9 for x 4 . We then calculate min( 5 3 , 3 2 )= 3 2 corresponding to w 2 . Thus we pivot with x 4 as the entering variable and w 2 as the leaving variable. Add 9 2 *row2 to the objective and - 3 2 *row2 to row1. The resulting dictionary is: 27 2 + 3 2 x 1 - 11 2 x 2 + 1 2 x 3 - 9 2 w 2 w 1 = 1 2 - 1 2 x 1 + 7 2 x 2 + 1 2 x 3 + 3 2 w 2 x 4 = 3 2 - 1 2 x 1 - 3 2 x 2 - 1 2 x 3 - 1 2 w 2 The rest of the pivots are shown using the pivot tool on the following page. The optimal solution is ( x 1 , x 2 , x 3 )=(2,0,1) and Z=17. 1

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pivot on w1, x1 pivot on x4, x2 pivot on w4,x1 2
Exercise 2. 2.2 Initialization: Write down the initial dictionary 0 +2 x 1 + x 2 w 1 = 4 +2 x 1 - x 2 w 2 = 3 - 2 x 1 - 3 x 2 w 3 = 5 - 4 x 1 - x 2 w 4 = 1 - x 1 - 5 x 2 The largest positive coeﬃcient in the objective is 2 for x 1 . We then calculate min(2, 3 2 , 5 4 ,1) = 1 corresponding to
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Unformatted text preview: w 4 . Thus we pivot with x 1 as the entering variable and w 4 as the leaving variable. Add 2*row4 to the objective, -2*row4 to row1, -2*row4 to row2, and-4*row4 to row3. The resulting dictionary is: 2-2 w 4-9 x 2 w 1 = 2-2 w 4 +9 x 2 w 2 = 1 +2 w 4 +7 x 2 w 3 = 1 +4 w 4 +19 x 2 x 1 = 1-w 4-5 x 2 Now notice the the coﬃcients in the objective are all negative so we have an optimal solution: ( x 1 , x 2 )=(1,0) and Z=2. You should also include a plot of the feasible region -a two dimensional plot of the problem constraints. An arrow should be drawn from the starting point (0,0) to the optimal point (1,0) which is reached after one pivot. 3...
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hw1solutions - w 4 Thus we pivot with x 1 as the entering...

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