in Computer Networks
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2 votes
2 votes

in Computer Networks
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2 Comments

932?
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I was doing this way,820 bytes of payload  got fragmented into 4parts of payload 248bytes so extra header would be 8*4bytes(So 852) and then we wont have problem as MTU is bigger,How u did it?Are those 8bytes frame headers?So payload should be 224 in each having 28bytes of [email protected]
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1 Answer

4 votes
4 votes
Best answer
820 bytes is the data which comes to network layer of P. at network layer 20 more bytes of IP header is added.
at P-R1 max MTU is 1024-DLL header = 1012 bytes. this includes IP header also. we need to fit 820 bytes in 1012-20 bytes
but, as 992>820 no need to do fragmetation
 R1-R2 contains 256-8 is max MTU= 248 B. this includes IP header also. max data it can support is 228 bytes
but 228 is not divisiible by 8
so divide 820 4 parts.. 224,224,224,148.. each of these packet gets 20 bytes IP header+ 8 bytes DLL header.
R2-Q max MTU is 500 bytes. no need to do fragmentation
1)224+20+8
2)224+20+8
3)224+20+8
4)148+20+8
total 932 bytes
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4 Comments

Answer should be 948,because when frames will travel through link R2-Q, DLL will remove header of last link and will add new header of 12 Byte.
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I think the last fragment should be 152+20+8

because in solution it is given that 148+20+8

and 148 is not divisible by 8 so we need to pad 4 bits then it will be 148 + 4 = 152.
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@Shubhanshu the Last fragment need not be divisible by 8 as we are not going to store this value as fragment offset anywhere. moreover, how can we pad more data bits or decrease the data bits, This is isn't possible. Correct me if I'm wrong
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