Question

Suppose two hosts (Host-A and Host-B) are connected through a shared Ethernet bus of 1 Mbps. The distance between two nodes is 5 meters. The signal propagation speed over physical media is 2.5 x 10^5 meters/second. Compute the propagation delay in this case. What will be the propagation delay for a shared bus of 10 Mbps instead Suppose both hosts sense idle channel at the same time and starts transmission. Will it result in collision (yes or no) If yes: How much time is needed to detect this collision at Host-A? If yes: How much time is needed to detect this collision at Host-B?

Answer 1

Answer:

Following are the responses to this question:

Explanation:

Given:

distance from nodes= 5 metres 

Levels of transmission $= 1 \ Mbps = 106 \ bps$

Speed of propagation = $\ 2.5 \times 10^5 \ \frac{m}{s}$

In case 1:

Calculating the propagation delay:

$\to Propagation \ delay = \frac{distance}{speed}$

                                  $=\frac{5}{(2.5 \times 10^5)}\\\\=\frac{50}{(25 \times 10^5)}\\\\= 2 \times 10^{-5} \ sec$

In case 2:

Calculating the delay be for a 10 Mbps shared bus?

$\to 10 \ Mbps = 10 \times 10^6 \ bps =10^7 \ bps$

Propagation delay $= 10^7 \times \text{(propagation delay for 1\ mbps)}$

                               $= 10^7 \times (2 \times 10^{-5}) \\\\= 10^2 \times 2 \\\\= 200 \ bits.$

In case 3:

When the hosts feel the channel is idle in CSMA/CD, they also will transfer it. It classifying as a collision since both hosts consider the channel to be idle.

In case 4:

The time required for this is 2T. Here is T's time of Propagation. So, Calculating the transmission time:

$=2\times (2\times 10^{-5})\\\\= 4\times 10^{-5}\\\\ = 0.00004 \ sec$