A production policy considering reworking of imperfect items and trade credit

Abstract

Making production decisions that will reduce total cost is a goal that most manufacturers pursue actively. However, the traditional production model development assumed that all products are perfect quality, which is far from reality. Since trade credit is a popular payment method in today’s business environment, this paper analyzes the production problem under trade credit and imperfect product reworking conditions. This work extends the traditional production model by considering reworking imperfect items and trade credit to cope with realistic situations. The objective of this study is to determine the optimal production lot size while minimizing the total cost. This paper develops an easy-to-use algorithm to solve the problem described, provides numerical examples to illustrate the proposed solution procedure, and discusses the impact of various system parameters.

Appendix 1: The proof of Proposition 1

• (1) Let $\Uppsi (Q) = {\frac{{dTVC_{1} }}{dQ}},\quad {\text{then}}\quad {\frac{d\Uppsi (Q)}{dQ}} = {\frac{2K\lambda }{{Q^{3} }}} - {\frac{{C_{p} I_{p} (P - \lambda )M^{2} \lambda }}{{Q^{3} }}} - {\frac{{M^{2} \lambda^{2} I_{e} S_{p} }}{{Q^{3} }}}.$ If $K\lambda - C_{p} I_{p} (P - \lambda )M^{2} \lambda - M^{2} \lambda^{2} I_{e} S_{p} > 0,$ we can know that $\Uppsi (Q)$ is an increasing function because ${\frac{d\Uppsi (Q)}{dQ}} > 0.$ When $Q \to \infty \quad {\text{or}}\quad 0^{ + } ,$ $\Uppsi (Q)$ can be obtained by

  $$ \begin{aligned} \mathop {\lim }\limits_{Q \to \infty } TVC_{1} (Q) = & {\frac{h\lambda (P - d - \lambda )}{{2P^{2} }}} + {\frac{{d\lambda h_{1} }}{{2P^{2} }}} + {\frac{{x^{2} \lambda h_{1} }}{{2P_{1} }}} + {\frac{{hx\lambda \left[ {P_{1} \left( {2P - 2\lambda - d} \right) - d\lambda } \right]}}{{2PP_{1}^{2} }}} \\ & + {\frac{{h\left[ {d\lambda + P_{1} \left( {\lambda - P} \right)} \right]^{2} }}{{2P^{2} P_{1}^{2} }}} + {\frac{{C_{p} I_{p} (P - \lambda )}}{P}} > 0 \\ \end{aligned} $$

  and $\mathop {\lim }\limits_{{Q \to 0^{ + } }} TVC_{1} (Q) = - \infty < 0.$ Therefore, there exists a unique solution Q (where 0 < Q < ∞) for $\Uppsi (Q) = 0$. This completes this proof.

• (2) Let $\Uppsi (Q) = {\frac{{dTVC_{2} }}{dQ}},$ then ${\frac{d\Uppsi (Q)}{dQ}} = {\frac{2K\lambda }{{Q^{3} }}} + {\frac{{C_{p} I_{p} M^{2} \lambda^{2} }}{{Q^{3} }}} - {\frac{{M^{2} \lambda^{2} I_{e} S_{p} }}{{Q^{3} }}}.$ If $2K\lambda + C_{p} I_{p} M^{2} \lambda^{2} - M^{2} \lambda^{2} I_{e} S_{p} > 0,$ we can know $\Uppsi (Q)$ is an increasing function. When $Q \to \infty \quad {\text{or}}\quad 0^{ + } ,$ $\Uppsi (Q)$ can be obtained by

  $$ \begin{aligned} \mathop {\lim }\limits_{Q \to \infty } \Uppsi (Q) = \, & {\frac{h\lambda (P - d - \lambda )}{{2P^{2} }}} + {\frac{{d\lambda h_{1} }}{{2P^{2} }}} + {\frac{{x^{2} \lambda h_{1} - P_{1} C_{p} I_{p} }}{{2P_{1} }}} \\ & + {\frac{{hx\lambda \left[ {P_{1} \left( {2P - 2\lambda - d} \right) - d\lambda } \right]}}{{2PP_{1}^{2} }}} + {\frac{{h\left[ {d\lambda + P_{1} \left( {\lambda - P} \right)} \right]^{2} }}{{2P^{2} P_{1}^{2} }}} > 0 \\ \end{aligned} $$

  and $\mathop {\lim }\limits_{{Q \to 0^{ + } }} \Uppsi (Q) = - \infty < 0$. Therefore, there exists a unique solution Q (where 0 < Q <∞) for $\Uppsi (Q) = 0$. This completes this proof.

• (3) Let $\Uppsi (Q) = {\frac{{dTVC_{3} }}{dQ}},$ from ${\frac{d\Uppsi (Q)}{dQ}} = {\frac{2K\lambda }{{Q^{3} }}} > 0,$, we can know $\Uppsi (Q)$ is an increasing function. When $Q \to \infty \quad {\text{or}}\quad 0^{ + } ,\quad \Uppsi (Q)$ can be obtained by

  $$ \begin{aligned} \mathop {\lim }\limits_{Q \to \infty } \Uppsi (Q) = & {\frac{h\lambda (P - d - \lambda )}{{2P^{2} }}} + {\frac{{d\lambda h_{1} }}{{2P^{2} }}} + {\frac{{x^{2} \lambda h_{1} }}{{2P_{1} }}} + {\frac{{hx\lambda \left[ {P_{1} \left( {2P - 2\lambda - d} \right) - d\lambda } \right]}}{{2PP_{1}^{2} }}} \\ & + {\frac{{h\left[ {d\lambda + P_{1} \left( {\lambda - P} \right)} \right]^{2} }}{{2P^{2} P_{1}^{2} }}} + {\frac{{S_{p} I_{e} }}{2}} > 0 \\ \end{aligned} $$

  and $\mathop {\lim }\limits_{{Q \to 0^{ + } }} \Uppsi (Q) = - \infty < 0.$ Therefore, there exists a unique solution Q (where 0 < Q <∞) for $\Uppsi \left( Q \right) = 0$. This completes this proof. □