BUS606: Operations and Supply Chain Management

This section contains a summary of the work of Benkherouf and Gilding needed to tackle the problem of this paper. Proofs of the results are omitted. Interested readers may consult Benkherouf and Gilding.

Consider the problem

$\mathbf{P}: \operatorname{TC}\left(t_{1}, \ldots, t_{n} ; n\right)=n K+\sum_{i=1}^{n} R_{i}\left(t_{i-1}, t_{\mathrm{i}}\right)$,   (3.1)

subject to

$0=t_{0} \text { < } t_{1} \text { < } \cdots \text { < } t_{n}=H$.   (3.2)

It was shown in Benkherouf and Gilding that, under some technical conditions, the optimization problem (P) has a unique optimal solution which can be found from solving a system of nonlinear equations derived from the first-order optimality condition. To be precise, $t_{0}=0$ and $t_{n}=H$ and ignore the rest of the constraints (3.2).

Write

$S_{n}:=\sum_{i=1}^{n} R_{i}\left(t_{i-1}, t_{i}\right)$.   (3.3)

Assuming that $R_{i}^{\prime} s$  are twice differentiable, then, for fixed $n$, the optimal solution in (P) subject to (3.2) reduces to minimizing $S_{n}$.

Use the notation $\nabla$ for the gradient, then setting $\nabla \mathrm{TC}\left(t_{1}, \ldots, t_{n} ; n\right)=0$ gives

$(\nabla \mathrm{TC})_{i}=\left(\partial R_{i}\right)_{y}\left(t_{i-1}, t_{i}\right)+\left(\partial R_{i+1}\right)_{x}\left(t_{i}, t_{i+1}\right)=0, \quad i=1, \ldots, n-1$.   (3.4)

Two sets of hypotheses were put forward.

Hypothesis 1. The functions $R_{i}$ satisfy, for $i=1, \ldots, n \text { and } y \text { > } x$,

(1) $R_{i}(x, y) \text { > } 0$

(2) $R_{i}(x, x) \text { = } 0$

(3) $\left(\partial R_{i}\right)_{x}(x, y) \text { < } 0 \text { < } \left(\partial R_{i}\right)_{y}(x, y)$

(4) $\left(\partial_{x} \partial_{y} R\right)_{i}(x, y) \text { < } 0$

Hypothesis 2. Define

$\begin{aligned} &\mathscr{L}_{x} z=\partial_{x}^{2} z+\partial_{x} \partial_{y} z+f(x) \partial_{x} z \\ &\mathscr{L}_{y} z=\partial_{y}^{2} z+\partial_{x} \partial_{y} z+f(y) \partial_{y} z \end{aligned}$   (3.5)

then there is a continuous function $f$ such that $\mathscr{L}_{x} R_{i} \geq 0, \mathscr{L}_{y} R_{i} \geq 0$ for all $i=1, \ldots, n$, and $\left(\partial R_{i}\right)_{y}+\left(\partial R_{i+1}\right)_{x}=0$ on the boundary of the feasible set.

The next theorem shows that under assumptions in Hypotheses 1 and 2, the function $S_{n}$ has a unique minimum.

Theorem 3.1.The system (3.4) has a unique solution subject to (3.2). Furthermore, this solution is the solution of (3.1) subject to (3.2).  Recall that a function $S_{n}$ is convex in $n$ if

$S_{n+2}-S_{n+1} \geq S_{n+1}-S_{n}$.   (3.6)

This is equivalent to

$\frac{1}{2}\left(S_{n}+S_{n+2}\right) \geq S_{n+1}$.   (3.7)

Theorem 3.2. If $s_{n}$ denotes the minimum objective value of (3.1) subject to (3.2) and $R_{i}(\mathrm{x}, y)=R(x, y)$ then $s_{n}$ is convex in $n$.

Based on the convexity property of $s_{n}$, the optimal number of cycles $n*$ is given by

$n^{*}=\min \left\{n \geq 1: s_{n+1}-s_{n} \text { > } 0\right\}$.   (3.8)

Now to solve (3.4) at $i=n-1$,

$\partial R_{n-1 y}\left(t_{n-2}, t_{n-1}\right)+\left(\partial R_{n}\right)_{x}\left(t_{n-1}, H\right)=0$.   (3.9)

Assume that $t_{n-1}$ is known, $t_{n-2}$ can be found uniquely as a function of $t_{n-1}$. Repeating this process for $i=n-2$, $i=1, t_{n-3}, \ldots, t_{n}$ are a function of $t_{n-1}$. So, the search for the optimal solution of (3.5) can be conducted using a univariate search method.