Linear Programming Duality

Suppose that we also suggested that the optimal solution might be attained at (x1, x2) = (1/2, 5/4), achieving an objective value equal to 2x1 + 3x2 = 19/4. We now construct a certificate for the optimality of the hint. The problem is asking us to maximize 2x1 + 3x2 (with x1, x2 non-negative), under some constraints. Let’s start by considering constraint 1 : since 4x1 + 8x2 ≤ 12, we definitely know that the optimal value of 2x1 + 3x2 is upper bounded by the value of 4x1 + 8x2; hence, 12 is an upper bound on the optimal value of the objective. Now, let’s consider 2 + 3 , i.e. the sum of constraints 2 and 3 : we have 5x1 + 3x2 ≤ 7 ; since definitely 2x1 + 3x2 ≤ 5x1 + 3x2 , a better upper bound on the optimal value of the objective is 7. Let’s now consider 12 1 , that is, let’s multiply constraint 1 by 1/2: we conclude that 2x1 + 4x2 ≤ 6; again, 2x1 + 4x2 is an upper bound on the objective 2x1+3x2 (remember that x1 and x2 are non-negative), and we improve our incumbent upper bound to 6. Now, consider instead 13 1 + 1 3 2 : we get 2x2 + 3x2 ≤ 5, improving the upper bound to 5. We still don’t know if 19/4 is optimal, but we definitely know that whatever the optimal value, it cannot be larger than 5. Finally, consider 5 16 1 + 1 4 3 ; we get 2x1 + 3x2 ≤ 19/4. This proves that an upper bound on the optimal value of the objective is 19/4. This means that the hint was correct, and the optimal value of the problem is indeed 19/4. In other terms, our optimality certificate is the triple (5/16, 0, 1/4), representing the multipliers that we applied to the constraints to obtain an upper bound matching the value we wanted to certify.