GATE CSE 2012 | Question: 44

Question

Consider a source computer $(S)$ transmitting a file of size $10^{6}$ bits to a destination computer $(D)$ over a network of two routers $(R_{1}\text{ and }R_{2})$ and three links $(L_{1},L_{2},\text{ and } L_{3})$. $L_{1}$ connects $S$ to $R_{1}$; $L_{2}$ connects $R_{1}$ to $R_{2}$; and $L_{3}$ connects $R_{2}$ to $D$. Let each link be of length $\text{100 km}$. Assume signals travel over each link at a speed of $10^{8}$ meters per second. Assume that the link bandwidth on each link is $\text{1 Mbps}$. Let the file be broken down into $1000$ packets each of size $1000$ bits. Find the total sum of transmission and propagation delays in transmitting the file from $S$ to $D$?

• A. $\text{1005 ms}$
• B. $\text{1010 ms}$
• C. $\text{3000 ms}$
• D. $\text{3003 ms}$

Comments

 @Divy Kala isn’t it the same thing that total transmission delay + propagation delay = total time is taken for the packet to reach the destination? 

and we are using pipelining since our data is fragmented into smaller fragments.

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The  question asks what the sum of all delays were, not how the entire file takes to be transferred. File transfer time is less than the sum of all delays because the former works on a pipeline. Inside the pipeline itself there are delays that happen over overlapping time durations. The question is either poorly worded, or the answer is wrong. Either case, the official answer key could have been challenged here.

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just a random foolish doubt
why aren't we considering aknowledgements
and why did we comeup with the pipelining technique for this caluculation

Answer 1

A common doubt is why after the first packet arrives, every other packet arrives in 1 ms when clearly there is transmission and propagation delay (1ms + 1ms) to consider.

To see why this is so, consider a single link connecting the source to destination. Lets say there is 1ms transmission time and 1ms propagation time. So the first packet reaches the destination in 2 ms. But does the next packet require 2ms too?

No, because the sender has already finished transmission of the second packet (at t = 2ms) when the first packet reached the destination. Hence there is only the 1ms propagation delay to consider now.

Apply that same logic here. The first packet takes ms to reach the destination. But after that, the following packet needs only 1ms propagation delay to reach the destination. This gives our answer of 6ms + 999 ms = 1005ms.

Comments

Here since both transmission and propagation time is the same, it may look like propagation time is controlling the time when the second packet reaches the destination. Take different values of T-t and T-p, and you will see it is taking max(T-t,T-p) time for 2^nd packet to reach destination 

Answer 2

Formula refer from here

https://www.geeksforgeeks.org/packet-switching-and-delays-in-computer-network/

 

 

Thanks

Comments

How did you assume $3 hops=3 links$? Aren’t hops==routers ?

Answer 3

Tt = 1ms, Tp = 1ms

Source~~Router1~~Router2~~Destination

Let’s think in terms of the last packet.

When the last packet reaches destination then all packets will reach and we will get the final answer. So we will count for last packet only. Now, the last packet will be transmitted on the first link L1 only when all previous packets have been transmitted.

So,

Time required = transmission time of 999 pkts from S + transmission time of last pkt From S + propagation from S to R  + transmission time of last pkt From R1 + propagation from R1 to R2 + transmission time of last pkt From R2 + propagation from R2 to D

= (999*1) + 1 + 1 + 1 + 1 + 1 +1

= 1005 ms

 

Answer 4

ok!

BW=10^3 Kbps

Tt=1ms

i.e., 1Kb (or 1 packet) is put on link in 1 ms

So, by the time 1st packet reaches R1, the IInd packet will already have been put on the 1st link.

(Please note Router processing time has been ignored here )

In other words, all the subsequent packets will be 1ms behind each other on the path

that's why after 6 second of packet 1's journey, it's +1ms+1ms ...    till 999 ms

Hope it helped :)