A hard disk with a transfer rate of 10 Mbytes/ second is constantly…
2004
A hard disk with a transfer rate of 10 Mbytes/ second is constantly transferring data to memory using DMA. The processor runs at 600 MHz, and takes 300 and 900 clock cycles to initiate and complete DMA transfer respectively. If the size of the transfer is 20 Kbytes, what is the percentage of processor time consumed for the transfer operation ?
- A.
5.0%
- B.
1.0%
- C.
0.5%
- D.
0.1%
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Correct answer: D
Step 1: Compute the DMA transfer time. Transfer time = transfer size / transfer rate = 20 KB / 10 MB/s ≈ 0.002 s (2 ms).
Step 2: Compute the processor busy time for the DMA operation. CPU cycles used = 300 (initiate) + 900 (complete) = 1200 cycles. At 600 MHz this is 1200 / 600,000,000 s = 2 × 10⁻6 s = 2 µs.
Step 3: Compute the fraction of processor time consumed. Fraction = CPU busy time / transfer time = 2 µs / 2 ms = 0.001 = 0.1%.
Final answer: The processor time consumed for the transfer operation is 0.1%.
Note: Using binary units (20 KiB and 10 MiB/s) gives a transfer time ≈ 1.95 ms and the same percentage to two significant figures (≈0.1%).
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