Process or Product Monitoring and Control
6.5.5. Principal Components
|Calculation of principal components example||A numerical example may clarify the mechanics of principal component analysis.|
|Sample data set||
Let us analyze the following 3-variate dataset with 10 observations.
Each observation consists of 3 measurements on a wafer: thickness,
horizontal displacement and vertical displacement.
|Compute the correlation matrix||
First compute the correlation matrix
|Solve for the roots of R||
Next solve for the roots of R, using software
|Compute the first column of the V matrix||
Substituting the first eigenvalue of 1.769 and R in the
appropriate equation we obtain
|Compute the remaining columns of the V matrix||
Repeating this procedure for the other 2 eigenvalues yields the matrix
|Compute the L1/2 matrix||
Now form the matrix L1/2, which is a diagonal
matrix whose elements are the square roots of the eigenvalues of
R. Then obtain S, the factor structure, using
S = V L1/2
|Compute the communality||
Next compute the communality, using the first two eigenvalues only
|Diagonal elements report how much of the variability is explained||
Communality consists of the diagonal elements.
This means that the first two principal components "explain" 86.62% of the first variable, 84.20 % of the second variable, and 98.76% of the third.
|Compute the coefficient matrix||
The coefficient matrix, B, is formed using the reciprocals of
the diagonals of L1/2
|Compute the principal factors||
Finally, we can compute the factor scores from ZB, where
Z is X converted to standard score form. These
columns are the principal factors.
|Principal factors control chart||These factors can be plotted against the indices, which could be times. If time is used, the resulting plot is an example of a principal factors control chart.|