Question

26) Given the following feasible solution, determine if the problem is degenerate and then find the optimal solution and its cost. Assume that capacity for source A is 10 and 30 for source B Destination A demands 10 units while destination B demands 30 units. Cost of shipping per unit is given as AA ($4), AB ($1), BA ($3), and BB ($2) Destination B Destination A Source A 10 Source B 30

Answer 1

Degeneracy test: m+n-1 = 2+2-1 = 3. The number of positive independent allocations (occupied cells) must be 3, but there are only 2 occupied cells in the given feasible solution. Therefore, the initial feasible solution is degenerate. In order to remove degeneracy, allocate a very small positive number (practically 0) to the empty cell with the lowest cost, which is AB.

Resulting feasible solution is:

	Destination A	Destination B
Source A	10 4	0   1
Source B	3	30 2

Cost = 10*4+0*1+30*2 = 100

Improvement index of empty cells is calculated by alternatively adding and subtracting the cost of cells which form a closed loop with the empty cell at one of its corner and non-empty cells at all the other corners. Closed loop for empty cell BA is formed by cells BA-AA+AB-BB

Improvement index of the empty cell BA = 3-4+1-2 = -2

Improvement index of empty cell is negative. which means the solution is not optimal.

Empty cell is allocated using stepping stone method.

Using the closed loop for this cell, allocate the minimum value of the non-empty cell with negative sign (as used in calculation of improvement index). That cell is AA and the minimum value is 10. So 10 is subtracted from AA, added to AB, subtracted from BB, and added to BA. Resulting tableau is following:

	Destination A	Destination B
Source A	4	10 1
Source B	10 3	20 2

Cost = 10*1+10*3+20*2 = 80

Again calculate the improvement index of empty cell AA, which is = 4-1+2-3 = 2

This is positive. Therefore, this solution is optimal.

Optimal solution is:

	Destination A	Destination B
Source A		10
Source B	10	20

Its cost = 80