SKEDSOFT

Introduction:In queuing model arrival rate consider as a input and service rate is consider as a output. In queuing model, two basic constituents are considered i.e. arrival rate and service rate. This model differs from the other model in the sense that the maximum number of customers in the system is limited to N. Finite Queue Length Model: (M / M / 1) : FCFS / N /∞

This model differs from the other model in the sense that the maximum number of customers in the system is limited to N. Therefore the equations of above model is valid for this model as long as n < N and arrivals will not exceed N under any circumstances. The various equations of the model is:

1)      p0 = (1− ρ) /(1− ρN 1),where ρ = λ /μ and λ /μ > 1 is allowed.

2)      pn = (1− ρ)ρn /(1− ρN 1)for all n = 0, 1, 2, ..N

3)      Average queue length E (n) = ρ[1− (1 N)ρN NρN 1]/ (1− ρ) (1− ρN 1).

4)      The average length of the waiting line = E (L) = [1− N ρN 1 (N −1)ρN ]ρ2 / (1− ρ) (1− ρN 1)

5)      Waiting time in the system = E (v) = E (n) / λ’ where λ’ = λ (1− ρN )

6)      Waiting time in the queue = E (w) = E (L) / λ’ = [(E (n) / λ’ / (1/μ)].

Example:

In a railway marshalling yard, good train arrives at the rate of 30 trains per day. Assume that the inter arrival time follows an exponential distribution and the service time is also to be assumed as exponential with a mean of 36 minutes.

 Calculate: (a) The probability that the yard is empty, (b) The average length assuming that the line capacity of the yard is 9 trains.

Solution

Data: λ = 30 / (60 × 24) = 1 / 48 trains per minute.   μ = 1 / 16 trains per minute.

Therefore ρ = (λ /μ) = 36 / 48 = 0.75.

The probability that the queue is empty is given by = p0 = (1− ρ) / (1− ρN 1), where N = 9.

{1 - 0.75) / [1- (0.75)9 1 = 0.25 / 0.90 = 0.28. i.e. 28 % of the time the line is empty.

Average queue length is =

= 3 trains.

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