Practice Question (Throughput)

Duration: 6 min

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AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This lecture reviews pure ALOHA throughput calculations, beginning with a GATE-2021 practice question (Q.22) and then working through textbook Examples 12.2 and 12.3 to establish the method before returning to solve Q.22. The instructor emphasizes identifying frame size, channel rate, and offered load G, computing transmission time Tf = L/B, then applying S = G e^{-2G}. For Example 12.3, a 200-bit frame on a 200 kbps channel gives Tf = 1 ms; at 1000 frames/s, G = 1 and S ≈ 0.135 (13.5%). The maximum pure ALOHA throughput is Smax = 1/(2e) ≈ 0.184 (18.4%) at G = 1/2. Finally, for Q.22 with a 1000-bit frame on a 1 Mbps channel and 1000 frames/s, Tf = 1 ms, G = 1, so S ≈ 13.5%.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens on slide Q.22, a GATE-2021 pure ALOHA throughput problem: 1,000-bit frames on a 1 Mbps (≈10^6 bps) channel with transmissions following a Poisson process at 1,000 frames/s. The instructor then shifts to Example 12.2, circling the given values '200-bit frames' and '200 kbps', underlining 'average number of frames generated by the system', and writing G= in the corner. The core formula S = G x e^-2G is underlined, and the maximum throughput result Smax = 1/(2e) = 0.184 when G = 1/2 is highlighted as the key pure ALOHA result.

  2. 2:00 – 5:00 02:00-05:00

    The instructor works the throughput calculation using S = G * e^(-2G), circling the formula and the 18.4 percent maximum value. Handwritten annotations on the right introduce G1, G2 and a total G to break down frame generation rates (e.g., G1 = 2, then G1 = 1 and G2 = 1/2), with a note '1000 -> 184' linking frame rate to throughput. The lesson moves to Example 12.3: a pure ALOHA network transmitting 200-bit frames on a shared channel of 200 kbps. The frame transmission time is computed as 200/200 kbps = 1 ms and circled. For a system creating 1000 frames per second (1 frame per millisecond), G = 1, so S = G * e^-2G = 0.135 (13.5 percent) for case a, while the maximum throughput 0.184 (18.4 percent) is circled for case b.

  3. 5:00 – 6:15 05:00-06:15

    The video returns to the GATE-2021 problem Q.22, with red underlines highlighting '1,000 bits', '1 Mbps (≈ 10^6 bits per second)', and '1,000 frames per second'. Red handwritten work below the text shows Tf = L/B = 1000 x 1000 ns / 10^6, with the result '1ms' circled and an added '= 1'. The instructor then writes the throughput formula S = G x e^-2G, substitutes to get 1 x e^-2, and concludes with '= 13.5%', applying the same method demonstrated in Example 12.3 to solve the practice question.

The lecture teaches a consistent three-step method for pure ALOHA throughput problems. First, extract the frame length L and channel rate B to find transmission time Tf = L/B (in seconds). Second, determine the offered load G from the frame generation rate: if frames are created at a rate such that one frame is generated every Tf, then G = 1; more generally G equals the average number of frames generated per frame transmission time. Third, apply S = G e^{-2G} to find throughput as a fraction of channel capacity. The instructor repeatedly circles the formula and the maximum value Smax = 1/(2e) ≈ 0.184 (18.4%) at G = 1/2, contrasting it with the operating point G = 1 giving S ≈ 0.135 (13.5%). This pattern is first illustrated with Example 12.2 and the worked Example 12.3 (200-bit frames, 200 kbps channel, Tf = 1 ms), then transferred to the GATE-2021 Q.22 (1,000-bit frames, 1 Mbps channel, Tf = 1 ms), where the same G = 1 yields S ≈ 13.5%. The central takeaway is that pure ALOHA throughput depends only on the dimensionless offered load G, and that the maximum achievable efficiency is about 18.4%.

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