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Consider a disk with seek time of 4 ms, rotation speed of 15,000 rpm and 512-byte sectors with 500 sectors per track. Suppose that we wish to read a file consistinh of 2500 sectors for a total of 1.28 Mbytes. We would like to estimate the total time for the transfer?
asked in Operating System by (75 points)   | 30 views
26 ms????
Answer will depend on how files are stored in the disk
Me too think 26ms..

2 Answers

+1 vote
We just want to find transfer time

15000 revolution in 60 sec

1 revolution in $\frac{60}{15000}$ or

1 revolution in 4 mili second

1 revolution = covering one entire track = 500 sector

500 sector = 4 mili second

1 sector = 8 micro second

Now there are 2 different things 1) 2500 sectors so time = 2500* 8 mili seconds = 20 mili second

                                                    2) 1.28 MB= 1342177.28 Bytes or 2621.44 sectors = 2622 sectors = 20.976 mili seconds

Total time  case 1) 4 + 2 +20= 26 ms or case 2) 4+2+20.976=26.976 ms
answered by Boss (6.6k points)  
0 votes
Given data
Seek time = 4ms (considering its average seek time)
Rotation Speed = 15,000 RPM
Number of Sector = 500
Size of each sector = 512 B
We have to read 2500 Sector which is approx 1.23 MB
We know Disk access time = Avg Seek time + Avg rotation latency + Transfer Time

15,000 Rotation require 60 Sec
1 Rotation we can do in 4 ms
So avg rotation latency = 2 ms

Case 1 : Data are store in continuous way

To read 2500 sector we need to read 5 track and in 1 rotation we are reading 1 track
So to read 5 track we need 5* 4 ms  = 20 ms
So transfer time will be 20ms
Since data is store in continuous manner we only have to deal with Avg seek time and Avg rotation latency for first time
So disk access time  = 4+2+20 ms  = 26 ms

Case 2 : Data store in random way

As we know 1 track or 500 sector can be read in 4 ms
So 2500 sector we can read in 20 ms or 0.02 sec
Since data is store in random way every time we have to deal with seek time = 2500 * 4ms = 10,000 ms or 10 sec
Similarly every time we have to deal with rotation latency  = 2500 * 2ms = 5,000 ms or 5 sec
So total disk access time = 0.02 sec + 10 sec + 5 sec = 15.02 sec
answered by Active (1.9k points)  

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