Station A is sending data to station B over a full duplex error free c...
Station A is sending data to station B over a full duplex error free channel. A sliding window protocol is being used for flow control. The send and receive window size is being used for flow control. The send and receive window sizes are 6 frames each. Each frame is 1200 bytes long and the transmission time for such a frame is 70 micro sec. Acknowledgment frames sent by B to A are very small and require negligible transmission time. The propagation delay over the link is 300 micro sec. What is the max achievable throughput in this communication?
plz provide the solution for this question
3Comments
Shamshad Hussain saifi @shamshadhussain
12 Nov 2019 06:43 pm

@rihan :Throughput is the amount of useful data sent per unit time. Useful data means payload.

Since the window size is 6, the sender can send a maximum of 6 frames till the first frame is acknowledged.

Time is taken by the first frame to get acknowledged= transmission time of original frame + propagation delay from src to destination + transmission time of ack frame + propagation delay from receiver to src.

Since ack transmission is to be neglected, we get total time=70+300+300=670microsec.

The data sent during this time =6frames=6∗1200bytes=7200bytes.

Hence, throughput achieved is 7200bytes/670microsec=85.97Mbps.
 

Adeema jain @adeema
12 Nov 2019 06:53 pm
Why are you considering the channel as half duplex??
Rohit Panwar @panwarrohit
12 Nov 2019 06:55 pm
@adeema jain Since ack transmission is to be neglected.we have to take it as half duplex.

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