Problem

A communication line capable of transmitting at a rate of 50 Kb/s will be used to accommodate 10 sessions each generating Poisson traffic at a rate 150 packets/min. Packet lengths are exponentially distributed with mean 1000 bits.

(a)

For each session, find the average number of packets in queue, the average number in the system, and the average delay per packet, when the line is allocated to sessions by using:

(i)

10 equal capacity frequency-division multiplexed channels.

(ii)

Statistical multiplexing.

(b)

Repeat part (a) for the case where: five of the sessions transmit at a rate of 250 packets/min, while the other five transmit at a rate of 50 packets/min.

Solution

(a)

(i)

Each channel has a transmission rate of 5 Kb/s.
Since the mean packet length is (b), the service rate is 5 packets/s.

The arrival rate is given in packets per minute, so we must turn it into packets per second to find that (packets/s).

Mean delay is (s). This includes time in the queue and transmission time.
Mean transmission time is (s). Mean queueing delay is (s).

The average number of packets in the queue is given by Little's Law, using the queuing delay as T and as calculated above; thus, packets in queue on average.
The average number of packets in the system, including both packets in the queue and the packet that is being transmitted, is given again by Little's law, using this time the total time in the system as T: packet, for each session.

(ii)

In this case, there is a single shared queue (we assume that no additional information is needed for demultiplexing). Since the combination of Poisson input streams is a Poisson input stream with a rate equal to the sum of all rates, we now have an input stream generating Poisson traffic with packets/s.
The service rate is (packets/s), since C=50 (Kb/s).

Using the same methods as above, we find that:
the transmission time is (s),
the total time for a packet in queue or transmission is (s),
the time spent in queue is (s).

The mean number of packets in the queue is and
the mean number of packets on the system is .

Note that:
The number of packets in queue and in the system is the same in the case of statistical multiplexing as for each of the 10 (sub)streams when handled separately. The time spent in queue or in total, is 1/10 th when using statistical multiplexing.

(b)

(i)

Using equal capacity channels, we have different results for each type of session. The heavy sessions have a rate of 250 packets/min, or packets/s, while the light sessions have a rate of packets/s. Since we have divided the channel as in (a.i), the transmission rate is again 5 packets/s and the average transmission time for a packet is 0.2 s.

For the heavy sessions:

The total time in the system for a packet is s,
and the time spent in the queue is 1 s.

The average number of packets in the queue for these sessions is
and the total number of packets in queue or in transmission is .

For the light sessions:

The total time in the system is s,
and the time in the queue is 0.04 s.

The average number of packets in the queue is and the average number of packets both in queue and being transmitted is .

Note how the static division of transmission capacity makes sessions with different input rates have different delays.

(ii)

Using statistical multiplexing, the whole channel is a single server transmitting 50 packets/s.
The total input rate is packets/min, or 25 packets/s.

Note that these numbers are exactly the same as in (a.ii) above, so the results for this system are exactly the same as there. Statistical multiplexing equalizes the delay across all sessions.

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