Lecture 8: 9/19/2002 CS170 Fall 2002 1

CS170 Computer Organization and Architecture I

Ayman Abdel-Hamid

Department of Computer Science

Old Dominion University

Lecture 8: 9/19/2002

Lecture 8: 9/19/2002 CS170 Fall 2002 2

Outline

•An example of geometric mean use to summarize performance

•Fallacies and Pitfalls

Amdahl’s Law

Using MIPS as a performance Metric

Should cover section 2.7

Lecture 8: 9/19/2002 CS170 Fall 2002 3

Performance Summary Example1/2

M/C A M/C B

P1 1482 139

P2 2266 254

P3 6206 690

Which machine is faster according to total execution time? And by how much?

Total Execution Time (A) = 1482 + 2266 + 6206 = 9954

Total Execution (B) = 139 + 254 + 690 = 1083

Machine B is fastest by 9954/1083 = 9.27 times

Lecture 8: 9/19/2002 CS170 Fall 2002 4

Performance Summary Example2/2

M/C A M/C B

P1 1482 139

P2 2266 254

P3 6206 690

Which machine is faster by the geometric mean measure?

Remember how SPEC reported performance?

•Normalize in reference to one machine

•Choose A as reference machine

•Obtain Execution time ratios (ET Ratio)

ET Ratio(P1) = ET(A)/ET(B) = 1482/139 = 10.66

ET Ratio (P2) = 2266/254 = 8.92

ET Ratio(P3) = 6206/690 = 8.99

Geometric Mean = (Ratio (P1) * Ratio(P2) * Ratio(P3))1/3

Geometric Mean = 9.49

Machine B is 9.49 times faster than A according to geometric mean measure

Lecture 8: 9/19/2002 CS170 Fall 2002 5

Amdahl’s Law1/3

Pitfall

Expecting the improvement of one aspect of a machine to increase performance by an amount proportional to the size of the improvement

Program runs in 100 sec on a machine

Multiply operations responsible for 80 sec of time

How much do we need to improve the speed of multiplication if program is to run 5 times faster?

Execution time after improvement =

(Execution time affected by improvement/Amount of improvement + Execution time unaffected)

Execution time after improvement = 80/n + (100-80) = 20 = (100/5)

20 = 80/n + 20 80/n = 0 no n can be found to achieve the requested improvement

Make the common case fast

Lecture 8: 9/19/2002 CS170 Fall 2002 6

Amdahl’s Law2/3

Another form of Amdahl’s Law (to yield Speedup)

Speedup = Performance after improvement/Performance before

Speedup = Execution time before/Execution time after improvement

Assume new hardware added to machine

f = fractions of all operations which use new hardware

s = speedup of those operations using new hardware

Execution time with new hardware is Tnew

Execution time without new hardware is Told

Tnew = f* Told/s + (1-f) * Told

Overall speedup S = Told/Tnew

Speedup = s / (s –f * (s-1))

fs 0.1 Speedup2 1.0526325 1.08695710 1.098901

s 0.5 Speedup2 1.3333335 1.66666710 1.818182

s 0.9 Speedup2 1.8181825 3.57142910 5.263158

s 0.99 Speedup2 1.9801985 4.80769210 9.174312

Lecture 8: 9/19/2002 CS170 Fall 2002 7

Amdahl’s Law3/3

Example of memory versus processor speedup

A = B op C

Assume memory access takes 4 cycles and a typical operation takes 2 cycles

Which of the following achieves the best increase in performance

•Increase memory speed by 50%

•Double operation speed

Calculate how many memory accesses are needed first?

1 to get instruction from memory

2 to get B and C from memory

1 to store result (A) back in memory

Then we need a total of 4 memory access operations

Memory access time = 4 (accesses) * 4 (cycles/access) = 16 cycles

Operation time = 1 (operation) * 2 (cycles/operation) = 2 cycles

Total number of cycles = 16 + 2 = 18

Option 1 increase memory speed by 50%

s1 = 1.5 (how?)

f1 = memory access time/ total time

= 16/18 = 0.889

S1 = 1.42

Option 2 double operation speed

s2 = 2

f2 = operation time/total time

= 2/18 = 0.111

S2 = 1.059

Lecture 8: 9/19/2002 CS170 Fall 2002 8

MIPS as a Performance MetricMIPS is million instructions per second

MIPS = instruction count / (Execution time * 106)

Instruction execution rate (instruction/sec)

Faster machines have a higher MIPS rating

Problems with MIPS

•Does not take into account capabilities of instructions

(can not compare computers with different ISA)

•Varies between programs on the same computer

(a machine can not have a single MIPS rating for all programs)

•Can vary inversely with performance

•Example page 78