DISK Numerical Ex.(2)

1. If the Disk has average rotational speed of 3000 rpm and seek latency of 16 ms find the Average access time.

Solution:-

Average Access Time or Latency = Seek latency + rotational delay + data transfer delay 

Rotational latency: If it is not given in question then we can get as

                                        1/2 (one rotation time)

So, 

3000 rotation in       1 minute

1 roatation in           =  (1/3000 ) * 60 sec

                                       =  1/50 sec.

So rotational latency = 1/2 *(1/50)= 100 sec. = 10 m-sec.

Average Access Latency = 16 + 10 = 26 m-sec. // here data transfer delay is not given , so neglect it.

 

2. For the following data are given to calculate Average Access time for transferring 250 bytes from disk.

 No. of tracks = 500

 No of sectors per track = 100

 Bytes per sector = 500 B

 Rotation time =600 rpm

Time taken to move from one track to adjacent track is  1 msec. 

Solution:

Average Access time = avg seek time + rotational delay + data transfer delay

let's divide the question into three parts:

1. Calculation of Avg .seek time: 

They are saying that no. of tracks = 500 and time taken to move from one track to adjacent track is  1 msec. 

Avg. time to move from track0 to track 499 = { (500 + 0) / 2} * 1msec

                                                                                    = 250 msec.

2. Calculation of Rotational delay= 1/2(one rotation time)

   600 rotation  in         1 minute

    1 roatation   in          (1/600) * 60 sec. = 100 m-sec

    Rotational delay = 100/2 = 50 msec.

 

3. Data tranfser rate : 

Capacity of track = 100 * 500 = 50000 B

50000 B  in       1 rotation or

50000 B  in        100 msec 

250 B in              (100 / 50000) * 250 = 0.5 msec.

Answer is

So Total latency =  250 + 50 + 0.5 = 300.5 msec 

 

 

                    

 

 

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DISK Numerical Ex. (1)

1. Consider a disk with following specifications:

no. of surfaces = 16

tracks/ surface = 128

sector / track = 256 and 

Bytes / sector = 512 B

 

A. Find the Capacity of Disk.

Solutions:  

The capacity of disk = (no. of surfaces * (tracks/ surface) * (sector / track) *(Bytes / sector)

= 16 * 128 * 256 * 512 = 228 B = 512 Bytes

 

B. Find the No. of bits required to address the sector.

Solution:  Total no. of sectors = 16 *128* 256  =219 

So we need log2(219 ) = 19 bits.

 

C. let the diameter of the innermost track is 21 cm. find the maximum recording density.

Solution: density is associated with tracks.

Density = track capacity /  circumference

So track capacity = (sector / track) *(Bytes / sector)

                                  = 256 * 512 = 217 = 128 KB

circmunference = 2 *π * r

                                 = 2 * (22/7)* (21/2)

                                = 66 cm

So density = 128 / 66 = 1.9 KB /cm.

 

D. let's suppose disk is rotating with 3600 rpm. What is the data transfer rate?

Solution: Please note that In one rotation, we access the data on one track.   what I mean to say is in one complete rotation we read the data equivalent to the capacity of one track. if no of surfaces are given we multiply with it.

So 

3600 rotation in 1 minute 

1 rotataion in 1/3600 minute = (1/3600)*60 sec.

                                                           = 1/60 sec.

now in 1/60 sec. , one rotation is completed and we know that in one rotation we are covering data equivalent to track capacity ie, 128 KB (we have calculated above)

So  

in 1/60 sec.   data covered is 128 KB

in 1-sec          data will be covered = 128 * 60 KB

but they have already, mentioned that there are 16 surfaces so we multiply with 16 

So =>   128 *60* 16 

          = 27 * 24 * 60 * 210 B

          = 120 * 220

         = 120 MBps

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Hard Disk Introduction(1)

Hey buddy, Before going into details let's know some basic terms. 

 

          

Basic Components of the disk:

1. Read/write head or Actuator arm: It is used to read or write data on disk. 

2. Platter: It means the complete circular disk, sometimes also called as plates.

3. Track: Circles are called track.there are a large number of the track on one plate.

  •     No of bytes in outer track = no of bytes in the inner track
  •     It all happens because of density variation.

As we go toward the outside track, density decreases.

let's suppose track capacity is Tc. Capacity means how much data it can store.

We all know that density means volume per distance.

So track density = Capacity of track / Circumference.

let us say radius is 'r' So cicrcumfernece= 2 π r

density=  Tc / 2 π r

now clearly we are seeing that

density  ∝  1/ r  .ie. on the increasing radius, density decrease and when we go outside, radius increases. So it is proved that As we go outside, density decreases.

4. Sector: Track is divided into equal partition called sectors. Normally the size of the sector is 512 KB.

Some latency associated with Disk:

Latency means time. Sometimes they mention latency and sometime time. both are the same thing.

1. Seek latency:

Normally when we have to read first we have to bring the read /write head to that track. The Time taken to move read/write head from one track to another is called Seek latency. 

To reduce seek time always we should read or write sequentially.

2. Rotational delay:

  The Time taken to wait for a particular sector to come under read/write head.

 Data transfer time mainly depend on Rotational delay.

Note : 

 Please remember this sequence it will help you to solve numerical.

Surface -> track -> sector -> byte 

Some formulas:- 

1. Size or capacity of hard disk =  no of surfaces *  tracks per surface  *  sector per track  * size of the sector.

2.Size or capacity of hard disk = Density  * circumference. 

3. No. of tracks  =    recording width  / inner track space.

4. Recording width = (outer disk diameter - inner disk diameter ) /

  

 

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