## M/M/1 (N/FIFO) System : Queuing Models

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 ——– 1 – 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 |

## Example: M/M/1 (N/FIFO) System

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 ——– 1 – 0.75 | – | (9 + 1)(0.75)^{9+1} ———————- 1 – (0.75)^{9 + 1} |

= 2.40 or 2 students. | |||

## M/M/1 Queuing System (∞/FIFO)

It is a queuing model where the arrivals follow a Poisson process, service times are exponentially distributed and there is only one server. In other words, it is a system with Poisson input, exponential waiting time and Poisson output with single channel.

Queue capacity of the system is infinite 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 **∞** for infinite capacity of the system.

#### Formulas

Probability of zero unit in the queue (P_{o}) = | 1 | − | λ —– μ |

Average queue length (L_{q} ) = | λ ^{2}——– μ (μ – λ ) | ||

Average number of units in the system (L_{s}) = | λ ——– μ – λ | ||

Average waiting time of an arrival (W_{q}) = | λ ———- μ(μ – λ ) | ||

Average waiting time of an arrival in the system (W_{s}) = | 1 ——— μ – λ |

#### Example 1

Students arrive at the head office of Universal Teacher Publications according to a Poisson input process with a mean rate of 40 per hour. The time required to serve a student has an exponential distribution with a mean of 50 per hour. Assume that the students are served by a single individual, find the average waiting time of a student.

Solution.

Given

λ = 40/hour, μ = 50/hour

Average waiting time of a student before receiving service (W_{q}) = | 40 ——— 50(50 – 40) | = | 4.8 minutes |

#### Example 2

New Delhi Railway Station has a single ticket counter. During the rush hours, customers arrive at the rate of 10 per hour. The average number of customers that can be served is 12 per hour. Find out the following:

- Probability that the ticket counter is free.
- Average number of customers in the queue.

Solution.

Given

λ = 10/hour, μ = 12/hour

Probability that the counter is free = | 1 – | 10 —– 12 | = | 1/6 |

Average number of customers in the queue (L_{q} ) = | (10)^{2}——– 12 (12 – 10) | = | 25/6 |

#### Example 3

At Bharat petrol pump, customers arrive according to a Poisson process with an average time of 5 minutes between arrivals. The service time is exponentially distributed with mean time = 2 minutes. On the basis of this information, find out

- What would be the average queue length?
- What would be the average number of customers in the queuing system?
- What is the average time spent by a car in the petrol pump?
- What is the average waiting time of a car before receiving petrol?

Solution.

Average inter arrival time = | 1 — λ | = 5minutes = | 1 — 12 | hour |

λ = 12/hour | ||||

Average service time = | 1 — μ | = 2 minutes = | 1 — 30 | hour |

μ = 30/hour | ||||

Average queue length, L_{q} = | (12)^{2}———–30(30 – 12) | = | 4 — 15 | |

Average number of customers, L_{s} = | 12 ——- 30 – 12 | = | 2 —- 3 | |

Average time spent at the petrol pump = | 1 ———- 30 – 12 | = | 3.33 minutes | |

Average waiting time of a car before receiving petrol = | 12 ——— 30(30 – 12) | = | 1.33 minutes |

#### Example 4

Universal Bank is considering opening a drive in window for customer service. Management estimates that customers will arrive at the rate of 15 per hour. The teller whom it is considering to staff the window can service customers at the rate of one every three minutes.

Assuming Poisson arrivals and exponential service find

- Average number in the waiting line.
- Average number in the system.
- Average waiting time in line.
- Average waiting time in the system.

Solution.

Given

λ = 15/hour,

μ = 3/60 hour

or 20/hour

Average number in the waiting line = | (15)^{2}———- 20(20 – 15) | = | 2.25 customers |

Average number in the system = | 15 ———- 20 – 15 | = | 3 customers |

Average waiting time in line = | 15 ———— 20(20 – 15) | = | 0.15 hours |

Average waiting time in the system = | 1 ——— 20 – 15 | = | 0.20 hours |

#### Example 5

Chhabra Saree Emporium has a single cashier. During the rush hours, customers arrive at the rate of 10 per hour. The average number of customers that can be processed by the cashier is 12 per hour. On the basis of this information, find the following:

- Probability that the cashier is idle
- Average number of customers in the queuing system
- Average time a customer spends in the system
- Average number of customers in the queue
- Average time a customer spends in the queue

Solution.

Given

λ = 10/hour, μ = 12/hour

P_{o} = | 1 – | 10 —– 12 | = | 1/6 |

L_{s} = | 10 ——- 12 – 10 | = | 5 customers | |

W_{s} = | 1 ———- 12 – 10 | = | 30 minutes | |

L_{q} = | (10)^{2}———–12(12 – 10) | = | 25/6 customers | |

W_{q} = | 10 ——— 12(12 – 10) | = | 25 minutes |

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