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Simulation Models |
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The Lajwaab Bakery Shop Problem
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Example| Daily demand | Probability |
|---|---|
| 0 | 0.01 |
| 15 | 0.15 |
| 25 | 0.20 |
| 35 | 0.50 |
| 45 | 0.12 |
| 50 | 0.02 |
Consider the following sequence of random numbers:
21, 27, 47, 54, 60, 39, 43, 91, 25, 20
Using this sequence, simulate the demand for the next 10 days. Find out the stock situation, if the owner of the bakery shop decides to make 30 cakes every day. Also estimate the daily average demand for the cakes on the basis of simulated data.
Solution.
Using the daily demand distribution, we obtain a probability distribution as shown in the following table.
Table 1
| Daily demand | Probability | Cumulative probability | Random Numbers |
|---|---|---|---|
| 0 | 0.01 | 0.01 | 0 |
| 15 | 0.15 | 0.16 | 1-15 |
| 25 | 0.20 | 0.36 | 16-35 |
| 35 | 0.50 | 0.86 | 36-85 |
| 45 | 0.12 | 0.98 | 86-97 |
| 50 | 0.02 | 1.00 | 98-99 |
At the start of simulation, the first random number 21 generates a demand of 25 cakes as shown in table 2. The demand is determined from the cumulative probability values in table 1. At the end of first day, the closing quantity is 5 (30-25) cakes.
Similarly, we can calculate the next demand for others.
Table 2
| Demand | Random Numbers | Next demand | Daily production = 30 cakes | |
|---|---|---|---|---|
| Left out | Shortage | |||
| 1 | 21 | 25 | 5 | |
| 2 | 27 | 25 | 10 | |
| 3 | 47 | 35 | 5 | |
| 4 | 54 | 35 | 0 | |
| 5 | 60 | 35 | 5 | |
| 6 | 39 | 35 | 10 | |
| 7 | 43 | 35 | 15 | |
| 8 | 91 | 45 | 30 | |
| 9 | 25 | 25 | 25 | |
| 10 | 20 | 25 | 20 | |
| Total | 320 | 10 | ||
Total demand = 320
Average demand = Total demand/no. of days
The daily average demand for the cakes = 320/10 = 32 cakes.
