University:
The University Of Western Ontario
Course:
MATH 3020A | Introduction to Abstract Algebra
Academic year:

2024
Views:

398

Pages:

2
Author:

Kyuwon J.

Question: Consider the following linear programming (LP) problem: 2x1 + 4x2 3x1 + 2x2 subject to −x1 −x2 Maximize 50x1 + 60x2 , ≤ 60, ≤ 40, ≤ 0, ≤ 0. At the point x̄ = (0, 15), determine the cone of the tight constraints. Specifically: • Identify which constraints are active (tight) at x̄ = (0, 15). • Represent the system of active constraints mathematically in matrix form. • Derive the cone of the tight constraints based on this representation. Solution: To determine the cone of the tight constraints, we analyze each constraint in the LP problem to check whether it is satisfied with equality at the given point x̄ = (0, 15). The constraints in the LP problem are as follows: 1. 2x1 + 4x2 ≤ 60, 2. 3x1 + 2x2 ≤ 40, 3. − x1 ≤ 0, 4. − x2 ≤ 0. Step 1: Evaluate Constraints at x̄ = (0, 15) Let us substitute x1 = 0 and x2 = 15 into each constraint to determine whether it is tight: • For 2x1 + 4x2 ≤ 60: Substituting x1 = 0 and x2 = 15, we get: 2(0) + 4(15) = 60. Since the left-hand side equals the right-hand side, this constraint is tight. • For 3x1 + 2x2 ≤ 40: Substituting x1 = 0 and x2 = 15, we get: 3(0) + 2(15) = 30 < 40. Since the inequality is strict, this constraint is not tight. • For −x1 ≤ 0: Substituting x1 = 0, we get: −0 = 0. This constraint is tight. • For −x2 ≤ 0: Substituting x2 = 15, we get: −15 < 0. Since the inequality is strict, this constraint is not tight. 1 Step 2: Identify Active Constraints Based on the evaluation above, the active (tight) constraints at x̄ = (0, 15) are: 2x1 + 4x2 = 60 and − x1 = 0. Step 3: Represent the Active Constraints in Matrix Form The tight constraints can be written as a system of linear equations: 2 4 x1 0 = . −1 0 x2 0 Here, the rows of the matrix represent the coefficients of the tight constraints. Step 4: Determine the Cone of the Tight Constraints The cone of the tight constraints is formed by the directions that satisfy this system of equations. Mathematically, the cone is spanned by the columns of the transpose of the coefficient matrix. Specifically, we compute: 2 −1 Cone: span , . 4 0 2 Step 5: Interpretation of the Cone The cone represents all linear combinations of the vectors and 4 −1 . These vectors indicate the directions in which we can move within the feasible region while 0 remaining along the boundaries defined by the active constraints. Thus, the cone of the tight constraints at x̄ = (0, 15) is: 2 −1 span , . 4 0 2