# Data-Driven Pavement Performance: Machine Learning-Based Predictive Models
## 6 Results
Comparisons of MSE and R\({}^{2}\) values for number of passes to fatigue damage have been shown in Table 2. As observed, CatBoost exhibits the least MSE of 3.755 \(\times\) 10\({}^{6}\) and maximum R\({}^{2}\) of 1.000, followed by LightGBM with MSE of 1.428 \(\times\) 10\({}^{8}\) and R\({}^{2}\) of 0.999. SVR, however, exhibits the highest MSE of 4.874 \(\times\) 10\({}^{10}\) and R\({}^{2}\) of 0.5375. Figure 9 shows the performance prediction in terms of number of passes to reach rut depth of 6 mm for different models. Figure 9a shows the prediction capability of the decision tree algorithm for maximum number of passes to reach rut depth of 6 mm for zero wander mode. Since the number of passes are in the range of a million in some instances, these graphs have been developed and scaled accordingly to exhibit higher MSE values. A total of nine folds for the decision tree algorithm have been developed, and Fold 3 exhibits the least MSE among other folds, with a magnitude of 454,458,547. Figure 9b shows the prediction of values in terms of number of passes to reach rut depth of 6 mm using gradient boosting. As observed, a total of 10 folds have been developed to further optimize the prediction model. Least MSE is observed at Fold 7, with a value of 22,056,971.7. The graph shows the least amount of data scatter among other algorithms used. Furthermore, the predictive performance of the gradient boosting model surpasses the decision tree algorithm. In Figure 9c, the predicted values have been plotted against the actual values using the KNN algorithm. Increased data scatter can be observed at a higher number of passes, where rut depth suddenly increases leading to increased prediction deficiency for this model. Least MSE has been observed by Fold 4, with an MSE of 6,335,127,849.05. \begin{table}
\begin{tabular}{c c c} \hline \hline
**Machine Learning** & **MSE** & **R\({}^{2}\)** \\
**Algorithm** & **MSE** & **R\({}^{2}\)** \\ \hline Linear Regression & 2.692 \(\times\) 10\({}^{9}\) & 0.9745 \\ Decision Tree & 4.161 \(\times\) 10\({}^{8}\) & 0.9961 \\ Random Forest & 3.453 \(\times\) 10\({}^{8}\) & 0.9967 \\ Gradient Boosting & 2.285 \(\times\) 10\({}^{7}\) & 0.9998 \\ KNN & 3.967 \(\times\) 10\({}^{9}\) & 0.9624 \\ Support Vector Regression & 4.874 \(\times\) 10\({}^{10}\) & 0.5375 \\ LightGBM & 1.428 \(\times\) 10\({}^{8}\) & 0.999 \\ CatBoost & 3.755 \(\times\) 10\({}^{6}\) & 1.000 \\ Linear Regression & 2.692 \(\times\) 10\({}^{9}\) & 0.9745 \\ Decision Tree & 4.161 \(\times\) 10\({}^{8}\) & 0.9961 \\ \hline \hline \end{tabular}
\end{table} 
Table 2: MSE and R\({}^{2}\) values of different models for number of passes to fatigue damage. Figure 9d shows linear regression being used to predict the values. The scatter pattern is the same as that of the KNN algorithm where scatter increases at a higher number of passes. However, the extent of scatter is less than that of KNN. A total of 10 folds have been developed for model optimization. Least MSE stays at 4,351,218,590.71, which outperforms the prediction capability of the KNN algorithm. In Figure 9e, predicted values for number of maximum passes to rut depth at 6 mm are shown using random forest. A total of 10 folds have been developed for model optimization. As observed, the random forest algorithm outperforms KNN and linear regression models, exhibiting an MSE of 350,754,482.9. In Figure 9f, SVR has been used to predict the values for maximum passes to rut depth at 6 mm. As observed, significant increase in data scatter occurs when compared to other algorithms. A total of 10 folds were developed. Fold 3 exhibits the least MSE of 29,456,518,554.68 among other folds. However, the MSE of SVR is significantly higher than other models. Figure 9g shows the comparison of predicted and actual values for decreased number of passes as a result of fatigue damage. As observed, CatBoost performs well with higher accuracy than other models, yielding the lowest MSE of 5,934,373.659 and R\({}^{2}\) of 1.000. All the data points stay on the perfect prediction line.