demand is exhausted ( b < a ), move one cell right horizontally tothe second column and allocate as much as possible.(i.e) x = min ( a – x , b ) If the supply is exhausted ( b > a ), move one cell down vertically to the second row and allocateas much as possible.(i.e) x = min ( a , b – x ) If both supply and demand are exhausted move one cell diagonally and allocate as much as possible. Step : Continue the above procedure until all the allocations are made Example . Obtain the initial solution for the following problem. Destination Supply Sources Demand Here total supply = + + + = , Total demand = + + = (i.e) Total supply =Total demand.
Therefore the given problem is balanced transportation problem. ∴ we can find an initial basic feasible solution to the given problem. From the above table we can choose the cell in the North West Corner. Here the cell is ( , A ) Allocate as much as possible in this cell so that either the capacity of first row is exhausted or the destination requirement of the first column is exhausted.
i.e. x = min ( , ) = Supply ( a i ) ( ) / Demand ( b j ) / XII Std - Business Maths & Stat EM Chapter - - Operations Research Reduced transportation table is a i b j Now the cell in the North west corner is ( , A ). Allocate as much as possible in the first cell so that either the capacity of second row is exhausted or the destination requirement of the first column is exhausted. i.e.
x = min ( , ) = a i ( ) / b j / Reduced transportation table is a i b j Here north west corner cell is ( ,B) Allocate as much as possible in the first cell so that either the capacity of second row is exhausted or the destination requirement of