Queuing Theory Questions and Answers

Q: A queueing system has one single server workstation that admits an infinitely long queue. The rate of arrival of jobs to the queueing system follows the Poisson distribution with a mean of 5 jobs/hour. The service time of the server is exponentially distributed with a mean of 6 minutes. In steady state operation of the queueing system, the probability that the server is not busy at any point in time is

The probability that the server is idle is 0.50.This is an M/M/1 queue. The arrival rate is \(\lambda = 5\) jobs/hour; a mean service time of 6 minutes gives a service rate \(\mu = 60/6 = 10\) jobs/hour.\[ \begin{aligned} \rho &= \frac{\lambda}{\mu} = \frac{5}{10} = 0.5 \\ P_{0} &= 1 - \rho = 0.50 \end{aligned} \]\(\rho\) is the fraction of time the server is busy, so the idle fraction is \(1 - \rho\). Here both happen to be 0.5. 0.20, 0.17 and 0.83 would each need a different arrival or service rate from those given; with \(\rho = 0.5\) the idle probability can only be 0.5.

Q: In usual notations of queueing systems, which of the following relationships is not true ?

The relationship that is not true is \(L_{s} = L_{q} + 1/\lambda\).The correct relation is \(L_{s} = L_{q} + \lambda/\mu\): the mean number in the system exceeds the mean number in the queue by the mean number in service, which for a single server is the utilisation \(\rho = \lambda/\mu\). It follows from multiplying \(W_{s} = W_{q} + 1/\mu\) by \(\lambda\) and applying Little's law.The other three are true. \(W_{s} = W_{q} + 1/\mu\) says time in system is waiting time plus mean service time. \(L_{s} = \lambda W_{s}\) and \(L_{q} = \lambda W_{q}\) are Little's law for the system and for the queue.

Q: Let \(L_{q}\) be the average number of customers in the queue, \(\lambda\) be the customer arrival rate and \(\mu\) be the average service rate. Then average waiting time for a customer in the queue for all infinite source queueing models is :

The average waiting time in the queue is \(L_{q}/\lambda\).Little's law applied to the queue alone states \(L_{q} = \lambda W_{q}\): the average number waiting equals the arrival rate times the average wait. It holds for any infinite-source queue in steady state, whatever the arrival and service distributions. So \(W_{q} = L_{q}/\lambda\).\(L_{q}/\mu\) divides by the wrong rate. \(\lambda/\mu\) is the traffic intensity \(\rho\), a pure number, and \(\mu/\lambda\) is its reciprocal; neither is a time.

Q: Consider the following two M/M/1 queueing systems where :In \(Q_{1}\) : arrival rate \(\lambda\), service rate \(\mu\)In \(Q_{2}\) : arrival rate \(2\lambda\), service rate \(2\mu\)Let \(W_{1}\) and \(W_{2}\) be the average waiting times of an arriving customer in \(Q_{1}\) and \(Q_{2}\) respectively. Then :

The answer is \(W_{1} \gt W_{2}\).In an M/M/1 queue the mean waiting time in the queue is \(W_{q} = \lambda/[\mu(\mu - \lambda)]\). Doubling both rates keeps the traffic intensity \(\rho = \lambda/\mu\) the same but halves every time:\[ \begin{aligned} W_{2} &= \frac{2\lambda}{2\mu(2\mu - 2\lambda)} \\ &= \frac{1}{2}\cdot\frac{\lambda}{\mu(\mu - \lambda)} \\ &= \frac{W_{1}}{2} \end{aligned} \]The same halving holds for the time in the system, \(1/(\mu - \lambda)\). The queue lengths are equal, but customers move through twice as fast, so they wait half as long. The waits are therefore not equal, and the relation can be decided.

Q: Which of the following is an example of the control of the queue discipline ?

The example is a separate server allocated for special category customers.Queue discipline is the rule deciding which waiting customer is served next: first come first served, priority, random order and so on. Serving a special category of customers through a separate server gives them priority, which is a control of the queue discipline.The other options act on other parts of the system: allowing customers to leave concerns customer behaviour (reneging), varying the number of servers controls the service mechanism, and adjusting the arrival rate controls the input process.

Q: Let \(D_{n}\) be the waiting time of the \(n\)th customer in G|M|1 queue. Consider the following recurrence equation :\[ D_{n+1} = \begin{cases} D_{n} + V_{n} - U_{n} & \text{if } D_{n} + V_{n} - U_{n} \ge 0 \\ 0 & \text{otherwise} \end{cases}. \]Which of the following interpretations of the variables in the above equation are valid ?(i) \(U_{n}\) is the interarrival time between \(n\)th and \((n + 1)\)th customer(ii) \(D_{n} + V_{n}\) is the departure time of the \(n\)th customer from the system

The keyed answer is both are valid.This is Lindley's recurrence. If the \(n\)th customer waits \(D_{n}\) and is served for \(V_{n}\), he leaves \(D_{n} + V_{n}\) time units after his own arrival. The next customer arrives \(U_{n}\) after him, the interarrival time between the \(n\)th and \((n + 1)\)th customers, so (i) is valid. He must wait \(D_{n} + V_{n} - U_{n}\) if that is positive, and 0 otherwise.So \(D_{n} + V_{n}\) is the departure time of the \(n\)th customer measured from his arrival, which is the sense in which (ii) is valid. The options rejecting either interpretation do not fit the recurrence.

Q: A canteen has a single cash counter. A customer has to buy coupons from the cash counter before getting the food items. Customers arrive at the counter according to a Poisson process with rate of 2 per 5 minutes. It takes 1.5 minutes on average to buy coupons. What is the long-run probability that the cash counter is free ?

The probability that the counter is free is \(1 - \dfrac{1.5}{2.5}\).Treat the counter as an M/M/1 (or M/G/1) queue. The arrival rate is \(\lambda = 2/5 = 0.4\) per minute and the mean service time is 1.5 minutes, so the traffic intensity is\[ \rho = \lambda \times 1.5 = 0.6 = \frac{1.5}{2.5} \]In the long run the server is busy a fraction \(\rho\) of the time, so it is free with probability \(1 - \rho = 0.4\). 2.5 minutes is the mean time between arrivals. \(1 - 1.5/5\) wrongly uses 5 minutes as the inter-arrival time, \(1 - 1.5/2\) uses the arrival count, and 0 would mean the counter is always busy.

Q: As the variance of service time increases, the waiting time of a customer in \(M|G|1\) queue :

As the service-time variance increases, the waiting time increases.The Pollaczek-Khinchine formula for the M/G/1 queue gives the mean waiting time in queue as\[ W_{q} = \frac{\lambda(\sigma^{2} + 1/\mu^{2})}{2(1 - \rho)} \]where \(\sigma^{2}\) is the variance of service time. With the arrival rate and mean service time fixed, \(W_{q}\) increases linearly with \(\sigma^{2}\). This is why the M/D/1 queue, with constant service, has half the waiting of the M/M/1 queue with the same mean.

A queueing system has one single server workstation that admits an infinitely long queue. The rate of arrival of jobs to the queueing system follows the Poisson distribution with a mean of 5 jobs/hour. The service time of the server is exponentially distributed with a mean of 6 minutes. In steady state operation of the queueing system, the probability that the server is not busy at any point in time is

A.0.20
B.0.17
C.0.50
D.0.83

Exam Year(s)

GATE ME - 2024,

Answer: C

The probability that the server is idle is 0.50.

This is an M/M/1 queue. The arrival rate is \(\lambda = 5\) jobs/hour; a mean service time of 6 minutes gives a service rate \(\mu = 60/6 = 10\) jobs/hour.

\[ \begin{aligned} \rho &= \frac{\lambda}{\mu} = \frac{5}{10} = 0.5 \\ P_{0} &= 1 - \rho = 0.50 \end{aligned} \]

\(\rho\) is the fraction of time the server is busy, so the idle fraction is \(1 - \rho\). Here both happen to be 0.5. 0.20, 0.17 and 0.83 would each need a different arrival or service rate from those given; with \(\rho = 0.5\) the idle probability can only be 0.5.

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In usual notations of queueing systems, which of the following relationships is not true ?

A.\(W_{s} = W_{q} + \dfrac{1}{\mu}\)
B.\(L_{s} = \lambda W_{s}\)
C.\(L_{q} = \lambda W_{q}\)
D.\(L_{s} = L_{q} + 1/\lambda\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The relationship that is not true is \(L_{s} = L_{q} + 1/\lambda\).

The correct relation is \(L_{s} = L_{q} + \lambda/\mu\): the mean number in the system exceeds the mean number in the queue by the mean number in service, which for a single server is the utilisation \(\rho = \lambda/\mu\). It follows from multiplying \(W_{s} = W_{q} + 1/\mu\) by \(\lambda\) and applying Little's law.

The other three are true. \(W_{s} = W_{q} + 1/\mu\) says time in system is waiting time plus mean service time. \(L_{s} = \lambda W_{s}\) and \(L_{q} = \lambda W_{q}\) are Little's law for the system and for the queue.

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Let \(L_{q}\) be the average number of customers in the queue, \(\lambda\) be the customer arrival rate and \(\mu\) be the average service rate. Then average waiting time for a customer in the queue for all infinite source queueing models is :

A.\(L_{q}/\mu\)
B.\(\mu/\lambda\)
C.\(\lambda/\mu\)
D.\(L_{q}/\lambda\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The average waiting time in the queue is \(L_{q}/\lambda\).

Little's law applied to the queue alone states \(L_{q} = \lambda W_{q}\): the average number waiting equals the arrival rate times the average wait. It holds for any infinite-source queue in steady state, whatever the arrival and service distributions. So \(W_{q} = L_{q}/\lambda\).

\(L_{q}/\mu\) divides by the wrong rate. \(\lambda/\mu\) is the traffic intensity \(\rho\), a pure number, and \(\mu/\lambda\) is its reciprocal; neither is a time.

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Consider the following two M/M/1 queueing systems where :

In \(Q_{1}\) : arrival rate \(\lambda\), service rate \(\mu\)
In \(Q_{2}\) : arrival rate \(2\lambda\), service rate \(2\mu\)

Let \(W_{1}\) and \(W_{2}\) be the average waiting times of an arriving customer in \(Q_{1}\) and \(Q_{2}\) respectively. Then :

A.\(W_{1} \lt W_{2}\)
B.\(W_{1} = W_{2}\)
C.\(W_{1} \gt W_{2}\)
D.The relation between \(W_{1}\) and \(W_{2}\) cannot be decided

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The answer is \(W_{1} \gt W_{2}\).

In an M/M/1 queue the mean waiting time in the queue is \(W_{q} = \lambda/[\mu(\mu - \lambda)]\). Doubling both rates keeps the traffic intensity \(\rho = \lambda/\mu\) the same but halves every time:

\[ \begin{aligned} W_{2} &= \frac{2\lambda}{2\mu(2\mu - 2\lambda)} \\ &= \frac{1}{2}\cdot\frac{\lambda}{\mu(\mu - \lambda)} \\ &= \frac{W_{1}}{2} \end{aligned} \]

The same halving holds for the time in the system, \(1/(\mu - \lambda)\). The queue lengths are equal, but customers move through twice as fast, so they wait half as long. The waits are therefore not equal, and the relation can be decided.

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Which of the following is an example of the control of the queue discipline ?

A.Customer is allowed to leave the facility
B.The number of servers are varied depending on the queue length
C.Separate server is allocated for special category customers
D.The mean arrival rate is adjusted

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The example is a separate server allocated for special category customers.

Queue discipline is the rule deciding which waiting customer is served next: first come first served, priority, random order and so on. Serving a special category of customers through a separate server gives them priority, which is a control of the queue discipline.

The other options act on other parts of the system: allowing customers to leave concerns customer behaviour (reneging), varying the number of servers controls the service mechanism, and adjusting the arrival rate controls the input process.

Discuss in Forum

Let \(D_{n}\) be the waiting time of the \(n\)th customer in G|M|1 queue. Consider the following recurrence equation :

\[ D_{n+1} = \begin{cases} D_{n} + V_{n} - U_{n} & \text{if } D_{n} + V_{n} - U_{n} \ge 0 \\ 0 & \text{otherwise} \end{cases}. \]

Which of the following interpretations of the variables in the above equation are valid ?

(i) \(U_{n}\) is the interarrival time between \(n\)th and \((n + 1)\)th customer
(ii) \(D_{n} + V_{n}\) is the departure time of the \(n\)th customer from the system

A.Both are not valid
B.Both are valid
C.Only (i) is valid
D.Only (ii) is valid

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The keyed answer is both are valid.

This is Lindley's recurrence. If the \(n\)th customer waits \(D_{n}\) and is served for \(V_{n}\), he leaves \(D_{n} + V_{n}\) time units after his own arrival. The next customer arrives \(U_{n}\) after him, the interarrival time between the \(n\)th and \((n + 1)\)th customers, so (i) is valid. He must wait \(D_{n} + V_{n} - U_{n}\) if that is positive, and 0 otherwise.

So \(D_{n} + V_{n}\) is the departure time of the \(n\)th customer measured from his arrival, which is the sense in which (ii) is valid. The options rejecting either interpretation do not fit the recurrence.

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A canteen has a single cash counter. A customer has to buy coupons from the cash counter before getting the food items. Customers arrive at the counter according to a Poisson process with rate of 2 per 5 minutes. It takes 1.5 minutes on average to buy coupons. What is the long-run probability that the cash counter is free ?

A.\(1 - \dfrac{1.5}{2}\)
B.\(1 - \dfrac{1.5}{5}\)
C.\(1 - \dfrac{1.5}{2.5}\)
D.0

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The probability that the counter is free is \(1 - \dfrac{1.5}{2.5}\).

Treat the counter as an M/M/1 (or M/G/1) queue. The arrival rate is \(\lambda = 2/5 = 0.4\) per minute and the mean service time is 1.5 minutes, so the traffic intensity is

\[ \rho = \lambda \times 1.5 = 0.6 = \frac{1.5}{2.5} \]

In the long run the server is busy a fraction \(\rho\) of the time, so it is free with probability \(1 - \rho = 0.4\). 2.5 minutes is the mean time between arrivals. \(1 - 1.5/5\) wrongly uses 5 minutes as the inter-arrival time, \(1 - 1.5/2\) uses the arrival count, and 0 would mean the counter is always busy.

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As the variance of service time increases, the waiting time of a customer in \(M|G|1\) queue :

A.increases
B.decreases
C.remains the same
D.cannot be determined

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

As the service-time variance increases, the waiting time increases.

The Pollaczek-Khinchine formula for the M/G/1 queue gives the mean waiting time in queue as

\[ W_{q} = \frac{\lambda(\sigma^{2} + 1/\mu^{2})}{2(1 - \rho)} \]

where \(\sigma^{2}\) is the variance of service time. With the arrival rate and mean service time fixed, \(W_{q}\) increases linearly with \(\sigma^{2}\). This is why the M/D/1 queue, with constant service, has half the waiting of the M/M/1 queue with the same mean.

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