I/O, Disks, and RAID.

I/O, Disks, and RAID

Goals for Today Review I/O Disks Prelim graded! RAID
How does a computer system interact with its environment? Disks How does a computer system permanently store data? Prelim graded! Discuss and pass back today RAID How to make storage both efficient and reliable?

The Requirements of I/O
So far in this course: We have learned how to manage CPU, memory What about I/O? Without I/O, computers are useless (disembodied brains?) But… thousands of devices, each slightly different How can we standardize the interfaces to these devices? Devices unreliable: media failures and transmission errors How can we make them reliable??? Devices unpredictable and/or slow How can we manage them if we don’t know what they will do or how they will perform? Some operational parameters: Byte/Block Some devices provide single byte at a time (e.g. keyboard) Others provide whole blocks (e.g. disks, networks, etc) Sequential/Random Some devices must be accessed sequentially (e.g. tape) Others can be accessed randomly (e.g. disk, cd, etc.) Polling/Interrupts Some devices require continual monitoring Others generate interrupts when they need service

Modern I/O Systems

Example Device-Transfer Rates (Sun Enterprise 6000)
Device Rates vary over many orders of magnitude System better be able to handle this wide range Better not have high overhead/byte for fast devices! Better not waste time waiting for slow devices

The Goal of the I/O Subsystem
Provide Uniform Interfaces, Despite Wide Range of Different Devices This code works on many different devices: int fd = open(“/dev/something”); for (int i = 0; i < 10; i++) { fprintf(fd,”Count %d\n”,i); } close(fd); Why? Because code that controls devices (“device driver”) implements standard interface. We will try to get a flavor for what is involved in actually controlling devices in rest of lecture Can only scratch surface!

Want Standard Interfaces to Devices
Block Devices: e.g. disk drives, tape drives, DVD-ROM Access blocks of data Commands include open(), read(), write(), seek() Raw I/O or file-system access Memory-mapped file access possible Character Devices: e.g. keyboards, mice, serial ports, some USB devices Single characters at a time Commands include get(), put() Libraries layered on top allow line editing Network Devices: e.g. Ethernet, Wireless, Bluetooth Different enough from block/character to have own interface Unix and Windows include socket interface Separates network protocol from network operation Includes select() functionality Usage: pipes, FIFOs, streams, queues, mailboxes

How Does User Deal with Timing?
Blocking Interface: “Wait” When request data (e.g. read() system call), put process to sleep until data is ready When write data (e.g. write() system call), put process to sleep until device is ready for data Non-blocking Interface: “Don’t Wait” Returns quickly from read or write request with count of bytes successfully transferred Read may return nothing, write may write nothing Asynchronous Interface: “Tell Me Later” When request data, take pointer to user’s buffer, return immediately; later kernel fills buffer and notifies user When send data, take pointer to user’s buffer, return immediately; later kernel takes data and notifies user

Life Cycle of An I/O Request
User Program Kernel I/O Subsystem Device Driver Top Half Device Driver Bottom Half Device Hardware

A Kernel I/O Structure

Device Drivers Device Driver: Device-specific code in the kernel that interacts directly with the device hardware Supports a standard, internal interface Same kernel I/O system can interact easily with different device drivers Special device-specific configuration supported with the ioctl() system call Device Drivers typically divided into two pieces: Top half: accessed in call path from system calls Implements a set of standard, cross-device calls like open(), close(), read(), write(), ioctl(), strategy() This is the kernel’s interface to the device driver Top half will start I/O to device, may put thread to sleep until finished Bottom half: run as interrupt routine Gets input or transfers next block of output May wake sleeping threads if I/O now complete

I/O Device Notifying the OS
The OS needs to know when: The I/O device has completed an operation The I/O operation has encountered an error I/O Interrupt: Device generates an interrupt whenever it needs service Pro: handles unpredictable events well Con: interrupts relatively high overhead Polling: OS periodically checks a device-specific status register I/O device puts completion information in status register Could use timer to invoke lower half of drivers occasionally Pro: low overhead Con: may waste many cycles on polling if infrequent or unpredictable I/O operations Some devices combine both polling and interrupts For instance: High-bandwidth network device: Interrupt for first incoming packet Poll for following packets until hardware empty After the OS has issued a command to the I/O device either via a special I/O instruction or by writing to a location in the I/O address space, the OS needs to be notified when: (a) The I/O device has completed the operation. (b) Or when the I/O device has encountered an error. This can be accomplished in two different ways: Polling and I/O interrupt. +1 = 58 min. (Y:38)

How does the processor actually talk to the device?
Controller read write control status Addressable Memory and/or Queues Registers (port 0x20) Hardware Memory Mapped Region: 0x8f008020 Bus Interface Processor Memory Bus CPU Regular Memory Interrupt Controller Bus Adaptor Address+ Data Interrupt Request Other Devices or Buses CPU interacts with a Controller Contains a set of registers that can be read and written May contain memory for request queues or bit-mapped images Regardless of the complexity of the connections and buses, processor accesses registers in two ways: I/O instructions: in/out instructions Example from the Intel architecture: out 0x21,AL Memory mapped I/O: load/store instructions Registers/memory appear in physical address space I/O accomplished with load and store instructions

Transfering Data To/From Controller
Programmed I/O: Each byte transferred via processor in/out or load/store Pro: Simple hardware, easy to program Con: Consumes processor cycles proportional to data size Direct Memory Access: Give controller access to memory bus Ask it to transfer data to/from memory directly Sample interaction with DMA controller (from book):

Main components of Intel Chipset: Pentium 4
Northbridge: Handles memory Graphics Southbridge: I/O PCI bus Disk controllers USB controllers Audio Serial I/O Interrupt controller Timers

The Memory Hierarchy Each level acts as a cache for the layer below it
CPU registers, L1 cache L2 cache primary memory disk storage (secondary memory) random access tape or optical storage (tertiary memory) sequential access

What does the disk look like?

Some parameters 2-30 heads (platters * 2) 700-20480 tracks per surface
diameter 14’’ to 2.5’’ tracks per surface sectors per track sector size: 64-8k bytes 512 for most PCs note: inter-sector gaps capacity: 20M-300G main adjectives: BIG, slow

Disk overheads To read from disk, we must specify:
cylinder #, surface #, sector #, transfer size, memory address Transfer time includes: Seek time: to get to the track Latency time: to get to the sector and Transfer time: get bits off the disk Track Sector Rotation Delay Seek Time

Modern disks Barracuda 180 Cheetah X15 36LP Capacity 181GB 36.7GB
Disk/Heads 12/24 4/8 Cylinders 24,247 18,479 Sectors/track ~609 ~485 Speed 7200RPM 15000RPM Latency (ms) 4.17 2.0 Avg seek (ms) 7.4/8.2 3.6/4.2 Track-2-track(ms) 0.8/1.1 0.3/0.4

52 years ago… On 13th September 1956, IBM 305 RAMAC computer system first to use disk storage 80000 times more data on the 8GB 1-inch drive in his right hand than on the 24-inch RAMAC one in his left…

Disks vs. Memory Smallest write: sector Atomic write = sector
Random access: 5ms not on a good curve Sequential access: 200MB/s Cost $.002MB Crash: doesn’t matter (“non-volatile”) (usually) bytes byte, word 50 ns faster all the time MB/s $.10MB contents gone (“volatile”)

Disk Structure Disk drives addressed as 1-dim arrays of logical blocks
the logical block is the smallest unit of transfer This array mapped sequentially onto disk sectors Address 0 is 1st sector of 1st track of the outermost cylinder Addresses incremented within track, then within tracks of the cylinder, then across cylinders, from innermost to outermost Translation is theoretically possible, but usually difficult Some sectors might be defective Number of sectors per track is not a constant

Non-uniform #sectors / track
Maintain same data rate with Constant Linear Velocity Approaches: Reduce bit density per track for outer layers Have more sectors per track on the outer layers (virtual geometry)

Disk Scheduling The operating system tries to use hardware efficiently
for disk drives  having fast access time, disk bandwidth Access time has two major components Seek time is time to move the heads to the cylinder containing the desired sector Rotational latency is additional time waiting to rotate the desired sector to the disk head. Minimize seek time Seek time  seek distance Disk bandwidth is total number of bytes transferred, divided by the total time between the first request for service and the completion of the last transfer.

Disk Scheduling (Cont.)
Several scheduling algos exist service disk I/O requests. We illustrate them with a request queue (0-199). 98, 183, 37, 122, 14, 124, 65, 67 Head pointer 53

Illustration shows total head movement of 640 cylinders.
FCFS Illustration shows total head movement of 640 cylinders.

SSTF Selects request with minimum seek time from current head position
SSTF scheduling is a form of SJF scheduling may cause starvation of some requests. Illustration shows total head movement of 236 cylinders.

SSTF (Cont.)

SCAN The disk arm starts at one end of the disk,
moves toward the other end, servicing requests head movement is reversed when it gets to the other end of disk servicing continues. Sometimes called the elevator algorithm. Illustration shows total head movement of 236 cylinders.

SCAN (Cont.)

C-SCAN Provides a more uniform wait time than SCAN.
The head moves from one end of the disk to the other. servicing requests as it goes. When it reaches the other end it immediately returns to beginning of the disk No requests serviced on the return trip. Treats the cylinders as a circular list that wraps around from the last cylinder to the first one.

C-SCAN (Cont.)

C-LOOK Version of C-SCAN
Arm only goes as far as last request in each direction, then reverses direction immediately, without first going all the way to the end of the disk.

C-LOOK (Cont.)

Selecting a Good Algorithm
SSTF is common and has a natural appeal SCAN and C-SCAN perform better under heavy load Performance depends on number and types of requests Requests for disk service can be influenced by the file-allocation method. Disk-scheduling algo should be a separate OS module allowing it to be replaced with a different algorithm if necessary. Either SSTF or LOOK is a reasonable default algo

Summary I/O Devices Types:
Many different speeds (0.1 bytes/sec to GBytes/sec) Different Access Patterns: Block Devices, Character Devices, Network Devices Different Access Timing: Blocking, Non-blocking, Asynchronous I/O Controllers: Hardware that controls actual device Processor Accesses through I/O instructions, load/store to special physical memory Report their results through either interrupts or a status register that processor looks at occasionally (polling) Device Driver: Device-specific code in kernel Disks: Latency Seek + Rotational + Transfer Also, queuing time Rotational latency: on average ½ rotation Improve performance (decrease queuing time) via scheduling

Announcements Homework 4 available later tonight Prelims graded
It is a programming assignment, so start early Prelims graded Mean 67.7 (Median 67), Stddev 14.2, High 96 out of 100! Good job! Re-grade policy Submit written re-grade request to Nazrul. Entire prelim will be re-graded. We were generous the first time… If still unhappy, submit another re-grade request. Nazrul will re-grade herself If still unhappy, submit a third re-grade request. I will re-grade. Final grade is law.

Grade distribution

Question #2 Algorithm Correctness Constraints (1) Pick up a knife
(2) Pick a fork (3) Cut out a slice of pizza and eat it (4) Return the knife and fork to the pile Correctness Constraints wait for a knife and then a fork, in that order! Key: Deadlock cannot occur since algorithm defines partial order thus, no circular waiting exists

Question #3 32 bit virtual address and 32-bit physical address, 8kB pages #bits for offset? #bits for index? Bytes required for PTE? Bytes required for page table? 3 bytes and 219*3=1.5 MB, respectively 13 and 19, respectively 32 bits Virtual Address 19 bits 13 bits Offset Virtual index Offset 13 bits 19 bits PageTablePtrA frame # VDRWE 1 2 3 … 219-1 Physical frame # 32 bits Physical Address 24 bits = 3 bytes PTE 5 bits 19 bits

Question #3 continued 32 bit virtual address and 24-bit physical address, 8kB pages #bits for offset? #bits for index? Bytes required for PTE? Bytes required for page table? 2 bytes and 219*2=1 MB, respectively 13 and 19, respectively 32 bits Virtual Address 19 bits 13 bits Offset Virtual index Offset 13 bits 11 bits PageTablePtrA frame # VDRWE 1 2 3 … 219-1 Physical frame # 24 bits Physical Address 16 bits = 2 bytes PTE 5 bits (24-13=) 11 bits

Question #4 Give a brief definition of the term “working set?”
Virtual memory pages touched within a window of time (or window of page references).

Question #5: CPU Scheduling
CPU Utilization w/ 10 I/O bound process and 1 CPU-bound I/O bound compute for 1ms, sleep for 10ms CPU bound computes indefinitely Context-switch overhead is 0.1ms CPU utilization w/ 1 ms quantum? scheduler incurs a 0.1ms context-switching cost for every context-switch, regardless of process type Cpu util = execTime/(execTime+contextSwitch) = 1/(1+0.1)=0.9090 CPU utilization w/ 10 ms quantum? #I/O*exI + #CPU*exC / (#I/O*(exI+cs) + #CPU*(exC+cs)) 10* *10 / (10*(1+0.1) *(10+0.1)) 20/( ) = 20/21.1 =

Question #5 continued What strategy can a process employ to maximize the amount of CPU time allocated to that process? Multilevel(-feedback) queue Use a large fraction of assigned quantum then relinquish the CPU before end of quantum thus, increasing the priority associated with the process Round robin Use entire quantum Or say no specific strategy Alternatively, use more threads

How is the disk formatted?
After manufacturing disk has no information Is stack of platters coated with magnetizable metal oxide Before use, each platter receives low-level format Format has series of concentric tracks Each track contains some sectors There is a short gap between sectors Preamble allows h/w to recognize start of sector Also contains cylinder and sector numbers Data is usually 512 bytes ECC field used to detect and recover from read errors

Cylinder Skew Why cylinder skew? How much skew? Example, if
10000 rpm Drive rotates in 6 ms Track has 300 sectors New sector every 20 µs If track seek time 800 µs 40 sectors pass on seek Cylinder skew: 40 sectors

Formatting and Performance
If 10K rpm, 300 sectors of 512 bytes per track 153,600 bytes every 6 ms  24.4 MB/sec transfer rate If disk controller buffer can store only one sector For 2 consecutive reads, 2nd sector flies past during memory transfer of 1st track Idea: Use single/double interleaving

Disk Partitioning Each partition is like a separate disk
Sector 0 is MBR Contains boot code + partition table Partition table has starting sector and size of each partition High-level formatting Done for each partition Specifies boot block, free list, root directory, empty file system What happens on boot? BIOS loads MBR, boot program checks to see active partition Reads boot sector from that partition that then loads OS kernel, etc.

Handling Errors A disk track with a bad sector Solutions:
Substitute a spare for the bad sector (sector sparing) Shift all sectors to bypass bad one (sector forwarding)

RAID Motivation Disks are improving, but not as fast as CPUs
1970s seek time: 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 100 disks have 1/100th mean time between failures of one disk So, we need Striping for performance, but we need something to help with reliability / availability 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
In industry, “I” is 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.

Some Raid Issues Granularity Redundancy
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 uniformly distribute redundancy info on disks: avoids load-balancing problems concentrate redundancy info on a small number of disks: partition the set into data disks and redundant disks

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 1 Stripe 2 Stripe 3 Stripe 7 Stripe 4 Stripe 5 Stripe 6 Stripe 8 Stripe 11 Stripe 9 Stripe 10 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 Expensive Stripe 0 Stripe 1 Stripe 2 Stripe 3 Stripe 0 Stripe 1 Stripe 2 Stripe 3 Stripe 7 Stripe 7 Stripe 4 Stripe 5 Stripe 6 Stripe 4 Stripe 5 Stripe 6 Stripe 8 Stripe 11 Stripe 8 Stripe 11 Stripe 9 Stripe 10 Stripe 9 Stripe 10 data disks mirror copies

Parity and Hamming Code
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

Parity and Hamming Code
Detection and correction require more work Hamming codes can detect double bit errors and detect & correct single bit errors 7/4 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> = < > If a bit is flipped, e.g. < > Hamming(<1111>) = <h2, h1, h0> = <111> compared to <010>, <101> are in error. Error occurred in bit 5.

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, but with no performance degradation Bit 0 Bit 1 Bit 2 Bit 3 Bit 4 Bit 5 Bit 6 data disks ECC disks

Raid Level 3 Use a parity disk A read accesses all the data disks
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 disk failure, read remaining disks plus parity disk to compute the missing data Single parity disk can be used to detect and correct errors Bit 0 Bit 1 Bit 2 Bit 3 Parity Parity disk data disks

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 Heavy load on the parity disk Stripe 0 Stripe 1 Stripe 2 Stripe 3 P0-3 Stripe 7 Stripe 4 Stripe 5 Stripe 6 P4-7 Stripe 8 Stripe 11 P8-11 Stripe 9 Stripe 10 Parity disk data disks

Raid Level 5 Block Interleaved Distributed Parity
Like parity scheme, but distribute the parity info over all disks (as well as data over all disks) Better read performance, large write performance Reads can outperform SLEDs and RAID-0 Stripe 0 Stripe 1 Stripe 2 Stripe 3 P0-3 Stripe 4 Stripe 5 Stripe 6 P4-7 Stripe 7 Stripe 8 Stripe 10 Stripe 11 Stripe 9 P8-11 data and parity disks

Raid Level 6 Level 5 with an extra parity bit
Can tolerate two failures What are the odds of having two concurrent failures ? May outperform Level-5 on reads, slower on writes

RAID 0+1 and 1+0

Stable Storage Handling disk write errors:
Write lays down bad data Crash during a write corrupts original data What we want to achieve? Stable Storage When a write is issued, the disk either correctly writes data, or it does nothing, leaving existing data intact Model: An incorrect disk write can be detected by looking at the ECC It is very rare that same sector goes bad on multiple disks CPU is fail-stop

Approach Use 2 identical disks 3 operations:
corresponding blocks on both drives are the same 3 operations: Stable write: retry on 1st until successful, then try 2nd disk Stable read: read from 1st. If ECC error, then try 2nd Crash recovery: scan corresponding blocks on both disks If one block is bad, replace with good one If both are good, replace block in 2nd with the one in 1st

CD-ROMs Spiral makes 22,188 revolutions around disk (approx 600/mm).
Will be 5.6 km long. Rotation rate: 530 rpm to 200 rpm

Logical data layout on a CD-ROM
CD-ROMs Logical data layout on a CD-ROM

I/O, Disks, and RAID.

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