Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) MCQ Quiz in मराठी - Objective Question with Answer for Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) - मोफत PDF डाउनलोड करा

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Latest Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) MCQ Objective Questions

Top Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) MCQ Objective Questions

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 1:

In the notation (a/b/c) : (d/e/f) for summarizing the characteristics of queueing situation, the letters ‘b’ and ‘d’ stand respectively for

  1. service time distribution and queue discipline
  2. number of servers and size of calling source
  3. number of servers and queue discipline
  4. service time distribution and maximum number allowed in system
  5. number of servers and service time distribution

Answer (Detailed Solution Below)

Option 1 : service time distribution and queue discipline

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 1 Detailed Solution

Kendall's notation for queues:

(a/b/c) : (d/e/f)

a = Inter-arrival rate of distribution, b = Service time distribution, c = Number of servers, 

d = System capacity (queue discipline), e = Populationn size, f = Service discipline

Arriving and service distributions can of the following type:

M: Exponential, D: Deterministic, Ek: Erlang with parameter k, Hk: Hyperexponential with parameter k, G: General

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 2:

Following data is obtained from a queuing model,

Arrival rate = 3.4/hr, Mean service time = 16 min, wage rate = Rs. 5/hr, Idle cost = Rs. 7/hr

What will be the total cost of arriving model for two hours?

  1. Rs. 190
  2. Rs. 156
  3. Rs. 146
  4. Rs. 210

Answer (Detailed Solution Below)

Option 3 : Rs. 146

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 2 Detailed Solution

Concept:

\({\rm{Length\;of\;system\;}} \Rightarrow {L_s} = \frac{\rho }{{1 - \rho }},\;P = \frac{\lambda }{\mu } = \frac{{Arrival}}{{service}}\)

Cost = length of system × idle cost/hr × Time + wage rate × time

Calculation:

\({L_s} = \frac{\rho }{{1 - \rho }} = \frac{{\frac{{3.4}}{{3.75}}}}{{1 - \frac{{3.4}}{{3.75}}}} = 9.7142\)

Cost = 9.714 × 7 × 2 + 5 × 2

∴ Cost = Rs. 146

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 3:

At a work station, 8 jobs arrive every minute. The mean time spent on each job in the work station is 6 seconds. The mean steady state number of jobs in the system will be

  1. 3
  2. 5
  3. 2
  4. 4

Answer (Detailed Solution Below)

Option 4 : 4

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 3 Detailed Solution

Concept:

λ = Arrival rate (customer/time)

μ = Service rate (customer/time).

Inter arrival rate is given by \( = \frac{1}{\lambda }\) (time/customer)

Inter service rate \(= \frac{1}{{\rm{\mu }}}\) (time/customer)

Now, \({\rm{\rho }} = \frac{{Arrival\;rate}}{{Service\;rate}} = \frac{\lambda }{{\rm{\mu }}}\) 

Some basic formula’s of Queueing Theory

Probability that system is idle PO = 1 – ρ

Probability of having exactly n customer in the system (Pn) = ρn PO

Average number of customers in the system (Ls) \( = \frac{\rho }{{1 - \rho }}\) 

Calculation:

λ = 8 jobs/min

\(\begin{array}{l} \frac{1}{\mu } = 6\;seconds/job\\ \therefore \mu = \frac{{60}}{{6l}} = 10\;jobs/min \end{array}\)

\(\rho = \frac{\lambda }{\mu } = \frac{8}{{10}} = 4/5\)

∴ steady state number of jobs \(\left( {{L_s}} \right) = \frac{\rho }{{1 - \rho }}\)

\( = \frac{{\frac{4}{5}}}{{1 - \frac{4}{5}}} = 4\)

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 4:

Which of the following is not an assumption of the simple single server queuing model ?

  1. Service times are Poisson distributed
  2. Queue discipline is first come first served
  3. Mean arrival rate is less than mean service rate
  4. Arrivals follow Poisson distribution

Answer (Detailed Solution Below)

Option 1 : Service times are Poisson distributed

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 4 Detailed Solution

The correct answer is Service times are Poisson distributed.

Key Points Service timings are not poison distributed in a simple single server queuing model. It is based on an exponential distribution system.

Important Points 

Queuing system refers to the two or more servers that provide service to the arrival customers. When two or more servers are busy, then the customers might choose to switch from one queue to another queue or the queuing systems. 

Component of queuing system:

  • Arrival system: Arrival of customers may emerge through the calling population. The calling population can be limited or not limited.
  • Service mechanism: The service mechanism is defined by the number of servers, each having its own queue and service distribution system.
  • Queue discipline: It is the rule that a server chooses to select the next customer from the queue.

Single server queueing model:

A single server queuing model serves customers one by one on a first come first served basis. After receiving service, the customer leaves the queue and the number of people in the queue reduces by 1.

Assumptions of single server queue model:

  • Customers are infinite and patient.
  • Customers' arrival follows an exponential distribution.
  • Service rates also follow an exponential distribution.
  • Customers are served on a first come first served basis.

 

Hence, the correct answer is  Service times are Poisson distributed.

 

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 5:

Consider the following statements:

In a single-server queuing model

1. The arrivals is a memory less process

2. The arrivals is described as a Poisson distribution

3. Uncertainty concerning the demand for service exists

Which of the above statements are correct?

  1. 1 and 2 only
  2. 1 and 3 only
  3. 1, 2 and 3
  4. 2 and 3 only

Answer (Detailed Solution Below)

Option 3 : 1, 2 and 3

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 5 Detailed Solution

Explanation:

  • We know that in a single-server queuing model, customer arrival is random or memory less process.
  • Generally, arrivals do not occur at fixed regular intervals of times but tend to be clustered or scattered in some fashion.
  • A Poisson distribution is a discrete probability distribution which predicts the number of arrivals in a given time. It assumes that arrivals are random and independent of all other operating conditions.
  • Uncertainty exists concerning the demand for service. In a single server model, the type of distribution used for service times is exponential distribution.
  • It involves the probability of completion of a service. It should be noted that Poisson distribution cannot be applied to servicing because of the possibility of the service facility remaining idle for some time. Under the condition of continuous service, the following characteristics of exponential distribution are written, without proof:

 

\({\rm{Probability\;of\;n\;complete\;services\;in\;time\;t}} = \frac{{{{\left( {\mu t} \right)}^n}.{e^{ - \mu t}}}}{{n!}}\)

Hence, all the statements are correct. So, option (3) is the correct option.

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 6:

In a telephone booth, the telephone calls arrive following Poisson distribution with average times of 9 minutes between two consecutive arrivals. The length of time call is exponentially distributed with a mean time of 3 minutes. The average waiting time of the queue in minutes is:

  1. 0.5
  2. 1.5
  3. 2
  4. 2.5

Answer (Detailed Solution Below)

Option 2 : 1.5

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 6 Detailed Solution

Concept:

In a queuing model:

The average waiting time in the queue is given by:

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

Where λ and μ are the arrival & service rates

Calculation:

In a telephone booth, the telephone calls arrive following Poisson distribution with average times of 9 minutes between two consecutive arrivals. The length of time call is exponentially distributed with a mean time of 3 minutes. The average waiting time of the queue in minutes is:

λ = 1/9 min, μ = 1/3 min

Putting the values:

\({W_Q} = \frac{{\frac{1}{9}}}{{\frac{1}{3}\left( {\frac{1}{3} - \frac{1}{9}} \right)}} = 1.5~min\)

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 7:

In single server queuing model if arrival rate is λ and service rate is μ, then what is the probability of system being idle?

  1. λ/μ
  2. μ/λ
  3. \(\left( {1 - \frac{\lambda }{\mu }} \right)\)
  4. \(\left( {\frac{{1 - \lambda }}{\mu }} \right)\)

Answer (Detailed Solution Below)

Option 3 : \(\left( {1 - \frac{\lambda }{\mu }} \right)\)

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 7 Detailed Solution

Concept:

λ = Arrival rate per unit time

μ = Service rate per unit time

ρ = Traffic intensity factor or utilization factor or channel efficiency (0 ≤ ρ ≤ 1)

ρ = λ/μ

Probability of utilization \(\rho = \frac{\lambda }{\mu }\)

(1 - ρ) shows the time factor during which the system remains idle. It shows non - utilization factor.

(1 - ρ) shows the probability that the system is empty, i.e. there is no customer. It also shows the probability that a new customer (just entered in the system) will be immediately get the service. It can be interpreted that if the system is idle, then it shows the probability of having no customer in the system.

∴ Probability of idleness = \(1\; - \;\rho = 1 - \frac{\lambda }{\mu } = \frac{{\mu - \lambda }}{\mu }\) 

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 8:

Consider a single server queuing model with Poisson arrivals (λ = 4/hour) and exponential service (μ = 4/hour). The number in the system is restricted to a maximum of 10. The probability that a person who comes in leaves without joining the queue is

  1. 1/11
  2. 1/10
  3. 1/9
  4. ½

Answer (Detailed Solution Below)

Option 1 : 1/11

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 8 Detailed Solution

Concept:

Traffic intensity \(ρ \; = \;\frac{\lambda }{\mu}\; \)

\(\mathop \sum \limits_{n\; = \;0}^{n} {P_n}\; = \;1\)

The probability that there are n customer in the system Pn = P0 × ρn

where, P0 = probability of zero customer in queue

Calculation:

Given: 

λ = 4 per hour

μ = 4 per hour

Traffic intensity \(ρ \; = \;\frac{\lambda }{\mu}\; = \;\frac{4}{4}\; = \;1\)

We know that,

\(\mathop \sum \limits_{n\; = \;0}^{10} {P_n}\; = \;1\)

∴ P0 + P1 + P3 + … + P10 = 1

P0 + ρP0 + ρ2P0 + ρ3P0 + … + ρ10P0 = 1

P0(1 + ρ + ρ2 + …. + ρ10) = 1

P0(1 + 1 + …. + 1) = 1

\({P_0}\; = \;\frac{1}{{11}}\)

Probability that a person who comes in leaves without joining the queue i.e.

\({P_{11}}\; = \;{ρ ^{11}}{P_0}\; = \;{\left( 1 \right)^{11}} × \frac{1}{{11}}\; = \;\frac{1}{{11}}\)

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 9:

A queuing system using Kendall’s notation is expressed in the symbolic form as (M/M/3); (FCFS/6). How many number of servers in the system?

  1. 6
  2. 3
  3. 2
  4. 1
  5. 4

Answer (Detailed Solution Below)

Option 2 : 3

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 9 Detailed Solution

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

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 10:

In a queuing theory, the number of arrivals and service rate follows respectively

  1. Exponential distribution, Poisson distribution
  2. Poisson distribution, Exponential distribution
  3. Poisson distribution, Binomial distribution
  4. Exponential distribution, Binomial distribution
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : Poisson distribution, Exponential distribution

Queueing Model 1 (M/M/1) : (Infinity/Infinity/FIFO) Question 10 Detailed Solution

In queuing theory,

  • Arrival rate or arrival pattern follows Poisson distribution service rate or service pattern follows Exponential distribution.

Poisson probability Distribution for Arrivals Pattern:

  • The number of customer arrivals in a specific time period follows the Poisson Probability Distribution pattern and is expressed as follows:
  • \(P\left( X \right) = \frac{{{\lambda ^x}{e^{ - \lambda }}}}{{X!}}\)
  • X = the number of arrivals in the time period
  • λ = the average or mean number of customers arrivals per unit of time period or mean arrival rate

Exponential Probability Distribution for Service Times

  • The service time for a unit or customer is variable and random. An exponential distribution function expresses the service time as follows:
  • f(t) = μ e-μt
  • μ is the rate that units or customers are served.
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