Acme Ltd is a producer of an item,
which is used as sports item. The demand forecast for this item
for the next 5 periods is 5000, 12,000, 11,000, 7,500 and 5,000
units. The production cost per unit made on regular time is
Mu 50 and Mu 54 for each unit produced on overtime. This item
can also be subcontracted at a cost of Mu 57 per unit. Inventory
holding costs are Mu 6 per unit per period and no backorders
are allowed. In each period, 9000 units can be produced on regular
time, 1800 units can be produced on overtime and up to 2500
units can be subcontracted. Set up an appropriate transportation
problem format to solve the above aggregate production-planning
problem.
| Production
Period |
Source* |
1 |
2 |
3 |
4 |
5 |
Capacity |
|
1
|
RT
|
50
|
56
|
62
|
68
|
74
|
9000
|
|
OT
|
54
|
60
|
66
|
72
|
78
|
1800
|
|
SC
|
57
|
63
|
69
|
75
|
81
|
2500
|
|
2
|
RT
|
x
|
50
|
56
|
62
|
68
|
9000
|
|
OT
|
x
|
54
|
60
|
66
|
72
|
1800
|
|
SC
|
x
|
57
|
63
|
69
|
75
|
2500
|
|
3
|
RT
|
x
|
x
|
50
|
56
|
62
|
9000
|
|
OT
|
x
|
x
|
54
|
60
|
66
|
1800
|
|
SC
|
x
|
x
|
57
|
63
|
69
|
2500
|
|
4
|
RT
|
x
|
x
|
x
|
50
|
56
|
9000
|
|
OT
|
x
|
x
|
x
|
54
|
60
|
1800
|
|
SC
|
x
|
x
|
x
|
57
|
63
|
2500
|
|
5
|
RT
|
x
|
x
|
x
|
x
|
50
|
9000
|
|
OT
|
x
|
x
|
x
|
x
|
54
|
1800
|
|
SC
|
x
|
x
|
x
|
x
|
57
|
2500
|
|
|
Demand
|
5000
|
12000
|
11000
|
7500
|
5000
|
|
|
* Here RT: Regular Time, OT: Overtime, SC: Subcontracted
The problem can now be
solved. Starting period 1, we attempt to satisfy demand using
the cheapest cost alternative available as long as capacity
is available. The cheapest source in column 1 is RT in period
1. Since there is capacity of 9000 units available and a demand
of 5000, we produce 5000 units. Once the demand is satisfied
in period 1, we move on to 2nd period. In period 2, the cheapest
source is RT, however the demand is 12,000 units and only
9000 units can be produced using RT. Thus we produce 9000
units and remaining units 1800 units using OT and 1200 units
using RT inn period 1. Next we move on to period 3 and so
on. The solution is given in the following table.
| Production
Period |
Source* |
1 |
2 |
3 |
4 |
5 |
Capacity |
| 1 |
RT |
5000 |
1200 |
|
|
|
9000 |
| OT |
|
|
|
|
|
1800 |
| SC |
|
|
|
|
|
2500 |
| 2 |
RT |
x |
9000 |
|
|
|
9000 |
| OT |
x |
1800 |
|
|
|
1800 |
| SC |
x |
|
|
|
|
2500 |
| 3 |
RT |
x |
x |
9000 |
|
|
9000 |
| OT |
x |
x |
1800 |
|
|
1800 |
| SC |
x |
x |
200 |
|
|
2500 |
| 4 |
RT |
x |
x |
x |
7500 |
|
9000 |
| OT |
x |
x |
x |
|
|
1800 |
| SC |
x |
x |
x |
|
|
2500 |
| 5 |
RT |
x |
x |
x |
x |
5000 |
9000 |
| OT |
x |
x |
x |
x |
|
1800 |
| SC |
x |
x |
x |
x |
|
2500 |
|
Demand |
5000 |
12000 |
11000 |
7500 |
5000 |
|
|
Cost computations
| Period |
Demand |
RT prodn
cost |
OT Prod cost
|
SC cost |
End of period
inventory cost |
| 1 |
5000 |
6200 x 50= 310,000 |
0 |
0 |
1200 x 6 = 7200 |
| 2 |
12000 |
9000 x 50 =450,000 |
1800x 54 =972,000 |
0 |
0 |
| 3 |
11000 |
9000 x 50 =450,000 |
1800x 54 =972,000 |
200 x 57 =11,400 |
0 |
| 4 |
7500 |
7500 x 50=375,000 |
0 |
0 |
0 |
| 5 |
5000 |
5000 x 50=250,000 |
0 |
0 |
0 |
| |
Total cost (in Rs) |
1835,000 |
194,400 |
11,400 |
7200 |
| |
Grand cost (in Rs) |
2048,000 |
|
|
|
|
|
|
