Information Technology Reference
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= (1 +
:2
estimated .2% seek time
33:3
10:5) ms
= 1:06 ms
Rotation time. Since we don't know the position of the disk when
the seek finishes and since sectors are scattered randomly, a
simple and reasonable estimate for the time after the seek fin-
ishes for the 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. Similar to the example on page 361, transfer time
for each sector is at most 0.0095 ms
Total time. 1.06 + 4.15 + .0095 = 5.22 ms per request, so 500
requests will take about 2.6 s. Notice that the time for the SCAN
scheduled time is less than half the 7.8 s time for the FIFO-
scheduled time for the example on page 361
Exercises
1. Discussion. Some high-end disks in the 1980s had multiple disk arm
assemblies per disk enclosure in order to allow them to achieve higher
performance. Today, high-performance server disks have a single arm
assembly per disk enclosure. Why do you think disks so seldom have
multiple disk arm assemblies today?
2. How many sectors does a track on the disk described in Figure 12.3 on
page 360 have?
3. For the disk in Figure 12.3 on page 360, estimate the distance from the
center of one track to the center of the next track.
4. A disk may have multiple surfaces, arms, and heads, but when you issue
a read or write, only one head is active at a time. It seems like one could
greatly increase disk bandwidth for large requests by reading or writing
with all of the heads at the same time. Given the physical characteristics
of disks, can you figure out why no one does this?
5. For the disk described in Figure 12.3 on page 360, consider a workload
consisting of 500 read requests, each of a randomly chosen sector on disk,
assuming that the disk head is on the outside track and that requests are
serviced in P-CSCAN order from outside to inside. How long will servicing
these requests take?
Note: Answering this question will require making some estimates.
