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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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8 Answers

3 votes
3 votes

We are storing 180 bytes of application data at a. now 20 bytes of transport layer data will be added to this and that will be header of transport layer after that network layer will also add its 20 bytes of data as you can see that is also known as network layer header in total vpp transmitting 220 bytes of data in which 200 words is our actual data for transport layer and 20 bytes is network layer had a we will be sending this particular data at a speed of 1 MBPS now when we will be sending this data at we will receive 220 bytes of data in which 20 points of data will be cut off because that was that is network layer header. now this 200 bytes of data will be further divided into three packets of 100 words each as our be will be having hundreds of packet length now in totality we will be seeing that in order to transmit 180 bytes of data we are actually transmitting 260 bytes of data that means efficiency is 9 by 13 into 512kbps because I eventually we will be sending this 512kbps link

1 votes
1 votes
$Max$ $Packet$ $Size$ $is$ $given$ $in$ $the$ $question$ $to$ $confuse$ $you$

A can send $180B$ of data to B.

As A can send 1000B data in one go so we will send 180B data + 20B TCP header + 20B IP header = 220B packet

B will receive this packet and removes off 20B IP as it is no longer required. It will now send this 200B data in 3 parts each containing max data of 80B as its limit is 100B of packet.

3 packets of 100B 100B and 60B will be made. = 260B

now data rate is original data sent per actual packet being sent.

In a packet of 260B made at B only 180B data is data rest is headers.

so

data rate = $\frac{180}{260}$

and now we will multiply this chunk by 512kbps which is given in the question.

data rate = $\frac{180}{260}*512kbps$

which is equal to Option B 354.5 kbps
0 votes
0 votes

Answer - 354.46

Data rate is = Actual Data or useful data / Time

Useful data is = 180 bytes

total data = 260 bytes.

Bandwidth = 512Kbps
That means

8/(512x1000) secs = for 1 byte of data

(8x260)/(512x1000) secs = for 260 bytes of data = 0.0040625

Data Rate = 180 / 0.0040625 = 44307.69230 bytes/sec

which is 354.46 Kb/s
 

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0 votes

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