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

Following are the responses to this question:

Step-by-step explanation:

Given:

distance from nodes= 5 metres

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

Speed of propagation =
`\ 2.5 * 10^5 \ (m)/(s)`

In case 1:

Calculating the propagation delay:

`\to Propagation \ delay = (distance)/(speed)`

`=(5)/((2.5 * 10^5))\\\\=(50)/((25 * 10^5))\\\\= 2 * 10^(-5) \ sec`

In case 2:

Calculating the delay be for a 10 Mbps shared bus?

`\to 10 \ Mbps = 10 * 10^6 \ bps =10^7 \ bps`

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

`= 10^7 * (2 * 10^(-5)) \\\\= 10^2 * 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* (2* 10^(-5))\\\\= 4* 10^(-5)\\\\ = 0.00004 \ sec`