GATE CS Mastery
Cheat sheets

GATE CS 2015 Set 1 · Computer Networks

Question 39

NATPrinted as CS Q.292 marksNo negative markingMedium (estimate)

Question

Consider a LAN with four nodes S1,S2,S3S_1, S_2, S_3 and S4S_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 transmit in the same slot. The probabilities of generation of a frame in a time slot by S1,S2,S3S_1, S_2, S_3 and S4S_4 are 0.1,0.2,0.30.1, 0.2, 0.3 and 0.40.4, respectively. The probability of sending a frame in the first slot without any collision by any of these four stations is _____.

Official answer and solution
Accepted answers
0.4 to 0.46
From
Official GATE 2015 answer key

Solution

Compute the sum of probabilities that exactly one station generates a frame while all other three remain idle.

Let p1=0.1,p2=0.2,p3=0.3,p4=0.4p_1 = 0.1, p_2 = 0.2, p_3 = 0.3, p_4 = 0.4 be the probabilities that stations S1,S2,S3,S4S_1, S_2, S_3, S_4 transmit in the slot.
A frame is transmitted without collision if and only if exactly one station transmits while the remaining three stations do not transmit.

Computing the probability for each station:

  1. Only S1S_1 transmits:
    P1=p1(1−p2)(1−p3)(1−p4)=0.1×0.8×0.7×0.6=0.0336.P_1 = p_1(1 - p_2)(1 - p_3)(1 - p_4) = 0.1 \times 0.8 \times 0.7 \times 0.6 = 0.0336.
  2. Only S2S_2 transmits:
    P2=(1−p1)p2(1−p3)(1−p4)=0.9×0.2×0.7×0.6=0.0756.P_2 = (1 - p_1)p_2(1 - p_3)(1 - p_4) = 0.9 \times 0.2 \times 0.7 \times 0.6 = 0.0756.
  3. Only S3S_3 transmits:
    P3=(1−p1)(1−p2)p3(1−p4)=0.9×0.8×0.3×0.6=0.1296.P_3 = (1 - p_1)(1 - p_2)p_3(1 - p_4) = 0.9 \times 0.8 \times 0.3 \times 0.6 = 0.1296.
  4. Only S4S_4 transmits:
    P4=(1−p1)(1−p2)(1−p3)p4=0.9×0.8×0.7×0.4=0.2016.P_4 = (1 - p_1)(1 - p_2)(1 - p_3)p_4 = 0.9 \times 0.8 \times 0.7 \times 0.4 = 0.2016.

Summing these mutually exclusive probabilities:

P(no collision)=0.0336+0.0756+0.1296+0.2016=0.4404.P(\text{no collision}) = 0.0336 + 0.0756 + 0.1296 + 0.2016 = 0.4404.

The value falls within the accepted range [0.40,0.46][0.40, 0.46].

Final answer

0.44

Common trap

Adding P(no station transmits)=(0.9)(0.8)(0.7)(0.6)=0.3024P(\text{no station transmits}) = (0.9)(0.8)(0.7)(0.6) = 0.3024, which corresponds to an idle slot where no frame is sent at all.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2015 Computer Science and Information Technology (CS) Session 1
Question
39 of 65, printed as CS Q.29
Section
Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.