# hw3sol - EGM6365 Homework#3 1 A linear programming problem...

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Unformatted text preview: EGM6365 Homework #3 1. A linear programming problem is given as Maximize f = 2 x1 + 5 x2 Subject to 3 x1 + 2 x2 ≤ 12 2 x1 + 3x2 ≥ 6 (1) x1 ≥ 0, x2 unrestricted (i) (ii) Solve Eq. (1) using MATLAB. Convert Eq. (1) into the standard form and solve the standards problem using MATLAB. Compare two results. (i) Original Problem >> f=[-2; -5]; >> A=[3 2;-2 -3]; >> b=[12; -6]; >> lb=[0; -inf]; >>[x,fval,exitflag,output,lambda]=linprog(f,A,b,,,lb) x = [0, 6] f = -30 exitflag = 1 output = iterations: 5 cgiterations: 0 algorithm: 'lipsol' Thus, the optimum point is x* = (0, 6) and the value of the objective function is f* = 30. (ii) Standard Problem Let new design variable X = [x1, y1, y2, s1, s2], where x2 = y1 – y2. >> >> >> >> >> f = [-2; -5; 5; 0; 0]; Aeq = [3 2 -2 1 0; 2 3 -3 0 -1]; beq = [12; 6]; lb = [0; 0; 0; 0; 0; 0]; [x, fval, exitflag, output, lambda] = ... linprog(f, , , Aeq, beq, lb) x =[0.0000, 136.7656, 130.7656, 0.0000, 12.0000] f = -30 exitflag = 1 output = iterations: 5 cgiterations: 0 algorithm: 'lipsol' Note that x2 = 136.7656 – 130.7656 = 6.0. Two problems converge to the same solution. 2. Solve the linear programming problem using MATLAB. Submit all your input and output data. Maximize f = 2 x1 + 5 x2 − 4.5 x3 + 1.4 x4 Subject to 5 x1 + 3 x2 + 1.5 x3 ≤ 8 −1.8 x1 − 6 x2 + 4 x3 + x4 ≥ 3 −3.6 x1 + 8.2 x2 + 7.5 x3 + 5 x4 = 15 xi ≥ 0, i = 1 to 4 >> f = [-2; -5; 4.5; -1.4]; >> A=[5 3 1.5 0; 1.8 6 -4 -1]; >> Aeq = [-3.6 8.2 7.5 5]; >> b = [8; -3]; >> beq = [15]; >> lb = zeros(4,1); >> [x, fval, exitflag, output] = ... linprog(f,A,b,Aeq,beq,lb,,,optimset('Display','iter')); Residuals: Primal Dual Duality Total Infeas Infeas Gap Rel A*x-b A'*y+z-f x'*z Error --------------------------------------------------Iter 0: 2.03e+03 1.31e+01 2.47e+03 3.90e+02 Iter 1: 2.15e+02 2.43e-15 2.42e+02 1.24e+01 Iter 2: 2.27e+00 5.62e-15 1.30e+01 6.05e-01 Iter 3: 1.96e-14 2.22e-16 6.46e+00 7.15e-01 Iter 4: 4.99e-12 1.09e-15 2.99e-01 6.65e-02 Iter 5: 3.69e-13 3.64e-15 1.52e-02 3.44e-03 Iter 6: 1.69e-14 3.67e-15 7.74e-07 1.75e-07 Iter 7: 1.02e-14 1.68e-15 7.74e-13 1.74e-13 x = [1.4164, 0.0000, 0.6119, 3.1020] fval = -4.4221 Thus, the optimum design is x* = [1.4164, 0.0, 0.6119, 3.1020] and the objective function is f* = 4,4221. All three constraints are active and corresponding Lagrange multipliers are [0.0278, -2.6567, 0.8113]. Note that the λ2 < 0 because the original constraint is greater than or equal to type. ...
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