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23-Ind-A5 Quality Planning, Control, and Assurance · May 2018

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

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams, May 2018. Closed-book examination. Any five of the six questions constitute a complete paper; all six are answered in full below. Relevant statistical tables (cumulative standard normal distribution, factors for constructing variables control charts, the F distribution, and MIL-STD-105E Tables 1 and II-A) are reproduced/applied from the paper's own attached appendices.

Reference texts: Montgomery, Introduction to Statistical Quality Control (8th ed.) — Ch. 1–2 (quality philosophy, cost of quality, Six Sigma/TQM, ISO 9000), Ch. 4–6 (magnificent seven SPC tools, process capability, X̄-S and attributes control charts), Ch. 9 (average run length), Ch. 13–14 (designed experiments/factorial designs, acceptance sampling and MIL-STD-105E).

Question 5: 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; 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, engineering-judgment stage); (2) parameter (robust) design — using statistically designed experiments to choose the nominal (target) settings of the controllable design factors so that the response is both on-target and minimally sensitive to uncontrollable "noise" factors, without necessarily requiring 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, guided by the Taguchi loss function's economic sensitivity $k$.

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 can set and will specify in the final design). The outer array is a second, separate orthogonal design over the noise factors (uncontrollable in the field — e.g. ambient temperature, supplier-to-supplier material variation, customer usage pattern) that are deliberately varied during the experiment to represent field variability. Every inner-array run is repeated across every outer-array combination (a "crossed array"), so a signal-to-noise ratio can be computed at each inner-array setting, directly measuring how sensitive that design choice is to the noise factors — the setting with the best S/N ratio is the one that is both on-target and robust to conditions the engineer cannot control after the product ships.

(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%.

Treatment totals (sum of the 2 replicates)
Combination(1)ababcacbcabc
Rep I285295306313290302310308
Rep II290282298325280290326302
Total575577604638570592636610

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$/8$): $$A=\frac{-575+577-604+638-570+592-636+610}{8}=4.00,\qquad B=\frac{-575-577+604+638-570-592+636+610}{8}=21.75$$ $$C=1.75,\qquad AB=-2.00,\qquad AC=-5.00,\qquad BC=-0.75,\qquad ABC=-10.00$$
  2. Sum of squares ($SS=\text{contrast}^2/16$, i.e. $(8\times\text{effect})^2/16$): $$SS_A=64.0,\ SS_B=1892.25,\ SS_C=12.25,\ SS_{AB}=16.0,\ SS_{AC}=100.0,\ SS_{BC}=2.25,\ SS_{ABC}=400.0$$
  3. Error term. $SSE=\sum(y_I-y_{II})^2/2=469.0$ over $df_E=8$, so $MSE=469.0/8=58.625$. Critical value $F_{0.05,1,8}=5.318$.
  4. F-test each effect ($F=SS/MSE$).
    EffectABCABACBCABC
    $F$1.0932.280.210.271.710.046.82
    vs. $F_{crit}=5.32$nsSIGnsnsnsnsSIG
    $$\boxed{\text{Significant at }\alpha=0.05:\ B\ (\text{Mo main effect}),\ ABC\ (\text{three-way interaction})}$$
B (Mo)21.75ABC-10.00AC-5.00A (Ni)4.00AB-2.00C (Ti)1.75BC-0.75Effect estimates, 2³ factorial on tensile strength (Q5b)
Fig. 5.1 — Effect estimates ranked by magnitude; $B$ (molybdenum) dominates, with the three-factor interaction $ABC$ the only other effect exceeding the $F_{0.05,1,8}$ significance threshold.

(c) Best combination and predicted mean response at Ni=17%, Mo=4%, Ti=0.6%

Given. Treatment means $\bar y=\text{total}/2$: (1)=287.5, a=288.5, b=302.0, ab=319.0, c=285.0, ac=296.0, bc=318.0, abc=305.0. Effect estimates from part (b); grand mean $\bar{\bar y}=300.125$.

Find. The factor-level combination with the highest mean tensile strength, and the fitted mean response at Ni=17%, Mo=4%, Ti=0.6%.

  1. Best combination. Scanning the 8 treatment means, ab (Ni=18%, Mo=5%, Ti=0.55%) has the highest mean, 319.0, narrowly ahead of bc (Ni=15%, Mo=5%, Ti=0.65%) at 318.0. $$\boxed{\text{Highest mean tensile strength: Ni=18\%, Mo=5\%, Ti=0.55\% (mean}=319.0)}$$
  2. Coded factor levels for the prediction point. With centre/half-range $(16.5,1.5)$ for Ni, $(4.5,0.5)$ for Mo, $(0.6,0.05)$ for Ti: $$x_1=\frac{17-16.5}{1.5}=0.333,\qquad x_2=\frac{4-4.5}{0.5}=-1.000,\qquad x_3=\frac{0.6-0.6}{0.05}=0$$
  3. Fitted response-surface model (coefficients = effect$/2$): $$\hat y=\bar{\bar y}+\tfrac{A}{2}x_1+\tfrac{B}{2}x_2+\tfrac{C}{2}x_3+\tfrac{AB}{2}x_1x_2+\tfrac{AC}{2}x_1x_3+\tfrac{BC}{2}x_2x_3+\tfrac{ABC}{2}x_1x_2x_3$$ Since $x_3=0$, every term containing $C$, $AC$, $BC$, or $ABC$ vanishes, leaving only the grand mean plus the $A$, $B$, and $AB$ contributions: $$\hat y=300.125+2.00(0.333)+(-10.875)(-1)\cdot(-1)+\ldots$$
  4. Evaluate. $$\hat y=300.125+\underbrace{2.00(0.333)}_{0.667}+\underbrace{10.875(-1)}_{-10.875}+\underbrace{(-1.00)(0.333)(-1)}_{0.333}=290.25$$ $$\boxed{\hat y(17,4,0.6)=290.25\text{ (units of Table 5's response scale)}}$$
Check — the prediction uses the full fitted model (all 7 estimated effects); because the evaluation point sits exactly at the Ti centre ($x_3=0$), the terms in $C$, $AC$, $BC$ and $ABC$ (including the statistically significant $ABC$) all vanish regardless, so a reduced model keeping only the significant $B$ and $ABC$ terms would differ from $\hat y=290.25$ only by whether the (individually non-significant) $AB$ term is retained — a $\pm1.0$ sensitivity, small next to the $290$ magnitude of the response.
Question 5 — final results
QuantityValue
Effects A / B / C4.00 / 21.75 / 1.75
Effects AB / AC / BC / ABC-2.00 / -5.00 / -0.75 / -10.00
Significant at $\alpha=0.05$B, ABC
Best combinationab: Ni=18%, Mo=5%, Ti=0.55% (mean 319.0)
$\hat y$ at Ni=17%,Mo=4%,Ti=0.6%290.25