Information Technology Reference
In-Depth Information
Rotation time. Once the disk head settles on the right track, it
must wait for the desired sector to rotate under it; since there is
no reason to expect the desired sector to be particularly near or
far from the disk head when it settles, a reasonable estimate for
rotation time is 4.15 ms, one half of the time that it takes a 7200
RPM disk to rotate once.
Transfer time. The disk's surface bandwidth is at least 54 MB/s,
so transferring 512 bytes takes at most 9.5 S (0.0095 ms).
Total time. 10.5 + 4.15 + .0095 = 14.65 ms per request, so 500
requests will take about 7.8 seconds.
Example: Sequential access workload.
Question: For the disk described in Figure 12.3, consider a workload con-
sisting of a read request for 500 sequential sectors on the same
track. How long will servicing these requests take?
Answer: Disk access time is seek time + rotation time + transfer time.
Seek time. Since we don't know which track we're starting with or
which track we're reading from, we'll use the average seek time,
10.5 ms, as an estimate for the seek time.
Rotation time. Since we don't know the position of the disk when
the request is issued, a simple and reasonable estimate for the
time for the first desired block to rotate to the disk head is 4.15
ms, one half of the time that it takes a 7200 RPM disk to rotate
once.
Transfer time. A simple answer is that 500 sectors can be trans-
ferred in 4.8 to 2.0 ms, depending on whether they are on the
inner or outer tracks.
500 sectors 512
bytes
second
54 10
6
bytes
sector
= 4:8 ms
500 sectors 512
bytes
second
128 10
6
bytes
sector
= 2 ms
(Too) simple answer. These three estimates give us a range from
10:5 + 4:15 + 2 = 16:7 ms
to
10:5 + 4:15 + 4:8 = 19:5 ms
More precise answer. However, this simple answer ignores the
track buffer. Since the transfer time is a large fraction of the ro-
tation time (about 1/4 to 1/2 of the time for a full rotation), we
