This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Math464 - HW 3 Due on Thursday, Feb 4 1 Linear Optimization (Spring 2010) Brief solutions to Homework 3 1. We rewrite the second constraint for cash availability in the LP formulation (given in the solu- tions to Homework 1) as follows: max 3 x 1 + 3 . 4 x 2 (net profit) s.t. 3 x 1 + 4 x 2 20 , 000 (max machine hours) . 3 x 1 + 0 . 38 x 2 4 , 000 (cash available, including re-financing) x 1 ,x 2 (non-negativity) The line (equation) of the second constraint lies entirely above the line of the first constraint, which connects (6666 . 67 , 0) and (0 , 5000). In other words, the cash availability constraint is redundant . The feasible region is a triangle with origin and the above two points as vertices. The optimal solution is (6666 . 67 , 0), i.e., produce 6666 . 67 units of product 1 and none of product 2. The corresponding maximum net profit is $20,000. To model the situation where the machine hours could be increased by 2000 hours at the cost of $400, change the right-hand-side value of the first constraint to 22000, and that of the second constraint to 3600. The second constraint still remains redundant, and the optimal solution isconstraint to 3600....
View Full Document