GeeksForGeeks AIM 2 - computer networks
in Computer Networks
57 views
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
by
57 views

2 Comments

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?

0
0

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

 

0
0

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