Waiting Line Models
In this section and the subsequent sections of this chapter, we explain several models. In presenting the models below, we start slowly and provide several examples, so that you can acquire a better feeling for waiting line models. Be patient and give yourself plenty of time to study these sections; otherwise, you may easily get confused.

"With time and patience the mulberry leaf becomes a silk gown." -Chinese Proverb

The M/M/1 (µ/FIFO) system

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 µ for infinite capacity of the system.

Formulas

Probability of zero unit in the queue (Po) = 1 - l
-----
m
       
Average queue length (Lq ) = l 2
--------
m (m - l )
   
Average number of units in the system (Ls) =

l
--------
m - l
   
Average waiting time of an arrival (Wq) = l
----------
m(m - l )
   
Average waiting time of an arrival in the system (Ws) = 1
---------
m - l

Example 1

Students arrive at the head office of www.universalteacher.com 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
l = 40/hour, m = 50/hour

Average waiting time of a student before receiving service (Wq) = 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
l = 10/hour, m = 12/hour

Probability that the counter is free = 1 - 10
-----
12		 = 1/6
Average number of customers in the queue (Lq ) = (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

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

Solution.

Average inter arrival time = 1
---
l 		= 5minutes = 1
---
12		 hour
l = 12/hour  
         
Average service time = 1
---
m 		= 2 minutes = 1
---
30		 hour
m = 30/hour  
   

Average queue length, Lq =

(12)2
-----------
30(30 - 12)		 = 4
---
15
   
Average number of customers, Ls =

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

  1. Average number in the waiting line.
  2. Average number in the system.
  3. Average waiting time in line.
  4. Average waiting time in the system.

Solution.

Given
l = 15/hour,
m = 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
l = 10/hour, m = 12/hour

Po = 1 - 10
-----
12		 = 1/6
Ls =

10
-------
12 - 10

 = 5 customers
       
Ws = 1
----------
12 - 10		 = 30 minutes
       

Lq =

(10)2
-----------
12(12 - 10)		 = 25/6 customers
   
   
Wq = 10
---------
12(12 - 10) 		= 25 minutes

 


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