compeng 701_2

7/23/2019 COMPENG 701_2

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Computer components

• CPU - Central Processor Unit(e.g., G3, Pentium III, RISC)

• RAM - Random Access Memory(generally lost when power cycled)

• VRAM - Video RAM (amount setsscreen size and color depth)

• ROM -Read Only Memory (usedfor boot)• IO Input/Output: IO through

interfaces such as SCSI (SmallComputer System Interface), USB(Universal Serial Bus), Firewire(video standard), Ethernet

• HD - Hard Disk Drive (permanentstorage); CDROM - Compact DiskRead-only-memory; CD-RW; DVDRead/Write

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Computer Architecture

• Computer Organization

– The von Neumann architecture

– Same storage device for both instructions and data

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Computer Architecture

• Device Controllers

– Memory mapped I/O – Direct Memory Access (DMA)

• Instruction Set

– Data transfer operations

– Arithmetic / logic operations

– Control flow instructions

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The architecture of a hypothetical

machine

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Computer basics

• Smallest thing a computer knows is a bit 0 or 1

(false/true)• CPU knows how to perform and, or, xor (exclusive or) operations

– And returns true if both same – Or returns true if either true

– Xor returns true if different

• CPU is a massive collection of and and or gates• A specific CPU has a set of instructions it can

execute (usually 50-100, machine language)

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Basics

• The number of instructions per seconds is set by

the “clock speed”, e.g., 500 MHz Pentium III• One clock tick is called a cycle, modern CPUscan often execute more than one instruction per

cycle• All programs end up as set of instructions to beexecuted by the CPU (compilers/linkers or

interpreters do this for you)• Floating point speed is measured in floating

point operations per seconds, or flops

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Bits/Bytes and words

• Bits are grouped into larger units

– 8 bits = 1 byte (still common) – 2/4/8 bytes = word (varies between CPU)

– Most desktop machines are 32-bit words but64 bits machines are becoming more common(set by data bus)

– Why important? Sets minimum size unit youcan access in program, and often precision

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Grouping words and bytes

• Number of unique values that can berepresented depends on number of bits

• With n bits one can have 2n unique values• For n=8 have (byte) = 256

• Grouped into larger units to represent differentdata• ASCII (American Standard Code for Information

Interchange) – Basic version is 7 bit (127 characters) – A-Z, a-z, 0-9 and special characters – Values <32 are “control characters”

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Numbers and bases

• Numbers represented in different base systems

– Binary base 2 (0-1) – Octal base 8 (0-7)

– Hexadecimal base 16 (0-15, with A-F representing

10-15) – E.g, 5410=3616=668=1101102

• Prefixes: kilo = 1024; mega=1048576;

giga=10737741824 (approximately 103,106,109)

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Program Execution

“The machine cycle”

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The composition of an instruction

for a machine

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Stored Program

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Performing the fetch step of the

machine cycle I

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Performing the fetch step of the

machine cycle II

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Decoding the instruction 35A7

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Mnemonics

• It is hard to remember commands as numbers

• Use words associated with the numbers

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Some Assembly language

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Components: OS

• The Operating System (OS)

– Controls everything in the way the computer works. – Not Specific to a CPU type but often some OSs are

associated with specific CPUs

• G3/4/5 68x series MacOS• Pentium, x86 DOS (Windows)

• SPARC Solaris (Unix)

– OS controls IO and memory management

• Program implementations are dependent on OS

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Programming interface to OS

• Depending on language used, OS

interface may or may not be important• For Fortran, C, C++ when program is

linked OS routines are needed – How to read from keyboard or file?

– How to write to screen or disk?

• In your program you do not need to go intothe low-level (OS) details

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Storage in memory

• Memory may be treated as a linear array ofbytes from 1 to <size of memory>

• The OS keeps track of which parts of memoryare being used and which parts are still free foruse by programs and data

• Some computers do “byte-swapping”, i.e., thebytes are not counted linearly but rather areswitched. The main (but not only) styles are Big

Endian (HP, Sun, Macs) and Little Endian (PC).• Affects ability to transfer binary data (TCP knowsthis and will accommodate this up to a certaindegree)

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Hard disks

• Hard disks contain thecomputers “file system” (allows

access through file names)• Directory structure points towhere files are located (reasonfor having less space available

than the size of disk + somecalibration tracks)• Actual content of HD and

directories depend on OS used

(different standards exist, e.g.,FAT16, FAT32, NTFS forWindows, EXT2 for Linux

• In general, OS can only use

their own file-system

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Accessing RAM vs. HDD

• The highest possiblebandwidth (peak

bandwidth) for the varioustypes of RAM

– However, RAM also has tomatch the motherboard,chipset and the CPUsystem bus

• HDD ~ only 80MB/s

• In MATLAB: try save,load, pack, clear

Module type Max.Transfer, MB/s

SD RAM, PC100 800

SD RAM, PC133 1064

Rambus, PC800 1600

Rambus, Dual PC800 3200

DDR 266 (PC2100) 2128

DDR 333 (PC2700) 2664

DDR 400 (PC3200) 3200

DUAL DDR PC3200 6400

DUAL DDR2-400 8600

DUAL DDR2-533 10600

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RAM and “fast” RAM/cache

• A CPU cache is a cache usedby the central processing unit of

a computer to reduce theaverage time to access memory

– Access time: roughly speaking“CPU speed against the bus speed”

http://en.wikipedia.org/wiki/Cache

http://en.wikipedia.org/wiki/Central_processing_unit

http://en.wikipedia.org/wiki/Computer

http://en.wikipedia.org/wiki/Computer_storage

http://en.wikipedia.org/wiki/Computer_storage

http://en.wikipedia.org/wiki/Computer

http://en.wikipedia.org/wiki/Central_processing_unit

http://en.wikipedia.org/wiki/Cache

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Integers

• Integer numbers can be represented exactly (up

to the range allowed by the number of bytes)• A 2-byte integer, unsigned 0-65535, signed

±32767 (sometimes called short)

• A 4-byte integer, unsigned 0-4294967295,signed ±2147483827

– (With a 32-bit address bus, can have 4Gbytes ofmemory—reason max memory is limited incomputers)

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Floating point

• Representations vary between machines (often

reason binary files can not be shared)

– Precise layout of bits depends on machine andformat; all formats are (mantissa)*2(exponent)

Th IEEE t d d f fl ti

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The IEEE standard for floating

point arithmetic• Single precision (32 bits=4 bytes)

S EEEEEEEE FFFFFFFFFFFFFFFFFFFFFF0 1 8 9 31

The value V represented by the word may be determined as follows:

• If E=255 and F is nonzero, then V=NaN ("Not a number")• If E=255 and F is zero and S is 1, then V=-Infinity• If E=255 and F is zero and S is 0, then V=Infinity• If 0<E<255 then V=(-1)S * 2(E-127) * (1.F) where "1.F" is intended to represent

the binary number created by prefixing F with an implicit leading 1 and a

binary point• If E=0 and F is nonzero, then V=(-1)S * 2 (-126) * (0.F). These are

"unnormalized" values.• If E=0 and F is zero and S is 1, then V=-0• If E=0 and F is zero and S is 0, then V=0

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Single precision floating point• In particular

0 00000000 00000000000000000000000 = 0

1 00000000 00000000000000000000000 = -0

0 11111111 00000000000000000000000 = Infinity1 11111111 00000000000000000000000 = -Infinity

0 11111111 00000100000000000000000 = NaN1 11111111 00100010001001010101010 = NaN

0 10000000 00000000000000000000000 = +1 * 2(128-127) * 1.0 = 20 10000001 10100000000000000000000 = +1 * 2(129-127) * 1.101 = 22*(23+22+1)/23 = 6.5

1 10000001 10100000000000000000000 = -1 * 2(129-127) * 1.101 = -6.5

0 00000001 00000000000000000000000 = +1 * 2(1-127) * 1.0 = 2(-126)

0 00000000 10000000000000000000000 = +1 * 2(-126) * 0.1 = 2(-127)

0 00000000 00000000000000000000001 = +1 * 2(-126) *

0.00000000000000000000001 =2(-149) (Smallest positive value)

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Double precision floating point

S EEEEEEEEEEE FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF

0 1 11 12 63

The value V represented by the word may be determined as follows:

• If E=2047 and F is nonzero, then V=NaN ("Not a number")

• If E=2047 and F is zero and S is 1, then V=-Infinity

• If E=2047 and F is zero and S is 0, then V=Infinity• If 0<E<2047 then V=(-1)S * 2(E-1023) * (1.F) where "1.F" is intended to represent

the binary number created by prefixing F with an implicit leading 1 and a binarypoint.

• If E=0 and F is nonzero, then V=(-1)S

* 2(-1022)

* (0.F) These are "unnormalized"values.

• If E=0 and F is zero and S is 1, then V=-0

• If E=0 and F is zero and S is 0, then V=0

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Is the finite precision an issue?

• An extended example: condition number of a

symmetric matrix.

Consider a system of linear equations

and the perturbed system

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

1997

1999

998999

9991000

y

x

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ 01.1997

99.1998

ˆ

ˆ

998999

9991000

y

x

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Example

Note

What went wrong?

Recallan n-by-n matrix A is symmetric iff A=A

T

.Fact (“spectral decomposition”):

If A is symmetric, it may be written as

A=UDUT,where D is the diagonal matrix,

U is unitary (i.e., UUT=I, the identity matrix).

99.18

97.20

ˆ

ˆ but

1

1

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

+=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =⎟⎟

⎠

⎞⎜⎜⎝

⎛ y

x

y

x

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Example

Fact: U is unitary is “almost the same” as

U is a rotation matrix(“almost the same” because U might includereflections).

Fact: For A=UDUT as before, the diagonalelements of D are the eigenvalues of A andcolumns of U are the right eigenvectors of A.

Recall t is an eigenvalue of A iff det(A-tI)=0,x is the corresponding right eigenvector iff

Ax=tx.

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Example

• How does A act on x, step-by-step: Ax=UDUTx=UD(UTx)=U(D(UTx)),

that is, “rotate, scale, rotate back”.

Define the condition number of A as

χ(A)=|t|max(A) / |t|min(A)where |t|max and |t|min are the largest and the smallest inabsolute value eigenvalues of A.

χ(A) shows how far off the solution to Ax=b may be fromthe solution to Ax=b+Δb (a measure of “relativesingularity”).

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Assignment 2MATLAB and C code are posted on the web. The MATLAB script is slightly modified in an

attempt to produce a more accurate graph: the C call is referred to several timesusing reps counter.

1. Using the MATLAB script from the class, try to identify the cache size on your

machine. Note that usually a number in MATLAB occupies 8 bytes (double precisionfloating point), and that the function requires roughly 2-times the memory needed tostore the x vector (x in the input, x_new in the output). Explain your results. As asanity check, you might want to use a CPU info tool.

2. Condition number I: for the class example, answer the following:a) find the matrix spectral decomposition and compute the matrix condition number,b) find perturbations Δb of the right-hand side of the equation with (the Euclidean norm) ||Δb||=1

that give the largest and the smallest errors in the solution vector; plot each of the twoperturbations, together with their image under UT, under D-1UT and, finally, under A-1,

c) given the equation for the curve ||Δb||=1, find its image under A-1 and plot it,d) find the explicit expression for the condition number χ(A-γ I) for γ > 0.

3. Condition number II: given a 2x2 matrix with double-precision entries, what is theworst condition number this matrix might have? Explain your answer.

4. Condition number III: for the question 2 of the first assignment, is there any ill-conditioning (ill-conditioning refers to a matrix having large condition number, henceoften resulting in numerical instability; if you happen to encounter complexeigenvalues, you are on the wrong track; for the definition in class, matrix must be

symmetric)? Explain your answer.5. Bonus question: find all the distinct spectral decompositions of the identity matrix.

compeng 701_2

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