CS170 Computer Organization and Architecture I

1
CS170 Computer Organization and Architecture I
Ayman Abdel-Hamid Department of Computer
Science Old Dominion University Lecture 8
9/19/2002
2
Outline

• An example of geometric mean use to summarize
  performance
• Fallacies and Pitfalls
• Amdahls Law
• Using MIPS as a performance Metric
• Should cover section 2.7

3
Performance Summary Example1/2
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
4
Performance Summary Example2/2
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

5
Amdahls 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
6
Amdahls Law2/3
Another form of Amdahls 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))
7
Amdahls 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

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
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
8
MIPS as a Performance Metric

• MIPS 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