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Official PYQ · 2021Set 2 · Q54Data Link Layer: Framing, Error Detection, MAC, Ethernet BridgingNAT2 marks

GATE CS 2021 Set 2 Q54 — Data Link Layer: Framing, Error Detection, MAC, Ethernet Bridging

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Consider a network using the pure ALOHA\text{ALOHA} medium access control protocol, where each frame is of length 1,0001,000 bits. The channel transmission rate is 11 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 1,0001,000 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 ______________
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Answer: 130 : 140

To find the throughput of the pure ALOHA\text{ALOHA} network in frames per second, we can proceed step-by-step:

1. Calculate the Frame Transmission Time (TfrT_{fr})

The time taken to transmit a single frame of length LL over a channel of transmission rate RR is given by:

Tfr=LR=1,000 bits106 bits/second=10−3 seconds=1 msT_{fr} = \frac{L}{R} = \frac{1,000 \text{ bits}}{10^6 \text{ bits/second}} = 10^{-3} \text{ seconds} = 1 \text{ ms}

2. Calculate the Offered Load (GG)

The offered load GG is the average number of transmission attempts (both new and retransmitted frames) per frame transmission time (TfrT_{fr}):

G=λ×TfrG = \lambda \times T_{fr}

Given that the aggregate transmission rate λ=1,000\lambda = 1,000 frames per second:

G=1,000 frames/second×0.001 seconds=1G = 1,000 \text{ frames/second} \times 0.001 \text{ seconds} = 1

3. Calculate the Throughput

In pure ALOHA\text{ALOHA}, the probability of a successful transmission is e−2Ge^{-2G}. Therefore, the throughput SS (in frames per frame time) is:

S=Ge−2GS = G e^{-2G}

Substituting G=1G = 1:

S=1×e−2≈0.135335 frames per frame timeS = 1 \times e^{-2} \approx 0.135335 \text{ frames per frame time}

To find the throughput in terms of successful frames per second (SsecS_{\text{sec}}):

Ssec=STfr=0.1353350.001≈135.34 frames/secondS_{\text{sec}} = \frac{S}{T_{fr}} = \frac{0.135335}{0.001} \approx 135.34 \text{ frames/second}

Rounding to the nearest integer, we get 135 (which lies within the official range of 130 to 140).

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