GATE CSE 2014 | Set 3 | Question: 25

Question

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

• A. $\text{i}$ only
• B. $\text{i}$ and $\text{ii}$ only
• C. $\text{ii}$ and $\text{iii}$ only
• D. $\text{i, ii}$ and $\text{iii}$

Comments

This is not at all ambiguous, as JASHAN said, this checks your aptitude about the subject

Destination is not exceptional, it changes the fragmentation offset, so yes, offset can be changed

so smart na

this is computer science,

here you will find the most beautiful questions framed by brilliant and sophisticated minds of IITs and IISC professors

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@JashanArora
The question says “When D reaches B”. Here, D is already defined as an IP datagram.
How can we say that a datagram (a network-layer entity) has reached its destination when one of its fragments is still at the physical layer and reassembly has not yet occurred?

We cannot introduce new interpretations to justify ambiguity in the question.
Doesnt D reaching the destination mean that it has reached the Network layer and has been reassembled? Or the question should have been more precise with the wording?
Please correct me if I am wrong.

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The official key for this question is "Option D"

You can check here : Official Key

Answer 1

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.

Comments

But in this question Host A datagram has been referred as Datagram(D). And it's mention there too that when D reaches B, so how can we assume that D has been fragmented in the route..? Can anyone explain me..

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Adityagto Dec 6

it is clearly mentioned that packet D reaches its destination,

so if fragmentation has occured during transmission,

after reaching to it destination all fragments combine it its original sequence, therefore fragment offset remains unchanged.

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Gate is game of words, it has stated as "may be"

Answer 2

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.

Comments

@meghna nicely explained

 

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yes ur intuition is correct
but they clearly mentioned 'may' so due to this we will consider both cases
with fragmentation and no fragmentation
and clearly in fragmentation the Fragment Offset changes
so, option D is correct
if they had given fields 'will' change
so we can say option b is right.

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The official key for this question is "Option D"

You can check here : Official Key

Answer 3

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.

Comments

But original packet offset will remain same, new generated fragments can have different FO.

Answer 4

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.

Answer 5

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 :)

Comments

@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

Answer 6

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.

Comments

it is clearly mentioned that D reaches its destination,

so if fragmentation occured during transmission, after reaching to it destination all fragments combine it its original sequence, therefore fragment offset remains unchanged.