Processor Sharing Queues

The M/M/1/PS Queue Single server of rate μ shared equally across all jobs (no job ever waits in queue) Jobs arrive according to Poisson process of rate λ When n jobs are present, they each receive a service rate of μ/n Markov chain for system: i is number of jobs in system Arrival rate of λ in each state Departure rate is constant and equal to μ (n jobs that each gets μ/n) The chain is identical to that of an M/M/1/FCFS queue 1 μ 2 3 λ

More on the M/M/1/PS Queue State probabilities are as with the M/M/1/FCFS queue πn = ρn(1- ρ) E[N] = ρ/(1- ρ) and using Little’s Law E[T] = 1/(μ – λ) So where is the difference with an FCFS server? Main impact is that system time is now dependent on a job’s own service time: T(x) = αx for job with service time x Taking averages: E[T] = αE[S]  α = E[T]/E[S] = 1/(1- ρ) Therefore E[T|x] = x/(1- ρ)

More on PS Response Time Let (x)=b(x)/[1 – B(x)] be the rate of completing service given an attained service time of x, where we recall (p. 208) our earlier failure rate result (b(x) and B(x) are the density and c.d.f. of the service time, respectively) Let n(x) be the average density of jobs in the system with an attained service rate of x Assume also that service is given in quanta of size x n[x+x] = n(x)[1–(x)x] + o(x) ; [1–(x)x] is the probability that a job that has received x amount of service requires more than x+x Lettings x→0, this gives dn(x)/dx = – (x)n(x), which has a solution of the form n(x) = n(0) exp[-{0 to x}(y)dy] But exp[-{0 to x}(y)dy] = ln[1 – B(x)] , so that n(x) = n(0)[1 – B(x)] We also have that n(x) = λ[1 – B(x)]dT(x)/dx; based on applying Little’s Law to a system with finite service quanta x and then letting x→0 This implies dT(x)/dx = n(0)/λ and therefore T(x) = xn(0)/λ, i.e., T(x) is proportional to x

Two-Stage Coxian Distribution μ1 μ2 p 1-p E[S]=1/μ1+ p/μ2 E[S2]=2/μ12+ 2p/μ22 C2=2(p μ12 + μ22)/(μ1μ2[pμ1+ μ2]) – 1 Can also emulate system with two classes of jobs Small & big jobs Small jobs arrive according to Poisson process of rate (1-p)λ and have an average service time of duration 1/μ1 Big jobs arrive according to Poisson process of rate pλ and have an average service time of duration 1/μ1+1/μ2 Recall that the two-stage Coxian distribution can be used to obtain distributions with both C2 > 1 and C2 < 1 μ1 = 1 and μ2 = 2 yields C2 = 2/(p+2) < 1 μ1 = 1 and μ2 = 1/2 yields C2 = (6p+1)/(2p+1) > 1

M/Cox/1/PS Queue Probability that system is empty is still π0 = (1- ρ), where ρ = λE[S] = λ(1/μ1+ p/μ2) Probability that the system has n1 jobs in phase 1 and n2 jobs in phase 2 follows a product form πn1,n2 = binom(n1+n2,n1) ρ1n1 ρ2n2 (1- ρ), where ρ1 = λ/μ1 and ρ2 = λp/μ2; (ρ = ρ1 + ρ2) Probability πn of n jobs in the system