ECE260: Fundamentals of Computer Engineering Technologies ... · ECE260: Fundamentals of Computer Engineering CPI Example • Given two Computers that implement some program using

ECE260: Fundamentals of Computer Engineering

James Moscola Dept. of Engineering & Computer Science York College of Pennsylvania


Based on Computer Organization and Design, 5th Edition by Patterson & Hennessy

Technologies & Performance


Technology Trends• Electronics technology continues to evolve rapidly

• Increased capacity and performance (i.e. larger and faster)

• Reduced cost

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DRAM Capacity


Semiconductor Technology• Silicon is a semiconductor

• Does not conduct electricity well

• Used as the base material for integrated circuits (ICs)

• Can add various materials to silicon to modify its properties

• Conductors (i.e. wires)

• Insulators

• Switches (i.e. transistors)

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ECE260: Fundamentals of Computer Engineering 4

Manufacturing Integrated Circuits (ICs)


Intel Core i7 Wafer• 300mm wafer, 280 processor dies, 32nm technology

• Each chip is 20.7 x 10.5 mm

5


Integrated Circuit Cost• Wafer cost and area are fixed

• Defect rate is determined by manufacturing process

• Die area determined by architecture, circuit design, and process technology

• Yield — proportion of working dies per wafer

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Cost per die = Cost per wafer

Dies per wafer × Yield

Yield = 1

(1 + (Defects per area × Die area/2))2

Dies per wafer ≈Wafer area

Die area


Defining Performance can be Challenging • Which airplane has the best performance?

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0 100 200 300 400 500

DouglasDC-8-50

BAC/SudConcorde

Boeing 747

Boeing 777

Passenger Capacity

0 2000 4000 6000 8000 10000

Douglas DC-8-50

BAC/SudConcorde

Boeing 747

Boeing 777

Cruising Range (miles)

0 500 1000 1500

DouglasDC-8-50

BAC/SudConcorde

Boeing 747

Boeing 777

Cruising Speed (mph)

0 100000 200000 300000 400000

Douglas DC-8-50

BAC/SudConcorde

Boeing 747

Boeing 777

Passengers x mph


Response Time and Throughput• Multiple different measures of computing performance

• Response time (a.k.a. execution time) <= this will be our primary focus for now

• Time to complete a task

• Throughput

• Total work done per unit time (e.g. tasks/hour)

• How are response time and throughput affected by:

• Replacing the processor with a faster version?

• Both are improved since each task can be done faster thus allowing more tasks to get completed

• Adding more processors?

• Only throughput increases since more tasks can be done simultaneously, but no single task is faster

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Defining Performance• Define performance in terms of execution time

• The smaller the execution time, the higher the performance

• Comparing performance between systems

• Performance of SystemX is greater than performance of SystemY

• Quantifying the relative performance

• SystemX is n times faster than SystemY

• Example: time taken to run a program

• 10s on SystemA, 15s on SystemB

• SystemA is 1.5 times faster than SystemB

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PerformanceX = 1ExecutionTimeX

PerformanceX > PerformanceY

1>

1ExecutionTimeX ExecutionTimeY

PerformanceX = n

PerformanceY

PerformanceA =

ExecutionTimeB= n

PerformanceB ExecutionTimeA

15s= 1.5

10s


Measuring Performance• Program time can be measure in different ways:

• Elapsed time

• Total response/execution time (as if measured with a stopwatch)

• Includes: processing, I/O, operating system overhead, idle time, etc.

• Not a great measure since your computer is typically busy with more than one task

• CPU time

• Time spent processing a specific task

• Does NOT include: I/O, operating system overhead, idle time, etc.

• Comprises User CPU time + System CPU time

• User CPU time is time spent in program

• System CPU time is time spent in operating system doing tasks related to user program

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CPU Clocking• Operation of digital hardware governed by a constant-rate clock

• Clock period: duration of a clock cycle as measured in seconds

• e.g., 250ps = 0.25ns = 250×10–12s

• Clock frequency (rate): cycles per second

• e.g., 4.0GHz = 4000MHz = 4.0×109Hz

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1= Clock Frequency

Clock Period

1= Clock Period

Clock Frequency


CPU Time• Relating CPU Time, CPU cycles for a program, and clock cycle time (a.k.a. clock period)

• To improve performance, reduce CPU Time

• Reduce CPU Time by:

• Reducing number of clock cycles

• Increasing clock rate

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CPU Time = CPU Clock Cycles × Clock Cycle

Time

CPU Time =CPU Clock Cycles

Clock Frequency (i.e. rate)


CPU Time Example• Existing computer called ComputerA has a 2GHz clock rate and completes a program in 10 seconds

• CPU TimeA = 10s, Clock RateA = 2GHz, Clock CyclesA = <not given, but easy to calculate>

• Goal: Design a new computer, ComputerB, that can run same program in 6 seconds

• Increasing clock rate will require 1.2 times as many clock cycles as ComputerA

• What clock rate should be specified for ComputerB?

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Clock RateB =Clock CyclesB

=1.2 × Clock CyclesA

CPU TimeB 6s

Clock CyclesA = CPU TimeA × Clock RateA

Clock CyclesA = 10s × 2×109 Hz

Clock CyclesA = 20×109 cycles

Clock RateB =Clock CyclesB

=1.2 × 20×109 cycles

= 4GHzCPU TimeB 6s


Instruction Count and Performance• Instruction count for a program is the number

of instructions executed to run the program

• Determined by program logic, the Instruction Set Architecture (ISA), and the compiler

• Different instructions may require different number of clock cycles

• ADD, SUBTRACT, SHIFT ==> 1 clock cycle

• MULTIPLY ==> 4 clock cycles

• Execution time can be computed as

• Average clock cycles per instruction (CPI)

• Determined by CPU hardware

• If different instructions have different CPI then average CPI is affected by instruction mix

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CPU Clock Cycles =

Instructions for program (i.e. count)

×Average Clock

Cycles per Instruction (CPI)


Instruction Count and Performance• Recall:

• Thus, using:

• We can rewrite CPU Time as:

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CPU Clock Cycles =


×Average Clock


CPU Time =Instructions for program (i.e. count)

×Average Clock


× Clock Cycle Time


Time


Instruction Count and Performance• Furthermore:

• Can also be written as:

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CPU Time =Instructions for program (i.e. count)

×Average Clock


× Clock Cycle Time

CPU Time =Instruction Count × CPI

Clock Frequency (i.e. rate)


CPI Example• Given two Computers that implement some program using the same ISA

• ComputerA: Cycle Time = 250ps, CPI = 2.0

• ComputerB: Cycle Time = 500ps, CPI = 1.2

• Which computer is faster, and by how much?

• First, determine the CPU Time for each computer using the following equation where

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Time

CPU Clock Cycles =


×Average Clock



CPI Example (Continued)• Compute CPU Time for each

• Which is faster?

• ComputerA is 1.2 times faster than ComputerB

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CPU TimeA = Instruction Count × CPIA × Clock Cycle TimeA

CPU TimeA = Instruction Count × 2.0 × 250ps

CPU TimeA = Instruction Count × 500ps

CPU TimeB = Instruction Count × CPIB × Clock Cycle TimeB

CPU TimeB = Instruction Count × 1.2 × 500ps

CPU TimeB = Instruction Count × 600ps

CPU PerformanceA=

CPU TimeB=

Instruction Count × 600ps= 1.2

CPU PerformanceB CPU TimeA Instruction Count × 500ps


Power Trends• Clock rate and power requirements increased rapidly for many years

• In PostPC era, we are interested in reducing power requirements to conserve battery life

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Power Trends• In modern integrated circuits, primary source of energy consumption is dynamic energy

• Dynamic energy is consumed when transistors switch states (i.e. 0 → 1 or 1 → 0)

• The dynamic power requirements are dependent on the dynamic energy and the frequency at which transistors switch states

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Power ∝ 1/2 × Capacitive Load × Voltage2 × Frequency Switched


Reducing Power Example• Developed a new CPU with the following features

• 85% of capacitive load of old CPU

• 15% voltage reduction and 15% frequency reduction

• What is the impact on dynamic power?

• Use the following power equation

• The new CPU uses about half the power of the old CPU

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PowerNew=

(CapacitiveLoadOld × 0.85) × (VoltageOld × 0.85)2 × (FrequencySwitchedOld × 0.85)= 0.854 = 0.52

PowerOld CapacitiveLoadOld × VoltageOld2 × FrequencySwitchedOld

Power ∝ 1/2 × Capacitive Load × Voltage2 × Frequency Switched

Documents

ECE260: Fundamentals of Computer Engineering Technologies ... · ECE260: Fundamentals of Computer Engineering CPI Example • Given two Computers that implement some program using