a9sol - C O350 L INEAR O PTIMIZATION S OLUTIONS 9 Exercise...

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CO350 L INEAR O PTIMIZATION - S OLUTIONS 9 Exercise 1. (1) A = " 1 1 2 0 2 2 3 3 1 2 # , b = " 3 7 # , c = [5 , 8 , 4 , 2 , 3] T . We use x 4 as basic variable. (2) Phase 1 is the LP A = " 1 1 2 0 2 1 2 3 3 1 2 0 # , b = " 3 7 # , c = [0 , 0 , 0 , 0 , 0 , - 1] T . Iteration 1, B = { 6 , 4 } and x * B = [3 , 7] T (a) Solve A T B y = c B " 1 0 0 1 # " y 1 y 2 # = " - 1 0 # y = " - 1 0 # (b) Choose entering variable c 1 = c 1 - A T 1 y = 0 - [1 , 2][ - 1 , 0] T = 1 > 0 So x 1 enters the basis. (c) Choose leaving variable - solve A B d = A 1 " 1 0 0 1 # " d 6 d 4 # = " 1 2 # " d 6 d 4 # = " 2 1 # t = min d i > 0 x * 6 d 6 , x * 4 d 4 = min 3 1 , 7 2 = 3 r = 6 so x 6 is leaving. (d) x * 1 := t = 3 and x * 4 := x * 4 - td 4 = 7 - 3 × 2 = 1 (e) B = { 1 , 4 } and all artificial variables are zero, hence the first phase is over. (3) For Phase II we have B = { 1 , 4 } and corresponding x * = [3 , 1] T together with A = " 1 1 2 0 2 2 3 3 1 2 # , b = " 3 7 # , c = [5 , 8 , 4 , 2 , 3] T . Iteration 1, (a) Solve A T B y = c B " 1 2 0 1 # " y 1 y 2 # = " 5 2 # y = " 1 2 # 1
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2 (b) Choose entering variable c 2 = c 2 - A T 2 y = 8 - [1 , 3][1 , 2] T = 1 > 0 So x 2 enters the basis. (c) Choose leaving variable - solve A B d = A 2 " 1 0 2 1 # " d 1 d 4 # = " 1 3 # d = " 1 1 # t = min d i > 0 x * 1 d 1 , x * 4 d 4 = min 3 1 , 1 1 = 1 hence x 4 is leaving.
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