Prediction of Crack Porosity from Other Easily Soil Properties Using Ridge Regression Analysis

Abstract

The development of accurate, easy, and low-cost method to determine soil cracks porosity (SCP) is important in the evaluation of ecosystem to manage the hydrological, erosional, and geochemical cycles. Indeed, these procedures are cumbersome, time-and energy-consuming, and costly. Accordingly, intensive efforts are being made to formulate a high-performance model to estimate SCP. Pedotransfer functions (PTFs) have often been developed using multiple linear regression models with no distinction about their involvement in multicollinearity. The paper was focused on ridge regression (RR) to overcome multicollinearity problems via regularizing the regression coefficients by imposing a penalty on their magnitudes. In the application of RR,choosing the ridge parameter (k) is important to control the amount of shrinkage of the regression coefficients. A total of 61 soil samples were analyzed from different land uses (forest, cropland, and pasture) inIraqi Kurdistan Region. Eighty-two percent of the soil sampleswere selected as the training set, and the rest of 11 soil samples (18%) were used as the testing set. Threemethods of modeling regression (simple linear regression SLR, multiple linear regression MLR, and ridge regression RR) were used to formulate accurate PTFs for predicting SCP. The results clearly showed multicollinearity problem (wrong sing and value of regression coefficient) in SMR and the most of MLR; therefore,they are not recommended to be used in predicting SCP. Soil crack porosity was positively correlated with each of clay (C), liquid limit (LL), plastic limit (PL), and percent of shrinkage limit (PLS). While, the most influential variable for predicting SCP is liquid limit (RMSE=126.194 cm2 m-2; adj.R2=0.884). The best model (RMSE=122.786 cm2 m-2, adj. R2=0.891) to predict SCP among all the models is formulated by RR technique when k=0.31395 which gave the lowest intercept value with very low VIF (0.582).