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the primal problem remains feasible. If A be increase in the first resource, it can be determined from the condition [215 -17515+41 f60 B 'b 1/5 275 8 + 2A 1075+ 2275-2751 5 5/5+4/5+4/5 9+A [215 215][*74) 20] As x and x, remain feasible (20) for all values of A 20, the first resource can be increased indefinitely while maintaining the condition that each additional unit will increase the objective function by 29/5 (@) The second resource should be decreased as each additional unit of the second resource decreases the objective function by 2/5. Let A be the decrease in the second resource. To find its extent, we make use of the condition that the current solution remains feasible so long as B 'b= [275 -1/5715 175 2/52- 8+A 10/5-275+4/57 5 575 +475-21/5) 9-24 20] 9-2A Evidently x, remains positive only so long as 20 or A S 9/2. 5 If A > 9/2, x; becomes negative and must leave the solution. EXAMPLE 6.6-1.2 Given the following L.P. problem: Maximize 2- x + 2x2x subject to 3x + x>- *s 10, - *, + 4x2 + x3 26, * + xy S4 X, X, X, 20, and the optimal solution given by table 6.59, find the separate ranges of b, b, and by consistent with the optimal solution TABLE 6.59 1 0 0 -M Basis X X4 A 0 1 1 0 3 4 -1 2 2 0 - 3 -2 -M Xy CB 0 2 0 X4 X2 Xs Z 3 0 1 0 -1 X2 0 1 0 2 1 0 0 0 0 Xy 0 0 1 0 b 6 4 10 0 0