Rec10_sol - Fall 2010 Optimization I (ORIE 3300/5300)...

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Fall 2010 Optimization I (ORIE 3300/5300) Recitation 10 Solution (a) The solution to the primal problem is: ampl: model Abc.mod; data Abc.dat; solve; MINOS 5.5: optimal solution found. 2 iterations, objective 19 ampl: display X; X [*] := 1 2 2 0 3 1 4 0 5 1 6 0 Model File (Primal Problem): param n > 0; # number of variables param m > 0; # number of constraints param A {1. .m, 1. .n}; # constraint matrix param b {1. .m}; # right-hand side param c {1. .n}; # objective function coefficients var X {1. .n} >= 0; maximize Objective: sum {j in 1. .n} c[j]*X[j]; subject to Constraint {i in 1. .m}: sum {j in 1. .n} A[i,j]*X[j] = b[i]; Data File: param m := 3; param n := 6; param A: 1 2 3 4 5 6 := 1 2 1 3 1 0 0 2 2 3 1 0 1 0 3 1 1 2 0 0 1 ; param b := 1 7.0 2 6.0 3 4.0; param c := 1 5.0 2 3.0 3 9.0 4 0.0 5 0.0 6 0.0; (b) Not shown. 1
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Fall 2010 Optimization I (ORIE 3300/5300) (c) The solution to the dual problem is: ampl: model dualAbc.mod; data Abc.dat; solve; MINOS 5.5: optimal solution found. 1 iterations, objective 19
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This note was uploaded on 01/25/2011 for the course ORIE 5300 taught by Professor Todd during the Fall '08 term at Cornell.

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Rec10_sol - Fall 2010 Optimization I (ORIE 3300/5300)...

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