In the network shown below, frames are generated at Node A and sent to Node C th

Question

In the network shown below, frames are generated at Node A and sent to Node C through Node B. Nodes B and C acknowledge each data frame immediately upon receipt to Nodes A and B, respectively. The channel data rate between Nodes A and B is 256 kbps and the data rate between nodes B and C is 1 Mbps. The lines between each pair of nodes are full-duplex, and the propagation delay for each line is 10 msec/mile. All data frames are 1000 bits long and ACKs are separate frames of negligible length. A sliding window protocol with a window size of W is used between Nodes A and B, and a stop-and-wait protocol is used between Nodes B and C. All lines are error-free and no frames can be lost.

What is the maximum window size Wmax that can be used between nodes A and B without flooding the buffers of node B?

Hint: The buffers of node B will not be flooded if the average number of frames entering and leaving Node B are equal over a long time interval (at steady-state).

Explanation / Answer

Answer:

Propagation delay=10msec/km

length of frame L=1000 bits

Consider A to B:

Distance between A and B=2000 miles

=3219 km (1 mile=1.6 km)

Data rate between A and B=256 kbps

Propagation time=distance*propagation delay

=3219*10 msec

=32190 msec

Transamition time=L/R

=1000*10-3 /256*103*10-3

=3.90 msec

Now consider B to C:

Distance between B and C=500miles

=805 km (1 mile=1.6 km)

Data rate between B and C=1 Mbps

=1024 kbps

Propagation time=distance*propagation delay

=805*10 msec

=8050 msec

Transamition time=L/R

=1000*10-3 /1024*103*10-3

=0.977 msec

A sliding window protocol with a window size of x is used to transmit between A and B.

The first frame takes 32190 msec to transmit 'x' frames in 32190(2x-1)msec.

B can transmit one frame to C at a time. It takes 10+0.977=10.977 msec fro the frame to be received at C

and additional 10msec for C's acknowledgment to return to B.

So, B can transmit one frame every 20.977 msec or 'X' frames every (20.977) *X msec

So, (20.977)*X=32190(2X-1)

20.977X=64380X-32190

64380X-20.977X=32190

64359.02X=32190

X=32190/64359.02

X=0.5

Hence, the maximum window size (Wmax) that can be used between A and B without flooding the buffer of mode B is 0.5

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