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# 12-sol - Mass Storage Structure CHAPTER 12 Practice...

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12 C H A P T E R Mass Storage Structure Practice Exercises 12.1 The accelerating seek described in Exercise 12.3 is typical of hard-disk drives. By contrast, floppy disks (and many hard disks manufactured before the mid-1980s) typically seek at a fixed rate. Suppose that the disk in Exercise 12.3 has a constant-rate seek rather than a constant- acceleration seek, so the seek time is of the form t = x + yL , where t is the time in milliseconds and L is the seek distance. Suppose that the time to seek to an adjacent cylinder is 1 millisecond, as before, and is 0.5 milliseconds for each additional cylinder. a. Write an equation for this seek time as a function of the seek distance. b. Using the seek-time function from part a, calculate the total seek time for each of the schedules in Exercise 12.2. Is your answer the same as it was for Exercise 12.3(c)? c. What is the percentage speedup of the fastest schedule over FCFS in this case? Answer: a. t = 0 . 95 + 0 . 05 L b. FCFS 362.60; SSTF 95.80; SCAN 497.95; LOOK 174.50; C-SCAN 500.15; (and C-LOOK 176.70). SSTF is still the winner, and LOOK is the runner-up. c. (362 . 60 - 95 . 80) / 362 . 60 = 0 . 74 The percentage speedup of SSTF over FCFS is 74%, with respect to the seek time. If we include the overhead of rotational latency and data transfer, the percentage speedup will be less. 43

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44 Chapter 12 Mass-Storage Structure 12.2 Is disk scheduling, other than FCFS scheduling, useful in a single-user environment? Explain your answer. Answer: In a single-user environment, the I/O queue usually is empty. Requests generally arrive from a single process for one block or for a sequence of consecutive blocks. In these cases, FCFS is an economical method of disk scheduling. But LOOK is nearly as easy to program and will give much better performance when multiple processes are performing concurrent I/O , such as when a Web browser retrieves data in the background while the operating system is paging and another application is active in the foreground. 12.3 Explain why SSTF scheduling tends to favor middle cylinders over the innermost and outermost cylinders. Answer: The center of the disk is the location having the smallest average distance to all other tracks. Thus the disk head tends to move away from the edges of the disk. Here is another way to think of it. The current location of the head divides the cylinders into two groups. If the head is not in the center of the disk and a new request arrives, the new request is more likely to be in the group that includes the center of the disk; thus, the head is more likely to move in that direction. 12.4 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
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