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

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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170 votes
170 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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46 votes
46 votes

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

6 votes
6 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)

 

Answer:

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