Two-echelon supply chain inventory model with controllable lead time

Abstract

In this study, we consider a two-echelon supply chain inventory problem consisting of a single vendor and a single buyer with controllable lead time. This paper presents an integrated a single vendor and a single buyer inventory model in order to minimize the sum of the ordering cost/setup cost, holding cost and crashing cost by simultaneously optimizing the optimal order quantity, lead time and number of deliveries. Here, we consider the lead time crashing cost is an exponentially function of lead time. The main contribution of this proposed model is to find minimizing the integrated total cost for the single vendor and single buyer. The main contribution of proposed model is an efficient iterative algorithm developed to minimize integrated total cost for a single vendor and a single buyer system with controllable lead time reduction. It can be obtained simultaneously by optimizing the optimal solution. Mathematical modelling and solution procedure are employed in this study for optimizing the order quantity, lead time and the number of deliveries from a single vendor and a single buyer in one production run with the objective of minimizing integrated total cost. Graphical representation is also presented to illustrate the proposed model. The result is illustrated with the help of numerical example. Finally, sensitivity analysis is carried out with respect to the key parameters and some managerial implications are also included. The computer flowchart of the algorithm to illustrated the model. Matlab coding is also developed to derive the optimal solution and present numerical examples to illustrate the model.

Appendix

We want to prove the Hessian Matrix of ITC(Q,L,n) at point $\left( {Q^{*} ,\,L^{*} } \right)$ for fixed n is positive definite. We first obtain the Hessian matrix H as follows

$$ H = \left[ \begin{gathered} \frac{{\partial^{2} ITC\left( {Q,\,L,\,n} \right)}}{{\partial Q^{2} }}\,\,\,\,\,\,\,\,\frac{{\partial^{2} ITC\left( {Q,\,L,\,n} \right)}}{\partial Q\partial L} \hfill \\ \frac{{\partial^{2} ITC\left( {Q,\,L,\,n} \right)}}{\partial L\partial Q}\,\,\,\,\,\,\,\,\frac{{\partial^{2} ITC\left( {Q,\,L,\,n} \right)}}{{\partial L^{2} }} \hfill \\ \end{gathered} \right] $$

where

$$ \frac{{\partial^{2} ITC\left( {Q,\,L,\,n} \right)}}{{\partial Q^{2} }} = \frac{2D}{{Q^{3} }}\left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right) > 0 $$

$$ \frac{{\partial^{2} ITC\left( {Q,\,L,\,n} \right)}}{\partial Q\partial L} = \frac{{\partial^{2} Tc\left( {Q,\,L,\,n} \right)}}{\partial L\partial Q} = \frac{{DCe^{\frac{C}{L}} }}{{Q^{2} L^{2} }} $$

$$ \frac{{\partial^{2} ITC\left( {Q,\,L,\,n} \right)}}{{\partial L^{2} }} = \frac{D}{Q}\left( {\frac{{2Ce^{\frac{C}{L}} }}{{L^{3} }} + \frac{{C^{2} e^{\frac{C}{L}} }}{{L^{4} }}} \right) - \left( {\frac{{rc_{b} k\sigma L^{{ - \frac{3}{2}}} }}{4}} \right) > 0 $$

We proceed by evaluating the principal minor determinant of the Hessian matrix H at point$\left( {Q^{*} ,\,L^{*} } \right)$. The first principal minor determinant of H then becomes

$$ \left| {H_{11} } \right| = \frac{2D}{{Q^{3} }}\left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right) > 0 $$

$\left| {H_{22} } \right| = \frac{2D}{{Q^{3} }}\left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right)\,\left( {\frac{D}{Q}\left( {\frac{{2Ce^{\frac{C}{L}} }}{{L^{3} }} + \frac{{C^{2} e^{\frac{C}{L}} }}{{L^{4} }}} \right) - \left( {\frac{{rc_{b} k\sigma L^{{ - \frac{3}{2}}} }}{4}} \right)} \right) - \left( {\frac{{DCe^{\frac{C}{L}} }}{{Q^{2} L^{2} }}} \right)^{2}$ $\,\, = \frac{2D}{{Q^{3} }}\left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right)\left( {\frac{{2DCe^{\frac{C}{L}} }}{{QL^{3} }} + \frac{{DC^{2} e^{\frac{C}{L}} }}{{QL^{4} }} - \left( {\frac{{rc_{b} k\sigma L^{{ - \frac{3}{2}}} }}{4}} \right)} \right) - \left( {\frac{{DCe^{\frac{C}{L}} }}{{Q^{2} L^{2} }}} \right)^{2}$ $= \frac{2D}{{Q^{3} }}\left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right)\left( {\frac{{2DCe^{\frac{C}{L}} }}{{QL^{3} }} + \frac{{DC^{2} e^{\frac{C}{L}} }}{{QL^{4} }}} \right) - \frac{2D}{{Q^{3} }}\left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right)\left( {\frac{{rc_{b} k\sigma L^{{ - \frac{3}{2}}} }}{4}} \right) - \left( {\frac{{DCe^{\frac{C}{L}} }}{{Q^{2} L^{2} }}} \right)^{2} > 0$ Since $\,\left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right)\left( {\frac{{4D^{2} Ce^{\frac{C}{L}} }}{{Q^{4} L^{3} }}} \right) + \left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right)\left( {\frac{{2D^{2} C^{2} e^{\frac{C}{L}} }}{{Q^{4} L^{4} }}} \right) > \left( {\frac{{Drc_{b} k\sigma L^{{ - \frac{3}{2}}} }}{{2Q^{3} }}} \right)\left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right) + \left( {\frac{{DCe^{\frac{C}{L}} }}{{Q^{2} L^{2} }}} \right)^{2}$ $\left( {\frac{D}{{Q^{4} }}} \right)\left( {\left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right)\left( {\frac{{4DCe^{\frac{C}{L}} }}{{L^{3} }}} \right) + \left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right)\left( {\frac{{2DC^{2} e^{\frac{C}{L}} }}{{L^{4} }}} \right)} \right) > \left( {\frac{D}{{Q^{4} }}} \right)\left( {\left( {\frac{{Qrc_{b} k\sigma L^{{ - \frac{3}{2}}} }}{2}} \right)\left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right) + \frac{{DC^{2} e^{\frac{C}{L}} }}{{L^{4} }}} \right)$ $\left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right)\left( {\frac{{4DCe^{\frac{C}{L}} }}{{L^{3} }}} \right) + \left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right)\left( {\frac{{2DC^{2} e^{\frac{C}{L}} }}{{L^{4} }}} \right) > \left( {A + \frac{S}{n} + e^{\frac{C}{L}} } \right)\left( {\frac{{Qc_{b} k\sigma L^{{ - \frac{3}{2}}} }}{2}} \right) + \left( {\frac{{DC^{2} e^{\frac{C}{L}} }}{{L^{4} }}} \right)$

Therefore $\left| {H_{22} } \right| > 0$.

Hence for fixed n, the Hessian matrix is positive and ITC(Q,L,n) is convex with respect to (Q,L).