ITPEC IP April 2018 Question 3
A PC plays back video data recorded at 500×400 pixels per frame, 30 frames per second, and 3 bytes per pixel while reading it from a peripheral device. When the data transfer rate between the peripheral device and the PC is 1×10^6 bytes per second, in order to play back the video without delay, the data must be compressed to one n-th (1/n) in advance. Which of the following is the minimum value of n? Here, the time to expand the compressed data and other impacts can be ignored.
Explanation
18 — compare the data rate the video needs with the rate the connection can deliver.
- •Bytes per frame: 500 × 400 × 3 = 600000 bytes
- •Bytes per second: 600000 × 30 = 18000000 = 18 × 106 bytes per second
- •Required compression: 18 × 106 ÷ (1 × 106) = 18
Compressed to 1/18, the video needs exactly 1 × 106 bytes per second, which the connection carries, so playback keeps up. Any smaller n needs more than the transfer rate.
Answer (a)
Why not others:
- •(b) 54 and (c) 72 — they also let the video play without delay, but compress more than needed; the question asks for the minimum n
- •(d) 144 — 18 × 8: it matches computing the video rate in bits (144 million bits per second) and dividing by a transfer rate given in bytes
Key rule: Uncompressed video rate = pixels per frame × bytes per pixel × frames per second; divide it by the transfer rate, in the same unit, to get the minimum compression factor.
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