A package express carrier is considering expanding the fleet of aircraft used to transport packages. Of primary importance is that there is a total of $350 million allocated for purchases. Two types of aircraft may be purchased - the C1A and the C1B

The C1A costs $25 million, while the C1B costs $18 million. The C1A can carry 60,000 pounds of packages, while the C1B can only carry 40,000 pounds of packages. Of secondary importance is that the company needs at least 10 new aircraft. It takes 150 hours per month to maintain the C1A, and 100 hours to maintain the C1B. The least level of importance is that there are a total of 1,200 hours of maintenance time available per month.

(a) First, formulate this as an integer programming problem to maximize the number of pounds that may be carried.

(b) Second, rework the problem differently than in part (a) to suppose the company decides that what is most important to them is that they keep the ratio of C1Bs to C1As in their fleet as close to 1.2 as possible to allow for flexibility in serving their routes. Formulate the goal programming representation of this problem, with the other three goals having priorities P2, P3, and P4, respectively.

(a) Let A = number of C1As to purchase
B = number of C1Bs to purchase

Maximize 60000A + 40000B
Subject to: 25A + 18B ? 350 (budget, in millions)
A + B ? 10 (minimum number of aircraft)
150A + 100B ? 1200 (maintenance)
A, B ? 0 and integer

(b) Let A = number of C1As to purchase
B = number of C1Bs to purchase
d1- = underachievement of ratio of C1B to C1A
d1+ = overachievement of ratio of C1B to C1A
d2- = underachievement of budget target
d2+ = overachievement of budget target
d3- = underachievement of planes
d3+ = overachievement of planes
d4- = underachievement of maintenance hours
d4+ = overachievement of maintenance hours

Minimize P1d1- + P1d1+ + P2d2+ + P3d3- + P4d4+
Subject to: -1.2A + B + d1- - d1+ = 0 (ratio of 1.2 C1Bs to C1As)
25A + 18B + d2- - d2+ = 350 (budget, in millions)
A + B + d3- - d3+ = 10 (minimum number of aircraft)
150A + 100B + d4- - d4+ = 1200 (maintenance)
all variables ? 0