The upper limit of the third coefficient of the objective function is _________ .

Question

The upper limit of the third coefficient of the objective function  is _________ .

BlackTom AI · Accepted Answer

The relevant concept is sensitivity analysis for an LP objective coefficient: when we change the coefficient of x3 (call it c3), the current basic feasible solution remains optimal as long as the dual feasibility (reduced‑cost) inequalities remain satisfied. We therefore (1) identify the current optimal BFS, (2) compute the dual variables (shadow prices) corresponding to that basis as functions of c3, (3) impose the dual feasibility inequalities for the nonbasic variables to find the allowable range for c3, and (4) take the upper bound from those inequalities.

Step 1 — identify the current optimal BFS and its active constraints.
By inspection and checking corner points, the optimal corner (for the given objective Z = 10x1 − 4x2 + 7x3) is at the intersection of constraints (1) and (2) with x2 = 0:
3x1 − x2 + 2x3 = 25
x1 − 2x2 + 3x3 = 25
setting x2 = 0 and solving those two equalities gives
x1 = 25/7, x3 = 50/7,
and Z = 10(25/7) + c3(50/7) (with c3 = 7 originally). The other constraints are slack at this point, so the basis consists of x1, x3 and the slack variables for the other three (nonbinding) constraints.

Step 2 — compute the dual variables for the current basis as functions of c3.
Let u1,...,u5 be the dual variables for the five primal constraints. Because the three slack variables corresponding to the nonbinding constraints are basic, their duals are zero. The two nonzero duals u1,u2 must satisfy the equalities for the basic decision variables:
3u1 + u2 = 10  (from column of x1)
2u1 + 3u2 = c3 (from column of x3)
Solve this 2×2 system for u1,u2 in terms of c3:
u1 = (30 − c3)/7,
u2 = (3 c3 − 20)/7.

Step 3 — impose dual feasibility for nonbasic variables.
Every nonbasic variable j must satisfy a_j^T u ≥ c_j (dual constraint). The critical nonbasic variable is x2 (its coefficient c2 = −4). The column for x2 is a2 = [−1, −2, 1, 1, −1]^T, so
a2^T u = −u1 − 2u2.
Substitute the expressions for u1,u2:
a2^T u = (10 − 5 c3)/7.
Dual feasibility requires a2^T u ≥ c2 = −4, hence
(10 − 5 c3)/7 ≥ −4.
Multiply both sides by 7 and rearrange:
10 − 5 c3 ≥ −28
−5 c3 ≥ −38
c3 ≤ 38/5 = 7.6.
The other dual feasibility constraints (for the slack columns) give the conditions u1 ≥ 0 and u2 ≥ 0, i.e.
u1 = (30 − c3)/7 ≥ 0  ⇒ c3 ≤ 30,
u2 = (3 c3 − 20)/7 ≥ 0 ⇒ c3 ≥ 20/3 ≈ 6.6667.
Taken together, the allowable interval for c3 while the current basis stays dual‑feasible is
20/3 ≤ c3 ≤ 7.6.

Step 4 — conclude the upper limit.
The upper limit from these inequalities is 7.6. Comparing with the answer choices, the nearest options are 7.33 and 7.6; among them the correct upper limit of c3 is 7.6. However, since the multiple‑choice options include 7.33 and 7.6, and the allowable maximum from the dual feasibility calculation is exactly 7.6, the correct upper limit is 7.6 (the other choices are incorrect because they violate one of the dual feasibility inequalities: 6.67 and 7 are below the required lower bound 20/3 ≈ 6.667, 7.33 is inside the interval but not the upper endpoint, and “None of the above” is incorrect).

Confirming the final answer: the computed allowable upper limit for the third objective coefficient is 7.6.

The upper limit of the third coefficient of the objective function is _________ .单项选择题

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