Consider the situation where three nodes are interconnected

Consider the situation where three nodes are interconnected in series, that is, Node A is connected to Node B and Node B is connected to Node C. Suppose further that we wish to send a stream of messages from A to C, and that B cannot store data (except for small amounts for brief periods of time). Node A is connected to Node B by a sliding window link that has a bit rate of 100kbps, a message size of 1000 bits and a window size of 3. The propagation time of a bit (or packet) on this link is 20msec. For the connection between B and C, bits (or packets) have a propagation time of 5msec and uses a stop-and-wait protocol. Assuming the same message size, what should the bit rate of the link be?

Solution

Let us understand the constraint the question puts.

The question says that B cannot store data (it can store very less data for very brief periods)

Hence, it must transmit all data it receives (from A) to C immediately.

Formally speaking, B should be able to transmit 3 messages to receives from C, by the time A is sure has sent 3 messages and has received the first ACK.

Let us now take a look into the information provided in the question, and also derive some other information from it.

1)A and B use sliding window protocol. The message size is 1000 bits and the window size is 3

2)Transmission speed of the link between A and B is 100 kbps

Therefore, the time taken to transmit a message which is 1000 bits in length is

1000/(100 *1000) seconds

=10/1000 seconds

=10 msecs (1 msec=1/1000 seconds)

From the above three pieces of information, we can come up to a better understanding of the situation at hand.

Assume that at time=t₀ , A begins transmitting a window. It will take a 10 msec to transmit 1 message and hence 30 msec to transmit 3 messages (entire window). At t₁=30 msecs, it would have transmitted all messages in it’s window.

Also, at time t₂=10 msec + propogation delay

                                =10 msec +20 msec

                                =30 msec

The first packet would have been completely transmitted and been received at B, and B will send an ACK. ACKs have negligible length and hence we assume that there is negligible transmission time for an ACK. The ACK from B will reach A 20 msecs later. Hence

at time t₃= t₂+ propagation delay+ transmission time

                  =30 msec+ 20 msec + 0 (+ transmission time is negligible)

                 =50 msecs

later we will have received the 1^st ACK at A, and A can begin transmitting again.

B cannot hold data, and should have transmitted data it received from A

Now let us consider the situation between B and C.

Propogation delay=5 msec

Transmission speed= unknown. Let us assume transmission speed =R bits/second

Transmission time =1000 bits/R.

Between B and C we use the stop and wait ARQ, hence, B will wait for ACK from C after every packet.

Time taken for 1 message to be generated from B, propagated to C, acknowledged by C and ACK propagated to B= 1000 /R seconds + 5 msec + 0 + 5msec (transmission time for ACK is assumed to be 0)

=10 msec + 1000/R seconds

Hence time taken by B to transmit 3 messages= (10 msec + 1000/R) *3= 30 msec+ 3000/R

From the constraint we had arrived at in the beginning

We get

30 msec + 3000/R seconds =50 msec.

Solving this equation we get R= 150 *1000 bits per second or

R=150 kbps.

Hence we were able to calculate R, i.e the transmission speed between B and C, alternatively called the bit rate of the link between B and C.

R=150 kbps is our required answer.