The remaining constraints are greater than

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The remaining constraints are “greater than” constraints because the corresponding primal variables x 2 , . . . , x 5 are nonnegative. The variable y 1 is nonnegative because the first con- straint in the primal is a “less than” constraint, while y 2 , y 3 are free because the second and third constraints in the primal are equality constraints. (c) Model file: #hw9dual.mod param m; param n; param A {1..m, 1..n}; param b {1..m}; param c {1..n}; var y {1..m}; minimize objective: sum{j in 1..m} b[j]*y[j]; subject to constrainteq: sum{i in 1..m}A[i, 1]*y[i] = c[1]; subject to constraints {j in 2..n}: sum{i in 1..m} A[i, j]*y[i] >= c[j]; subject to nonnegativity: y[1] >= 0; 2
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Fall 2010 Optimization I (ORIE 3300/5300) Data file #hw9.dat param m = 3; param n = 5; param A: 1 2 3 4 5:= 1 -1 3 -2 4 2 2 -2 -1 3 2 1 3 3 -1 2 -2 -3; param b:= 1 10 2 6 3 -7; param c:= 1 11 2 -3 3 2 4 -8 5 -12; Output: ampl: reset; model hw9dual.mod; data hw9.dat; solve; MINOS 5.5: optimal solution found. 1 iterations, objective -27 ampl: display y; y [*] := 1 0 2 -1 3 3 ; (d) Consider the primal solution x * = ( - 1 , 0 , 0 , 2 , 0) T from part (a) and the dual solution y * = (0 , - 1 , - 3) T from part (c). We first need to show that x * satisfies all the primal constraints and that y * satisfies all the dual constraints. Next, we check that the primal objective value corresponding to x * is - 27 and that the dual objective value corresponding to y * is als o - 27. Since the objective values of x * and y * are equal and they are both feasible for the primal and dual, respectively, then by Corollary 17.7 (Certificate of optimality), x * is an optimal solution of the primal problem and y * is an optimal solution of the dual problem. 3
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Fall 2010 Optimization I (ORIE 3300/5300) Grading scheme: (a) 2 points; (b) 3 points); (c) 2 points; (d) 3 points. (Note: there are various ways to code the LPs in AMPL and the one presented in this solution is only one of many; any correct formulation is accepted.) 4
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