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Homework Solutions 01

# Homework Solutions 01 - IEOR E4007 G Iyengar Sept 24th 2008...

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IEOR E4007 G. Iyengar Sept. 24th, 2008 Solutions to Homework #1 1. Simple linear program (a) The feasible region of the LP is shown below (0,0) (0,2) (0.5,0) (b) The optimal solution is the corner (0 , 2) with an optimal value 2. (c) The slacks in the main constraints are as follows: 12 x 1 +3 x 2 6s l a c k= 0 3 x 1 + x 2 7s l a c 5 x 2 10 slack = 8 (d) The dual linear program is given by min 6 u 1 +7 u 2 +10 u 3 subject to 12 u 1 3 u 2 2 3 u 1 + u 2 + u 3 1 , u 1 ,u 2 3 0 . Since the second and third main constraints are slack, complementary slack- ness implies that u 2 = u 3 =0 . Since x 2 > 0, complementary slackness implies that 3 u 1 + u 2 + u 3 =1 . Thus , u 1 / 3. (e) The answer depends on your interpretation of the question. If you simply Fnd the RHS range from AMPL or Excel, that gives you the range for each coeﬃcient assuming the other coeﬃcients are held constant. Under this assumption, we can solve the problem as following. If we perturb the Frst component by c 1 , the solution (0 , 2) is optimal as long as the objective vector lies within the cone generated by the active constraints. The 1

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active constraint at (0 , 2) is the frst constraint and x 1 0. The outer normal vector at these two points are (12 , 3) and ( 1 , 0). So (0 , 2) is optimal iF there exists non-negative λ 1 2 such that (2 + r 1 , 1) = λ 1 (12 , 3) + λ 2 ( 1 , 0) This implies 1 = 3 λ 1 λ 1 = 1 3 .A l so 2+ r 1 =12 λ 1 λ 2 r 1 =4 λ 2 2 r 1 = λ 2 r 1 2 So we can increase c 1 by 2 and decrease it by infnity. Similarly, we can perturb the second component (2 , 1+ r 2 )= λ 1 (12 , 3) + λ 2 ( 1 , 0) This implies 2= 1 2 λ 1 λ 2 r 2 =3 λ 1 The frst equation implies an additional constraint on λ 1 12 λ 1 =2+ λ 2 λ 1 = 1 6 + λ 2 12 λ 1 1 6 So now the second equation implies r 2 λ 1 1 2 r 2 ≥− 1 2 So we can decrease c 1 by a halF and increase it by infnity. The other interpretation is what is the range oF the objective Function iF all coor- dinates oF the objective vector can change simultaneously. ( c 1 ,c 2 λ 1 (12 , 3) + λ 2 ( 1 , 0) This means c 2 λ 1 c 2 0. Substitute this into the frst component gives you c 1 λ 1 λ 2 c 1 c 2 λ 2 4 c 2 c 1 = λ 2 4 c 2 c 1 Which gives you the fnal answer oF 4 c 2 c 1 and c 2 0. There is a more general method using duality. The corner (0 , 2) will remain optimal, as long as there exists a dual Feasible u which satisfes complementary slackness. 2
Since the second and third main constraints are slack, complementary slack- ness implies that u 2 = u 3 =0 . Since x 2 > 0, complementary slackness implies that 3 u 1 + u 2 + u 3 = c 2 .Thu s , u 1 = c 2 / 3 c 2 0. This value of u 1 must satisfy the Frst constraint in the dual, i.e. 12 u 1 c 1 ,o r equivalently 4 c 2 c 1 . Thus, the current corner (0 , 2) remains optimal as long as 4 c 2 c 1 and c 2 0.

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Homework Solutions 01 - IEOR E4007 G Iyengar Sept 24th 2008...

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