Interactive end-of-chapter exercises


Quantitative Comparison of Packet Switching and Circuit Switching

This question requires a little bit of background in probability (but we'll try to help you though it in the solutions). Consider the two scenarios below:




Round your answer to two decimals after leading zeros



Question List


1. When circuit switching is used, what is the maximum number of users that can be supported?

2. Suppose packet switching is used. If there are 29 packet-switching users, can this many users be supported under circuit-switching? Yes or No.

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

4. Suppose packet switching is used. What is the probability that one user (any one among the 29 users) is transmitting, and the remaining users are not transmitting?

5. When one user is transmitting, what fraction of the link capacity will be used by this user? Write your answer as a decimal.

6. What is the probability that any 11 users (of the total 29 users) are transmitting and the remaining users are not transmitting?

7. What is the probability that more than 15 users are transmitting?




Solution


1. When circuit switching is used, at most 15 users can be supported. This is because each circuit-switched user must be allocated its 10 Mbps bandwidth, and there is 150 Mbps of link capacity that can be allocated.

2. No. Under circuit switching, the 29 users would each need to be allocated 10 Mbps, for an aggregate of 290 Mbps - more than the 150 Mbps of link capacity available.

3. The probability that a given (specific) user is busy transmitting, which we'll denote p, is just the fraction of time it is transmitting, i.e. 0.2. The probability that one specific other user is not busy is (1-p), and so the probability that all of the other Nps-1 users are not transmitting is (1-p)Nps-1. Thus the probability that one specific user is transmitting and the remaining users are not transmitting is p*(1-p)Nps-1, which has the numerical value of 0.00039.

4. The probability that exactly one (any one) of the Nps users is transmitting is Nps times the probability that a given specific user is transmitting and the remaining users are not transmitting. The answer is thus Nps * p * (1-p)Nps-1, which has the numerical value of 0.011.

5. This user will be transmitting at a rate of 10 Mbps over the 150 Mbps link, using a fraction 0.067 of the link's capacity when busy.

6. The probability that 11 specific users of the total 29 users are transmitting and the other 18 users are idle is p11(1-p)18. Thus the probability that any 4 of the 7 users are busy is choose(29, 11) * p11(1-p)18, where choose(29, 11) is the (29, 11) coefficient of the binomial distribution). The numerical value of this probability is 0.013.

7. The probability that more than 15 users of the total 29 users are transmitting is Σ i=16,29 choose(29, i) * pi(1-p)29 - i. The numerical value of this probability is 3.00E-5. Note that 15 is the maximum number of users that can be supported using circuit switching. With packet switching, nearly twice as many users (29) are supported with a small probability that more than 15 of these packet-switching users are busy at the same time.



That's incorrect

That's correct

The answer was: 15

Question 1 of 7

The answer was: No

Question 2 of 7

The answer was: 0.00039

Question 3 of 7

The answer was: 0.011

Question 4 of 7

The answer was: 0.067

Question 5 of 7

The answer was: 0.013

Question 6 of 7

The answer was: 3.00E-5

Question 7 of 7

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