Start with the same assumptions as in Exercise 7.8...

Start with the same assumptions as in Exercise 7.8. Nowcalculate the performance in both I/Os per second and megabytes persecond of these two disk organizations assuming the request patternis reads of 4 KB of sequential sectors where the average seekdistance is 10 tracks. Assume the 4KB are aligned under the samearm on each disk in the array.

Answers ( 1 )

  1. marcel thiele
    marcel thiele
       Dear user,
               Fallacy states that seek time orseek Distance,
         Average seek distance = 10Tracks = 5280 sectors
              Average seek Time = 0.9 msec.
            Disk   rotationallow   access to = (5280*10000)/60
                                                            =88000 sectors/sec
                  Time needed to read 4 KB = 0.09 msec.
       Disk access Time ( for singledisk)
               = Average seek Time + Average rotational Delay +Transfer Time + Controller overhead
               = 0.9+ ( 0.5/10000RPM) +0.3+0.09
              =1.29msec
             1 IO(single)= 1/1.29 = 0.77 IO/sec almost equals to 1 IO/sec
          you can calculate for diskarrays
            The answer for diskarrays is 4.23msec and
                     IO/sec for arrays is 0.24 IO/sec.
            If you get anydoubt post the question.

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