# excess- k notation

DESCRIPTION

Excess- k notation. A method for representing signed integers add excess amount k to every number to obtain a positive integer (smallest number that can be represented is k ) most significant bit represents sign (1 = positive, 0 = negative) Examples (excess-128): - PowerPoint PPT PresentationTRANSCRIPT

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Excess-k notation

A method for representing signed integers— add excess amount k to every number to obtain a positive integer

(smallest number that can be represented is k)— most significant bit represents sign (1 = positive, 0 = negative)

Examples (excess-128):

1 0 1 0 0 0 0 1 = +330 1 0 1 1 1 1 1 = 331 0 0 0 0 0 0 0 = 0 unique representation of 0!

If bias k = 2n-1, then excess-k is almost identical to two’s complement— sign-bit reversed (1 means positive)— comparing integers identical to the unsigned case

Big endian (MSB first) vs. little endian (LSB first)

128 64 32 16 8 4 2 1

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Floating Point Numbers

How can we represent 3.14 ?— What’s wrong with: (int_part, frac_part)— 3.14 and 3.014 have the same representation!

The leading-zeroes problem can be solved if numbers are normalized— write the number in the form d.f 10e , d is a single non-zero digit— normalized(3.14) = 3.14 100, normalized(0.314) = 3.14 101

In binary, the “d” part will always be 1 (zero is a special case)— this implicit 1 can be ignored

Ideal representation scheme has these features:— can represent positive and negative, low and high magnitude— it is easy to compare two numbers— it is easy to do basic math

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IEEE 754 standard

Format for single-precision (32-bit) and double-precision (64-bit) reals

The normalized (non-zero) binary number 1.f 2e is stored as

Comparison of floats almost identical to comparison of ints!

MIPS has separate floating point registers and instructions

23-bit fraction f8-bit exponent eexcess-127

notation

1 sign bit1 =

negative0 =

positive

single precision float

52-bit fraction f11-bit exponent eexcess-1023

notation

1 sign bit1 =

negative0 =

positive

double precision double

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Two notions of performance

Which has higher performance?

From a passenger’s viewpoint: latency (time to do the task)— hours per flight, execution time, response time

From an airline’s viewpoint: throughput (tasks per unit time)— passengers per hour, bandwidth

Latency and throughput are often in opposition

Aircraft DC to Paris Passengers

747 6 hours 500

Concorde 3 hours 125

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Some Definitions

Latency is time per task (e.g. hours per flight)

If we are primarily concerned with latency,

Performance(x) = 1 execution_time(x)

Throughput is number of tasks per unit time (e.g. passengers per hour)

Performance(x) = throughput(x) Again, bigger is better

Relative performance: “x is N times faster than y”

N = Performance(x) Performance(y)

Bigger is better

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CPU performance

The obvious metric: how long does it take to run a test program?— Aircraft analogy: how long does it take to transport 1000

passengers?

Our vocabulary Aircraft analogy

N instructions N passengers

c cycles per instruction (1/c) passengers per flight

t seconds per cycle t hours per flight

Time = N c t seconds Time = N c t hours

Cycles Per Instruction

CPU timeX,P = Instructions executedP * CPIX,P * Clock cycle timeX

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Instructions executed:—the dynamic instruction count (#instructions actually executed)—not the (static) number of lines of code

Cycles per instruction:—average number of clock cycles per instruction—function of the machine and program

• CPI(floating-point operations) CPI(integer operations)• Pentium executes same instructions as an 80486, but faster

—Single-cycle machine: each instruction takes 1 cycle (CPI = 1)• CPI can be 1 due to memory stalls and slow instructions• CPI can be 1 on superscalar machines

Clock cycle time: 1 cycle = minimum time it takes the CPU to do any work—clock cycle time = 1/ clock frequency—500MHz processor has a cycle time of 2ns (nanoseconds)—2GHz (2000MHz) CPU has a cycle time of just 0.5ns—higher frequency is usually better

The three components of CPU performance

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CPU timeX,P = Instructions executedP * CPIX,P * Clock cycle timeX

The easiest way to remember this is match up the units:

Make things faster by making any component smaller!

Often easy to reduce one component by increasing another

Execution time, again

Seconds

=

Instructions

*

Clock cycles

*

Seconds

Program

Program Instructions Clock cycle

Program Compiler ISA Organization

Technology

InstructionExecuted

CPI

Clock Cycle Time

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Let’s compare the performances two x86-based processors—An 800MHz AMD Duron, with a CPI of 1.2 for an MP3

compressor—A 1GHz Pentium III with a CPI of 1.5 for the same program

Compatible processors implement identical instruction sets and will use the same executable files, with the same number of instructions

But they implement the ISA differently, which leads to different CPIs

CPU timeAMD,P= InstructionsP * CPIAMD,P * Cycle timeAMD

=

CPU timeP3,P = InstructionsP * CPIP3,P * Cycle timeP3

=

Example: ISA-compatible processors

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Another Example: Comparing across ISAs

Intel’s Itanium (IA-64) ISA is designed facilitate executing multiple instructions per cycle. If it achieves an average of 3 instructions per cycle, how much faster is it than a Pentium4 (which uses the x86 ISA) with an average CPI of 1?

a) Itanium is three times fasterb) Itanium is one third as fastc) Not enough information

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Improving CPI

Some processor design techniques improve CPI— Often they only improve CPI for certain types of instructions

where Fi = fraction of instructions of type i

First Law of Performance:

Make the common case fast

i

n

ii FCPICPI

1

Second Law of Performance:

Make the fast case common

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Example: CPI improvements

Base Machine:

How much faster would the machine be if:— we added a cache to reduce average load time to 3 cycles?— we added a branch predictor to reduce branch time by 1

cycle?— we could do two ALU operations in parallel?

Op Type Freq (Fi) CPIi contribution to CPI

ALU 50% 3

Load 20% 6

Store 20% 3

Branch 10% 2

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Amdahl’s Law states that optimizations are limited in their effectiveness

Example: Suppose we double the speed of floating-point operations—If only 10% of the program execution time T involves floating-

point code, then the overall performance improves by just 5%

Amdahl’s Law

Execution time after

improvement =

Time affected by improvement

+Time

unaffected by improvement

Amount of improvement

Execution time after

improvement

=

0.10 T

+0.90

T=

0.95 T

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