In the last article, we learned how to determine the coefficients of a predictive model for 2-level screening designs. It is more complex to determine model coefficients for multi-level experiments so for those, we rely on statistical methods software.

In this article, we look at using the model to develop solutions. So that we learn the basics, we first use some simple algebra to find a solution. Then, in the next article, we will explore some common tools that are found in DOE software programs to help uncover solutions.

Let’s revisit the camera battery life example from the last article. We ended up with the following model, where y-hat I the predicted battery life and the factors are:

- X1 – Wall thickness
- X3 – Material Type
- X4 – Ambient Temperature

$$ \displaystyle \hat{Y}=46.5-2.5X_{1}-3.5X_{4}-3.5X_{3}X_{4} $$

This model is the “coded” model, which is based on factor levels ranging from “-1” to “+1”. The coded model has some advantages in that the factors and interactions are all orthogonal to each other. Essentially this allows us to have independent (clean) estimates of the main and interaction effects. We can work with the coded model and then convert solutions to real units.

So, we have one equation and 3 unknowns. Suppose our objective is to find a solution (levels of the factors) that would result in a predicted life of 45. Obviously, we could find an infinite set of values for wall thickness, material type, and temperature that would produce a given result. So, where should we start? It is often helpful to interpret the main and interaction effects. For the interaction effect, we use a 2-factor interaction plot. We have:

When lithium (X_{3} low) is used, changes in temperature have no effect on battery life. This property is desirable.

This plot shows us that when the material type is “Lithium”, then the Ambient Temperature has no effect on the average battery life. That may be a desirable situation since we cannot control the ambient temperature. So, we will force X_{3} to the low value, which is “-1” on the coded scale, and then simplify the equation as shown below.

Starting with the model

$$ \displaystyle \hat{Y}=46.5-2.5X_{1}-3.5X_{4}-3.5X_{3}X_{4} $$

Suppose the target value is 45.0. We plan to use lithium so we set X_{3} = -1.

$$ \displaystyle 45=46.5-2.5X_{1}-3.5X_{4}-3.5\left(-1\right)X_{4} $$

Simplifying, we have:

$$ \displaystyle \begin{align*}

45 & =46.5-2.5X_{1}-3.5X_{4}+3.5X\\

45 & =46.5-2.5X_{1}

\end{align*} $$

As might be expected, the Temperature factor (X_{4}), simply falls out of the equation when we force the material type to be -1 (Lithium). We are left with a simple equation with only one unknown (X_{4} = Wall Thickness). Finally, we can solve for the value of X_{1} that produces a response of 45.

Since we are left with 1 equation and 1 unknown, we can solve for X_{1}.

$$ \displaystyle \begin{align*}

45-46.5 & =-2.5X_{1}\\

-1.5 & =-2.5X_{1}\\

0.6 & =X_{1}

\end{align*} $$

The solution is to set wall thickness equal to 0.6 (on the coded scale).

Note: each factor must fall between -1 and 1 for the solution to be feasible. We should not extrapolate beyond the region over which we experimented because we do not know of the nonlinearities or interactions that may exist beyond the design space we selected.

A key requirement is that the solution for the factor level must be within the range of which we varied the factor in the experiment (-1 to +1 on the coded scale). Here, we got 0.6 which is a “feasible solution” since it is in the design space. Solving for a solution that is outside of our design space is risky since we did not collect data in that region.

Finally, how do we know what the coded solution of 0.6 is in real units? We must de-code this solution, so we know the actual wall thickness to use. As shown in the graphic below the actual levels for wall thickness were, low = 0.8 mm and high = 1.6 mm.

If our coded solution were 0, it would be obvious that the uncoded solution would be halfway between 0.8 mm and 1.6 mm and would be equal to 1.2 mm. But what about our coded solution of 0.6? We need to find the value on the uncoded scale that is 60% of the way between the middle value and high value.

We take 60% of the actual distance on the uncoded scale (0.6*0.4mm) = 0.24 mm

Then add this to the middle value of 1.2 mm. We get 1.44 mm which is the uncoded solution.

In this example, things worked out rather nicely once we realized that we wanted the solution to be insensitive to ambient temperature changes. By using Lithium, the temperature factor became insignificant and we were left with only one factor to solve for. The real world is not this simple. Let us look at one more example which will introduce an important graphical tool to help us find solutions.

In this example, we have a model which predicts “run out” of a key characteristic. The desired value is 0. The predictive model is developed and contains 3 significant factors (through three main effects and one 2-factor interaction)

An experiment was performed to understand the factors affecting “run out”. Tool coating, temperature, cycle time, and the temp/cycle time interaction were significatnt.

$$ \displaystyle \hat{Y}=2.2+2.5X_{1}-3.5X_{2}+3X_{3}+2X_{2}X_{3} $$

Target run out = 0 and to minimize cost, tool coating (X_{1}) is set at -1. We then have:

$$ \displaystyle 0=2.2-2.5-3.5X_{2}+3X_{3}+2X_{2}X_{3} $$

After forcing one of the factors to be low for cost reason, we are left with one equation and two unknowns. Many solutions are possible as illustrated below. If one solution exists, an infinite number of solutions will exist. Suppose we set X_{1} = 0.

$$ \displaystyle \begin{align*}

0.3 & =-3.5X_{2}+3X_{3}+2X_{2}X_{3}\\

0.3 & =-3.5\left(0\right)+3X_{3}+2\left(0\right)X_{3}\\

0.3 & =3X_{3}\\

X_{3} & =0.1

\end{align*} $$

So, one solution to the problem is:

$$ \displaystyle \begin{align*}

X_{1} & =-1\\

X_{2} & =0\\

X_{3} & =0.1

\end{align*} $$

Another solution, starting with:

$$ \displaystyle \begin{align*}

X_{1} & =-1\\

X_{3} & =\frac{1}{3}

\end{align*} $$

So, another solution to the problem is:

$$ \displaystyle \begin{align*}

\hat{Y} & =2.2+2.5X_{1}-3.5X_{2}+3X_{3}+2X_{2}X_{3}\\

0 & =2.2+2.5\left(-1\right)-3.5X_{2}+3\left(\frac{1}{3}\right)+2X_{2}\left(\frac{1}{3}\right)\\

0 & =2.2-2.5-3.5X_{2}+1+\frac{2}{3}X_{2}\\

-0.7 & =-3.5X_{2}+1+\frac{2}{3}X_{2}\\

-0.7 & =-2.83X_{2}\\

X_{2} & =0.247

\end{align*} $$

Clearly, given that X_{1} is held low, we could keep changing X_{2} and calculate a value for X_{3} that solves the equation. A computer is good at quickly finding and displaying all the solutions graphically. A contour plot allows us to view combinations of two different factors that provide a constant response (along a contour line). The illustration below shows combinations of X_{2} and X_{3} that solve the equation for responses of -1.2, 0, 1.2. So, the bold curve (contour) is our desired response and is the contour that we want. All of the possible solutions may be illustrated on a contour plot.

3 contours of Y given X_{1} = -1 with

$$ \displaystyle \begin{align*}

X_{2} & \in\left[-1,1\right]\\

\text{and}\\

X_{3} & \in\left[-1,1\right]

\end{align*} $$

We’ll explore the use of contour plots and other tools in the next article to help us quickly find solutions to our models.

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