It is a **queuing model** where the arrivals follow a Poisson process,
service times are exponentially distributed and there is only one server.
Capacity of the system is limited to N with first in first out mode.

The first M in the notation stands for Poisson input, second M for Poisson output, 1 for the number of servers and N for capacity of the system.

ρ = λ/μ | |||

P_{o} = |
1 − ρ -------- 1 − ρ^{N
+ 1} |
||

L_{s} = |
ρ |
− | (N + 1)ρ^{N+1} ----------- 1 − ρ^{N
+ 1} |

L_{q} = |
L_{s} - λ/μ |
||

W_{q} = |
L_{q}---- λ |
||

W_{s} = |
L_{s}---- λ |

Students arrive at the head office of Universal Teacher Publications according to a Poisson input process with a mean rate of 30 per day. The time required to serve a student has an exponential distribution with a mean of 36 minutes. Assume that the students are served by a single individual, and queue capacity is 9. On the basis of this information, find the following:

- The probability of zero unit in the queue.
- The average line length.

Solution.

λ = | 30 --------- 60 X 24 |
||

= 1/48 students per minute | |||

μ = 1/36 students per minute | |||

ρ
= 36/48 = 0.75 N = 9 |
|||

P_{o} = |
1- 0.75------------- 1- (0.75) ^{9 + 1} |
||

= 0.26 | |||

L_{s} = |
0.75 |
- | (9 + 1)(0.75)^{9+1} ---------------------- 1 - (0.75)^{9 + 1} |

= 2.40 or 2 students. | |||