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

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, E_k: Erlang with parameter k, H_k: Hyperexponential with parameter k, G: General

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

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 P_O = 1 – ρ

Probability of having exactly n customer in the system (P_n) = ρⁿ P_O

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

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 \(\left( {{L_s}} \right) = \frac{\rho }{{1 - \rho }}\)

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

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.

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.

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:

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

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.

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.

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

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

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.