EE132B-HW Set #2 - Sol.
Prof. Izhak Rubin
Problem 1 Suppose users share a 1 Mbps link. Also suppose that the traffic generation process of each user alternates (independently of the other users) between periods of activity (active modes), when the user generates data at a constant rate of 100 Kbps and periods of inactivity (inactive mode) when the user generates no data. Suppose further that the user is active (independent of other users) only 10 percent of the time. (1) When circuit switching is used to allocate resources on the shared link, how many users can be supported? (2) Suppose now that users are supported by using accordance with a packet switching method, so that the communications channel is shared on a statistical multiplexing manner. Assume now that the communications channel is used to support a number of users that is 4 times larger than those supported under the circuit switching operation. Use normal distribution to approximate the probability that more than 10 users are simultaneously active. (Hint: Central Limit Theorem. You may find Φ(3.16) = 0.99921 useful, where Φ(x) is the CDF of a normal random variable.) Ans: 1.
1 Mbps 100 Kbps
= 10 users.
2. Number of user sources is 40. Let {X1 , ..., X40 } be i.i.d Bernoulli random variables with P r{Xi = 1} = 0.1 = 1 − P r{Xi = 0} ∀ i = 1...40.
By Central Limit Theorem, Z = X1 + ... + X40 can be approximated by a Gaussian random variable with mean µ = 0.1 × 40 = 4 and variance σ 2 = 0.1 × 0.9 × 40 = 3.6. Z−4 Hence, P r{Z ≤ 10} = P r{ √ ≤ 3.6
10−4 √ } 3.6
≈ 0.99921.
P r{Z > 10} ≈ 1 − 0.99921 = 0.00079.
Problem 2 For the Gaussian distribution with mean µ and variance σ 2 , find the moment generating function. Using the moment generating function, calculate the mean and the variance. (Hint: The probability density function for a Gaussian random variable X with mean µ and variance σ 2 is given by 1 (x − µ)2 fX (x) = √ . exp − 2σ 2 2πσ 2 !
1
EE132B-HW Set #2 - Sol.
Prof. Izhak Rubin
Ans: By using the laplace transform, we obtain h
ΦX (s) = E e−sX
i
2 1 1 e− 2σ2 (x−µ) dx −∞ 2πσ 2 Z ∞ 1 2 2 2 1 √ e− 2σ2 [x −2(µ−σ s)x+µ ] dx = −∞ 2πσ 2 i h Z ∞ 2 − 12 (x−(µ−σ2 s)) +2(µσ2 s)−σ4 s2 1 √ dx = e 2σ −∞ 2πσ 2 Z ∞ 2 1 1 2 1 2 2 √ = e− 2 (2µs−σ s ) e− 2σ2 (x−(µ−σ s)) dx −∞ 2πσ 2 1 2 2 = e− 2 (2µs−σ s ) .
=
Z
∞
e−sx √
We can obtain the first and second moment of X, which is given by 1 d ΦX (s) = − lim − 2µ − 2σ 2 s ΦX (s) = µ. s→0 s→0 ds 2 2 d E[X 2 ] = lim 2 ΦX (s) s→0 d s " # 2 1 2 2 = lim σ ΦX (s) + 2µ − 2σ s ΦX (s) s→0 2 = µ2 + σ 2 .
E[X] = − lim
Thus, the variance is given as V ar[X] = E[X 2 ] − E[X]2 = σ 2 .
Problem 3 Consider the following probability density function: fX (x) =
λe−λ|x| , ∀x ∈ (−∞, ∞), 2
where λ > 0. (a) Calculate the mean directly. (b) Calculate the variance directly. (c) Find the moment generating function. (d) Find the mean and the variance from the moment generating function. Ans: 2
(1)
EE132B-HW Set #2 - Sol.
Prof. Izhak Rubin
∞ 0 (a) E [X] = −∞ x f (x) dx = −∞ x λ eλx dx + R ∞ λ −λx R ∞ λ −λx2 = − 0 x 2 e dx + 0 x 2 e dx = 0
R
R
(b) V ar [X] = E [X 2 ] R ∞ 2 λ −λ|x| dx = −∞ x e R ∞ 22λ −λx = 2 0 x 2 e dx = h
R∞ 0
x λ2 e−λx dx
2 λ2
i
(c) ΦX (s) = E e−sX R ∞ −sx λ −λ|x| = −∞ e 2e dx R 0 λ (λ−s)x R ∞ λ −(λ+s)x = −∞ e dx + dx 0 2e 2 λ 1 1 = 2 λ−s + λ+s h
i
1 1 + λ+s λ−s s→0 h i 1 1 d2 λ + E [X 2 ] = lim ds 2 2 λ−s λ+s s→0 2 V ar [X] = λ2
d λ (4) E [X] = − lim ds 2
=0 =
2 λ2
Problem 4 A coin is flipped until at least a head occur and at least two tails occur. Define three random variables X, Y , Z to be the trial numbers at which the first head, the first tail, and the second tail are observed, respectively. Assume that at any trial, the probability that a head occurs is p ∈ (0, 1). Let q = 1 − p. (a) Find the joint probability P (X = m, Y = n, Z = r). (b) Verify that your answer in (a) is valid by checking r) = 1.
P
m,n,r
P (X = m, Y = n, Z =
(c) Calculate P (Z < X) using the joint probability derived in (a). Ans: (a) P (X = m, Y = n, Z = r) = P (Y = n, Z = r | X = m)P (X = m) = P (Z = r | X = m, Y = n)P (Y = n | X = m)P (X = m). Note that P (X = m) = pq m−1 ∀m ≥ 1. Next, we evaluate P (Y = n | X = m):
1 , m ≥ 2, n = 1 n−2 p q , m = 1, n ≥ 2 P (Y = n | X = m) = 0 , otherwise.
3
EE132B-HW Set #2 - Sol.
Prof. Izhak Rubin
Finally we evaluate P (Z = r | X = m, Y = n). We can only evaluate m, n values where P (Y = n | X = m) 6= 0:
1 , r = 2, m > 2, n = 1 r−3 , r > 2, m = 2, n = 1 P (Z = r | X = m, Y = n) = p q r−n−1 p q , r > n, m = 1, n ≥ 2
We can thus calculate P (X = m, Y = n, Z = r) as:
P (X = m, Y = n, Z = r) =
pq m−1 pr−3 q × pq m−1 pr−n−1 q × pn−2 q × pq m−1 0
=
(b) ∞ X
pq m−1 +
m=3
∞ X
pr−2 q 2 +
r=3
∞ X
∞ X
pq m−1 pr−2 q 2 pr−2 q 2 0
pr−2 q 2 =
n=2 r=n+1
, r = 2, m > 2, n = 1 , r > 2, m = 2, n = 1 , r > n, m = 1, n ≥ 2 , otherwise.
, r = 2, m > 2, n = 1 , r > 2, m = 2, n = 1 , r > n, m = 1, n ≥ 2 , otherwise.
∞ n−1 2 X pq 2 pq 2 p q + + 1 − q 1 − p n=2 1 − p
p(1 − p)2 p(1 − p)2 p(1 − p)2 + + p 1−p (1 − p)2 = (1 − p)2 + p(1 − p) + p =1 =
(c) P (Z < X) =
∞ X
pq m−1 =
m=3
pq 2 = q2 1−q
Problem 5 Two DTEs are communicating via modems and RS-232C interfaces over a half duplex link. Data transmission is one directional with each message of length M. After a message is transmitted, the sender waits for a CTS message of length na before sending another message. Assume error-free operation and the following parameter values: • R (line bit rate) = 9600 bps • M (message length) = 1000 bits (all data bits) 4
EE132B-HW Set #2 - Sol.
Prof. Izhak Rubin
• nh (overhead in each message packet) = 24 bits • na (acknowledgement length) = 24 bits • tp (propagation time in one direction) = 1 msec Minimum transition time between states (1,3) and (1,1) (CTS delay) is equal to 10 msec at message transmitting end, and 20 msec at message receiving end (CTS transmitting end). All other delays during message transmission and reception are assumed to be negligible. For the stated conditions, determine the effective data rate in bps. Ans:
H
Data message
ACK
tp
Data message
Data message t
CTS delay in TX
tp
H
H
ACK
H
CTS delay in RX
Data message
tp
t
Figure 1: The waveforms seen by transmitter and receiver of the data message The Figure 1 illustrates waveforms seen by transmitter and receiver of the data message. As the figure indicates, the receiver completes reception of the data message h (plus overhead) at a time equal to tp plus the time M +n (which requires to transmit R the data message) and is ready to transmit after a delay equal to the CTS delay at the receiver. Reception of the CTS at the transmitter is then complete at time tp plus nRa after the receiver begins transmission. After another delay, equal to the CTS delay at the transmitter, the transmitter can begin to send the next message. Thus, the time required per data message of M bits is 24 + 1000 + 24 + 0.001 × 2 + 0.020 + 0.010 = 0.14117 (seconds) 9600 As a result, the effective data rate is
1000 0.14117 (seconds)
5
= 7.0837 × 103 [bps].
