empty cells to see whether the use of any of them would reduce total cost. If we find such a route, then we will allocate as much as possible to it. First, let us consider allocating one ton of wheat to cell 1A. If one ton is allocated to cell 1A, cost will be increased by $6—the transportation cost for cell 1A. However, by allocating one ton to cell 1A, we increase the supply in row 1 to 151 tons, as shown in Table B-12.
A requirement of this solution method is that units can only be added to and subtracted from cells that already have allocations. That is why one ton was added to cell 3B and not to cell 2B. It is from this requirement that the method derives its name. The process of adding and subtracting units from allocated cells is analogous to crossing a pond by stepping on stones (i.e., only allocated-to cells).
By allocating one extra ton to cell 3B we have increased cost by $5, the transportation cost for that cell. However, we have also increased the supply in row 3 to 276 tons, a viola- tion of the supply constraint for this source. As before, this violation can be remedied by subtracting one ton from cell 3A, which contains an allocation of 200 tons. This satisfies the supply constraint again for row 3, and it also reduces the total cost by $4, the transportation cost for cell 3A. These allocations and deletions are shown in Table B-14.