Question 5 of 6: Product Design Stages, Robust Design, and a $2^3$ Factorial Experiment on Tensile Strength

Question 5: Product Design Stages, Robust Design, 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) Three stages of product design; parameter-design objectives; inner and outer arrays

Taguchi's three stages of product/process design are: (1) system design — using engineering/scientific knowledge to select the basic technology, architecture, and configuration that will meet the functional requirements (a largely qualitative stage); (2) parameter (robust) design — using statistically designed experiments to choose the nominal settings of the controllable design factors so the response is both on-target and minimally sensitive to uncontrollable noise, without requiring more expensive components; (3) tolerance design — only after parameter design has minimized sensitivity to noise, selectively tightening tolerances (at extra cost) on just the factors whose variability most affects the response, per the Taguchi loss function's economic sensitivity discussed in Q1(a).

Usual objectives of the parameter-design stage: (i) choose factor-level settings that put the mean response on target; (ii) simultaneously minimize the response's sensitivity (variance) to noise factors that cannot be controlled once the product is in the field, i.e. maximize an appropriate signal-to-noise ratio; and (iii) achieve both without resorting to costlier components or tighter tolerances — the improvement comes from the choice of nominal settings, not from spending more.

Inner and outer arrays: in a robust-design (parameter-design) experiment, the inner array is an orthogonal design over the controllable design factors (the ones the engineer sets and specifies in the final design). The outer array is a second, separate orthogonal design over the noise factors (uncontrollable in the field — ambient conditions, material lot-to-lot variation, customer usage pattern) that are deliberately varied to represent field variability. Every inner-array run is crossed with every outer-array combination, so a signal-to-noise ratio can be computed at each inner-array setting, directly measuring how sensitive that design choice is to noise — the setting with the best S/N ratio is both on-target and robust to conditions the engineer cannot control after shipment.

(b) Effect estimates and significance testing for the $2^3$ factorial

Given. $2^3$ factorial ($k=3$ factors A=Ni, B=Mo, C=Ti), $r=2$ replicates; low/high levels Ni: 15%/18%, Mo: 4%/5%, Ti: 0.55%/0.65% (source table headers Ni/Mo/Ti; "manganese" in the question stem is a wording slip against the table's own Mo column).

Find. Main effects $A,B,C$; interaction effects $AB,AC,BC,ABC$; which are significant at $\alpha=0.05$.

Approach. Standard $2^3$ sign-table contrasts, each effect $=\text{contrast}/(4r)$; sum-of-squares $SS=\text{contrast}^2/(8r)$; pooled replicate error $SSE=\sum(y_I-y_{II})^2/2$ with $df_E=8(r-1)=8$; compare each $F=SS/MSE$ against $F_{0.05,1,8}$.

1. Effect estimates (contrast$/(4r)=$contrast$/8$): $$A=\frac{-540+593-554+586-502+522-578+596}{8}=15.375,\qquad B=\frac{-540-593+554+586-502-522+578+596}{8}=19.625$$ $$C=-9.375,\qquad AB=-2.875,\qquad AC=-5.875,\qquad BC=17.875,\qquad ABC=2.375$$
2. Sum of squares ($SS=\text{contrast}^2/16$, i.e. $(8\times\text{effect})^2/16$): $$SS_A=945.56,\ SS_B=1540.56,\ SS_C=351.56,\ SS_{AB}=33.06,\ SS_{AC}=138.06,\ SS_{BC}=1278.06,\ SS_{ABC}=22.56$$
3. Error term. Each replicate pair differs by only 0–8 units (e.g. $(1)$: 270 vs. 270; $a$: 295 vs. 298), giving $SSE=\sum(y_I-y_{II})^2/2=22.5$ over $df_E=8$, so $MSE=22.5/8=2.8125$. Critical value $F_{0.05,1,8}=5.318$.
4. F-test each effect ($F=SS/MSE$).

   Effect	A	B	C	AB	AC	BC	ABC
   $F$	336.2	547.8	125.0	11.8	49.1	454.4	8.0
   vs. $F_{crit}=5.32$	SIG	SIG	SIG	SIG	SIG	SIG	SIG

   $$\boxed{\text{All seven effects (three main, four interaction) are statistically significant at }\alpha=0.05}$$

(c) Best combination and predicted mean response at Ni=17%, Mo=4%, Ti=0.62%

Given. Treatment means $\bar y=\text{total}/2$: (1)=270.0, a=296.5, b=277.0, ab=293.0, c=251.0, ac=261.0, bc=289.0, abc=298.0. Effect estimates from part (b); grand mean $\bar{\bar y}=279.4375$.

Find. The factor-level combination with the highest mean tensile strength, and the fitted mean response at Ni=17%, Mo=4%, Ti=0.62%.

1. Best combination. Scanning the 8 treatment means, abc (Ni=18%, Mo=5%, Ti=0.65% — all three factors at their high level) has the highest mean, 298.0. $$\boxed{\text{Highest mean tensile strength: Ni=18\%, Mo=5\%, Ti=0.65\% (mean}=298.0)}$$
2. Coded factor levels for the prediction point. With centre/half-range $(16.5,1.5)$ for Ni, $(4.5,0.5)$ for Mo, $(0.60,0.05)$ for Ti: $$x_1=\frac{17-16.5}{1.5}=0.333,\qquad x_2=\frac{4-4.5}{0.5}=-1.000,\qquad x_3=\frac{0.62-0.60}{0.05}=0.400$$
3. Fitted response-surface model (coefficients = effect$/2$, all seven terms retained since all seven effects are significant): $$\hat y=\bar{\bar y}+\tfrac{A}{2}x_1+\tfrac{B}{2}x_2+\tfrac{C}{2}x_3+\tfrac{AB}{2}x_1x_2+\tfrac{AC}{2}x_1x_3+\tfrac{BC}{2}x_2x_3+\tfrac{ABC}{2}x_1x_2x_3$$
4. Evaluate. $$\hat y=279.4375+7.6875(0.333)+9.8125(-1)+(-4.6875)(0.4)+(-1.4375)(0.333)(-1)+(-2.9375)(0.333)(0.4)+8.9375(-1)(0.4)+1.1875(0.333)(-1)(0.4)$$ $$\boxed{\hat y(17,4,0.62)=266.67\text{ (tensile-strength response units)}}$$