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Problem 1 a)
maximize 3y 1 + 4y 2
subject to
y 1 + 2y 2 ≤ − 2 y 1 + y 2 ≤ − 1 y 1 , y 2 ≤ 0.
b) x 1 > 0 implies that e 1 = 0 in the dual. s∗ 1 = 3 − x∗ 1 − x∗ 2 = 1 > 0 implies that y 1 = 0 in the dual. Therefore, we must have that 2y 2 = −2 implying y∗ 2 = −1 and y 1 ∗ = 0.
c) Recall that in a minimization problem, the negative of the dual variable corresponding to the primal constraint is the constraint’s shadow price. Therefore, it is −y 2 = 1.
Problem 2 a) The solution x 1 = 1, x 2 = 3 is optimal to the new problem. This is due to the fact that 2(1) + 2(3) = 8 < 10, or, the current optimal solution satisfies the new constraint.
b) First, just add the new constraint/its slack variable as is to the tableau:
Set new row 3 = -(row 2) + row 3:
We have our starting tableau. The third row is our pivot row since its r.h.s is negative. There is only one entry in the third row that is negative in applying the dual min-ratio test. So, s 3 will leave the basis and s 2 will enter. Performing EROS, we obtain 0 0 2 0 1 12 1 0 .5 0 -.5 1. 0 1 0 0 1 2 0 0 -.5 1 -.5.
This tableau is our final tableau since all entries in the r.h.s are non-negative.
Problem 3
a) B−^1 =
[
−^2316
5 3 −
1 6
]
. It appears in the columns of the tableau corresponding to the initial
basis.
b) Need B−^1 b ≥ 0. Or
[
−^2316
5 3 −
1 6
] [
]
[
]
≥ 0. Rearranging 2 + ∆ 6 ≥ 0,
we obtain that ∆ ≥ −12 and rearranging 5 − ∆ 6 ≥ 0 we obtain ∆ ≤ 30. This implies that 28 ≤ b 2 ≤ 70, where b 2 are the number of labor hours, for the current basis to remain optimal.
c) We will insert the functions x 1 = 2 + ∆ 6 and x 2 = 5 − ∆ 6 into the objective function. As a function of ∆, the objective function becomes 2700 + 400∆ 6 + 380(−∆ 6 ) = 2700 + 206 ∆. This implies the shadow price of the second constraint is 206 = 103 = 3.33. Since the objective function is to maximize profits (revenues - costs), the shadow price acts as a premium, so we are willing to pay (cost of labor hour + shadow price of labor hour) = 8.33.
d) In this case, ∆ = −20. Therefore, the tableau for the current basis is: 0 0 11003 103 2700 1 0 −^2316 -^43 0 1 53 -^16253
The first row is our pivot row since the r.h.s is negative. There is only a single entry that is negative and, therefore, s 1 will enter the basis while x 1 leaves. Performing EROS, we end up with our optimal tableau: (^550 0 0 ) −^32 0 1 -^14 5 2 1 0
1 4 5
Problem 4 a) When we increase the r.h.s of a ≤ constraint, we enlarge the feasible region. Therefore, the current optimal solution remains feasible to the new problem and, therefore, we can only improve the objective function. When we increase the r.h.s of a ≥ constraint, we are shrinking the feasible region. Therefore, it is possible that we cut out the optimal solution to the old problem so we would need to move to a worse solution.
b) By way of contradiction, suppose that (D) is feasible. Therefore, there exists a y that is a solution to (D). By weak duality, we have c>x ≥ b>y for all feasible x to (P). This derives our contradiction since we can make the objective function of (P) arbitrarily small (since it is unbounded).
c) Given any x such that Ax ≤ b, we must have that Ax ≤ ¯b since b ≤ ¯b. This implies that x^1 is feasible to (LP2) or, equivalently, c>x^1 provides an upper bound on the optimal solution value to (LP2). Therefore, c>x^1 ≥ c>x^2.
d) The primal problem is infeasible.
Problem 5 a) 2 ≤ c 1 ≤ 4, 1. 5 ≤ c 2 ≤ 3, and −∞ < c 3 ≤ 4.
