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I am facing serious troubles while calculating number of fragments. I read a formula #fragments = ceil (payload/ mtu-header)  but this isn’t giving correct answer in all the cases. 

Before I show my approach, I would like to ask what does PDU imply? Does it mean header is already added? Because in questions like “TL PDU is X bytes”, I assume that according to NL, X is payload, to which it adds IP header. So NL PDU is X bytes or (X+ ip header) bytes? 

For question 26:

Assuming NL PDU means it includes header(20B) , then useful data (payload) = 17056 bytes which is to be fragmented. MTU is 200B out of which 20B is for header. so it can carry 180 of useful data. So for the formula I read till now(calling it formula 1), #frags= ceil (17056/180)=95

But the fragments can’t carry 180B of data, it has to be multiple of 8. So it carries 176B. (calling this formula 2). So #frags=ceil(17056/176)=97 

No answers matched. 

Now assuming NL PDU means only useful data, ie. payload is 17076B which is to be fragmented. Using the formula 1 #frags= ceil(17076/180)=95

Using formula 2, #frags= ceil(17076/176)=98

No match either. Where am I going wrong?

in Computer Networks by (288 points) | 17 views

1 Answer

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You are wrong to think that these test series give right solutions.

PDU is a generic name for what we call packets at different Layer. For example PDU for Network Layer is a "Datagram".

So here PDU for NL is given as = 17076 (TL segment + NL header). Therefore payload here is simply 17076-20 = 17056.

Now we divide 17056 B into chunks of 176 B. And we simply need 97 of these chunks (Fragmented datagrams) of which last one may not be full.

So option C is right answer. 

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