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Question

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Problem 3-02 (Algorithmic) Consider the following linear program Max 3A2B 1A 1B S 3A 1B S 1A 2B S A, B 2 0 8 21 16 The value of the optimal solution is 24. Suppose that the right-hand side of the constraint 1 is increased from 8 to 9 a. Use the graphical solution procedure to find the new optimal solution (i) 26 (ii) B 26 24 24 20 18 16 14 12 10 20 18 16 12 10 Optimal Solution A- 6,B 3 3A 2B 24 Optimal Solution A-5.2, B 5.4 3A+2B 26.4 2 4 6 8 10 12 14 16 18 20 2 4 6 8 10 12 14 16 18 20

Explanation / Answer

a) The correct graph is (i)

b) Dual value of constraint 1 = 1.5

(if right-hand-side (RHS) of constraint 1 is changed by 1 unit, objective function will change in the same direction by 1.5 units)

c) The right-hand-side range for constraint 1 is = RHS - Allowable decrease = 8 - 1 = 7

to  RHS value + Allowable increase = 8+2.6 = 10.6

d) The improvement in the value of the optimal solution will be 0.5   for every unit increase in the RHS of constraint 2 as long as the RHS is between = 21 - 13 = 8 to 21+3 = 24

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