1. A disk has a average seek time of 4ms (milliseconds), a rotational speed of 5
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Question
1. A disk has a average seek time of 4ms (milliseconds), a rotational speed of 5400 rotations per minute. Each track on the disk has 16 sectors, each of 1Kbyte, and the transfer from disk to main memory proceeds at the rate of 20Mbytes/second.
a. What is the average access time of this disk for reading one sector?
b. A 1.2Ghz computer containing the above disk loads a program of size 10Kb stored in sequential sectors on the same track. No disk cache is present. Thereafter it executes the program for 600 million cycles. What percentage of the total (disk + CPU) time is spent accessing the disk? Assume an average seek time and rotational latency.
Explanation / Answer
A disk has an average seek time of 4ms
A rotational speed of 5400 rotations per minute that 900 rotations per second
10000/900 = 11.11111 ms
T_rotation = 11.111111/2 = 5.555556
Each track on the disk has 16 sectors and each of 1Kbyte
The transfer from disk to main memory proceeds at the rate of 20Mbytes/second
Rotation (R) Time is 1/R, or the inverse of the Rotation Speed.
If the disk spins at 900 rotations per second, the rotation time is 1/900 seconds that nearly 0.001111.
Latency (L) is average latency is one half the time it takes to rotate the disk (Rotation Time).
If the rotation time is 0.001111 seconds, the average latency is 0.000556 seconds.
Sectors to Read: the number of sectors to read is the total number of bytes needed divided by the number of bytes per sector.
If 1KB for each sector that there are 16 sectors 16KB.
Transfer Time: time to transfer one sector, times the number of sectors needed.
Avg time to read=T seek+ T_rotational+ T_Transfer+ T_Controller
so the avg time to read 0.139 msec
average seek time of 4ms
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