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# disk scheduling -FCFS, SCAN, CSCAN, SSTF

I know how these method works, but I'm woundering,
if the request queue is like this: 15, 33, 45, 8, 10, 75, 15, 33
So if 50% of the request are for the fixed cylinders, which method{fcfs, scan, cscan, sstf} is likely to have a better performance for this case?

I think the sstf method will have a better performance, because if we know the requests, then It is no point to use cscan, and scan method, because they both perfrom unnecssary checking after the cylinder 75.
SCAN: 30->33->33->45->75->99(end of platter)->15->15->10->8(total :166)
CSCAN:30->33->33->45->75->99->0->8->10->15->15(total :258)

can anyone give me some ideas cos I'm not too sure...
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redsar
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2 Solutions

Commented:
Hi redsar,

You are right  ... in this case, sstf will give the best performance

but the best strategy perhaps would have been
30->8-> ... ->75 total 22+67=89
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Author Commented:
so , in general, SSTF will give us the best performance among the those strategy(sstf, scan, csan, fcfs) if we know 50% of the request are for a small, fixed number of cylinders?

so the best strategy by taking advantage of this "hot spot" on the disk would be the LOOK scheduling
as you mentioned?
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Commented:
In what I mentioned, the idea was
if ( current < max && current > min )
get current - min ; get max-current ... which ever value is smaller, start with that end (min/max)

>SSTF will give us the best performance among the those strategy(sstf, scan, csan, fcfs) if we know
>50% of the request are for a small, fixed number of cylinders?
Generally sstf is best but there can be degenerate cases where other algorithms will perform better or atleast the same ...e.g head at 80 and requests for 99 and 0
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Commented:
Hi redsar,

There have been variants of SCAN and CSCAN which dont seek to end of platter.They consider the last request on the queue and then scan back.

But yes,SSTF does give you the best performance most of the times.
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Author Commented:
thanks , got the idea
and If this is the linked list for the request:
33->72->47->8->90->NULL
and I wanna perform SCAN algorithum :
72->90->99->47->33->8
what I think is using the disk head to divide the linked list into two sub linked list,
one is sorted: 72->90
another: 47->33->8
and then combine that....and then take the first from the linked list and then free that process,
but it seems pretty messy to do that, any simpler or any other ideas,
thanks again?
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Commented:
1. maintain doubly linked list ...
2. sort it
3. start from smallest request > disk head ... proceed till end of disk
4. back to original (starting position of step 3) ... traverse the list in other direction
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Author Commented:
well....i supposed to use a single linked list, rather than a double linked list..
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Commented:
In that case there is no other way except the one you listed
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Author Commented:
thanks...
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