The RTT between a router in Bangalore and my office is about 405 ms. If a stop-a

Question

The RTT between a router in Bangalore and my office is about 405 ms. If a stop-and-wait ARQ protocol is used to communicate 1500B packets at 2 Mb/s to that router, what is the utilization? Answer within 0.001

ANSWER: Utilization = 0.01459854014

Using the Information above answer the following:

Given the above utilization, what is the average number of bits sent per second between the two endpoints if stop-and-wait is used? Answer to the nearest 1000 b/s.

ANS: _________

Now suppose two nodes near each other communicate 1500B packets over a 2 Mb/s link using stop and wait with a RTT of 6 ms. What is the utilization in this case? Answer to the nearest 0.01.

ANS: _________

Under this second set of assumptions, how many bits can be communicated from one to the other per second? Answer to the nearest 1000 b/s.

ANS: _________

Explanation / Answer

Given

RTT between a router in Bangalore and the office is about 405 ms.

Stop-and-wait ARQ protocol is used to communicate

Packet size(L)=1500B

Bandwidth(BW)=2 Mb/s

Let TT be the transmission time and TP be the propagation time.

now let us first calculate TT= L/BW

=> TT=(1500*8)/(2*106)

=6ms.

( For calculating the utilization intially the formula used is : Utilization=TT/(TT+2TP)

=TT/(TT+RTT) as RTT=2TP

=6/(6+405)

=6/411

=0.01459854014 )

Average number of bits sent per second between the two endpoints is the throughput

Throughput=Utilization * Bandwidth

= 0.01459854014 * 2*106

=29197 b/s

Now when the two nodes near each other communicate 1500B packets over a 2 Mb/s link using stop and wait with a RTT of 6 ms.

then

Utilization=TT/(TT+2TP)

=TT/(TT+RTT) as RTT=2TP

=6/(6+6)

=0.50

Now for total bits communicated from one end to the other end per second ,

Throughput=Utilization * Bandwidth

Throughput=0.50 * 2 * 106

=1000,000 b/s or 1000 Kb/s

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