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52 votes
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Consider three IP networks $A, B$ and $C$. Host $H_A$ in network $A$ sends messages each containing $180$ $bytes$ of application data to a host $H_C$ in network $C$. The TCP layer prefixes $20$ byte header to the message. This passes through an intermediate network $B$. The maximum packet size, including $20$ byte IP header, in each network, is:

  • $A:$ $1000$ $\text{bytes}$
  • $B: 100$   $\text{bytes}$
  • $C: 1000$ $\text{bytes}$

The network $A$ and $B$ are connected through a $1$ $Mbps$ link, while $B$ and $C$ are connected by a $512$ $Kbps$ link (bps = bits per second).

What is the rate at which application data is transferred to host $H_C$? Ignore errors, acknowledgments, and other overheads.

  1. $325.5$ $\text{Kbps}$
  2. $354.5$ $\text{Kbps}$
  3. $409.6$ $\text{Kbps}$
  4. $512.0$ $\text{Kbps}$
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This is indeed a confusing question because at one hand they say "find rate at which data is transferred and on other hand they say ignore other overheads". Although here overheads means delay of any kind but tcp+ ip header also considered as overhead in that case option B&D both are possible according to comprehension of one individual. They should have been more specific.
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@Arjun Sir

What if there is another host D and there is a link of $1Mbps$ between $H_C$ to $H_D$ and network $C$ also support MTU of $100bytes$ and we need to find the data rate.

will Data rate be = $\frac{180*8 bits}{1.76ms + 4.0625ms + 2.080ms}$

                              = $208.62 Kbps$

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edited by

Part 1 of the question :- https://gateoverflow.in/1052/gate2004-56

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8 Answers

71 votes
71 votes
Best answer
$\text{Packet A sends an IP packet of 180 bytes of data + 20 bytes of TCP header}$
$\text{+ 20 bytes of IP header to B}$.

IP layer of $B$ now removes $20\text{ bytes}$ of IP header and has $200\text{ bytes}$ of data. So, it makes $3$ IP packets - $[80 + 20,  80 + 20 , 40 + 20]$ and sends to $C$ as the IP packet size of $B$ is $100$. So, $C$ receives $260$ bytes of data which
includes $60\text{ bytes}$ of IP headers and $20\text{ bytes}$ of TCP header.

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).

So, here $180\text{ bytes}$ of application data are transferred from $A$ to $C$ and this causes $260\text{ bytes}$ to be transferred from $B$ to $C$.

Time to transfer $260\text{ bytes}$ from $\text{B-C}=\dfrac{260\times 8}{(512\times 1000)}$
$=\dfrac{65}{16000}=\dfrac{13}{3200}$.

So, data rate $=\dfrac{180\times 3200}{13}=44.3\text{ kBps}= 44.3 \times 8 = 354.46\text{ kbps}.$

Correct Answer: $B$
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4 Comments

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Throughput = Efficiency * Bandwidth right then how can we calculate efficiency at TCP layer?
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Why you guys not considering datalink  layer header +preamble+sfd??. Is it. because of this words in the question " Ignore errors, acknowledgments, and other overheads."??

260+78 = 338

=> (180/338)*512Kbps
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156 votes
156 votes
Over all bytes received by NETWORK $C =260\text{ bytes}$ (this include tcp header and ip header).

EFFECTIVE DATA TRANSFERRED FROM NETWORK $B$ TO NETWORK $C=180\text{ Byte}.$

HENCE EFFECTIVE EFFICIENCY $= \dfrac{180}{260}\times 512\text{ Kbps}=354.46\text{ Kbps}$
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Kya baat hai Kya baat hai Kya baat hai Sahi hai, Thanks buddy you made my life easier :-)
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why this is not best answer ?
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thanks best ans
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42 votes
42 votes

(B) is correct option, after the delivery of first packet, Host C will receive a new packet for each 4.0625 milliseconds.

4 Comments

ohh okay so u mean 100 bytes will be transmitted as soon as rececived from A by B

then 100 bytes agn and then 60 bytes thus a total time of 4.06msec.

 

just got a little confused here then exactly how  fragmentation of packet at B..take place?
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"ohh okay so u mean 100 bytes will be transmitted as soon as rececived from A by B"

not 100 bytes 220 bytes ...fragmentation is done at B not at A...

AND

"just got a little confused here then exactly how  fragmentation of packet at B..take place?"

actually practiclly definite there are some delay for Fragmentation and queuing delay...

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This is the actual correct explanation because application data rate is the data/time it took to transfer.

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3 votes
3 votes

the rate at which application data is transferred to host HC

It means Question is asking for Throughput.

We know that, Efficiency = $\frac{Useful Data Bytes Transferred}{Total data Bytes Transferred}$

Here, bottleneck is created by B to C link i.e @ 512 kbps. Effective bandwidth BW = 512 kbps.

$Throughput = \eta * BW$$= \frac{180}{260} * 512 kbps = 354.46 kbps.$

Correct Ans: (B)

 

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nice:)
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