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Consider three IP networks $A, B$ and $C$. Host $H_A$ in network $A$ sends messages each containing $180$ bytes of application data to a host $H_C$ in network $C$. The $\text{TCP}$ layer prefixes $20$ byte header to the message.

This passes through an intermediate network $B$.The maximum packet size, including $20$ byte IP header, in each network is:

• A: $\text{1000 bytes}$
• B: $\text{100 bytes}$
• C: $\text{1000 bytes}$

The network $A$ and $B$ are connected through a $1$ Mbps link, while $B$ and $C$
are connected by a $512$ Kbps link (bps = bits per second).

Assuming that the packets are correctly delivered, how many bytes, including headers, are delivered to the $IP$ layer at the destination for one application message, in the best case? Consider only data packets.

1. $200$
2. $220$
3. $240$
4. $260$

data is not being fragmented at network A as it’s mss is 1000
if we do defragmentation at every node, the message you are sending on whatsapp will get delivered tomorrow.
Right

Packet $A$ sends an $IP$ packet of $180$ bytes of data $+ 20$ bytes of TCP header $+ 20$ $bytes$ of IP header to $B$.

$IP$ layer of $B$ now removes $20$ $bytes$ of $IP$ header and has $200$ bytes of data. So, it makes $3$ IP packets - $[80 + 20, 80 + 20 , 40 + 20]$ and sends to $C$ as the Ip packet size of $B$ is $100$. So, $C$ receives $260$ bytes of data which includes $60$ bytes of $IP$ headers and $20$ bytes of TCP header.

For data rate, we need to consider only the slowest part of the network as data will be getting accumulated at that sender (data rate till that slowest part, we need to add time if a faster part follows a slower part).

So, here $180$ bytes of application data are transferred from $A$ to $C$ and this causes $260$ $bytes$ to be transferred from $B$ to $C$.

Correct Answer: $D$
by

"For data rate, we need to consider only the slowest part of the network as data will be getting accumulated at that sender (data rate till that slowest part, we need to add time if a faster part follows a slower part)."

hey someone tell me what is this??? What if 1mb follows 512kb

@RamaSivaSubrahmanyam It simply means that if you have many networks with different bandwidths, then take the slowest bandwidth as the bottle neck bandwidth

Hey! I am asking this" data rate till that slowest part, we need to add time if a faster part follows a slower part"  .

it's well known that if two buckets maintains two different rates slowest rate will be the output rate...

but i need above one...

Hence, total of 100 + 100 + 60 bytes = 260 bytes would be received by the destination.

Why IP network B removes the header but dont put it back
fragmentation and de-fragmentation both must be done by network B.

but here why is it done only at the destination?
Why the defragmentation is not done when packet is leaving the network B. Why the defragmentation is done only at sender.
nice explanation!!
180B application data + 20B TCP header = 200B
For network layer, 200B data is to be sent with 20B header in each packet.

No problem with A, C.
B has maximum packet size 100.

We will split  200 as  80+80+40
(80+20), (80+20), (40+20)= 260  (Ans)
by

@Ahwan

Can you please let me know that why defragmentation is not done at b itself. Why we are doing the degragmentation at the destination.
Defragmentation can't be done at middle of the path because multiple packets may take different paths but finally reaches to the destination, that's the reason reassembly algorithm will be applied at destination only.
Why are we removing the IP header at B?

Here actually network A sends packet with one ip header to B but cause fragmentation at network B total 3 headers (in which 2 headers are overhead) added ( because each fragment going to have ip header cause of datagram service) that is an overhead. that’s why even 180 bytes of application data are transferred from A to C but network C receives ( 80+20+80+20+40+20 =260 bytes ) because of overhead that arises cause of fragmentation