M/Ek/1 (∞/FIFO) System: Queuing Theory
It is a queuing model where the arrivals follow a Poisson process, service time follows an Erlang (k) probability distribution and the number of server is one.
Queue capacity of the system is infinite with first in first out mode. The first M in the notation stands for Poisson input, k for number of phases, 1 for the number of servers and ∞ for infinite capacity of the system.
An Erlang Family (Ek)
Ek of a probability distribution is the probability distribution of a random variable, which can be expressed as the sum k independently, identically distributed exponential variables.
| The expected numbers of customers in the queue, Lq = | (1 + k) -------- 2k |
X | λ2 ------- μ (μ - λ ) |
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| The expected waiting time before being served, Wq = | (1 + k) |
X | λ |
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| The expected time spent in the system, Ws = | Wq | + | 1 ----- μ |
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| The expected numbers of customers in the system, Ls = | λ Ws | ||||
Example 1
The registration of a student at Universal Teacher Publications requires three steps to be completed sequentially. The time taken to perform each step follows an exponential distribution with mean 30/3 minutes and is independent of each other. Students arrive at the head office according to a Poisson input process with a mean rate of 25 per hour. Assuming that there is only one person for registration. On the basis of this information, find the following:
- expected waiting time
- expected numbers of students in the queue.
Solution.
This is an M/Ek/1 system.
Here k = 3, λ = 25 per hour.
| Service time per phase = | 1 --- 3μ |
= | 30 ---- 3 |
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| Therefore, μ = 30 per hour. | |||||||
| The expected numbers of students in the queue, Lq = | 1 + 3 ------ 2 x 3 |
X | (25)2 ------- 30(30 - 25) |
= | 2.78 students or 3 students | ||
| The expected waiting time before being served, Wq = | 1 + 3 |
X | 25 -------- 30(30 - 25) |
= | 1/9 hour or 6.67 minutes | ||
Example 2: M/Ek/1 (∞/FIFO) System (Queuing Models)
Repair of a certain type of machine requires three steps to be completed sequentially. The time taken to perform each step follows an exponential distribution with mean 20/3 minutes and is independent of each other. The machine breakdown follows a Poisson process with rate of 1per 2 hours. Assuming that there is only one repairman, find out
- The expected idle time of a machine.
- The average waiting time of a broken down machine in a queue.
- The expected number of broken down machines in the queue.
- The average number of machines which are not in operation
Solution.
This is an M/Ek/1 system.
Here k =3, λ =1/2 per hour.
| Service time per phase = | 1 --- 3μ |
= | 20 ---- 3 |
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| Therefore, μ = 3 per hour. | |||||||
| The expected numbers of customers in the queue, Lq = | 1 + 3 ------ 2 X 3 |
X | (1/2)2 ------- 3(3 - 1/2) |
= | 1.33 minutes | ||
| The expected waiting time before being served, Wq = | 1 + 3 |
X | 1/2 -------- 3(3 - 1/2) |
= | 2 minutes 40 seconds | ||
| The expected time spent in the system, Ws = | 2 ---- 45 |
+ | 1 ---- 3 |
= | 22 minutes 40 seconds | ||
| The expected numbers of customers in the system, Ls = | 1 ---- 2 |
X | 17 ----- 45 |
= | 11.33 minutes | ||
As k approaches to infinity, the expressions of Lq, Wq , Ws and Ls are given by
| Lq = | λ2 -------- 2μ (μ - λ ) |
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| Wq = | λ |
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| Ws = | Wq | + | 1 ---- μ |
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| Ls = | Lq | + | λ ---- μ |
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Example 3
At Indira Gandhi airport, it takes exactly 6 minutes to land an aeroplane, once it is given the signal to land. Although incoming planes have scheduled arrival times, the wide variability in arrival times produces an effect which makes the incoming planes appear to arrive in a Poisson fashion at an average rate of 6 per hour. This produces occasional stack-ups at the airport, which can be dangerous and costly. Under these circumstances, how much time will a pilot expect to spend circling the field waiting to land?
Solution.
Here, service time is fixed being equal to 6 minutes.
The service distribution is last member of Erlang family, i.e., for
k = ∞
Mean arrival rate of aeroplanes, λ =
6 per hour
Mean landing rate of planes, μ = (1/6)
X 60 = 10 per hour
| Wq = | 6 |
= | 3/40 hours | = | 4.5 minutes | ||
