# Queuing theory question [closed]

At a computerized Rail Reservation window, the customer arrivals are considered to be Poisson with an average inter-arrival time of ten minutes. The railway clerk's time for serving the customer (examination of reservation slip, making reservation, collecting cash and issuing reservation ticket) may be assumed to be distributed negative exponentially with an average of 6 minutes.

(a)What is the chance the customer will straightaway get a ticket without having to wait in line?

(b)For what part of time will the railway clerk be busy?

(c)Find the average queue length, average number of customers in the system, average waiting time of a customer and the average time a customer spends in the system.

(d)There is a complaint the waiting time of an arrival exceeds 45 minutes and an additional clerk is required. Determine how large the new arrival rate should be to justify a second clerk?

How can we answer these questions? I am working on these questions. If any member knowing the correct answers to all of these question, may reply.

• Here is another queueing theory question: How long must you wait until a random poster on the internet comes along and does your school assignment for you. Commented Mar 17 at 11:41
• Here is how you can ask homework or homework-adjacent questions here: or.meta.stackexchange.com/questions/229/…
– EhsanK
Commented Mar 17 at 17:04
• I’m voting to close this question because it violates our homework policy by failing to show work toward the answer.
– prubin
Commented Mar 28 at 22:14

This is $$\ M/M/1/\infty/\infty\$$ queue model.

The symbols in turn represent:

• The distribution of the time between successive customer arrivals is negatively exponential
• The distribution of hours of service is negatively exponential
• Number of service desks is 1
• The system capacity limit is infinity
• The number of customer sources is infinity
• (Default) FCFS, first-come, first-served

# Solution

Write $$\rho=\lambda /\mu$$ for the utilization of the buffer.

Assuming $$\rho=\lambda /\mu<1$$ to ensure the convergence of the subsequent derivation.

We have $$\sum\limits_{i=0}^{\infty} \pi_i =1 \quad \text{ and }\quad \pi_i=\rho^i \pi_0\text{ for } i=1,2,\cdots$$

$$\quad \text{ thus }\quad \sum\limits_{i=0}^{\infty} \pi_0 \rho^i =1$$

The probability that the stationary process is in state $$i$$ (contains $$i$$ customers, including those in service) is $$\pi_i = (1-\rho) \rho^i$$

(a) The probability that the customer will straightaway get a ticket without having to wait in line, $$\pi_0=1-\rho=1-\frac{\lambda}{\mu}$$

(b) For what part of time will the railway clerk be busy? $$1-\pi_0=\frac{\lambda}{\mu}$$

(c)

The average queue length, $$L_q= \sum\limits_{n=c}^{\infty} (n-c)\pi_n = \sum\limits_{n=1}^{\infty} (n-1)\pi_n = \sum\limits_{n=1}^{\infty}(n-1) (1-\rho)\rho^n =\frac{{\rho}^2}{1-{\rho}}=\frac{\lambda ^2}{\mu (\mu -\lambda)}$$

The average number of customers in the system, $$L_s=\sum\limits_{n=0}^{\infty} n \pi _n = \sum\limits_{n=0}^{\infty} n (1-\rho) \rho^n=\frac{\rho}{1-\rho}= \frac{\lambda}{\mu -\lambda}$$

The time $$T$$ that a customer spends in the system, is distributed accroding to negatively exponential distribution with parameter $$\mu-\lambda$$:

$$P(T>t) =e^{-(\mu-\lambda)t} , \quad t\geq 0$$

The average time a customer spends in the system, $$W_s=\mathbb{E} [T]=\frac{1}{\mu-\lambda}$$

The average waiting time of a customer, $$W_q=W_s-\frac{1}{\mu}= \frac{\lambda}{ \mu (\mu -\lambda)}$$

(d)

MeanQueueSize of $$M/M/c/\infty/\infty$$ is

$$W_q= \frac{c \lambda \mu \Gamma (c) \left(\frac{\lambda }{\mu }\right)^c}{c! e^{\lambda /\mu } (\lambda -c \mu )^2 \Gamma \left(c,\frac{\lambda }{\mu }\right)+c \mu \Gamma (c) (c \mu -\lambda ) \left(\frac{\lambda }{\mu }\right)^c}$$

When $$c=2$$, MeanQueueSize of $$M/M/2/\infty/\infty$$ is

$$W_q=\frac{\lambda ^3}{4 \mu ^3-\lambda ^2 \mu }$$

We want to solve the equation $$W_q\operatorname{|}\limits_{\mu \rightarrow \frac{1}{6}}=45$$ for variable $$\lambda$$

\[ScriptCapitalQ] = QueueingProcess[\[Lambda], \[Mu], 2];
\[Lambda] /.
NSolve[QueueProperties[\[ScriptCapitalQ], "MeanQueueSize"] ==
45 /. {\[Mu] -> 1/6}, \[Lambda]] // Select[#, # > 0 &] &


The new arrival rate should be 0.32631 services per minute to justify a second clerk.

# Mathematica code

some Mathematica APIs store the famous properties.

(* M/M/2/∞/∞ queue*)
\[ScriptCapitalQ] = QueueingProcess[\[Lambda], \[Mu], 2];
QueueProperties[\[ScriptCapitalQ]]

• customer arrivals follows poisson distribution. In our question $\lambda=0.1, \mu =0.1667$ where $\lambda$ is customer arrival time per minute whereas $\mu$ is service time per minute Commented Apr 2 at 15:29