disk management

Question

my approach 
seek time 30
rotationn latency=10
tranfer time :
64 kb prog page size is 4kb so in 16 pages this prog wiil be accesed through memory since track size is given 32 kb
so in 32kb is read in one rtotaion in 20msec
so 4kb --------------------------------(20/32)*4 
=2.5
for 1 page it take transfer time as 2.5msec as program is spread in pages so total acces 16 pages we need totoal time as :
(30+10+2.5)*16=680 msec
correct me !! if wrogn thanks..given ans by them is 120.

Answer 1

one track size = 32 MB

number of tracks required to store the data = 64MB/32 MB = 2 Track.

Seek time : time require to move the R/W Header to desired track

Rotational Latency : placing the R/W from data is beginning

Now Total time taken for the first track =  Seek Time + Rotation latency+ Transfer time

                                                                =  30 ms  +  10 ms+ 20 ms

                                                                =  60 ms

After reading the complete one track again, there will a need to place R/W header at the track in which next 32MB data present. So for this,it will again perform seek operation and find the location from where the next data is present.

Total time taken to transfer the second track =  Seek Time   +  Rotation latency +  Transfer time

                                                                         =   30 ms  + 10 ms  +  20 ms

                                                                         =   60 ms

Total time = 60 + 60 = 120 ms

Comments

question has said about pages as program is spread in pages,that thing why we are not including during solving.??

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It may be you confusion is, "each pages are stored in different tracks i.e only 1 page memory is contiguous. Thats why you taking transfer of one page and multiplying it with total number of pages in the program."

See the Question says, "Pages of the programs are contiguously placed on Disk." & One complete page always stored in the memory contiguously by paging scheme only. So all pages of the program  means i.e 16 pages are stored contiguously.  I have taken a all possible pages in a track but for pages which are there in the second track we will have to search it again means it will take seek time + rotational latency.

PS : complete page always stored in the memory contiguously by paging scheme only.

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Since they said Continuously so why to consider seek time again ???