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HW5_solution

# HW5_solution - IEOR 4004 HW#5 solution 1 Simplex Form Max z...

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Unformatted text preview: IEOR 4004 HW #5 solution 1. Simplex Form: Max z s.t. x 4x x 5x 3x x 3x x 0 x 2x 3x s s 4 2 2x x ,x ,x ,x ,s ,s Table 1 Z 1 Table 2 Z 1 Table 3 X1 1/3 4/3 2/3 X1 1 4 2 X2 1 5 3 X3 1 3 1 X4 1 2 3 S1 1 S2 1 RHS 0 4 2 BV Z=0 S1=4 S2=2 Ratio 4/3 2 X2 2/3 5/3 4/3 X3 0 1 0 X4 1/3 2/3 7/3 S1 1/3 1/3 1/3 S2 1 RHS 4/3 4/3 2/3 BV Z=4/3 S1=4/3 S2=2/3 Ratio 2 2/7 X2 X3 X4 S1 Z X1 1 3/7 6/7 0 0 2/7 8/9 9/7 1 0 3/7 2/7 4/7 0 1 1/7 Optimal solution: z=10/7, x3=8/7, x4=2/7, x1=x2=s1=s2=0; S2 1/7 2/7 3/7 RHS 10/7 8/7 2/7 BV Z=10/7 S1=8/7 S2=2/7 Ratio 2. Weak Duality Theorem: If (x1,x2...xn) is feasible for the primal and (y1,y2...yn) is feasible for the dual, then: ; So according to the weak duality theorem, if x*and y* are feasible, Zp(x*) ZD(y*). If Zp is unbounded, Zp is infinitely large. Thus, ZD cannot be larger than infinite, and if X* is feasible (but unbounded), y* will be infeasible. 3. Pg 313 #5 Primal: Dual: Max z=4x1+ x2 Min w=6y1+10y2 s.t. 3y1+ 6y2 4 s.t. 3x1+2x2 6 6x1+3x2 10 2y1+ 3y2 1 x1, x2 0 y1, y2 0 Optimal z+2x2+s2=20/3 *From Dual theorem: if primal is a normal maximum question, the optimal value of 1th dual variable is the coefficient of S1 in row 0 of the optimal primal; Since the optimal value for y1=0 and y2=1, optimal solution of w=10 z=20/3 The primal and dual should has the same optimal solution, so the computation is incorrect. 4. Pg328 #3 Max z=5x1+3x2+ x3 Y2 s.t. 2x1+ x2+ x3 6 x1+2x2+ x3 7 x1, x2, x3 0 Dual: Min w=6y1+7y2 1 s.t. 2y1+ y2 5 (1) y1+2y2 3 (2) 2 y1+ y2 1 (3) Y1 3 y1, y2, y3 0 Optimal for dual: y1=7/3, y2=1/3, w=49/3 For primal: Max z=5x1+3x2+ x3 s.t. 2x1+ x2+ x3 + S1 = 6 x1+2x2+ x3 + S2= 7 x1, x2, x3, s1, s2 0 According to the complementary slackness: Since y1=7/3>0 => s1=0; y2=1/3>0 => s2=0; Thus w=z=49/3, x1 and x2 are the basic variable in the primal: 5x1 +3x2+x3=49/3 2x1+x2=6 => Optimal for primal: z=49/3, x1=5/3, x2=8/3, x3=0. X1+2x2=7 ...
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