Question

Suppose users share a 10 Mbps link. Also suppose each user transmits continuously at 2 Mbps when transmitting, but each user transmits only 20 percent of the time. (See the discussion of statistical multiplexing in Section 1.3.) a. When circuit switching is used, how many users can be supported? b. For the remainder of this problem, suppose packet switching is used. Why will there be essentially no queuing delay before the link if two or fewer users transmit at the same time? Why will there be a queuing delay if three users transmit at the same time? c. Find the probability that a given user is transmitting. d. Suppose now there are three users. Find the probability that at any given time, all three users are transmitting simultaneously. Find the fraction of time during which the queue grows.Previous question

Answer 1

a. When circuit switching is used, only 1 user can be supported because circuit switching dedicates the entire bandwidth to a single user's connection, so the bandwidth cannot be shared among multiple users.

b. If two or fewer users transmit at the same time, the link capacity is sufficient to handle the transmission rates, and there will be no queuing delay before the link. If three users transmit at the same time, the link capacity of 10 Mbps is exceeded, and a queuing delay will occur because packets will have to wait in a queue before being transmitted.

c. The probability that a given user is transmitting is 0.2 because each user transmits only 20 percent of the time.

d. If there are three users, the probability that all three users are transmitting simultaneously is 0.2 * 0.2 * 0.2 = 0.008. The fraction of time during which the queue grows is equal to the probability that all three users are transmitting simultaneously because that is when the link capacity is exceeded and packets must wait in a queue.