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L22_Iterative computations of the Transportation algorithm

# L22_Iterative computations of the Transportation algorithm...

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Iterative computations of the   Transportation algorithm

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Iterative computations of the Transportation algorithm After determining the starting BFS by any one of the three methods discussed earlier, we use the following algorithm to determine the optimum solution Step1: Use the Simplex optimality condition to determine the entering variable as a current non-basic variable that can improve the solution. If the optimality condition is satisfied by all non-basic variables, the current solution is optimal and we stop. Otherwise we go to Step 2.
Step 2. Determine the leaving variable using the Simplex feasibility condition. Change the basis and go to Step 1. The determination of the entering variable from among the current non-basic variables is done by the method of multipliers . In the method of multipliers, we associate with each row a dual variable (also called a multiplier) u i and with each column we associate a dual variable (also called a multiplier) v j .

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Noting that each row corresponds to a constraint and each column corresponds to a constraint we recall from duality theory that At any simplex iteration , Primal z-equation Left hand side Right hand side coefficient of = of corresponding - of corresponding variable x j dual constraint dual constraint That is ij j i ij ij c v u c z - + = - " " (Verify this by taking m=3 and n=4 !)
Since there are m+n-1 basic variables and since 0 = - ij ij c z ij j i c v u = + for all such basic variables, we have m+n-1 equations to determine the m+n variables j i v u , We arbitrarily choose one of them and equate to zero and determine the remaining m+n-1 of them. Then we calculate ij j i ij ij c v u c z - + = - for all non-basic variables x ij . Then the entering variable is that one for which ij j i c v u - + is most positive.

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We do all this on the transportation tableau itself (and NOT separately) as the following example shows.
3 7 6 4     5 2 4 3 2     2 4 3 8 5     3     3    3     2    2 Destination Supply S o u r c e Demand 3 2 1 1 1 2 Thus x 32 enters the basis. u 1 =0 v 1 =3 v 2 =7 u 2 = -3 v 3 =6 u 3 = 2 v 4 =3 Total Cost =48 Starting Tableau 0 -1 -2 -2 1 6

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Determining the leaving variable We first construct a closed loop that starts and ends at the entering variable cell. The loop consists of connected horizontal and vertical segments only (no diagonals are allowed). Except for the entering variable cell, each vertex (or corner) of the closed loop must correspond to a basic variable cell. The loop can cross itself and bypass one or more basic variables . The amount θ to be allocated to the entering variable cell is such that it satisfies all the demand and supply restrictions and must be non-negative.
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