Queueing Theory MCQ Quiz - Objective Question with Answer for Queueing Theory - Download Free PDF

Explanation:

Given the data:

• Arrival rate (λ) = 120 customers per hour
• Service rate (μ) = 150 customers per hour

The utilization factor (ρ) is the ratio of the arrival rate to the service rate. It represents the proportion of time the server is busy. The formula for the utilization factor is:

ρ = λ / μ

Substituting the given values:

ρ = 120 / 150

Simplifying this fraction:

ρ = 0.8

The utilization factor (ρ) indicates that the server is busy 80% of the time. Therefore, the proportion of time the server is idle (1 - ρ) can be calculated as:

Idle Time Proportion = 1 - ρ

Substituting the value of ρ:

Idle Time Proportion = 1 - 0.8

Idle Time Proportion = 0.2

Therefore, the proportion of time the server is idle is 0.2

Concept:

The given problem follows the M/M/1 queuing model, where arrivals follow a Poisson distribution, and the service times are exponentially distributed.

The expected queue length L_q is given by:

\( L_q = \frac{\lambda^2}{\mu (\mu - \lambda)} \)

where:

• \( \lambda \) = Mean arrival rate (customers per hour)
• \( \mu \) = Mean service rate (customers per hour)

Calculation:

Step 1: Compute the Service Rate \( \mu \)

Given that the mean service time per customer is **3 minutes**, we convert it to an hourly rate:

\( \mu = \frac{60}{\text{Service Time}} = \frac{60}{3} = 20 \text{ customers per hour} \)

Step 2: Compute the Expected Queue Length

Substituting values into the formula:

\( L_q = \frac{8^2}{20 (20 - 8)} \)

\( L_q = \frac{64}{20 \times 12} \)

\( L_q = \frac{64}{240} = 0.267 \)

Explanation:

Given: λ = 5/hr, μ = 10/hr

∴ \(\rho=\frac{\lambda}{\mu}=\frac{5}{10}=\frac{1}{2}\)

Now, probability for the system to be idle,

P₀ = 1 - ρ = 1 - 0.5 = 0.5

Concept:

In an M/M/1 queuing system, we deal with a single server queue where arrivals follow a Poisson process, and service times are exponentially distributed. The key parameters are the arrival rate \(\lambda\) and the service rate \(\mu\).

The average number of customers in the system, denoted by \(L\), can be determined using the formula:

\( L = \frac{\lambda}{\mu - \lambda} \)

where:

• \(\lambda\) is the arrival rate of customers per unit time.
• \(\mu\) is the service rate of customers per unit time.

Given:

• Arrival rate, \(\lambda = 6\) customers per hour
• Service rate, \(\mu = 10\) customers per hour

Calculation:

Using the formula for the average number of customers in the system:

\( L = \frac{\lambda}{\mu - \lambda} \)

Substitute the given values:

\( L = \frac{6}{10 - 6} = \frac{6}{4} = 1.5 \)

Therefore, the average number of customers in the system is:

\( \boxed{1.5 \text{ customers}} \)

Thus, the correct answer is option 2: 1.5 customers.

The correct answer is Shifting from one queue to another parallel queue.

Key Points

• Queuing theory deals with problems which involve queuing (or waiting). Typical examples might be: banks/supermarkets - waiting for service, computers - waiting for a response, public transport - waiting for a train or a bus.
• Jockeying can be described as the movement of a waiting customer from one queue to another (of shorter length or which appears to be moving faster, etc.) in anticipation of a shorter delay.

Additional Information

•  As explained, queuing theory is the study of the movement of people, objects, or information through a line.

• Balking is when customers deciding not to join the queue if it is too long and reneging is where customers leave the queue if they have waited too long for service.

Hence, the correct answer is Shifting from one queue to another parallel queue.

Concept:

No. of customers in the system, \({L_S} = \frac{\lambda }{{\mu - \lambda }}\)

No. of customers in the queue, \({L_q} = \frac{{{\lambda ^2}}}{{\mu \left( {\mu - \lambda } \right)}}\)

Calculation:

Arrival rate λ = 10 minute/customer = 6 customers/hour

Service rate, μ = 6 minute/customer = 10 customers/hour

The average length of the queue, \({L_q} = \frac{{{\lambda ^2}}}{{\mu \left( {\mu - \lambda } \right)}}\)

\(\;{L_q} = \frac{{{6^2}}}{{10\;\left( {10\; - \;6} \right)}} = \;0.9 = \frac{9}{{10}}\)

Points to remember

• Average arrival time and the time spent in the system (Waiting time in system) = \({W_s} = \frac{1}{{\mu - \lambda }}\)
• Average arrival time and the time spent in the queue (before being served) (Waiting time in queue) = \({W_q} = \frac{\lambda }{\mu }.\frac{1}{{\mu - \lambda }}\)
• Average number of customers in the system = \({L_s} = \lambda {W_s} = \lambda .\frac{1}{{\mu - \lambda }}\)
• Average number of customers in the queue = Average queue length = \({L_q} = \lambda {W_q} = \lambda .\frac{\lambda }{\mu }.\frac{1}{{\mu - \lambda }} = \frac{{{\lambda ^2}}}{\mu }\left( {\frac{1}{{\mu - \lambda }}} \right)\)
• Probability that there are k customers in the system = (ρ)k(1 - ρ) = \(\frac{\lambda }{\mu }\left( {\frac{1}{{\mu - \lambda }}} \right)\)

• Probability that there are more than k customers = \({\left( {\frac{\lambda }{\mu }} \right)^{k + 1}}\)

The correct answer is \(\mathbf{\frac{6}{36}}\)

Key Points

•  The probability of a particular event E occurring is given by \(P(E)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes or sample space.}}\)
• The sample space of this problem, where two dice are thrown together is 36
• The sum of values on the dice resulting in a total of 7 is given by the following table:

  Serial number	Value of observation in dice 1	Value of observation in dice 2
  1	1	6
  2	6	1
  3	3	4
  4	4	3
  5	2	5
  6	5	2

• The number of favourable outcomes, in this case, is 6.
• Let F be the event of getting a 7 when two dice are thrown together.
• Therefore P(F)=\(\frac{6}{36}\)

Important Points

•  Sample space is the set of all the possible outcomes in the given experiment. In this case, there are 36 outcomes possible when 2 dice are thrown together.
• Event denotes the probability of getting a single outcome of an experiment. In this problem, the event is to get the value of 7.
• The outcome is the result of the experiment.

Hence, the probability of getting a total of 7 on two dice thrown together is \(\frac{6}{36}\)

Concept:

There are four types of customer behaviours.

1. Jockeying- When a customer keeps changing the queue just in hope to get fast service.
2. Balking-Customer does not join the queue and leaves the system as the queue is very long
3. Reneging- Customer joins the queue for a short period then leave the system as the queue is moving very slowly.
4. Cheaters- Customer takes illegal means like bribing, fighting etc just in hope to get service faster.

Concept:

Queuing Theory

Given:

Mean arrival rate (λ) = 10/hr, Mean service rate (μ) = 15/hr

∴ Traffic intensity, \(\rho = \frac{\lambda }{\mu } = \frac{2}{3}\)

Queue length = \(\frac{{{\rho ^2}}}{{1 - \rho }}\)

Calculation:

Queue length = \(\frac{{4/9}}{{1/3}}\) = 4/3

Concept:

The mean waiting time of a patient, needing medical checkup facility in the queue, is,

\({W_q} = \frac{\lambda }{{\mu \left( {\mu - \lambda } \right)}}\)

Calculation:

λ = 32 per hour

\(\mu = \frac{{60\; \times \;60}}{{90}} = 40\;{\rm{per\;hour}}\)

\(\therefore {W_q} = \frac{{32}}{{40\left( {40 - 32} \right)}}\)

Wq = 0.1 hour

∴ Wq = 6 minute

Explanation:-

Queuing models are represented by Kendell and Lee notation whose general form is (a/b/c) : (d/e/f)

where,

a =  Probability distribution for arrival pattern, b  = Probability distribution for service pattern, c =  No of servers in the system, d =  Service rule or service order, e = Size or capacity of  the system, f = Size or capacity of calling population

Therefore, No of server in the system (M/M/3) : (FCFS/6) is  = 3

Explanation:

• The ratio of arrival to the service rate indicates the percentage of time sever is busy, it is known as utilization factor, system utilization channel efficiency, and clearing ratio. It indicates the probability that a new customer has to wait.
• The arrival rate of customer = λ  (Follows poisons distribution)
• Service rate  = μ  (Follows exponential distribution)
• If the ratio of  mean arrival to mean service rate is increased
• λ > μ
• The customer arrival rate is more than the service rate, hence the customer moves slower in the system and the queue length will keep on increasing and after a certain time, the incoming population will not get service.

Average number of customers are given as

λ = 30/hr  = 0.5 /min

Probability of customer arrival in 1 minute or less, is given as

P(≤t) = 1 - e^-λt

P(≤1) = 1 - e^{-0.5 × 1}

p(≤t) = 0.393

Probability of customers arrival in 3 minutes or less, is given as

P(≤t) = 1 - e^-λt

P(≤3) = 1 - e-0.5 × 3

p(≤t) = 0.7768

Probability of customer arrival between 1 and 3 minute, is given as

P(≤3) - P(≤1) = 0.7768 - 0.393

P(≤3) - P(≤1) = 0.38

Concept:

For the Poisson process of rate λ, and for any t > 0, the Probability mass function for N(t) (i.e., the number of arrivals in (0,t]) is given by the Poisson PMF

\({P_{N\left( t \right)}}\left( n \right) = \frac{{{{\left( {\lambda t} \right)}^n}\exp \left( { - \lambda t} \right)}}{{n!}}\)

If the arrivals of a Poisson process are split into two new arrival processes, each new process is independent.

Calculation:

Given:

Customers arrive at a shop according to the Poisson distribution with a mean of 10 customers/hour.

⇒ λ = 10 customers/60 min = 1 customer/6 min = 1/6 customers per minute;

The manager notes that no customer arrives for the first 3 minutes after the shop opens,

⇒ t = 3 min and n = 0;

From the Poisson’s PMF,

The probability of zero customers in 3 minutes will be

\({P_{N\left( 3 \right)}}\left( 0 \right) = \frac{{{{\left( {\frac{1}{6} \cdot 3} \right)}^0}\exp \left( { - \frac{1}{6} \cdot 3} \right)}}{{0!}} = {e^{ - 0.5}} = 0.606\)

Now,

The probability that the customer arrives in the next 3 minutes = 1 - P(0)

∴ The probability that the customer arrives in the next 3 minutes = 1 - 0.606

∴ The probability that the customer arrives in the next 3 minutes = 0.39

Concept:

Waiting time in the queue \(W_q=\frac{L_q}{λ}\) and \(L_q=\frac{ρ^2}{1-ρ}\)

\(\rho=\frac{\lambda}{μ}\)

where, L_q = length of queue, λ = arrival rate, μ = service rate

Calculation:

Given: 

λ = 5 cars per hour, μ = 1 car per 10 minute = 6 cars per hour

⇒ \(\rho =\frac{5}{6}\)

 \(L_q= \frac{(5/6)^2}{1-(5/6)}=\frac{25}{6}\)

 \(W_q=\frac{25/6}{5}=\frac{5}{6}\) hours = \(\frac{5}{6}\times60=50~min\)