운영체제 연습문제 답

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운영체제 연습문제 답에 대한 보고서 자료입니다.

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chedule is 143, 913, 948, 1022, 1470, 1509, 1750,
1774, 130, 86. The total seek distance is 3319.
e. The C-SCAN schedule is 143, 913, 948, 1022, 1470, 1509,
1750, 1774, 4999, 0, 86, 130. The total seek distance is
9985.
f. The C-LOOK schedule is 143, 913, 948, 1022, 1470, 1509,
1750, 1774, 86, 130. The total seek distance is 3363.
14.10 Requests are not usually uniformly distributed. For example, a
cylinder containing the file system FAT or inodes can be
expected to be accessed more frequently than a cylinder that
only contains files. Suppose you know that 50 percent of the
requests are for a small, fixed number of cylinders.
a. Would any of the scheduling algorithms discussed in this
chapter be particularly good for this case? Explain your
answer.
b. Propose a disk-scheduling algorithm that gives even better
performance by taking advantage of this "hot spot" on the
disk.
c. File systems typically find data blocks via an indirection
table, such as a FAT in DOS or inodes in UNIX. Describe
one or more ways to take advantage of this indirection to
improve the disk performance.
Answer:
a. SSTF would take greatest advantage of the situation. FCFS
could cause unnecessary head movement if references to the
"high-demand" cylinders were interspersed with references
to cylinders far away.
b. Here are some ideas. Place the hot data near the middle of
the disk. Modify SSTF to prevent starvation. Add the
policy that if the disk becomes idle for more than, say,
50 ms, the operating system generates an anticipatory seek
to the hot region, since the next request is more likely
to be there.
c. Cache the metadata in primary memory, and locate a file's
data and metadata in close physical proximity on the
disk. (UNIX accomplishes the latter goal by allocat- ing
data and metadata in regions called cylinder groups.)
14.11 Why is rotational latency usually not considered in disk
scheduling? How would you modify SSTF, SCAN, and C-SCAN to
include latency optimization?
Answer: Most disks do not export their rotational position
information to the host. Even if they did, the time for this
information to reach the scheduler would be subject to
imprecision and the time consumed by the scheduler is variable,
so the rotational position information would become
incorrect. Further, the disk requests are usually given in terms
of logical block numbers, and the mapping between logical blocks
and physical locations is very complex.
14.14 What are the tradeoffs involved in rereading code pages from the
file system versus using swap space to store them?
Answer: If code pages are stored in swap space, they can be
transferred more quickly to main memory (because swap space
allocation is tuned for faster performance than general file
system allocation). Using swap space can require startup time if
the pages are copied there at process invocation rather than
just being paged out to swap space on demand. Also, more swap
space must be allocated if it is used for both code and data
pages.

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운영체제,   Operating system concepts,   연습문제