A warpage optimization method for injection molding using artificial neural network with parametric sampling evaluation strategy

Abstract

A sequential optimization design method based on artificial neural network (ANN) surrogate model with parametric sampling evaluation (PSE) strategy is proposed in this paper. The quality index, such as warpage deformations, thickness uniformity, and so on, is a nonlinear, implicit function of the process conditions, which are typically evaluated by the solution of finite element (FE) equations, a complicated task which often involves huge computational effort. The ANN model can build an approximate function relationship between the design variables and quality index, replacing the expensive FE reanalysis of the quality index in the optimization. Moldflow Corporation’s Plastics Insight software is used to analyze the quality index of the injection-molded parts. The optimization process is performed by a Parametric Sampling Evaluation (PSE) function. PSE is an infilling sampling criterion. Although the design of experiment size is small, this criterion can take the relatively unexpected space into consideration to improve the accuracy of the ANN model and quickly tend to the global optimization solution in the design space. As examples, a scanner, a TV cover, and a plastic lens are investigated. The results show that the sequential optimization method based on PSE sampling criterion can converge faster and effectively approach to the global optimization solution.

Appendix 1

Definition 1

If p is a positive real variable, and E = {e _j (I), j = 1, 2,…, m} is a set of sampling evaluation functions, then

$$ S(E) = \frac{1}{p}\ln \sum\limits_{{j = 1}}^m {\exp \left( {p{e_j}(I)} \right)} $$

(1)

is a parametric sampling evaluation function, where p is a positive real variable and the sampling evaluation function at sampling point j is

$$ {e_j}{(}I{)} = \sigma {(}x{)[}{u_j}\Phi {(}{u_j}{)} + \phi {(}{u_j}{)]}\;{u_j} = \frac{{{Y_j} - \hat{y}(x)}}{{\sigma (x)}} $$

(2)

Definition 2

If, for any $E(I) = \left\{ {{e_1}(I),{e_2}(I), \cdot \cdot \cdot, {e_m}(I)} \right\}$, and $\overline E (I) = \left\{ {{{\overline e }_1}(I),{{\overline e }_2}(I), \cdots, {{\overline e }_m}(I)} \right\}$, $E(I),\overline E (I) \in {E^m}$ with ${e_j}(I) \leqslant {\overline e_j}(I)$ (1 ≤ j ≤ m) and there exists at least one ${{{j {_0}}}}(1 \leqslant{{j_0} } \leqslant m)$ such that ${e_{j_{0}}} (I) < {\overline e_{j_{0}}}(I)$, then $E(I) \leqslant \overline E (I)$ or simply $E \leqslant \overline E$.

Definition 3

If, for any $E,\overline E \in {E^m}$ with $E \leqslant \overline E$, $S(E) < S(\overline E )$, then S(E) is a strictly monotone increasing function of E.

Theorem 1

The PSE function S(E) is a strictly monotone increasing function of E, and if p → ∞ then

$$ S(E) = \frac{1}{p}\ln \sum\limits_{{j = 1}}^m {\exp \left( {p{e_j}(I)} \right)} = \max {e_j}(I) $$

(3)

Proof. Let

$$ E = \left\{ {{e_j}(I)} \right\} \leqslant \overline E = \left\{ {{{\overline E }_j}(I)} \right\},{ }j = 1,2, \cdot \cdot \cdot, m $$

(4)

by Definition 2

$$ {e_j}(I) \leqslant {\overline e_j}(I),{ }j = 1,2, \cdots, m $$

(5)

and there exists at least one j ₀(1 ≤ j ₀ ≤ m) such that

$$ {e_{{{j_0}}}}(I) < {\overline e_{{{j_0}}}}(I) $$

(6)

Then for p > 0,

$$ p{e_{{{j_0}}}}(I) < p{\overline e_{{{j_0}}}}(I) $$

(7)

$$ \exp (p{e_{{{j_0}}}}(I)) < \exp (p{\overline e_{{{j_0}}}}(I)) $$

(8)

Hence

$$ \sum\limits_{{j = 1}}^m {\exp (p{e_j}(I)) < \sum\limits_{{j = 1}}^m {\exp (p{{\overline e }_j}(I))} } $$

(9)

Taking logarithms on both sides and dividing by p

$$ S(E) = (1/p)\ln \sum\limits_{{j = 1}}^m {\exp (p{e_j}(I))} < (1/p)\ln \sum\limits_{{j = 1}}^m {\exp (p{{\overline e }_j}(I))} $$

(10)

i.e., S(E) is a strictly monotone increasing function of increasing function of E.

The p norm of the q-dimensional vector

$$ {E_e} = {\left\{ {{e^{{{e_1}(I)}}},{e^{{{e_2}(I)}}}, \cdots, {e^{{{e_m}(I)}}}} \right\}^T} $$

(11)

is given by

$$ {\left\| {{E_e}} \right\|_p} = {\left( {\sum\limits_{{j = 1}}^m {{e^{{p{e_j}(I)}}}} } \right)^{{(1/p)}}} $$

(12)

The uniform norm, also called the maximum norm, is defined by

$$ {\left\| {E_e} \right\|}_{\infty } = {\mathop {\lim }\limits_{p \to \infty } }{\left[ {{\sum\limits_{j = 1}^m {e^{{pe_{j} {\left( I \right)}}} } }} \right]}^{{{\left( {1/p} \right)}}} $$

(13)

Since ${{e}^{{{{e}_{j}}(I)}}} > 0$ by Jensen’s inequality, the norm is a strictly monotone decreasing function of its order, i.e.

$$ {S_p} < {S_r}{\text{ for }}r < p $$

(14)

The importance of this inequality is that it holds also in the limit as p → ∞. Thus, we have

$$ {S_p}\left( {{E_e}} \right) = \mathop{{\lim }}\limits_{{p \to \infty }} {\left( {\sum\limits_{{j = 1}}^m {{e^{{p{e_j}(I)}}}} } \right)^{{\tfrac{1}{p}}}} = {S_{\infty }}({E_e}) = \max \left( {{e^{{{e_j}(I)}}}} \right) $$

(15)

Taking logarithms on both side of equation gives

$$ \mathop{{\lim }}\limits_{{p \to \infty }} (1/p)\ln \sum\limits_{{j = 1}}^m {\exp (p{e_j}(I))} = \max { }({e_j}(I)) $$

(16)

and the proof is completed.