In: Operations Management
G and P Manufacturing would like to minimize the labor cost of producing dishwasher motors for major
appliance manufacturer. Two models of motors exist and the time in hours required for each model in each
production area is tabled here, along with the labor cost.
Model 1 |
Model 2 |
|
Area A (hrs) |
12 |
16 |
Area B (hrs) |
10 |
8 |
Area C (hrs) |
14 |
12 |
Cost ($) |
100 |
120 |
Currently, labor assignments provide for 38,000 hours in each of Area A, 25,000 hours in Area B, and 27,000 hours in area C.
2,000 hours are available to be transferred from Area B to Area C and a combined total of 4,000 hours are available to be transferred from Area A to Areas B and C.
We would like to develop the linear programming model to minimize the labor cost, whose solution would tell G&P how many of each model to produce and how to allocate the workforce.
Let P1 = the number of model 1 motors to produce
P2 = the number of model 2 motors to produce
TAC = the number of hours transferred from A to C
TAB = the number of hours transferred from A to B
TBC = the number of hours transferred from B to C
What is the objective function?
Correct answer is c.Min 100 P1 + 120 P2
Which of the following represents the resource availability constraint for Area B?
Correct answer is d. 10P1 +8P2 <= 25,000 -TBC + TAB
25000 hours have been allocated to area B and the number of hours transferred from A to B(TAB) along with accommodate the number of hours transferred from B to C
No matter what the resource allocation is, Area A will always have the highest resource availability.
Correct answer is a.True
reason : As we can see there are 38000 hours in Area A and even if all the 4000 hours are transferred from Area A to Areas B and C , Area A will always have the highest resource availability
Let Pij = the production of product I in period j. To specify that production of product 2 in period 4 and in period 5 differs by no more than 80 units, we need to add which pair of constraints?
Correct answer is b. P24 – P25 <= 80; P25 – P24 <= 80