Problem 3 Suppose users share a 9 Mbps link. Each user consumes 1.5 Mbps when tr

Question

Problem 3 Suppose users share a 9 Mbps link. Each user consumes 1.5 Mbps when transmitting, and only transmits with probability 0.2 How many users can be supported using circuit switching? Assume packet switching for the rest of the problem. What is the probability that 7 out of 10 users are transmitting? What is the maximum number of users, such that the probability of exceeding the link capacity is no more than 0.01? Hint: Gradually increase the number of users and evaluate the probability of exceeding the link capacity for each value of this number. a) b) c)

Explanation / Answer

section (a) 6 users can be supported because each user requires 1.5 of the link bandwidth

so, 1.5*6=9

section (b) Probability that 7 out of 10 user uses the link is

10C7 P7 (1-P)3 = 120 * (0.2)7 * (0.8)3

=0.000786

section (c) we have values of P and (1-P) and n, we have to find value of r, means no. of users for which probability less then 0.1

10Cr Pr (1-P)(10-r) <= 0.01

now increase the no. of users (i.e. r) from 1 to 10

10C1 (0.2) (0.8)9 = 0.2

10C2 (0.2)2 (0.8)8 = 0.3

10C3 (0.2)3 (0.8)7 = 0.2

.......

10C5 (0.2)5 (0.8)5 = 0.02

10C6 (0.2)6 (0.8)4 = 0.003

which is less then 0.01

so, maximum number of users can be 6

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