Please Give Formulas on how to get answers. I cant find it in my book and need t
ID: 3568710 • Letter: P
Question
Please Give Formulas on how to get answers. I cant find it in my book and need to know how to do it for future classes.
Suppose that a certain disk drive has the following characteristics:
Rotational speed: 5400 rpm, Average seek time: 5 ms, Minimum seek time: 0.5 ms
500 sectors per track, with each sector holding 2048 bytes of data
1. Assuming that a 50 MB file is stored contiguously on the disk, how long would it take to read the entire file into memory?
2. Assuming that a 50 MB file is stored on random sectors scattered all over the disk, how long would it take to read the entire file into memory?
Explanation / Answer
a)
The rational speed is 7200rpm=120rps.
Each track contains=500 sectors,
Each sector is of size=2048 bytes.
Average seeking time =5ms.
Minimum seeking time=0.5 ms.
Inorder to read the 2048 bytes of data, the accurate time to transfer the data is given as,
Average Disk Access Time=average rotation time+average seek time+data transfer time+
controller overhead time.
Ignore the controller overhead time or estimate it to 1ms.
The actual seeking time =average seeking time*2+minimum seeking time*2
=5ms*2+0.5*2
=10+1ms
=11ms.
Each track contains 500 sectors, total data contained in each track is
=number of sectors*size of each sector
=500*2048
=1024000bytes
=1024kB
Therefore, each tack contains 1024kB of data.
Data transfer rate per second is,
=rational speed*1 track data
=120rps*1024kB
=122,880kB/s
=122.88MB/s
So, in order to read the 50MB data the number tracks would be,
1 track contains 1024 kB=1.024MB
? tracks contains 50MB data=> 50/1.024= approximately 50 tracks
Rotation time = number of tracks/120rps=50/120=0.416s.=416ms
Therefore, access time to read bytes of data is,
=416ms+5ms+50MB/122.8MB*0.5ms
=416+5+0.23ms
=421ms
b)
Since, the access time for 50MB is 421ms,
Therefore, the average access time is 210.5ms
Random access time is=seek time+average access time+1(120*500)
=11m+210.5ms+0.000016ms
=221.5
The data transfer rate per second is=1/(0.2215)=4.514
So each transfer per second is 122.8MB
4.514*122.8
554MB/s