Economic model predictive inventory routing and control

Abstract

The paper proposes an economic model predictive control (EMPC) strategy for the inventory routing problem under demand uncertainty. The strategy is illustrated using an application on industrial gas distribution systems, where the product is transported to customers in small tanks and the inventory levels at the customers’ sites are monitored and controlled by the supplier following a vendor managed inventory approach. The objective is to produce balanced decisions for the joint routing and the inventory control problem over the planning horizon with respect to the decision maker’s perspective against stock-out risk. The proposed EMPC strategy makes use of a mixed integer mathematical programming optimization model that describes the deterministic inventory routing problem with simultaneous pickups and deliveries over a specific planning horizon. A time series decomposition forecasting model is used for predicting future demand and an exact linearization of the quadratic term of the objective function guarantees optimality of the solutions. The proposed methodology is illustrated using two examples featuring a single distribution centre, and three customers with simple and complex demand profiles. It is shown that EMPC offers a useful tool for producing balanced decisions between transportation and inventory costs and tracking of the safety inventory levels.

Appendices

Appendix 1: Prioritization of objectives in the steady state case

The total traveling and inventory cost $C_{tot}$ cannot exceed the value $C_{\max } =({K_{avl} +n} )c_{\max } + \textit{IC}_{\max }$, where $c_{max}$ is the maximum cost of traveling between any pair of vertices of the graph and $\textit{IC}_{\max }$ is the maximum total inventory holding cost that cannot exceed; ${\textit{IC}}_{\max }=\sum \nolimits _{i\in N} \sum \nolimits _{t\in H} {h_i d_{i,t}}$.

Then $a^{*}$ is computed as follows:

$$\begin{aligned} a^{*}=\mathop {\arg }\limits _a \left\{ { \min \left( { 10^{a-1}} \right) , 10^{a-1}>C_{\max } +IC_{\mathrm{max}} } \right\} , \quad a \in \mathbf{Z}^{+} \end{aligned}$$

(24)

This way $C_{tot}$ coexists with the number of vehicles K in a joint objective function (1), named from now on obj, i.e. the number of used vehicles is $K=\lfloor {obj/10^{a^{*}}} \rfloor$ and the total traveling and inventory holding cost is $C_{tot} =obj-K\cdot 10^{a^{*}}$. The coefficient $10^{a^{*}}$ works as a weight multiplier and $UB=10^{a^{*}}({K+1})$ is an upper bound to the candidate solutions.

The following example illustrates the role of the first term of the objective function. Assuming that $c_{\max } = 34, K_{avl} = 3$ and $n = 3$, then $C_{\max } = 204$. Assuming also $\textit{IC}_{\max } = 100$, from Eq. (24) it follows that $a^{*} = 4$, because $10^{a^{*}-1} = 10^{3}$ is the minimum value for which $10^{a-1} > C_{\max } + \textit{IC}_{\max }$. The first two summations of the objective function i.e. $\sum \nolimits _{k\in V} {\sum \nolimits _{(0,i) \in A} {z_{k,0,i,t} } }$ represent the number of vehicles. Without loss of generality, assuming a single period horizon ($T = 1$), this sum multiplied by $10^{a^{*}}$ will incur a cost that is always higher than the cost of the remaining terms of the objective function. Therefore, the solution algorithm will show preference to solutions with less number of vehicles even if the sum of transportation and inventory costs are lower. Based on the previous assumptions a solution value obj = 20195 implies that the number of vehicles is $K= \left\lfloor {obj/10^{a^{*}}} \right\rfloor =2$ and the remaining cost is $C_{tot} =obj-K\cdot 10^{a^{*}}=195$. This solution will be preferred over another with $\textit{obj}^{*}= 30160$ for which $K^{*}=\left\lfloor {obj^{*}/10^{a^{*}}} \right\rfloor =3$ even though $C^{*}_{tot} =obj^{*}-K\cdot 10^{a^{*}}=160 < C_{tot}$.

Appendix 2: Subtour elimination constraints

To illustrate the function of the sub-tour elimination constraints, an example is considered that involves one depot denoted by $\{0, 4\}$ and three customers $\{1, 2, 3\}$. For brevity and without loss of generality, the period and vehicle indexes can be dropped. Then constraint set (5) becomes:

$$\begin{aligned} \sum _{\begin{array}{c} l=0 \\ l\ne i \end{array}}^{N\cup \{0\}} {Y_{l,i} } -\sum _{\begin{array}{c} m=0 \\ m\ne i \end{array}}^{N\cup \{4\}} {Y_{i,m} } \ge 0 \quad \forall i\in \{ {1,2,3}\} \end{aligned}$$

(25)

i.e.; the number of full tanks delivered to each customer i cannot receive a negative value. Figure 8 presents subtour 1–3–2–1, where $Y_{1,3}, Y_{3,2}$ and $Y_{2,1}$ represent the truck loads of full tanks along the links 1–3, 3–2 and 2–1 respectively.

Fig. 8

Subtour for an example with three customers

The numbers of full tanks delivered to customers 1, 2, and 3 are respectively $u_{1}=Y_{2,1} - Y_{1,3}, u_{2}=Y_{3,2} - Y_{2,1}$ and $u_{3}=Y_{1,3}-Y_{3,2}$. Then, unless $Y_{1,3}, Y_{3,2}$ and $Y_{2,1} = 0$, at least one of $u_{1}, u_{2}$ and $u_{3}$ must be negative and thus constraint (25) is violated. Therefore, the presence of this set of constraints eliminates possible subtours and allows only routes that start and end at the depot. The same applies for any other subtour; 1–2–1, 2–3–2 etc. Besides, a trip with empty loads ($Y_{1,3}, Y_{3,2}$ and $Y_{2,1} = 0$) is not efficient because unnecessary transportation cost is added. The same rationale can be followed to demonstrate the function of the subtour elimination constraints (7) for the part of empty tanks transportation and generalized for all periods and vehicles and any number of customers.