PAGING

Question

A Program of size 64MB is stored on disk which supports an average seek time of 30ms and rotation time of 20ms.Page size is 4MB and track size is 32MB. if the pages of program are contiguosly placed on disk then what would be the total time required to load the program from disk?

Answer 1

Tseek + Trot + Ttrans

2(30  + 10 )+ 2.5*16=120ms

Comments

why you have multiplied (30+10) by 2??

Answer 2

Seek time = 30 ms
Rotation Time = 20 ms
Rotation Latency = $\frac{20}{2}$

Transfer rate = 20ms---------------32MB(1 track)
                       1ms                        ?                   
          
                  = $\frac{32}{20}$ = 1.6MB/ms
For one page 1ms ------------------1.6MB
                       ?                        4MB

                  =   $\frac{4}{1.6}$ = 2.5ms
Total Time for one page = 30 + 10 + 2.5 = 42.5ms

Total number of page = $\frac{64}{4}$ = 16

Total time for 16 page = 16*42.5 = 680ms

Comments

i dont know its right or wrong
but i did it like time taken to transfer one page Xms
then time to take transfer total page of program?
which is mentioned above.

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I think they mention "contiguously" so, this fact should be taken into consideration

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https://gateoverflow.in/28495/calculating-disk-access-time