MadeEasy Test Series: Computer Networks - Csma Cd

Question

A)340 bits B) 240bits

Answer 1

The maximum distance between any two nodes is between $D_{10}$ and $D_{9}$ i.e (1000+900)

So delay to cover this distance = $\frac{1000+900}{2\times 10^8}=9.5\,microsec$

Since the hub produces delay of $2.5\,microsec$, so delay due to this $=2\times2.5=5$

Total delay $=9.5\times 2+5=24\,microsec$

Let minimum size $=x$

$\frac{x}{10\times10^6}=24\times10^{-6}$

$x=240\,bits$

Correct me if I'm wrong

Comments

What is the reason of taking maximum distance?