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Performance (metrics
for measuring the performance of a
computer system)
MehranMehran RezaeiRezaei
2
Overview
• Review of the previous lecture
• Introduction (how to measure performance)
• Metrics
• Relationships among the metrics
3
Two notions of “performance”
° Time to do the task
– execution time, elapsed time, latency
° Tasks per day, hour, week, sec, ns. ..
– throughput, bandwidth
Elapsed time and throughput often are in opposition
Plane
Boeing 747
BAD/Sud Concorde
Speed
610 mph
1350 mph
DC to Paris
6.5 hours
3 hours
Passengers
470
132
Throughput (pmph)
286,700
178,200
Which has higher performance?
4
Example
• Time of Concorde vs. Boeing 747? – Concord is 1350 mph / 610 mph = 2.2 times faster
– = 6.5 hours / 3 hours
• Throughput of Concorde vs. Boeing 747 ? – Concord is 178,200 pmph / 286,700 pmph = 0.62 “times
faster”
– Boeing is 286,700 pmph / 178,200 pmph = 1.6 “times faster”
• Boeing is 1.6 times (“60%”)faster in terms of throughput
• Concord is 2.2 times (“120%”) faster in terms of flying time
• We will focus primarily on execution time for a single job
5
Definitions
• Performance is in units of things-per-second
–higher is better
• If we are primarily concerned with response
time
performance(x) = 1
execution_time(x)
" X is n times faster than Y" means
Performance(X) n = ----------------- Performance(Y)
6
Example
• If machine A runs a program in 10
seconds and machine B runs the same
program in 15 seconds, which machine
is faster and by how much?
Increase performance = Decrease execution
time
To me is a bit confusing
Improve performance = Improve execution
time
9
• For the first few chapters we focus on
CPU time
– User CPU time
– System CPU time
• Time Utility of UNIX 140.3u 39.5s 7:23 41%
0.41=(140.3+39.5)/443
About 59% of the elapsed time was used for
I/O and some other stuff (I/O bound).
Execution time
10
System CPU time vs. User CPU time
• Which one do you think reflects the
performance of the system (better)?
• From now on:
If system performance
Elapsed time
If CPU performance
User CPU time
11
Finally
• Clock rate and clock period
I call them CF and CT
CF=1/CT
• CPU Execution time=
# of Cycles for program * CT =
# of Cycles for program/CF
12
Example
• A program runs in 10 seconds on computer
A, which has 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 ask
the designer to target for machine B?
13
The final Formula
Execution time =
Total number of cycles for program * CT =
Total number of Instruction *
Average number of cycles per Instruction *
CT =
Instruction Count*
Average Cycles per Instruction*
CT =
IC * CPI * CT
16
Example
• 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 program p.
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?
17
How could you measure them?
Execution time =
Total number of cycles for program * CT =
Total number of Instruction *
Average number of cycles per Instruction *
CT =
Instruction Count*
Average Cycles per Instruction*
CT =
IC * CPI * CT
18
Do not forget that,
• Clock rate and clock period
I call CF and CT
CF=1/CT
• CPU Execution time=
# of Cycles for program * CT =
# of Cycles for program/CF
19
Clock Cycles
Instead of reporting execution time in seconds, we often use
cycles
Clock “ticks” indicate when to start activities (one abstraction):
cycle time = time between ticks = seconds per cycle
clock rate (frequency) = cycles per second (1 Hz. = 1 cycle/sec)
A 200 Mhz. clock has a cycle
time
time
seconds
program
cycles
program
seconds
cycle
1
200 106 109 5 nanoseconds
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Example
IClass IC CPI Mix
Load 766 2 9.7 %
Store 3538 2 45.11 %
Ubranch 182 3 2.32 %
Cbranch 947 4 12.1 %
Int Op 2409 1 30.77 %
Exec. Time = (CPIi * ICi)*T=
(766*2+3538*2+182*3+947*4+2409*1)*T=
=Total Ins. * (CPIi * Mixi)*T=
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Averaging
Program Exe. Time
P1 5 seconds
P2 10 seconds
P3 20 seconds
P4 3 seconds
Arithmetic Mean=(1/n) Exei=9.5
22
Averaging
Weights Program Exe. Time
20 % P1 5 seconds
30 % P2 10 seconds
5 % P3 20 seconds
45 % P4 3 seconds
Arithmetic Mean = 0.2*5 + 0.3*10 + 0.05*20 + 0.45*3=
6.35
23
Amdahl’s Law
• If you enhance an aspect of a machine
that improves a portion of the execution
time say by a factor n, how much will
be the overall improvement?
Execution time
24
Amdahl’s Law (Cont’d)
• We have optimized the system so that
50 % of execution time is improved by
factor of 10. How much is the overall
speedup?
25
MIPS
• Is Million Instructions Per Second Million Instructions Per Second a
good performance metric?
How many Instruction per second = IC/Exec.
… … …
code
10 s 10
billion
Ins
t.
MIPS = IC/(Exe. *106) =1/(IPC*T* 106)
