8.7
(a) The BFS are: (0, 0), (175, 0), (0, 150, 0, 0, 0), (400/3, 250/3), (100, 100), (150, 50).
(b) The optimal solutions are elements of the infinite set {(x1 , x2 ) : (x1 , x2 ) = λ1 (400/3, 250/3)+λ2 (100, 100)+
λ3 (175, 0) where λ1 + λ2 + λ3 = 1, λ1 , λ2 , λ3 ≥ 0}.
(c) Note that the canonical form of the problem has three additional slack variables and is:
max
s.t.
25x1 + 50x2
x1 + 2x2 + s1 = 300
2x1 + x2 + s2 = 350
x1 + x2 + s3 = 200
x1 , x2 , s1 , s2 , s3 ≥ 0.
Iteration 1
Starting Simplex from the basic feasible solution x0 = (0, 0, 300, 350, 200) and basis B 0 = {s1 , s2 , s3 }, we
compute the two directions dx1 = (1, 0, −1, −2, −1) and dx2 = (0, 1, −2, −1, −1). Because 50 = c> dx2 >
c> dx1 = 25, we choose to enter dx2 . The step size is:
λ = min{300/ − (−2), 350/ − (−1), 200/ − (−1)} = 150.
So, the new BFS is x1 = (0, 150, 0, 200, 50) and the new basis is B 1 = {x2 , s2 , s3 }.
Iteration 2
The new directions are
dx1 = (1, −1/2, 0, −3/2, −1/2) and ds1 = (0, −1/2, 1, 1/2, 1/2).
The reduced costs are cx1 = 0 and cs1 = −25. Thus, the B 1 is an optimal basis and x1 is an optimal
BFS.
(d) To use Simplex to see if there are multiple optimal solutions, we must verify that the step size is positive
if dx1 is chosen. The minimum ratio test indicates that the step size is
λ = min{150/(1/2), 200/(3/2), 50/(1/2)} = 100.
Because λ > 0 we can conclude there are infinitely many optimal solutions.
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