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M/M/1 : FIFO/N Model

In this queueing model, arrivals and departures are Poisson with rates λ and µ respectively. There is one server and the capacity of the system is infinity i.e., very large. We shall derive the steady -state probabilities and other characteristics.

 

Let                   Pn(t) = Probability of n arrivals during time interval t

 

If h > 0 and small then

 

Pn(t + h) = P (n arrivals during t and none during h)

or

P (n - 1 arrivals during t and one during h)

Or                                  P (n + 1 arrivals during t and one departure during h)

P (zero arrival in h) = e- λh = 1- λ

P (one arrival in h) = 1 - e- λh = λh

P (zero departure in h) = h e- µh 1- µh

P {one departure in h) = 1 - e- µh = µh.

Then for n > 0, we can write

Taking h -> 0. we obtain

These are called difference-differential equations.

 

The solution of ( 1) and (2) will give the transient-state probabilities, Pn(t). But the solution procedure is complex. So with certain assumptions we shall obtain the steady state solution.

 

For steady state, Jet us consider

t -> ∞, λ< µ

 

P'n(t) -> ∞, Pn(t) ->Pn for n = 0, 1, 2, ....

 

(Here λ = µ=> No queue and λ > µ => explosive state).

 

From (1) and (2) we obtain

λ .Pn - 1 + µ Pn + 1 - (λ+ µ) Pn = 0, n > 0

 

- λP0 + µP1 = 0, n = 0

P1 = (λ/ µ) P0.

From (3),                                                                     for n = 1,

= Pn (1 - ρ), ρ < 1, n :2: 0.

 

Now Ls = Expected number of customers in the system.

Ws = Expected waiting time in the system

 

= Ls / λ (By Little's formula) = 1 / (µ - λ).

 

Lq = Average queue length

 

= Ls – λ / µ = ρ2 (1- ρ)

Wq = Expected waiting time in queue

 

=Lq / λ(By Little's formula)

P (at least n customers in the system)

 

=

Let m = No. of customers in the queue

 

P (m>O)          = P (n > l) = 1- P (n ≤ l )

=1-{P (n=O) + P (n=l)}

=1-{(1 - p) + p (1- p)} = p2.

 

Therefore, Average length of non-empty queue

 

= E [m|m > 0]

Variance of system length/Fluctuation of queue

(a) Waiting Time Distributions

 

Let the time spent by a customer in the system be given as follows

 

Ts = t'l + t2 + .... + tn + tn + 1

where t’1 is the additional time taken by the customer in service, t2, .... , tn are the service times of other customers ahead of him and tn + 1 is the service time of arriving customer. Here T5 is the sum of (n + 1 ) independently identically exponentially distributed random variables and follows a gamma  distribution with parameters !l and n + I . The conditional pdf w(t | n + 1 ) of T5 is given by

w(t | n + 1 ) = µ / n! (µt)n. e-µt , t > 0.

 

Then the pdf of T5 i s obtained by first multiplying the expression w(t | n + 1 ) with the probability that there are n customers in the system and then summing over all values of n from 0 to ∞ and is given below :

 

pdf of T5 = (u – λ) e- (µ- λ)t, t > 0

 

which is an exponential distribution with parameter (!l - A.). We can also compute the pdf Tq of waiting time of an incoming customer before he receives the service following a similar line of argument. Thus

The second component means the customer starts receiving service immediately after the arrival if there is no customer in the system.

 

Also we can obtain

Example l. At a public telephone booth arrivals are considered to be Poisson with an average interval time of 10 minutes. The length of a phone call may be treated as service, assumed to be distributed exponentially with mean = 2.5 minutes. Calculate the following :

 

(a)    Average number of customers in the booth

(b)   Probability that a fresh arrival will have to wait for a phone call.

(c)    Probability that a customer completes the phone call in less than 1 0 minutes and leave.

(d)   Probability that queue size exceeds at least 5.

 

Solution. Here

 

λ= 1/10to customers per minute.

µ=1/ 2.5 customers per minute.

(a) Average number of customers in the booth

(b) P (a fresh arrival will have to wait)

 

= 1 - P (a fresh arrival will not have to wait)

= 1 - P (no customers in the booth)

= 1- P[ X = 0]

= 1 - (1 - ρ) = ρ= 0.25

 

(c) P (phone call completes in less than 10 min.)

= P [ Ts < l0]

Example 2. At a one-man barber shop, customers arrive according to the Poisson distribution with a mean arrival rate of 4 per hour and his hair cutting time was exponentially distributed with an average hair-cut taking 12 minutes. There is no restriction in queue length. Calculate the following :

 

(a)    Expected time in minutes that a customer has to spend in the queue.

(b)   Fluctuations of the queue length.

(c)    Probability that there is at least 5 customers in the system.

(d)   Percentage of time the barber is idle in 8-hr. day.

 

Solution.

λ = 4 per hour = 1/15 per minute.

µ= 1/12 per minute.

Ρ = λ/ µ=  12/15 =0.8 < 1(b) Fluctuations of queue length

(c) P (at least 5 customers in the system) = p5 = (0.8)5 = 0.33

 

(d) P (barber is idle) = P (no customers in the shop)

= 1 - ρ = 1 - 0.8 = 0.2

 

Percentage of time barber is idle = 8 * .2 = 1 .6.

 

Example 3. In the Central Railway station 15 computerised reservation counters are available. A customer can book the ticket in any train on any day in any one of these counters. The average time spent per customer by each clerk is 5 minutes. Average arrivals per hour during three types of activity periods have been calculated and customers have been surveyed to determine how long they are willing to wait during each type of period.

 

Type of period                        Arrivals per hr.                        Customer s acceptable

waiting time

Peak                            110                                                     15 Minutes

Normal                        60                                                        10 Minutes

Low                             30                                                        5 Minutes

 

Making suitable assumptions on this queueing process, determine how many counters should be kept open during each type of period.

 

Solution. Assumptions are as follows :

(i)                 Arrivals follow Poisson distribution with average arrival rate λ

(ii)               Service time follows exponential distribution with average service rate µ

(iii)             The given system can be considered as m different single server queueing system (m ≤ 15) and there is no jockeying and balking.

 

Then the arrival rate λ (per hour) = 100/m (peak period)

= 60/m (normal period)

= 30/m (low period)

The service rate at all the periods = 60 / 5= 1 2 per hour

 

Peak period :               Wq = 15/60 = (100/m) / 12(12-100/m), =>m =  12.22

 

Hence 13 counters must be kept open to ensure that the average waiting time does not exceed 15 minutes.

 

Normal period :           Wq = 10/60 = (60 / m) / 12 ( 1 2 - 60 / m) , => m= 7.5

 

Hence 8 counters must be kept open during the normal period so that the waiting time does not exceed 10 minutes.

 

Low period :

Wq = 5/60= (30/m) / 12 (1 2 - 30 / m), =>m= 5

 

Hence 5 counters must be kept open during the low period so that the waiting time does not exceed 5 minutes.