A system transfers data in blocks of 8 KB. Each block takes 2 ms to transfer and 3 ms to process. Assuming transfer and processing use separate resources, double buffering is used, and startup and shutdown overhead are ignored, what is the approximate steady-state throughput?

Options

  • A. 1600 KB/s
  • B. 2667 KB/s
  • C. 4000 KB/s
  • D. More than one of the above
  • E. None of the above

Correct Answer (Detailed Explanation is Below)

B. 2667 KB/s

Detailed Explanation

Correct Answer: Approximately 2667 KB/s.

Explanation: With double buffering and separate transfer and processing resources, the steady-state time per block is the larger of the two times: max(2 ms, 3 ms) = 3 ms.
Throughput: 8 KB / 0.003 s = approximately 2666.67 KB/s.
Without overlapping transfer and processing, each block would take 2 + 3 = 5 ms, giving 1600 KB/s. The calculation assumes the resources can operate concurrently without additional contention.
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BPSC TRE 4 operating system practice set 3

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