In this problem, Total supply = 5 +2 +3 = 10 Total demand = 3 + 3 + 2 + 2 = 10 =

Question

In this problem,

Total supply = 5 +2 +3 = 10

Total demand = 3 + 3 + 2 + 2 = 10 = Total supply

Since Total supply = Total demand, the problem is balanced.

a) The North-West Corner Method :-

The north-west corner method generates an initial allocation according to the following procedure:

1. Allocate the maximum amount allowable by the supply and demand constraints to the variable x11 (i.e. the cell in the top left corner of the transportation tableau).

2. If a column (or row) is satisfied, cross it out. The remaining decision variables in that column (or row) are non-basic and are set equal to zero. If a row and column are satisfied simultaneously, cross only one out (it does not matter which).

3. Adjust supply and demand for the non-crossed out rows and columns.

4. Allocate the maximum feasible amount to the first available non-crossed out element in the next column (or row).

5. When exactly one row or column is left, all the remaining variables are basic and are assigned the only feasible allocation.

Note that there are 3 Sources and 4 Destinations and so m = 3, n = 4

m + n 1 = 6 basic variables

x11 = 3. Cross out column 1. The amount left in row 1 is 2.
x12 = 2. Cross out row 1. 1 unit is left in column 2.
x22 = 1. Cross out column 2. 1 unit is left in row 2.
x23 = 1. Cross out row 2. 1 unit is left in column 3.
x33 = 1. 2 units are left in column 3 and row 3.
x34 = 2.
This provides the initial basic feasible solution x11 = 3, x12 = 2, x22 = 1, x23 = 1, x33 = 1, x34 = 2. The remaining variables are non-basic and therefore equal to zero. The solution has m + n 1 = 6 basic variables as required.

b ) The Least-Cost Method :-

This method usually provides a better initial basic feasible solution than the North-West Corner method since it takes into account the cost variables in the problem.

1. Assign as much as possible to the cell with the smallest unit cost in the entire tableau. If there is a tie then choose arbitrarily.

2. Cross out the row or column which has satisfied supply or demand. If a row and column are both satisfied then cross out only one of them.

3. Adjust the supply and demand for those rows and columns which are not crossed out.

4. When exactly one row or column is left, all the remaining variables are basic and are assigned the only feasible allocation.

For the above example:

• Cells (2, 1) and (2,4) both have 2 cost so we arbitrarily choose the first and assign x21 = 2. Cross out row 2. The amount left in column 1 is 1.

• x11 = 1. Cross out column 1. The amount left in row 1 is 4.

• x32 = 3. Cross out column 2 and row 3.

Explanation / Answer

For the following transportation problem: (a) Identify the initial basic feasible solution using three algorithms (Northwest corner, least cost and penalty cost methods) and (b) Solve the problem to optimality using the algorithm you learnt (stepping stone method) 2 20 10 12 4 10 29 15 20 14 19 16 19 18 15

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