MAE 552 Heuristic Optimization

Instructor: John Eddy

Lecture #17

3/4/02

Taguchi’s Orthogonal Arrays

S/N Ratio

•Why use the signal / noise ratio?– Given a product or process with a target

performance, deviation from that performance can typically be expressed in terms of (i.t.o) statistics (Taguchi – Quality loss).

– Two things you want are for your mean to be “on target” and for your variance to be low. Take the example of a machine that throws darts.

S/N Ratio•Here are the results of 4 such machines:

Good Mean – Bad Var.

Bad Mean – Good Var.

Good Mean – Good Var.

Bad Mean – Bad Var.

S/N Ratio

•Noise can be interpreted as the variance observed in the process (since variance is commonly a direct result of noise).

•The signal can be interpreted as the desired value (the value you would like your mean to take).

•The S/N ratio is then 2y

S/N Ratio

•Another means of calculating the S/N ratio is to take -10*log(d) where d represents the mean squared deviation from the target.

•So in our example, the target is 0 defects, and the squared deviation for each count is the square of the count itself.

S/N Ratio

•Should we always use an S/N ratio?

– Taguchi says yes because it accounts for both the mean and standard deviation.

•Will it always be a “maximize” situation?– Yes it will, assuming that it is always desirable

to achieve the target value with very little deviation.

Back to our ExampleA B C D

1 1 1 1 1

2 1 2 2 2

3 1 3 3 3

4 2 1 2 3

5 2 2 3 1

6 2 3 1 2

7 3 1 3 2

8 3 2 1 3

9 3 3 2 1

Taguchi Example

•Continuing with our example:

FactorLevels

1 2 3

A: Temperature -20 -45 -60

B: Pressure -30 -40 -55

C: Settling Time -50 -35 -40

D: Clean Method -45 -40 -40

All the level averages for our example problem.

Example

•Using the values for the level averages, we can plot the factor effects and see visually which factors have the greatest influence on the performance of our product or process.

Example

Overall Mean

Example

•Considering the effects visualized in the previous figure, which levels do you think are optimal?

•Recall that we want to maximize our S/N ratio, so we choose the levels that cause the greatest positive deviation from our mean. Since our deviation was:

mfactor, level – m

Example

•We clearly want to choose those levels with the largest values because we have a negative overall mean and we will therefore be adding that value to our level means.

•Consider Factor A, our deviation from the mean cause by factor A at level 1 is:

-20 – (-41.67) = 21.67

Example

•And our deviation from the mean cause by factors 2 and three are:

-45 – (-41.67) = -3.33

-60 – (-41.67) = -18.33

•So by comparison, level 1 causes the greatest positive deviation from the mean, and will be chosen as our optimal setting.

Example

•Likewise, for the remaining factors, we will choose the following levels as optimal:

B) Pressure - Level 1

C) Settling Time - Level 2

D) Cleaning Method - Level 2 or 3

•The resulting equivalent experiment is then:1 1 2 2/3 - does not appear in our array!!

Example•The final step would then be to run a validation experiment to see that the optimal solution is actually the best thus far.

•If it is not, then you would perhaps do one of two things:

– Choose the best configuration of any of the experiments you did run.

– Refine your system and re-run the experiments. (could mean changing level settings to “hone in” on good regions, etc.

Comparison

•What if we didn’t use the Taguchi approach, what else might we do to estimate factor effects?

– One approach is a one-at-a-time method in which all but one factor are held constant while the other is varied throughout its levels. How many experiments does this require?

Comparison• Answer:

For N factors each having L levels, we would need:

L x N experiments

That would be 12 in our example (and it turns out that we would not achieve the same accuracy in our estimate of the level means).

Comparison

•Another approach is the brute-force full factorial method. That is, conduct every possible experiment for the factors and levels.

•How many experiments would this require?

Comparison• Answer:

For N factors each having L levels, we would need:

LN experiments

That would be 81 experiments in our example. The only good thing is that we would be guaranteed to find the best configuration for our levels.

ANOM

•The approach we have taken with this problem is called the “analysis of means”.

•This approach relies on 2 assumptions for its validity:

– Use of an additive model is appropriate– Use of an orthogonal array is appropriate.

Additive Model

•What do we mean by additive model?

•First, let’s define a short hand for our factor effects:

– Let ai represent the deviation from the mean caused by setting factor a to level i. So i.t.o. our previous example,

b2 = mb2 - m

Additive Model

•We are going to show that the procedure we applied to our example is equivalent to applying an additive model of the individual factor effects to determine performance.

•This means that by using such an additive model, we should be able to predict the performance of our product or process based on the factor effects.

Additive Model

•Our additive model is stated as follows:

η(Ai, Bj, Ck, Dl) = m + ai + bj + ck + dl + e

•Where e is an error term that accounts for the error incurred by using the additive model and the error incurred by any lack of repeatability of measuring η for any given experiment.

Additive Model• As in most subject areas, an additive model is also sometimes called a superposition model or a variables separable model.

• It is possible for our factor effects to be non-linear (quadratic, cubic, etc.) but cross product terms (between factors) are not allowed.

Cross correlation would be like: “The effect of factor A at level 2 while factor C is at level 3 is …”

Additive Model• According to our previous definition of factor level effects, we can show that the sum of the effects of all the levels of a given factor is equal to 0. That is (in our example):

a1 + a2 + a3 = 0

b1 + b2 + b3 = 0

c1 + c2 + c3 = 0

d1 + d2 + d3 = 0

We will show this for factor a.

Additive Model

•Recall that ai = mAi – m;

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Additive Model

•Now, recall our representation of the observation value i.t.o the overall mean, factor effects, and error term:

η(Ai, Bj, Ck, Dl) = m + ai + bj + ck + dl + e

•We can represent factor effects i.t.o this equation as shown on the following slide.

Additive Model

Additive Model

•So from the previous result, we see that:

mA3 = m + a3 + some error term

This is sensible enough, recall that our representation of ai was originally mAi – m.

So mA3 as we have derived it here, is an estimate of m + a3 and thus our approach was in fact the use of an additive model.

Additive Model

•A note on the error term.

We have shown that the error is actually the average of 3 error terms (one corresponding to each experiment).

We typically treat each of the individual experiment error terms as having zero mean and some variance.

Additive Model•Define Replication Number nr:

–The replication number is the number of times a particular factor level is repeated in an orthogonal array.

•It can be shown that the error variance of the average effect of a factor level is smaller than the error variance of a single experiment by a factor equal to its nr.

Additive Model

•So, to obtain the same accuracy in our factor level averages using a one-factor-at-a-time approach, we would have to conduct 3 experiments at each of the 3 levels of each factor for a total of 36 experiments.