GATE CSE 2015 | Set 1 | Question: 29

Question

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__________________.

Comments

@ankitgupta.1729 sir, can you pls retagged this question properly. The tag should be slotted Aloha under MAC Protocol.

---

@avik_ghosh it is already done by @Hira Thakur

---

Reference : Kurose & Ross

prerequisite for this Question  : We must know the some basic concept of probability ..

1) Independent Event ==> The probability of one station transmitting does not affect or depend on the others.

2) Multiplicative rule of Independent Event ==> One specific station transmitting while all other specific stations remain silent, the individual probabilities of each event are multiplied together.

3) Mutually Exclusive Events ==> Only one station can be the sole transmitter in a given slot.

4) Addition rule of Disjoint Events ==> Because these success are mutually exclusive, the total probability of a successful transmission is the sum of the individual probabilities.

==> A successful transmission in the first slot occurs if exactly one station transmits but if two or more transmit, there is a collision.

==> From the independent random-access method, the total success probability ($P_{\text{success}}$) is the sum of the probabilities of each individual station transmitting alone while all other stations remain idle -->

$$P_{\text{success}} = \sum_{i=1}^{4} p_i \prod_{j \neq i} (1 - p_j)$$

 

$$P_{\text{success}} = p_1(1-p_2)(1-p_3)(1-p_4) + (1-p_1)p_2(1-p_3)(1-p_4) + (1-p_1)(1-p_2)p_3(1-p_4) + (1-p_1)(1-p_2)(1-p_3)p_4$$

$$P_{\text{success}} = (0.1 \times 0.8 \times 0.7 \times 0.6) + (0.9 \times 0.2 \times 0.7 \times 0.6) + (0.9 \times 0.8 \times 0.3 \times 0.6) + (0.9 \times 0.8 \times 0.7 \times 0.4)$$

$$P_{\text{success}} = 0.0336 + 0.0756 + 0.1296 + 0.2016 = 0.4404$$

Hence we can say that the probability of sending a frame without any collisions in the first slot is 0.4404.

Answer 1

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

Comments

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 $S_1$.

---

@toxicdesire S1 ,S2,S3,S4 are not slots they are nodes.

---

same

Answer 2

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

Comments

nice divya mam ji

---

good explanation

Answer 3

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

Answer 4

Pi is the probability of frame being generated by station si

Answer 5

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

Answer 6

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