MGMT 1050 Lecture 22: MGMT 1050 Lecture 22 Notes
MGMT 1050 Lecture 22 Notes – Revolutions per Minute
Introduction
• For a typical hard disk rotating at 3600 revolutions per minute, or 60 revolutions per
second, the average latency is average latency = 1 2 × 1 60 = 8.33 milliseconds Once the
sector is reached, the transfer of data can begin.
• Since the disk is rotating at a fixed speed, the time required to transfer the block, known
as transfer time, is defined by the number of sectors on a track.
• Since this establishes the percentage of the track that is used by a single data block.
• The transfer time is defined by transfer time = 1 number of sectors × rotational speed
• If the hard drive in the example contains 30 sectors per track, the transfer time for a
single block would be transfer time = 1 30 × 60 = 0.55 milliseconds
• A table of typical disks of different types, comparing various characteristics of the disks.
• Since the total time required to access a disk block is approximately the sum of these
three numbers, a typical disk access might require 20 to 25 msec.
• To put these speeds in perspective, consider that the typical modern computer can
execute an instruction in less than 1 nanosecond.
• Thus, the CPU is capable of executing millions of instructions in the time required for a
single disk access.
• This should make it very clear to you that disk I/O is a major bottleneck in processing
• Also that it is desirable to find other work that the CPU can be doing while a program is
waiting for disk I/O to take place.
• An expansion of part of a track to show a single data block is shown
• The block consists of a header, 512 bytes of data, and footer.
• An inter block gap separates the block from neighboring blocks.
• The layout of the header for a Windows-based disk.
• The track positions, blocks, and headers must be established before the disk can be
used.
• The process to do this is known as formatting the disk.
