In this section and the subsequent sections of this chapter, we explain several Queuing 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.
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.
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 

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 
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 
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:
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 
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
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 
=  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 
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
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 
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:
Solution.
Given
λ = 10/hour, μ = 12/hour
P_{o} =  1   10  12 
=  1/6 
L_{s} =  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 