TCP Data Transfer Rate [modified Gate Question]

Question

Consider three IP networks A,B and C. Host $H_A$ in network A sends messages each containing 200 bytes of application data to a host $H_C$ in network C. This passes through an intermediate network B . The maximum packet size, including 20 byte IP header, in each network is:

• A: 1000 bytes
• B: 100 bytes
• C: 200 bytes

The network A and B are connected through a 512Kps link, while B and C are connected by a 1 Mbps link (bps = bits per second). What is the rate at which application data is transferred to host $H_C$ ? Ignore errors, acknowledgements, and other overheads.

Comments

message = 200

TCP datagram = 200 + 20

IP datagram = 220 + 20

fragmentation will take place at network B and 220 ip packet will be fragmented in 3 ip datagram. 80+20 , 80+20 , 60 + 4[ to make it  multiple of 8] + 20.

So, total data send = 280 insted of 200.

Hence, rate at which application data will be send = (200/280)*512 Kbps.

Answer 1

Here MTU is 100 which is required to be consider.

Which means 20+80.

Now, 200B data from application layer is coming, at TCP 20B of header is Added.

At Network Layer Also ip header is added, so, it becomes [20+220]B.

Fragmentation has to be done in Network B, three fragments are

[20+80]   [20+80]  [20+60]

efficiency = 200B/220+20*3 =0.76

Data Rate = 0.76* 512 Kbps (consider bottleneck BW)

=393.84Kbps