New learning functions for active learning Kriging reliability analysis using a probabilistic approach: KO and WKO functions

Abstract

Reducing the cost of calculation without compromising the accuracy of the solution is a recognized challenge for optimizing the reliability analysis, which became possible using surrogate models trained with robust techniques, such as active learning Kriging (AK) reliability methods. In the AK reliability method, a Kriging predictor is built with a small size of design of experiments (DoE) and becomes more accurate in the vicinity of the limit state function (LSF) in a stepwise manner, called the learning process, until a stopping criterion is met. The motivation of the current study is to enhance the accuracy and efficiency of AK reliability analysis by developing new learning functions, new stopping criteria, and a new method of selection of the next candidate for updating the DoE in the learning process. In this paper, two new learning functions named Kriging occurrence (KO) and weighted KO (WKO) are proposed based on a probability-based approach. A hybrid selection for the next candidate is introduced which simultaneously considers the probability of improvement and the density of DoE and a new stopping criterion is recommended based on the relative mean of the learning functions. A thorough study of the literature is conducted where 12 learning functions are summarized and their performances are compared to that of newly developed learning functions through five comparative examples. The result of the study shows that the new learning function can enhance the accuracy and efficiency of the learning process.

Appendix 1: Hc learning function—A revisit to H learning function

During a check on the derivation of the learning function H in the literature (Lv et al. 2015), a mistake was observed. Therefore, the authors independently derived the equation for the H learning function which is presented in the following as H_c or corrected H learning function. Also, at the end of this appendix, in Table 13, the sources of inaccuracy in the derivation of the mistakes in the source for H learning function derivation are presented. Eq. (A1) shows the information entropy of the surrogate response estimator (i.e., $\widehat{G}\left({\varvec{x}}\right)$), whose solution is the H_c learning function.

$${\mathrm{H}}_{\mathrm{c}}\left(\widehat{G}\left({\varvec{x}}\right)\right)=\left|-{\int }_{\overline{G }\left(x\right)-\varepsilon }^{\overline{G }\left(x\right)+\varepsilon }\mathrm{ln}\left[f\left(\widehat{G}\left({\varvec{x}}\right)\right)\right]f\left(\widehat{G}\left({\varvec{x}}\right)\right)d\left(\widehat{G}\left({\varvec{x}}\right)\right) \right|=\left|-{\int }_{{G}^{-}}^{{G}^{+}}f\left(\widehat{G}\right)\mathrm{ln}\left[f\left(\widehat{G}\right)\right]d\widehat{G}\right|,$$

(A1)

where $f\left(\widehat{G}\left({\varvec{x}}\right)\right)$ is the PDF of $\widehat{G}\left({\varvec{x}}\right)$ which is assumed to be a normal distribution, $d\left(\widehat{G}\left({\varvec{x}}\right)\right)$ is the differential of $\widehat{G}\left({\varvec{x}}\right)$, $\varepsilon$ corresponds to $2{\sigma }_{\widehat{G}\left(x\right)}$, ${G}^{+}$ is the upper bound of the integration (i.e., ${G}^{+}=\overline{G }\left(x\right)+\varepsilon$), ${G}^{-}$ is the lower bound of the integration (i.e., ${G}^{-}=\overline{G }\left(x\right)-\varepsilon$), and x is the desired design site, which is not presented in the rest of the equations for simplicity. By substituting the PDF of $\widehat{G}\left({\varvec{x}}\right)$ in Eq. (A1) and simplifying the integral equation, the H_c learning function can be written down in terms of two other integrals (i.e., I₁ and I₂), as presented in Eq. (A2).

$${\mathrm{H}}_{\mathrm{c}}\left(\widehat{G}\right)=\left|-{\int }_{{G}^{-}}^{{G}^{+}}\frac{1}{\sqrt{2\pi }{\sigma }_{\widehat{G}}}\mathrm{exp}\left\{-\frac{{\left(\widehat{G}-{\mu }_{\widehat{G}}\right)}^{2}}{2{\sigma }_{\widehat{G}}^{2}}\right\}\mathrm{ln}\left[\frac{1}{\sqrt{2\pi }{\sigma }_{\widehat{G}}}\mathrm{exp}\left\{-\frac{{\left(\widehat{G}-{\mu }_{\widehat{G}}\right)}^{2}}{2{\sigma }_{\widehat{G}}^{2}}\right\}\right]d\widehat{G} \right|=\left|-{\int }_{{G}^{-}}^{{G}^{+}}\frac{1}{\sqrt{2\pi }{\sigma }_{\widehat{G}}}\mathrm{exp}\left\{-\frac{{\left(\widehat{G}-{\mu }_{\widehat{G}}\right)}^{2}}{2{\sigma }_{\widehat{G}}^{2}}\right\}\left[\mathrm{ln}\left(\frac{1}{\sqrt{2\pi }{\sigma }_{\widehat{G}}}\right)+\mathrm{ln}\left(\mathrm{exp}\left\{-\frac{{\left(\widehat{G}-{\mu }_{\widehat{G}}\right)}^{2}}{2{\sigma }_{\widehat{G}}^{2}}\right\}\right)\right]d\widehat{G} \right|=\left|-{\int }_{{G}^{-}}^{{G}^{+}}\frac{1}{\sqrt{2\pi }{\sigma }_{\widehat{G}}}\mathrm{exp}\left\{-\frac{{\left(\widehat{G}-{\mu }_{\widehat{G}}\right)}^{2}}{2{\sigma }_{\widehat{G}}^{2}}\right\}\left[-\mathrm{ln}\left(\sqrt{2\pi }{\sigma }_{\widehat{G}}\right)-\frac{{\left(\widehat{G}-{\mu }_{\widehat{G}}\right)}^{2}}{2{\sigma }_{\widehat{G}}^{2}}\right]d\widehat{G} \right|=\left|\mathrm{ln}\left(\sqrt{2\pi }{\sigma }_{\widehat{G}}\right){\int }_{{G}^{-}}^{{G}^{+}}\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left\{-\frac{{\left(\widehat{G}-{\mu }_{\widehat{G}}\right)}^{2}}{2{\sigma }_{\widehat{G}}^{2}}\right\}\frac{d\widehat{G}}{{\sigma }_{\widehat{G}}}+\frac{1}{\sqrt{2\pi }{\sigma }_{\widehat{G}}}{\int }_{{G}^{-}}^{{G}^{+}}\frac{{\left(\widehat{G}-{\mu }_{\widehat{G}}\right)}^{2}}{2{\sigma }_{\widehat{G}}^{2}}\mathrm{exp}\left\{-\frac{{\left(\widehat{G}-{\mu }_{\widehat{G}}\right)}^{2}}{2{\sigma }_{\widehat{G}}^{2}}\right\}d\widehat{G}\right|=\left|{I}_{1}+{I}_{2}\right|.$$

(A2)

The solution to the integral I₁ is straightforward and is presented in Eq. (A3).

$${I}_{1}=\mathrm{ln}\left(\sqrt{2\pi }{\sigma }_{\widehat{G}}\right){\int }_{{G}^{-}}^{{G}^{+}}\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left\{-\frac{{\left(\widehat{G}-{\mu }_{\widehat{G}}\right)}^{2}}{2{\sigma }_{\widehat{G}}^{2}}\right\}\frac{d\widehat{G}}{{\sigma }_{\widehat{G}}}=\mathrm{ln}\left(\sqrt{2\pi }{\sigma }_{\widehat{G}}\right){\int }_{{G}^{-}}^{{G}^{+}}\phi \left(\frac{\widehat{G}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)d\left(\frac{\widehat{G}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)=\mathrm{ln}\left(\sqrt{2\pi }{\sigma }_{\widehat{G}}\right){\int }_{{G}^{-}}^{{G}^{+}}\phi \left(\frac{\widehat{G}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)d\left(\frac{\widehat{G}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)=\mathrm{ln}\left(\sqrt{2\pi }{\sigma }_{\widehat{G}}\right)\left({\left.\Phi \left(\frac{\widehat{G}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\right|}_{{G}^{-}}^{{G}^{+}}\right.=\mathrm{ln}\left(\sqrt{2\pi }{\sigma }_{\widehat{G}}\right)\left[\Phi \left(\frac{{G}^{+}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)-\Phi \left(\frac{{G}^{-}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\right].$$

(A3)

The solution to integral I₂ requires a change of the variables and integral by parts. Eq. (A4) shows the integral part I₂.

$${I}_{2}=\frac{1}{\sqrt{2\pi }{\sigma }_{\widehat{G}}}{\int }_{{G}^{-}}^{{G}^{+}}\frac{{\left(\widehat{G}-{\mu }_{\widehat{G}}\right)}^{2}}{2{\sigma }_{\widehat{G}}^{2}}\mathrm{exp}\left\{-\frac{{\left(\widehat{G}-{\mu }_{\widehat{G}}\right)}^{2}}{2{\sigma }_{\widehat{G}}^{2}}\right\}d\widehat{G}.$$

(A4)

To solve I₂ a variable z is defined as an auxiliary variable which is the standard normal formulation $\widehat{G}$ (i.e., $z=\frac{\widehat{G}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}$ and $dz=\frac{d\widehat{G}}{{\sigma }_{\widehat{G}}}$), and the corresponding integral boundaries were built (i.e., ${z}^{+}=\frac{{G}^{+}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}$ and ${z}^{-}=\frac{{G}^{-}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}$), so that Eq. (A4) can be written down in the form of Eq. (A5).

$${I}_{2}=\frac{1}{2\sqrt{2\pi }}{\int }_{{z}^{-}}^{{z}^{+}}{z}^{2}\mathrm{exp}\left\{-\frac{{z}^{2}}{2}\right\}dz.$$

(A5)

To solve Eq. (A5), integration by parts was used (i.e., $\int udv=uv-\int vdu$), where $u=z$ and $dv=z\mathrm{exp}\left\{-\frac{{z}^{2}}{2}\right\}dz$ were considered (i.e., $v= -\mathrm{exp}\left\{-\frac{{z}^{2}}{2}\right\}$ and $du= dz$). Thus, the solution to Eq. (A5) can be presented as Eq. (A6), by applying the integration by parts.

$${I}_{2}=\frac{1}{2\sqrt{2\pi }}\left({\left.-z\mathrm{exp}\left\{-\frac{{z}^{2}}{2}\right\}\right|}_{{z}^{-}}^{{z}^{+}}\right.+\frac{1}{2\sqrt{2\pi }}{\int }_{{z}^{-}}^{{z}^{+}}\mathrm{exp}\left\{-\frac{{z}^{2}}{2}\right\}dz=-\frac{1}{2}\left({\left.z\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left\{-\frac{{z}^{2}}{2}\right\}\right|}_{{z}^{-}}^{{z}^{+}}\right.+\frac{1}{2}{\int }_{{z}^{-}}^{{z}^{+}}\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left\{-\frac{{z}^{2}}{2}\right\}dz=-\frac{1}{2}\left({\left.z\phi \left(z\right)\right|}_{{z}^{-}}^{{z}^{+}}\right.+\frac{1}{2}{\int }_{{z}^{-}}^{{z}^{+}}\phi \left(z\right)dz=-\frac{1}{2}\left({\left.z\phi \left(z\right)\right|}_{{z}^{-}}^{{z}^{+}}\right.+\frac{1}{2}\left({\left.\Phi \left(z\right)\right|}_{{z}^{-}}^{{z}^{+}}\right.=-\frac{1}{2}\left[{z}^{+}\phi \left({z}^{+}\right)-{z}^{-}\phi \left({z}^{-}\right)\right]+\frac{1}{2}\left[\Phi \left({z}^{+}\right)-\Phi \left({z}^{-}\right)\right]=-\frac{1}{2}\left[\left(\frac{{G}^{+}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\phi \left(\frac{{G}^{+}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)-\left(\frac{{G}^{-}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\phi \left(\frac{{G}^{-}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\right]+\frac{1}{2}\left[\Phi \left(\frac{{G}^{+}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)-\Phi \left(\frac{{G}^{-}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\right].$$

(A6)

By substituting the solutions to I₁ and I₂ integrals, from Eq. (A5) and Eq. (A6), into Eq. (A2), the solution for learning function H_c can be presented in Eq. (A7) .

$${\mathrm{H}}_{\mathrm{c}}\left(\widehat{G}\right)=\left|{I}_{1}+{I}_{2}\right|=\left|\mathrm{ln}\left(\sqrt{2\pi }{\sigma }_{\widehat{G}}\right)\left[\Phi \left(\frac{{G}^{+}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)-\Phi \left(\frac{{G}^{-}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\right]+\frac{1}{2}\left[\Phi \left(\frac{{G}^{+}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)-\Phi \left(\frac{{G}^{-}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\right]-\frac{1}{2}\left[\left(\frac{{G}^{+}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\phi \left(\frac{{G}^{+}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)-\left(\frac{{G}^{-}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\phi \left(\frac{{G}^{-}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\right]\right|.$$

(A7)

By factoring in the same statements, Eq. (A7) can be written down in the form of Eq. (A8).

$${\mathrm{H}}_{\mathrm{c}}\left(\widehat{G}\right)=\left|\left(\mathrm{ln}\left(\sqrt{2\pi }{\sigma }_{\widehat{G}}\right)+\frac{1}{2}\right)\left[\Phi \left(\frac{{G}^{+}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)-\Phi \left(\frac{{G}^{-}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\right]-\frac{1}{2}\left[\left(\frac{{G}^{+}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\phi \left(\frac{{G}^{+}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)-\left(\frac{{G}^{-}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\phi \left(\frac{{G}^{-}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\right]\right|.$$

(A8)

By substituting the value of G⁺ and G⁻ boundaries (i.e., ${G}^{-}=\overline{G }-\varepsilon =-2{\sigma }_{\widehat{G}}$ and ${G}^{+}=\overline{G }+\varepsilon =2{\sigma }_{\widehat{G}}$), using $\varepsilon =2{\sigma }_{\widehat{G}}$ and $\overline{G }=0$, Eq. (A8) can be written as Eq. (A9).

$${\mathrm{H}}_{\mathrm{c}}\left(\widehat{G}\right)=\left|\left(\mathrm{ln}\left(\sqrt{2\pi }{\sigma }_{\widehat{G}}\right)+\frac{1}{2}\right)\left[\Phi \left(\frac{2{\sigma }_{\widehat{G}}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)-\Phi \left(\frac{-2{\sigma }_{\widehat{G}}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\right]-\frac{1}{2}\left[\left(\frac{2{\sigma }_{\widehat{G}}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\phi \left(\frac{2{\sigma }_{\widehat{G}}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)-\left(\frac{-2{\sigma }_{\widehat{G}}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\phi \left(\frac{-2{\sigma }_{\widehat{G}}-{\mu }_{\widehat{G}}}{{\sigma }_{\widehat{G}}}\right)\right]\right|.$$

(A9)

Therefore, using Eq. (A9) as the learning function H_c for the desired design site of x, the full format of the equation can be presented as Eq. (A10).

$${\mathrm{H}}_{\mathrm{c}}\left(\widehat{G}(x)\right)=\left|\left(\mathrm{ln}\left(\sqrt{2\pi }{\sigma }_{\widehat{G}}\left(x\right)\right)+\frac{1}{2}\right)\left[\Phi \left(\frac{2{\sigma }_{\widehat{G}}\left(x\right)-{\mu }_{\widehat{G}}\left(x\right)}{{\sigma }_{\widehat{G}}\left(x\right)}\right)-\Phi \left(\frac{-2{\sigma }_{\widehat{G}}\left(x\right)-{\mu }_{\widehat{G}}\left(x\right)}{{\sigma }_{\widehat{G}}\left(x\right)}\right)\right]-\frac{1}{2}\left[\left(\frac{2{\sigma }_{\widehat{G}}\left(x\right)-{\mu }_{\widehat{G}}\left(x\right)}{{\sigma }_{\widehat{G}}\left(x\right)}\right)\phi \left(\frac{2{\sigma }_{\widehat{G}}\left(x\right)-{\mu }_{\widehat{G}}\left(x\right)}{{\sigma }_{\widehat{G}}\left(x\right)}\right)+\left(\frac{2{\sigma }_{\widehat{G}}\left(x\right)+{\mu }_{\widehat{G}}\left(x\right)}{{\sigma }_{\widehat{G}}\left(x\right)}\right)\phi \left(\frac{-2{\sigma }_{\widehat{G}}\left(x\right)-{\mu }_{\widehat{G}}\left(x\right)}{{\sigma }_{\widehat{G}}\left(x\right)}\right)\right]\right|.$$

(A10)

It should be noted that Eq. (A10) is different from the one shown in the referenced study (Lv et al. 2015). Therefore, in the current study, the performance of both learning functions was examined. To further investigate, the mistakes in the referenced study (Lv et al. 2015) are presented in Table 13.

Table 13 Mistakes in the H learning function

which is compatible with the finding in Eq. (A10).