Respuesta :
Complete Question:
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?
Answer:
a) 51.2 msec. b) 5.12 msec
Explanation:
If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:
- Tw = K*512* bit time
The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:
a) BW = 1 Mbps = 10⁶ bps
⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec
b) BW = 10 Mbps = 10⁷ bps
⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec
Answer:
a) 51.2 msec. b) 5.12 msec
Explanation:
Let [tex]K=100,[/tex]
- The time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression
- The time that the adapter should delay until detecting a channel subsequent to distinguishing a first impact, is given by the following expression:
[tex]Tw = K\times 512\times bit time[/tex]
- The bit time, is just the inverse of the channel bandwidth,
- which is expressed in bits per second, so for the two instances posed by the question, we have:
a) [tex]BW=1 Mbps = 10 ^ 6 bps[/tex]
[tex]Tw = 100\times 512\times(1/10^{6} ) bps[/tex]
[tex]= 51.2\times10^{-3} sec[/tex].
[tex]= 51.2 msec[/tex]
b) [tex]BW = 10 Mbps = 10^{7} bps[/tex]
[tex]Tw = 100\times512\times(1/10^{7} ) bps[/tex]
[tex]= 5.12\times10^{3} sec.[/tex]
[tex]= 5.12 msec[/tex]
To learn more, refer:
brainly.com/question/14292191
