For a 7200 RPM disk, use the parameters as shown in the textbook (5ms average se

Question

For a 7200 RPM disk, use the parameters as shown in the textbook (5ms average seek time; 1Gbit/sec bandwidth; 0.1ms controller overhead). Assume the block size is 512 bytes.

a) How long does it take to access 512 bytes of contiguous data on average? What is the average throughput (bytes per second) if the user accesses 512 bytes for each I/O? Assume the data is perfectly aligned on the disk.

b) How long does it take to access 1MB of contiguous data on average? What is the average throughput (bytes per second) if the user accesses 1MB for each I/O? Assume the data is perfectly aligned on the disk.

c) If the user expects his/her average throughput to reach 90% of the disk max bandwidth, what is the minimal size of his/her I/O?

d) How long does it take to access 10 bytes of contiguous data in average? Assume the 10 bytes reside in one block.

Explanation / Answer

The average access time is the sum of the seek time, rotational latency, and transfer time. The seek time is given as 5ms. Once the head is in the right place, on average we will need to wait for half a rotation of the disk for the correct sector to come under the head. Thus, on average, the rotational latency is half the time it takes the disk to make a complete revolution. The disk spins at 7200 RPM, so it takes 1/7200 of a minute to make one revolution. Equivalently, (1000 ms/sec × 60 sec/minute) / 10000 RPM = 8.33 ms to make one revolution. So rotational latency is 4ms

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