Chapter 11I/O Management and Disk

Scheduling

Patricia RoyManatee Community College, Venice, FL

©2008, Prentice Hall

Operating Systems:Internals and Design Principles, 6/E

William Stallings

GIVEN CREDIT WHERE IT IS DUE

Some of the lecture notes are borrowed from Dr. Mark Temte at Indiana University – Purdue University Fort Wayne and from Dr. Einar Vollset at Cornell University

I have modified them and added new slides

I/O MANAGEMENT AND DISK SCHEDULING

We have already discussed I/O techniques Programmed I/O Interrupt-driven I/O Direct memory access (DMA)

Also mentioned logical I/O where the OS hides most of the details of device I/O in system service routines. Then processes see devices in general terms such as read, write, open, close, lock, unlock

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OTHER ISSUES . . .

I/O buffering Physical disk organization Need for efficient disk access Disk scheduling policies RAID

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I/O BUFFERING

Can be block-oriented or stream-oriented Block-oriented buffering

Information is stored in fixed-size blocks Transfers are made a block at a time Used for disks and tapes

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I/O BUFFERING Stream-oriented

Transfer information as a stream of bytes Used for terminals, printers, communication ports,

mouse and other pointing devices, and most other devices that are not secondary storage

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I/O BUFFERING I/O is a problem under paged virtual memory

The target page of any I/O operation must be present in a page frame until the transfer is complete

Otherwise, there can be single-process deadlock

Example Suppose process P is blocked waiting for I/O event to

complete Then suppose the target page for the I/O is swapped

out to disk The I/O operation is subsequently blocked waiting

for the target page to be swapped in, but the rules of suspension dictate that it stays suspended on the disk until the I/O event completes!

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I/O BUFFERING Resulting OS complications

The target page of any I/O operation must be locked in memory until the transfer is complete

A process with pending I/O on any page may not be suspended

Solution: Do I/O through a system I/O buffer in main memory assigned to the I/O request System buffer is locked in memory frame Input transfer is made to the buffer Block moved to user space when needed

This decouples the I/O transfer from the address space of the application

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Physical Disk Organization

read-write head

track

sectors

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Physical Disk Organization

boom

cylinder

PHYSICAL DISK ORGANIZATION When the disk drive is operating, the disk is

rotating at constant speed To read or write, the disk head must be positioned

on the desired track and at the beginning of the desired sector

Once the head is in position, the read or write operation is then performed as the sector moves under the head

Seek time is the time it takes to position the head on the desired track

Rotational delay or rotational latency is the additional time it takes for the beginning of the sector to reach the head once the head is in position

Transfer time is the time for the sector to pass under the head

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PHYSICAL DISK ORGANIZATION Access time

= seek time + rotational latency + transfer time

Efficiency of a sequence of disk accesses strongly depends on the order of the requests

Adjacent requests on the same track avoid additional seek and rotational latency times

Loading a file as a unit is efficient when the file has been stored on consecutive sectors on the same track/cylinder of the disk

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TRANSFER TIME

Depend on the rotation speed of the disk: T = b/(rN) Where T: transfer time

b: number of bytes to be transferred N: number of bytes on a track/cylinder r: rotation speed, in revolutions per second

Thus, the total average access time is Ts + 0.5/r + b/(rN), where Ts is the average seek time.

A TIME COMPARISON Consider a disk with an advertised average seek

time of 4ms, rotation speed of 7500 rpm, and 512-byte sectors with 500 sectors per track. Suppose that we wish to read a file consisting of 2500 sectors for a total of 1.28 Mbytes. We would like to estimate the total time for the transfer. 1) Sequential organization: the file is stored on

5 adjacent tracks (5 tracks * 500 sectors/track = 2500 sectors).

2) The sectors are distributed randomly over the disk.

DISK SCHEDULING POLICIES

Seek time is the reason for differences in performance

For a single disk there will be a number of I/O requests

If requests are selected randomly, get poor performance

DISK SCHEDULING POLICIES

First-in, first-out (FIFO) Process request sequentially Fair to all processes Approaches random scheduling in performance if

there are many processes Example: 55, 58, 39, 18, 90, 160, 150, 38, 184

DISK SCHEDULING POLICIES

Shortest Service/Seek Time First Select the disk I/O request that requires the least

movement of the disk arm from its current position

Always choose the minimum seek time Example: 55, 58, 39, 18, 90, 160, 150, 38, 184 Requests for tracks far away from the current

position may never be served, if requests for closer tracks are issued continuously

DISK SCHEDULING POLICIES

SCAN (aka Elevator Algorithm) Arm moves in one direction only, satisfying all

outstanding requests until it reaches the last track in that direction

Direction is reversed Example: 55, 58, 39, 18, 90, 160, 150, 38, 184

DISK SCHEDULING POLICIES

C-SCAN Restricts scanning to one direction only When the last track has been visited in one

direction, the arm is returned to the opposite end of the disk and the scan begins again

In case of repetitive requests to one track, we will see “arm stickiness” in SSTF, SCAN, C-SCAN

DISK SCHEDULING POLICIES

N-step-SCAN Segment the disk request queue into subqueues

of length N Subqueues are processed one at a time, using

SCAN New requests added to other queue when a

queue is processed

DISK SCHEDULING POLICIES

FSCAN Two subqueues One queue is empty for new requests

Disk Scheduling Algorithms

IN-CLASS EXERCISE

Prob. 11.3 a) Perform the same type of analysis as that of

the previous table for the following sequence of disk track requests: 27, 129, 110, 186, 147, 41, 10, 64, 120. Assume that the disk head is initially positioned over track 100 and is moving in the direction of decreasing track number.

b) Do the same analysis, but now assume that the disk head is moving in the direction of the increasing track number.

RAID MOTIVATION

Disks are improving, but not as fast as CPUs 1970s seek time: 50-100 ms. 2000s seek time: <5 ms. Factor of 20 improvement in 3 decades

We can use multiple disks for improving performance By striping files across multiple disks (placing parts of each

file on a different disk), parallel I/O can improve access time Striping reduces reliability

What’s the mean time between failures of 100 disks, assuming T as the mean time between failures of one disk?

The mean time between failures of 100 disks = 1/100 times of the mean time between failures of one disk

So, we need striping for performance, but we need something to help with reliability

To improve reliability, we can add redundant data to the disks, in addition to striping

RAID A RAID is a Redundant Array of Inexpensive Disks

“I” is also for “Independent” The alternative is SLED, single large expensive disk

Disks are small and cheap, so it’s easy to put lots of disks (10s to 100s) in one box for increased storage, performance, and availability

The RAID box with a RAID controller looks just like a SLED to the computer

Data plus some redundant information is striped across the disks in some way

How that striping is done is key to performance and reliability.

RAID LEVEL 0 Level 0 is nonredundant disk array Files are striped across disks, no redundant info High read throughput Best write throughput (no redundant info to write) Any disk failure results in data loss

Reliability worse than SLED

Stripe 0

Stripe 4

Stripe 3Stripe 1 Stripe 2

Stripe 8 Stripe 10 Stripe 11

Stripe 7Stripe 6Stripe 5

Stripe 9

data disks

RAID LEVEL 1

Mirrored Disks Data is written to two places

On failure, just use surviving disk On read, choose fastest to read

Write performance is same as single drive, read performance is 2x better

Replication redundancy is expensive

data disks mirror copies

Stripe 0

Stripe 4

Stripe 3Stripe 1 Stripe 2

Stripe 8 Stripe 10 Stripe 11

Stripe 7Stripe 6Stripe 5

Stripe 9

Stripe 0

Stripe 4

Stripe 3Stripe 1 Stripe 2

Stripe 8 Stripe 10 Stripe 11

Stripe 7Stripe 6Stripe 5

Stripe 9

PARITY AND HAMMING CODES

What do you need to do in order to detect and correct a one-bit error ? Suppose you have a binary number, represented as a

collection of bits: <b3, b2, b1, b0>, e.g. 0110 Detection is easy Parity:

Count the number of bits that are on, see if it’s odd or even EVEN parity is 0 if the number of 1 bits is even

Parity(<b3, b2, b1, b0 >) = P0 = b0 b1 b2 b3 Parity(<b3, b2, b1, b0, p0>) = 0 if all bits are intact Parity(0110) = 0, Parity(01100) = 0 Parity(11100) = 1 => ERROR! Parity can detect a single error, but can’t tell you which of

the bits got flipped

HAMMING CODE HISTORY

In the late 1940’s Richard Hamming recognized that the further evolution of computers required greater reliability, in particular the ability to not only detect errors, but correct them. His search for error-correcting codes led to the Hamming Codes, perfect 1-error correcting codes, and the extended Hamming Codes, 1-error correcting and 2-error detecting codes.

PARITY AND HAMMING CODES

Detection and correction require more work Hamming codes can detect & correct single bit errors [7,4] binary Hamming Code

h0 = b0 b1 b3 h1 = b0 b2 b3 h2 = b1 b2 b3 H0(<1101>) = 0 H1(<1101>) = 1 H2(<1101>) = 0 Hamming(<1101>) = <b3, b2, b1, h2, b0, h1, h0> = <1100110> If a bit is flipped, e.g. <1110110> a = h0 b0 b1 b3 = 1 b = h1 b0 b2 b3 = 0 c = h2 b1 b2 b3 = 1 abc = <101>, the 5th bit is in error and switch it

EXTENDED [8,4] BINARY HAMM. CODE As with the [7,4] binary Hamming Code:

h0 = b0 b1 b3h1 = b0 b2 b3h2 = b1 b2 b3

Add a new bit p such that p = b0 b1 b2 b3 h0 h1 h2 . i.e., the new bit

makes the XOR of all 8 bits zero. p is called a parity check bit.

Assume at most 2 errors: If parity check passes and abc = 000, the system is correct; If parity check passes and abc ≠ 000, the system has two

errors; If parity check fails, there is one error and abc indicates

which bit is in error.

RAID LEVEL 2 Bit-level striping with Hamming (ECC) codes for error correction All 7 disk arms are synchronized and move in unison Complicated controller Single access at a time Tolerates only one error

data disks

Bit 0 Bit 3Bit 1 Bit 2 Bit 4 Bit 5 Bit 6

ECC disks

RAID LEVEL 3 Use a parity disk

Each bit on the parity disk is a parity function of the corresponding bits on all the other disks

A read accesses all the data disks A write accesses all data disks plus the parity disk On a disk failure, read remaining disks plus parity disk

to compute the missing data

data disksParity disk

Bit 0 Bit 3Bit 1 Bit 2 Parity

RAID 3

X4(i) = X3(i) X2(i) X1(i) X0(i)

If drive X1 has failed. How to recover it?

RAID LEVEL 4 Combines Level 0 and 3 – block-level parity with

Stripes A read accesses all the data disks A write accesses all data disks plus the parity disk

data disksParity disk

Stripe 0 Stripe 3Stripe 1 Stripe 2 P0-3

Stripe 4

Stripe 8

Stripe 10 Stripe 11

Stripe 7Stripe 6Stripe 5

Stripe 9

P4-7

P8-11

RAID 4 (BLOCK-LEVEL PARITY)

X4(i) = X3(i) X2(i) X1(i) X0(i)How to execute a write operation, for instance on drive X1?

Heavy load on the parity disk

RAID 5 (block-level distributed parity)

RAID 6 (dual redundancy)

• Level 5 with an extra parity bit, which is generated with a different and independent data check algorithm

• Can tolerate two failures

I/O BUFFERING FOR THROUGHPUT Perform input transfers in advance of requests being

made With I/O buffers, an application can process the data

from one I/O request while awaiting another Time needed to process a block of data . . .

without buffering = C + Twith buffering = M + max{ C, T }

where:C = computation timeT = I/O memory/disk transfer timeM = memory/memory transfer time (buffer to user)

Perform output transfers sometime after the request is made

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DOUBLE BUFFERING Use two system buffers instead of one A process can transfer data to or from one

buffer while the operating system empties or fills the other buffer

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CIRCULAR BUFFERING More than two buffers are used List of system buffers are arranged in a circle Appropriate in applications where there are bursts

of I/O requests

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SOME RAID ISSUES Granularity

fine-grained: Stripe each file over all disks. This gives high throughput for the file, but limits to transfer of 1 file at a time

coarse-grained: stripe each file over only a few disks. This limits throughput for 1 file but allows more parallel file access

Redundancy concentrate redundancy info on a small number of disks:

partition the set into data disks and redundant disks uniformly distribute redundancy info on disks: avoids

load-balancing problems