1. Note: base 2 numbers. 1 MB = 1024 KB SSD controller overhead is 0.1 ms per fi

Question

1. Note: base 2 numbers. 1 MB = 1024 KB

SSD controller overhead is 0.1 ms per file
HD controller overhead is 0.1 ms per file
HD average seek time is 5 ms
HD average rotational delay is 4 ms
Transfer time (both HD & SSD) for a 4kb block of data requires 0.26 ms

a) What is the time required to transfer a single 4 megabyte file on this hard drive? (in ms, round to a whole number)

b) What is the time required to transfer 1000 4 kylobyte files on this hard drive? (in ms, round to a whole number)

c) What is the time (in ms) required to transfer a single 4 megabyte file on this SSD? (in ms, round to a whole number)

d) What is the time required to transfer 1000 4 kylobyte files on this SSD? (in ms, round to a whole number)

Explanation / Answer

Block Size =4Kb

Transfer Time =0.26 ms

a) Transfer Time= HD seek Time + HD rotational delay + transfer time + SSD controller overhead + HD controller Overhead

=5+4+(4*1024 KB(File Size)*0.26/4KB(Size of Block)+0.1+0.1

=9.2+(1024*0.26)

=275 ms

b) Transfer time for 1 block of 4Kb=5+4+0.26+0.1+0.1

=9.46 ms

So total time to transfer 1000 blocks=1000*9.46

=9460 ms

c)Transfer rate for SSD=0.26 ms/4Kb

hence transfer time for 4 Mb file= Transfer Time+SSD overhead

4*1024 kb*(0.26ms/4Kb)+0.1

=266.34ms

d)Transfer time for one block of 4Kb= 0.26 ms(Transfer Time) +0.1 ms (SSD overhead)=0.36 ms

Transfer time to 1000 blocks =1000*0.36 =360 ms

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