networking

Question

Suppose a TCP message that contains 2048 bytes of data and 20 bytes of TCP
header is passed to IP for delivery across two networks of the Internet (i.e., from
the source host to a router to the destination host). The first network uses 14-byte
headers and has an MTU of 1024 bytes; the second uses 8-byte headers with an
MTU of 512 bytes. Each network’s MTU gives the size of the largest IP datagram
that can be carried in a link-layer frame. Give the sizes and offsets of the sequence
of fragments delivered to the network layer at the destination host. Assume all IP
headers are 20 bytes.

Comments

check is this correct 

Answer 1

Data size is 2048B and header size is 20B.

so total size of datagram is 2068B.

MTU size is 1024B it includes 14B header and 20B ip header .so allowable data size is 1024-14-20=990B.

2048B data is passed through 990B MTU data must be fragmented.and fragmented chunks are 984,984,80B.

and offset values are (0-122) (123-245) (246-255).

2nd MTU size is 512B it includes 8Bheader and 20B ip header.data size is 512-8-20=484B.

if 984B is passed through MTU data must be fragmented.chunks are 480,480,24B

offset values are (256-315) (316-375) (376-378).

2nd 984B also fragmented 480,480,24B.

offset values are (379-438) (439-498) (499-501).

last 80B if passed no fragmentation .offset is (502-509).

so total fragments are 9

Comments

Why we are considering TCP header in fragments ?  It is similar to Gate 2015 question with UDP header. There we were not considering UDP header in each fragments.

https://gateoverflow.in/8255/gate2015-2-52

Answer 2

Total data to fragment is 2048+20(TL-Header)= 2068 B.

We have to remember that Transport Layer Header will now also be considered as part of main data. Please see the diagram, you'll understand.

Answer 3

 

So No of Fragments received by destination = 7 

Offset value of 6th fragment =247

Answer (b)