Question

How long does it take to load a 64-KB program from a disk whose average seek time is 10 msec., whose rotation time is 20 msecs., and whose track holds 32-KB for a 2-KB page size?(How to solve this?)

Comments

@shivani i also think seek time ,rotational latency should be considered again.

according to me  contiguous mean pages will be stored in the contiguous manner in one track.After reading the one track completely .It has to place R/W header in next track so that it can R/W .So seek time and rotational latency should considered again.

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Ok fine so for this question if contiguous is mentioned then access time is 80msec right

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yes i think so.

Answer 1

1) continous pages

2) random pages

continous pages

seek time = 10ms 

rotational latency= half of the 1 rotation time = 20/2 = 10

transfer time => 1 rotation time we can transfer------> total size of track

                                ?------------------------------------------------> required data

                             20------------------------------->32KB

                             X--------------------------------->2KB* 32  (as continous pages are there we need not to send them 1 by 1...in one                                                                                                                 attempt one by one we can send 32 pages)          

X=   40           

                                so total transfer time = (seek time+rotational latency + transfer time)

                                                          =(10+10+40)= 60

now...u r getting...80 ///i will tell why...u r adding seek time+ rotational latency twice....like

10+10+20////for first track & 10+10+20 for second track and in two track whole data will be transfered...

but seek time+ rotational latency it will be added only once....bcoz once u reach the position from where ur data starts....the work of  seek time+ rotational latency ends there only...need not to be considered twice..

now random)....all answered correct for random///keep it up i will copy it from akriti's  answer below..

seek time = 10msec

rotation latency = 20msec/2=10msec

now one track contains 32/2 i.e 16 pages..and we have to load 64/2 i.e 32 pages..

16 pages i.e  1 track needs 20msec

16  pages-> 20msec

1 page ->20/16 msec

to tranfer 32 pages,we need 32(10+10+20/16) i.e 680 msec

Comments

shivani see the images of track & sectors...first then ask ur doubt...///i knw wat ur question is but first see it...and read the defination of seek time & rotational latency

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@Akriti Check out these 2 questions once

https://gateoverflow.in//30488/the-total-time-reuired-to-load-the-program-from-disk

https://gateoverflow.in//33540/disk-management

They considered as what I said 2 times seek time and rotational latency

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i too have same doubt… can you plzz help if you are now clear with it?? @shivanisrivarshini

Answer 2

IT IS NOT MENTIONED THAT THE SECTORS ARE CONTINIOUSLY DISTRIBUTED IN THE TRACK ..SO IT MAY BE THE CASE THAT THE SECTORS ARE RANDOMLY DISTRIBUTED

SO TO ACCESS A SECTOR WE NEED 

1.MOVE ARM HEAD TO THE TRACK(ST)

2.PLACE THE SECTOR UNDER THE HEAD(RL)

3.READ OR WRITE THE SECTOR

IN ONE ROTATION =1 TRACK=32KB

20MS=32KB

1KB=20/32MS

2KB=(20/32)*2=1.25 MS

SO TO ACCESS ONE SECTOR WE NEED =10+10+1.25=21.25MS

NUMBER OF SECTOR=64KB/2KB=32 NUMBER OF SECTORS

THEREFORE THE COMPLETE TIME= 32*21.25=680MS

 

Comments

what do you exactly  mean by continous distribution of pages

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read this

http://www.utdallas.edu/~muratk/courses/db08files/parameters_of_disks.pdf

 

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How long does it take to load a 64−KB64−KB program from a disk whose average seek time is 5msec,5msec, whose rotation time is 5msec,5msec, and whose tracks hold 1MB1MB

1. for a 2−KB2−KB page size?
2. for a 4−KB4−KB page size?

The pages are spread randomly around the disk and the number of cylinders is so large that the chance of two pages being on the same cylinder is negligible.

Could you please answer this QST

Answer 3

64 KB program will be organized into 2 tracks because of each track capacity is 32KB.

To load entire track we require 20msec. To load 2KB we require 1.25 msec.

I/O time =seek time+avg.rotation latency+transfer time

                 10msec+10msec+1.25msec=21.25msec

                Since 64KB program is organized into 2 tracks then I/O time will be 2(21.25)=42.5 msec