In: Math
Georgia Cabinets manufactures kitchen cabinets that are sold to local dealers throughout the Southeast. Because of a large backlog of orders for oak and cherry cabinets, the company decided to contract with three smaller cabinetmakers to do the final finishing operation. For the three cabinetmakers, the number of hours required to complete all the oak cabinets, the number of hours required to complete all the cherry cabinets, the number of hours available for the final finishing operation, and the cost per hour to perform the work are shown here:
Cabinetmaker 1 | Cabinetmaker 2 | Cabinetmaker 3 | |
Hours required to complete all the oak cabinets | 47 | 40 | 27 |
Hours required to complete all the cherry cabinets | 64 | 51 | 36 |
Hours available | 40 | 30 | 35 |
Cost per hour | $34 | $41 | $52 |
For example, Cabinetmaker 1 estimates it will take 47 hours to complete all the oak cabinets and 64 hours to complete all the cherry cabinets. However, Cabinetmaker 1 only has 40 hours available for the final finishing operation. Thus, Cabinetmaker 1 can only complete 40/47 = 0.85,or 85%, of the oak cabinets if it worked only on oak cabinets. Similarly, Cabinetmaker 1 can only complete 40/64 = 0.63, or 63%, of the cherry cabinets if it worked only on cherry cabinets.
Formulate a linear programming model that can be used to determine the percentage of the oak cabinets and the percentage of the cherry cabinets that should be given to each of the three cabinetmakers in order to minimize the total cost of completing both projects. If the constant is "1" it must be entered in the box.
Let | O1 = percentage of Oak cabinets assigned to cabinetmaker 1 |
O2 = percentage of Oak cabinets assigned to cabinetmaker 2 | |
O3 = percentage of Oak cabinets assigned to cabinetmaker 3 | |
C1 = percentage of Cherry cabinets assigned to cabinetmaker 1 | |
C2 = percentage of Cherry cabinets assigned to cabinetmaker 2 | |
C3 = percentage of Cherry cabinets assigned to cabinetmaker 3 |
Min | O1 | + | O2 | + | O3 | + | C1 | + | C2 | + | C3 | |||
s.t. | ||||||||||||||
O1 | + | C1 | ≤ | Hours avail. 1 | ||||||||||
O2 | + | C2 | ≤ | Hours avail. 2 | ||||||||||
O3 | + | C3 | ≤ | Hours avail. 3 | ||||||||||
O1 | + | O2 | + | O3 | = | Oak | ||||||||
C1 | + | C2 | + | C3 | = | Cherry | ||||||||
O1, O2, O3, C1, C2, C3 ≥ 0 |
Solve the model formulated in part (a). What percentage of the oak
cabinets and what percentage of the cherry cabinets should be
assigned to each cabinetmaker? If required, round your answers to
three decimal places. If your answer is zero, enter "0".
Cabinetmaker 1 | Cabinetmaker 2 | Cabinetmaker 3 | |
---|---|---|---|
Oak | O1 = | O2 = | O3 = |
Cherry | C1 = | C2 = | C3 = |
What is the total cost of completing both projects? If required,
round your answer to the nearest dollar.
Total Cost = $
If Cabinetmaker 1 has additional hours available, would the
optimal solution change? If required, round your answers to three
decimal places. If your answer is zero, enter "0". Explain.
because Cabinetmaker 1 has of hours. Alternatively, the
dual value is which means that adding one hour to this constraint
will decrease total cost by $.
If Cabinetmaker 2 has additional hours available, would the
optimal solution change? If required, round your answers to three
decimal places. If your answer is zero, enter "0". Use a minus sign
to indicate the negative figure. Explain.
because Cabinetmaker 2 has a of . Therefore, each
additional hour of time for cabinetmaker 2 will reduce total cost
by $ per hour, up to a maximum of hours.
Suppose Cabinetmaker 2 reduced its cost to $38 per hour. What effect would this change have on the optimal solution? If required, round your answers to three decimal places. If your answer is zero, enter "0".
Cabinetmaker 1 | Cabinetmaker 2 | Cabinetmaker 3 | |
---|---|---|---|
Oak | O1 = | O2 = | O3 = |
Cherry | C1 = | C2 = | C3 = |
What is the total cost of completing both projects? If required,
round your answer to the nearest dollar.
Total Cost = $
The change in Cabinetmaker 2’s cost per hour leads to changing
objective function coefficients. This means that the linear
program
The new optimal solution the one above but with a total cost of
$ .