Solve Transportation Problem: NW Corner & Stepping Stone

[mathmari]

Hey!

Suppose ee have a transportation problem and we get a first solution by the North-West-Corner method. Then we want to optimize the solution by the stepping stone algorithm.

We pick an empty cell and we want to see how the cost changes when we would add at this cell an unit.
Then we pick the cell with the most negative result.
The total result/cost always be optimized after that step, right? (Wondering)

Suppose we have the following tableau at one step:

	G1	G2	G3	capacity
L1	20 / 20	18 / 0	40 / 20	40
L2	50 / 0	28 / 30	60 / 0	30
L3	19 / 0	38 / 0	15 / 20	20
Required Amount	20	30	40	90

The amounts after the / is the currect optimal solution.

Is it possible to add an unit at the cell $L_1G_2$ ? When we add there an unit there we delete one from $L_2G_2$, add one at $L_3G_3$ and delete one from $L_1G_3$. Is this correct ? (Wondering)

Only the sum of the elements of one column must always be equal to the required amounts, but the sum of the elements of the rows must be $\leq $ the capacity, or not? (Wondering)

But happens when we add an unit at $L_2G_1$ ? (Wondering)

 

[jvicens]

Hello!

Yes, it is possible to add an unit at the cell $L_1G_2$. When you add an unit there, you will delete one from $L_2G_2$, add one at $L_3G_3$, and delete one from $L_1G_3$. This is correct as it follows the stepping stone algorithm, which involves finding the cell with the most negative result and making adjustments to optimize the solution.

You are correct in saying that the sum of the elements in each column must be equal to the required amounts, and the sum of the elements in each row must be less than or equal to the capacity. This ensures that all the demands are met and the capacity of each source is not exceeded.

When you add an unit at $L_2G_1$, you will delete one from $L_1G_1$, add one at $L_3G_2$, and delete one from $L_2G_3$. This is also correct according to the stepping stone algorithm.

It's important to note that the stepping stone algorithm is an iterative process, so you may need to make multiple adjustments before reaching the optimal solution. I hope this helps clarify the process for you! Let me know if you have any other questions.