16.3 Estimation of Preference Orderings

On the basis of the reported preference values for each stimulus conjoint analysis determines the part-worths. Conjoint analysis uses an additive model of the form


\begin{displaymath}
Y_k=\sum_{j=1}^J\sum_{l=1}^{L_j}\beta_{jl}I(X_j=x_{jl})+\mu,...
...,K
\textrm{ and } \forall \;j \; \sum_{l=1}^{L_j}\beta_{jl}=0.
\end{displaymath} (16.1)

$X_j$ ($j=1,\ldots,J$) denote the factors, $x_{jl}$ ( $l=1,\ldots ,L_j$) are the levels of each factor $X_j$ and the coefficients $\beta_{jl}$ are the part-worths. The constant $\mu$ denotes an overall level and $Y_k$ is the observed preference for each stimulus and the total number of stimuli are:

\begin{displaymath}K = \prod_{j=1}^J L_j.\end{displaymath}

Equation (16.1) is without an error term for the moment. In order to explain how (16.1) may be written in the standard linear model form we first concentrate on $J=2$ factors. Suppose that the factors engine power and airbag safety equipment have been ranked as follows:
      airbag
      1 2
  50 kW 1 1 3
engine 70 kW 2 2 6
  90 kW 3 4 5

There are $K=6$ preferences altogether. Suppose that the stimuli have been sorted so that $Y_1$ corresponds to engine level 1 and airbag level 1, $Y_2$ corresponds to engine level 1 and airbag level 2, and so on. Then model (16.1) reads:

\begin{displaymath}\begin{array}{ll}
Y_1 = \beta_{11} + \beta_{21} + \mu\\
Y_2 ...
..._{21} + \mu\\
Y_6 = \beta_{13} + \beta_{22} + \mu.
\end{array}\end{displaymath}

Now we would like to estimate the part-worths $\beta_{jl}$.

EXAMPLE 16.3   In the margarine example let us consider the part-worths of $X_1 =$ usage and $X_2 =$ calories. We have $x_{11}=1$, $x_{12}=2$, $x_{13}=3$, $x_{21}=1$ and $x_{22}=2$. (We momentarily re-labeled the factors: $X_3$ became $X_1$). Hence $L_1=3$ and $L_2=2$. Suppose that a person has ranked the six different products as in Table 16.5.

Table 16.5: Ranked products.
      $X_2$ (calories)
      low high
      1 2
  bread 1 2 1
$X_1$ (usage) cooking 2 3 4
  universal 3 6 5


If we order the stimuli as follows:

\begin{displaymath}\begin{array}{ll}
Y_1 = &\textrm{Utility}\left(X_1=1 \wedge X...
...m{Utility}\left(X_1=3 \wedge X_2=2\right),\nonumber
\end{array}\end{displaymath}

we obtain from equation (16.1) the same decomposition as above:

\begin{displaymath}\begin{array}{ll}
Y_1 = \beta_{11} + \beta_{21} + \mu\\
Y_2 ...
..._{21} + \mu\\
Y_6 = \beta_{13} + \beta_{22} + \mu.
\end{array}\end{displaymath}

Our aim is to estimate the part-worths $\beta_{jl}$ as well as possible from a collection of tables like Table 16.5 that have been generated by a sample of test persons. First, the so-called metric solution to this problem is discussed and then a non-metric solution.

Metric Solution

The problem of conjoint measurement analysis can be solved by the technique of Analysis of Variance. An important assumption underlying this technique is that the ``distance'' between any two adjacent preference orderings corresponds to the same difference in utility. That is, the difference in utility between the products ranked 1st and 2nd is the same as the difference in utility between the products ranked 4th and 5th. Put differently, we treat the ranking of the products--which is a cardinal variable--as if it were a metric variable.

Introducing a mean utility $\mu$ equation (16.1) can be rewritten. The mean utility in the above Example 16.3 is $\mu = (1+2+3+4+5+6)/6 =21/6=3.5$. In order to check the deviations of the utilities from this mean, we enlarge Table 16.5 by the mean utility $\bar{p}_{x_{j\bullet}}$, given a certain level of the other factor. The metric solution for the car example is given in Table 16.6:

Table 16.6: Metric solution for car example.
  $X_2$ (airbags)    
  1 2 $\bar{p}_{x_{1\bullet}}$ $\beta_{1l}$
  50 kW 1 1 3 2 $-1.5$
$X_1$ (engine) 70 kW 2 2 6 4 $-0.5$
  90 kW 3 4 5 4.5 1.5
$\bar{p}_{x_{2\bullet}}$ 2.33 4.66 3.5  
$\beta_{2l}$ $-1.16$ 1.16    



Table 16.7: Metric solution for Table 16.5.
  $X_2$ (calories)    
  low high    
  1 2 $\bar{p}_{x_{1\bullet}}$ $\beta_{1l}$
  bread 1 2 1 1.5 $-$2
$X_1$ (usage) cooking 2 3 4 3.5 0
  universal 3 6 5 5.5 2
$\bar{p}_{x_{2\bullet}}$ 3.66 3.33 3.5  
$\beta_{2l}$ 0.16 $-$0.16    


EXAMPLE 16.4  

In the margarine example the resulting part-worths for $\mu =3.5$ are

\begin{displaymath}
\begin{array}{rcrllllr}
\beta_{11} & = & -2 & & \beta_{21} &...
...eta_{22} & = &-0.16\\
\beta_{13} & = & 2 & & & &
\end{array}.
\end{displaymath}

Note that $\sum\limits_{l=1}^{L_j}\beta_{jl}=0$ ($j=1,\ldots,J$). The estimated utility $\hat{Y}_1$ for the product with low calories and usage of bread, for example, is:

\begin{displaymath}\hat{Y}_1=\beta_{11}+\beta_{21}+\mu= -2+0.16+3.5=1.66.\end{displaymath}

The estimated utility $\hat{Y}_4$ for product 4 (cooking ($X_1=2$) and high calories ($X_2=2$)) is:

\begin{displaymath}\hat{Y}_4=\beta_{12}+\beta_{22}+\mu= 0-0.16+3.5=3.33.\end{displaymath}

The coefficients $\beta_{jl}$ are computed as $\bar{p}_{x_{jl}}-\mu$, where $\bar{p}_{x_{jl}}$ is the average preference ordering for each factor level. For instance, $\bar{p}_{x_{11}}=1/2*(2+1)=1.5$.

The fit can be evaluated by calculating the deviations of the fitted values to the observed preference orderings. In the rightmost column of Table 16.8 the quadratic deviations between the observed rankings (utilities) $Y_k$ and the estimated utilities $\hat{Y}_k$ are listed.


Table 16.8: Deviations between model and data.
Stimulus $Y_k$ $\hat{Y}_k$ $Y_k-\hat{Y}_k$ $(Y_k-\hat{Y}_k)^2$
1 2 1.66 0.33 0.11
2 1 1.33 $-$0.33 0.11
3 3 3.66 $-$0.66 0.44
4 4 3.33 0.66 0.44
5 6 5.66 0.33 0.11
6 5 5.33 $-$0.33 0.11
$\sum$ 21 21 0 1.33


The technique described that generated Table 16.7 is in fact the solution to a least squares problem. The conjoint measurement problem (16.1) may be rewritten as a linear regression model (with error $\varepsilon=0$):

\begin{displaymath}
Y=\data{X}\beta +\varepsilon
\end{displaymath} (16.2)

with $\data{X}$ being a design matrix with dummy variables. $\data{X}$ has the row dimension $K=\prod\limits_{j=1}^{J}L_j$ (the number of stimuli) and the column dimension $D= \sum\limits_{j=1}^J L_j-J$. The reason for the reduced column number is that per factor only ($L_j-1$) vectors are linearly independent. Without loss of generality we may standardize the problem so that the last coefficient of each factor is omitted. The error term $\varepsilon$ is introduced since even for one person the preference orderings may not fit the model (16.1).

EXAMPLE 16.5   If we rewrite the $\beta$ coefficients in the form
\begin{displaymath}
\left(\begin{array}{c}
\beta_1\\
\beta_2\\
\beta_3\\
...
...- \beta_{13}\\
\beta_{21}- \beta_{22}\\
\end{array}\right)
\end{displaymath} (16.3)

and define the design matrix $\data{X}$ as

\begin{displaymath}
\data{X}=\left(\begin{array}{cccc}
1\ \vline & 1 & 0\ \vline...
...ne & 1\\
1\ \vline & 0 & 0\ \vline & 0\\
\end{array}\right),
\end{displaymath} (16.4)

then equation (16.1) leads to the linear model (with error $\varepsilon=0$):
\begin{displaymath}
Y = \data{X} \beta + \varepsilon.
\end{displaymath} (16.5)

The least squares solution to this problem is the technique used for Table 16.7.

In practice we have more than one person to answer the utility rank question for the different factor levels. The design matrix is then obtained by stacking the above design matrix $n$ times. Hence, for $n$ persons we have as a final design matrix:

\begin{displaymath}
\data{X}^*=1_n \otimes \data{X}=\left. \left(
\begin{array}{...
...\vdots\\
\data{X}
\end{array}\right)\right\} n-\textrm{times}
\end{displaymath}

which has dimension $(nK)(L-J)$ (where $L=\sum\limits_{j=1}^JL_j$ ) and $Y^* =(Y_1^{\top},...,Y_n^{\top})^{\top}$. The linear model (16.5) can now be written as:
\begin{displaymath}
Y^* = \data{X}^* \beta + \varepsilon^*.
\end{displaymath} (16.6)

Given that the test people assign different rankings, the error term $\varepsilon^*$ is a necessary part of the model.

EXAMPLE 16.6   If we take the $\beta$ vector as defined in (16.3) and the design matrix $\data{X}$ from (16.4), we obtain the coefficients:
\begin{displaymath}
\begin{array}{llrll}
\hat{\beta}_1 & = & \phantom{-}5.33 & =...
...imits_{l=1}^{L_j}\hat{\beta}_{jl}& = &\phantom{-}0.
\end{array}\end{displaymath} (16.7)

Solving (16.7) we have:

\begin{displaymath}
\begin{array}{lclcl}
\hat{\beta}_{11}&=&\hat{\beta}_2-\frac{...
...ht)+\frac{1}{2}(\hat{\beta}_4)&=&\phantom{-}3.5.\\
\end{array}\end{displaymath} (16.8)

In fact, we obtain the same estimated part-worths as in Table 16.7. The stimulus $k=2$ corresponds to adding up $\beta_{11},\beta_{22},$ and $\mu$ (see (16.3)). Adding $\hat{\beta}_1$ and $\hat{\beta}_2$ gives:

\begin{displaymath}\hat{Y}_2 = 5.33 - 4 = 1.33 .\end{displaymath}


Nonmetric solution

If we drop the assumption that utilities are measured on a metric scale, we have to use (16.1) to estimate the coefficients from an adjusted set of estimated utilities. More precisely, we may use the monotone ANOVA as developed by Kruskal (1965). The procedure works as follows. First, one estimates model (16.1) with the ANOVA technique described above. Then one applies a monotone transformation $\hat{Z}=f(\hat{Y})$ to the estimated stimulus utilities. The monotone transformation $f$ is used because the fitted values $\hat{Y}_k$ from (16.2) of the reported preference orderings $Y_k$ may not be monotone. The transformation $\hat{Z}_k=f(\hat{Y}_k)$ is introduced to guarantee monotonicity of preference orderings. For the car example the reported $Y_k$ values were $Y=(1,\; 3,\; 2,\; 6,\; 4,\; 5)^{\top}$. The estimated values are computed as:

\begin{displaymath}
\begin{array}{lclcl}
\hat{Y}_1 &=&-1.5 -1.16 +3.5&=0.84&\\
...
...
\hat{Y}_6 &=&\phantom{-}1.5 +1.16 +3.5&=6.16.&\\
\end{array}\end{displaymath}

If we make a plot of the estimated preference orderings against the revealed ones, we obtain Figure 16.1.

Figure 16.1: Plot of estimated preference orderings vs. revealed rankings and PAV fit. 48517 MVAcarrankings.xpl
\includegraphics[width=1.1\defpicwidth]{carrankings.ps}

We see that the estimated $\hat{Y}_4=4.16$ is below the estimated $\hat{Y}_6=6.16$ and thus an inconsistency in ranking the utilities occurrs. The monotone transformation $\hat{Z}_k=f(\hat{Y}_k)$ is introduced to make the relationship in Figure 16.1 monotone. A very simple procedure consists of averaging the ``violators'' $\hat{Y}_4$ and $\hat{Y}_6$ to obtain $5.16$. The relationship is then monotone but the model (16.1) may now be violated. The idea is therefore to iterate these two steps. This procedure is iterated until the stress measure (see Chapter 15)

\begin{displaymath}
\textrm{STRESS} = \frac{\sum\limits_{k=1}^K(\hat{Z}_k-\hat{Y}_k)^2}{\sum\limits_{k=1}^K(\hat{Y}_k-\bar{\hat{Y}})^2}
\end{displaymath} (16.9)

is minimized over $\beta$ and the monotone transformation $f$. The monotone transformation can be computed by the so called pool-adjacent-violators (PAV) algorithm.

Summary
$\ast$
The part-worths are estimated via the least squares method.
$\ast$
The metric solution corresponds to analysis of variance in a linear model.
$\ast$
The non-metric solution iterates between a monotone regression curve fitting and determining the part-worths by ANOVA methodology.
$\ast$
The fitting of data to a monotone function is done via the PAV algorithm.