Hart manufacturing makes three products. each product requires manufacturing operations in three departments: a, b and c. the labor-hour requirements, by department, are:

2 department product 1 product 2 product 3

a 1.50 3.00 2.00

b 2.00 1.00 2.50

c 0.25 0.25 0.25

during the next production period, the labor-hours available are 450 in department a, 350 in department b, and 50 in department c. the profit contributions per unit are $25 for product 1, $28 for product 2, and $30 for product 3. a- (10 points) formulate a linear programming model for maximizing total profit contribution. provide the formulation.

and then

- (10 points) management realized that the optimal product mix, taking setup costs into account, might be different from the one recommended in part (b). formulate a mixed integer linear program that takes setup cost into account. management has also stated that we should not consider making more than 175 units of product 1, 150 of product 2, or 140 units of product 3. provide the formulation. setup costs are $400 for product 1, $550 for product 2, and $600 for product 3

Consider the following calculation

Explanation:

Let Pi  = units of product i produced

Max 25P1 + 28P2 + 30P3

s.t.

1.5P1 + 3P2 + 2P3 = 450

2P1 + 1P2 + 2.5P3 = 350

.25P1 + .25P2 + .25P3 = 50

P1, P2, P3 = 0

b. The optimal solution is

P1 = 60

P2 = 80 Value = 5540

P3 = 60

This solution provides a profit of $5540.

c. Since the solution in part (b) calls for producing all three products, the total setup cost is

$1550 = $400 + $550 + $600.

Subtracting the total setup cost from the profit in part (b), we see that

Profit = $5540 - 1550 = $3990

d. We introduce a 0-1 variable yi

that is one if any quantity of product i is produced and zero

otherwise.

With the maximum production quantities provided by management, we obtain 3 new constraints:

P1 = 175y1

P2 = 150y2

P3 = 140y3

Bringing the variables to the left-hand side of the constraints, we obtain the following fixed charge

formulation of the Hart problem.

Max 25P1 + 28P2 + 30P3 - 400y1 - 550y2 - 600y3

s.t.

1.5P1 + 3P2 + 2P3 = 450

2P1 + 1P2 + 2.5P3 = 350

.25P1 + .25P2 + .25P3 = 50

P1 - 175y1 = 0

P2 - 150y2 = 0

P3 - 140y3 = 0

P1, P2, P3 = 0; y1, y2, y3 = 0, 1

P1 = 100 y1 = 1

P2 = 100 y2 = 1 Value = 4350

P3 = 0 y3 = 0

The profit associated with this solution is $4350. This is an improvement of $360 over the solution  in part (c).

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