Kind of **Slotted Aloha** protocol. :)

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+32 votes

Consider a LAN with four nodes $S_1, S_2, S_3,$ and $S_4$. Time is divided into fixed-size slots, and a node can begin its transmission only at the beginning of a slot. A collision is said to have occurred if more than one node transmits in the same slot. The probabilities of generation of a frame in a time slot by $S_1, S_2, S_3,$ and $S_4$ are $0.1, 0.2, 0.3$ and $0.4$ respectively. The probability of sending a frame in the first slot without any collision by any of these four stations is__________________.

+61 votes

Best answer

$P= P(S1) P(\neg S2) P(\neg S3) P(\neg S4)$

$\qquad + P(\neg S1) P(S2) P(\neg S3) P(\neg S4)$

$\qquad + P(\neg S1) P(\neg S2) P(S3) P(\neg S4) $

$\qquad + P(\neg S1) P(\neg S2) P(\neg S3) P(S4) $

$ \quad = $ $ 0.1 * 0.8 * 0.7 * 0.6 \\ + 0.9 * 0.2 * 0.7 * 0.6 \\ + 0.9 * 0.8 * 0.3 * 0.6 \\ + 0.9 * 0.8 * 0.7 * 0.4$

$ \quad = \textbf{0.4404}$

$\qquad + P(\neg S1) P(S2) P(\neg S3) P(\neg S4)$

$\qquad + P(\neg S1) P(\neg S2) P(S3) P(\neg S4) $

$\qquad + P(\neg S1) P(\neg S2) P(\neg S3) P(S4) $

$ \quad = $ $ 0.1 * 0.8 * 0.7 * 0.6 \\ + 0.9 * 0.2 * 0.7 * 0.6 \\ + 0.9 * 0.8 * 0.3 * 0.6 \\ + 0.9 * 0.8 * 0.7 * 0.4$

$ \quad = \textbf{0.4404}$

+2

why u didnot consider that how many ways a node can be choosen before transmitting...since there are 4 node anyone can be choosen..for example s1 can be choosen in 4c1 ways,...

0

S1 is a fixed node- not the first node that transmits. I have considered the case for each of S1-S4.

–1

I THINK IT IS FINITE POPULATION ALOHA QUESTION........... AM I RIGHT @ARJUN SIR

I HAVE SEEN SUCH TYPE OF QUESTION IN THAT TOPIC.

I HAVE SEEN SUCH TYPE OF QUESTION IN THAT TOPIC.

+1

@arjun Sir , isn't first slot is S1 ?

so the probability would be simply P = P(S1) P(~S2) P(~S3) P(~S4)

so the probability would be simply P = P(S1) P(~S2) P(~S3) P(~S4)

+11

@pC , we have to find

"The probability of sending a frame in the first slot without any collision** by any of these four stations** is___ "

It means , In first slot ,any of these 4 stations can send their frames **without collision..**

**So , **Possibilities are :- 1) S1 will send and s2,s3,s4 will not send

2) S2 will send and s1,s3,s4 will not send

3) S3 will send and s1,s2,s4 will not send

4) S4 will send and s1,s2,s3 will not send

+1

PLease help me with this that why are we not considering ^{n}c1 _{or }1/4 in each case.

I know my assumption is not correct but its confusing me whether to take any of above terms.

0

@Manu Thakur @Arjun i answered it as 0.44 and it is considered as wrong.. will this be like in gate exam also?

+13 votes

The probability of sending a frame in the first slot

without any collision by any of these four stations is

sum of following 4 probabilities

Probability that S1 sends a frame and no one else does +

Probability thatS2 sends a frame and no one else does +

Probability thatS3 sends a frame and no one else does +

Probability thatS4 sends a frame and no one else does

= 0.1 * (1 - 0.2) * (1 - 0.3) *(1 - 0.4) +

(1 -0.1) * 0.2 * (1 - 0.3) *(1 - 0.4) +

(1 -0.1) * (1 - 0.2) * 0.3 *(1 - 0.4) +

(1 -0.1) * (1 - 0.2) * (1 - 0.3) * 0.4

= 0.4404

without any collision by any of these four stations is

sum of following 4 probabilities

Probability that S1 sends a frame and no one else does +

Probability thatS2 sends a frame and no one else does +

Probability thatS3 sends a frame and no one else does +

Probability thatS4 sends a frame and no one else does

= 0.1 * (1 - 0.2) * (1 - 0.3) *(1 - 0.4) +

(1 -0.1) * 0.2 * (1 - 0.3) *(1 - 0.4) +

(1 -0.1) * (1 - 0.2) * 0.3 *(1 - 0.4) +

(1 -0.1) * (1 - 0.2) * (1 - 0.3) * 0.4

= 0.4404

+7 votes

**Required probability**

=P(only one frame can transmit in a given time slot)

=P(only S1 sends)+ P(only S2 sends)+ P(only S3 sends)+ P(only S4 sends)

=(**0.1***0.8*0.7*0.6)+(0.9***0.2***0.7*0.6)+(0.9*0.8***0.3***0.6)+(0.9*0.8*0.7***0.4**)

=0.0336+ 0.0756 + 0.1296 + 0.2016

=**0.4404**

0 votes

probablity=0.1*(1-0.2)*(1-0.3)*(1-0.4)+(1-0.1)*0.2*(1-0.3)*(1-0.4)+(1-0.1)*(1-0.2)*).3*(1-0.4)+(1-0.1)*(1-0.2)*(1-0.3)*0.4

=>0.0336+0.0756+0.1296+0.2016=.4404

=>0.0336+0.0756+0.1296+0.2016=.4404

0 votes

Required probability=

=P(only one frame can transmit in a given time slot)

=P(only S1 sends)+ P(only S2 sends)+ P(only S3 sends)+ P(only S4 sends)

=(**0.1***0.8*0.7*0.6)+(0.9***0.2***0.7*0.6)+(0.9*0.8***0.3***0.6)+(0.9*0.8*0.7***0.4**)

So that which can not send than do minus from 1-(which cannot send on this slot of time expect which are sending in that slot of time we get probability of sending )

=0.0336+ 0.0756 + 0.1296 + 0.2016

=0.4404

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