a)Total available bandwidth = 1 Mbps = 1000 KbpsEach user requires send data at the rate of = 500 kbpsAs it is circuit switched network we have to dedicate the bandwidthSo 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 accommodatedSo total data that may be produced by all users when using simultaneously = 500N kbpsBut each user is active for 10% of time so at any given time the maximum demand for bandwidthwould be = 50N kbpsSo we have50N = 1000Or N= 20Thus 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)= 0.0009Similarly P4 = 0.14 * (1-0.1) (4-4)= 0.0001Thus probability of link congestion = P3+P4 = 0.0009 + 0.0001 = 0.001