Disk Management


#1

Consider a disk which has an average seek time of 30 ns and rotational rate of 360 rpm each track of disk 512 sectors each of size 512 bytes . What is the time taken to transfer the four successive sectors…?


#2

Given Data,
Average seek time = 30ns
Rotation rate = 360 RPM
Number of sector = 512
Sector capacity = 512 Bytes

Here we have to find disk access time of 4 sector i.e 4 * 512B = 2048 B
We know data transfer rate is 1536KB/sec
i.e 1 B data can be transferred in 11536×10311536×103 sec
i.e 2048 B can be transferred in 20481536×103sec=0.00133sec20481536×103sec=0.00133sec

Now 1 rotation is done in 1616 sec, so Avg rotation latency is 112112 sec i.e 0.0833 sec

As we know Disk access time = Avg seek time + Avg rotation latency + Data transfer rate
DAT = 30ns + 0.0833 sec + 0.00133 sec
we can neglect avg seek time as its too small
Therefore Disk Access time = 0.0846 secs