Madeeasy CBT

Question

Ethernet frame may carry data upto 512 byte (i.e. mtu 512B ).If total number of fragments are denoted by x and payload bytes in last fragment is y then value of y*2^x will be?

Comments

I am not able to view picture properly, it gets blurred on zooming . but I am getting no. Of fragments as 3 and last fragments is of size 230 B. I am not able go view equation to be computed. Plz type it in question. And verify is my answer correct??

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Sorry. Edited now.

And I don't know. The answer given is 1632. I don't think its correct though.

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Are you sure??? Maybe 1696 will be given.  I am getting 1696 as answer. Payload in last is 212 and this makes 212×8. Post the solution given by made easy.

Answer 1

Total length=Data+Header

1200=Data+20

Data=1180. 

MTU => 512 bytes => 492 excluding IP header which is mandatory. 

Since 492 is not a multiple of 8, maximum Payload that can be sent in one segment is 488.

So there will be 3 segments =>

488+20 

488+20

204+20

So answer is 204*(2³) => 1632

Comments

Why are you subtracting 20? It's Ethernet frame, so the ip header should be enclosed within the frame,  isn't it?

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Because each fragment with carry it's own IP header. It's mandatory.

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Mutiple of 8 comstraint is in IP header.

Why are you implying it in Ethernet frame???? Ehternet allows min. Payload of 46B which is also not a multiple of 8.

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Similar question!!

https://gateoverflow.in/8255/gate2015-2_52

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They are not same. In that question a UDP message was there which had to pass thorugh network layer. Here we already have IP packet. So no need to pass from network layer. So multiple of 8 constraint does not imply and also header size should be 18 B

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Yes...but that IP packet can't pass as a single unit. So router will fragment it. Each segment will have its own IP header.

In the above question also.... situation is same. Only difference is that Payload is 'UDP Header + Data' for network layer.

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But ethernet will also add header right??? I know it will not affect answer. Still what is ethernet header size??

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It's upto 18 bytes.

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This is where my doubt is since we are given MTU therfore we can use only 494 B for payload. But only IP header is included in each frame and ethernet header is not.

And MTU= paylod + header.

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That's correct. But then the question will mention about Ethernet header size. We can solve accordingly after that. :)

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@Akash.

Payload for the last fragment =204 = 204

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rest of answer is correct.

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@sushant why did you add 20 B in last fragment???

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What's that extra 20B for?

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Yes, it should be 204. I consider last fragment as last frame. My mistake.