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78 votes
78 votes

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 $D$?

  1. $\text{1005 ms}$
  2. $\text{1010 ms}$
  3. $\text{3000 ms}$
  4. $\text{3003 ms}$
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 @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
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8 Answers

130 votes
130 votes
Best answer
Routers are store and forward devices.

Propagation time $=\dfrac{100\;\text{km}}{10^8\;\text{m/s}}=1\;\text{milli second}$

Transmission time for a packet $=\dfrac{1000 \text{ bits}}{10^6 \text{ bits/sec}}=1\;\text{milli second}$

Packets will be forwarded in a pipelined manner after the first packet reaches the receiver, in every $1\;\text{ms}$ a new one arrives.

Now time taken by packet no $1$ to reach destination is :

$\text{1 ms ($T_x$ at sender) + 1 ms ($Tp$ from sender to R1) + 1ms  ($T_x$ at R1)}$

$\text{+ 1 ms($T_p$ from R1 to R2) + 1ms ($T_x$ at R2) + 1 ms ( $Tp$ from R2 to destination) = 6ms}$

So, time for packet, $1000=6\;\text{ms} + 999\;\text{ms} =1005\;\text{ms}$

Correct Answer: $A$
edited by

4 Comments

After first packet reaches the destination, transmission time and propagation time overlaps.
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just a random doubt, suppose that Tt=2ms and Tp=3ms, what will be the time for rest of the 999 packets?

will it be 999*2 ms or 999*3 ms, after going through the comments I got a little confused wheter to take the Tp or max(Tp, Tt). please clarify...
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edited by

@Abjineet Singh It will be Tt for remaining 999 packets as long as transmission time is same for all the Links because second packet will be transmitted after Tt of first packet got transmitted .Rest of journey from Source to destination is same for every packet.  so 

At t=0 start

At t=Tt 1st packet got transmitted…...then rest of journey whatever tp it is

At t=2Tt 2nd packet transmitted ..then .rest of journey...so it all depends on Tt ... 

At t=3Tt 3rd packet and so on…

=======================================================================

But note that unlike this question   if

suppose there are two links and Transmission  for Link1=15    and transmission time for Link2= 30  and tp=10

Then 

At t=15 first packet transmitted  over Link1           At t=30  second packet transmitted over Link1

At t=25 first packet reached   at R1                              At t=40  second packet reached at R1

At  t=55  first packet  transmitted over Link2 …………..note (second packet reached  at t=40 but waiting for its turn hence data getting accumulated at second router  )

At t=65  first packet  reached at destination

At t=55+30=85 second packet transmitted over Link2

At t=95 second packet reached at destination …..which is (time taken to reached first packet +slowest link) i.e. 65+30=95 .……….so here bottleneck is slowest link L2 which is 30

to understand this concept

refer this question 

https://gateoverflow.in/43572/gate2004-57

 and  make sure you understand @arjun sir’s comment on this question 

For data rate, we need to consider only the slowest part of the network as data will be getting accumulated at that sender (data rate till that slowest part, we need to add time if a faster part follows a slower part).

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91 votes
91 votes
First all data needs to be transmitted from source and after all packets transmission from source, just focus on last packet jorney, and u will get it.

Transmission time for all packets from Source:= $\frac{10^6}{1\times 10^6}$ = 1 sec = $1000$ms.
(Now the last packet is our main focus, bcoz the moment last packet reaches, all previous are already reached to Destination)

Last packet time = $3\times \text{propagation time of link} + 2 \times \text{tranmsion time of router} $ (Transmission time for source is already included in above $1000$ msec)
= $(3 \times 1) + (2 \times 1)$ = 5ms

Total time= $1000+5 = 1005 \text{ms}$

4 Comments

Whenever you see a big explanation of any question, @Sachin Mittal 1 Sir is always there with his version of the cleanest and beautiful answer like these :) 

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BEST ANSWER
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Yup understood it perfectly now.
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15 votes
15 votes

We have been given following info:

Signal Speed = 10^8 m/s

Bandwidth, BW = 1 Mbps = 10^6 bps

Propagation time, Tp = Link Distance/ Link Speed = (100x10^3)/(10^8) = 10^-3 sec = 1 ms.

[Propagation time is the time taken by packet to travel through the link]

Transmission time for 1 packet, Tt = Packet Size / BW = 1000 / 10^6 = 10^-3 sec = 1 ms.

[Transmission time is the time taken by source or router to process or put the data on Link]

Link filling time can be calculated as:

Pipeline Filling time
Time L1 R1 L2 R2 L3 D
t=0            
t=1 P1          
t=2 P2 P1        
t=3 P3 P2 P1      
t=4 P4 P3 P2 P1    
t=5 P5 P4 P3 P2 P1  
t=6 P6 P5 P4 P3 P2 P1

This way time taken by packet P1 to reach Destination (D) = 6 ms

Now, as you can see, at every ms, 1 packet will reach D.

Therefore, time taken for remaining 999 packets = 999 ms.

So, total time taken to transmit 1000 packets = 999 + 6 = 1005 ms.

Hence Option (A) is correct answer.

 

4 Comments

somehow I found it the most intuitive
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Great explation ... this question based on just coa pipelining question ..thnks
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Better explanation!👍
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For this question, the transmission time for every packet on each link is the same, so we are taking that at every Tp, we are getting one new packet at the receiver, what can we say about the same if transmission time hadn’t been constant at every link ??

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3 votes
3 votes
S------R1--------R2--------D        here we will use the concept of packet switching first of all ,transmission time = L/B which is equal to 1mili second from S to D  total  Tt= s to R1(1milisec) +R1to R2(1milisec) +R2 to D(1milisec)=3milisec  ,now for remaining 999 packets it will be 999*1milisec=999milisec   (We send 1 packet and now we are sending remaining 999 packets)

so total Tt=3+999=1002

now the propogation delay Tp=D/V , here it is mentioned that (Let each link be of length 100 km) Total distance from S To D is  300km    so Tp=300*10^3/10^8=3milisec.

now the sum of Tt + Tp=1002+3=1005 milisecond. ANSWER IS A.

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edited by
can u solve this question  for Tt =3 sec and Tp =6??????
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Answer:

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