In: Operations Management
The Ace Manufacturing Company has orders for three similar products.
Product | Orders (units) |
---|---|
AA | 2,200 |
BB | 700 |
CC | 1,400 |
Three machines are available for the manufacturing operations. All three machines can produce all the products at the same production rate. However, due to varying defect percentages of each product on each machine, the unit costs of the products vary depending on the machine used. Machine capacities for the next week and the unit costs are shown below.
Machine | Capacity (units) |
---|---|
11 | 1,800 |
22 | 1,800 |
33 | 1,100 |
Machine | |||
---|---|---|---|
Product | 1 | 2 | 3 |
A | $1.00 | $1.30 | $1.10 |
B | $1.20 | $1.40 | $1.00 |
C | $0.90 | $1.20 | $1.20 |
Use the transportation model to develop the minimum cost production schedule for the products and machines.
1. Show the production schedule.
(xA1, xA2, xA3, xB1, xB2, xB3, xC1, xC2, xC3) = ??????
2. Determine the cost (in dollars) of the production schedule.
Total = $ ????
This is a demand supply transportation problem. Let us represent the problem tabularly:
Machine | |||||
Product | 1 | 2 | 3 | Orders | |
A | 1 | 1.3 | 1.1 | 2200 | |
B | 1.2 | 1.4 | 1 | 700 | |
C | 0.9 | 1.2 | 1.2 | 1400 | |
Capacity | 1800 | 1800 | 1100 | 4700 | |
4300 | Total |
Here we can clearly see that there is mismatch between supply and demand, this is unbalanced
Machine | |||||
Product | 1 | 2 | 3 | Orders | |
A | 1 | 1.3 | 1.1 | 2200 | |
B | 1.2 | 1.4 | 1 | 700 | |
C | 0.9 | 1.2 | 1.2 | 1400 | |
Dummy | 0 | 0 | 0 | 400 | |
Capacity | 1800 | 1800 | 1100 | 4700 | |
4700 | Total |
We have tried to balance it by adding a row.
Now lets solve this with Northwest corner method:
Machine | ||||
Product | 1 | 2 | 3 | Orders |
A | 1800 | 400 | 0 | 2200 |
B | 0 | 700 | 0 | 700 |
C | 0 | 700 | 700 | 1400 |
Dummy | 0 | 0 | 400 | 400 |
Capacity | 1800 | 1800 | 1100 |
Now lets multiply the value and cost table to get the final answer
1800*1 + 400*1.3 + 700*1.4 + 700*1.2 + 700*1.2 + 400*0
= 4980
Hence part a) production schedule is shown
& part b) the final cost is found = $4980