R E F E R E N C E : D A V I D A . P A T T E R S O N & J O H N L . H E N N E S S Y – C O M P U T E R O R G A N I Z A T I O N A N D D E S I G N

Module 1a: Organization & Architecture:

Structure & Function

MODULE 1 (Overview & Computer Performance)

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Computer Architecture Computer Organization

Refers as a set of attributes of a system as seen by programmer

Deals with all physical components of computer systems that interacts with each other to perform various functionalities

The lower level of computer organization is known as microarchitecture which is more detailed and concrete.

Comp. Architecture vs Comp. Organization 2

The instruction set

The number of bits to

represent data types

I/O mechanisms

memory addressing techniques

Control signals

Interfaces between computer and

peripherals

The memory

technology being

used

Computer Architecture Computer Organization

The difference between architecture and organization is best described by a non-computer example.

“Is the gear level in a motorcycle part of it is architecture or organization?

The architecture of a motorcycle is simple; it transports you from A to B. The gear level belongs to the

motorcycle's organization because it implements the function of a motorcycle but is not part of that function”

Comp. Architecture vs Comp. Organization

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Computer Architecture Computer Organization

Refers to those attributes visible to the programmer

Refers to the how features are implemented

Comp. Architecture vs Comp. Organization

Can we multiply 2 numbers?

Yes, we can multiply

How to multiply.

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The Computer Family

Many computer manufacturers offer a family of computer models, all with the same architecture but with differences in organization.

All Intel x86 family share the same basic architecture

The IBM System/370 architecture first introduced in 1970 included a number of models that share the same basic architecture and has survived to this day as the architecture of IBM’s mainframe product line.

The newer models retained the same architecture so that the customer’s software investment was protected (code compatibility)

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6

Register different

growing fast….

same

architecture

differences in

organization

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Structure and Function

A computer is a complex system with a hierarchical system of interrelated subsystems with different levels.

At each level, the designer is concerned with structure and function:

Structure: The way in which the components are interrelated.

Function: The operation of each individual component as part of the structure.

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Structure

4 main structural

components

Central processing unit (CPU)

Main memory

I/O

System interconnection

Controls the operation of the

computer and performs its data

processing functions

Stores data

Moves data between the

computer and its external

environment

Mechanism for

communication among

CPU, main memory, and

I/O

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Structure: CPU

CPU

Control unit

Arithmetic and logic unit

(ALU)

Registers

CPU interconnection

Controls the operation

of the CPU

Performs the

computer’s data

processing

functions Provides storage

internal to the

CPU

Mechanism for

communication among the

control unit, ALU, and

registers

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THE COMPUTER: TOP-LEVEL STRUCTURE

Computer

Main

Memory

Input

Output

Systems

Interconnection

Peripherals

Communication

lines

Central

Processing

Unit

Computer

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FUNCTION

Functions

• process data in variety of forms and requirements Data Processing

• short and long term data storage for retrieval and update Data storage

• move data between computer and outside world. Data movement

• control of process, move and store data using instruction.

Control

There are only four functions

How are functions performed?

Through PROGRAMS

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Program

A sequence of steps

For each step, a computer function is executed

For each operation, a different/new set of control signals is needed

For each operation a unique code (instruction) is provided e.g. ADD, MOVE

A hardware segment accepts the code and issues the control signals

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Executing A Program

Approach 1: Hardwired program

connecting/combining various logic components to store data and perform arithmetic and logic operations

Hardwired systems are inflexible

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Executing A Program

Approach 2: Software

General purpose hardware can do different tasks, given correct control signals

Instead of re-wiring, supply a new set of control signals through instruction codes

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R E F E R E N C E : W I L L I A M S T A L L I N G S – C O M P U T E R O R G A N I Z A T I O N & A R C H I T E C T U R E ( C H : P E N T I U M E V O L U T I O N )

Computer Evolution

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Von Neumann Machine

1945: stored-program concept first implemented for EDVAC (Electronic Discrete Variable Computer).

Key concepts:

Data and instructions are stored in a single read-write memory.

The contents of this memory are addressable by location, without regard to the type of data contained there

Execution occurs in a sequential fashion from one instruction to the next

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Structure of von Neumann machine 18

Microprocessors (µP) Intel

Microprocessor : all CPU components on a single chip

1971 - 4004

First microprocessor

4 bit

Followed in 1972 by 8008

8 bit

Both designed for specific applications

1974 - 8080

Intel’s first general purpose microprocessor

Designed to be the CPU of a general purpose microcomputer

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Intel µP Evolution ..

8080 first general purpose microprocessor

8 bit data path

Used in first personal computer – Altair

8086 much more powerful

16 bit

instruction cache, prefetch few instructions

8088 (8 bit external bus) used in first IBM PC

80286 16 MB memory addressable

80386 First 32 bit design

Support for multitasking- run multiple programs at the same time

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.. Intel µP Evolution ..

80486 sophisticated powerful cache and instruction pipelining

built in maths co-processor

Pentium Superscalar technique - multiple instructions executed in parallel

Pentium Pro Increased superscalar organization

Aggressive register renaming

branch prediction

data flow analysis

speculative execution

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.. Intel µP Evolution

Pentium II MMX technology graphics, video & audio processing

Pentium III Additional floating point instructions for 3D graphics

Pentium 4 Further floating point and multimedia enhancements

Itanium 64 bit

Core Duo starts of a multi core processor

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Intel Evolution

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R E F E R E N C E : D A V I D A . P A T T E R S O N & J O H N L . H E N N E S S Y – C O M P U T E R O R G A N I Z A T I O N A N D D E S I G N

Module 1b: Understanding & Measuring

Performance

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Introduction

Hardware performance is often key to the effectiveness of an entire system of hardware and software.

For different types of applications, different performance metrics may be appropriate, and different aspects of a computer systems may be the most significant factor in determining overall performance.

Understanding how best to measure performance and limitations of performance is important when selecting a computer system

To understand the issues of assessing performance. Why a piece of software performs as it does?

Why one instruction set can be implemented to perform better than another?

How some hardware feature affects performance?

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Why measure performance?

Performance is important!

Identify HW/SW performance problems

Comparisons:

Which machine is faster?

Which ISA is better?

Which implementation (of an ISA) is faster?

Expose significant performance issues (enable us to ignore unimportant issues)

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More than one way to measure performance

Performance is evaluated differently by different entity.

Better performance means faster processing speed (e.g. faster completion of a task/job)

Better performance means higher throughput (doing more jobs in a time given)

Better performance means doing more jobs at a smaller cost

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Which plane has better performance?

If higher throughput (transporting

more passengers) is better

performance If higher speed

is better

performance

If better performance

means having a long

range

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Understanding terminology

Execution time (a.k.a response time) :The total time it takes from start to completion of a task

Throughput :The total amount of tasks completed in a given time interval

CPU execution time (a.k.a CPU time) :The actual time CPU spends on a specific task

User CPU time: time the CPU spends on running the actual program

System CPU time: time the CPU spends on OS overhead on behalf of the program

Clock cycle (a.k.a ticks, cycle) :Discrete time intervals (the processor clock which runs at a constant rate). Usually in nanoseconds (ns) or picoseconds (ps)

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Understanding terminology

Clock period (a.k.a clock cycle time): the duration of one clock cycle. In sec, or msec

Clock rate (or frequency) : the speed that the microprocessor executes each instruction or each vibration of the clock. In MHz/GHz. Frequency = 1/clock period

1 MHz representing 1 million cycles per second,

1 GHz representing 1 thousand million cycles per second (109)

Clock cycles per instruction (CPI) : The average number of clock cycles each instruction takes to execute

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Figure 1

Figure 2

1 cycle time =

how length of

this clock cycle

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Common performance metrics

MB/s, Mb/s: Megabytes, Megabits Per Second

MIPS: Millions of Instructions Per Second

CPI: Clock Cycles Per Instruction

IPC: Instructions Per Clock cycle

Hz: (processor clock frequency) cycles Per Second

LIPS: Logical Interference Per Second

FLOPS: Floating-Point arithmetic Operations Per Second

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Computer performance measures

Performance is related to execution time.

To maximize performance, we want to minimize the execution time

If performance of Computer A is 10 times better than Computer B, what is the relation between their execution times?

This shows that CompB

needs 10x more time than

CompA to execute a given

task.

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CPU Execution Time 34

Clock period = 1

frequency

If a processor has frequency, 320 MHz:

Clock period = 1 = 3.125ns

320 000 000

Clock rate = frequency (Hz)

“the frequency at which a CPU is

running. It is measured in Hz unit”

Example 1: Improving Performance

Our favorite program runs in 10 seconds on computer A, which has a 4 GHz clock. Computer B will run this program in 6 seconds, given that computer B requires 1.2 times as many clock cycles as computer A for this program. What is computer B’s clock rate? Answer: 8Ghz

What do we know?

Computer A

CPU Execution Time = 10s

Clock rate (CR) = 4GHz = 4 x 109 Hz

Computer B

CPU Execution Time = 6s

Clock cycle (CC) = 1.2 x clock cycle Computer A

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Example 1: Improving Performance

What do we know?

Computer A

CPU Execution Time = 10s

Clock rate (CR) = 4GHz = 4 x 109 Hz

Computer B

CPU Execution Time = 6s

Clock cycle (CC) = 1.2 x clock cycle Computer A

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Clock Cycles per Instruction (CPI)

Previously, our calculations of Execution time did not include the number of instructions needed for the program.

Different instructions may take different amounts of time to execute, depending on what they do

Example: The MOV (Move) instruction – moving data from one place to another

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The MOV instruction : Analogy

Analyze Conrad’s movement of putting the red balls into the container.

Balls from prime storage:

-walk

-fetch ball

-walk (halfway)

-walk

-put ball in container

Total = 5

Balls from sub storage:

-fetch ball

-walk

-put ball in container

Total = 3

To do 5 movements takes

longer to execute than 3

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CPU Execution Time

a.k.a Instruction count

CPU clock cycle

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Example 2: Using Performance Equation

Suppose we have two implementations of the same instruction set architecture (ISA) and for the same program. Which computer is faster and by how much?

Computer A: clock cycle time=250 ps and CPI=2.0

Computer B: clock cycle time=500 ps and CPI=1.2

Note: because both computer uses the same program, and the Instruction Count is not given, we can assume it to be a variable I

Remember the formula

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Example 2: Using Performance Equation

Remember: the lower the

execution time, the better the

performance.

Computer A is faster

How much faster is Computer A?

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Example 2 (continued)…

We can conclude, A is 1.2

times faster than B for

this program

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Measuring the CPI

Sometimes it is possible to compute the CPU clock cycles by looking at the different types of instructions and using their individual clock cycle counts

Ci = count of the number of instructions of class i executed

CPIi = average number of cycles per instruction for that instruction class

n = number of instruction classes

Remember that overall CPI for a program will depend on both the number of cycles for each instruction type and the frequency of each instruction type in the program execution

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Sample: Calculate CPI

You are on the design team for a new processor. The clock of the processor runs at 200 MHz. The following table gives instruction frequencies for Benchmark B, as well as how many cycles the instructions take, for the different classes of instructions. For this problem, we assume that (unlike many of today's computers) the processor only executes one instruction at a time.

If we say that there are 100 instructions, then:

30 of them will be loads and stores.

50 of them will be arithmetic instructions.

20 of them will be all others.

Formula: (30 * 6) + (50 * 4) + (20 * 3) = 440 cycles/100 instructions = 4.4 cycles per instruction

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Instruction Type Frequency Cycles

Loads & Stores 30% 6 cycles

Arithmetic Instructions 50% 4 cycles

All Others 20% 3 cycles

Factors Affecting the CPU Performance 45

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Example 3 : Comparing Code Segments

A compiler designer is trying to decide between two code sequences for a particular computer. The hardware designers have supplied the following facts:

For a particular high-level-language statement, the compiler writer is considering two code sequence that require the following instruction counts:

a) Which code sequence executes the most instructions?

b) Which will be faster?

c) What is the CPI for each sequence?

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Example: code segments

Example 3 : Part (a)

Sequence 1 executes 2 + 1+ 2 = 5 Instructions

Sequence 2 executes 4 + 1+ 1 = 6 Instructions

Seq 2 executes THE MOST instructions

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Example 3 : Part (b)

Using this equation

Takes 10 cycles to execute

5 instructions

Takes 9 cycles to execute

6 instructions Seq 2 is FASTER

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Example 3 : Part (c)

Code SEQ2 uses fewer

clock cycles, it must

have a lower CPI

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Example 4 : Comparing Code Segments

A processor has 3 classes of instructions:

Which code sequence is faster?

Instruction CPI Code SEQ1

Code SEQ2

Clock cycles SEQ1

Clock cycles SEQ2

A 1 5 3 5 3

B 2 3 2 6 4

C 5 1 2 5 10

9 ins. 16 clock

cycles

7 ins. 17 clock

cycles

Code SEQ1 Takes 16 cycles

to execute 9 instructions

Code SEQ2 Takes 17 cycles

to execute 7 instructions

Code SEQ1 is FASTER

Recall 51

Example 4a: Calculating with CPI

The ADD instruction takes 1 clock cycle to execute, while the MUL instruction takes 3 clock cycles. If a program consists of 20 ADD and 10 MUL instructions, what is the average CPI?

What do we know?

Instruction Clock cycles

Instruction count

ADD 1 20

MUL 3 10

There are 2 instructions

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Example 4a: Calculating average CPI

Instruction Clock cycles

Instruction count

ADD 1 20

MUL 3 10

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Homework (to make you cleverer )

Instruction Instructions

count

Clock

Cycles

a) CPI b) Execution time

A 20 3

B 25 1

C 10 2

D 30 2

E 10 3

F 5 4

CPU X runs a program/code sequence Y which consists of 100 instructions. Calculate and fill in the table below: a) The CPI for each instruction class given below. b) The execution time for each instruction class, given a clock

cycle time is 0.25miliseconds. c) The CPU X’s execution time d) The CPU X’s clock rate

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Did you get the same???

Instruction Instruction

count Clock Cycles

a) CPI b) Execution

time

clock cycle time

0.25

A 20 3 0.15 0.75

B 25 1 0.04 0.25

C 10 2 0.2 0.5

D 30 2 0.07 0.5

E 10 3 0.3 0.75

F 5 4 0.8 1

15 3.75

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(a) (b)

(c)

(d) Clock rate = ∑clock cycles = 15 = 4 ∑ execution time 3.75

Increasing the CPU Performance

Decreasing the clock cycle time

Datapath organization leading to lower CPI

Reduction in the number of executed instructions.

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Example 5: Improve Performance

Our favourite program runs in 20 seconds on Computer P, which has 8 GHz clock. We are trying to help a computer designer build Computer Q that will run this program in 5 seconds. The designer has determined that the substantial increase in the clock rate is possible, but this will affect the rest of the CPU design, causing computer Q to require 1.5 times as many clock cycles as computer P for this program. What clock rate should we tell the designer to target?

What do we know?

Computer P

CPU Execution Time = 20s

Clock rate (CR) = 8GHz = 8 x 109 Hz

Computer Q

CPU Execution Time = 5s

Clock cycle (CC) = 1.5 x clock cycle Computer P

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What do we know?

Computer P

CPU Execution Time = 20s

Clock rate (CR) = 8GHz = 8 x 109 Hz

Computer Q

CPU Execution Time = 5s

Clock cycle (CC) = 1.5 x clock cycle Computer P

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Mandatory Homework

Do Tutorial Module 1 (e-learning).. It is COMPULSORY for my class!

Submission date will be announced.

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Understanding the Units

CPU execution time for a program = Seconds for the program (S/P)

Clock cycle = clock cycles per program (C/P)

Clock cycle time = Seconds per clock cycle (S/C)

Clock rate = clock cycle per second (C/S)

Instruction count = Instructions executed for the program (I/P)

Clock cycle per instruction = Average number of clock cycles per instructions (C/I)

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Understanding the Units

It cancels each other to

give the unit.

Example:

10s = 20cycle/ clock rate

Clock rate = 20/10 cycle per seconds = 2Hz

1 Hz is 1 cycle per second

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