Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
Answer:
Following are the responses to this question:
Explanation:
Given:
distance from nodes= 5 metres
Levels of transmission [tex]= 1 \ Mbps = 106 \ bps[/tex]
Speed of propagation = [tex]\ 2.5 \times 10^5 \ \frac{m}{s}[/tex]
In case 1:
Calculating the propagation delay:
[tex]\to Propagation \ delay = \frac{distance}{speed}[/tex]
[tex]=\frac{5}{(2.5 \times 10^5)}\\\\=\frac{50}{(25 \times 10^5)}\\\\= 2 \times 10^{-5} \ sec[/tex]
In case 2:
Calculating the delay be for a 10 Mbps shared bus?
[tex]\to 10 \ Mbps = 10 \times 10^6 \ bps =10^7 \ bps[/tex]
Propagation delay [tex]= 10^7 \times \text{(propagation delay for 1\ mbps)}[/tex]
[tex]= 10^7 \times (2 \times 10^{-5}) \\\\= 10^2 \times 2 \\\\= 200 \ bits.[/tex]
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:
[tex]=2\times (2\times 10^{-5})\\\\= 4\times 10^{-5}\\\\ = 0.00004 \ sec[/tex]
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
