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.in queue we have first come first out model.
Poisson Arrival / Poisson output / Number of channels / Infinite capacity / FIFO Model M / M / 1 / (∞ / FIFO): Formulae used:
1) Average number of arrivals per unit of time = λ
2) Average number of units served per unit of time = μ
3) Traffic intensity or utility ratio = ρ= λ /μ
4) Probability that the system is empty = P0 = (1- ρ )
5) Probability that there are .n. units in the system = Pn = ρn P0
6) Average number of units in the system = E (n)= ρ/(1- ρ0)= λ/ (μ –λ)=Lq λ /μ
7) Average number of units in the waiting line = EL= ρ2/(1- ρ0)= λ2/ μ (μ –λ)
8) Average waiting length (mean time in the system) = E (L / L > 0)= 1/ (μ –λ)= 1/(1- ρ)
9) Average length of waiting line with the condition that it is always greater than zero= ρ/(1- ρ0)2= λ/ (μ –λ)2
10) Average time an arrival spends in the system = E (v)= 1/ μ (1- ρ)= 1/ (μ –λ)
11) P (w > 0) = System is busy = ρ
12) Idle time = (1− ρ)
13) Probability distribution of waiting time = P (w) dw = μ ρ (1- ρ)e−μw(1−ρ)
14) Probability that a consumer has to wait on arrival = (P (w > 0)= ρ
15) Probability that a new arrival stays in the system =P(v)dv = μ(1− ρ) e−μv(1−ρ) dv,
Example: A T.V. Repairman finds that the time spent on his jobs have an exponential distribution with meanof 30 minutes. If he repairs sets in the order in which they come in, and if the arrival of sets isapproximately Poisson with an average rate of 10 per 8 hour day, what is repairman’s expected idletime each day? How many jobs are ahead of the average set just brought in?
Solution:This problem is Poisson arrival/Negative exponential service / single channel /infinite capacity/ FIFO type problem.
Data: λ = 10 sets per 8 hour day = 10 / 8 = 5/4 sets per hour. Given 1/μ = 30 minutes, hence μ = (1/30) × 60 = 2 sets per hour. Hence, Utility ratio = ρ = (λ /μ) = (5/4) / 2 = = 5 / 8. = 0.625. This means out of 8 hours 5 hours the system is busy i.e. repairman is busy. Probability that there is no queue = The system is idle = (1− ρ) = 1 . (5 / 8) = 3 / 8 = That is out of 8 hours the repairman will be idle for 3 hours. Number of sets ahead of the set just entered = Average number of sets in system = λ / (μ − λ) = ρ / (1− ρ) = 0.625 / (1 . 0.625) = 5 / 3 ahead of jobs just came in.
