HW-7%20Solution

# HW-7%20Solution - 4.5.3 Z 1 0 0 0 Step 2 1 0 0 0 Step 3 1 0...

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Unformatted text preview: 4.5.3 Z 1 0 0 0 Step 2 1 0 0 0 Step 3 1 0 0 0 x1 ‐2 3 1 1 0 0 1 0 0 0 1 0 x2 1 1 ‐1 1 1 ‐ 4 ‐1 2 0 0 0 1 1‐ 3 5‐ 2 ‐3 1.5 1 0.5 ‐1.5 x3 ‐1 1 2 0 0 1 0 0 0 1 0 0 s1 0 1 0 0 2 ‐3 1 ‐1 1.5 ‐1 0.5 ‐0.5 s2 0 0 1 1 0 0 s3 0 0 0 20 20 30 0 1 0.5 ‐2 0.5 0.5 RHS 0 60 10 20 Ratio 20 10 15/2 10 10 25 10 15 5 None 5 This is an optimal tableau with optimal solution z = 25, s1 = 10, x1 = 15, x2 = 5, s2 = s3 = 0. 4.6.2 Z ‐1 0 0 x1 ‐1 1 x2 ‐1 ‐1 s1 0 1 0 s2 0 1 0 RHS 0 1qÀ Ratio 0 Step 3 ‐1 0 0 0 0 1 0 2 0 0 1 ‐1 0 0.5 ‐0.5 1 1 0.5 0.5 1 2 1.5 0.5 0.5 This is an optimal tableau with optimal solution z = - 2, x1 = 1.5, x2 = 0.5, s1 = s2 = 0 4.7.2 Z 1 0 0 Step 2 1 0 0 x1 3 5 ‐1 0 8.5 0.5 x2 ‐6 7 2 0 0 1 s1 0 1 0 0 1 0 s2 0 0 1 3 ‐3.5 0.5 RHS 0 35 2 6 28 1 Ratio 5 1 This is an optimal tableau with optimal solution z = 6, s1 = 28, x2 = 1, s2 = x1 = 0. Since the nonbasic variable x1 has a zero coefficient in Row 0 we can enter x1 into the basis to obtain the alternative optimal solution z = 6, x1 = 56/17, x2 = 45/17. By averaging these two optimal solutions, a third optimal solution may be obtained. This yields the optimal solution z = 6, x1 = 28/17, x2 = 31/17. 4.8.1 Z 1 0 0 Step 2 1 0 0 x1 0 1 ‐1 ‐2 0 ‐1 x2 ‐2 ‐1 1 0 0 1 s1 0 1 0 0 1 0 s2 0 0 1 2 1 1 RHS Ratio 0 4 None 1 1 2 5 1 Since x1 has a negative coefficient in Row 0 and a non-positive coefficient in each constraint, we have an unbounded LP. From the final tableau we find that (holding s2 = 0) z = 2 + 2x1 s1 = 5 x2 = 1 + x1 s2 = 0 Thus if 2 + 2x1 = 10,000 or x1 = 4,999 we can find a point in the feasible region with z = 10,000. s1 = 5, x1 = 4,999, x2 = 5,000, s2= 0 is a point in the feasible region having Z=10,000. ...
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## This note was uploaded on 10/19/2009 for the course IEOR 160 taught by Professor Hochbaum during the Fall '07 term at Berkeley.

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HW-7%20Solution - 4.5.3 Z 1 0 0 0 Step 2 1 0 0 0 Step 3 1 0...

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