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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]
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