# GE330_lect12 - Lecture 12 Transportation Algorithm March 9...

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Lecture 12 Transportation Algorithm March 9, 2009

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Last Lecture The transportation model - The transportation problem - The LP formulation Variants of the transportation model - The assignment model - The transhipment model - General minimum cost ﬂow problems 2
The Transportation Model The general transportation model: min m X i =1 n X j =1 c ij x ij s.t. n X j =1 x ij = a i , i = 1 , 2 , ··· ,m m X i =1 x ij = b j , j = 1 , 2 , ··· ,n x ij 0 , i = 1 , 2 , ··· ,m ; j = 1 , 2 , ··· ,n (1) It is in the standard form with m + n constraints and m × n variables. One of the equations is redundant. A BFS has m + n - 1 basic variables. 3

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The Transportation Model For the example we talked about in the last lecture min z = 80 x 11 + 215 x 12 + 100 x 21 + 108 x 22 + 102 x 31 + 68 x 32 s.t. x 11 + x 12 = 1000 x 21 + x 22 = 1500 x 31 + x 32 = 1200 x 11 + x 21 + x 31 = 2300 x 12 + x 22 + x 32 = 1400 x ij 0 , The A matrix is x 11 x 12 x 21 x 22 x 31 x 32 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 1 0 1 0 1 0 0 1 0 1 0 1 4
Simplex Method: General Form (for Min) 1. We have a basic columns A B (1) , . . . , A B ( m ) and basic solution x B = B - 1 b , B = { B (1) , . . . , B ( m ) } . 2. Compute the reduced cost for nonbasic x j , j / B , by ¯ c j = c T B B - 1 A j - c j . If ¯ c j 0 , stop - we have an optimal solution. Otherwise, select an index j with ¯ c j > 0 and go to Step 3. 3. Compute direction d = B - 1 A j ( j will enter the basis, this is corre- sponding to the ratio test). If d 0 (all coordinates), the optimal value is -∞ . Otherwise, ﬁnd θ * such that θ * = min { ` | d ` > 0 } x B ( ` ) d ` . 4. Let l * be the index where the above min is attained. Form a new basis by bringing x j in the basis and x ` leaving the basis. The new basic solution has x new B ( i ) = x B ( i ) - θ * d i for i 6 = ` , and x new j = θ * .

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The Dual of the Transportation Model The general transportation model: min m X i =1 n X j =1 c ij x ij s.t. n X j =1 x ij = a i , i = 1 , 2 , ··· ,m ······ u i m X i =1 x ij = b j , j = 1 , 2 , ··· ,n ······ v j max m X i =1 a i u i + n X j =1 b j v j s.t. u i + v j c ij , i = 1 , 2 , ··· ,m ; j = 1 , 2 , ··· ,n 5
Primal-Dual Relationship Revisit Primal min c T x s.t. Ax = b x 0 Dual max b T y s.t. A T y c If x * is an optimal solution of the primal problem with corre- sponding basis

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GE330_lect12 - Lecture 12 Transportation Algorithm March 9...

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