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

@Sidd1425 one point i would like to add. The pipeline formula can only be used here when $T_t=T_p$.

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@rhl More like when $T_{t} <= T_{p}$ right ? Because in case of queuing delay, we need to add that also ?

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packet1  takes total 6msec to reach D  (3Tt + 3Tp)
packet2 is just behind packet1 so,  at 7msec it reaches D
similarily rest all 999 packets (packet no. 2  - 1000)  will take  999*1msec  
so Total Time  =  6 + 999 = 1005

Answer 1

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: $A$

Comments

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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@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…

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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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For calculation of Tt and Tp  refer to this answer itself, 

Tt = 1ms

Tp = 1ms

then we can directly use Store and Forward formula,

                  T = (N + k - 1) Tt  +   k x Tp

here, N = no. of packets   and     k = no. of links 

                  T = (1000+3−1) × 1 +  3×1

                     =  1002 + 3

                     =  1005

Answer 2

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

Comments

@Markzuck I don't think there is a valid reason to consider transmission delay at the destination.

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\[
T_{\text{end-to-end}} = 1000T_t + T_p + 2(T_t + T_p)
\]
 

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sir,where the question says the time to send the whole file ....there are asking the sum of transmission and propagation delays although the processing can be parallel but still the delays are counted ex: suppose we perform two multiplication operation on multiprocessor machine though the processing is parallel and done in single multiplication time but the operations will be 2...coming to question one packet td and pd add to 6msec so for 1000 it should be 6000msec which is not the time to send the file as it's not asked

Can you please Clarify..

Answer 3

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.

 

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 ??

Answer 4

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&rsquo;t hops==routers ?

Answer 5

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.

Comments

can u solve this question  for Tt =3 sec and Tp =6??????

Answer 6

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