From Schaum's OS (7)

[wood]

This one is simple but interesting... so I'm posting.

7.20 A disk has 8 sectors per track and spins at 600 rpm. It takes the controller 10 ms from the end of one I/O operation before it can issue a subsequent one. How long does it take to read all 8 sectors using the following interleaving systems?

(a) No interleaving
(b) Single interleaving
(c) Double interleaving

[ambreenjava]

Okay,
Interleaving is when sectors are arranged in alternate order.

(a) when there is no interleaving

then

sectors would be like this

s1|s2|s3|s4|s5|s6|s7|s8

Since the disk revolves at 600 rpm then
it means it revolves at 10 revolutions per second and 1 revolutions per milli sec.

That means 100/8 = 12.5 millisec is spent on each of the 8 sectors

and the time it takes to load another request is 10ms

thats less than 12.5

In first request it read sector s1
then while the disk is spinning the controller went to fetch another request. And 10msec are waster out of 12.5 that means more than half of the s2 had already passed therefore the heads to rotate again to fetch s2. So will be the case of all 8 sectors meaning it will take

8 revolutions while each revolution takes 100 millisec and that would be 800 msec

(b)

the sectors will be arranged in the order

s1|s5|s2|s6|s3|s7|s4|s8

therefore in 1 rev 4 sectors will be read

and in 2 rev 8 sectors will be read

100 msec in 1 sec therefore 200 msec in 2 sec

(c)

sectors would be

s1|s4|s7|s2|s5|s8|s3|s6

here only 3 sectors will be read in 1 rev.

that means 300 msec for a total of 3 rev.

but 360/8 is 45 degrees
thats is only 3/4 of the disk rev is required in the 3rd rev
so we need to multiply 100*3/4 to get the actual time needed to read the last sector s8. since it comprises only 3/4

That means 275 msec is required

[wood]

Good job ambreenjava! You got it all right!

[The_Wonderful_Vipul]

Originally posted by ambreenjava

Okay,
Interleaving is when sectors are arranged in alternate order.

(a) when there is no interleaving

then

sectors would be like this

s1|s2|s3|s4|s5|s6|s7|s8

Since the disk revolves at 600 rpm then
it means it revolves at 10 revolutions per second and 1 revolutions per milli sec.

That means 100/8 = 12.5 millisec is spent on each of the 8 sectors

where did this 100 ms come from ?? so that you could say 100/8 ms ??

and the time it takes to load another request is 10ms

thats less than 12.5

In first request it read sector s1
then while the disk is spinning the controller went to fetch another request. And 10msec are waster out of 12.5 that means more than half of the s2 had already passed therefore the heads to rotate again to fetch s2. So will be the case of all 8 sectors meaning it will take

8 revolutions while each revolution takes 100 millisec and that would be 800 msec

(b)

the sectors will be arranged in the order

s1|s5|s2|s6|s3|s7|s4|s8

therefore in 1 rev 4 sectors will be read

and in 2 rev 8 sectors will be read

100 msec in 1 sec therefore 200 msec in 2 sec

(c)

sectors would be

s1|s4|s7|s2|s5|s8|s3|s6

here only 3 sectors will be read in 1 rev.

that means 300 msec for a total of 3 rev.

but 360/8 is 45 degrees
thats is only 3/4 of the disk rev is required in the 3rd rev
so we need to multiply 100*3/4 to get the actual time needed to read the last sector s8. since it comprises only 3/4

That means 275 msec is required

Vipul.

[AlbaLed]

> where did this 100 ms come from ?? so that you could say 100/8 ms ??

10 revs per second => 100 ms per revolution therefore time to read one sector or 1/8th of the track = 100/8