In this method, we suppose that a unit of quantity is taken from a used square and placed in an unused square. We see whether there is any reduction in total cost by this. The cost of squares which are now provided by an additional unit would be added and the costs of the squares from which a unit is taken of are subtracted.

The path thus traced out by the exercise of addition and subtraction is known as close path. The result of the addition and subtraction is known as improved index for that square.

Let us take an example.

We transfer one truck load from W to B. So one truck load would be provided less to A from W. Now to fulfill the rim requirement one more truck should be provided to A from X. This would satisfy all rim requirements.

Let us see the change in cost by this.

Addition to costs:

From W to B Rs 10

From X to A Rs 20

Total Rs 30

Deduction in costs:

From W to A Rs 5

From X to B Rs 30

Rs 35

Net change in cost = Rs 30 – 35 = — 5

This means that this business of additions and subtractions of quantities would reduce your cost by Rs 5 per truckload

-5 is known as improvement index for square WB

We can also do this directly WB – XA+ XA – XA

10 – 30 + 10 – 5 = — 5

Now, let us write the closed paths and improvements indices of all unused squares of our problem.

Unused square Closed Path Improvement Index

WB WB –XB + XA – WA –5

10 – 30 + 20 –5

WC WC – YC + YB – XB +XA –WA –9

10 – 12 + 8 – 30 + 20

XC XC – YC + YB – XB –14

20 – 12 + 8 – 30

YA YA – XA + XB – YB +7

5 – 20 + 30 – 8

Now, we should select an unused square with highest negative improvement index, and make it a used square. Here, XC is such a square. We would add to and subtract from squares mentioned in closed path a fixed quantity. This quantity would be the minimum quantity from all the squares bearing negative signs.

In Figure the squares on the closed path are drawn with their respective positive and negative signs. The inspection of these four squares would show that 20 truckloads are to be added and subtracted. The figure shows the quantity in the four squares after the addition and subtraction operations are carried out. A new matrix would be found with these changes in quantities in different squares. Figure would be said to be first, improved solution.

Let us see the reduction in cost by this improvement

Cost schedule

From plant To destination Quantity Cost Total Cost

(truckloads) (Rs/ Truckload)

W A 35 5 175

X A 10 20 200

X B 10 30 300

X C 20 20 400

Y B 40 8 320

1,395

So, now the cost has reduced from Rs 1,675 to Rs 1,395 i.e. reduction of Rs 280.

Now let us see whether further reduction in cost is possible. Till we get negative improvement index for any unused square, we must assume that further reduction in cost is possible. So, now let us find out closed paths and improvement indices of each of the unused squares first improved solution.

Unused Square Closed Path Improvement paths

WB WB – WA + XA – XB –5

10 – 5 + 20 – 30

WC WC – WA + XA – XB +5

10 – 5 + 20 – 20

YA YA – YB + XB – XA +7

5 – 8 + 30 – 20

YC YC – XC + XB – YB +7

12 – 20 + 30 – 8 + 14

Now we would select the unused square WB (as it has highest negative improvement index) and do additions and subtractions a we had done while obtaining first improved solution. As a result second improved solution would be obtained.

Again, we would have to calculate the total cost

From plant To destination Quantity Cost Total cost

W A 25 5 125

W B 10 10 100

X A 20 20 400

X C 20 20 400

Y B 40 8 320

1,345

So, here we have reduced the total transportation cost.

Now, let us calculate improvement indices or unused square to determine whether any further improvement can be done.

Unused square Closed path Improved Index

WC WC – WA + XA – XC +5

10 – 5 + 20 – 20

XB XB – WB + WA – XA

30 – 10 + 5 – 20 + 5

YA YA – YB + WB – WA

5 – 8 + 10 – 5 +2

YC YC – XC +XA – WA + WB – YB

12 – 20 + 20 – 5 + 10 + 8 + 9

Here, none of the unused squares has negative improvements index. So our second improved solution is the optimum solution. Minimum transportation cost would be involved if the combination of plants and destinations is done according to that solution. In this way we can solve transportation problem by stepping stone method.

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