Question

Suppose users share a 3 Mbps link. Also suppose each user requires 150 kbps when transmitting, but each user transmits only 10 percent of the time. (See the discussion of packet switching versus circuit switching 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. Find the probability that a given user is transmitting.
c. Suppose there are 120 users. Find the probability that at any given time, exactly n users are transmitting simultaneously. (Hint: Use the binomial distribution.)
d. Find the probability that there are 21 or more users transmitting simultaneously.

Answer 1

a) 20 users

b) 0.1 or 1/10

c) P ( X = n ) =
`^(120)C_(n) ( 0.10 )^n (0.9)^(120-n)`

d) 0.008

Step-by-step explanation:

Given data:

Link shared by users = 3 Mbps

each user requires ; 150 kbps

percentage transmitted by each user = 10%

a) Determine the number of users can be supported

number of users = 3 Mbps / 150 kbps = 3000 / 150 = 20 users

b) P( user is transmitting )

given that percentage transmitted by each user = 10%

since packet switching is used ; P ( user is transmitting ) = 10% = 0.1

c) P ( n users are transmitting simultaneously )

let the number of users ( i.e. N ) = 120

number transmitting = n

P ( user is transmitting ) = 0.10

using Binomial distribution

P ( X = n ) =
`^(N)`Cₙ ( P )ⁿ ( 1 - P )^
`^(N-n)`

=
`^(120)C_(n) ( 0.10 )^n (0.9)^(120-n)`

d) P( X ≥ 21 transmitting simultaneously )

P ( X = n ) =
`^(120)C_(n) ( 0.10 )^n (0.9)^(120-n)`

Therefore ; P( X ≥ 21 ) = 1 - P( X ≤ 20 )

= 1 - BINOMDIST(20,120,0.1,1) ( excel function )

∴ P( X ≥ 21 ) = 1 - 0.992059 ≈ 0.008