Linear Programming-Transportation Simplex

  • Thread starter tobinator250
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In summary, the student is trying to solve for a basic solution for a homework problem, but is stuck because two previously-basic variables have become zero simultaneously.
  • #1
tobinator250
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I don't know if I've posted this in the right place but I thought I'd give it a go anyway.

Homework Statement


See Attatchment


Homework Equations





The Attempt at a Solution



So for part a) max z:150X11+350X12+300X13+100X21+500X22+400X23

s.t. X11+X12+X13≤40
X21+X22+X23≤90
X11+X21≥30
X12+X22≥40
X13+X23≥60

For (b) eliminate first constraint and label constraints: u2,v1,v2,v3

Then solve for us and vs and i get u1=0 u2=100 v1=0 v2=400 and v3=300

Then c*12=50 so not optimal

(c) Enter X12 into basis create loop etc. but that give new solution as X12=40, X21=30 and X23=60. But then when you go through us and vs again to test for optimality, you've got 5 variables to solve for and only four equations so I'm stuck. Help please!
 

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  • #2
tobinator250 said:
I don't know if I've posted this in the right place but I thought I'd give it a go anyway.

Homework Statement


See Attatchment


Homework Equations





The Attempt at a Solution



So for part a) max z:150X11+350X12+300X13+100X21+500X22+400X23

s.t. X11+X12+X13≤40
X21+X22+X23≤90
X11+X21≥30
X12+X22≥40
X13+X23≥60

For (b) eliminate first constraint and label constraints: u2,v1,v2,v3

Then solve for us and vs and i get u1=0 u2=100 v1=0 v2=400 and v3=300

Then c*12=50 so not optimal

(c) Enter X12 into basis create loop etc. but that give new solution as X12=40, X21=30 and X23=60. But then when you go through us and vs again to test for optimality, you've got 5 variables to solve for and only four equations so I'm stuck. Help please!

Your new solution is not basic; you need a basis, and since two previously-basic variables were driven to zero simultaneously, you need to drop one and keep the other at level 0; that is, your new basic solution will be degenerate. Either x13 or x22 should drop out of the basis, leaving the other one in at zero value. One of these choices will lead to optimality. Try one, and if it does not work, then try the other.

RGV
 
  • #3
Thanks a lot :)
 

Related to Linear Programming-Transportation Simplex

1. What is Linear Programming?

Linear Programming is a mathematical method used to optimize resources or achieve a certain goal, given a set of constraints and a linear objective function. It involves using a system of linear inequalities to find the optimal solution.

2. What is the Transportation Simplex method in Linear Programming?

The Transportation Simplex method is a technique used to solve transportation problems in Linear Programming. It involves finding the optimal solution for transporting goods from multiple sources to multiple destinations, while minimizing transportation costs.

3. How does the Transportation Simplex method work?

The Transportation Simplex method works by starting with an initial feasible solution and iteratively improving it until the optimal solution is reached. This is done by identifying and resolving any unmet supply or demand requirements, and then applying the simplex algorithm to find the optimal solution.

4. What are the key assumptions in the Transportation Simplex method?

The key assumptions in the Transportation Simplex method include: 1) all supply and demand values are known and constant, 2) the transportation costs are linear and constant, 3) there are no restrictions on the number of units that can be transported, and 4) there is a single optimal solution.

5. What are the advantages of using the Transportation Simplex method?

The Transportation Simplex method offers several advantages, including: 1) it is a proven and well-established technique, 2) it is efficient and can handle large transportation problems, 3) it provides a feasible and optimal solution, and 4) it allows for sensitivity analysis to explore different scenarios and evaluate the impact of changes in input values.

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