a)
Total available bandwidth = 1 Mbps = 1000 Kbps
Each user requires send data at the rate of = 500 kbps
As it is circuit switched network we have to dedicate the bandwidth
So the link may support = 1000/ 500 = 2 users

b)
Each user is active for 10% of time.
We assume total N number of users can be accommodated
So total data that may be produced by all users when using simultaneously = 500N kbps
But each user is active for 10% of time so at any given time the maximum demand for bandwidth
would be = 50N kbps
So we have
50N = 1000
Or N= 20
Thus the link will be able to accommodate 20 users.

c) As the link is now connected to 4 host and there will be no congestion in the link when only 1 or two user is sending data. So the congestion will be seen when number of simultaneous user (k) >2 And probability of 3 users coming at a time P3= p3 * (1-p) (4-3) Where p= probability of activity of individual station which is in this case 0.1

So P3 = 0.13 * (1-0.1) (4-3)
= 0.0009
Similarly P4 = 0.14 * (1-0.1) (4-4)
= 0.0001
Thus probability of link congestion = P3+P4 = 0.0009 + 0.0001 = 0.001

Posted Date: 3/29/2013 3:35:27 AM | Location : United States

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