TABLE 3.176 2 0 0 f34 1 0 Cell evaluation matrix Since all cell values are non-negative, i.b.fs. is optimal. Total interest as per the above allocation = [100 0.18 + 200 x 0.16 + 50 * 0.16 + 150 0.16 + 50 * 0.15 + 125 * 0.13 +75 x 0.14] 1,000 => [18+ 32 +8 + 24 +7.5 + 16.25 +10.5] * 1,000 = + 1,16,250. Further, since some of the cell evaluations are zero, alternate optimal solutions exist. To get one such solution, we include, say, cell (P, T) as the basic cell and reallocate as indicated below: TABLE 3.177 TABLE 3.178 100 25 75 20050 150 200125 75 50125 75 125 125 Initial optimal solution with Alternate optimal solution reallocation Total interest as per the alternate solution = 3 (25 * 0.18 + 200 x 0.16 + 125 x 0.16 + 75 x 0,16 + 125 * 0.15 + 125 * 0.13 +75 0.17] 1,000 = 3 [4.5 + 32 + 20 + 12 + 18.75 + 16.25 + 12.75] 1,000 = 1,16,250. Since the total interest remains same, the alternate solution is also optimal EXAMPLE 3.7-5 A complay has factories at A, B and C which supply worehouses at P, Q, R and S. Weekly factory capacities are 250, 300 and 400 units respectively for normal shift. Factory A and B are capable of overtime production of 50 and 75 units per week with an incremental cost of 4 and 5 respectively. The current warehouse requirements are 200, 275, 275 and 300 units. The positions of the points A, B, C, P, Q, R and S are as shown below: 0 B R Fig. 3.2 Roads PQ, OS SR and RP form a square of 10 km side, whereas roads AB, BC and CA form an equilateral triangle. A lies midway between P and Q. Transportation cost is estimated at 2.50 per kilometer. Determine the optimum distribution for minimum cost