We're back again. In our last lecture, we talked about the mechanics of the canonical analysis. I think, now, we need to illustrate this with an example. This is example 11.2 for the book, it's a continuation of that chemical process example that we started back in example 11.1, and where we used steepest ascent to arrive at a neighborhood of the optimum. Now, we've decided that around to that point of about 85 minutes of reaction tab and 175 degrees, we probably need to think about filling a second-order model because we saw curvature in that second first-order model that we fit around that point. In table 11.6, here's the design that we're going to use to do this. Now, this is the original, two square design with center points around 85 minutes and 175 degrees. That's the center. What we've done is, we've added axial runs to this experiment to create a central composite design. In coded units, we used alpha values for our axial runs of plus or minus 1.414. That's both in the x_1 and the x_2 direction. This is the complete second-order design now with the factorial portion, that's the first four runs, then the five center runs, and in the four axial runs shown at the bottom of the table. Here are the responses, now, this is kind of interesting. Look at this. So far we've only been working with yield, well, when this experiment was run, the experimenters also, were measuring two other responses, viscosity, that's y_2, and molecular weight, that's y_3. Obviously, we want to maximize yield, but we have to be concerned about viscosity and molecular weight because those are physical properties of this material that customers are interested in. So in this second phase of the study, we probably want to look at all of these responses, and incidentally, this is very typical of response surface work. You very rarely have only a single response that you're interested in, particularly, when you get to the optimization phase. But for the moment, we're going to concentrate just on yield. This is what our central composite design looks like now in coded units. Remember, all we've done here is we've added the axial runs to the original four runs at the corners of the square and the five runs at the center. Here is some computer output for this experiment. What you really need to pay attention to is this analysis of variance table down at the bottom. That's the key here, and you'll notice that the linear terms are highly significant. These p-values are quite small. The pure quadratic terms are highly significant. These p-values are quite small. The interaction term is only significant at about the 10 percent level. But I would leave it in. I think, fitting the full quadratic here is usually a pretty good idea unless you have a lot of non-significant terms. I'm going to leave that term in its significant at about the 10 percent level. There's no indication of lack of fit here. The F statistic for lack of fit would be very, very small. So there's really no reason to believe that there's really anything to worry about here in terms of the second-order model not be an adequate fit to the data. So here is an another table of output display. This gives you the regression coefficient. This is actually the model coefficient in terms of the coded units, and this is the model in terms of natural units. Finally, here's a summary table that shows you the fitted value, the predicted value, the residuals, and other diagnostics. If you conduct a normal probability plot of the residuals, everything is fine. So this model is a very good fit to the data. This slide shows you contour plots and response surface plots. This is a contour plot of yield, and you notice that the surface appears to be a pure maximum. It looks like that somewhere in here, there's a point of maximum response, and it's a pretty dramatic shape. You can see that by looking at the response surface plot in the figure on the right. But this would be a good place to illustrate the canonical analysis, so that you can see how this is actually done. First of all, let's find the precise mathematical location to the stationary point. Vector b is just the vector of the first-order regression coefficients, and matrix B is the two-by-two matrix that has the pure quadratic coefficients on the main diagonal and one half of the interaction coefficient on the off-diagonal. So both of these matrices are shown here. Now, the stationary point is just going to be calculated as minus a half times the inverse of that B matrix times vector b. The stationary point turns out to be encoded units x_1 as would be 0.389, and x_2 as would be 0.305. We can now, actually convert those into the natural units. Here's the conversion into the natural units. When you solve those equations, you get about 87 minutes of reaction time in about 176.5 degrees, and that's pretty close to what we would have probably picked out from just looking at this picture, we would have probably picked something somewhere around there, and that's very close to what we actually found as the mathematical solution for the stationary point. We probably ought to look at the canonical model form as well. We are sure this is a maximum, but let's be sure that we understand how to do the canonical analysis to calculate or to obtain the canonical form of the model, we need the eigenvalues of the B matrix. The eigenvalues of B, are the roots of this determinant equation that you see here. The determinant of B minus lambda I equal to 0. Now, when you form that quantity and calculate the determinant, that reduces to the quadratic equation in lambda that you see here. The roots of that quadratic equation are going to be the eigenvalues lambda one and lambda two. The roots of that quadratic equation turned out to be lambda one is minus 0.964 and lambda two is minus 1.4147. So this is the canonical form. Both eigenvalues are negative. That's a strong indication that the stationary point is a point of maximum response. Lambda two is larger in absolute value than lambda one. That indicates that the surface is probably more sensitive in the w_2 direction. We've only talked about minimum and maximum in saddle systems. There are other kinds of response surfaces that we sometimes encounter. Typically, they'll be Ridge Systems and Ridge Systems are typically indicated when you have one or more of your eigenvalues that are very, very small. Theoretically, an eigenvalue of zero would indicate a Ridge System of some sort, because that would be an enormous amount of elongation in one direction. But you rarely ever find an eigenvalue that's actually exactly zero. What you'll find is eigenvalues that are much smaller than others, and the rule of thumb I use, if an eigenvalue is smaller than the next one by an order of magnitude, you've probably got a ridge. This is what I'm sort of trying to illustrate here. Here's a situation where lambda one is essentially zero. It's very, very small, so the surface is very elongated in this direction. Here's another case, where we have an elongated system where lambda one is very, very small, so the system is highly elongated along the w_1 direction and not elongated at all in the w_2 direction. These two Ridge Systems look different, don't they? Yeah. This one is called a stationary ridge, and at the stationary Ridge System, we have this ridge behavior, but the stationary point is inside the region of experimentation. So this is our region of experimentation and the stationary point is inside that region. This is an example of a rising Ridge System. This is our experimental region, but the stationary point is somewhere out here beyond our region of experimentation. So in this kind of system, we would want to try to run further experiments up this ridge to see if we can improve our response. Whereas, in this system, we can probably operate anywhere along that ridge, and get pretty good results, or along this ridge and get pretty good results. It's much more flexibility in the stationary ridge system. So that's sort of an overview of the canonical analysis and a quick introduction to how ridge systems work. We'll pick up with this material next time.