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 Wmaxthat 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

Given data:

Propagation Delay = 10ms/mile = 10*ms/1.6km = 6.25 ms/km

Length of the frame L = 1000 bits

Consider A --> B:

Distance A and B (d) = 2000 miles ( 1mile=1.6 km)

= 3219 km

Data Rate A and B (R) = 256 kbps

Propagation time = Distance * propagation Delay

                             = 3219 * 6.25 * 10-3

                             = 20.11 msec

Transmission Time = L/R

                               = 1000 * 103/256*103

                               = 3.9 msec

Consider B --> C

Distance B and C (d) = 500 miles ( 1mile=1.6 km)

         = 805 km

Data Rate B and C = 1 mbps

Propagation time = Distance * propagation Delay

                             = 805 * 6.25 * 10-3

                             = 5.03 msec

Transmission Time = L/R

                               = 1000 * 103/1024*103

                               = 0.977 msec

To calculate the window size, need to calculate window size between A and B first.

A sliding window protocol is used to transmit the frames over A to B. Assume it as x.

The first frame takes 20.11 msec to transmit ‘x’frames in 20.11(2x-1) msec.

B can transmit the frames by using stop and wait protocol.

Total time taken to receive the frames at C = 6.25 + 0.977 = 7.22 msec

Additional time of acknowledgement received by the sender B = 6.25 msec.

So, B can transmit after each 13.47 msec

or ‘x’frames every x*13.47 msec

Thus, window size can be by equating both calculate x

13.47* x = 20.11(2x-1)

13.47 x = 40.22 x – 20.11

20.11 = 26.75 x

x=0.75

Hence, the window size without flooding the buffer that can be used between A and B is 0.75.

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