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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 $\text{iii}$
in Computer Networks by Veteran (105k points)
edited by | 4.9k views
0
option(A )
+15

Therefore, (D) is correct option!

0

If checksum is applicable for all fields including TTL, then we know that TTL is changed(decrement by1) in each networking device like router,so checksum value is changed that time,So receiver receive different checksum than what sender calculate.

Then what will happen? @Manu Thakur

0
checksum calculated at source(s) and sent to router(R1)... R1 verifies checksum first, then TTL value changes and R1 computes new checksum and sends it to R2... Now R2 verifies checksum, then TTL changes and new checksum computed.. so on till destination..

So, checksum values can change in between though there are no errors..

If there are any errors, checksum can detect it while checksum verification at each router or destination..

8 Answers

+62 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.

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?
+4
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.
+8
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..!!!

+5

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 ?

+6

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.

0

@MiNiPanda same doubt

@ankitgupta.1729 If we consider the case at B after reassembly procedure is done(given D reaches B), then the offset value will be the same as the value when D was at A. 

Please help with option 3.

0

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.

WRONG.

TTL doesn't change on every hop. TTL changes for every hop through the router and at the destination.

Check: https://gateoverflow.in/1983/gate2014-2-25

0

@MiNiPanda

"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? "

even I have the same doubt.did u find the answer for it?

+10 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.
by Active (3.4k 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
+1
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.
by Boss (10k points)
0
But original packet offset will remain same, new generated fragments can have different FO.
+2 votes
D option
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 :)

by (75 points)
+1
@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.
by Junior (743 points)
+1 vote

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)

by (161 points)
+1 vote
TTL will change (No doubt)

Due to change in TTL it will cause change in the header and since CHECKSUM will find some changes (due to change in TTL value) CHECKSUM value will be changing at each spot.

Now coming to the FRAGMENT OFFSET part, since question has asked about options which MAY be different from original Datagram then, YES FRAGMENT OFFSET ''MAY" also change if the MTU of other network is smaller.

Hope this helps.
by (33 points)

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