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- Problem 5.
(1) Since B is a basis matrix, its determinant is nonzero. Consider the deter- minant of B +δE as a function of δ. It is a polynomial in δ, which is nonzero for δ = 0. Therefore, it is also nonzero for all δ sufficiently close to zero. Hence, B + δE is nonsingular for small δ. (2) We have:
xB = (B + δE)−^1 b = (B(I + δB−^1 E))−^1 b = (I + δB−^1 E)−^1 B−^1 b.
(3) By nondegeneracy of x∗, we have B−^1 b > 0 , which implies (B + δE)−^1 b > 0 for sufficiently small δ (because we have a continuous function of δ as long as B + δE is invertible), and therefore primal feasibility is preserved. Because of our dual nondegeneracy assumption, the nonbasic components of c′^ −c′ B B−^1 A are also positive. Using the same continuity argument, they remain positive when B is replaced by B + δE and δ is sufficiently small. We conclude that the optimality conditions are also preserved, for small δ. (4) We have:
c′ B xB = c′ B (B + δE)−^1 b = c′ B (I + δB−^1 E)−^1 B−^1 b ≈ c′ B (I − δB−^1 E)−^1 B−^1 b = c′ B B−^1 b − δc′ B B−^1 EB−^1 b = c′x∗^ − δp 1 x∗ 1.
- Problem 5.
(1) ¯c 3 ≥ 0 and ¯c 5 ≥ 0. (2) Let x 3 enter the basis. Then x 1 exits. The resulting optimal solution is x 3 = 3/ 4 , x 4 = 1/ 2 , x 2 = 1. 75 , x 1 = 0, x 5 = 0. (3) Since 2x 3 + x 4 + γx 5 = 2, x 3 , x 5 ≥ 0 and γ > 0 , x 3 and x 5 are bounded. Thus, even if ¯c 3 , c¯ 5 < 0 , the optimal cost is bounded. (4) We need B−^1 (b + e 1 ) ≥ 0. Now B−^1 b = (1, 2 , 3). Also, B−^1 e 1 = (− 1 , 2 , 4). Hence, we need, 1− ≥ 0 , 2+2 ≥ 0 , 3+4 ≥ 0 , which leads to − 3 / 4 ≤ ≤ 1. (5) Using the formula in page 209 of the text, we need 4 ≤ c¯ 3 and δ ≤ ¯c 5.
- Problem 5.
(1) Let xi be the number of lamps produced in month i, i = 1, 2 , 3 , 4. Let si be the number of lamps purchased from company C in month i, i = 1, 2 , 3 , 4. Let Ii be the inventory in the end of the month i, i = 1, 2 , 3 , 4. The problem company A is facing is as follows: 1
min 35
∑^4
i=
xi + 50
∑^4
i=
si + 5
∑^4
i=
Ii
s.t. x 1 + s 1 − I 1 = 150 x 2 + s 2 + I 1 − I 2 = 160 x 3 + s 3 + I 2 − I 3 = 225 x 4 + s 4 + I 3 − I 4 = 180 xi ≤ 160 , i = 1, 2 , 3 , 4 xi, si, Ii ≥ 0 , i = 1, 2 , 3 , 4
(2) The optimal solution is x∗ 1 = 160, x∗ 2 = 160, x∗ 3 = 160, x∗ 4 = 160, I 1 ∗ = 10 , I 2 ∗ = 10, I 3 ∗ = 0, I 4 ∗ = 0, s∗ 1 = 0, s∗ 2 = 0, s∗ 3 = 55, s∗ 4 = 20, and the minimum cost is $26, 250. (3) The dual prices for the constraints x 1 ≤ 160 , x 2 ≤ 160 , x 3 ≤ 160 are 5,10, respectively. The allowable decrease on the RHS of 160 is 10,10,160 re- spectively. The proposed preventive maintenance will decrease capacity to 151,153,155 respectively. Since these changes are within the allowable ranges, the dual prices are valid. The cost of the maintenance is thus 5 × 9 = 45 for January, 10 × 7 = 70 for February, and 15 × 5 = 75 for March. So, it is profitable to do the maintenance in January. (4) Since the dual prices during the months of January, February, March are 5,10,15, which are smaller than the cost of 45, company A should not buy lamps from company D. (5) Currently s∗ 2 = 0. The reduced cost of this variable is 5. So, if company C gives a discount more than $5/lamp, company A should start buying lamps from company C in February. (6) Since the allowable increase for the cost of I 2 is 5, and the proposed increase is 3, the current solution is optimal. Thus, the optimal solution remains the same, but the cost increases by 3 × 10 = $30. (7) The proposed decrease is 60, whereas the allowable decrease is 55. So, the current basis will change. The optimal cost as a function of the RHS is convex. It is also increasing as the more lamps we produce the higher the cost. The dual price for this constraint is 40, i.e., a lower bound on the cost is thus 26, 250 − 60 × 40 = 23, 850. An upper bound is 26, 250 − 55 × 40 = 24 , 050. If we resolve the problem with 90 instead of 150, we find that the new cost is 23,875.
- Problem 5.
(1) There exist x^1 = B−^1 (b − 10 f ) ≥ 0 and x^2 = B−^1 (b + 10f ) ≥ 0 such that:
Ax^1 = b − 10 f, Ax^2 = b + 10f. Take x = (3/4)x^2 + (1/4)x^1. Then x satisfies: Ax = b + 5f, x ≥ 0. But x = B−^1 (b + 5f ), so for θ = 5, the current basis is feasible. By the optimality of this basis for θ = − 10 , 10 , we have (c − 10 d)′^ − (c − 10 d)B B−^1 A ≥ 0
