Mark M.

asked • 05/23/20# Not sure how to determine PMF

In random access communication protocols a population of users is contending for the use of a common channel to transmit their packets, without the control and supervision of a master station. In the sequel we assume that time is divided into timeslots. The communication channel can handle at most two packets in a timeslot. Because all users transmit packets when they themselves find appropriate, it will be possible of course that multiple users transmit a packet during the same time. If one or two users transmit a packet over the channel in a certain timeslot, then these transmissions are successfull. If three users transmit a packet over the channel in the timeslot, a small collision occurs and only one of the three packets is transmitted successfully. If four or more users transmit a packet in the timeslot, then a big collision occurs and all the packets transmitted in the timeslot are lost.

Suppose now that we have n users that always have packets to transmit over the channel. At the beginning of each timeslot each user, independently of all the other users, decides to transmit a packet with probability p = α/n for some α > 0.

Give the probability mass function PK(k; α, n) of the number of users, K, that transmit a packet over the communication channel in a certain timeslot in this case and evaluate it for n = 5, 10 and 100 and for α = 2.

Not sure how to determine the PMF here, could anyone help? Thanks!

## 1 Expert Answer

P(k = 1) = P(x=1)+P(x=3);P(k=2)=P(x = 2);P(k=0) = 1 - P(k = 1) - P(k = 2)

a)n = 5;p = 2/5

P(k=1) = P(x = 1 or 3) = binom.dist(1,5,2/5,0)+binom.dist(3,5,2/5,0) = .4896

P(k=2) = P(x = 2) = binom.dist(2,5,2/5,0) = .3456

P(k = 0) = 1 - P(k=1) - P(k=2) = 1 - .4896 - .3456 = .1648

b)n=10;p=2/10=1/5

P(k=1) = P(x = 1 or 3) = binom.dist(1,10,1/5,0)+binom.dist(3,10,1/5,0) = 0.469762048

P(k=2) = P(x = 2) = binom.dist(2,10,1/5,0) = 0.301989888

P(k = 0) = 1 - P(k=1) - P(k=2) = 1 - 0.469762048 - 0.301989888 = 0.228248064

c)n = 100;p=2/100=1/50

P(k=1) = P(x = 1 or 3) = binom.dist(1,100,1/50,0)+binom.dist(3,100,1/50,0) = 0.452928095933768

P(k=2) = P(x = 2) = binom.dist(2,100,1/50,0) = 0.273413911569774

P(k = 0) = 1 - P(k=1) - P(k=2) = 1 - 0.452928095933768 - 0.273413911569774 = 0.273657992496458

Tom K.

05/24/20

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Stanton D.

You'd probably start with a Poisson distribution, and modify for losses incurred when n>=3 users attempt transmission. (Or you might, as my Verizon service likes to do, simply permit some cross-channel bleed when the network gets full!)05/24/20