Question

Consider a circuit-switching scenario in which Ncs users, must share a link of capacity 150 Mbps. Each user alternates between periods of activity when the user generates data at a constant rate of 20 Mbps and periods of inactivity when he generates no data. Suppose further that he is active only 20% of the time. Assume a circuit switched TDM. What is the maximum number of circuit-switched users that can be supported? Assume packet switching is used for the remaining parts of this question. Suppose there are 13 packet- switching users (i.e., Nps = 13). Can this many users be supported under circuit-switching? Explain. What is the probability that a given (specific) user is transmitting, and the remaining users are not transmitting? What is the probability that one user (any one among the 13 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? What is the probability that any 7 users (of the total 13 users) are transmitting and the remaining users are not transmitting? (Hint: you will need to use the binomial distribution.) What is the probability that more than 7 users are transmitting? Comment on what this implies about the number of users supportable under circuit switching and packet switching. What is the probability that more than 7 users are transmitting? Comment on what this implies about the number of users supportable under circuit switching and packet switching.

Answer 1

Lets, answer each question one-by-one.

1. What is the maximum number of circuit-switched users that can be supported?

Ans: Given that, each user alternates between periods of activity when the user generates data at a constant rate of 20 Mbps and periods of inactivity when he generates no data. Because of this each user should be allocated 20 Mbps bandwidth from shared link capacity of 150 Mbps.

Hence,maximum number of circuit-switched users that can be supported = 150/20 = 7.5

2. Assume packet switching is used for the remaining parts of this question. Suppose there are 13 packet- switching users (i.e., Nps = 13). Can this many users be supported under circuit-switching? Explain.

Ans: No, these many users can not be supported under circuit-switching because, among 13 packet-switching users, each one requires bandwidth of 20 Mbps. This results in aggregate bandwidth of 20*13 = 260 Mbps which way more than given shared link capacity of 150 Mbps.

3.What is the probability that a given (specific) user is transmitting, and the remaining users are not transmitting?

Ans: Lets say p is the probability that a given (specific) user is transmitting and he is active for 20% of the time. Hence, p=0.20 So, probability of one specific user not transmitting is (1-p). Hence, probability of all of the other (N_ps - 1) users are not transmitting is (1-p)^{N_ps - 1}Therefore the probability of that a given (specific) user is transmitting, and the remaining users are not transmitting is p¹ * (1-p)^{N_ps - 1} = (0.20)¹ * (1-0.20)^13-1 = (0.20)¹ * (0.8)¹² = 0.01374

4. What is the probability that one user (any one among the 13 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?

Ans: the probability that one user (any one among the 13 users) is transmitting, and the remaining users are not transmitting is N_ps * p¹ * (1-p)^{N_ps - 1} = 13 * (0.20)¹ * (1-0.20)^13-1 = 13 * (0.20)¹ * (0.8)¹² = 13 * 0.01374 = 0.17862

fraction of the link capacity will be used by this use is = 20/150 = 0.133

5. What is the probability that any 7 users (of the total 13 users) are transmitting and the remaining users are not transmitting?

Ans: The probability that any 7 users (of the total 13 users) are transmitting and the remaining users are not transmitting is p⁷* (1-p)⁶. Hence, probability of 7 users out of 13 users transmitting (using Binomial Theorem) is : (13 choose 7) * p⁷ * (1-p)⁶ = [13! / ((13-7)! *(7!))] * (0.20)⁷ * (0.8)⁶ = 1716 * 0.0000128 * 0.262144 = 0.0057579

6. What is the probability that more than 7 users are transmitting? Comment on what this implies about the number of users supportable under circuit switching and packet switching.

Ans: The probability that more than 7 users are transmitting is = (13 choose i) * pⁱ * (1-p)^13-i

= [(13 choose 8) * p⁸ * (1-p)^13-8 ] + [(13 choose 9) * p⁹ * (1-p)^13-9 ] + [(13 choose 10) * p¹⁰ * (1-p)^13-10 ] + [(13 choose 11) * p¹¹ * (1-p)^13-11 ] + [(13 choose 12) * p¹² * (1-p)^13-12 ] + [(13 choose 13) * p¹³ * (1-p)^13-13 ]

= [1287 * 0.2⁸ * (0.8)⁵ ] + [715 * 0.2⁹ * (0.8)⁴ ] + [286 * 0.2¹⁰ * (0.8)³ ] + [78 * 0.2¹¹ * (0.8)² ] + [13 * 0.2¹² * (0.8)¹ ] + [1 * 0.2¹³ * (0.8)⁰ ]

= [1287 * 0.00000256 * 0.32768] + [715 * 0.000000512 * 0.4096] + [286 * 0.0000001024 * 0.512] + [78 * 0.00000002048 * 0.64] + [13 * 0.000000004096 * 0.8] + [1 * 0.0000000008192 * 1]

= 0.00179613 + 0.0001499 + 0.0000149 + 0.0000010 + 0.0000000425 + 0.0000000008192

=0.0019619733

Maximum number of users supported using circuit switching are 7. With packet switching, almost twice the users (13) are supported with a probability that more than 7 packet switching users are transmitting at the same time.