On coalition formation in a non-convex multi-agent inventory problem

Abstract

A multi-agent inventory problem is a situation in which several agents face individual inventory problems and can coordinate their orders to reduce costs. This paper analyses a multi-agent inventory problem in which each agent faces a continuous-review inventory problem, with a deterministic linear demand, no holding costs and a limited capacity warehouse. In the case under study, shortages are allowed as follows. Goods are bought from an external supplier and then they are stored in each agent’s warehouse. These stored goods usually satisfy demand. However, each agent may alternative produce their own goods, which are less costly but of lower quality. When a shortage of the purchased goods occurs, demand is satisfied with the produced goods until a new order arrives. The problem under study herein arises in a farming community, and is a variation of a problem addressed by Fiestras-Janeiro et al. (2015). However, the existence of two acquisition costs makes it substantially different from Fiestras-Janeiro et al.’s problem and significantly complicates its analysis since the resulting cost functions may now be non-convex. This paper establishes the optimal inventory policies for our problem and obtains a stable order structure when agents allocate the joint costs using a proportional rule. In addition, it illustrates the performance of our model and results in an example.

Appendix

In this appendix, we detail the steps to obtain an optimization problem equivalent to Problem P:

$$\begin{aligned} \begin{array}{lll} \mathbf P &{} &{}\\ &{}\min &{} \sum _{j=1}^{2^n-1}\left( \frac{c^p(S_j)}{\sum _{i\in S_j}c^p(i)}\right) V_j\\ &{}\text {subject to}&{} \\ &{}&{}\sum _{j=1}^{2^n-1}V_j=1,\\ &{}&{}V_j \in \{0,1\}, \text{ for } \text{ all } j=1,\dots ,2^n-1. \end{array} \end{aligned}$$

Using the explicit expression for the cost of any coalition given in (5) and denoting $W=1/x$, we rewrite problem P above as

$$\begin{aligned} \begin{array}{lll} \mathbf P1 &{} &{}\\ &{}\min &{} \left( \frac{\sum _{i=1}^{n}b_id_iY_iW+c_1\sum _{i=1}^nd_iX_i+\sum _{i=1}^n(c_2d_i-b_iK_i)Y_i}{\sum _{i\in N}c^p(i)(X_i+Y_i)}\right) \\ &{}\text {subject to}&{} \\ &{}&{}X_i+Y_i\le 1, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}\sum _{j=1}^nY_j\ge 1,\\ &{}&{} WY_i\ge \frac{K_i}{d_i}Y_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} X_iW\le \frac{K_i}{d_i}X_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}\sum _{i=1}^nb_id_iY_iW^2=2a+2(c_1-c_2)\sum _{i=1}^nK_iY_i+\sum _{i=1}^nb_i\frac{K^2_i}{d_i}Y_i,\\ &{}&{}W\ge 0,\ X_i,\ Y_i \in \{0,1\}, \text{ for } \text{ all } i=1,\dots ,n. \end{array} \end{aligned}$$

In what follows, the most efficient coalition will be determined by the set of index $i=1,\dots n$, such that $X_i+Y_i=1$. For every $i=1,\dots ,n$, we denote by $Y_i$ the binary variable that represents the membership (or not) of i to I(S), and by $X_i$ the other binary variable associated to the membership (or not) of i to $S{\setminus } I(S)$, respectively.

Thus, we can transform this problem into a separable problem using a well-known variable transformation (see, for instance, Williams 2013, p. 28). If we define

$$\begin{aligned} U=\frac{1}{\sum _{i\in N}c^p(i)(X_i+Y_i)}, \end{aligned}$$

then we can rewrite the problem as

$$\begin{aligned} \begin{array}{lll} \mathbf P2 &{}&{}\\ &{}\min &{}\sum _{i=1}^{n}b_id_iY_iUW+c_1\sum _{i=1}^nd_iX_iU+\sum _{i=1}^n(c_2d_i-b_iK_i)Y_iU\\ &{}\text {subject to}&{} \\ &{}&{}X_i+Y_i\le 1, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}\sum _{j=1}^nY_j\ge 1,\\ &{}&{}\sum _{i\in N}c^p(i)(X_i+Y_i)U=1,\\ &{}&{} WY_i\ge \frac{K_i}{d_i}Y_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} X_iW\le \frac{K_i}{d_i}X_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}\sum _{i=1}^nb_id_iY_iW^2=2a+2(c_1-c_2)\sum _{i=1}^nK_iY_i+\sum _{i=1}^nb_i\frac{K^2_i}{d_i}Y_i,\\ &{}&{}U\ge 0,\ W\ge 0,\ X_i,\ Y_i \in \{0,1\}, \text{ for } \text{ all } i=1,\dots ,n. \end{array} \end{aligned}$$

Constraints four and five of Problem P2 can be linearized using the variables $Z_i^X=X_iW,$ $Z_i^Y=Y_iW$ and the results in Tawarmalani et al. (2002). In addition, it is necessary to include the following set of constraints associated to this transformation:

$$\begin{aligned} \begin{array}{ll} &{} Z^Y_i\ge \frac{K_i}{d_i}Y_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{} Z^X_i\le \frac{K_i}{d_i}X_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}0\le W-Z_i^X\le (1-X_i)M, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}0\le W-Z_i^Y\le (1-Y_i)M, \text{ for } \text{ all } i=1,\dots ,n,\\ &{} Z_i^X\le MX_i,\ Z_i^Y\le MY_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{} Z_i^X\ge 0,\ Z_i^Y\ge 0, \text{ for } \text{ all } i=1,\dots ,n, \end{array} \end{aligned}$$

with M being a sufficiently large constant. Then, we obtain the following problem P3:

$$\begin{aligned} \begin{array}{lll} \mathbf P3 &{}&{}\\ &{}\min &{}\sum _{i=1}^{n}b_id_iZ_i^YU+c_1\sum _{i=1}^nd_iX_iU+\sum _{i=1}^n(c_2d_i-b_iK_i)Y_iU\\ &{}\text {subject to}&{} \\ &{}&{}X_i+Y_i\le 1, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}\sum _{j=1}^nY_j\ge 1,\\ &{}&{}\sum _{i\in N}c^p(i)(X_i+Y_i)U=1,\\ &{}&{} Z^Y_i\ge \frac{K_i}{d_i}Y_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} Z^X_i\le \frac{K_i}{d_i}X_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}0\le W-Z_i^X\le (1-X_i)M, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}0\le W-Z_i^Y\le (1-Y_i)M, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} Z_i^X\le MX_i,\ Z_i^Y\le MY_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}\sum _{i=1}^nb_id_i(Z_i^Y)^2=2a+2(c_1-c_2)\sum _{i=1}^nK_iY_i+\sum _{i=1}^nb_i\frac{K^2_i}{d_i}Y_i,\\ &{}&{}U\ge 0,\ W\ge 0,\ X_i,\ Y_i \in \{0,1\}, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} Z_i^X\ge 0,\ Z_i^Y\ge 0, \text{ for } \text{ all } i=1,\dots ,n, \end{array} \end{aligned}$$

with M being a sufficiently large constant.

In this formulation, the last two addends of the objective function can be linearized using the variables $R_i^X=X_iU,$ $R_i^Y=Y_iU$ and, again, the results in Tawarmalani et al. (2002). Then we obtain a new formulation by introducing the following constraints:

$$\begin{aligned} \begin{array}{ll} &{}0\le U-R_i^X\le (1-X_i)M, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}0\le U-R_i^Y\le (1-Y_i)M, \text{ for } \text{ all } i=1,\dots ,n,\\ &{} R_i^X\le MX_i,\ R_i^Y\le MY_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{} R_i^X\ge 0,\ R_i^Y\ge 0, \text{ for } \text{ all } i=1,\dots ,n, \end{array} \end{aligned}$$

with M being a sufficiently large constant. Then we obtain the following problem P4:

$$\begin{aligned} \begin{array}{lll} \mathbf P4 &{}&{}\\ &{}\min &{}\sum _{i=1}^{n}b_id_iZ_i^YU+c_1\sum _{i=1}^nd_iR_i^X+\sum _{i=1}^n(c_2d_i-b_iK_i)R_i^Y\\ &{}\text {subject to}&{} \\ &{}&{}X_i+Y_i\le 1, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}\sum _{j=1}^nY_j\ge 1,\\ &{}&{}\sum _{i\in N}c^p(i)(R_i^X+R^Y_i)=1,\\ &{}&{}0\le U-R_i^X\le (1-X_i)M, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}0\le U-R_i^Y\le (1-Y_i)M, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} R_i^X\le MX_i,\ R_i^Y\le MY_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} Z^Y_i\ge \frac{K_i}{d_i}Y_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} Z^X_i\le \frac{K_i}{d_i}X_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}0\le W-Z_i^X\le (1-X_i)M, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}0\le W-Z_i^Y\le (1-Y_i)M, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} Z_i^X\le MX_i,\ Z_i^Y\le MY_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}\sum _{i=1}^nb_id_i(Z_i^Y)^2=2a+2(c_1-c_2)\sum _{i=1}^nK_iY_i+\sum _{i=1}^nb_i\frac{K^2_i}{d_i}Y_i,\\ &{}&{}U\ge 0,\ W\ge 0,\ X_i,\ Y_i \in \{0,1\}, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} Z_i^X\ge 0,\ Z_i^Y\ge 0, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} R_i^X\ge 0,\ R_i^Y\ge 0, \text{ for } \text{ all } i=1,\dots ,n, \end{array} \end{aligned}$$

with M being a sufficiently large constant. Finally, for every $i=1,\ldots ,n$, we transform the terms $Z_i^YU$ into a difference of two new terms as follows:

$$\begin{aligned} \alpha _i=\frac{1}{2}(Z_i^Y+U),\quad \beta _i=\frac{1}{2}(Z_i^Y-U), \end{aligned}$$

and, thus, $Z_i^YU=\alpha _i^2-\beta _i^2$. Then, the final formulation of our problem is

$$\begin{aligned} \begin{array}{lll} \mathbf P ^{\prime }&{}&{}\\ &{}\min &{}\sum _{i=1}^{n}b_id_i(\alpha _i^2-\beta _i^2)+c_1\sum _{i=1}^nd_iR_i^X+\sum _{i=1}^n(c_2d_i-b_iK_i)R_i^Y\\ &{}\text {subject to}&{} \\ &{}&{}X_i+Y_i\le 1, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}\sum _{j=1}^nY_j\ge 1,\\ &{}&{}\sum _{i\in N}c^p(i)(R_i^X+R^Y_i)=1,\\ &{}&{}\alpha _i=\frac{1}{2}(Z_i^Y+U),\quad \beta _i=\frac{1}{2}(Z_i^Y-U), \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}0\le U-R_i^X\le (1-X_i)M, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}0\le U-R_i^Y\le (1-Y_i)M, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} R_i^X\le MX_i,\ R_i^Y\le MY_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} Z^Y_i\ge \frac{K_i}{d_i}Y_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} Z^X_i\le \frac{K_i}{d_i}X_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}0\le W-Z_i^X\le (1-X_i)M, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}0\le W-Z_i^Y\le (1-Y_i)M, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} Z_i^X\le MX_i,\ Z_i^Y\le MY_i, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}\sum _{i=1}^nb_id_i(Z_i^Y)^2=2a+2(c_1-c_2)\sum _{i=1}^nK_iY_i+\sum _{i=1}^nb_i\frac{K^2_i}{d_i}Y_i,\\ &{}&{}U\ge 0,\ W\ge 0,\ X_i,\ Y_i \in \{0,1\}, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} Z_i^X\ge 0,\ Z_i^Y\ge 0, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{} R_i^X\ge 0,\ R_i^Y\ge 0, \text{ for } \text{ all } i=1,\dots ,n,\\ &{}&{}\alpha _i\ge 0, \text{ for } \text{ all } i=1,\dots ,n. \end{array} \end{aligned}$$

Thus, the minimization problem defined in Step 3 of Procedure 4.1 can be expressed as a quadratic problem with some binary variables.