C is correct..
Please vote my answer branliest! Thanks.
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:
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: in solution.
Explanation:
It is basically 194 divided by 11 since we are evenly grouping 194 seeds into 11 pots. This gives 17.636363…
This means that the best estimate is around that number.
Answer:
Two computers can safely have the same IP address in certain cases. In most cases, if those two computers are on the same local network, it breaks connectivity for one or both of them. Internet protocols work by sending small, individually addressed messages. Each message can be routed differently.
Explanation:
