Harddisks: Seek, Read, Write, Read, Write, Slow ?

[Skybuck Flying]

How big a file? Does it read all the blocks in order from beginning to
end(sequential)?

100 MB, Yes.

How do you decide that?

The program has 1 seek instruction.

What 1MB cache handled where?

Officialy it's called 'User Space'. But let's just call it 'Program Space'.
Memory allocated by the program.

What says there is a seek to where here?

It does a seek into the file at a random location.

Huh?

Because it seeks more.

Loop:

Seek somewhere in file.

Write 4 KB

Write 4 KB

Write 4 KB

etc up to 1 MB / 4 KB.

go back to loop.

So after it has written 1 MB to the file... (in pieces of 4KB)... it goes
back to loop and does a seek again.

That's why it does more seeks than the read test.... at least for big files


I thought 1 MB cache was enough to reach maximum harddisk speed... or 'good
enough' speed...

But 1 MByte/Sec is not really satisfactory for a 100 Megabit Network.

So I would prefer higher write speeds of say 4 MByte/Sec to 8 MByte/Sec

A bigger cache and less seeking might be required to achieve that... I am
not sure about though

Bye,
Skybuck.

 

[Skybuck Flying]

<snip>

1 seek that you instructed it to complete. This is a multitasking operating
system.

So what's your point ?

During testing I have no other applications open to minimize any other disk
activity.

 

[Bruce Roberts]

I will read some documents about that and try to understand... then I will
get back on this

www.storagereview.com

 

[Skybuck Flying]

Nope, calculate the theoretical rate for the data group to be transferred
at the rate of one per revolution.

I don't understand yet how to calculate the data transfer rate

I will read some documents about that and try to understand... then I will
get back on this

Each time you switch from read to write
you likely lose the rest of the current disk rotation.

Sounds reasonable.

What is the relation of the location of each 4KB block of data? Are they
sequential in a file? Where is a write located compared to the preceeding
read?.

In concept 3 yes, they should be, but might not be because of file
fragmentation.

Seeking has to do with moving the head to a different track. Are you doing
random I/O of these blocks in a big file or to multiple files?

1 File.

It sounds like you haven't well thought out or at least described what you
are doing.

Well I might come back on that

Anyway here are two new concepts... actually the real concept behind concept
3.

Concept 4:

Seek

Skip 4 KByte block
Skip 4 KByte block
Write 4 KByte block
Skip 4 KByte block
Write 4 KByte block
Skip 4 KByte block
Skip 4 KByte block
Write 4 KByte block
etc

That's what I would like to do with a file...

I only want to update some parts of a file... not the complete file... etc.

The number of seeks I can use is limited so I have to try and do as many
writes as possible per seek.

Get it now ?

Concept 5:

Skip 4 KByte block
Skip 4 KByte block
Read 4 KByte block
Skip 4 KByte block
Read 4 KByte block
Skip 4 KByte block
Skip 4 KByte block
Read 4 KByte block
etc

I don't think harddisks have a 'skip x number of bytes' command.

 

[Skybuck Flying]

Here is what needs to be done as efficiently as possible:

Transferring a file of 100 MByte from Computer A to Computer B.

The file is split into packets which are send across a network.

The packets are 512 bytes.

Sometimes some packets might go lost.

This creates gaps in the file.

These gaps have to be filled in later

Seeking to all these gaps could be very slow.

So I see some possible solutions.

Problem:

Packet 1, 5, 9, 13 are re-transmitted and have to be stored to file.

Possibly slow solution 1:

Seek Packet 1, Write Packet 1
Seek Packet 5, Write Packet 5
Seek Packet 9, Write Packet 9
Seek Packet 13, Write Packet 13.

Possibly faster solution 2:

Seek Packet 1,
Read Packet 1,2,3,4,5,6,7,8,9,10,11,12,13 into memory.

( Packet 1,5,9,13 are empty )

Overwrite packet 1,5,9,13 with new data into memory.

Seek Packet 1
Write Packet 1,2,3,4,5,6,7,8,9,10,11,12,13 from memory to harddisk.

Possibly bad solution 3:

( because of harddisk design, switching from read to write to read to write
is slow )

Seek Packet 1
Write Packet 1
Skip or Read Packet 2
Skip or Read Packet 3,
Skip or Read Packet 4,
Write Packet 5
Skip or Read Packet 6,
Skip or Read Packet 7,
Skip or Read Packet 8,
Write Packet 9
Skip or Read Packet 10,
Skip or Read Packet 11,
Skip or Read Packet 12,
Write Packet 13.

If the harddisk and software had a skip command this solution might be very
fast.

But because it doesn't have a skip command... I tried a read command
instead.

But funny enough that gives very bad performance.

So currently I think solution 2 is best at the moment.

At this point I am out of ideas

Bye,
Skybuck.

 

[Jim Macklin]

Try a CD-R



| Here is what needs to be done as efficiently as possible:
|
| Transferring a file of 100 MByte from Computer A to
Computer B.
|
| The file is split into packets which are send across a
network.
|
| The packets are 512 bytes.
|
| Sometimes some packets might go lost.
|
| This creates gaps in the file.
|
| These gaps have to be filled in later
|
| Seeking to all these gaps could be very slow.
|
| So I see some possible solutions.
|
| Problem:
|
| Packet 1, 5, 9, 13 are re-transmitted and have to be
stored to file.
|
| Possibly slow solution 1:
|
| Seek Packet 1, Write Packet 1
| Seek Packet 5, Write Packet 5
| Seek Packet 9, Write Packet 9
| Seek Packet 13, Write Packet 13.
|
| Possibly faster solution 2:
|
| Seek Packet 1,
| Read Packet 1,2,3,4,5,6,7,8,9,10,11,12,13 into memory.
|
| ( Packet 1,5,9,13 are empty )
|
| Overwrite packet 1,5,9,13 with new data into memory.
|
| Seek Packet 1
| Write Packet 1,2,3,4,5,6,7,8,9,10,11,12,13 from memory to
harddisk.
|
| Possibly bad solution 3:
|
| ( because of harddisk design, switching from read to write
to read to write
| is slow )
|
| Seek Packet 1
| Write Packet 1
| Skip or Read Packet 2
| Skip or Read Packet 3,
| Skip or Read Packet 4,
| Write Packet 5
| Skip or Read Packet 6,
| Skip or Read Packet 7,
| Skip or Read Packet 8,
| Write Packet 9
| Skip or Read Packet 10,
| Skip or Read Packet 11,
| Skip or Read Packet 12,
| Write Packet 13.
|
| If the harddisk and software had a skip command this
solution might be very
| fast.
|
| But because it doesn't have a skip command... I tried a
read command
| instead.
|
| But funny enough that gives very bad performance.
|
| So currently I think solution 2 is best at the moment.
|
| At this point I am out of ideas
|
| Bye,
| Skybuck.
|
|

 

[Skybuck Flying]

Anyway...

With my 'Possibly faster solution 2' mention somewhere else in this thread.

The big question is:

How many reads/writes per seek ?

A secondary question would be how big the buffer size (block size to
transfer in one read or write ).

The answer to this secondary question might be simply 512 bytes... or 4 KB
which should give best results because of cpu page size, ram page size and
possible harddisk page size etc. So that answer is fixed.

The remaining question would be:

How many reads of 4 KB per seek ?

How many seeks per second are ok ? (possible only 1 )

The desired speed is 4 MByte/Sec

The desired speed could be 8 Mbyte/Sec.

4 MByte/Sec divided by 4 KB buffer = 1000 reads.

That's a lot !

The second desired speed is 2000 reads that s twice as much

Seeing these specification talk about 1 to 2 milliseconds for a command,
read, switch whatever... it doesn't take a lot of imagination to see how
exceeing 1 second is easily achieved and ofcourse not desired.

Suppose each read command take 1 millisecond. 2000x1= 2 seconds... <- oops.

Or 2 Seeks = 30 Milliseconds + 1000 reads <- 1030 milliseconds <- oops.

Etc...



So the big question is

1. How to achieve MByte/Sec writing... since that seems most difficult.

Timing is the key here...

The bad news is that every harddisk has different timing.

One could assume worst case timing for all factors etc.

Only thing left to do is to come up with a formula to find out how many
reads, how many seeks and how big the transfer buffer should be...

the transfer buffer might be fixed to 4 KB so that might not be variable
since 4 KB achieves best performance... so it can't do better than that...

So that leaves... seeks vs reads....

StorageReview doesn't seem to help with the formula's... (or least I didn't
see any page about it ) so I ll dive back into my IBM/Hitachi Deskstar 14
GXP document and try to make sense of it

Skybuck.

 

[Skybuck Flying]

On the other hand...

There are only 3 variables which I can control with my
software/windows/delphi/TFileStream

1. TFileStream.Seek <- seek command

2. TFileStream.Read,
TFileStream.Write <- read command or write command

3. The Read/Write Buffer Size

But then there are ofcourse more factors involved:

4. Offset of seek is also a factor.
5. Switching between read and write is a factor.

6. Switching between seek, read and write might be a factor.
7. Even pausing for a while might be a factor.

And then there are ofcourse external factors... OS, Other Hardware etc.

So at least 5 factors amount into 6 requirements

1. The program should do sequential seeks or random seeks. Randomize Offset,
Or increment Offset. <- Sequential seeks is pretty much bullshit. But not
quite... since you could seek ahead only.
2. The program should do, read only, write only, or mixed, possibly with a
read:write ratio or randomizer.
3. The program should do seeks per second.
4. The program should do writes per second. Writes per seek should therefore
be calculated.
5. The program should do reads per second. Reads per seek should therefore
be calculated.
6. The program should do buffer size variable.

5 Causes 2 to be partially done... except for the mixed and randomizer.
Read/Write Randomizer is less interesting. Mixed could be interesting but is
also automatically handled.
Ratio is also handled by 2.

So all this stuff would give a single result

Speed Per Second

( Data Transfer Speed From/To Harddisk in Bytes/Sec )

So instead of trying to understand a document about how a single harddisk
works, it's charactistics, it's timing factors and terms and how it is
related to performance and unapplyable formula's for application software
because of missing information like current cylinder number and track and
sector number, a test program could be made to find the sweet spot in
practice

To top things off some artificial intelligence mechanisme should try out all
kinds of settings and find the sweet spot hehehehe.

Bye,
Skybuck.

 

[Folkert Rienstra]

Not really.

Yes, really. How much depends per cylinder. Not that it matters.

What limitation is that exactly?

You need to ask? Not that it matters.

That's false.

Nope. Not that it matters.

Not necessarily.

Indeed, as it's automatically done by the Servo system.

 

[Folkert Rienstra]

I don't understand yet how to calculate the data transfer rate

I will read some documents about that and try to understand... then I will
get back on this


Sounds reasonable.

But it's not.
In order to write to the correct sector the drive has to know
if it is the right sector, therefor needs to read the header.
If it can't, writes would go one sector per rev.
Obviously not.

And this is how it used to be when Sector-IDs were still in use
and obviously things improve with time, not worsen, so if it could
do it then it certainly can do it now.

[snip]

 

[Ron Reaugh]

So does moving to the next track etc.

That's an electronic headswitch or an adjacent cylinder seek.

 

[Ron Reaugh]

You need to describe in more detail your program and settings. Why do you
assume that your program statements have any first order relation to what's
happening at the physical disk I/O level?

 

[Ron Reaugh]

During testing I have no other applications open to minimize any other disk
activity.

There's all sorts of possible blocking and other issues related to OS
scheduling that could be at work even in an idle system.

 

[Ron Reaugh]

I don't understand yet how to calculate the data transfer rate

Let's assume that your programatic 4K block has a direct relation to the
actual disk I/O size(it probably does NOT.) Now assume that the disk spins
at 7200 RPM or 120 revs per second. So the throughput of one 4K block per
rev is 120 x 4,000 = bytes per second = 480,000. That's the max speed as
any seeking will slow things down more.

 

[Ron Reaugh]

Go back under the bridge where you belong.

news:[email protected]...

 

[Skybuck Flying]

Let's assume that your programatic 4K block has a direct relation to the
actual disk I/O size(it probably does NOT.) Now assume that the disk spins
at 7200 RPM or 120 revs per second. So the throughput of one 4K block per
rev is 120 x 4,000 = bytes per second = 480,000. That's the max speed as
any seeking will slow things down more.

The 4KB size is probably because of the way CPU's and RAM work... CPU's and
RAM like moving 4 KBytes at a time

Hmmm...

Does that mean a harddisk is only able to read 1 block of data per
revolution ?

That can't be correct since I have another program that reads/writes block
of roughly 512 bytes...

That would mean max: 512 * 120 = 61 KB...

I am pretty sure the program was able to achieve higher speeds then that.

That could mean a couple of things:

1. Windows XP 'groups' multiple 'seeks + read' pairs together if they are
for example sequantial... and turn it into one big read to the harddisk...
and then give it back in multiple reads to my program.

or

2. The harddisk has some read ahead cache... so that multiple reads can
occur without having to wait for another revolution.

There are a couple of things I wonder about:

1. How long does it take to read 1 sector from the moment it is found
(seeked, so not including seek time) to the moment it is completely read.

Maybe this is a weird question because harddisk's like to 'cache' as much
sectors as possible during a read... I think this is what they call re-ahead
cache ?... or just harddisk read cache ?

Anyway I did some calculations...

And my drive is able to have a sustained speed of 8 Million bytes per
second.

So that is roughly:

( 8.000.000 Bytes / 512 Bytes Per Sector ) / Sec = 15625 Sectors / Sec.

So to find the time per sector in seconds:

1 second / 15625 Sectors = 0.000064 Seconds Per Sector

So that means roughly 0.064 milliseconds per sector.

0.000064 * 1000 = 0.064 Millisecond per Sector.

Maybe I can take this as an average... so it means inclusive of command
overhead, switch times, all that stuff.

So I could try to use this in calculations to find throughput.

The remaining factor is seek time...

The document about my harddisk states:

Full Stroke Read Seek Time 18 MSec
Full Stroke Write Seek Time 19 MSec

This seems about right since I like to think seek time is mostly 20
milliseconds anyway in practice. Unless many sequantial seeks are done...

But for randomly seeking this seems about right.

So now I might be able to do some calculations like this:

Suppose I have a write cache like in the write program I posted the results
from.

1 Megabyte write cache.

This cache consists out of 250 blocks of 4 KB = 1 MByte.

( Maybe the 4 KB is not that much relevant anyway )

First a seek is done:

20 Milliseconds gone.

1 Megabyte / 512 sector size = 2048 Sectors have to be written.

2048 * 0.064 = 131.072 milliseconds.

That's 151 Milliseconds gone.

1000 milliseconds / 151 = 6.6.

So the program should roughly be able to do this 6.6 times.

6.6 x 1 MByte = 6.6 Mbyte / Sec write speed.

The weird thing is my program does not achieve this speed....

It sticks around at 1 MByte/Sec.

Only the first second it achieved 6 or 7 MByte/Sec. Which can probably be
ignored since the program has to still start etc...

But still... the program did not update the stats every second...

So this time I took a look at the task manager and the process list, it
shows the I/O Write Bytes.

I notice that the harddisk is spinning a lot and making noise a lot.

Sometimes it only updates with 20 KB and sometimes with 1 or 2 MByte.

Could this be because of fragmentation I wonder ?!

I have a pretty fragmented file system / drive !

So it could be that it has to seek a lot etc...

BUTT

It remains weird... because when I use the other program... to do sequential
read of the file... from start to end... it goes much faster !

So that's a bit weird.

Maybe it's because the write program... seeks randomly into the file.... so
it could start writing anywhere... maybe half way a cylinder or track etc...
so then at first has to wait a bit or so until a revolution is past... but
from that moment on it should write sequantially and that should go fast ?!

Or maybe windows xp has to do all kinds of updates somewhere in the ntfs
system etc...

Oh well...

Skybuck.

 

[Ron Reaugh]

The 4KB size is probably because of the way CPU's and RAM work... CPU's and
RAM like moving 4 KBytes at a time

No, you said 4KB writes. The CPU/RAM aren't relevant.

Hmmm...

Does that mean a harddisk is only able to read 1 block of data per
revolution ?

Unless the I/O is structured to meet streaming requirements, yes one block
per rev is often the result.

That can't be correct since I have another program that reads/writes block
of roughly 512 bytes...

That would mean max: 512 * 120 = 61 KB...

I am pretty sure the program was able to achieve higher speeds then that.

Yes, you achieved some streaming probably via intelligent buffering in the
language I/O routines, OS or onboard the drive.

That could mean a couple of things:

1. Windows XP 'groups' multiple 'seeks + read' pairs together if they are
for example sequantial...

Exclude seek from the statement.

and turn it into one big read to the harddisk...
and then give it back in multiple reads to my program.

Exactly and the drive itself will read into its onboard buffer even more
physically sequential data than you asked for which puts it in a position to
send that additional data back to you with no further actual disk surface
I/O. Keep doing that fast and the drive will try and stay ahead.......and
your streaming and full disk read rate which is the number of user sectors
per track at revolution speed. There are equivalent mechanisms for write
streaming

or

2. The harddisk has some read ahead cache... so that multiple reads can
occur without having to wait for another revolution.

There are a couple of things I wonder about:

1. How long does it take to read 1 sector from the moment it is found
(seeked, so not including seek time) to the moment it is completely read.

Generally speaking it is the fraction of a revolution derived from the
number of sectors per track and the RPM.

Maybe this is a weird question because harddisk's like to 'cache' as much
sectors as possible during a read...

Well, there's an algorithm "as much" is too strong.

I think this is what they call re-ahead
cache ?... or just harddisk read cache ?

There's RAM onboard a HD. It's used for write caching, read caching and
microcode etc.

Anyway I did some calculations...

And my drive is able to have a sustained speed of 8 Million bytes per
second.

That's very slow for any recent drive. Current drive are generally in the
30-60MB/sec. vicinity.

 

[Skybuck Flying]

As I suspected the 100 MB file I was using to test... is pretty heavily
fragmented

I found this nice little too to find out if it's fragmented.

http://www.sysinternals.com/ntw2k/info/defrag.shtml

This was the output

: n \test files\bufferlog.txt

Clusters for file: C:\test files\bufferlog.txt
VCN: 0 LCN: 1300746 LEN: 16
VCN: 16 LCN: 1346392 LEN: 32
VCN: 48 LCN: 1345381 LEN: 64
VCN: 112 LCN: 2295765 LEN: 128
VCN: 240 LCN: 2510809 LEN: 219
VCN: 459 LCN: 2619481 LEN: 85
VCN: 544 LCN: 2467607 LEN: 218
VCN: 762 LCN: 2326190 LEN: 102
VCN: 864 LCN: 1866455 LEN: 218
VCN: 1082 LCN: 2778070 LEN: 102
VCN: 1184 LCN: 241057 LEN: 225
VCN: 1409 LCN: 2642790 LEN: 95
VCN: 1504 LCN: 2765237 LEN: 217
VCN: 1721 LCN: 2053603 LEN: 119
VCN: 1840 LCN: 403797 LEN: 217
VCN: 2057 LCN: 1280178 LEN: 119
VCN: 2176 LCN: 1870824 LEN: 225
VCN: 2401 LCN: 2483382 LEN: 111
VCN: 2512 LCN: 1184381 LEN: 227
VCN: 2739 LCN: 1280035 LEN: 125
VCN: 2864 LCN: 2559019 LEN: 229
VCN: 3093 LCN: 2764708 LEN: 123
VCN: 3216 LCN: 2058491 LEN: 233
more ('q' to quit):
VCN: 3449 LCN: 1382934 LEN: 119
VCN: 3568 LCN: 404503 LEN: 243
VCN: 3811 LCN: 2522295 LEN: 125
VCN: 3936 LCN: 857266 LEN: 243
VCN: 4179 LCN: 1179615 LEN: 126
VCN: 4305 LCN: 2176930 LEN: 243
VCN: 4548 LCN: 1281750 LEN: 126
VCN: 4674 LCN: 1465119 LEN: 213
VCN: 4887 LCN: 1184609 LEN: 173
VCN: 5060 LCN: 1866681 LEN: 250
VCN: 5310 LCN: 700707 LEN: 130
VCN: 5440 LCN: 700422 LEN: 252
VCN: 5692 LCN: 1822219 LEN: 132
VCN: 5824 LCN: 1284880 LEN: 255
VCN: 6079 LCN: 2514075 LEN: 146
VCN: 6225 LCN: 2456167 LEN: 259
VCN: 6484 LCN: 2559526 LEN: 144
VCN: 6628 LCN: 2178713 LEN: 260
VCN: 6888 LCN: 2313913 LEN: 153
VCN: 7041 LCN: 240583 LEN: 263
VCN: 7304 LCN: 2501987 LEN: 153
VCN: 7457 LCN: 2204221 LEN: 271
VCN: 7728 LCN: 1871065 LEN: 147
VCN: 7875 LCN: 1158481 LEN: 292
more ('q' to quit):
VCN: 8167 LCN: 2391566 LEN: 126
VCN: 8293 LCN: 2284425 LEN: 304
VCN: 8597 LCN: 2179820 LEN: 110
VCN: 8707 LCN: 2560656 LEN: 322
VCN: 9029 LCN: 2812066 LEN: 93
VCN: 9122 LCN: 2166757 LEN: 328
VCN: 9450 LCN: 2177179 LEN: 88
VCN: 9538 LCN: 90173 LEN: 335
VCN: 9873 LCN: 543314 LEN: 80
VCN: 9953 LCN: 2559672 LEN: 338
VCN: 10291 LCN: 2791169 LEN: 78
VCN: 10369 LCN: 1871243 LEN: 342
VCN: 10711 LCN: 1792079 LEN: 74
VCN: 10785 LCN: 180051 LEN: 346
VCN: 11131 LCN: 1598333 LEN: 69
VCN: 11200 LCN: 2188595 LEN: 351
VCN: 11551 LCN: 2209458 LEN: 65
VCN: 11616 LCN: 2179948 LEN: 352
VCN: 11968 LCN: 2063382 LEN: 63
VCN: 12031 LCN: 2218857 LEN: 17
VCN: 12048 LCN: 2167213 LEN: 353
VCN: 12401 LCN: 2068155 LEN: 78
VCN: 12479 LCN: 2325918 LEN: 17
VCN: 12496 LCN: 2177854 LEN: 365
more ('q' to quit):
VCN: 12861 LCN: 2479980 LEN: 65
VCN: 12926 LCN: 2589648 LEN: 34
VCN: 12960 LCN: 2168480 LEN: 383
VCN: 13343 LCN: 2481639 LEN: 63
VCN: 13406 LCN: 2600003 LEN: 34
VCN: 13440 LCN: 2392009 LEN: 384
VCN: 13824 LCN: 2458964 LEN: 62
VCN: 13886 LCN: 2609991 LEN: 34
VCN: 13920 LCN: 2059631 LEN: 205
VCN: 14125 LCN: 1643586 LEN: 51
VCN: 14176 LCN: 2293193 LEN: 389
VCN: 14565 LCN: 2167968 LEN: 72
VCN: 14637 LCN: 2591060 LEN: 51
VCN: 14688 LCN: 2171942 LEN: 391
VCN: 15079 LCN: 2480746 LEN: 69
VCN: 15148 LCN: 2502200 LEN: 68
VCN: 15216 LCN: 2219938 LEN: 394
VCN: 15610 LCN: 2204502 LEN: 82
VCN: 15692 LCN: 2582807 LEN: 68
VCN: 15760 LCN: 1158805 LEN: 399
VCN: 16159 LCN: 2168041 LEN: 76
VCN: 16235 LCN: 2798382 LEN: 85
VCN: 16320 LCN: 2170575 LEN: 203
VCN: 16523 LCN: 2799079 LEN: 85
more ('q' to quit):
VCN: 16608 LCN: 1708210 LEN: 203
VCN: 16811 LCN: 2815815 LEN: 85
VCN: 16896 LCN: 157273 LEN: 404
VCN: 17300 LCN: 854129 LEN: 102
VCN: 17402 LCN: 2829296 LEN: 102
VCN: 17504 LCN: 2883531 LEN: 419
VCN: 17923 LCN: 2809129 LEN: 87
VCN: 18010 LCN: 2766550 LEN: 105
VCN: 18115 LCN: 2287803 LEN: 199
VCN: 18314 LCN: 2456794 LEN: 118
VCN: 18432 LCN: 2176267 LEN: 423
VCN: 18855 LCN: 2306638 LEN: 98
VCN: 18953 LCN: 1994350 LEN: 120
VCN: 19073 LCN: 1280453 LEN: 200
VCN: 19273 LCN: 2168119 LEN: 138
VCN: 19411 LCN: 2284221 LEN: 197
VCN: 19608 LCN: 1868690 LEN: 150
VCN: 19758 LCN: 2491310 LEN: 187
VCN: 19945 LCN: 2483073 LEN: 151
VCN: 20096 LCN: 2594851 LEN: 183
VCN: 20279 LCN: 1868527 LEN: 156
VCN: 20435 LCN: 2172345 LEN: 200
VCN: 20635 LCN: 2623014 LEN: 151
VCN: 20786 LCN: 2479269 LEN: 200
more ('q' to quit):
VCN: 20986 LCN: 2300299 LEN: 156
VCN: 21142 LCN: 2186850 LEN: 192
VCN: 21334 LCN: 2512740 LEN: 175
VCN: 21509 LCN: 1753435 LEN: 193
VCN: 21702 LCN: 1281897 LEN: 177
VCN: 21879 LCN: 2558254 LEN: 194
VCN: 22073 LCN: 1530465 LEN: 177
VCN: 22250 LCN: 1283329 LEN: 188
VCN: 22438 LCN: 2209694 LEN: 429
VCN: 22867 LCN: 242236 LEN: 164
VCN: 23031 LCN: 240867 LEN: 178
VCN: 23209 LCN: 2869945 LEN: 436
VCN: 23645 LCN: 2479534 LEN: 178
VCN: 23823 LCN: 2789327 LEN: 178
VCN: 24001 LCN: 1282077 LEN: 179
VCN: 24180 LCN: 2314613 LEN: 180
VCN: 24360 LCN: 2871359 LEN: 442
VCN: 24802 LCN: 2513850 LEN: 129
VCN: 24931 LCN: 1556977 LEN: 465
VCN: 25396 LCN: 2176692 LEN: 110
VCN: 25506 LCN: 89547 LEN: 467
VCN: 25973 LCN: 241313 LEN: 4
Enumerate file clusters: STATUS_SUCCESS

The BufferLog.TXT was extracted from a WinZip file...

So I see two possibilities:

1. WinZip causes lot's of fragmentation

or

2. The free space was already fragmented...

I think case 2 was the case

Since I first deleted a compressed file etc...

And I know I deleted many files before making the compressed file etc...

So in other words... the free space was already pretty much fragmented

It would be nice to have a simple little tool... just to defragment a single
file for testing purposes...

Also I wonder about the Logic Cluster Numbers...

How does that work... performance wise ?

Are sequential Logic Cluster Numbers always garanteed to achieve maximum
performance ?

That would be amazing... so I don't think that's necessarily the case...

I read something about harddisk liking to writing/filling as many clustors
in a single 'cylinder'... a hollow cylinder that is... only writing the
rings of the 'cylinder'

At least that's how I imagine it could work:

For example:

Platter 1. Inner track 0 is filled.

Platter 2. Inner track 0 is filled

Platter 3. Inner track 0 is filled.

/\
| this is what they call a cylinder i think...
\/

Then when all tracks 'vertically' are filled.

Platter 1. Inner track 1 is filled

Platter 2. Inner track 1 is filled

Platter 3. Inner track 1 is filled.

Etc.

This should be done to minimize track to track movement etc...

So I wonder how windows xp 'distributes' the logic cluster numbers across
the disk...

Does it first fill up all tracks on a platter before proceeding to the next
platter...

Like so:

Case 1:

Platter 1, Track 1, Track 2, Track 3, Track 4, Track 5

Or does it use the 'cylinder' performance trick

Case 2:

Platter 1, Track 1, Track 4

Platter 2, Track 2, Track 5

Platter 3, Track 3, Track 6

Etc

That would be something... that would mean Windows XP optimizes
cluster/track placement for each disk differently... since each disk has
different geometry properties...

Whatever the case is...

I will assume placing all logical cluster numbers sequantially will still
give much better performance than randomly.

Skybuck.

 

[Skybuck Flying]

Hmm...

Here is a little link about harddisk's layout and performance optimizations
though it's kinda old.

http://home.no.net/tkos/info/hd.html

 

[Skybuck Flying]

So what I am interested in...

Is a little bit of information about

Harddisk numbering in general...

Like sector numbering, track numbering, platter numbering, head numbering,
ring numbering, etc.

How it corresponds to the physical location...

And ofcourse how harddisk read these