Answer:
Given that the time to read data off a 7200 rpm disk drive will be roughly 75% of 5400 rpm disk, at 30% idle time of 7200 rpm disk will the power consumption be equal, on average for the two disk
Explanation:
Let the 7200-disk busy time = x ( we need to calculate it)
As given in the question the in the given time to read data off a 7200 rpm disk drive will be roughly 75% of 5400 rpm disk. This mean that we equalize the equations of both disk power
7200-disk power = 5400-disk power
here we apply the formula to calculate the 7200 rpm disk busy time
4.0 * (1-x) + 7.9 * x = 2.9 * (1-x/0.75) + 7.0 * x/0.75
4.0 + 3.9 x = 2.9 – 3.87 x + 9.33 x
1.1 = 5.47 x – 3.9 x
we calculate the value for x
Hence x = 0.70
Applying formula to calculate the idle time.
Idle time = 1.0 – 0.70 = 0.30
