MCS-284 Homework 8 (Fall 1999)
Due: December 6, 1999
Do exercise 8.3 on page 701. Either disk layout might in principle be
faster depending on the specific numerical parameters of the disk
drive. Each has 8 sectors per track. One has sector number 8 located
directly in from sector number 0. The other has the sectors in the
inner track skewed somewhat, so that sector number 8 is positioned in
from some non-zero sector. That non-zero sector happens to be 2 in
the illustration. You should not worry too much about the exact
amount of skew -- whether it is 1/4 of the way around, or 1/3 of the
way, or what. The correct amount of skew varies from disk drive to
disk drive. But, for any given disk drive, there is some amount of
skew that will produce optimal performance, generally significantly
better than the neatly aligned arrangement where sector 8 is aligned
with sector 0. The question you should be addressing is: why?
Do exercise 8.4 on page 701. Be careful about the
distinction between microseconds and seconds, and check that your
answer is of a plausible order of magnitude.
Do exercise 8.5 on page 701. Again, you need to be careful
about orders of magnitude.
- Do exercise 8.17 on page 703. This refers back to
the calculation on page 686, which has one little mystery in it.
The last step in the calculation divides 64 KB by 22.8 ms
and arrives at approximately 2.74 MB/sec. In order to make the
arithmetic work here, you have to assume that KB is the 210
kind of K and MB is the 220 kind of M, rather than
103 and 106. For the size of the I/O transfers
(64 KB) it is indeed reasonable that the K would be of the
210 kind. For the I/O rate (2.74 MB/sec), however, it
would have been more traditional to use the 106 version of
M, as our authors remark on page 642. In working your own
version of this calculation on the homework, you may use either
version of K and M, but be sure to explicitly state which. Also,
remember that there are two issues that limit how many drives you put
on a SCSI bus.
Instructor: Max Hailperin