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Host A (on TCP/IP v4 network A) sends an IP datagram D to host B (also on TCP/IP v4 network B). Assume that no error occurred during the transmission of D. When D reaches B, which of the following IP header field(s) may be different from that of the original datagram D?

  1. TTL
  2. Checksum
  3. Fragment Offset
  1. $\text{i}$ only
  2. $\text{i}$ and $\text{ii}$ only
  3. $\text{ii}$ and $\text{iii}$ only
  4. $\text{i, ii}$ and $\text{iii}$
asked in Computer Networks by Veteran (106k points)
edited by | 3.6k views
0
option(A )
+12

Therefore, (D) is correct option!

7 Answers

+54 votes
Best answer

The answer is OPTION D.

Whenever an IP packet is transmitted, the value in Time to Live (TTL) field will be decremented on every single hop. Hence, TTL is changed on every hop.

Now, since TTL changes, hence the Checksum of the packet will also change.

For the Fragmentation offset, A packet will be fragmented if the packet has a size greater than the Maximum Transmission Unit (MTU) of the network. Hence, Fragmentation offset can also be changed.

answered by Active (1.4k points)
edited by
0
I also saw at few places where the answer is (b) and not (d).

Did any one check in the official 2014 gate key??

Any one knows anything about this?
+3
If there is NO ERROR (given in que) how can we come to conclusion that checksum field will change?

Checksum will recompute and stored again at each router. (but after computation we get same checksum because see below)

If checksum not match then datagram is discared.

But Here Datagram reaches to Destination, It is only possible if checksum match.

If Checksum Matching, How it can change?

SO please give more insight for claim that checksum field change.

Thanks.
0
@parthbkgadoya i agreed with you.. can anyone please tell what was the answer in the official key? bcz how check sum can be changed?

@Arjun sir please help to varify
+1
Official Answer is D) Header checksum is recomputed and verified in all intermediate routers as some header fields change like(TTL). So after the packet reaches the destination the header checksum will be surely different from the one computer at the source. And as mentioned in the answer Fragmentation may occur at any point when packet size is larger than the MTU of the interface.
+5
Checksum is recomputed at every router.
0

i know the checksum is recomputed every time but in que they have said 

"Assume that there is no error occured during the transmission of D"

what is significance of this statement?

+3
They just meant that Data is correctly transmitted so no other field like ip addresses etc will get changed or corrupted.
+1

the answer for the above question is (D) for sure..although even I didnt get the answer yet but some1 is asking 

Official question and answers for 2014..

https://www.qualifygate.com/download/qna2014/Key_CS_9.pdf

https://www.qualifygate.com/download/qna2014/CS03_2014.pdf

0

since packet reached without error..hence checksum should not be changed....and some people are saying with change in TTL value makes change in checksum but whats the relation in them..plz explain..!!!

+3

dhruvkc123

in IP datagram checksum is calculated on header part .... and TTL is the part of IP header ,if TTL changes then definite value of checksum is being changed.... 

0

@hs_yadav fine...got it now..thnx 

0

What if there is no fragmentation at all ?

0

I understand i and ii but I am having problem with iii.

The question says "When D reaches B,"

If it had said "when fragment reaches B" then I accept that the offset would be different. But until all the fragments don't reach B how can we say that "D" has reached B? And when all fragments have reached B then they are merged and then only we can say it as "D". This D doesn't differ in fragmentation offset. Isn't it? 

Please correct me.

+6 votes
Answer should be B and not D

Reason :- Since, in the question it clearly says that the Datagram D reaches the destination, if there was fragmentation done along the way, the Datagram D would have been divided into different datagrams say A, B, C. Obviously if the same datagram D is reaching, there was no fragmentation hence no change in Frag. offset.

So, only TTL and checksum will be different.
answered by Active (3.2k points)
0
Did you check in the official 2014 gate key?? I also saw a few places where the answer is (b) and not (d).

Any one knows anything about this?
0
But same ip packet can reach sdestination via multiple frames
0
Ans should be D, TTl will definitely change and other fields which "may" change during journey of the packet is FO (fragment offset), MF(more fragment field) and TL(total length), in case fragmentation occurs. Also "options" field may also change, and if so then header length will also change.

So overall, TTL change (compulsory) and Fragment offset (optional) does change checksum field,which makes the IP header field(s) different from the original datagram. hence answer should be D.
+4 votes
Correct Option Will be -D

TTL:: It is obvious that TTL value will be change at each router so checksum will also.

Fragment::It may change or not depend upon MTU of forwarding device as router here.But in worst it must be change.
answered by Loyal (9.1k points)
+2 votes
D option
answered by Boss (16k points)
+1 vote

Answer is D

If due to MTU restriction we need to fragment the packets the Checksum value changes/recomputed wether error has occured or not...

So, TTL changes in every hop, Fragment Offset can change hence Checksum changes

http://stackoverflow.com/questions/27111459/how-does-the-udp-checksum-change-for-ip-fragments  In this answer see last bullet point in box :)

answered by (77 points)
0
@arjun sir why would offset feild change

And if fragmentation is done there is no D left it will be fragmented so from which fragment header we will compare

If we will compare from first fragment offset as it's identification no. And offset both are matching so fragment offset didn't change.

Length feild is getting changed but not offset
+1 vote
D option is correct bcoz at each inetermediate node ttl value will be change so  checksum also and each node fragmention may also be possible so all are correct.
answered by (399 points)
0 votes

so TTL definitely changes in each hop so check sum would also change.suppose the MTU of network is less than frame may get fragmented so fragement offset may also change .so answer must be (d)

answered by (83 points)
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