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Unformatted text preview: CO350 LINEAR OPTIMIZATION  SOLUTIONS 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 , , , , , 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 # ⇒ y = " 1 # (b) Choose entering variable c 1 = c 1 A T 1 y = 0 [1 , 2][ 1 , 0] T = 1 > 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 > 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 ....
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This note was uploaded on 01/05/2012 for the course CO 350 taught by Professor S.furino,b.guenin during the Spring '07 term at Waterloo.
 Spring '07
 S.Furino,B.Guenin

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