The Role of Performance

Chapter - 2
Why examining performance is important?

• Hardware performance is often key to

the effectiveness of an entire system.
Why assessing the performance is challenging?

• The scale and intricacy of modern software

systems, together with the wide range of
performance improvement techniques employed
by hardware designers have made performance
assessment much more difficult.

• For different types of applications, different

performance metrics may be appropriate and
different aspects of a computer system may be
the most significant in determining overall
performance.
 Defining Performance
• How subtle the question of performance can be:
• Example:
Airplane Passenger Cruising Cruising Passenger
Capacity range speed throughput
(miles) (m.p.h.) (passengers
* m.p.h.)
Boeing 777 375 4630 610 2,28,750
Boeing 747 470 4150 610 2,86,700
Concorde 132 4000 1350 1,78,200
DC-8-50 146 8720 544 79,424
Which of the planes in this table had the best performance?
 Defining Performance

• Which of the planes in this table had the best

performance?
• What is performance?
• If highest cruising speed – Concorde
• If longest range – DC-8
• If largest capacity – 747
• If we want to transport 450 passengers –
which one will be the fastest one?
– 747
 Defining Performance
Running a program on two different workstations

Better
 Defining Performance

• Response Time or Execution Time :

the time between the start and
completion of a task.

• It includes the time for disk

accesses, memory accesses, I/O
activities, operating system
overhead, CPU execution time, and
so on.
 Defining Performance

Better
Time
1 sec/op 1 sec/op (V)
Shared
2 sec/op (C)
Computer
 Defining Performance

• Throughput or bandwidth:
• the total amount of work done in a
given time.
 Throughput and Response Time
• Do the following changes to a computer
system increase throughput, decrease
response time, or both?
– Replacing the processor in a computer with a
faster version – both.
– Adding additional processors to a system that
uses processors for separate tasks –
throughput (also response time).

Changing either one often affect the other.

 Relative performance
1
PerformanceX =
Execution timeX

Performance of X is greater than the performance of Y

PerformanceX > PerformanceY

1 1
>
Execution timeX Execution timeY
Execution timeY > Execution timeX
X is faster than Y
 Relative performance

• X is n times faster than Y, it means,

PerformanceX
= n
PerformanceY

PerformanceX Execution timeY

= =n
PerformanceY Execution timeX
 Relative performance

• Example: If machine A runs a program in 10

seconds and machine B runs the same program
in 15 seconds, how faster is A than B?
– A is n times faster than B if
PerformanceA
= n
PerformanceB
Execution timeB 15
=n⇒ = 1.5
Execution timeA 10
– A is 1.5 times faster than B
 Relative performance

• We could also say that – Machine B is 1.5 times

slower than machine A. since

PerformanceA
= n
PerformanceB

PerformanceA
PerformanceB =
n
 Measuring Performance
• Time is the measure of computer
performance.
• Program execution time is measured in
seconds per program.
• Wall-clock time / response time / elapsed
time / execution time – total time to
complete a task, including - disk
accesses, memory access, I/O activity,
OS overhead.
 Measuring Performance

• CPU execution time or CPU time

• is the time the CPU spends computing
for a task and does not include time
spent waiting for I/O or running other
programs.

CPU execution time or CPU time ≤ Response time

 Measuring Performance

User CPU time

CPU time
System CPU time
• User CPU time – the CPU time spent in
the program

• System CPU time – the CPU time spent

in the OS performing tasks on behalf of
the program
 Measuring Performance

Execution Time
CPU time

For I/O User CPU System

and Others time CPU time
 Measuring Performance

• Example:
• Unix time command –
• 90.7u 12.9s 2:39 65%

User CPU time System CPU time Elapsed time

(90.7 seconds) (12.9 seconds) 2*60 + 39 =
(159 seconds)

90.7 + 12.9
= 0.65
159
 Measuring Performance

• System Performance – considering

elapsed time on an unloaded system

• CPU Performance – considering user

√ CPU time.
Measuring Performance

• Clock cycle – Almost all computers

are constructed using a clock that
determines when events take place.
These discrete time intervals are
called clock cycles (ticks / clock ticks
/ clock periods / clocks / cycles).

• Clock rate – Inverse of clock period.

 Relating the Metrics

CPU execution time CPU clock cycle Clock cycle

= ×
for a program for a program time

CPU execution time CPU clock cycle for a program

=
for a program Clock rate

Hardware designer can improve performance by

reducing either the length of the clock cycle or
the number of clock cycles required for a
program.
 Improving Performance
Our favorite program runs in 10 seconds on
computer A, which has a 400 MHz clock. We
are trying to help a computer designer build a
machine, B, that will run this program in 6
seconds. The designer has determined that a
substantial increase in the clock rate is possible,
but this increase will affect the rest of the CPU
design, causing machine B to require 1.2 times
as many clock cycles as machine A for this
program. What clock rate should we tell the
designer to target?
 Improving Performance (Cont.)
CPU clock cycleA
CPU timeA =
Clock rateA
CPU clock cycleA
10 Seconds =
400 × 106 cycles/sec
CPU clock cycleA = 10 seconds × 400 × 106 cycles/sec
= 4000 × 106 cycles
CPU clock cycleB
CPU timeB =
Clock rateB
1.2 × CPU clock cycleA
CPU timeB =
Clock rateB
 Improving Performance (Cont.)
1.2 × 4000 × 106 cycles
6 seconds =
Clock rateB
1.2 × 4000 × 106 cycles
Clock rateB =
6 seconds
= 800 MHz

Machine B must therefore have twice the clock

rate of A to run the program in 6 seconds.
 Hardware Software Interface

• Since Machine had to execute the

instructions to run the program, the
execution time must depend on the
number of instructions in a program.
Average clock
CPU clock cycles Instructions
= × cycles per
(for a program) for a program
instruction

CPI
Using the Performance Equation

• Suppose we have two implementations

of the same instruction set architecture.
Machine A has a clock cycle time of 1 ns
and a CPI of 2.0 for some program, and
machine B has a clock cycle time of 2 ns
and a CPI of 1.2 for the same program.
Which machine is faster for this program,
and by how much?
 Continuation
Let the number of instructions of the program be I
CPU clock cyclesA = I × 2.0
CPU clock cyclesB = I × 1.2
CPU timeA = CPU clock cyclesA × Clock cycle timeA
= I × 2.0 × 1 ns = 2I ns
CPU timeB = I × 1.2 × 2 ns = 2.4I ns

CPU performanceA Execution timeB 2.4I ns

= = = 1.2
CPU performanceB Execution timeA 2I ns
A is 1.2 times faster than B
 Continuation
• Basic performance equation

CPU time = Instruction count × CPI × clock cycle time

Instruction count × CPI

CPU time =
Clock rate
 Continuation
• 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.

n
• CPU clock cycle = iΣ= 1 (CPIi × Ci)
• Ci – No. of instructions of class i
• CPIi – CPI for instruction class i
 Comparing Code Segments
• Example
– The hardware designer supplied:
Instruction Class CPI for this class
A 1
B 2
C 3
– Two code sequences requires the following:
Code Sequence Instruction Counts for instruction class
A B C
1 2 1 2
2 4 1 1
– Which code sequence executes the most instructions?
– Which will be faster?
– What is the CPI for each sequence?
 Solution

• Sequence 1 executes 2 + 1 + 2 = 5
instructions.
• Sequence 2 executes 4 + 1 + 1 = 6
instructions.
• So sequence 2 executes most
instructions.
 Solution

• CPU clock cycles1 = (2×1) + (1×2) +

(2×3) = 2 + 2 + 6 = 10 cycles
• CPU clock cycles2 = (4×1) + (1×2) +
(1×3) = 4 + 2 + 3 = 9 cycles
• So code sequence 2 is faster.
 Solution
CPU clock cycles1 10
CPI1 = = = 2
Instruction count1 5

CPU clock cycles2 9

CPI2 = = = 1.5
Instruction count2 6

When comparing two machines, we must look at all three

components, which combine to form execution time.
To Be Continued …