Q. The accelerating search for described in subsequent exercise is typical of hard-disk drives. By contrast floppy disks and several hard disks manufactured before the mid-1980s typically seek at a fixed rate. Presume that the disk in subsequent exercise has a constant-rate seek rather than a constant acceleration seek consequently seek time is of the form t = x + yL where t is the time in milliseconds and L is the seek distance. Presume that the time to seek to an adjacent cylinder is 1 millisecond as before and is 0.5 milliseconds for every additional cylinder.

a. Write an equation for this seek time like a function of the seek distance.

b. Utilizing the seek-time function from a part calculate the total seek time for each of the schedules in Exercise 12.2. Is your answer the similar as it was for Exercise 12.3(c)?

c. What is the percentage accelerate of the fastest schedule over FCFS in this case?

Answer:

a. t = 0.95 + 0.05L

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 as well as LOOK is the runner-up.

c. (362.60 - 95.80)/362.60 = 0.74 the percentage accelerate of SSTF over FCFS is 74% with respect to the seek time. If we comprise the overhead of rotational latency and data transfer the percentage speedup will be less.

Posted Date: 7/29/2013 1:43:39 AM | Location : United States

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