Consider a disk with an average seek time of 5ms, an average rotational latency

Question

Consider a disk with an average seek time of 5ms, an average rotational latency of 5ms, and a transfer rate of 40MB/sec.

If the block size is 1KB, how long would it take to read a block?

If the block size is 100KB, how long would it take to read a block?

If the goal is to read 1K, a 1KB block size is better as the remaining 99KB are wasted. If the goal is to read 100KB, the 100KB block size is better since the 1KB block size needs 100 seeks and 100 rotational latencies. What is the minimum size request for which a disk with a 100KB block size would complete faster than one with a 1KB block size?

Explanation / Answer

Transfer Rate = 40MB/s = 40*1000 KB/s = 40 KB/ms
Transfer Time = Transfer Rate * Block Size = 40 * 1 ms = 40 ms

Time to read the block = Average Seek Time + Average Rotational Latency + Transfer Time = 5 + 5 + 40 = 50 ms.
Answer - It will take 50 ms to read 1 KB block size.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.