written 5.0 years ago by
teamques10
★ 64k
|
• modified 4.0 years ago |
Calculate the output statistics for the queueing system whose inter-arrival and service times for ten arrivals are given below:
| Inter-arrival time | - | 8 | 6 | 1 | 8 | 3 | 8 | 7 | 2 | 3 |
|---|---|---|---|---|---|---|---|---|---|---|
| Service time | 4 | 1 | 4 | 3 | 2 | 4 | 5 | 4 | 5 | 3 |
|
written 5.0 years ago by
teamques10
★ 64k
|
• modified 5.0 years ago |
Solution:
| Customer | IAT | AT | ST | Time Service Begin |
Time Cust waits in queue |
Time Service ends |
Time Cust spends in s/m |
Idle time server |
|---|---|---|---|---|---|---|---|---|
| 1 | - | 0 | 4 | 0 | 0 | 4 | 4 | 0 |
| 2 | 8 | 8 | 1 | 8 | 0 | 9 | 1 | 4 |
| 3 | 6 | 14 | 4 | 14 | 0 | 18 | 4 | 5 |
| 4 | 1 | 15 | 3 | 18 | 3 | 21 | 6 | 0 |
| 5 | 8 | 23 | 2 | 23 | 0 | 25 | 2 | 2 |
| 6 | 3 | 26 | 4 | 26 | 0 | 30 | 4 | 1 |
| 7 | 8 | 34 | 5 | 34 | 0 | 39 | 5 | 4 |
| 8 | 7 | 41 | 4 | 41 | 0 | 45 | 4 | 2 |
| 9 | 2 | 43 | 5 | 45 | 2 | 50 | 7 | 0 |
| 10 | 3 | 46 | 3 | 50 | 4 | 53 | 7 | 0 |
| - | $\sum$ 46 | - | $\sum$ 35 | - | $\sum$ 9 | - | $\sum$ 44 | $\sum$ 18 |
AT: Arrival time
IAT: Interarrival time
ST: Service Time
o/p statistics
Avg waiting time for a customer = $\frac{9}{10} = 0.9$
Probability of Idle server = $\frac{18}{53} = 0.34$
Probability customer has to wait = $\frac{3}{10} = 0.3$
Avg service time = $\frac{35}{10} = 3.5$
Avg time between arrival = $\frac{46}{9} = 5.11$
Avg waiting time for customer who waits = $\frac{9}{3} = 3$
Avg time customer spent in the system = $\frac{44}{10} = 4.4$
Max queue length = 1
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