Testbook test series - Computer Networks TCP

Question

Question: TCP opens a connection using an initial sequence number of 3500 and sends data at a rate of 5MBps. The other party opens a connection with a sequence number of 1200. Wrap around time for both sequence number differs by 12562.77 sec. Calculate the data rate (in KBps) for the second party.

The answer given is 320-321 but I calculated the answer to be 366.9.

In the solution, they calculated WAT as $\frac{2^{_{32}}-Initial Sequence}{Bandwidth}$

Whereas I calculated WAT as $\frac{2^{_{32}}}{Bandwidth}$

Which one is correct?

Answer 1

Let bandwidth we have to find be y
According to question
$\frac{2^{32}}{y} -\frac{2^{32}}{5*10^{6}}=12562.77$
2^32/5*(10^6) is 858.9 (calculate using calculator)
now
$\frac{2^{32}}{y} =12562.77 + 858.9$

$\frac{2^{32}}{12562.77 + 858.9 } = y$
$y=\frac{2^{32}}{12562.77 + 858.9 }$
$y=320018.426$
$y=320.018426 Kbps$

Comments

Thanks! I realized my mistake. I was subtracting 858.9 from 12562.77 (instead of adding it). I took 12562.77 as the sum of both WATs and not the difference. I guess you shouldn't rush while solving numerical questions.