Using disk A specified above, how long will it take to read a file that is 26 KB

Question

Using disk A specified above, how long will it take to read a file that is 26 KB in length located contiguously on one cylinder? Using disk B specified above, how long will it take to read a file that is 26 KB in length located contiguously on one cylinder? During the boot process the computer reads many files. How much faster would the I/O be for a system with disk A be compared to a system with disk B if 500 files of 26KB are read? A 1X CD drive can read 75 blocks/second where each block contains2KB of data. It spins at 500 revolutions per minute (rpm). Faster drives read and spin proportional to their X rating Assume the seek time is 90 ms. How long does it take to read a 26 KB file with a 16X CD reader? If you upgraded to a 48X CD reader with the a 90 ms seek time, how long would it take to read a 26 KB file? How long does it take to read an 800 KB file with the same 16X CD reader?

Explanation / Answer

4. For 16X CD drive, read rate = 75*16 blocks/second = 1200 blocks/second

Spin rate = 500*16 rpm = 8000rpm = 133.33 rotation per second

Rotational latency (i.e. half rotation time) = 1/(2*8000) min = 60/16000 sec = 3/800 sec = 3.75 ms

So, to read a 26KB file, it needs to access 13 blocks as each block contains 2KB file.

Time taken to read one block = 1000/1200 ms = 0.833 ms

So, total time taken to read this file = (number of blocks to be read) * (seek time + rotational latency + read time) = 13*(90ms + 3.75 ms + 0.833 ms) = 1229.5833 ms = 1.23 sec

5. For 48X CD drive, read rate = 75*48 blocks/second = 3600 blocks/second

Spin rate = 500*48 rpm = 24000rpm = 400 rotation per second

So, to read a 26KB file, it needs to access 13 blocks.

Rotational latency (i.e. half rotation time) = 1/(2*400) sec = 1/800 sec = 1.25 ms

So, to read a 26KB file, it needs to access 13 blocks as each block contains 2KB file.

Read time (time taken to read 1 blocks) = (1000)/3600 ms = 0.2778 ms

So, time taken to read this file = (number of blocks to be read) * (seek time + rotational latency + read time) = 13*(90ms + 1.25 ms + 0.2778 ms) = 1189.8614 ms = 1.19 sec

6. For 16X CD drive, read rate = 75*16 blocks/second = 1200 blocks/second

Spin rate = 500*16 rpm = 8000rpm = 133.33 rotation per second

Rotational latency (i.e. half rotation time) = 1/(2*8000) min = 60/16000 sec = 3/800 sec = 3.75 ms

So, to read a 800KB file, it needs to access 400 blocks as each block contains 2KB file.

Read time (time taken to read 1 blocks) = (1000)/1200 ms = 0.833 ms

So, time taken to read this file = (number of blocks to be read) * (seek time + rotational latency + read time) = 400*(90ms + 3.75 ms + 0.833 ms) = 37833.2 ms = 37.83 sec

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