Folie a: Name
Exercise 1
Advanced Computer Architecture
Exercise 1
Prof. Dr.-Ing. Axel Hunger
Department of Electrical Engineering and Information Technology
Institute for Computer Engineering
g
g
Dipl.-Ing. M.A. Lebedev
Institute for Computer Engineering
University of Duisburg-Essen
Faculty of Engineering
Computer Engineering
Prof. Dr.-Ing. Axel Hunger
BB 321, Tel: 0203 379 - 1019
Exercise Advanced Computer Architektur
E-mail: michail.lebedev@uni-due.de Lecturer: Dipl. -Ing. M.A. Lebedev
Link for downloading ACA foil
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Exercise 1
http://www.fb9dv.uni-duisburg.de/ti/en/education/teaching/ss09/advra/index.php
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Faculty of Engineering
Computer Engineering
Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
2 of 37
Table of content
Exercise 1
1. Speedup, throughput, efficiency
2. Amdahls Law
3. Scalability
4. Speedup by pipelining
5. Multiple choice (examination preparation)
University of Duisburg-Essen
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Prof. Dr.-Ing. Axel Hunger
Literature
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
3 of 37
Exercise 1
Books
David E. Culler, Jasweinder Pal Singh, Anoop Gupta: Parallel
Computer Architecture: A Hardware/Software Approach, Morgan
Kaufmann, 1999, ISBN 1-55860-343-3
J. Hennessy, D. Patterson: Computer Architecture: A
quantitative approach, Morgan Kaufmann Publishers, Inc., 3rd
edition, 2002
Andrew S. Tanenbaum: Computerarchitektur: Strukturen-KonzepteGrundlagen, ISBN 3-8273-7151-1
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Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
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1. Multiple Choice: General Questions
Exercise 1
1. Who is regarded as the founder of Computer Architecture?
a)
b)
c)
d)
e)
Alan Turing
Konrad Zuse
John von Neumann
John William Mauchly
None of the answers above is correct
2. What is characteristic for the organization of a computer
architecture?
a)
b)
c)
d)
e)
Size
Dynamic behaviour
Static behaviour
Speed
None of the answers above is correct
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Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
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1. Multiple Choice
Exercise 1
3. What is usually regarded as the von Neumann Bottleneck?
a)
b)
c))
d)
e)
Processor/memory interface
Control unit
A ith ti llogical
Arithmetic
i l unitit
Instruction set
None of the answers above is correct
4. How does the number of transistors per chip increase
according to Moores law?
a)
b)
c)
d)
e)
Quadratically
Li
Linearly
l
Cubicly
Exponentially
None of the answers above is correct
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Computer Engineering
Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
6 of 37
1. Multiple Choice: General Questions
Exercise 1
5. Who is regarded as the founder of Computer Science?
a)
b)
c)
d)
e)
Alan Turing
Konrad Zuse
J. Presper Eckert
John William Mauchly
None of the answers above is correct
6. Which is the fastest storage unit in a usual memory
hierarchy?
a) Cache
b) Main memory
c) Hard disk
d) Register
e) None of the answers above is correct
University of Duisburg-Essen
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Computer Engineering
Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
7 of 37
1. Multiple Choice: General Questions
Exercise 1
7. Which cache miss does not occur in case of a fully
associative cache?
a)
b)
c)
d)
e)
Conflict miss
Capacity miss
Compulsory miss
Cold start miss
None of the answers above is correct
8. Which miss even occurs in infinite caches?
a)
b)
c)
d)
e)
Coherence miss
Capacity
C
it miss
i
Conflict miss
Cold start miss
None of the answers above is correct
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Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
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1. Multiple Choice: General Questions
Exercise 1
9. What is stored in a Translation Lookaside Buffer?
a)
b)
c)
d)
e)
System dumps
Physical addresses
Program data
Operating system log files
None of the answers above is correct
University of Duisburg-Essen
Faculty of Engineering
Computer Engineering
Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
9 of 37
Execution time, Throughput, Speedup
Question not precise!
What is better?
Exercise 1
Aeroplane
NY to
Paris
Speed
Passengers
Throughput
(Persons/h)
Boeing 747
6.5 h
610 mph
470
72.3
Concorde
3h
1350 mph
132
44.0
Execution time T (response time, latency) [sec], [h], ...
Throughput X (bandwidth) [1/sec], [1/h], ....
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Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
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Definition of Speedup
Exercise 1
Speedup S (Acceleration):
A is S times faster than B
T(B)
S =
= 6.5h / 3h = 2.167
T(A)
Speedup is a measure for the judgement of the
processing of a single task (passenger).
Throughput is a measure for the judgement of the
processing of the whole work load (with what
aeroplane type can an airline transport more passengers?).
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Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
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Performance Goal
Speedup
speedup( p processors) =
performance( p processors)
performance(1 processor )
Scientific computing: performance=1/time
speedup( p processors) =
Exercise 1
time(1 processor )
time( p processors)
Efficiency
efficiency( p processors) =
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Prof. Dr.-Ing. Axel Hunger
speedup( p processors)
p
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
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Speedup based on Throughput
Exercise 1
Performance = throughput = transactions / minute
speedup( p processors) =
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Prof. Dr.-Ing. Axel Hunger
Amdahls Law (1)
tpm( p processor )
tpm(1 processors)
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
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Exercise 1
The performance gain that can be obtained by improving some portion of a
computer can be calculated using Amdahls Law. Amdahl Law states that the
performance improvement to be gained from using some faster mode of
execution is limited by the fraction of the time the faster mode can be used.
A d hl L
Amdahls
Law d
defines
fi
th
the speedup
d th
thatt can b
be gained
i db
by using
i a particular
ti l
feature. What is speedup? Suppose that we can make an enhancement to a
machine that will improve performance when it is used.
Speedup is the ratio
Performance for entire task using the enhancement when possible
Speedup =
Performance for entire task without using the enhancement
Alternatively,
y,
Execution time for entire task without using the enhancement
Speedup =
Execution time for entire task using the enhancement when possible
Universitytells
of Duisburg-Essen
Exercise
Computer
Architektur
Speed
us how much faster a task will run using
theAdvanced
machine
with
the
Faculty of Engineering
Lecturer: Dipl. -Ing. M.A. Lebedev
Computer Engineeringas opposed to the original machine.
14 of 37
enhancement
Prof. Dr.-Ing. Axel Hunger
Amdahls Law (2)
Exercise 1
Amdahls Law gives us a quick way to find the speedup from some
enhancement, which depends on two factors:
1. The fraction of the computation time in the original machine that can be
converted to take advantage of the enhancement
2 The
2.
Th improvement
i
t gained
i d by
b the
th enhanced
h
d execution
ti mode;
d that
th t is,
i how
h
much faster the task would run if the enhanced mode were used for the
entire program
The execution time using the original machine with the enhanced mode will be
the time spent using the unenhanced portion of the machine plus the time
spent using the enhancement:
F
Fraction
i enhanced
Execution timenew = Execution timeold (1 Fractionenhanced ) +
Speedupenhanced
The overall speedup is the ratio of the execution times:
Speedupoverall =
Execution timeold
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Execution timenew
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Computer Engineering
Prof. Dr.-Ing. Axel Hunger
1
Fraction
Speedupenhanced
Advanced
Computer Architektur
enhanced
+
(1 Fractionenhanced ) Exercise
Lecturer: Dipl. -Ing. M.A. Lebedev
Amdahls Law (3)
15 of 37
Exercise 1
In 1967, Gene Amdahl (developer of the IBM 360/xx computer)
defined the performance increase of a program with fixed problem
size for parallel processing as:
Ts
Speedup S(p) =
Sequential execution time
=
f * Ts + (1 f) * Ts/p
Exe. time (seq.+ parallel)
with Ts : Execution time for sequential processing of the whole task
f : Fraction of the execution time for program segments which
cannot run in parallel (f = 0..1)
p : Number of parallel processing elements (processors)
for p : S(p) = 1 / f , or for f 0: S(p) = p
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Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
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Acceleration of Programs (1)
Exercise 1
For efficient parallel processing it is necessary to achieve
speedups that are close to the number of processors used.
S(p)
S(p) = p (ideal)
S(p) < p (real)
Speedup
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Prof. Dr.-Ing. Axel Hunger
Acceleration of Programs (2)
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Prof. Dr.-Ing. Axel Hunger
Number of processors
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
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Exercise 1
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
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Definition of Efficiency
Exercise 1
Efficiency defines the ratio of speedup and number of processors
used.
Efficiency indicates which share of the processor performance
can be utilized.
T(sequential)
=
Efficiency
E(p) = S(p) / p
p * T(parallel)
with 0 < E(p) 1
Number of processors p
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Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
19 of 37
2. Multiple Choice: Speedup and Amdahls law
Exercise 1
1. Which value has the speedup of a parallel program that
achieves an efficiency of 75% on 32 processors?
a)
b)
c)
d)
e)
18
24
16
20
None of the answers above is correct
2. Which speedup could be achieved according to Amdahls law for
infinite number of processors if 2.5% of a program is sequential
and the remaining part is ideally parallel?
a) Infinite speedup
b) 40
c) 4
d) 400
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e) None of the answers above is correct
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Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
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10
2. Multiple Choice: Speedup and Amdahls law
Exercise 1
3. Which speedup could be achieved according to Amdahls law for 9
processors if 10% of a program is sequential and the remaining
part is ideally parallel?
a)
b)
c)
d)
5 Ts
5
1,098
None of the answers above is correct
4. Which is the speedup that can be obtained on 100 processors if
93% of the program is ideally parallel, the remaining part is
sequential and the sequential execution time is 10000 s?
a)
b)
c)
d)
12.6
10.6
1.075
None of the answers above is correct
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Computer Engineering
Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
21 of 37
2. Multiple Choice: Speedup and Amdahls law
Exercise 1
5. Suppose that we are considering an enhancement to the processor
of a server system used for Web serving. The new CPU is 10 times
faster on computation in the Web serving application than the
original processor. Assuming that the original CPU is busy with
computation 40% of the time and is waiting for I/O 60% of the time,
what is the overall speedup gained by incorporating the
enhancement?
a)
b)
c)
d)
1
2
1.56
None of the answers above is correct
Fractionenhanced = 0.4
Speedupenhanced = 10
Speedupoverall =
University of Duisburg-Essen
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Prof. Dr.-Ing. Axel Hunger
1
0.6 +
0.4
10
1
1.56
0.64Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
22 of 37
11
2. Multiple Choice: Speedup and Amdahls law
Exercise 1
5. A common transformation required in graphics engines is square root.
Implementions of floating-point (FP) square root vary significantly in
performance, especially among processors designed for graphics. Suppose
FP square root (FPSQR) is responsible for 20% of the execution time of a
critical graphics benchmark. One proposal is to enhance the FPSQR
hardware and speed up this operation by a factor of 10. The other
alternative is just to try to make all FP instructions in graphics processor run
faster by a factor of 1.6; FP instructions are responsible for a total of 50% of
the execution time for the application. The design team believes that they
can make all FP instructions run 1.6 times faster with the same effort as
required for the fast square root. Compare these two design alternatives.
We can compare these two alternatives by comparing the speedups:
SpeedupFPSQR =
1
0.2
(1 0.2 ) +
10
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1
= 1.22
0.82
SpeedupFP =
1
0.5
(1 0.5 ) +
1.6
1
= 1.23
0.8125
Exercise Advanced Computer Architektur
Improving
the performance of the FP operations overall is
slightly
better
Faculty of Engineering
Lecturer:
Dipl. -Ing.
M.A. Lebedev
Computer Engineering
23 of 37
because
of
the
higher
frequency.
Prof. Dr.-Ing. Axel Hunger
Summary of the benefits of Amdahls Law (1)
Exercise 1
Amdahls Law expresses the law of diminishing returns: The incremental
improvement in speedup gained by an additional improvement in the
performance of just a portion of the computation diminishes as
i
improvements
t are added.
dd d An
A important
i
t t corollary
ll
off Amdahls
A d hl Law
L is
i that
th t
if an enhancement is only usable for a fraction of a task, we cant speed
up the task by more than the reciprocal of 1 minus that fraction.
A common mistake in applying Amdahls Law is to confuse fraction of
time converted to use an enhancement and fraction of time after
enhancement is in use. If, instead of measuring the time that we could
use the enhancement in a computation,
computation we measure the time after the
enhancement is in use, the results will be incorrect!
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Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
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12
Summary of the benefits of Amdahls Law (2)
Exercise 1
Amdahls Law can serve as a guide to how much an enhancement will
improve performance and how to distribute resources to improve costperformance. The goal, clearly, is to spend resources proportional to
where
h time
ti is
i spend.
d Amdahls
A d hl Law
L is
i particularly
ti l l useful
f l for
f comparing
i
the overall system performance of two alternatives, but it can also be
applied to compare two CPU design alternatives.
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Scalability
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
25 of 37
Exercise 1
A computer architecture or a program is scalable if the efficiency of
program processing remains consant for increasing processor number
In general, this is only possible for a simultaneous increase of the
problem
bl size
i
A program (an algorithm) is perfectly scalable if a linear increase of n
is enough in case of a linear increase of p to achieve constant efficiency
S
Speedup
d is
i usually
ll reduced
d
d by
b additional
dditi
l parallel
ll l overhead:
h d
V(p) = p*T(p) T(seq)
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Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
26 of 37
13
.... Scalability
Exercise 1
Definition of mean parallel overhead:
V(p) = V(p) / p
Causes:
Startup costs of an event (process or communication start)
Costs for the distribution/administration of shared data
Costs for synchronization
What is better?
Less communication by bigger work packages for fewer
processors (from fine to coarse granularity)
Smaller work p
packages
g distributed to more p
processors
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Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
27 of 37
3. Multiple Choice: Scalability
Exercise 1
1. Which is the sequential execution time of a program with
mean parallel overhead 2 s and parallel execution time
7 s on 100 processors?
a)
b)
c)
d)
603 s
797 s
500 s
None of the answers above is correct
2. The sequential execution time of a program is 500 s, its parallel
execution time is 8 s and the mean parallel overhead of the system
is 3 s. How many processors are in the system?
a)
b)
c)
d)
163
63
100
None of the answers above is correct
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Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
28 of 37
14
Problem 6: Scalability
Exercise 1
3. Which is the sequential execution time of a program with mean
parallel overhead 3s and parallel execution time 8s if 10% of the
program is sequential and the remaining part is ideally parallel?
(The speedup is considered as 5)
a) 65 s
b) 45 s
c) 42.5 s
d) None of the answers above is correct
4. Which is the parallel execution time of a program with mean
parallel overhead 1 s and sequential execution time 90 sec for 10
processors?
p
a) 10 s
b) 9 s
c) 11 s
d) 8 s
e) None
of the answers above is correct
University
of Duisburg-Essen
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Computer Engineering
Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
29 of 37
Invention of the Assembly Line (Pipeline Principle)
Exercise 1
In 1913, Henry Ford was the first to introduce the pipeline principle
into industrial fabrication with production start of Tin Lizzy.
Principle:
A complex task is distributed into
a sequence of simple partial tasks.
Advantage:
Higher throughput (Lizzies / h)
Simplification (and thus faster
processing) of partial tasks
(CISC RISC)
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Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
30 of 37
15
Sequential Laundry
6
Exercise 1
10
11
12
Time
30 40 20 30 40 20 30 40 20 30 40 20
T
a
s
k
o
r
d
e
r
A
B
C
D
The sequential laundry needs 6 hours for 4 tasks.
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Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
31 of 37
Pipeline Laundry
6
Exercise 1
10
11
12
Time
30 40 40 40 40 20
T
a
s
k
o
r
d
e
r
A
B
The pipeline laundry
needs 3.5 hours
for 4 tasks.
C
D
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Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
32 of 37
16
What does Pipelining Teach Us?
Exercise 1
Pipelining does not improve the execution time of a single task; it improves
the throughput of the total work load.
Pipeline speed is limited by the slowest pipeline stage.
Several tasks are processed simultaneously.
Possible speedup = Number of pipeline (pipe) stages
Speedup reduction by imbalanced lengths of the pipe stages
Time for filling and emptying of the pipeline must be considered.
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Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
33 of 37
Speedup for Pipelining
Exercise 1
For efficient pipeline use, the n number of tasks (processes) to be
processed should be as high as possible
For a single task, a pipeline works purely sequentially
For sequential processing of a task, a pipeline with k equal stages
For
and execution time T per stage needs
For a single task:
T1 = k * T
For n tasks without pipelining: Tseq = n * T1 = n * k * T
For n tasks and pipelining:
Tpipe = (k - 1) * T + n * T = (k - 1 + n) * T
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Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
34 of 37
17
.... Speedup for Pipelining
Tseq
S=
Exercise 1
n*k
=
Tpipe
k1+n
For n = 1 (only one task) S = 1, i.e. same behaviour as sequential
processing
For n >> k S = k, i.e. speedup grows for pipes which are always well
filled with the number of stages
If we define a mean overhead of a pipe on the analogy of V(p) we obtain:
V(k) = (k * Tpipe Tseq) / k = (k 1) * T Filling!
University of Duisburg-Essen
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Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
35 of 37
3. Multiple Choice: Speedup by Pipelining
Exercise 1
1. Which is the speedup of pipelining that can be obtained with 7
tasks if the mean overhead of a pipeline is considered as 5 and an
execution time per stage of 1 cycle?
a)
b)
c)
d)
35
3.5
3
6
None of the answers above is correct
2. Which is the mean overhead of a pipeline with 5 stages and an
execution time per stage of 1 cycle?
a))
b)
c)
d)
2 cycles
l
3 cycles
4 cycles
None of the answers above is correct
University of Duisburg-Essen
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Computer Engineering
Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
36 of 37
18
3. Multiple Choice: Speedup by Pipelining
Exercise 1
3. What is the execution time per stage of a pipeline that has
5 equal stages and a mean overhead of 12 cycles?
a)
b)
c)
d)
2 cycles
3 cycles
4 cycles
None of the answers above is correct
University of Duisburg-Essen
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Prof. Dr.-Ing. Axel Hunger
Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
37 of 37
Summary
Exercise 1
1. Speedup, throughput, efficiency
2. Amdahls Law
3. Scalability
4. Speedup by pipelining
5. Multiple choice (examination preparation)
Next Lecture: Pipelines
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Exercise Advanced Computer Architektur
Lecturer: Dipl. -Ing. M.A. Lebedev
38 of 37
19
