Generalization Of Model (M/m/1):(fcfs/∞/ ∞)
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.
Generalization of model (M /M / 1) : (FCFS/∞/ ∞) : (Birth – Death process)
- In waiting line system each arrival can be considered to be a birth i.e. if the system is in the state E_{n}, i.e. there are n units in the system and there is an arrival then the state of the system changes to the state E_{n}_{ 1}.
- Similarly when there is a departure from the system the state of the system becomes E_{n}_{-1}.
- ., then it will mean that some person interested in joining the queue may not join due to long queue. Similarly, if μ is also dependent on the queue length it may affect the service rate.
- In this case both λ and μ cannot be taken to be fixed.
- Three cases may occur, which are described below.
- In this model, arrival rate and service rate i.e. λ and μ do not remain constant during the queuing phenomenon and vary to λ_{1}, λ_{2} . λ_{n}and μ_{1, }μ_{2}.. μnrespectively. Then:
p_{1}= (λ0 /μ_{1} ) p_{0}
p_{2}= (λ0 /μ_{1}) (λ_{1} /μ_{2} ) p_{0}
............
............ pn = (λ0 /μ_{1}) (λ_{1} /μ_{2} )...(λn−2 /μn−1 )× (λn−1 /μn ) p0
But there are some special cases when:
1.λn = λ and μn = μ then, p0 = 1− (λ /μ), = (λ /μ)n ×[1− (λ /μ)] pn
2.When λn = λ /(n 1) and μn = μ p_{0} = e^{−ρ} pn = [ρn /(n!)]× e^{−ρ} , where ρ = (λ /μ)
3.When λn = λ and μn = n ×μ then p0 = e−ρ and p = (ρn / n!) x e^{−ρ. n}
Example:
A transport company has a single unloading berth with vehicles arriving in a Poisson fashion at an average rate of three per day. The unloading time distribution for a vehicle with .n. unloading workers is found to be exponentially with an average unloading time (1/2) xn days. The company has a large labour supply without regular working hours, and to avoid long waiting lines, the company has a policy of using as many unloading group of workers in a vehicle as there are vehicles waiting in line or being unloaded. Under these conditions find
(a) What will be the average number of unloading group of workers working at any time?
(b) What is the probability that more than 4 groups of workers are needed?
Solution Let us assume that there are .n. vehicles waiting in line at any time. Now service rate is dependent on waiting length hence μn = 2n vehicles per day (when there are .n. groups of workers in the system). Now λ = 3 vehicles per day and μ = 2 vehicles per day. (With one unloading labour group) Hence, pn = (ρn / n!)× e−ρ for n ≥ 0 Therefore, expected number of group of workers working any specified instant is
The probability that the vehicle entering in service will require more than four groups of workers is==0.019.