GeeksForGeeks AIM 2 - computer networks
in Computer Networks
0 votes
0 votes

Consider a message with 60000 bits long, that is to be sent from a source to a destination, there are two routers between source to destination. 

Each of link in the path has bandwidth 1 Mbps, each packet is 1000 bit long, total time taken (in msec) to reach the last bit to the destination, propagation delay is 10 msec ________.

in Computer Networks


my solution:

total packets = 60000 / 1000 = 60

The packets will be transmitted 3 times, at A, R1 and R2. They’ll have propagation delay on 3 links A – R1, R1 – R2 and R2 – B.

Also, 1st packet will take full time from A to B and rest 59 packets will follow in a pipeline. 

Transmission time = 1 ms, propagation delay = 10 ms (given)

Therefore, total time = 1 * (3 * 1 + 3 * 10) ms + 59 * (3 * 10) ms = 623 ms

answer key’s solution:

Why did they count transmission delay for 59 packets? Shouldn’t it be propagation delay?


No, your solution is incorrect.Packet used to transfer in pipeline manner.It is not like untill first packet will reach to the destination, all remaining packet will wait. For better understanding I will suggest you to solve this previous year question.

GATE CSE 2012 | Question: 44 - GATE Overflow



Subscribe to GO Classes for GATE CSE 2022

1 Answer

2 votes
2 votes

when first packet is transmitted on last link see the scenario

At t=22 it reaches second router
At t=23 it is transmitted from second router
At t=33 it reaches destination
At t=24 second packet is transmitted from second router
At t=34 second packet reaches destination
So every 1 second packet reaches destination.

That means 1st packet reaching destination at 33ms

remaining (60-1=59 packets) can reach the destination with= 59+33 =92ms.

So your analysis is correct Until last step. That means we will not consider the Propagation time for every packet. It should be Transmission time(59*Tt). Hope this helps.

Related questions