Information Technology Reference
In-Depth Information
So, even a fairly large request (500 sectors or 250KB in this case) can incur
significant overheads from seek and rotational latency.
Example: Efficient access.
Question: For the disk described in Figure 12.3, how large must a request
that begins on a random disk sector be to ensure that the disk
gets at least 80% of its advertised maximum surface transfer
bandwidth?
Answer: When reading a long sequence of logically sequential blocks, the
disk will read an entire track, then do a 1 track seek (or a head
switch and resettle, which amounts to the same thing) and then
read the next track. Notice that track buffering allows the disk to
read an entire track in one rotation regardless of which sector the
head is over when it settles on the track and starts successfully
reading. So, for the outer tracks, it reads for one rotation (8.4 ms)
and then does a minimum seek (1 ms).
So, to achieve 80% of peak bandwidth after a random seek (10.5
ms), we need to read enough rotations worth of data to ensure
that we spend 80% of the total time reading. If x is the number
of rotations we will read, then we have
0:8 totalTime
= x rotationTime
0:8(10:5 ms + (1 + 8:4)xms)
=
8:4x ms
x =
9:09
So, we need to read at least 9.09 rotations worth of data to reach
an efficiency of 80%. Since each rotation takes 8.4 ms and trans-
fers data at 128 MB/s, 9.09 rotations transfers 9.77 MB of data,
or about 19,089 sectors.
12.1.2
Disk scheduling
Because moving the disk arm and waiting for the platter to rotate are so expen-
sive, performance can be significantly improved by optimizing the order in which
pending requests are serviced. Disk scheduling can be done by the operating
system, by the disk's rmware, or both.
FIFO. The simplest thing to do is to process requests in rst-in-rst-out
(FIFO) order. Unfortunately, a FIFO scheduler can yield poor performance.
For example, a sequence of requests that alternate between the outer and inner
tracks of a disk will result in many long seeks.
