Question 5 of 6: Taguchi Methods, and a $2^3$ Factorial Experiment on Tensile Strength

Question 5: Taguchi Methods, and a $2^3$ Factorial Experiment on Tensile Strength (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Taguchi methods, parameter-design objectives, inner/outer arrays, and robust design

Taguchi's methods apply statistically designed experiments (fractional-factorial and orthogonal-array designs) across three stages of product/process design. System design selects the basic technology/concept. Parameter design — Taguchi's central contribution — uses orthogonal-array experiments to choose the NOMINAL levels of the CONTROLLABLE design/process parameters that make performance insensitive to uncontrollable "noise," using inexpensive, ordinary-grade components wherever possible. Tolerance design is used only as a last resort, after parameter design: it tightens tolerances (or specifies higher-grade components) on the few parameters found, from the parameter-design experiment, to still transmit unacceptable variation — because tightening a tolerance costs money on every unit made forever, while parameter design is often free (it only changes a target setting, not a component grade).

The usual objectives in parameter design are to find the combination of control-factor levels that (i) brings the mean response onto target, (ii) minimizes the variance transmitted to the response from uncontrollable noise (maximizing an appropriate signal-to-noise ratio, chosen according to whether the response is nominal-the-best, smaller-the-better, or larger-the-better), and (iii) does both using the least expensive components/settings available, rather than by controlling or eliminating the noise itself.

The inner array is the orthogonal array of CONTROLLABLE design parameters — the variables the engineer is free to set (analogous to Ni, Mo, Ti in part (b), if this experiment were being run for robustness rather than mean-effect estimation). The outer array is a second orthogonal array of NOISE factors — variables that are uncontrollable in the field (ambient temperature, load variation, part-to-part material variation, customer usage pattern) but which CAN be deliberately varied in the experiment to simulate that real-world variability. Every inner-array run is crossed with every outer-array combination, so each inner-array setting produces a whole DISTRIBUTION of outcomes (one per noise combination) from which a signal-to-noise ratio, not just a single mean, is computed for that setting.

A robust design is one whose performance is insensitive to variation in the uncontrollable noise factors — achieved by deliberately exploiting NONLINEARITY or INTERACTION between a control factor and a noise factor (choosing the control-factor level at which the response curve is locally flat with respect to the noise factor), rather than by the far more expensive route of trying to control, shield, or eliminate the noise itself.

(b) $2^3$ factorial: main effects, interaction effects, and significance at $\alpha=0.05$

Given. $2^3$ factorial (Ni, Mo, Ti) with 2 replicates ($N=16$ runs total), tensile-strength data in the table above.

Find. Estimates of the 3 main effects and 4 interaction effects, and which are statistically significant at $\alpha=0.05$.

Approach. Use the standard signed contrast method: for each effect, form the contrast by summing every observation with the sign it carries in that effect's column of the $2^3$ sign table, divide by $4n$ (half the runs, times replicates) for the effect and square-and-divide by $8n$ for its sum of squares, then compare each effect's mean square against the pooled error mean square via an $F$-test with $(1,\ 8n-8)$ degrees of freedom.

1. Treatment totals and grand mean. With $n=2$ replicates per the 8 corners, $N=16$, grand mean $\bar{\bar y}=283.125$.
2. Effects (contrast$/4n$, $n=2$).

   Effect	Contrast	Effect estimate	$SS$
   Ni	30	3.75	56.25
   Mo	62	7.75	240.25
   Ti	88	11.00	484.00
   Ni×Mo	30	3.75	56.25
   Ni×Ti	4	0.50	1.00
   Mo×Ti	−8	−1.00	4.00
   Ni×Mo×Ti	48	6.00	144.00

3. Error term and $F$-test. $SS_T=1021.75$ (about $\bar{\bar y}$); summing the seven effect $SS$ above gives $SS_{\text{factors}}=985.75$, so $SSE=SS_T-SS_{\text{factors}}=36.0$ with $df_E=16-8=8$ (one degree of freedom "used up" by each of the 8 treatment means, leaving the within-cell replicate variation as pure error), giving $MSE=36.0/8=4.5$. The critical value is $F_{0.05,1,8}=5.32$. $$F=\frac{SS_{\text{effect}}}{MSE}$$
4. Significance decisions.

   Effect	$F=SS/4.5$	Significant at $\alpha=0.05$?
   Ni	12.50	Yes
   Mo	53.39	Yes
   Ti	107.56	Yes
   Ni×Mo	12.50	Yes
   Ni×Ti	0.22	No
   Mo×Ti	0.89	No
   Ni×Mo×Ti	32.00	Yes

   All three main effects, the Ni×Mo two-factor interaction, and the full Ni×Mo×Ti three-factor interaction are $\boxed{\text{statistically significant}}$ at $\alpha=0.05$; the Ni×Ti and Mo×Ti interactions are not.

(c) Best factor combination, and predicted mean response at Ni=16%, Mo=5%, Ti=0.55%

Given. The eight treatment means from part (b); significant-effects model from part (b).

Find. Which corner of the design gives the highest mean tensile strength, and the predicted mean response at Ni=16%, Mo=5%, Ti=0.55% using the significant terms only.

Approach. Since all three main effects are positive AND the significant Ni×Mo and Ni×Mo×Ti interactions are also positive, the effects reinforce rather than cancel at the all-high corner; confirm directly from the eight treatment means, then build a coded first-order-plus-interaction prediction equation from the significant effects only and evaluate it at the coded target levels.

1. Treatment means.

   Ni	Mo	Ti	Mean
   10	3	0.5	270.5
   20	3	0.5	276.0
   10	6	0.5	281.5
   20	6	0.5	282.5
   10	3	0.6	288.0
   20	3	0.6	282.5
   10	6	0.6	285.0
   20	6	0.6	​299.0

   The maximum is at Ni=20%, Mo=6%, Ti=0.6% (all three factors at their HIGH level), $\bar y=\boxed{299.0}$ — consistent with all three main effects and both surviving interactions (Ni×Mo, Ni×Mo×Ti) being positive, so no factor benefits from being run low.
2. Coded prediction equation (significant terms only). With $x_1,x_2,x_3$ the coded (−1/+1 at the low/high levels) variables for Ni, Mo, Ti, dropping the non-significant Ni×Ti and Mo×Ti terms: $$\hat y=\bar{\bar y}+\frac{\text{Ni}}{2}x_1+\frac{\text{Mo}}{2}x_2+\frac{\text{Ti}}{2}x_3+\frac{\text{Ni}{\times}\text{Mo}}{2}x_1x_2+\frac{\text{Ni}{\times}\text{Mo}{\times}\text{Ti}}{2}x_1x_2x_3$$ $$=283.125+1.875x_1+3.875x_2+5.5x_3+1.875x_1x_2+3.0x_1x_2x_3$$
3. Code the target levels. $x=(actual-\text{mid})/(\text{half-range})$: Ni mid=15,half=5 → $x_1=(16-15)/5=0.2$; Mo mid=4.5,half=1.5 → $x_2=(5-4.5)/1.5=0.333$; Ti mid=0.55,half=0.05 → $x_3=(0.55-0.55)/0.05=0$.
4. Evaluate. Since $x_3=0$, both the Ti main-effect term and the three-factor-interaction term (which carries $x_3$) vanish identically: $$\hat y=283.125+1.875(0.2)+3.875(0.333)+0+1.875(0.2)(0.333)+0=283.125+0.375+1.292+0.125=\boxed{284.9\text{ (psi}\times10)}$$