A communication system capable of transmitting at a rate of 50[Kbps] will be use

Question

A communication system capable of transmitting at a rate of 50[Kbps] will be used to accommodate 10 sessions, each generatingPoisson traffic at a rate of 150 [packets/minute]. Packets lengthare exponentially distributed with mean of 1000 [bits].
(a) For each session, find the average number of packets in thequeue, the average number of packets in the system, and the averagedelay per packet in the system, when the line is allocated to thesessions using:
(1) 10 equal-capacity time-division multiplexed channels
(2) Statistical multiplexing
(b) Repeat part (a) for the case where five of the sessionstransmit at a rate of 250 [packets/minute], while the other fivetransmit at a rate of 50 [packets/minute].

Explanation / Answer

(a)For each session the arrival rate is lambda = 150/60 = 2.5 packets/sec.When the line is divided into 10 lines of capacity 5 Kbps, thenmu is 5 packets/sec for each line and the averagepacket transmission time is 1/mu = 0.2 sec.The corresponding utilization factor is rho = lambda/mu = 0.5.We have for each sessionThe average number in each queue is Nq = rho^2/(1-ro)=0.5,The average number is each line is N = rho/(1-rho)=1,The average delay is T = N/lambda = 0.4 sec.For all 10 sessions collectively, Nq and N must be multiplied by 10to give Nq = 5 and N = 10.When statistical multiplexing is used, all sessions are merged intoa single session with 10 times larger lambda and mu; lambda = 25,1/mu = 0.02. We obtain rho = 0.5, Nq=0.5, N=1, and T = 0.04 sec.Therefore Nq, N and T have been reduced by a factor of 10 over theTDM case.(b)For sessions transmitting at 250 packets/min we haverho=(250/60)*0.2 = 0.833,and we haveNq=(0.833)^2/(1-0.833)=4.158,N = 4.99,T= N/lambda = 4.99/(250/60) = 1.197 secFor the sessions transmitting at 50 packets/min we haverho = (50/60)*0.2 = 0.166,Nq = 0.033,N = 0.199T = 0.199/(50/60) = 0.239The corresponding averages over all sessions are:Nq = 5 * (4.158 + 0.033) = 20.95N = 5 *(4.99 +0.199) = 25.95T = N/lambda = N/(5*lambda1 + 5*lambda2)= 25.95/(5*(250/60)+5*(50/60))= 1.038 secWhen statistical multiplexing is used, the arrivalrate of the combined session is5*(250+50) = 150 packets/sec and the same valuesfor Nq, N and T as in (a) are obtained.

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