Question

PLEASE SOLVE ONLY 9.13

9.12 IS ONLY GIVEN FOR THE FULL EXPLANATION OF 9.13

9.12 Elementary physics states that when an object is subjected to a constant acceleration a, the relationship between distance d and time t is given by d at2. Suppose that, during a seek, the disk in Exercise 9.11 accelerates the disk arm at a constant rate for the first half of the seek then decelerates the disk arm at the same rate for the second half of the seek. Assume that the disk can perform a seek to an adjacent cylinder in 1 millisecond and a full-stroke seek over all 5,000 cylinders in 18 milliseconds The distance of a seek is the number of cylinders over which the head moves. Explain why the seek time is proportional to the square root of the seek distance a. function of the seek b. Write an equation for the seek time as a distance. This equation should be of the form t = x + yvZ, where t is the time in milliseconds and L is the seek distance in cylinders. Calculate the total seek time for each of the schedules in Exercise C. 9.11. Determine which schedule is the fastest (has the smallest total seek time) d. The percentage speedup is the time saved divided by the original time. What is the percentage speedup of the fastest schedule over FCFS? 9.13 Suppose that the disk in Exercise 9.12 rotates at 7,200 RPM. What is the average rotational latency of this disk drive? a. b. What seek distance can be covered in the time that you found for part a?

Answer 1

a) The average rotational latency: Half of  is the time it takes for the disk to make one revolution.

7200 rpm=120 rotations per second

The average rotational latency for disk is one half the amount of time.

The average rotation latency =1/(2x120)=1/240=4.167ms

b) distance covered:

We know, d = 1/2 at² . Hence we get t = √ (2d/a).

We have two hints above:
For 1 seek between 2 ( the difference, L=1)cylinders , t=1 ms

For seeks between 5000 (L=4999) cylinders, t=18ms.

FINDING THE EQUATION BY GETTING VALUES FOR X and Y ( pure mathematics here)

Solve the simultaneous equations t = x+y √ L putting L = 1 ,t = 1, and L = 4999, t = 18,

we get the values of x and y and our equation becomes:  t = 0.756+ 0.244√ L.

In equation:

t = 0.756 + 0.244√ L , put value of t = 4.167 to get L = 195.58,

Conclusion: During the time for average rotational latency. we seek over 195 tracks