ITPEC FE Morning April 2022 Question 31
Audio data of 2.4 Mbytes encoded at 192 kbit/s is to be downloaded and played back without interruptions over a network with a communication speed of 128 kbit/s. What is the minimum number of seconds required as the data buffering time before the start of playback?
Explanation
50 seconds — the download is slower than playback, so playback may start only as early as lets it finish no sooner than the download.
- •Data size: 2.4 Mbytes = 2.4 × 8 = 19.2 Mbit = 19,200 kbit
- •Playback time: 19,200 ÷ 192 = 100 s
- •Download time: 19,200 ÷ 128 = 150 s
Playback that starts after T seconds of buffering ends at T + 100, and it must not end before the download does: T + 100 ≥ 150, so T ≥ 50. With T = 50, at t seconds after the download starts the data received is 128t kbit and the data played is 192(t − 50) kbit; the difference 9,600 − 64t stays positive until both reach 19,200 kbit at t = 150, so playback never stalls.
Answer (a)
Why not others:
- •(b) 100 — the playback time of the whole file, not the waiting time before it.
- •(c) 150 — the time to download the entire file before playing it; safe, but not the minimum, because playback can start while the download continues.
- •(d) 250 — download time plus playback time, the moment playback ends if the whole file is downloaded first.
Key rule: When a stream downloads more slowly than it plays, the minimum buffering time is the total download time minus the playback time.
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