1. Statistics
CSE 807

2. Experimental Design and Analysis
How to:
Design a proper set of experiments for measurement or simulation.
Develop a model that best describes the data obtained.
Estimate the contribution of each alternative to the performance.
Isolate the measurement errors.
Estimate confidence intervals for model parameters.
Check if the alternatives are significantly different.
Check if the model is adequate.

3. Example Personal workstation design. Processor:68000, Z80, or 8086.
Memory size: 512K, 2M, or 8M bytes.
Number of Disks: One, two, three, or four.
Workload: Secretarial, managerial, or scientific.
User education: High school, college, or Post-graduate level.

4. Terminology Response Variable: Outcome.
E.g., throughput, response time.
Factors: Variables that affect the response variable.
E.g., CPU type, memory size, number of disk drivers, workload used, and user’s educational level.
Also called predictor variables or predictors.
Levels: The value that a factor can assume.
E.g., the CPU type has three levels:
68000, 8080, or Z80.
# of disk drives has four levels.
Also called treatment.

5. Terminology (cont’d)
Primary Factors: The factors whose effects need to be quantified.
E.g., CPU type, memory size only, and number of disk drives.
Secondary Factors: “Factors whose impact need not be quantified.
E.g., the work loads.
Replication: Repetition of all or some experiments.

6. Terminology (cont’d)
Design: The number of experiments, the factor level and number of replications for each experiment.
E.g., Full Factorial design with 5 replications:
3 X 3 X 4 X 3 X 3 or 324 experiments, each repeated five times.
Experimental Unit: Any entity that is used for experiments.
E.g., users. Generally, no interest in comparing the units.
Goal - minimize the impact of variation among the units.

7. Terminology (cont’d)
Interaction => Effect of one factor depends upon the level of the other.
Non-interacting Factors
Interacting Factors

8. Common Mistakes in Experimentation
1. The variation due to experimental error is ignored.
2. Important parameters are not controlled.
3. Effects of different factors are not isolated.
4. Simple one-factor-at-a-time designs are used
5. Interactions are ignored.
6. Too many experiments are conducted.
Better: two phases.

9. Types of Experimental Designs
Simple Designs: Vary one factor at a time
#of Experiments =
Not statistically efficient.
Wrong conclusions if the factors have interaction.
Not recommended.

10. Types of Experimental Designs (cont’d)
Full Factorial Design: All combinations.
# of Experiments =
Can find the effect of all factors.
Too much time and money.
May try 2k design first

11. Types of Experimental Designs (cont’d)
Fractional Factorial Designs: Save time and expense.
Less information.
May not get all interactions.
Not a problem if negligible interactions.

12. A Sample Fractional Factorial Design.

13. Exercise
The performance of a System being designed depends upon the following three factors:
a. CPU type: 68000, 8086, 80286
b. Operating System type: CPM, MS-DOS, UNIX
c. Disk drive type: A, B, C
How many experiments are required to analyze the performance if
a. There is significant interaction among factors.
b. There is no interaction among factors
c. The interactions are small compared to main effects.

14. 2k Factorial Designs k factors, each at two levels. Easy to analyze.
Helps in sorting out impact of factors.
Good at the beginning of study.
Valid only if the effect is unidirectional.
E.g., memory size, the number of disk drives

15. 22 Factorial Designs xA= xB= Two factors, each at two levels
Performance in MIPS
Cache
Size
Memory size
4M Bytes
16M Bytes 45 75 15 25 1K 2K
-1 if 4M bytes memory
1 if 16M bytes memory
-1 if 1M bytes cache
1 if 2M bytes cache
xA=
xB=

16. Model y = q0 + qAxA + qBxB +qABxAxB 15= q0 - qA - qB + qAB
y = xA + 10xB + 5xAxB
Interpretation: Mean performance = 40 MIPS
Effect of memory = 20 MIPS
Effect cache = 10 MIPS
Interaction between memory and cache = 5 MIPS

17. Computation of Effects
Model: y = q0 + qAxA + qBxB +qABxAxB
Substitution:
y1 = q0 - qA - qB + qAB
y2 = q0 + qA - qB - qAB
y3 = q0 - qA + qB - qAB
y4 = q0 + qA + qB + qAB

18. Computation of Effects (cont’d)
Solution:
q0 =1/4 (y1 + y2 + y3 + y4)
qA =1/4 (-y1 + y2 - y3 + y4)
qB =1/4 (-y1 - y2 + y3 + y4)
qAB =1/4 (y1 - y2 - y3 + y4)
Notice that effects are linear combinations of responses.
Sum of the coefficients is zero => contrasts.
Notice: qA = Column A x Column y
qB = Column B x Column y
qAB = Column A x Column B x Column y

19. Sign Table Method

20. Allocation of Variation
Importance of a factor = proportion of the variation explained
Sample variance of
Variation of y  Numerator
= sum of squares total (SST)

21. Allocation of Variation (cont’d)
For a 22 design:
Variation due to
Variation due to interaction
SST = SSA + SSB + SSAB
Fraction explained by
Variation  Variance

22. Derivation Model: yi = q0 + qAxAi + qBxBi +qABxAixBi Notice
1. The sum of entries in each column is zero:
2. The sum of the squares of entries in each column is 4:

23. Derivation (cont’d)
3. The columns are orthogonal (inner product of any two columns is zero):

24. Derivation (cont’d)
Sample mean

25. Derivation (cont’d)
Variation of y
Product terms

26. Example Memory-cache study: Total Variation Total variation = 2100
Variation due to memory = 1600 (76%)
Variation due to cache = 400 (19%)
Variation due to interaction = 100 (5%)

27. Case Study: Interconnection Net
Memory interconnection networks:
Omega and Crossbar.
Memory reference patterns:
random and Matrix
Fixed factors:
1. Number of processors was fixed at 16.
2. Queued requests were not buffered but blocked.
3. Circuit switching instead of packet switching.
4. Random arbitration instead of round robin.
5. Infinite interleaving of memory => no memory back contention.

28. 22 Design for Interconnection Networks
Factors Used in the Interconnection Network Study
Level
Response

29. Interconnection Network Study (cont’d)
Para-
meter
Mean Estimate
Variation Explained q0 qA qB
qAB
0.5725
0.0595
3.5
-0.5
1.0
0.0
1.871
-0.145
0.413
0.051
17.2%
77.0%
5.8%
20%
80% 0%
10.9%
87.8%
1.3% T N R

30. Interpretation of Results
Average throughput =
Most effective factor = B = reference pattern => The address patterns chosen are very different.
Reference pattern explains  (77%) of variation
Effect of network type =
Omega networks = Average
Crossbar networks = Average
Difference between the two = 0.119
Slight interaction (0.0346) between reference pattern and network type.

31. General 2k Factorial Designs
k factors at two levels each.
2k experiments.
2k effects:
k main effects
Two factor interactions
Three factor interactions...

32. 2k Design Example Three factors in designing a machine: Cache size
Memory size
Number of processors

33. 2k Design Example (cont’d)
Cache
Size
4M Bytes
16M Bytes
1K Byte
2K Byte
1 Proc 14 10
2 Proc 46 50 22 34 58 86

34. Analysis =18%+4%+71%+4%+1%+2%+0% =100%
Number of Processors (C) is the most important factor

35. Exercise Analyze the 23 design:B1 B2 C1
100 40 C2 15 30
120 20 10 50
a. Quantify main effects and all interactions.
b. Quantify percentages of variation explained.
c. Sort the variables in the order of decreasing importance