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 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 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.