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**Unformatted text preview: **-y 1 + 2 y 4 =-3 , 2 y 1 + y 4 = 4. Solving this gives the dual solution ( 11 5 , , ,-2 5 ) T , which is feasible. Hence the solution is optimal. 2. (a) (D) min { y T b : A T y ≥ c, y ≥ } (D’) min { 0 : A T y ≥ c, y ≥ } (b) Since (P) is unbounded, (D) is infeasible. Since both (D) and (D’) have the same set of constraints, (D’) is infeasible. (c) Since (D’) is infeasible, (P’) is either infeasible or unbounded. However, y = 0 is a feasible solution to (P’). Therefore, (P’) must be unbounded. (d) Since (P’) is unbounded, there exists a series of feasible solution d such that c T d → ∞ . This means that there exists a feasible solution for which c T d > 0. Such a solution satsifies Ad ≤ , d ≥ , c T d > 0....

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