In: Operations Management
A craftsman builds two kinds of birdhouses, one for wrens and one for bluebirds. Each wren birdhouse takes 3 hours of labor and 4 units of lumber. Each bluebird house requires 2 hours of labor and 10 units of lumber. The craftsman has available 80 hours of labor and 100 units of lumber, and he wants to build at least 6 wren houses. Wren houses profit $8 each and bluebird houses profit $16 each. How many of each kind of birdhouses should be built in order to maximize total profit? Formulate this as a linear programming problem (i.e., DO NOT solve it.)
Answer: Maximize the objective value, because its the profit
No. of wren houses to make is 25 units
No. of bluebird houses to make is 0 units
Maximum profit is $200
Explanation:
W is no. of wren houses
B is no. of bluebird houses
Labor hours: 3*W + 2*B ≤ 80---- Eqn1
lumber units: 4*W + 10*B ≤ 100---- Eqn2
W >= 6---- Eqn3
Profit of Objective, maximize 8*W + 6*B = z
Cell no. | B | C | D | E | F | G | H |
6 | W | B | Total | Max capacity | |||
7 | Decision variable | ||||||
8 | Maximize | 8 | 6 | 0 | 0.00 | ||
9 | Subject to | ||||||
10 | equation 1 | 3 | 2 | 0 | 0.00 | <= | 80 |
11 | equation 2 | 4 | 10 | 0 | 0.00 | <= | 100 |
12 | equation 3 | 1 | 0 | 0 | 0.00 | >= | 6 |
formulae
Cell no. | B | C | D | E | F | G | H |
6 | W | B | Total | Max capacity | |||
7 | Decision variable | ||||||
8 | Maximize | 8 | 6 | 0 | =C8*$C$7 + D8*$D$7 | ||
9 | Subject to | ||||||
10 | equation 1 | 3 | 2 | 0 | =C10*$C$7 + D10*$D$7 | <= | 80 |
11 | equation 2 | 4 | 10 | 0 | =C11*$C$7 + D11*$D$7 | <= | 100 |
12 | equation 3 | 1 | 0 | 0 | =C12*$C$7 + D12*$D$7 | >= | 6 |
solver
Solution
Cell no. | B | C | D | E | F | G | H |
6 | W | B | Total | Max capacity | |||
7 | Decision variable | 25.00 | 0.00 | ||||
8 | Maximize | 8 | 6 | 0 | 200.00 | ||
9 | Subject to | ||||||
10 | equation 1 | 3 | 2 | 0 | 75.00 | <= | 80 |
11 | equation 2 | 4 | 10 | 0 | 100.00 | <= | 100 |
12 | equation 3 | 1 | 0 | 0 | 25.00 | >= | 6 |
W= 25
B= 0
Optimum= $200