Using the probability of demand, we can assign random digits to demand, similarly, we can assign random digits to lead time by using the probability to lead time.
Table A: Random Digit Assignment for Daily Demand
Daily Demand |
Probability |
Cumulative Prob |
Random Digit Assignment |
3 |
0.20 |
0.20 |
1 - 20 |
4 |
0.35 |
0.55 |
21 - 55 |
5 |
0.30 |
0.85 |
56 - 85 |
6 |
0.15 |
1.00 |
86 - 00 |
Table B: Random Digit Assignment for Lead Time
| Lead Time | Probability | Cumulative Prob | Random Digit Assignment |
| 1 | 0.36 | 0.36 | 1- 36 |
| 2 | 0.42 | 0.78 | 37 - 78 |
| 3 | 0.22 | 1.00 | 79 - 00 |
The random digit for lead time for first cycle is given by 46. Mapping it (46) to table B gives a lead time of 2 days. Hence two random digits are required for demand (4,5)
R.D. for lead time |
46 |
75 |
86 |
27 |
63 |
- |
- |
- |
- |
- |
R.D. for demand |
4 |
5 |
4 |
5 |
6 |
3 |
4 |
4 |
6 |
4 |
After mapping, these R.D. for demand from table A, if gives demand of '3' for both random digits. So, lead time demand for first cycle is b(3+3). In this way, lead time demand is calculated for remaining cycles and given Table 'C'.
cycle |
Random Digit for Lead Time |
Lead Time |
Random Digit Demand |
Daily Demand |
Lead Time Demand |
1 |
46 |
2 |
4 5 |
3 3 |
6 |
2 |
75 |
2 |
4 5 |
3 3 |
6 |
3 |
86 |
3 |
6 3 4 |
3 3 3 |
9 |
4 |
27 |
1 |
4 |
3 |
3 |
5 |
63 |
2 |
6 4 |
3 3 |
6 |
Table 'C': Simulation table for Lead Time Demand