Truck synchronization at single door cross-docking terminals

Abstract

We address the problem of synchronizing the loading and discharging operations of trucks at a particular cross-docking center, with one door at both the inbound and outbound sides, aiming at minimizing the makespan of the whole process. We propose a mixed integer linear model and a Lagrangian decomposition scheme. We derive conditions for optimally solving both the Lagrangian relaxation and the dual problems. Based on the theoretical results, we propose a Lagrangian heuristic for computing many feasible solutions and gathering the best one among them. An extensive computational experience validates our Lagrangian heuristic, also in comparison with a state-of-the-art benchmark solver and a heuristic algorithm from the literature.

Appendices

Appendix 1: Proof of Proposition 4

Proof

For a given set of multipliers, taking into account Proposition 3 and Theorem 1, we have

$$\begin{aligned} Z_{LR}^O(\sigma )=C_{\max }-\sum _{h \in H}\sum _{j \in J}h \sigma _{j}y_{jh}= {\left\{ \begin{array}{ll} m-\displaystyle \sum _{h=1}^mh\sigma _{[h]} &{} \text{ if } \ \displaystyle \sum _{j=1}^m \sigma _j \le 1\\ H-\displaystyle \sum _{H-m+1}^Hh\sigma _{[h]} &{} \text{ if } \ \displaystyle \sum _{j=1}^m \sigma _j > 1 \end{array}\right. } \end{aligned}$$

Let $\delta >0$ be an increase in any component ${\hat{\jmath }}$ of $\sigma$, ${\hat{\sigma }}=\sigma \delta e_{{\hat{\jmath }}}$ and let ${\hat{h}}$ the time slot associated with ${\hat{\jmath }}$, that is $[{\hat{h}}]={\hat{\jmath }}$. We indicate by ${\varDelta } Z^O$ the variation $Z_{LR}^O({\hat{\sigma }})-Z_{LR}^O(\sigma )$. Three cases can occur

• c1) $\ \ \displaystyle \sum _{j\in j}\sigma _j <1$ and $\displaystyle \sum _{j \in J \setminus \{ {\hat{\jmath }}\}}\sigma _j + \sigma _{{\hat{\jmath }}}+\delta \le 1;$

• c2) $\ \ \displaystyle \sum _{j\in J}\sigma _j > 1$ , (of course it is also $\displaystyle \sum _{j \in J \setminus \{ {\hat{\jmath }}\}}\sigma _j + \sigma _{{\hat{\jmath }}}+\delta > 1 );$

• c3) $\ \ \displaystyle \sum _{j\in J}\sigma _j \le 1$ and $\displaystyle \sum _{j \in J \setminus \{ {\hat{\jmath }}\}}\sigma _j + \sigma _{{\hat{\jmath }}}+\delta > 1.$

Case c1. If the time slot associated with ${\hat{\jmath }}$ does not change, we simply get

$$\begin{aligned} {\varDelta } Z^O =Z^O_{LR}({\hat{\sigma }})-Z^O_{LR}(\sigma )= -{\hat{h}}\delta < 0; \end{aligned}$$

conversely, let $h_M$, ${\hat{h}}+1 \le h_M \le m$, be the newest time slot associated with ${\hat{\jmath }}$. Note that it is

$$\begin{aligned} \sigma _{[{\hat{h}}]} + \delta > \sigma _{[h_{M}]}. \end{aligned}$$

(50)

We get

$$\begin{aligned} Z_{LR}^O({\hat{\sigma }})&= m - \sum _{h=1}^{{\hat{h}}-1}h\sigma _{[h]} - \sum _{h={\hat{h}}}^{h_{M}-1}h\sigma _{[h+1]}-h_{M}(\sigma _{[{\hat{h}}]}+\delta ) - \sum _{h=h_{M}+1}^{m} h\sigma _{[h]}\\ Z_{LR}^O(\sigma )&= m - \sum _{h=1}^{m}h\sigma _{[h]} \end{aligned}$$

Therefore, from Proposition 3 and (50), it holds

$$\begin{aligned} {\varDelta } Z^O= & {} Z^O_{LR}({\hat{\sigma }})-Z^O_{LR}(\sigma )\\= & {} \sum _{h = {\hat{h}}}^{h_{M}-1}h(\sigma _{[h]}-\sigma _{[h]+1]})+h_{M}(\sigma _{[h_{M}]}-\sigma _{[{\hat{h}}]}-\delta ) <0 \end{aligned}$$

Case c2. The proof is similar to that of Case c1 and we omit it.

Case c3. Assume without loss of generality that the trucks are initially scheduled at the time slots $(1,\ldots ,m)$. By effect of the increase in $\sigma _{{\hat{j}}}$ they will be rescheduled at the time slots $(H-m+1,\ldots ,H)$. Again we have two subcases. If the ordering of the trucks does not change, noting that it is:

$$\begin{aligned} {\hat{\sigma }}_{[H-m+h]}=\left\{ \begin{array}{ll} \sigma _{[h]}&{}\,\,\,\text{ for } \,\,h=1,\ldots ,m, \,\, h \ne {\hat{h}}\\ \sigma _{[{\hat{h}}]}+ \delta &{}\,\,\,\text{ for } \,h={\hat{h}}, \\ \end{array} \right. \end{aligned}$$

we have

$$\begin{aligned} Z_{LR}^O({\hat{\sigma }})= & {} H-\displaystyle \sum _{h=H-m+1}^{H}h{\hat{\sigma }}_{[h]}=H-\displaystyle \sum _{h=1}^{{\hat{h}}-1}(H-m+h) \sigma _{[h]}+\\&- (H-m+{\hat{h}})(\sigma _{[{\hat{h}}]} + \delta ) - \sum _{h={\hat{h}}+1}^{m}(H-m+h) \sigma _{[h]} \end{aligned}$$

while

$$\begin{aligned} Z_{LR}^O(\sigma )= m-\sum _{h=1}^m h \sigma _{[h]}=m-\displaystyle \sum _{h=1}^{{\hat{h}}-1}h \sigma _{[h]} - {\hat{h}}\sigma _{[{\hat{h}}]} - \sum _{h={\hat{h}}+1}^{m}h \sigma _{[h]}. \end{aligned}$$

Therefore, taking into account $1-\sum _{h=1}^m \sigma _{[h]}-\delta <0$, we obtain

$$\begin{aligned} {\varDelta } Z^O= & {} Z^O_{LR}({\hat{\sigma }})-Z^O_{LR}(\sigma )\\= & {} (H-m)-(H-m)\sum _{h=1}^m \sigma _{[h]} - (H-m+{\hat{h}})\delta \\= & {} (H-m)\left( 1-\sum _{h=1}^m \sigma _{[h]}-\delta \right) -{\hat{h}}\delta <0 \end{aligned}$$

Finally, if the increase in $\sigma _{[{\hat{h}}]}$ changes the ordering of the trucks, let $h_{M}$ be the largest index such that $\sigma _{[{\hat{h}}]} + \delta > \sigma _{[h_{M}]}$. Then, resorting the new sequence and taking into account that it is:

$$\begin{aligned}{\hat{\sigma }}_{[H-m+h]}=\left\{ \begin{array}{ll} \sigma _{[h]}&{}\,\,\,\text{ for } \,h=1,\ldots ,({\hat{h}}-1), \,\, \\ \sigma _{[h+1]} &{}\,\,\,\text{ for } \,h={\hat{h}},\ldots ,(h_M-1), \,\, \\ \sigma _{[{\hat{h}}]}+ \delta &{}\,\,\,\text{ for } \,h=h_M \,\, \\ \sigma _{[h]}&{}\,\,\,\text{ for } \, h=(h_M+1),\ldots ,m, \,\, \end{array} \right. \end{aligned}$$

we have:

$$\begin{aligned} Z_{LR}^O({\hat{\sigma }})= & {} H-\displaystyle \sum _{h=H-m+1}^{H}h{\hat{\sigma }}_{[h]} \\= & {} H-\sum _{h=1}^{{\hat{h}}-1}(H-m+h) \sigma _{[h]} - \sum _{h={\hat{h}}}^{h_{M}-1}(H-m+h)\sigma _{[h+1]}\\&-(H-m+h_{M})(\sigma _{[{\hat{h}}]}+\delta )+\sum _{h=h_{M} + 1}^{m}(H-m+h) \sigma _{[h]} \end{aligned}$$

and

$$\begin{aligned} Z_{LR}^O(\sigma )= & {} m-\sum _{h=1}^m h \sigma _{[h]}\\= & {} m-\sum _{h=1}^{{\hat{h}}-1}h\sigma _{[h]} - \sum _{h={\hat{h}}}^{h_{M}-1}h\sigma _{[h]}-h_{M}\sigma _{[h_{M}]}-\sum _{h=h_{M}+1}^mh\sigma _{[h]} \end{aligned}$$

Therefore,

$$\begin{aligned} {\varDelta } Z^O= & {} Z^O_{LR}({\hat{\sigma }})-Z^O_{LR}(\sigma ) \\= & {} (H-m)-(H-m)\sum _{h=1}^{m} \sigma _{[h]} \\&-\sum _{h={\bar{h}}}^{h_{M}-1}h(\sigma _{[h+1]}-\sigma _{[h]})-h_{M}(\sigma _{[{\hat{h}}]}-\sigma _{h_{[M]}})-(H-m+h_{M})\delta \\= & {} (H-m)\left( 1-\left( \sum _{h=1}^{m} \sigma _{[h]} +\delta \right) \right) - \sum _{h={\hat{h}}}^{h_{M}-1}h(\sigma _{[h+1]} -\sigma _{[h]})\\&-h_{M}(\sigma _{[{\hat{h}}]}+\delta -\sigma _{h_{[M]}})\\= & {} (H-m)\left( 1-\left( \sum _{h=1}^{m} {\hat{\sigma }}_{[h]}\right) \right) \\&- \sum _{h={\hat{h}}}^{h_{M}-1}h(\sigma _{[h+1]} -\sigma _{[h]})-h_{M}(\sigma _{[{\hat{h}}]}+\delta -\sigma _{h_{[M]}}) < 0 \end{aligned}$$

$\square$

Appendix 2: The instance generator Algorithm