Question

A program must write 4 sectors onto a disk. Each sector is 0.5KB. The disk drive rotates at 10,000 rpm. The average seek time is 6 ms and the disk controller overhead is 1 ms. The transfer rate is 1 MB/sec. The current location of the R/W head is not known. What is the time to write this data if a. Sectors are on the same track and are contiguous b. Sectors are on the adjacent cylinders. The sectors can appear anywhere on the cylinder.

Answer 1

To transfer one sector to the computer, we must wait the equivalent Seek Time + Rotational Latency + Transfer Time + overhead

Rotational latency = 60/10000 seconds = 0.006 seconds = 6 ms

Seek time = 6ms

Transfer rate = 1 MB/sec

to transfer 0.5 KB

Transfer time = 0.5/1000 seconds = 0.5 ms

Overhead = 1 ms

1. When sectors are contiguos = Seek Time + Rotational Latency + overhead+ 4*transfer time = 6 + 6 + 1 + 4*0.5 = 15 ms

2. Sectors can appears anywhere = 4 * (Seek Time + Rotational Latency + overhead+transfer time) = 4 *(6+6+1+0.5) = 4 * 13.5 = 54 ms