RELIABILITY IMPROVEMENT THROUGH ROBUST DESIGN
5.11 DESIGN OPTIMIZATION
5.11.5 Analysis of Variance
Graphical response analysis is an intuitive method for identifying significant factors and interactions. The method is easy to understand and use when the number of factors is small. However, the analysis may become tedious if a
TABLE 5.11 One-Factor Experimental Layout
Factor Level Observation Total Average 1 y11 y12 · · · y1n y1. y1.
2 y21 y22 . . . y2n y2. y2.
... ... ... . . . ... ... ...
p yp1 yp2 . . . ypn yp. yp.
relatively large number of factors are involved. In these situations, analysis of variance (ANOVA) is more efficient.
ANOVA for One-Factor Experiments To understand the concept and procedure for ANOVA, we first consider a one-factor experiment designed to determine the effect of factor A. Factor A has p levels, each of which contains n replicates.
Let yij be the j th observation of quality characteristic y taken at level i. The experimental layout is shown in Table 5.11. Statistically, the experiment is to test the hypothesis that the mean responses at all levels are equal. Let yi. and yi. denote the total and average of the observations at leveli, respectively. Also let y.. andy.. be the grand total and grand average of all observations. Then we have
where N= pn is the total number of observations in the experiment.
We define the total corrected sum of squares as SST =
to measure the overall variability in the data. SST can be written as SST =
DESIGN OPTIMIZATION 167
SSA is called the sum of squares of the factor, and SSE is the sum of squares of the error. Equation (5.38) indicates that the total corrected sum of squares can be partitioned into these two portions.
Factor A hasp levels; thus, SSA hasp− 1 degrees of freedom. There are N observations in the experiment, so SST has N− 1 degrees of freedom. Because there aren observations in each of p levels providing n− 1 degrees of freedom for estimating the experimental error, SSE has p(n− 1) = N − p degrees of freedom. Note that the degrees of freedom for SST equals the sum of degrees of freedom for SSAand SSE. Dividing the sum of squares by its respective degrees of freedom gives the mean square MS: namely,
MSx = SSx dfx
, (5.39)
where x denotes A or E and dfx is the number of degrees of freedom for SSx. TheF statistic for testing the hypothesis that the mean responses for all levels are equal is
F0= MSA
MSE, (5.40)
which has an F distribution with p− 1 and N − p degrees of freedom. We conclude that factor A has a statistically significant effect at 100α% significance level if
F0 > Fα,p−1,N−p.
For the convenience of numerical calculation, the sums of squares may be rewritten as
The procedure for the analysis of variance can be tabulated as shown in Table 5.12. The table is called an ANOVA table.
TABLE 5.12 One-Factor ANOVA Table Source of
Variation Sum of Squares
Degrees of
Freedom Mean Square F0
Factor SSA p− 1 MSA MSA/MSE
Error SSE N− p MSE
Total SST N− 1
TABLE 5.13 Temperature Data from the Engine Testing
A/F Ratio Temperature (◦C) Total Average 10.6 701 713 722 716 2852 713.00 11.6 745 738 751 761 2995 748.75 12.6 773 782 776 768 3099 774.75
Example 5.8 An experiment was designed to investigate the effects of the air-to-fuel (A/F) ratio on the temperature of the exhaust valve in an automobile engine. The experiment was replicated with four samples at each A/F ratio. The experimental data are summarized in Table 5.13. Determine whether A/F ratio has a strong influence at the 5% significance level.
SOLUTION The total and average of the temperature observations are com-puted and summarized in Table 5.13. The grand total and grand average are
y..=
The sums of squares are SST =
The calculation of mean squares and F0 is straightforward. The values are sum-marized in the ANOVA table as shown in Table 5.14. Because F0= 55.64 >
F0.05,2,9= 4.26, we conclude that the A/F ratio has a strong effect on the exhaust valve temperature at the 5% significance level.
ANOVA for Orthogonal Inner Arrays In the design of experiment, the purpose of an outer array is to expose samples to noise factors. After the experimental data TABLE 5.14 ANOVA Table for the Exhaust Temperature Data
Source of
Variation Sum of Squares
Degrees of
Freedom Mean Square F0
Factor 7689.5 2 3844.75 55.67
Error 621.5 9 69.06
Total 8311 11
DESIGN OPTIMIZATION 169
are collected, the outer array has completed its role. The outer array usually is not involved in subsequent ANOVA for design optimization unless we are interested in understanding the effects of noise factors on the quality characteristic. The optimal levels of design parameters are determined using analysis of variance for the inner array.
A column of an inner array may be assigned to a factor, an interaction, or an error (empty column). An I-level column in LN(IP) can be considered as an I-level factor, each I-level havingn= N/I replicates. Thus, (5.41) can be employed to calculate the total corrected sum of squares of an inner array, and (5.42) applies to a column of an inner array. LetT be the total of observations: namely,
T =
N i=1
yi,
whereyi represents ˆηi oryi as shown in, for example, Table 5.9. Then the total corrected sum of squares can be written as
SST =
N i=1
yi2−T2
N. (5.44)
Also, let Tj denote the total of observations taken at level j in a column. The sum of squares of columni having I levels is
SSi = I
For a two-level column, (5.45) reduces to SSi = (T0− T1)2
The sum of squares of column 2 of the array is
SS2= 3
In an inner array, some columns may be empty and are treated as error columns. The sum of squares for an error column is computed with (5.45). Then
the sums of squares for all error columns are added together. If an assigned col-umn has a small sum of squares, it may be treated as an error colcol-umn, and the sum of squares should be pooled into the error term. The total corrected sum of squares equals the total of the sums of squares for factor columns, interaction columns, and error columns. Recall that the number of degrees of freedom is I− 1 for an I-level column and is N − 1 for LN(IP). The number of degrees of freedom for error is the sum of the degrees of freedom for error columns. The mean square andF statistic for a factor or interaction are computed from (5.39) and (5.40), respectively. It is concluded that the factor or interaction is important at the 100α% significance level if
F0> Fα,I−1,dfe,
where dfe is the number of degrees of freedom for error.
The computation for ANOVA may be burdensome, especially when a large number of factors and interactions are involved. There are several commercial software packages, such as Minitab, which can perform the calculation.
If the quality characteristic is smaller-the-better or larger-the-better, the ANOVA for signal-to-noise ratio data determines the significance of the factors and interactions. The next step is to select the optimum levels of the significant factors and interactions. The selection method was discussed in Section 5.11.4.
ANOVA should be performed for both signal-to-noise ratio and mean response data if the quality characteristic belongs to the nominal-the-best type. In such cases, the levels of dispersion factors are selected to maximize the signal-to-noise ratio, while the mean adjustment factors are set at the levels that bring the response on target. The procedure for choosing the optimal setting is the same as that for the graphical analysis.
Once the optimal setting is specified, the average response at the optimal setting should be predicted by using (5.37). A confirmation test is run to verify that the predicted value is achieved.
Example 5.9 Refer to Examples 5.4 and 5.5. The design of experiment for the rear spade has four control factors and two interactions. L8(27) is used as an inner array to accommodate the control factors. The outer array is filled with two combinations of noise levels. Two test units of the same setting of control factors were run at each of the two noise combinations. The fatigue life data (in 1000 cycles) are shown in Table 5.15.
Fatigue life is a larger-the-better characteristic. The signal-to-noise ratio for each row of the inner array is computed using (5.35) and is shown in Table 5.15.
For example, the value of the ratio for the first row is ˆη1 = −10 log
DESIGN OPTIMIZATION 171
TABLE 5.15 Cross Array and Fatigue Life Data for the Rear Spade Inner Array
D B D× B C D× C A Outer Array
Run 1 2 3 4 5 6 7 z11 z12 ˆη
1 0 0 0 0 0 0 0 7.6 8.2 6.2 6.9 17.03
2 0 0 0 1 1 1 1 7.1 6.7 4.9 4.2 14.55
3 0 1 1 0 0 1 1 4.8 6.3 5.2 3.9 13.68
4 0 1 1 1 1 0 0 6.2 5.2 4.4 5.1 14.17
5 1 0 1 0 1 0 1 3.9 4.3 3.6 4.7 12.18
6 1 0 1 1 0 1 0 5.7 5.1 4.7 3.8 13.38
7 1 1 0 0 1 1 0 6.4 5.9 5.7 6.4 15.67
8 1 1 0 1 0 0 1 6.8 6.2 4.3 5.5 14.72
The total of the values of the signal-to-noise ratio is
T =
8 i=1
ˆηi= 17.03 + 14.55 + · · · + 14.72 = 115.38.
The sum of squares for each column is calculated from (5.46). For example, the sum of squares for column 2 (factor B) is
SS2 =1
8× (17.03 + 14.55 + 12.18 + 13.38
− 13.68 − 14.17 − 15.67 − 14.72)2= 0.15.
The sums of squares of the factors, interactions, and error are given in Table 5.16.
Because each column has two levels, the degrees of freedom for each factor, interaction, and error is 1. Note that in the table the sum of squares for factor B
TABLE 5.16 ANOVA Table for the Fatigue Life Data of the Rear Spade
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square F0
A 3.28 1 3.28 27.33
B 0.15 1 0.15
C 0.38 1 0.38 3.17
D 1.51 1 1.51 12.58
D× B 9.17 1 9.17 76.42
D× C 0.63 1 0.63 5.25
e 0.09 1 0.09
(e) (0.24) (2) (0.12)
Total 15.21 7
is pooled into error e to give the new error term (e). Thus, the new error term has 2 degrees of freedom.
The critical value of the F statistic is F0.1,1,2= 8.53 at the 10% significance level. By comparing the critical value with theF0values for the factors and inter-actions in the ANOVA table, we conclude that A, D, and D× B have significant effects, whereas B, C, and D× C are not statistically important.
For comparison, we use the graphical response method described in Section 5.11.4 to generate the main effect plots and interaction plots, as shown Figure 5.24. Figure 5.24a indicates that the slopes for factors A and D are steep, and thus these factors have strong effects, whereas factor B is clearly insignificant.
From the figure it is difficult to judge the importance of factor C because of the marginal slope. This suggests that ANOVA should be performed. Figure 5.24b shows that the interaction between factors B and D is very strong, although factor B itself is not significant. As indicated in Figure 5.24c, there appears an interaction between factors C and D because of the lack of parallelism of the two line segments. The interaction, however, is considerably less severe than that between B and D. ANOVA shows that this interaction is statistically insignificant, but the value ofF0 is close to the critical value.
Once the significant factors and interactions are identified, the optimal levels should be selected. Because the interaction D× B has a strong effect, the levels of B and D are determined by the interaction effect. From the interaction plot, we choose B0D0. Figure 5.24a orT0 and T1 of factor A calculated for ANOVA suggests that A0 be selected. Because factor C is deemed insignificant, C0 is chosen to maintain the current manufacturing process. In summary, the design should use material type 1, forging thickness 7.5 mm, and bend radius 5 mm with normal shot peening in manufacturing.
The value of the signal-to-noise ratio predicted at the optimal setting A0B0C0
D0 is obtained from (5.37) as
ˆη = T + (A0− T ) + (D0− T ) + [(D0B0− T ) − (D0− T ) − (B0− T )]
= A0+ D0B0− B0= 15.06 + 15.49 − 14.28 = 16.27.
A confirmation test should be run to verify that the signal-to-noise ratio predicted is achieved.
The estimated average fatigue life at the optimal setting A0B0C0D0 is ˆy = 10ˆη/20× 1000 = 6509 cycles. This life estimate is the average of life data over the noise levels and unit-to-unit variability.
5.12 ROBUST RELIABILITY DESIGN OF DIAGNOSTIC SYSTEMS