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GATE CS 2021 Set 2 · Computer Networks

Question 64

NATPrinted as CS Q.542 marksNo negative markingMedium (estimate)

Question

Consider a network using the pure ALOHA medium access control protocol, where each frame is of length 1,000 bits. The channel transmission rate is 1 Mbps (=106= 10^6 bits per second). The aggregate number of transmissions across all the nodes (including new frame transmissions and retransmitted frames due to collisions) is modelled as a Poisson process with a rate of 1,000 frames per second. Throughput is defined as the average number of frames successfully transmitted per second. The throughput of the network (rounded to the nearest integer) is ______.

Official answer and solution
Accepted answers
130 to 140
From
Official GATE 2021 answer key

Solution

Compute the normalized offered traffic GG and evaluate the pure ALOHA throughput formula S=Ge−2GS = G e^{-2G}, then scale by frame duration.

Given parameters:

  • Frame size L=1000 bitsL = 1000\text{ bits}
  • Channel transmission rate R=1 Mbps=106 bits/secondR = 1\text{ Mbps} = 10^6\text{ bits/second}
  • Frame transmission time: Tfr=LR=1000106=10−3 seconds=1 msT_{\text{fr}} = \frac{L}{R} = \frac{1000}{10^6} = 10^{-3}\text{ seconds} = 1\text{ ms}

The aggregate transmission arrival rate is λ=1000 frames/second\lambda = 1000\text{ frames/second}. The normalized channel load (offered traffic per frame transmission time) is:

G=λ×Tfr=1000×10−3=1G = \lambda \times T_{\text{fr}} = 1000 \times 10^{-3} = 1

In pure ALOHA, the vulnerable period for a packet is 2Tfr2 T_{\text{fr}}. The probability of successful transmission without collision is:

Psucc=e−2G=e−2P_{\text{succ}} = e^{-2G} = e^{-2}

The throughput SS in frames per frame transmission time is:

S=Ge−2G=1⋅e−2=e−2≈0.135335 frames per TfrS = G e^{-2G} = 1 \cdot e^{-2} = e^{-2} \approx 0.135335\text{ frames per } T_{\text{fr}}

Convert to average number of successfully transmitted frames per second:

Throughput=STfr=0.13533510−3=135.335 frames/second\text{Throughput} = \frac{S}{T_{\text{fr}}} = \frac{0.135335}{10^{-3}} = 135.335\text{ frames/second}

Rounding to the nearest integer gives 135 (official key range: 130 to 140).

Final answer

135

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

Source

Paper
GATE 2021 Computer Science and Information Technology - Session 2 (CS2)
Question
64 of 65, printed as CS Q.54
Section
Computer Science and Information Technology

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