edited by
17,535 views
44 votes
44 votes
Consider a typical disk that rotates at $15000$ rotations per minute (RPM) and has a transfer rate of $50 \times 10^6$ bytes/sec. If the average seek time of the disk is twice the average rotational delay and the controller's transfer time is $10$ times the disk transfer time, the average time (in milliseconds) to read or write a $512$-byte sector of the disk is _____
edited by

4 Answers

Best answer
83 votes
83 votes

Average time to read/write $=$ Avg. seek time $+$ Avg. rotational delay $+$ Effective transfer time

Rotational delay $= \frac{60}{15}$ = $4$ ms

Avg. rotational delay $= \frac{1}{2} \times 4$ = $2$ ms

Avg. seek time $= 2 \times 2$ = $4$ ms

Disk transfer time $= \frac{512 \text{ Bytes}}{50*10^6 \text{ Bytes/sec}}$ = $0.0102$ ms

Effective transfer time $= 10 \times $  disk transfer time $=$ $0.102$ ms

So, avg. time to read/write $=$ $4 + 2 +0.0102 +0.102$ $= 6.11$ ms  $\bf{\approx \;\; 6.1}$ ms 

Reference: http://www.csc.villanova.edu/~japaridz/8400/sld012.htm

edited by
4 votes
4 votes

Avg rotation latency :

$15,000\ rot\rightarrow60sec$

$1\ rot\rightarrow\ ?$

$1\ rot=4ms$

$\dfrac{1}{2}rot\rightarrow2ms$


$Avg\ Seek\ time=2\times Avg\ rotation\ latency$

$Avg\ Seek\ time=4ms$


$Data\ transfer\ time:$

$1\ sec\leftarrow 50\times 10^6B$

$?\leftarrow 512B$

$0.0102ms$


$Controller's\ transfer\ time=10\times data\ transfer\ time$

$Controller's\ transfer\ time=0.102ms$


Total$=6.1ms$

Answer:

Related questions

38 votes
38 votes
5 answers
3
36 votes
36 votes
6 answers
4
go_editor asked Feb 13, 2015
15,523 views
The number of states in the minimal deterministic finite automaton corresponding to the regular expression $(0+1)^* (10)$ is _____.