Passage to India

From Elementary physics, we know that, when an object is subjected to a constant acceleration a, the relationship between distance d and time t is given by d = Ð...at2. Suppose that, during a seek, the disk in Exercise 13.2 accelerates the disk arm at a constant rate for the first half of the seek, then decelerates the disk arm at the seek rate for the second half of the seek. Assume that the disk can perform a seek the an adjacent cylinder in 1 millisecond, and a fill-stoke seek over all 5000 cylinders in 18 milliseconds.

a. The distance of a seek is the number of cylinders that the head moves. Explain why the

seek time is proportional to the square root of the seek distance.

The head accelerates and decelerates and the same constant for equal amounts of time.

Therefore we can rewrite the equations d = Ð... t2, or ((d*2)1/2) = t.

b. Write and equation for time as a function of distance. ((d*2)1/2) = t.

c. Calculate the total seek time for each of the schedules in Exercise 13.2. Determine

which schedule is the fastest (has the smallest total seek time).

86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130

a.) FCFS = 327.8

b.) SSTF = 137.2

c.) SCAN = 148.6 - SCAN goes to zero which adds time

d.) LOOK = 137.2 - These are the same because the

e.) C-SCAN = 137.2 - starting point is the left most value

d. The percentage speedup is the time saved divided by the original time. What is the

percentage speedup of the fastest schedule over FCFS?

327.8-137.2 = 190.6

190.6/327.8 = .58 or 58%