max x � 13 x 1 7 x 2 12 x 3 s t 2 x 1 3 x 2 x 3 w 4 5 4 x 1 7 x 2 2 x 3 w 5 11

# Max x ? 13 x 1 7 x 2 12 x 3 s t 2 x 1 3 x 2 x 3 w 4 5

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max x ζ = 13 x 1 + 7 x 2 - 12 x 3 s . t . 2 x 1 + 3 x 2 - x 3 + w 4 = 5 - 4 x 1 - 7 x 2 + 2 x 3 + w 5 = - 11 3 x 1 - 4 x 2 - 2 x 3 + w 6 = - 8 x 1 , x 2 , x 3 , w 4 , w 5 , w 6 0 . Consider the following auxiliary problem min t 7 + t 8 s . t . 2 x 1 + 3 x 2 - x 3 + w 4 = 5 - 4 x 1 - 7 x 2 + 2 x 3 + w 5 - t 7 = - 11 3 x 1 - 4 x 2 - 2 x 3 + w 6 - t 8 = - 8 x 1 , x 2 , x 3 , w 4 , w 5 , w 6 , t 7 , t 8 0 . Lizhi Wang ([email protected]) IE 534 Linear Programming September 21, 2012 8 / 12 The auxiliary problem Define K = { i : b i < 0 , i = 1 , ..., m } , and let k = |K| . Define LP1 as min x,w, t ( k X i =1 t i : Ax + w + H t = b ; x, w, t 0 ) . I t R k × 1 is a new variable called the artificial variable. I H R m × k is defined as H i,j = - 1 , if i = K ( j ) ; 0 , otherwise. , i = 1 , ..., m ; j = 1 , ..., k. Lizhi Wang ([email protected]) IE 534 Linear Programming September 21, 2012 9 / 12 Starting fbp for LP1 The dimensions of the partition for LP1 are |N| = n + k and |B| = m. The following basic partition is a fbp for LP1 . ( N 1 = { 1 , ..., n } ∪ { n + K} , B 1 = { 1 , ..., n + m + k }\N 1 ) The fbp ( B 1 , N 1 ) uniquely determines the following fbs. ( x B 1 = A - 1 B 1 b = | b | , x N 1 = 0) Lizhi Wang ([email protected]) IE 534 Linear Programming September 21, 2012 10 / 12 Relationship between LP0 and LP1 LP0 : max x,w { ζ = c > x : Ax + w = b, x 0 , w 0 } . LP1 : min x,w, t ( k X i =1 t i : Ax + w + H t = b ; x, w, t 0 ) . Let ( x * , w * , t * ) be an optimal solution to LP1 . If t * = 0 , then ( x * , w * ) can be used as a fbs (and its corresponding partition as a fbp) to continue with the Simplex iterations.  #### You've reached the end of your free preview.

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