Question

Packets arrive at an infinite buffer at a Poisson rate of 100 packets/sec, and are transmitted over a link of rate 1 Mbps in an FCFS manner. An arriving packet is a 100-bit packet with probability 0.2, a 1000-bit packet with probability 0.5, and a 10000-bit packet with probability 0.3 . Find (a) the average waiting time for a packet, and (b) the average number of packets in the buffer, in steady state.

Answer 1

Solution

Assuming that the transmission over link follows an exponential distribution, the given problem is a direct application of M/M/1 Queue Sysstem.

Back-up Theory

An M/M/1 queue system is characterized by arrivals following Poisson pattern with average rate λ, [this is also the same as exponential arrival with average inter-arrival time = 1/ λ] service time following Exponential Distribution with average service time of (1/µ) [this is also the same as Poisson service with average service rate = µ] and single service channel.

Let n = number of customers in the system and m = number of customers in the queue.

[Trivially, n = m + number of customers under service.]

Let (λ/µ) = ρ

The steady-state probability of n customers in the system is given by P_n = ρⁿ(1 - ρ) …………(1)

The steady-state probability of no customers in the system is given by P₀ = (1 - ρ) …….……(2)

Average queue length = E(m) = (λ²)/{µ(µ - λ)} ........…………………………………………..…..(3)

Average number of customers in the system = E(n) = (λ)/(µ - λ).....……………………………..(4)

Average waiting time = E(w) = (λ)/{µ(µ - λ)} …………………………...........……………………..(5)

Average time spent in the system = E(v) = {1/(µ - λ)}………………………........………………..(6)

Now to work out the solution,

An arriving packet is a 100-bit packet with probability 0.2, a 1000-bit packet with probability 0.5, and

a 10000-bit packet with probability 0.3 =>

average packet size = (100 x 0.2) + (1000 x 0.5) +(10000 x 0.3) = 3520.

This, in conjunction with, ‘Packets arrive at a Poisson rate of 100 packets/sec’ =>

average arrival rate in terms of bits = 3520 x 100 = 0.352 x 10⁶ per sec.

‘transmitted over a link of rate 1 Mbps’ => service rate is 10⁶ per sec.

Thus, given data in the above terminology would be: λ = 0.352 x 10⁶ per second and µ = 10⁶ per second …….........................................................................................................….. (7)

Part (a)

Average waiting time for a packet = E(w) = (λ)/{µ(µ - λ)} [vide (5) above]

= (0.352 x 10⁶)/{10⁶ x (0.648 x 10⁶)}

= 0.543 x 10^-6 sec. ANSWER

Part (b)

Average number of packets in the buffer = E(n) = (λ)/(µ - λ) [vide (4) above]

= (0.352 x 10⁶)/(0.648 x 10⁶)

= 0.543 ANSWER