Transcript I/O - Thayer School of Engineering at Dartmouth

ENGS 116 Lecture 18
1
I/O Interfaces,
A Little Queueing Theory
RAID
Vincent H. Berk
November 11, 2005
Reading for Today: Sections 7.1 – 7.4
Reading for Monday: Sections 7.5 – 7.9, 7.14
Homework for Friday Nov 18: 5.4, 5.17,
6.4, 6.10, 7.3, 7.21, 8.9/8.10, 8.17
ENGS 116 Lecture 18
2
Common Bus Standards
• ISA
• PCI
• AGP
• PCMCIA
• USB
• FireWire/IEEE 1394
• IDE
• SCSI
ENGS 116 Lecture 18
3
Programmed I/O (Polling)
CPU
Is the
data
ready?
Memory
IOC
no
yes
read
data
Device
However, checks for
I/O completion can be
dispersed among
computationally
intensive code.
store
data
done?
yes
Busy wait loop
not an efficient
way to use the CPU
unless the device
is very fast!
no
ENGS 116 Lecture 18
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Interrupt-Driven Data Transfer
CPU
add
sub
and
or
nop
(1) I/O
interrupt
Memory
IOC
(2) save PC
device
(3) interrupt
service addr
User program progress only halted during
actual transfer
(4)
read
store
...
rti
user
program
interrupt
service
routine
memory
1000 transfers at 1 ms each:
1000 interrupts @ 2 µ sec per interrupt
1000 interrupt service @ 98 µ sec each = 0.1 CPU seconds
 0.1 x 10-6 sec/byte  0.1 µsec/byte
 1000 bytes = 100 µsec
1000 transfers  100 µsecs = 100 ms = 0.1 CPU seconds
Device transfer rate = 10 MBytes/sec
Still far from device transfer rate! 1/2 time in interrupt overhead
ENGS 116 Lecture 18
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Direct Memory Access (DMA)
Time to do 1000 transfers at 1 msec each:
1 DMA set-up sequence @ 50 µ sec
1 interrupt @ 2 µ sec
1 interrupt service sequence @ 48 µ sec
CPU sends a starting address,
direction, and length count to
DMAC. Then issues "start".
.0001 second of CPU time
CPU
Memory
DMAC
0
IOC
Memory
Mapped I/O
device
DMAC provides handshake signals for peripheral
controller and memory addresses and handshake
signals for memory.
ROM
RAM
Peripherals
n
DMAC
ENGS 116 Lecture 18
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Input/Output Processors
CPU
D1
IOP
D2
main memory
bus
Mem
. . .
I/O
bus
(1)
Dn
CPU
(4) issues instruction to IOP
IOP
(2) interrupts when done
(3)
target device
where commands are
OP Device Address
looks in memory for commands
memory
Device to/from memory
transfers are controlled
by the IOP directly.
IOP steals memory cycles.
OP Addr Cnt Other
what
to do
where
to put
data
how
much
special
requests
ENGS 116 Lecture 18
7
Summary
• Disk industry growing rapidly, improves bandwidth and areal
density
• Time = queue + controller + seek + rotate + transfer
• Advertised average seek time much greater than average seek
time in practice
• Response time vs. bandwidth tradeoffs
• Processor interface: today peripheral processors, DMA, I/O
bus, interrupts
ENGS 116 Lecture 18
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A Little Queueing Theory
System
Arrivals
Departures
• More interested in long-term, steady-state behavior
 Arrivals = Departures
• Little’s Law: mean number of tasks in system
= arrival rate  mean response time
– Observed by many, Little was first to prove
– Applies to any system in equilibrium as long as nothing
inside the system (black box) is creating or destroying tasks
• Queueing models assume state of equilibrium:
input rate = output rate
ENGS 116 Lecture 18
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A Little Queueing Theory
Queue
Arrivals
Server
Departures
Avg. arrival rate 
Avg. service rate 
Avg. number in system N
Avg. system time per customer T = avg. waiting time + avg.
service time
• Little’s Law: N =   T
• Service utilization  =  / 
•
•
•
•
ENGS 116 Lecture 18
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A Little Queueing Theory
• Server spends a variable amount of time with customers
– Service distribution characterized by mean, variance, squared
coefficient of variance (C)
– Squared coefficient of variance: C = variance/mean2, unitless
measure
• Exponential distribution: C = 1; most short relative to average
• Hypoexponential distribution: C < 1; most close to average
• Hyperexponential distribution: C > 1; most further from average
• Disk response times: C ≈ 1.5, but usually pick C = 1 for simplicity
ENGS 116 Lecture 18
11
Average Residual Service Time
How long does a new customer wait for the current customer to finish
service?
Average residual time =
1
 Mean service time 1 C 
2
If C = 0, average residual service time = 1/2 mean service time
ENGS 116 Lecture 18
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Average Wait Time in Queue
• If something at server, it takes average residual service time to
complete
• Probability that server is busy is 
• All customers in line must complete, each averaging Tservice
Tse rvi ce  1  C  
Tq 
2  1 
• If exponential distribution, C = 1 and
Tq  Tse rvi ce 

1   
ENGS 116 Lecture 18
13
M/M/1 and M/G/1
• Assumptions so far:
– System in equilibrium
– Time between two successive arrivals in line are random
– Server can start on next customer immediately after prior customer
finishes
– No limit to the queue, FIFO service
– All customers in line must complete, each averaging Tservice
• Exponential distribution (C = 1) is “memoryless” or Markovian,
denoted by M
• Queueing system with exponential arrivals, exponential service times,
and 1 server: M/M/1
• General distribution is denoted by G: can have M/G/1 queue
ENGS 116 Lecture 18
14
An Example
Processor sends 10 8-KB disk I/Os per second, requests and service
times exponentially distributed, avg. disk service = 20 ms.
ENGS 116 Lecture 18
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Another Example
Processor sends 20 8-KB disk I/Os per second, requests and service
times exponentially distributed, avg. disk service = 12 ms.
ENGS 116 Lecture 18
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Yet Another Example
Processor sends 10 8-KB disk I/Os per second, C = 1.5, avg. disk
service = 20 ms.
ENGS 116 Lecture 18
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Manufacturing Advantages of Disk Arrays
Disk Product Families
Conventional:
4 disk designs
3.5”
5.25”
Low End
Disk Array:
1 disk design
3.5”
10”
14”
14”
High End
ENGS 116 Lecture 18
18
Replace Small # of Large Disks with
Large # of Small Disks!
Data Capacity
Volume
Power
Data Rate
I/O Rate
MTBF
Cost
IBM 3390 (K)
20 GBytes
97 cu. ft.
3 KW
15 MB/s
600 I/Os/s
250 KHrs
$250K
IBM 3.5" 0061
320 MBytes
0.1 cu. ft.
11 W
x70
23 GBytes
11 cu. ft.
1 KW
1.5 MB/s
55 I/Os/s
50 KHrs
$2K
120 MB/s
3900 IOs/s
??? Hrs
$150K
large data and I/O rates
Disk Arrays have potential for
high MB per cu. ft., high MB per KW
awful reliability
ENGS 116 Lecture 18
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Array Reliability
• Reliability of N disks = Reliability of 1 Disk ÷ N
50,000 Hours ÷ 70 disks = 700 hours
Disk system MTTF: Drops from 6 years to 1 month!
• Arrays without redundancy too unreliable to be useful!
Hot spares support reconstruction in parallel with
access: very high media availability can be achieved
ENGS 116 Lecture 18
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Redundant Arrays of Disks
• Files are "striped" across multiple spindles
• Redundancy yields high data availability
Disks will fail
Contents reconstructed from data redundantly stored in the array
 Capacity penalty to store it
 Bandwidth penalty to update
Mirroring/Shadowing (high capacity cost)
Techniques:
Horizontal Hamming Codes (overkill)
Parity & Reed-Solomon Codes
Failure Prediction (no capacity overhead!)
VaxSimPlus — Technique is controversial
ENGS 116 Lecture 18
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Redundant Arrays of Disks (RAID) Techniques
1
0
0
1
0
0
1
1
• Disk Mirroring, Shadowing
Each disk is fully duplicated onto its "shadow"
Logical write = two physical writes
100% capacity overhead
• Parity Data Bandwidth Array
Parity computed horizontally
Logically a single high data bandwidth disk
• High I/O Rate Parity Array
Interleaved parity blocks
Independent reads and writes
Logical write = 2 reads + 2 writes
Parity + Reed-Solomon codes
1
0
0
1
0
0
1
1
1
1
0
0
1
1
0
1
1
0
0
1
0
0
1
1
Parity
disk
1
0
0
1
0
0
1
1
0
0
1
1
0
0
1
0
ENGS 116 Lecture 18
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RAID 0: Disk Striping
• Data is distributed over disks
• Improved bandwidth and seek time on read and write
• Larger virtual disk
• No redundancy
ENGS 116 Lecture 18
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RAID 1: Disk Mirroring/Shadowing
recovery
group
• Each disk is fully duplicated onto its "shadow"
Very high availability can be achieved
• Bandwidth sacrifice on write:
Logical write = two physical writes
• Half seek time on reads
• Reads may be optimized
• Most expensive solution: 100% capacity overhead
Targeted for high I/O rate, high availability environments
ENGS 116 Lecture 18
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RAID 3: Parity Disk
10010011
11001101
10010011
...
logical record
Striped physical
records
P
1
0
0
1
0
0
1
1
1
1
0
0
1
1
0
1
1
0
0
1
0
0
1
1
0
0
1
1
0
0
0
0
• Parity computed across recovery group to protect against hard disk failures
33% capacity cost for parity in this configuration
wider arrays reduce capacity costs, decrease expected availability,
increase reconstruction time
• Arms logically synchronized, spindles rotationally synchronized
logically a single high capacity, high transfer rate disk
Targeted for high bandwidth applications
ENGS 116 Lecture 18
RAID 4 & 5: Block-Interleaved Parity and
Distributed Block-Interleaved Parity
• Similar to RAID 3, requiring same number of disks.
• Parity is computed over blocks and stored in blocks.
• RAID 4 places parity on last disk
• RAID 5 places parity blocks distributed over all disks:
Advantage: parity block is always accesses on read/writes
• Parity is updated by reading the old-block and parity-block, and writing
the new-block and the new parity-block. (2 reads, 2 writes)
25
ENGS 116 Lecture 18
Disk access advantage RAID 4/5 over RAID 3
26
ENGS 116 Lecture 18
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Problems of Disk Arrays:
Small Writes
RAID-5: Small Write Algorithm
1 Logical Write = 2 Physical Reads + 2 Physical Writes
D0'
new
data
D0
D1
D2
D3
old (1. Read)
data
+ XOR
P
old
(2. Read)
parity
+ XOR
(3. Write)
D0'
D1
(4. Write)
D2
D3
P'
ENGS 116 Lecture 18
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I/O Benchmarks: Transaction Processing
• Transaction Processing (TP) (or On-line TP = OLTP)
– Changes to a large body of shared information from many terminals, with
the TP system guaranteeing proper behavior on a failure
– If a bank’s computer fails when a customer withdraws money, the TP
system would guarantee that the account is debited if the customer
received the money and that the account is unchanged if the money was
not received
– Airline reservation systems & banks use TP
• Atomic transactions makes this work
• Each transaction  2 to 10 disk I/Os and 5,000 to 20,000 CPU
instructions per disk I/O
– Efficient TP software, avoid disks accesses by keeping information in main
memory
• Classic metric is Transactions Per Second (TPS)
– Under what workload? How is machine configured?
ENGS 116 Lecture 18
I/O Benchmarks: Transaction Processing
• Early 1980s great interest in OLTP
– Expecting demand for high TPS (e.g., ATM machines, credit cards)
– Tandem’s success implied medium range OLTP expands
– Each vendor picked own conditions for TPS claims, reported only CPU
times with widely different I/O
– Conflicting claims led to disbelief of all benchmarks  chaos
• 1984 Jim Gray of Tandem distributed paper to Tandem employees
and 19 in other industries to propose standard benchmark
• Published “A measure of transaction processing power,” Datamation,
1985 by Anonymous et. al
– To indicate that this was effort of large group
– To avoid delays of legal department of each author’s firm
– Author still gets mail at Tandem
29
ENGS 116 Lecture 18
I/O Benchmarks: TP by Anon et. al
• Proposed 3 standard tests to characterize commercial OLTP
– TP1: OLTP test, DebitCredit, simulates ATMs (TP1)
– Batch sort
– Batch scan
• Debit/Credit:
– One type of transaction: 100 bytes each
– Recorded 3 places: account file, branch file, teller file; events
recorded in history file (90 days)
• 15% requests for different branches
– Under what conditions, how to report results?
30
ENGS 116 Lecture 18
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I/O Benchmarks: TP1 by Anon et. al
• DebitCredit Scalability: size of account, branch, teller, history function
of throughput
TPS
Number of ATMs
Account-file size
10
1,000
0.1 GB
100
10,000
1.0 GB
1,000
100,000
10.0 GB
10,000
1,000,000
100.0 GB
– Each input TPS  100,000 account records, 10 branches, 100 ATMs
– Accounts must grow since a person is not likely to use the bank more
frequently just because the bank has a faster computer!
• Response time: 95% transactions take ≤ 1 second
• Configuration control: just report price (initial purchase price + 5 year
maintenance = cost of ownership)
• By publishing, in public domain