Question

Question requires a little bit of background in probability:

Consider the two scenarios below:
▪ A circuit-switching scenario in which Ncs users, each requiring a bandwidth of 25 Mbps, must share a link of capacity 200 Mbps.
▪ A packet-switching scenario with Nps users sharing a 200 Mbps link, where each user again requires 25 Mbps when transmitting, but only needs to transmit 20 percent of the time.

Answer the following:
a. When circuit switching is used, what is the maximum number of circuit-switched users that can be supported? Explain your answer.
b. For the remainder of this problem, suppose packet switching is used. Suppose there are 35 packet-switching users (i.e., Nps = 35). Can this many users be supported under circuit-switching? Explain.
c. What is the probability that a given (specific) user is transmitting, and the remaining users are not transmitting?
d. What is the probability that one user (any one among the 35 users) is transmitting, and the remaining users are not transmitting? When one user is transmitting, what fraction of the link capacity will be used by this user?
e. What is the probability that any 15 users (of the total 35 users) are transmitting and the remaining users are not transmitting?

Answer 1

PLEASE RATE AND LET ME KNOW IF YOU HAVE DOUBT AND NEED EXPLANATIONS.

SOLUTION:

Solution! Tatal Bandwidth avallable capacity) = 200 Mbps Bandwidth requirement of each user = 25 Mbps (a) When Circuit Switching is used, a part of bandwidth is restul reserved for a particullar user and that part of bandwidth can't be used by other data packets (foo from other users) even though of the user is not using the link. That part of bandwidth is available for other users when the requirement of first user is fulfilled, it, there is no contagious contrat un circuit switching. No. of users that can be supported = Total Bandwidth available - Bandwidth requirement of I user = 200 Mbps 25 Mbps 28 users. (Aus) Thus, Only 8 users can be supported for Circuit switching over the link at a time. (b) Suppose there are 35 packet switching users. # The link will not support 35 users at a time for circuit switching because the guien link can support mocinum of 8 users at a time. [solved above] © for Packet Switching, Total no.ey users 35 Probability that user is transmitting = 20% of time = 20 / 0 = 012 . .. Probability that user is not transmitting = 1-0.2=0.8 Now, the probability that 'n' users out of 35 are active = P(n) : : Pln] = 35cm (0.2)^(0.8735-n (sms) . Ates Also, the probability that no users is active die,n=o. Plo) = 356 (0-2)9 (0.8)35-0 = 1 x 1 x 0.00040567 = 0.0004 (Aus) no

(a) Probability that one out of 35 user is active, lie, nat :: P(1) = 350,.(0:2)*(0.9 = 35 x (0-2) x 5.0706 x 1024 = 3.5494 X10-3 = 0.0035 (Aus) Since, each uses requires 25 Mbps. fruction of Bandwith used by one user = 20 = 5 = 0.125 (Aus) (l) Probability that is users out of 35 are active, ie n=15. .: P(15) = 3566.CO.2365.co.2325-15 = 1.2270 X10-3 = 0.0012. (Ames) Remaining in-active users =35-15 = 20 non-active usou. Probability that 20 unts are not transmitting = 1-P(15) =) - 0.0012 20.9988 (Aus)