{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

L22_Iterative computations of the Transportation algorithm

L22_Iterative computations of the Transportation algorithm...

Info icon This preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Iterative computations of the   Transportation algorithm
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 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 .
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 !)
Image of page 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.
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
We do all this on the transportation tableau itself (and NOT separately) as the following example shows.
Image of page 6
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
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern