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Consider a disk with a sector size of $512$ bytes, $50$ sectors per track, $2000$ tracks per surface and five double-sided platters (i.e. $10$ surfaces). The disk platters rotate at $5400$ rpm. The average seek time is $10$ msec. A block size is chosen as $1024$ bytes.

A file containing $100,000$ records of $100$ bytes each is to be stored on this disk, and no record is allowed to span two blocks. Then, the rotational latency of the disk is ________ seconds.

@Niharika 1   and  @rajesh170293

If the disk platters rotate at 5400rpm, the time required for one complete rotation, which is the maximum rotational delay, is 1/5400 * 60 = 0.011 seconds

The average rotational delay is half of the rotation time 0.011/2 =  0.0055

Formula: max rotational delay = (1/p) min = (60/p) sec

rotational delay  = ½ * cost of 1 revolution = ½ * (60/p) sec

Sir how A block size is chosen as 1024 bytes. I know that block size is equal to sector size and sector size is already mention as 512 byte.(I know these data not neccesary to answer this question but full question does not make any sence please comment)

Block is a group of sectors that the operating system can address (point to).

A block might be one sector, or it might be several sectors (2,4,8, or even 16). The bigger the drive, the more sectors that a block will hold.

Block is like a group of bytes handled, stored, and accessed as a logical data unit, such as an individual file record.  e. g. the first 1024 bytes of a file.

Sector : When a disk is formatted, tracks are defined (concentric rings from inside to the outside of the disk platter ) . Each track is divided into a slice, which is a sector. On hard drives and floppies, each sector can hold 512 bytes of data.

A sector is the smallest addressable unit of storage on a disk. Tracks are divided into sectors, with each sector 512 bytes long.

They contain data, but also contain information as to where the data is located, among other useful bits of information.

Hope now it is clear Why sector size is 512 B and Block size is 1024 B . In this question a Block have 2 sectors.

see the clear picture in reference [ 1 ] .

Reference:

Formula

maximum rotational delay = (1/p) min = (60/p) sec

rotational delay  = ½ * cost of 1 revolution = ½ * (60/p) sec

If the disk platters rotate at 5400rpm, the time required for one complete rotation, which is the maximum rotational delay, is 1/5400 * 60 = 0.011 seconds

The average rotational delay is half of the rotation time 0.011/2 =  0.0055

by

@Vaishali Jhalani

just click in this link http://www.utdallas.edu/~muratk/courses/db08files/parameters_of_disks.pdf

you will find that it always consider average rotational delay ..

you can also   CLICK here..

@Kaluti

see below slide ,

Average disk access time = Average seek time + Average Rotational delay + Transfer time + controller overhead + Queing delay

http://www.csc.villanova.edu/~japaridz/8400/sld012.htm

http://cssimplified.com/assignments/what-is-rotational-latency-in-the-context-of-disk-access-time-ignou-mca-assignment-2014-15

when calculate access time it is always average case Not total.

total access time = 3333.549 sec chech it