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Single channel queuing model

Single channel queuing model

(Arrival – Possion and Service time Exponential)

This is the simplest queuing model and is commonly used. It makes the following assumptions

(a)             Arriving customers are served on First Come First Serve (FCFS) basis.

(b)             There is no Balking or Reneging. All the customers wait the queue to be served, no one jumps the queue and no one leaves it.

(c)             Arrival rate is constant and does not change with time.

(d)             New customers arrival is independent of the earlier arrivals.

(e)             Arrivals are not of infinite population and follow Poisson’s distribution.

(f)               Rate of serving is known.

(g)             All customers have different service time requirements and are independent of each other.

(h)             Service time can be described by negative exponential probability distribution.

(i)               Average service rate is higher than the average arrival rate and over a period of time the queue keeps reducing.

Example 9.1 Assume a single channel service system of a library in a school. From past experiences it is known that on an average every hour 8 students come for issue of the books on an average rate of 10 per hour. Determine the following.

(a)             Probability of the assistant librarian being idle.

(b)             Probability that there are at least 3 students in system.

(c)             Expected time that a student is in queue.


(a) Probability that server is idle 

Example 9.2 At a garage, car owners arrive at the rate of 6 per hour and are served at the rate of 8 per hour. It is assumed that the arrival follows Poisson’s distribution and the service pattern is exponentially distributed. Determine.

(a)             Average Queue length.

(b)              Average waiting time.

Sol. Average arrival (mean arrival rate) A = 6 per hour.

Average (mean) service rate μ = 8 per hour.

Example 9.3 Customers arrive at a sales counter managed by a single person, according to a Poisson’s process with a mean rate of 20 per hour. The time required to serve a customer has an exponential distribution with mean of 100 seconds. Find the average waiting time of a customer.


Mean arrival rate λ = 20 per hour

Average or Mean service rate = 36 per hour as in 100 seconds one customer is served in I hour = 60 × 60 = 3600 seconds, 36 will be served Average waiting time of a customer in queue

Example 9.4 Self-help canteen employs one cashier at its counter. 8 customers arrive every 10 minutes all an average. The cashier can serve at the rate of one customer per minute. Assume Poisson’s distribution for arrival and exponential distribution for service patterns. Determine


(a)             Average number of customers in the system.


(b)             Average queue length


(c)             Average time a customer spends in the system


(d)             Average waiting time of each customer. 

Service rate μ = 1 customer/minute


(a) Average number of customers in the system 

Example 9.5 A branch library of State Library has only one clerk. The clerk is expected to perform various duties of issuing the books which are randomly distributed and call be approximated with Poisson’s distribution. He is able to issue 12 books per hour. The readers arrive at the rate of 10 per hour during the 10 hours the library is open. Determine


(a) Idle rate of the clerk.


(b) % time that the student has to wait.


(c) Average system time


Sol. Arrival rate λ  = 10 per hours


Service rate μ = 12 per hour

Example 9.6 An electrician repairs geysers. presses etc. He finds that the time he spends all repair of a geyser is exponentially distributed with mean 20 minutes. The geysers are repaired in the order in which these are received and their arrival approximates Poisson’s distribution with an average rate 16 per 8 hours day. Determine

(a) Electrician’s idle time each day

(b) How many geysers are ahead of the geyser just brought for repairs?

Sol. Arrival rate A = 2 / hour

Service rate μ = 3/hour (One geyser is repaired in 20 minutes three will be repaired in one hour).

(a) Electrician idle time = 8 – utilisation time.

Example 9.7 Arrival rate of telephone calls-at telephone booth are according to Poisson distribution, with an average time of 12 minutes between two consecutive calls arrival. The length of telephone calls is assumed to be exponentially distributed with mean 4 minutes.

a)     Determine the probability that person arriving at the booth will have to wait.

b)    Find the average queue length that is formed from time to time.

c)     The telephone company will install second booth when convinced that all arrival would expect to have to wait at least 5 minutes for the phone. Find the increase in flows or arrivals which will justify a second booth.

d)    What is the probability that an arrival will have to wait for more than 15 minutes before the phones is free?

e)     Find the fraction of a day that the phone will be in use.

Sol. Arrival rate λ = 1/12 minutes

Service rate μ = 1/4 minutes.

(a) Probability that a person will have to wait

Example 9.8. An emergency facility ill a hospital where only one patient call be attended to at any one time receives 96 patients in 24 hours. Based all past experiences, the hospital knows that one such patient, on an average needs 10 minutes of attention and this time would cost Rs. 20 per patient treated. The hospital wants to reduce the queue of patients from the present number to ½  patients. How much will it cost the hospital?

Thus to decrease the queue from 4/3 to ½, the budget per patient will have to be increased from Rs 100 to Rs. I5O.

Example 9.9 Customers arrive at the executive class air ticketing at the rate of 10 per hour. There is only one airlines clerk serving the customer at the rate of 20 per hour. If the conditions of single channel quelling model apply to this problem i.e., arrival rate and service rate probability distribution are approximated to Poisson’s and Exponential respectively; determine

(a) System being idle probability.

(h) The probability that there is not customer waiting to buy the ticket.

(c) The probability that the customer is being served and nobody is waiting.

Sol. -λ= 10 per hour

μ = 20 per hour

pn = Probability that there are n customer in the system 

= 0.5 (0.5)n for values of n = 1, 2, 3 …


(a) System being idle probability or o customer at the counter = p0 = 1 – pn = 1 – (0.5) × (0.5)0


= 1- 0.5 = 0.5

(b) Probability that there are more than 3 customers at the counter p (>3) = 

(c) Probability that there is no customer waiting = Probability that at the most 1 customer is waiting

= p0 + p1

= 0.5 + 0.5 × 0.5

= 0.5 + 0.25

= 0.75.

(d) Probability of customers being served and no body is waiting

p1 = 0.5 × 0.5

= 0.25.

Example 9.10 An electricity bill receiving window in a small town has only one cashier who handles and issues receipts to the customers. He takes on an average 5 per customer. It has been estimated that the persons coming for bill payment have no set pattern in a on an average 8 persons come per hour

The management receives a lot of complaints regarding customers waiting for long in queue and so decided to find out.

(a) What is the average length of queue?

(h) What time on an average, the cashier is idle?

(c) What is the average time for which a person has to wait to pay his bill?

(d) What is the probability that a person would have to wait jot at least 10 minutes

Sol. Making use of the usual notations

λ = 8 persons 1 hours

μ = 10 persons 1 hours 

Example 9.11 A small town has only one bus stand where the bus comes every 10 minutes. The commuters arrive in a random manner to use the bus facility. The commuters have complained that they have 10 wait for a long times in a queue to board the bus. Average rate of arrival of commuters is 4 per hours. Calculate

(a) The probability that a commuter has to wait.

(b) The wailing time of the commuter

Sol. λ= 4 per hour

μ= 6 per hour

(a) Probability that a commuter has to wait

Example 9.12 A bank plans to open a single server derive-in banking facility at a particular centre. It is estimated that 28 customers will arrive each hour on an average. If on an average it requires 2 minutes to process a customer transaction, determine

(a) The probability of time that the system will be idle.

(b) On the average how long the customer will have to wait before receiving the server.

(c) The length of the drive way required 10 accommodate all the arrivals. On the average 20 feet of derive way is required for each car that is waiting for service.

Sol. λ= 28 per hour

= 13

Length of drive way = 13 x 20 = 260 feet.

Example 9.13 A factory manufacturing tanks for military use has a separate tool room where Special Maintenance Tools (SMTs) are stored. The average time between requirement of a tool from tool room is 10 minutes and this follow the Poisson’s distribution. Average service time of the storekeeper is 9 minutes. Determine.

(a) Average queue length.

(b) Average length of non-empty queues.

(c) Average number of mechanics in the system including one who is being attended to.

(d) Mean waiting time of a mechanic.

(e) Average waiting time of mechanic who waits and

(f) Whether there is a need of employing another storekeeper so that cost of storekeeper idle time and mechanics waiting is reduced to the minimum. Assuming that a skilled mechanics cost Rs. 10 per hour and the storekeeper cost Rs. 1 per hour.

Sol. Using the usual notations

(b) Average number of workers/mechanics in the system.

It can be seen that the time cost of mechanics is much higher than the idle time cost, it is reasonable to use another storekeeper.