iki10230 pengantar organisasi komputer bab 5.3: on-line storage

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1 IKI10230 Pengantar Organisasi Komputer Bab 5.3: On-line Storage 23 April 2003 Bobby Nazief ([email protected]) Qonita Shahab ([email protected]) bahan kuliah: http://www.cs.ui.ac.id/kuliah/iki10230/ Sumber : 1. Hamacher. Computer Organization, ed-5. 2. Materi kuliah CS152/1997, UCB.

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IKI10230 Pengantar Organisasi Komputer Bab 5.3: On-line Storage. Sumber : 1. Hamacher. Computer Organization , ed-5. 2. Materi kuliah CS152/1997, UCB. 23 April 2003 Bobby Nazief ([email protected]) Qonita Shahab ([email protected]) bahan kuliah: http://www.cs.ui.ac.id/kuliah/iki10230/. - PowerPoint PPT Presentation

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AT90S8515*
Typical numbers (depending on the disk size):
500 to 2,000 tracks per surface
32 to 128 sectors per track
A sector is the smallest unit that can be read or written
Traditionally all tracks have the same number of sectors:
Constant bit density: record more sectors on the outer tracks
Recently relaxed: constant bit size, speed varies with track location
Platters
Track
Sector
Here is a primitive picture showing you how a disk drive can have multiple platters.
Each surface on the platter are divided into tracks and each track is further divided into sectors. A sector is the smallest unit that can be read or written.
By simple geometry you know the outer track have more area and you would thing the outer tack will have more sectors.
This, however, is not the case in traditional disk design where all tracks have the same number of sectors. Well, you will say, this is dumb but dumb is the reason they do it .
By keeping the number of sectors the same, the disk controller hardware and software can be dumb and does not have to know which track has how many sectors.
With more intelligent disk controller hardware and software, it is getting more popular to record more sectors on the outer tracks. This is referred to as constant bit density.
+2 = 32 min. (Y:12)
Seek time: position the arm over the proper track
Rotational latency: wait for the desired sector
to rotate under the read/write head
Transfer time: transfer a block of bits (sector)
under the read-write head
Typically in the range of 8 ms to 12 ms
(Sum of the time for all possible seek) / (total # of possible seeks)
Due to locality of disk reference, actual average seek time may:
Only be 25% to 33% of the advertised number
Sector
Track
Cylinder
Head
Platter
Arm
To read write information into a sector, a movable arm containing a read/write head is located over each surface.
The term cylinder is used to refer to all the tracks under the read/write head at a given point on all surfaces.
To access data, the operating system must direct the disk through a 3-stage process.
(a) The first step is to position the arm over the proper track. This is the seek operation and
the time to complete this operation is called the seek time.
(b) Once the head has reached the correct track, we must wait for the desired sector to
rotate under the read/write head. This is referred to as the rotational latency.
(c) Finally, once the desired sector is under the read/write head, the data transfer can begin.
The average seek time as reported by the manufacturer is in the range of 12 ms to 20ms and is calculated as the sum of the time for all possible seeks divided by the number of possible seeks.
This number is usually on the pessimistic side because due to locality of disk reference, the actual average seek time may only be 25 to 33% of the number published.
+2 = 34 min. (Y:14)
Rotational Latency:
Approximately 16 ms to 8 ms
per revolution, respectively
8 ms at 3600 RPM, 4 ms at 7200 RPM
Transfer Time is a function of :
Transfer size (usually a sector): 1 KB / sector
Rotation speed: 3600 RPM to 7200 RPM
Recording density: bits per inch on a track
Diameter typical diameter ranges from 2.5 to 5.25 in
Typical values: 2 to 12 MB per second
Sector
Track
Cylinder
Head
Platter
As far as rotational latency is concerned, most disks rotate at 3,600 RPM or approximately 16 ms per revolution.
Since on average, the information you desired is half way around the disk, the average rotational latency will be 8ms.
The transfer time is a function of transfer size, rotation speed, and recording density.
The typical transfer speed is 2 to 4 MB per second.
Notice that the transfer time is much faster than the rotational latency and seek time.
This is similar to the DRAM situation where the DRAM access time is much shorter than the DRAM cycle time.
***** Do anybody remember what we did to take advantage of the short access time versus cycle time? Well, we interleave!
+2 = 36 min. (Y:16)
Accessed sequentially
7 or 9 bits are recorded in parallel across the width of the tape
Capacity: 2 – 5 GB
Transfer rate: few hundreds KB/sec
Another challenger to magnetic disk as secondary storage device is optical disks or CDs.
The drawback of using CDs as secondary storage is that it is read-only.
The advantages of optical compact disk is that it is removable, inexpensive to manufacture, and some of them are write-once, which means you can make one reliable write to them.
The write-once feature gives CD the potential to compete with new tape technologies for archival storage.
+1 = 49 min. (Y:29)
Advantages of Optical Compact Disk:
It is removable
Free of EM interference
Have the potential to compete with new tape technologies for archival storage
Another challenger to magnetic disk as secondary storage device is optical disks or CDs.
The drawback of using CDs as secondary storage is that it is read-only.
The advantages of optical compact disk is that it is removable, inexpensive to manufacture, and some of them are write-once, which means you can make one reliable write to them.
The write-once feature gives CD the potential to compete with new tape technologies for archival storage.
+1 = 49 min. (Y:29)
7200 Revolutions Per Minute 120 Rev/sec
1 revolution = 1/120 sec 8.33 milliseconds
1/2 rotation (revolution) 4.16 ms
Average no. tracks move arm?
Sum all possible seek distances
from all possible tracks / # possible
Assumes average seek distance is random
Disk industry standard benchmark
73.4 GB, 3.5 inch disk
2¢/MB
17 watts (idle)
average seek time + average rotational delay + transfer time + controller overhead
= 5.3 ms + 0.5 * 1/(10000 RPM) + 0.5 KB / (50 MB/s) + 0.15 ms
= 5.3 ms + 3.0 ms + 0.5 KB / (50 KB/ms) + 0.15 ms
= 5.3 + 3.0 + 0.10 + 0.15 ms = 8.55 ms
source: www.ibm.com;
www.pricewatch.com; 2/14/00
Fallacy: Use Data Sheet “Average Seek” Time
Manufacturers needed standard for fair comparison (“benchmark”)
Calculate all seeks from all tracks, divide by number of seeks => “average”
Real average would be based on how data laid out on disk, where seek in real applications, then measure performance
Usually, tend to seek to tracks nearby, not to random track
Rule of Thumb: observed average seek time is typically about 1/4 to 1/3 of quoted seek time (i.e., 3X-4X faster)
UltraStar 72 avg. seek: 5.3 ms 1.7 ms
*
Fallacy: Use Data Sheet Transfer Rate
Manufacturers quote the speed off the data rate off the surface of the disk
Sectors contain an error detection and correction field (can be 20% of sector size) plus sector number as well as data
There are gaps between sectors on track
Rule of Thumb: disks deliver about 3/4 of internal media rate (1.3X slower) for data
For example, UlstraStar 72 quotes
50 MB/s internal media rate
Expect 37 MB/s user data rate
*
Disk Performance Example (revised)
Calculate time to read 1 sector for UltraStar 72 again, this time using 1/3 quoted seek time, 3/4 of internal outer track bandwidth; (8.55 ms before)
Disk latency = average seek time + average rotational delay + transfer time + controller overhead
= (0.33 * 5.3 ms) + 0.5 * 1/(10000 RPM)
+ 0.5 KB / (0.75 * 50 MB/s) + 0.15 ms
= 1.77 ms + 0.5 /(10000 RPM/(60000ms/M))
+ 0.5 KB / (37 KB/ms) + 0.15 ms
= 1.73 + 3.0 + 0.14 + 0.15 ms = 5.02 ms
*
Continued advance in capacity (60%/yr) and bandwidth (40%/yr)
Slow improvement in seek, rotation (8%/yr)
Time to read whole disk
Year Sequentially Randomly
3.5” form factor make sense in 5-7 yrs?
*
1970s: Mainframes 14 inch diameter disks
1980s: Minicomputers, Servers
Laptops, notebooks 2.5 inch disks
Palmtops didn’t use disks,
so 1.8 inch diameter disks didn’t make it
*
1 GB, 3600 RPM,
Digital camera, PalmPC?
in a successful product
Assuming past trends continue
14”
10”
5.25”
3.5”
3.5”
Conventional: 4 disk designs
Katz and Patterson asked in 1987:
*
Replace Small Number of Large Disks with Large Number of Small Disks! (1988 Disks)
Capacity
Volume
Power
??? Hrs
$150K
Disk Arrays have potential for large data and I/O rates, high MB per cu. ft., high MB per KW, but what about reliability?
9X
3X
8X
6X
measured as Mean Time To Failure (MTTF)
Reliability of N disks
(assuming failures independent)
Disk system MTTF:
Arrays too unreliable to be useful!
*
RAID 0
Redundancy yields high data availability
Availability: service still provided to user, even if some components failed
Disks will still fail
Capacity penalty to store redundant info
Bandwidth penalty to update redundant info
Redundant Arrays of (Independent) Disks
*
RAID 1: Disk Mirroring/Shadowing
Very high availability can be achieved
• Bandwidth sacrifice on write:
• Reads may be optimized
(RAID 2 – use Hamming Code – not interesting, so skip)
recovery
group
P contains sum of
other disks per stripe
P
10010011
11001101
10010011
on Read
But every sector has an error detection field
Rely on error detection field to catch errors on read, not on the parity disk
Allows independent reads to different disks simultaneously
*
Redundant Arrays of Inexpensive Disks RAID 4: High I/O Rate Parity
D0
D1
D2
D3
P
D4
D5
D6
P
D7
D8
D9
P
D10
D11
D12
P
D13
D14
D15
P
D16
D17
D18
D19
D20
D21
D22
D23
P
Small writes (write to one disk):
Option 1: read other data disks, create new sum and write to Parity Disk
Option 2: since P has old sum, compare old data to new data, add the difference to P
Small writes are limited by Parity Disk: Write to D0, D5 both also write to P disk
D0
D1
D2
D3
P
D4
D5
D6
P
D7
*
Redundant Arrays of Inexpensive Disks RAID 5: High I/O Rate Interleaved Parity
Independent writes
Increasing
Logical
Disk
Addresses