Step 1: Add to the transportation table a column on the E? HS titled u and row in the bottom of it labelled v.
Step 2:
a. Assign any value arbitrarily to a row or column variable u i or vj. Generally a value o ( zero) is assigned to the first row i,e, u 1 = 0
b. Consider every occupied cell in the first row individually and assign the column value vj( when the occupied cell is in the column of the row) which is such that the sum of the row and the column values is equal to the unit cost value in the occupied cell. With the help of these values consider other occupied cells one by one and determine the appropriate values of taking in each case u1+ vj= cij. Thus if ui is the row value of the row and vj is the column value of the column and cij is the unit cost of the cell in the row and column then the row and column values are obtained using the followings equation.
U i + Vj = Cij
For this solution let U1 =0 using equation given earlier we have for the occupied cell( 1 ,1) U1+ V1= C11, OR 0 + V1= 6 OR V1= 6
Similar U1 + V2 = C12 OR 0 + V2 = 4, OR 4 With V2 = 4 ,we get U2= 4( : U2= + 4 = 8) and U3= 0. From U3= 0 we get V3 = 2. The is how Ui values of 0, 4,and o and Vj values of 6 4and 2 are determined.
Step 3: Having determined all Ui and Vj values calculate for each unoccupied cell Δij = Ui + Vj - Cij. The Δ ij represent the opportunity costs of various cells. After obtaining the opportunity costs proceed in the same way as in the stepping stone method. If all the empty cells have negative opportunity costs the solution is optimal and unique. If some empty cell(S) has a zero opportunity costs but if non of the other empty cells have positive opportunity cost then it implies that the given solution is optimal but that it is not unique there exists other solution that would be as good as this solution.