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Consider a network having a host P connected to another host Q via two routers RI and R2. P is connected to RI through a link that can support a maximum frame length of $1024 bytes$ including header of size $12 bytes$. RI is connected to R2 through a link that supports a maximum frame length of $256 bytes$ including an $8 bytes$ header. R2 is connected to the host Q through the link that can support a maximum frame length of $512 bytes$ including the frame header Of $12 bytes$. Suppose a TCP message having $800 bytes$ Of data and $20 bytes$ of TCP header is passed to the IP at host P for delivery to Q. Find the total number of bytes that are sent from host P to Q (including all the headers)


Shouldn’t the ans be $948B?$

Here is the link to the question

https://gateoverflow.in/101500/ip-fragmentation

ago in Computer Networks by (613 points) | 13 views
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I feel the question is ambiguous. 

total number of bytes that are sent from host P to Q (including all the headers)

Does this mean:

  1. How many bytes P sends, period ? In this case it is 820(data)+20(IP-h)+12(DLL-h)
  2. How many bytes Q receives, period ? In this case it is 820(data) + 20*4(IP-h due to fragmentation at R1) + 12 * 4 (DLL-h since R2 has a different frame header) = 948B
  3. How many bytes are sent along the link totally, to get the data from P-Q. In this case it is (820 + 20 + 12)(at P) + (20*4 + 8 *4)(IP-h and DLL-h at R1) + (12*4)(DLL-h at R2)

I suppose you interepreted it as item(2) ? I felt item(3) also makes sense since the question makes a point to specify "including all the headers".

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It means how many Bytes of data is received by the receiver including all the headers

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