Integrated decision making model for pricing and locating the customer order decoupling point of a newsvendor supply chain

Abstract

A multi-product two echelon supply chain within a newsvendor framework is studied, in which semi-finished products are produced by a supplier and customized according to specific customer orders. The focus of this paper is to investigate a situation where the manufacturer wishes to determine the fraction of production performed by the supplier, its optimal semi-finished product order size, and price for each product type. In order to make the problem more practical, capacity and budget constraints are considered. The computability of the presented model is explored by a numerical example with Poisson arrival demands.

Appendices

Appendix 1

As $\lambda_{i}$ denotes the demand rate for product $i$, it is obvious that it only takes the nonnegative values. By assuming $\lambda_{i} \ge 0$, we will have (13) and (14):

$$\alpha_{i} - \beta_{P} P_{i} + \gamma_{P} (P_{j} - P_{i} ) - \beta_{K} K_{i} + \gamma_{K}^{{}} (K_{j} - K_{i} ) \ge 0$$

(13)

$$P_{j} \ge \frac{{(\beta_{P} + \gamma_{P} )P_{i} }}{{\gamma_{P} }} + \frac{{\beta_{K} K_{i} - \gamma_{K}^{{}} (K_{j} - K_{i} ) - \alpha_{i} }}{{\gamma_{P} }}$$

(14)

If we consider $\frac{{(\beta_{P} + \gamma_{P} )}}{{\gamma_{P} }} = a_{1}$ and $\frac{{\beta_{K} K_{i} - \gamma_{K}^{{}} (K_{j} - K_{i} ) - \alpha_{i} }}{{\gamma_{P} }} = b_{1}$, we can write (13) as (14):

$$P_{j} \ge a_{1} P_{i} + b_{1}$$

(15)

By the same assumption for $\lambda_{j} \ge 0$, we will have (16) and (17):

$$\alpha_{j} - \beta_{P} P_{j} + \gamma_{P} (P_{i} - P_{j} ) - \beta_{K} K_{j} + \gamma_{K}^{{}} (K_{i} - K_{j} ) \ge 0$$

(16)

$$P_{i} \ge \frac{{(\beta_{P} + \gamma_{P} )P_{j} }}{{\gamma_{P} }} + \frac{{\beta_{K} K_{j} - \gamma_{K}^{{}} (K_{i} - K_{j} ) - \alpha_{j} }}{{\gamma_{P} }}$$

(17)

If we consider $\frac{{(\beta_{P} + \gamma_{P} )}}{{\gamma_{P} }} = a_{2}$ and $\frac{{\beta_{K} K_{j} - \gamma_{K}^{{}} (K_{i} - K_{j} ) - \alpha_{j} }}{{\gamma_{P} }} = b_{2}$, we can write (17) as (18):

$$P_{i} \ge a_{2} P_{j} + b_{2} \, \Rightarrow \,\frac{{P_{i} - b_{2} }}{{a_{2} }} \ge P_{j}$$

(18)

By replacing (18) in (15), we will find the upper bound of $P_{i}$ as (19):

$$\frac{{P_{i} - b_{2} }}{{a_{2} }} \ge a_{1} P_{i} + b_{1} \, \Rightarrow \,\frac{{b_{1} a_{2} + b_{2} }}{{1 - a_{1} a_{2} }} \ge P_{i}$$

(19)

By using the same procedure and replacing (18) in (15), we will find the upper bound of $P_{j}$ as (20):

$$\frac{{P_{j} - b_{1} }}{{a_{1} }} \ge a_{2} P_{j} + b_{2} \, \Rightarrow \,\frac{{b_{2} a_{1} + b_{1} }}{{1 - a_{1} a_{2} }} \ge P_{j}$$

(20)

Appendix 2

In the sensitivity analysis section, both completion percentages are assumed to be equal and increase simultaneously, so we can assume $\theta_{1} = \theta_{2} = \theta$, and for the first product, we have (21):

$$\begin{aligned} \lambda_{1} &= 14 - 0.15P_{1} + 0.1(P_{2} - P_{1} ) - 0.12K_{1} \hfill \\ &\quad+\,0.1(K_{2} - K_{1} ) \hfill \\ \end{aligned}$$

(21)

By inserting $K_{1} (\theta_{1} ) = 1 - \theta$ and $K_{2} (\theta_{2} ) = 1.2(1 - \theta )$ to 21, we will have (22):

$$\begin{aligned} \lambda_{1} &= 14 - 0.15P_{1} + 0.1(P_{2} - P_{1} ) - 0.12(1 - \theta ) \hfill \\ &\quad+\,0.1(1.2(1 - \theta ) - (1 - \theta )) \hfill \\ \end{aligned}$$

(22)

Considering $\beta = 14 - 0.15P_{1} + 0.1(P_{2} - P_{1} )$ as a constant value, we will have (23):

$$\lambda_{1} = \beta - 0.1 + 0.1\theta$$

(23)

(23) shows that $\lambda_{1}$ is an increasing function of $\theta$. The same procedure can be applied for $\lambda_{2}$.