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Unformatted text preview: IEOR 162 4.11.1
Z 1 0 0 0 step 1 1 0 0 0 x1 5 4 4 1 0 0 1 0 x2 3 2 1 1 1.75 1 0.25 0.75 s1 0 1 0 0 Spring 2009 Homework 8 s2 0 0 1 0 s3 0 0 0 1 0 0 0 1 rhs 0 12 10 4 12.5 2 2.5 1.5 ratio 3 2.5 0 1.25 1 1 0 0.25 0 0.25 2 10 2 The tie in the ration test indicates that the next b.f.s will be degenerate.
step 2 Z x1 x2 s1 s2 s3 rhs ratio 1 0 0 1.75 0.5 0 16 0 0 1 1 1 0 2 0 1 0 0.25 0.5 0 2 4 0 0 0 0.75 0.5 1 0 0 step 3 1 0 0 1 0 1 16 0 0 1 0.5 0 2 2 0 1 0 0.5 0 1 2 0 0 0 1.5 1 2 0 This is an optimal tableau with the optimal solution being z = 16, x1 = x2 = 2, s2 = s3 = s1 = 0. This bfs corresponds to the following three sets of basic variables: {x1, x2, s1}, {x1, x2, s2}, and {x1, x2, s3}. The degeneracy is due to the fact that the 3 = 2 + 1 (problem has two decision variables!) constraints meet in a single point (point A = (2, 2) in the figure). IEOR 162 4.12.1 Spring 2009 Homework 8 After adding slack, excess and artificial variables we obtain min z = 4x1 + 4x2 + x3 + Ma3 s.t. x1 + x2 + x3 + s1 = 2 2x1 + x2 + s2 = 3 2x1 + x2 + 3x3  e3 + a3 = 3 Eliminating the basic variable a3 from z  4x1  4x2  x3  Ma3 = 0 we obtain z + (2M  4)x1 + (M  4)x2 + (3M  1)x3  Me3 = 3M. The simplex now yields
z 1 0 0 0 z 1 0 0 0 This is an optimal tableau with optimal solution z = 1, s1 = 1, s2 = 3, x3 = 1, x2 = x1 = e3 = 0. x1 2M4 1 2 2 x1 10/3 1/3 2 2/3 x2 M4 1 1 1 x2 11/3 2/3 1 1/3 x3 3M1 1 0 3 x3 0 0 0 1 s1 0 1 0 0 s1 0 1 0 0 s2 0 0 1 0 s2 0 0 1 0 e3 M 0 0 1 e3 1/3 1/3 0 1/3 a3 0 0 0 1 a3 1/3 M 1/3 0 1/3 rhs 3M 2 3 3 rhs 1 1 3 1 IEOR 162 4.12.4 Spring 2009 Homework 8 Adding excess and artificial variables we obtain min z = 3x1 + Ma1 + Ma2 s.t. 2x1 + x2e1 + a1 = 6 3x1 + 2x2 + a2 = 4 Eliminating a1 and a2 from z3x1Ma1Ma2 = 0 yields z + (5M3)x1 + 3Mx2Me1 = 10M. The simplex now yields Z 1 0 0 x1 5M3 2 3 x2 3M 1 2 e1 M 1 0 a1 0 1 0 a2 0 0 1 rhs 10M 6 4 1 0 2M/3 M 0 15M/3 10M/3+4 0 0 1/3 1 1 2/3 10/3 0 1 2/3 0 0 1/3 4/3 This is an optimal tableau. Note, however, that the artificial variable a1 is positive (a1 = 10/3) Thus original problem has no feasible solution. 4.14.2
Let x2 = x2+ x2 , where x2+ >= 0 , x2 >= 0. Applying the simplex we obtain Z 1 0 0 1 0 0 1 0 0 X1 2 3 1 0 1 0 0 1 0 x2+ 1 1 1 1/3 1/3 2/3 0 0 1 x 2 1 1 1 1/3 1/3 2/3 0 0 1 S1 0 1 0 2/3 1/3 1/3 1/2 1/2 1/2 S2 0 0 1 0 0 1 1/2 1/2 3/2 rhs 0 6 4 4 2 2 5 1 3 This is an optimal tableau with optimal solution z = 5, x1 = 1, x2+ = 3, x2 = 0, s1 = s2 = 0. Thus the optimal solution has x2 = 30 = 3. IEOR 162 4.16.2 Spring 2009 Homework 8 Let xi = Number of lots purchased from supplier i. min z = 10s1 + 6s2 + 4s3 + s4+ s.t 60x1 + 50x2 + 40x3 + s1  s1+ = 5,000 20x1 + 35x2 + 20x3 + s2  s2+ = 3,000 20x1 + 15x2 + 40x3 + s3  s3+ = 1,000 400x1 + 300x2 + 250x3 + s4 s4 + = 28,000 All variables nonnegative If we assign priorities in the following order: budget, excellent, good, and mediocre, the goal programming algorithm yields the following optimal solution x1 = 0, x2 = 280/3, x3 = 0. All goals are met except for the excellent chip goal (within the budget constraints there is no way to meet the excellent chip goal). (Excellent Chips) (Good Chips) (Mediocre Chips) (Budget Constraint) ...
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This note was uploaded on 09/06/2009 for the course IEOR 162 taught by Professor Zhang during the Spring '07 term at University of California, Berkeley.
 Spring '07
 Zhang

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