Question

In a Wi-Fi network, data packets are transmitted between a laptop computer and the access point. Since the transmission is wireless, errors can occur that may render the packet useless, so, when received in error, the packet is discarded. As a protection for errors, an acknowledgement is expected, and if the packet is received in error, it is retransmitted, until correctly received. Consider that the probability that a packet is discarded is p=0.6.

(a) What is the probability that the packet needs to be transmitted 3 times or more?

(b) If you could use more powerful coding to reduce the value of p, what should be p, such that the probability of having no more than two transmissions for the same packet is 0.8?

Answer 1

Probability of the packet is discarded = p = 0.6

therefore, Probability of the packet is transmitted susscefully = q = 0.4

X be a random variable.

X: Number of times packet needs to be discarded. ; 0 X

A)   

here we will use geometric distribution.

Packet need to be trasmitted 3 times or more that mean it has to be discard at least 2 times

P(X 2)

= 1 - P(X < 2)

= 1 - p^x q

= 1 - (0.4 + (0.6*0.4))

= 1 - (0.4 + 0.24)

= 1 - 0.64

= 0.46

B)

Probability of having no more than 2 transmission (we have to discard the transmission at most only once) = 0.8

therefore,

P(X 1) = 0.8

= p^x q = 0.8

= (q + p*q) = 0.8

= q (1+p) = 0.8

= (1-p) (1+p) = 0.8

= 1 - p² = 0.8

= p² = 1- 0.8 = 0.2

= p =

= p = 0.4472