The Gateway to Computer Science Excellence

+37 votes

Consider a disk pack with a seek time of $4$ milliseconds and rotational speed of $10000$ rotations per minute (RPM). It has $600$ sectors per track and each sector can store $512$ bytes of data. Consider a file stored in the disk. The file contains $2000$ sectors. Assume that every sector access necessitates a seek, and the average rotational latency for accessing each sector is half of the time for one complete rotation. The total time (in milliseconds) needed to read the entire file is__________________

+61 votes

Best answer

Since each sector requires a seek,

Total time $= 2000$ (seek time $+$ avg. rotational latency $+$ data transfer time)

Since data transfer rate is not given, we can take that in $1$ rotation, all data in a track is read. i.e., in $60/10000 = 6$ ms, $600 \times 512$ bytes are read. So, time to read $512$ bytes $= 6/600$ ms $= 0.01$ ms

$= 2000 \times (4 \ ms + 60 \times 1000 /2 \times 10000 + 0.01) $

$= 2000 \times (7.01 \ ms)$

$= 14020 \ ms. $

0

i am nt getting that part of ur calculation 60 * 1000 /2* 10000 .... plz reply as fast as possible...

0

@Arjun sir to calculate total time all time we have to be consider data transfer time ??. weather it is mention or not . but in case of seek time if it is not given we assume it as 0 why is it so ??

+9

@puja just see that

1track → 600 sectors

6ms←600 sectors (1 rotation means 600 sectors (or) 1 t

2000 sector → (2000) 0.01 = 20 ms

∴total time needed to read the entire file is:

= 2000 (4+3) +20= 14020 ms . hope it will help u

1track → 600 sectors

6ms←600 sectors (1 rotation means 600 sectors (or) 1 t

2000 sector → (2000) 0.01 = 20 ms

∴total time needed to read the entire file is:

= 2000 (4+3) +20= 14020 ms . hope it will help u

0

I have a doubt

for data transfer time

600 sector time taken 6ms

1 " " " .01ms

**then why r we not calculating**

**2000 sector time taken 20ms?**

why we r calculating with respect to total **2000 * (seek time + avg. rotational latency + data transfer time) **

0

if am doing in track wise ... 1 track ---> 6ms then 4 tracks (2000/600~= 4 ) then ans is 14024 , was there range for this ?

sresth as you know 6ms is for 1 track means 600 sectors and we have to find total time for 2000 sectors

sresth as you know 6ms is for 1 track means 600 sectors and we have to find total time for 2000 sectors

0

https://gateoverflow.in/3479/gate2007-it-44-isro2015-34

https://gateoverflow.in/761/gate2001-20

Confusing me so much :( :(

+1

@sid1221

your answer is differing because you are assuming that 2000 sectors are tightly filled in 4 tracks...but it can be case that all 2000 sectors are stored in 2000 different tracks.

your answer is differing because you are assuming that 2000 sectors are tightly filled in 4 tracks...but it can be case that all 2000 sectors are stored in 2000 different tracks.

0

yes that i know as nothing is given , then you can do this method also ....that why i was asking there was range for this

srestha i was also not satisfied for these 2 questions ..avg seek time :(

srestha i was also not satisfied for these 2 questions ..avg seek time :(

+2

Assume that every sector access necessitates a seek, and the average rotational latency

Means we randomly accessing different sectors. So every time seek and rotational latency will come into picture.

Hi @srestha ji,

What is your doubt ?

+1

@Arjun Sir,

Can we do like this below

Since T_{avg}= T_{seek} + T_{rotational} +T_{Data}_{ Transfer}

For every sector we require a seek and rotational delay is half the time to complete one revolution.

so for 2000 sectors the total seek and rotational delay is

2000*(4+3) =14000ms.

Now, Since a track consists of 600 sectors,

so

600 sectors are read in --------> 6msec

So, 2000 sectors of the file will be read in 20msec.

So, total time =14000+20=14020 ms.

0

Because we have to consider data transfer time while calculating total time.

You can calculate DTR using given RPM and sector size

You can calculate DTR using given RPM and sector size

0

Assume that every sector access necessitates a seek, and the average rotational latency for accessing each sector is half of the time for one complete rotation.

If this condition not given then ..how we will calculate time get data for 2000 sectors of file

+32 votes

Seek time (given) = 4ms

RPM = 10000 rotation in 1 min [60 sec]

So, 1 rotation will be =60/10000 =6ms [rotation speed]

Rotation latency= 1/2 * 6ms=3ms

# To access a file, total time includes =seek time + rot. latency +transfer time

TO calc. transfer time, find transfer rate Transfer rate = bytes on track /rotation speed

so, transfer rate = 600*512/6ms =51200 B/ms

transfer time= total bytes to be transferred/ transfer rate

so, Transfer time =2000*512/51200 = 20ms

Given as each sector requires seek tim + rot. latency = 4ms+3ms =7ms

Total 2000 sector takes = 2000*7 ms =14000 ms To read entire file ,total time = 14000 + 20(transfer time) = 14020 ms

RPM = 10000 rotation in 1 min [60 sec]

So, 1 rotation will be =60/10000 =6ms [rotation speed]

Rotation latency= 1/2 * 6ms=3ms

# To access a file, total time includes =seek time + rot. latency +transfer time

TO calc. transfer time, find transfer rate Transfer rate = bytes on track /rotation speed

so, transfer rate = 600*512/6ms =51200 B/ms

transfer time= total bytes to be transferred/ transfer rate

so, Transfer time =2000*512/51200 = 20ms

Given as each sector requires seek tim + rot. latency = 4ms+3ms =7ms

Total 2000 sector takes = 2000*7 ms =14000 ms To read entire file ,total time = 14000 + 20(transfer time) = 14020 ms

+1 vote

Seek time = 4ms

60s → 10000 rotations

Rotation Time :- (60 / 10000) = 6 ms (in 1- rotation)

Rotational latency :- (1 / 2 )6ms = 3ms

1track → 600 sectors

6ms←600 sectors (1 rotation means 600 sectors (or) 1 t

2000 sector → (2000) 0.01 = 20 ms

∴total time needed to read the entire file is:

= 2000 (4+3) +20

=8000+6000+20 = **14020**

0 votes

**In the question, we have to Find the time required to read / transfer the data (2000 sector /file ) data from disk **

**So we have to find the time required to read /transfer for 1 sector data than we find 2000(time required for one sector).**

so time for one sector is Total time =2000 (seek time +avg. rotational latency + data transfer time)

0 votes

Seek time $ =4ms$

In 60 seconds 1000 rotations.

Rotation Time$ = (60 / 10000) = 6 ms $ (Time to read one Track)

Rotational latency$ = (1 / 2 ) * 6ms = 3ms $

Transfer time = No. of tracks * time to read one track

$= $(Sectors in file/Sectors in one track) * (Time to read one track)

$= (2000/600)*6

= 20$ ms

total time needed to read the entire file is:

$= 2000*(4+3) + 20

= 8000 + 6000 + 20 = 14020$ ms

So, correct answer is **14020**.

–5 votes

SeekTime = 4ms

OneRotationalLatency = (60/10000) = 0.006s = 6ms

AverageRotationalLatency = 0.5 * 6 = 3ms

1 TrackAccessTime = 1 Rotation

4 TrackAccessTime = 4 * OneRotationalLatency = 24ms

600 SectorsAccessTime = 1 Rotation

600 SectorsAccessTime = 6ms

1 SectorAccessTime = 6/600 = 0.01ms = 10 ms

FullTracksToBeAccessed = floor(2000/600) = 3

ExtraSectorsToBeAccessed = 2000 - (600*3) = 200

TotalTime =

(4 * TrackAccessTime) + (200 * SectorAccessTime)

= (4 * 24) + (10*200) = 2096ms

OneRotationalLatency = (60/10000) = 0.006s = 6ms

AverageRotationalLatency = 0.5 * 6 = 3ms

1 TrackAccessTime = 1 Rotation

4 TrackAccessTime = 4 * OneRotationalLatency = 24ms

600 SectorsAccessTime = 1 Rotation

600 SectorsAccessTime = 6ms

1 SectorAccessTime = 6/600 = 0.01ms = 10 ms

FullTracksToBeAccessed = floor(2000/600) = 3

ExtraSectorsToBeAccessed = 2000 - (600*3) = 200

TotalTime =

(4 * TrackAccessTime) + (200 * SectorAccessTime)

= (4 * 24) + (10*200) = 2096ms

- All categories
- General Aptitude 1.9k
- Engineering Mathematics 7.5k
- Digital Logic 2.9k
- Programming and DS 4.9k
- Algorithms 4.4k
- Theory of Computation 6.2k
- Compiler Design 2.1k
- Databases 4.1k
- CO and Architecture 3.4k
- Computer Networks 4.2k
- Non GATE 1.4k
- Others 1.4k
- Admissions 595
- Exam Queries 573
- Tier 1 Placement Questions 23
- Job Queries 72
- Projects 18

50,737 questions

57,322 answers

198,393 comments

105,143 users