4.2k views
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__________________.

edited | 4.2k views
+1

Kind of Slotted Aloha protocol. :)

0
Someone please change the answer in key. It should be a range. Not exact $0.4404$. Official gate key also have range $0.4- 0.46$

I answered $0.404$ and it is giving wrong answer.

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

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

$\quad = \textbf{0.4404}$
by Veteran (431k points)
edited
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this is the right answer to this ques na??
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As far as I know. When official key comes we can confirm :)
+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,...
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S1 is a fixed node- not the first node that transmits. I have considered the case for each of S1-S4.
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ans is 0.4656 calculate it manually
–1
I THINK IT IS FINITE POPULATION ALOHA QUESTION........... AM I RIGHT @ARJUN SIR

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)
+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  nc1 or  1/4 in each case.

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

+1
if in the question they did not fix the slot as first slot then answer will be 1/4*0.4404
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@Arjun

if someone answers it as 0.44, will it be considered as wrong attempt?
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nice arjun sir
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@Manu Thakur @Arjun i answered it as 0.44 and it is considered as wrong.. will this be like in gate exam also?

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So, it's NOT the case that only Node $1$ transmits in time slot $S_1$.

Any one of the nodes can transmit in first time slot $S_1$.

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
by Loyal (9.9k points)
0
nice divya mam ji

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

by Boss (24k points)
edited by

Pi is the probability of frame being generated by station si

by Junior (687 points)
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
by Active (4k points)

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

by Active (1.8k points)

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

by Boss (24k points)
–1 vote
by Active (1.4k points)
+1
how?
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iska ans 0.4404 h aa rha hai ... but kaafi places pe 0.462 haai but i dnt get to dis ans anyhow.
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please explain this with the ans 0.462.... all the coachings have given ans but not explanation