5.a.

Dual formulation: Min 4 y1 + 5 y2 + 7 y3
Subject to, y1 + 2 y2 + 2 y3  3 y1 + 0 y2 + y3  2 2 y1 + 3 y2 + 3 y3  4 y1, y2, y3  0 Final dictionary of primal (verified by tableau obtained by simplex solver),
Z = 10.5 – 1.5 x3 – 2 x4 – 0.5 x5.
X2 = 1.5 – 0.5 x3 – x4 + 0.5 x5
X1 = 1 – 1.5 x3 – 0.5 x5
X6 = 0.5 + 0.5 x3 + x4 + 0.5 x5

We see that in the optimal dictionary for primal, the slack variables have coefficients:
-2 at x4, -0.5 at x5, 0 at x6
Then by applying , we obtain, y1 = 2, y2 = 0.5, y3 = 0

Check optimality and feasibility:
Plug into dual: w = 4 (2) + 5 (0.5) + 7 (0) = 8 + 2.5 + 0 = 10.5, which is equal to z* (= 10.5)­. (as expected by strong duality theorem)
We also see that all the constraints of the dual are satisfied.
So, solution is feasible and optimal.
5.b.

Dual formulation: Min 5 y1 + 3 y2
Subject to, y1 + y2  5 2 y1 + y2  6 3 y1 + 2 y2  9 y1 + 3 y2  8 y1, y2  0
Final dictionary of primal (verified by tableau obtained by simplex solver),
X2 = 2 – x3 + 2 x4 – x5 + x6
X1 = 1 – x3 – 5 x4 + x5 – 2 x6
Z = 17 – 2 x3 – 5 x4 – x5 – 4 x6. We see that in the optimal dictionary for primal, the slack variables have coefficients:
-1 at x5 and -4 at x6
Then by applying , we obtain, y1 = 1, y2 = 4
Check optimality and feasibility:
Plug into dual: w = 5 (1) + 3 (4) = 5 + 12 = 17, which is equal to z*­ (= 17). (as expected by strong duality theorem)
We also see that all the constraints of the dual are