Queuing Model [M/M/1:∞,FIFO] Single Server and Infinite Capacity with Solved Problems

[M/M/1:∞, FIFO] is one of the Queuing Model with Single Server and Infinite Capacity.

In this model, customers arrive in Poisson fashion and the service follows exponential fashion.

Jump to Probability and Queuing Theory Index Page

Important Formulae of [M/M/1:∞, FIFO]: (Operating Characteristics)

Read also:

Little's Formula

1. Traffic Intensity:

It is the ratio of the average arrival rate to the average departure rate. It is also the ratio of mean service time to mean arrival time. It is also called the Utilization Factor. It gives the proportion of the time server is busy. It is denoted by rho(ρ).

2. Steady State Probability (Limiting probability): 

It is also the probability of Queue being in a state 'n'. It is denoted by 'Pn'.

3. The probability that the Server is idle (no arriving customers):

 It is denoted by 'Po'.

Also, the probability that the server is busy is (1-P0) which is equal to ρ.

4. The average number of Customer in the System: 

It is denoted by Ls. 

5. The average number of Customer Waiting in the Queue.

Using Little's formula

6. The amount of Time Customer Spend in the System

Using Little's Formula

7. The Amount of Time Customer spend waiting in Queue

Using Little's Formula

8. The probability that the arriving customer has to wait in Queue for service:

9. The probability that the number of Customer in Queue exceed by K Customers = 

10. The probability that the number of Customers in System exceed by K Customers is:

11. The probability that the waiting time of a customer in the system exceeds by 't'

Solved Problems of (M/M/1: infinity, FIFO):

Solved Problem 1: A railway reservation center has a single Booking counter. During rush hour, Customers arrive at the rate of 30 customers per hour. The average number of customers that can be attended by the booking operator is 40 per hour. 

Assume that the arrival is Poisson and the service time is exponentially distributed. Find:

a. The probability that the operator is idle. 

b. The average number of Customer in the System.

c. The average time a customer spends in the System. 

d. The average number of Customer in the Queue. 

e. The average time a customer spends in the queue waiting for service. 

=> Solution:

Here Given, 

the arrival rate ( λ) = 30 customers per hour

the service rate (μ) = 40 customers per hour

traffic intensity (ρ) = λ/μ = 3/4

a. Probability that the operator is idle = Po = 1 - ρ = 1 - 3/4 = 1/4
b. The average number of Customer in the System is Ls 
We have, 

           Ls = ρ/(1-ρ) = 3

c. The average time a customer spends in the System is Ws

We have, 

                Ws = Ls / λ = 1/10 hours

d. The average number of Customer in the Queue is Lq

We have, 

Lq = Ls - λ/μ [Using Little's Formula]

       =  9/4 

e. The average time a customer spends in the queue waiting for service is Wq

We have, 

Wq = Lq / λ = 3/40 hours

---

Solved Problem 2: What is the probability that the customer has to wait15 minutes more to get served. Given that the arrival rate is 6 per hour and the service rate is 10 per hour. Assume that the arrival is Poisson and the service time is exponentially distributed.
=> Solution:
Here Given,
the arrival rate ( λ) = 6 per hour = 6/60 per minute = 1/10 per minute
the service rate (μ) = 10 per hour = 10/60 per minute = 1/6 per minute

The probability that the waiting time of a customer in the system exceeds by 't' is
P(Ts>t) = e^-(μ-λ)t 
              Here, t = 15 min
                  = e^{-(1/6-1/10)15
            = 1/e}

---

Solved Problem 3: A Barber takes half an hour on average for 1 haircut to complete. Customer arrives at the rate of 1 every 45 minutes in Poisson fashion. Find the
a. the number of customer in the shop
b. average time spent by the customer in service.
c. average time spent by a customer waiting in queue to get service.
d. idle time for the barber.
=> Solution:
Here Given,
the arrival rate ( λ) = 1/45 per minute = 60/45 per hour = 4/3 per hour
the service rate (μ) = 1/30 per minute = 60/30 per hour = 2 per hour

traffic intensity (ρ) = λ/μ = (4/3)/2 = 2/3

a. the number of customer in the shop is
Ls = ρ/(1-ρ) = 2
b. the average time spent by the customer in service is
Ws = Ls / λ =  3/2 = 1.5 hour
c. the average time spent by a customer waiting in queue to get service is
Wq = Lq / λ = 1 hour
d. Idle time for the barber is
Po = 1 - ρ = 1 - 2/3 = 1/3

---

Solved Problem 4: On average 96 patients per 24 hours, day requires service. 1 patient require an average of 10 minutes of service. Reduce the average number of patient waiting in Queue  to 1/2.

=> Solution:

Here Given, 

the arrival rate ( λ) = 96/24 = 4 per hour

the service rate (μ) =  1/10 per minute = 60/10 per hour = 6 per hour

traffic intensity (ρ) = λ/μ = 4/6 = 2/3

The average number of patient waiting in the Queue is Lq

We have,
Lq = Ls - λ/μ [Using Little's Formula]
         = ρ/(1-ρ) - ρ
      = 4/3

 To reduce the average number of patient waiting in Queue form 4/3 to 1/2, Service rate must be improved. Let μ1 be the new service rate. 

           Lq = 1/2
or,    λ² / μ1(μ1-λ) = 0.5
or,      μ1 = 8 patient per hour

So the average time required for service of 1 patient = 1/μ1 = 1/8 per hour = 7.5 per minute
Therefore, Inter-service time must be decreased by 10-7.5 = 2.5 minutes.

---

So these are the complete theory and Solved problems related to the Queuing Model [M/M/1:∞, FIFO].

Related Posts:

• Introduction and Features of Queuing System and Queuing Theory
• Queuing Model [M/M/1:∞,FIFO] Single Server and Infinite Capacity with Solved Problems
• Queuing Model [M/M/S:∞,FIFO] Multi-Server and Infinite Capacity with Solved Problems
• Queuing Model [M/M/1:K,FIFO] Single Server and finite queuing Capacity with Solved Problems
• Erlang Queuing Model [M/Ek/1: infinity, FIFO] single server with infinite queuing capacity solved problems