Avg. disk access time.

Question

What is the average access time for transferring 512 bytes of data with the following specification :

Avg. seek time = 5ms

Disk Rotation = 6000rpm, 40kbps

Controller Overhead = 0.1ms

Answer 1

Tavg=seektime+rotational latency+data transfer time+controller overhead

DISC ROTATION=6000RPM

                        =>6000 RORATION=1MIN=60 SEC

                        =>1 ROTATION=60/6000=10MSEC

SO RL=5 MSEC(ON AVG WE TAKE HALF )

DATA TRANSFER=40Kbps

40 kb=1sec

1b=1/40k sec

512 b=1/40k*512=12.8msec

THEREFORE Tavg=5+5+12.8+0.1=22.9msec

Comments

thnx, got it

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In the questioon 512 Byte is given, not 512 b. So, the answer will be 110.1

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The speed is given in Kilo bits and the data is in Bytes. So transfer time should be (512*8)/40 milli seconds.

Answer 2

Time taken for one full rotation = 60/6000 sec

                                                        = 1/100 sec = 0.01 sec = 10 msec

Average rotational delay = ½ x Time taken for one full rotation

                                            = ½ x 10 msec = 5 msec

Transfer time = (512bytes / 40 kb) sec

                         = 0.0125 sec = 12.5 msec

Average access time = Avg. seek time + Avg. rotational delay + Transfer time + Controlled overhead + Queing delay

                                      = 5 msec + 5 msec + 12.5 msec + 0.1 msec + 0

                                      = 22.6 msec