Scheduling shipments in closed-loop sortation conveyors

Abstract

At the very core of most automated sorting systems— for example, at airports for baggage handling and in parcel distribution centers for sorting mail—we find closed-loop tilt tray sortation conveyors. In such a system, trays are loaded with cargo as they pass through loading stations, and are later tilted upon reaching the outbound container dedicated to a shipment’s destination. This paper addresses the question of whether the simple decision rules typically applied in the real world when deciding which parcel should be loaded onto what tray are, indeed, a good choice. We formulate a short-term deterministic scheduling problem where a finite set of shipments must be loaded onto trays such that the makespan is minimized. We consider different levels of flexibility in how to arrange shipments on the feeding conveyors, and distinguish between unidirectional and bidirectional systems. In a comprehensive computational study, we compare these sophisticated optimization procedures with widespread rules of thumb, and find that the latter perform surprisingly well. For almost all problem settings, some priority rule can be identified which leads to a low-single-digit optimality gap. In addition, we systematically evaluate the performance gains promised by different sorter layouts.

Appendices

Appendix 1: Proof of Theorem 1

Proof

For simplicity, we say that the predecessor of the first shipment in l starts in period 0. For each shipment s, $s\in \mathcal{S}_l$, starting its travel in period p, we consider

• the immediate predecessor of s in l starting in period $p'$,

• the set $D_s$ of shipments in loading station $l'$, $l'\ne l$, such that s travels by $l'$, s, and shipments in $D_s$ travel in the same direction, and for each $s'\in D_s$ there is a period $p''$, $p'<p''< p$, such that s and $s'$ would be on the same tray if s would start in $p''$, and

• the set $D'_s$ of shipments in loading station $l'$, $l'\ne l$, traveling by l in a period $p''$, $p'<p''<p$.

Note that s must be loaded later than its predecessor and that $D_s$ and $D'_s$ prevent us from simply loading s in an earlier period but still after its predecessor. We refer to $D_s$ and $D'_s$ as delaying loaded shipments and delaying passing shipments, respectively, of s. Note that for shipments s and $s'$, $s,s'\in \mathcal{S}_l$, $s\ne s'$, we have $D_s\cap D_{s'}=\emptyset$ and $D'_s\cap D'_{s'}=\emptyset$. Intuitively speaking, this means that each shipment in loading station $l'$ can be a delaying shipment for any shipment in l at most twice: once as a delaying loaded shipment and once as a delaying passing shipment.

Now, assume that $\sigma _l(|\mathcal{S}_l|)$ is loaded onto the SC in period p, $p>2\cdot |\mathcal{S}|-|\mathcal{S}_l|$. There must then be a shipment s in l starting in p such that its predecessor starts in $p'$ with $p-p'>|D_s|+|D'_s|$, since

$$\begin{aligned} \sum _{s\in \mathcal{S}_l}\left( |D_s|+|D'_s|\right) \le 2\cdot \left( |\mathcal{S}|-|\mathcal{S}_l|\right) . \end{aligned}$$

Shipment s, then, can be loaded earlier in a period $p''$, $p'<p''<p$, for which no delaying shipment of s exists. Clearly, the objective value is not increased by this modification. Note that each shipment can be scheduled earlier only a finite number of times. Therefore, finally, the last shipment will be loaded earlier than in period $2\cdot |\mathcal{S}|-|\mathcal{S}_l|+1$. $\square$

Appendix 2: Proof of Lemma 2

Proof

First, it is not difficult to see by an interchange argument that shipments starting from l on the upper level are loaded in decreasing order of their indices (and therefore in non-increasing order of their travel times), and shipments starting from l on the lower level are loaded in increasing order of their indices (and therefore in non-increasing order of their travel times).

Now, assume that for the first shipment s loaded on the upper level in period p, and the first shipment $s'$ loaded on the lower level in $p'$, we have $s>s'$. Exchanging s and $s'$, i.e., loading s on the lower level in $p'$ and loading $s'$ on the upper level in p, cannot increase the objective value. Afterwards, we can sort shipments in both sequences according to their indices. This can be repeated until for the first shipment s loaded on the upper level in period p and the first shipment $s'$ loaded on the lower level in $p'$, we have $s<s'$. We can now choose $s^*$ from one of the two sequences; if one is empty, we choose $s^*$ from the other one. $\square$

Appendix 3: Proof of Lemma 5

Proof

Consider an arbitrary solution where $\chi$, $\chi >0$, is the number of times a shipment travels by a loading station. Consider the last period where a shipment s travels by a loading station l. Assume w. l. o. g. that s is traveling toward $l+1$ and that s starts in loading station $l'$, $l'<l$. Clearly, s is occupying the tray located at l in p, and no shipment from l can be loaded on this tray. We distinguish between two cases in the following.

First, if no shipment from l is loaded in p going in the other direction, we can remove s from the sequence at loading station $l'$, insert it into the sequence at loading station l such that it is ready to be loaded in period p, load it onto the same tray it occupies in the original solution (thus load it at l in period p), and obtain a solution where no shipment reaches its destination container later than in the original solution, and $\chi$ has been decreased.

Second, if a shipment $s'$ from l is loaded in p going in the other direction, then let $p'=p-(l-l')T/L$ be the period where s is loaded at $l'$. We then remove s and $s'$ from the sequence at loading stations $l'$ and l, respectively, and insert s and $s'$ into the sequence at loading stations l and $l'$, such that it is ready to be loaded in p and $p'$. Note that there is no conflict with any other shipments loaded at l or $l'$, since now s has the position formerly held by $s'$, and vice versa. We load s onto the same tray it occupies in the original solution (thus load it at l in period p), and load $s'$ this way as well (thus load it at $l'$ in period $p'$). Note that the destination container of $s'$ must be located between $l-1$ and l, since otherwise, s would not be the last shipment traveling by a loading station. Hence, sending $s'$ from $l'$ in the same direction as s is sent in the original solution is feasible, since this tray is not occupied when it reaches $l'$. Furthermore, $s'$ reaches its destination container earlier than does s, and s reaches its destination container during the same period as in the original solution. Therefore, we obtain a feasible solution without increasing the objective value. Finally, s travels by no loading station in the new solution, and $s'$ travels by one fewer loading station than does s in the original solution. Therefore, we decrease the number $\chi$.

Repeating the step described above, we can obtain a solution with $\chi =0$ which has an objective value not exceeding that of the original solution. $\square$