Queuing Simulation
Here we provide another example of how simulation is used in queuing theory.
Example 2
Students arrive at the head office of Universal Teacher Publications according to the following arrival and service time probability distributions:
| Inter-arrival Time (Min.) |
Frequency | Service Time (Min.) |
Frequency |
|---|---|---|---|
| 1 | 1 | 1 | 1 |
| 2 | 4 | 2 | 4 |
| 3 | 7 | 3 | 7 |
| 4 | 17 | 4 | 17 |
| 5 | 31 | 5 | 31 |
| 6 | 23 | 6 | 23 |
| 7 | 7 | 7 | 7 |
| 8 | 5 | 8 | 5 |
| 9 | 3 | 9 | 3 |
| 10 | 2 | 10 | 2 |
Using Monte Carlo Simulation, determine the following:
(a) The average waiting time before service
(b) The average time a student spends in the system
Solution.
From the given distribution of arrivals, the random numbers can be assigned to the arrival times as shown in table 1.
Table 1
| Inter-arrival Time (Min.) |
Frequency | Probability | Cumulative Probability |
R.No. |
|---|---|---|---|---|
| 1 | 1 | 0.01 | 0.01 | 0 |
| 2 | 4 | 0.04 | 0.05 | 1 - 4 |
| 3 | 7 | 0.07 | 0.12 | 5 - 11 |
| 4 | 17 | 0.17 | 0.29 | 12 - 28 |
| 5 | 31 | 0.31 | 0.60 | 29 - 59 |
| 6 | 23 | 0.23 | 0.83 | 60 - 82 |
| 7 | 7 | 0.07 | 0.90 | 83 - 89 |
| 8 | 5 | 0.05 | 0.95 | 90 - 94 |
| 9 | 3 | 0.03 | 0.98 | 95 - 97 |
| 10 | 2 | 0.02 | 1.00 | 97 - 99 |
Similarly, the random numbers can be assigned to the service times as shown in table 2.
Table 2
| Service Time (Min.) |
Frequency | Probability | Cumulative Probability |
R.No. |
|---|---|---|---|---|
| 1 | 1 | 0.01 | 0.01 | 0 |
| 2 | 4 | 0.04 | 0.05 | 1 - 4 |
| 3 | 7 | 0.07 | 0.12 | 5 - 11 |
| 4 | 17 | 0.17 | 0.29 | 12 - 28 |
| 5 | 31 | 0.31 | 0.60 | 29 - 59 |
| 6 | 23 | 0.23 | 0.83 | 60 - 82 |
| 7 | 7 | 0.07 | 0.90 | 83 - 89 |
| 8 | 5 | 0.05 | 0.95 | 90 - 94 |
| 9 | 3 | 0.03 | 0.98 | 95 - 97 |
| 10 | 2 | 0.02 | 1.00 | 97 - 99 |
The simulation worksheet as shown in table 3 is developed in the following way:
The first random number for inter-arrival time is 17, which corresponds to an inter-arrival time of 4 minutes. This implies that the first student arrives 4 minutes after the head office opens. Since the first student arrives at 10.04 a.m., therefore, the server has to wait for 4 minutes. The random number for service time is 90, which corresponds to a service time of 8 minutes. Thus, the first student leaves the system at 10.12 a.m. (10.04 + .08).
The next random number for inter-arrival time is 86, which corresponds to an inter-arrival time of 7 minutes. Since the second student arrives at 10.11 a.m., but the service can be started at 10.12 a.m. Therefore, he has to wait for 1 minute. Likewise, other values can be calculated.
Table 3
| S.No. | R.No. | Inter-arrival time | Arrival Time |
Service starts |
R.No. | Service | Waiting Time | ||
|---|---|---|---|---|---|---|---|---|---|
| Time | Ends | Server | Student | ||||||
| 1 | 17 | 4 | 10.04 | 10.04 | 90 | 8 | 10.12 | 4 | - |
| 2 | 86 | 7 | 10.11 | 10.12 | 59 | 5 | 10.17 | - | 1 |
| 3 | 84 | 7 | 10.18 | 10.18 | 95 | 9 | 10.27 | 1 | - |
| 4 | 79 | 6 | 10.24 | 10.27 | 68 | 6 | 10.33 | - | 3 |
| 5 | 33 | 5 | 10.29 | 10.33 | 51 | 5 | 10.38 | - | 4 |
| 6 | 55 | 5 | 10.34 | 10.38 | 82 | 6 | 10.44 | - | 4 |
| 7 | 6 | 3 | 10.37 | 10.44 | 72 | 6 | 10.50 | - | 7 |
| 8 | 42 | 5 | 10.42 | 10.50 | 1 | 2 | 10.52 | - | 8 |
| 9 | 93 | 8 | 10.50 | 10.52 | 77 | 6 | 10.58 | - | 2 |
| 10 | 38 | 5 | 10.55 | 10.58 | 80 | 6 | 11.04 | - | 3 |
| Total | 59 | 32 | |||||||
Average waiting time before service.
= 32/10 = 3.2 minutes.
Average service time.
= 59/10 = 5.9 minutes
Average time a student spends in the system.
5.9 + 3.2 = 9.1 minutes.
