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 }+ V_{j} = C_{ij}
For this solution let U_{1} =0 using equation given earlier we have for the occupied cell( 1 ,1) U_{1}+ V_{1}= C_{11}, OR _{0 }+ V_{1}= 6 OR V_{1}= 6
Similar U_{1} + V_{2} = C_{12} OR 0 + V_{2} = 4, OR 4 With V_{2} = 4 ,we get U_{2}= 4( : U_{2}= + 4 = 8) and U_{3}= 0. From U_{3}= 0 we get V_{3} = 2. The is how U^{i} values of 0, 4,and o and V_{j} values of 6 4and 2 are determined.
Step 3: Having determined all U_{i} and V_{j } values calculate for each unoccupied cell Δ_{ij }= U_{i} + V_{j} - C_{ij}. 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.