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COMPUTER ARCHITECTURE

Prof. Basel MahafzahLesson 1 - The role of performance - Part 1

Copyright Universit Telematica Internazionale UNINETTUNO

The Role ofPerformanceThe Role of

Performance

Part 1Part 1

OutlineOutline

Computer Performance Computer Performance

Execution Time Execution Time

Performance Performance

Definition of Performance Definition of Performance

Clock Cycles Clock Cycles

PerformancePerformance

What factors of systemperformance are hardwarerelated?

What factors of systemperformance are hardwarerelated?

Why is some hardwarebetter than others fordifferent programs?(example: G4 vs. pent. 4)

Why is some hardwarebetter than others fordifferent programs?(example: G4 vs. pent. 4)

(e.g., Do we need a new machine,or a new operating system?)

(e.g., Do we need a new machine,or a new operating system?)

How does the machinesinstruction set affectperformance?

How does the machinesinstruction set affectperformance?

Airplane Passengers Range SpeedAirplane Passengers Range Speed

Boeing 737 101 630 598

Boeing 747 470 4150 610

Concorde 132 4000 1350

Douglas 146 8720 544

Boeing 737 101 630 598

Boeing 747 470 4150 610

Concorde 132 4000 1350

Douglas 146 8720 544

(mi) (mph)(mi) (mph)

Example on performanceExample on performance

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Which of theseairplanes has the best

performance?

Which of theseairplanes has the best

performance?

The plane with the highestcruising speed is theConcorde

The plane with the highestcruising speed is theConcorde

The plane with the longestrange is the Douglas

The plane with the longestrange is the Douglas

The plane with the largestcapacity is the Boeing 747

The plane with the largestcapacity is the Boeing 747

How much faster is theConcorde compared to the 747?

How much faster is theConcorde compared to the 747?

How much bigger is the 747than the Douglas?

How much bigger is the 747than the Douglas?

Airplane Passengers SpeedAirplane Passengers Speed

Boeing 737 101 598

Boeing 747 470 610

Concorde 132 1350

Douglas 146 544

Boeing 737 101 598

Boeing 747 470 610

Concorde 132 1350

Douglas 146 544

(mph)(mph)ThroughputThroughput

228,750228,750

286,700286,700

178,200178,200

79,42479,424

If we define performance interms of speed, this leaves twopossible definitions:

If we define performance interms of speed, this leaves twopossible definitions:

1. Define the fastest plane

as the one with the highestspeed, taking a singlepassenger from one point toanother in the least time

1. Define the fastest plane

as the one with the highestspeed, taking a singlepassenger from one point toanother in the least time

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The Concorde would clearlybe the fastest

The Concorde would clearlybe the fastest

2. Define the fastest plane asthe one with the highest speed,taking 450 passengers fromone point to another in theleast time

2. Define the fastest plane asthe one with the highest speed,taking 450 passengers fromone point to another in theleast time

The Boeing 747 would clearlybe the fastest.Check the following table

The Boeing 747 would clearlybe the fastest.Check the following table

Airplane Passengers SpeedAirplane Passengers Speed

Boeing 737 101 598

Boeing 747 470 610

Concorde 132 1350

Douglas 146 544

Boeing 737 101 598

Boeing 747 470 610

Concorde 132 1350

Douglas 146 544

(mph)(mph)ThroughputThroughput

228,750228,750

286,700286,700

178,200178,200

79,42479,424

Computer Performance:TIME, TIME, TIMEComputer Performance:TIME, TIME, TIME

Response Time (latency)Response Time (latency)

How long does it takefor my job to run?

How long does it takefor my job to run?

How long does it taketo execute a job?

How long does it taketo execute a job?

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How long must I waitfor the database query?

How long must I waitfor the database query?

The time between the startand completion of a job

The time between the startand completion of a job

Throughput Throughput

What is the averageexecution rate?

What is the averageexecution rate?

How many jobs canthe machine run at once?

How many jobs canthe machine run at once?

How much work is gettingdone?

How much work is gettingdone?

The total amount of workdone in a given time

The total amount of workdone in a given time

ExampleExample

If we upgrade a machine witha new processor what do we

increase?

If we upgrade a machine witha new processor what do we

increase?

If we add a new machine tothe lab what do we increase?

If we add a new machine tothe lab what do we increase?

Answer:Answer:

Case 1: Both response timeand throughput are improved

Case 1: Both response timeand throughput are improved

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Case 2: No one task gets workdone faster, so onlythroughput increases

Case 2: No one task gets workdone faster, so onlythroughput increases

Answer:Answer:

But by increasing the numberof processors would reducethe waiting time in the queue,therefore response time couldimprove

But by increasing the numberof processors would reducethe waiting time in the queue,therefore response time couldimprove

Answer:Answer:

Execution TimeExecution Time

Elapsed Time Elapsed Time

counts everything (disk andmemory accesses, I/O, etc.)

counts everything (disk andmemory accesses, I/O, etc.)

a useful number, but oftennot good for comparisonpurposes

a useful number, but oftennot good for comparisonpurposes

CPU time CPU time

Doesnt count I/O or

time spent running otherprograms

Doesnt count I/O or

time spent running otherprograms

Can be broken up intosystem time, and user time

Can be broken up intosystem time, and user time

Our focus: user CPU time Our focus: user CPU time

time spent executingthe lines of code that arein our program

time spent executingthe lines of code that areinour program

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Example: Unix timeCommand

Example: Unix timeCommand

90.7u 12.9s 2:39 65%90.7u 12.9s 2:39 65%90.7u: User CPU Time (seconds)90.7u: User CPU Time (seconds)

12.9s: System CPU Time (seconds)12.9s: System CPU Time (seconds)

2:39: Elapsed Time (159 seconds)2:39: Elapsed Time (159 seconds)

65%: Percentage of Elapsed Time

that is CPU Time is

(90.7 + 12.9) / 159 = 0.65

65%: Percentage of Elapsed Time

that is CPU Time is

(90.7 + 12.9) / 159 = 0.65

More than a third of the elapsedtime in this example was spentwaiting for I/O, running otherprograms, or both

More than a third of the elapsedtime in this example was spentwaiting for I/O, running otherprograms, or both

Definition ofPerformanceDefinition ofPerformance

computer X is n timesfaster than computer YPerformanceX/PerformanceY=n

computer X is n timesfaster than computer YPerformanceX/PerformanceY=n

For some program runningon computer X, PerformanceX=1/Execution TimeX

For some program runningon computer X, PerformanceX=1/Execution TimeX

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Answer: A is 1.5 times fasterthan B

Answer:A is 1.5 times fasterthan B

Example: Computer A runs a

program in 10 seconds,Computer B runs the sameprogram in 15 seconds,how much faster is A than B?

Example: Computer A runs a

program in 10 seconds,Computer B runs the sameprogram in 15 seconds,how much faster is A than B?

Clock CyclesClock Cycles

Instead of reporting executiontime in seconds, we often usecycles

Instead of reporting executiontime in seconds, we often usecycles

seconds cycles seconds=

program program cycle



A simple formula relates the mostbasic metrics (clock cycles andclock cycle time) to CPU time:

A simple formula relates the mostbasic metrics (clock cycles andclock cycle time) to CPU time:

CPU execution time for aprogram = CPU clock cyclesfor a program X Clock cycletime

CPU execution time for aprogram = CPU clock cyclesfor a program X Clock cycletime

Or, because clock rate andclock cycle time are inverses:

Or, because clock rate andclock cycle time are inverses:

CPU execution time =CPU clock cycles fora program / Clock rate

CPU execution time =CPU clock cycles fora program / Clock rate

Clock ticks indicate whento start activities

(one abstraction):

Clockticksindicate whento start activities

(one abstraction):

timetime

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cycle time = time betweenticks = seconds per cycle cycle time = time betweenticks = seconds per cycle

clock rate (frequency)= cycles per second(1 Hz.= 1 cycle/sec)

clock rate (frequency)= cycles per second(1 Hz.= 1 cycle/sec)

A 200 MHz. clock has aA 200 MHz. clock has a

1 9 10 = 5 nsec6200 10

1 9 10 = 5 nsec6200 10

CycletimeCycletime

How to ImprovePerformance

How to ImprovePerformance





So, hardware designer can improveperformance by reducing eitherthe length of the clock cycle or thenumber of clock cycles required fora program.

So, hardware designer can improveperformance by reducing eitherthe length of the clock cycle or thenumber of clock cycles required fora program.

ExampleExample

Our favorite program runs in 10seconds on computer A, which has

a 400 MHz. clock. We are trying tohelp a computer designer build anew computer B, that will run thisprogram in 6 seconds

Our favorite program runs in 10seconds on computer A, which has

a 400 MHz. clock. We are trying tohelp a computer designer build anew computer B, that will run thisprogram in 6 seconds

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The designer can use newtechnology to substantiallyincrease the clock rate,but has informed us that thisincrease will affect the rest ofthe CPU design,

The designer can use newtechnology to substantiallyincrease the clock rate,but has informed us that thisincrease will affect the rest ofthe CPU design,

causing computer B to require1.2 times as many clock cyclesas computer A for the sameprogram. What clock rateshould we tell the designer totarget?

causing computer B to require1.2 times as many clock cyclesas computer A for the sameprogram. What clock rateshould we tell the designer totarget?

AnswerAnswer

First: find the number of clockcycles required for the programon computer A:

First: find the number of clockcycles required for the programon computer A:

CPU timeA = (CPU ClockCyclesA) / Clock RateA

CPU timeA = (CPU ClockCyclesA) / Clock RateA

10 seconds =

(CPU clock cyclesA) / 400 x 106

cycles/sec

10 seconds =

(CPU clock cyclesA) / 400 x 106

cycles/sec

CPU clock cyclesA =

10 sec X 400 X 106 cycles/sec= 4000 X 106 cycles

CPU clock cyclesA =

10 sec X 400 X 106 cycles/sec= 4000 X 106 cycles

CPU time for computer B canbe found using this equation:

CPU time for computer B canbe found using this equation:

CPU timeB = (1.2 X CPUclock cyclesA) / Clock rateB

CPU timeB = (1.2 X CPUclock cyclesA) / Clock rateB

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6 seconds = (1.2 X 4000 X 10

6

cycles) / Clock rateB

6 seconds = (1.2 X 4000 X 106

cycles) / Clock rateB

Clock rateB = (1.2 X 4000 X 106

cycles) / 6 seconds = (800 X 106

cycles) / second = 800 MHz

Clock rateB = (1.2 X 4000 X 106

cycles) / 6 seconds = (800 X 106

cycles) / second = 800 MHz

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