P9. Consider the discussion in Section 1.3 of packet switching versus circuit switching in which an example is provided with a 1 Mbps link. Users are generating data at a rate of 100 kbps when busy, but are busy generating data only with probability . Suppose that the 1 Mbps link is prop trans t=0 t= trans prop trans t=dtrans prop trans to trans s=2.5⋅108 L=120 bits R=56 kbps. prop trans p=0.1 replaced by a 1 Gbps link. a. What is N, the maximum number of users that can be supported simultaneously under circuit switching?

hello your question is not properly structured  below is the complete question

. Consider the discussion in Section 1.3 of packet switching versus circuit switching in which an example is provided with a 1 Mbps link. Users are generating data at a rate of 100 kbps when busy, but are busy generating data only with probability . Suppose that the 1 Mbps link is  replaced by a 1 Gbps link. a. What is N, the maximum number of users that can be supported simultaneously under circuit switching?

answer : N = 1000 users

Explanation:

The maximum number of users (n) that can be supported simultaneously under circuit switching

N =    Total transmission  / rate of data generation by users

Total transmission  = 1 Gbps = 1 * 10^9 bps

rate of data generation by user = 100 kbps = 10 * 10^4 bps

hence the maximum number of users that can be supported simultaneously

= ( (1* 10^9) / (10 * 10^4) )

= 10000 users

what is the question because that is not a question.

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered