Extended question to find the fragment sizes GATE-CS-2004

Question

Consider three IP networks A, B and C. Host HA in networks ‘A’ sends message each containing 180 B of application data to a host HC in network HC. The TCP layer prefixes 20 Bytes header to
the message. This passes through on intermediate network ‘B’. The maximum packet size, including 20B IP headers in each network is:
A. 500 Bytes
B. 100 Bytes
C.1000 Bytes
The network A and B are connected through 512 Kbps link, while B and C are connected by a 256 Kbps link.

1. Assuming that the packets are correctly delivered, how many Bytes including headers, are
delivered to IP layer at destination for one application message in the best case? Consider only data packets.
A. 220 B.240 C. 260 D. 280

 

2. What is the rate at which application data is transferred to host HC? Ignore errors,
acknowledgements and other overheads?
A. 196 Kbps B. 177.23 Kbps C. 354.5 Kbps D. 325.5 Kbps

 

3. What is the extra overhead caused by fragmentation?
A. 40 Bytes B. 20 Bytes C. 0 Bytes D. 60 Bytes

Answer 1

1.Application layer can recieve any amount of data ..in the given question A recieves 180 Byte of data and sends it to transport layer.

2.transport layer on recieving it adds a header of 20 Bytes to it total frame size=180+20=200B and sends it to network layer.

3.network layer on recieving 200 B of data adds 20 Bytes of header total length=220 Bytes..as maximum  packet size=1000 B so no need of fragmentation and then it sends to network B.

4 As B can recieve 100 byte of data only but A is sending 220 byte of data...first of all 20 byte of  is removed by network layer of B and then divide it into 3 fragments (80+20),(80+20),(40+20) ...and then send it to network C.

5.so network layer will recieve 260 Byte of data...(ANSWER)

PART B..

rate of transfer=(180(useful data)/260(total data))*240(bottleneck bandwidth)

                   =166.15KBps

part C->

without fragmentation -:headers required=220

with fragmentation=260

extra overhead=260-220=40 Bytes

Comments

Thanks a lot @saurav for the answer, can you plz have a look at the below:

1. Why is "20 byte of  is removed by network layer of B"? Does it happen in all cases? Do all intermediate networks remove the IP of the sender? If the inbetweeen networks(in this case B) remove the IP of sender(here A), how will the reciever(here C) come to know that the packet has come from A and not B?

2. Why is the bottleneck bandwidth 512, when B and C are connected via 256kbps bandwidth?

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the answer to your question lies that where your destination is!if the destination is not in your network then we need a router and in the router we have network layer which removes the previous IP header BUT in the new header the source address will be of the sender only.

part B-:bottleneck bandwidth =minimum bandwidth in rough language ...

i have taken wrong bandwidth earlier ...updated it please check.,

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Why is bottleneck bandwidth 240Kbps ?  Why not 256 Kbps ?

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yes it will be 256kbps