Retrieval scheduling in crane-based 3D automated retrieval and storage systems with shuttles

Abstract

Retrieval task scheduling has been extensively studied for 2D automated retrieval and storage systems (AS/RS). A good schedule can significantly reduce the makespan for finishing a given group of retrieval tasks. However, the task scheduling problem has never been studied for crane-based 3D AS/RS with shuttle-based depth movement mechanisms (DMMs), which has become increasingly popular in practice. This study considered how to schedule a group of retrieval requests in a crane-based 3D AS/RS with shuttle-based DMMs with the objective to minimize the makespan. A mixed-integer programing model was developed to represent the problem, and the problem was proven to be NP-hard. Four heuristics were investigated for their computational performance. First-Come-First-Serve is the current practice while the Percentage Priority to Shuttle Reallocation with the Shortest Leg rule was developed based on the existing rule for scheduling storage and retrieval tasks in 3D AS/RS with conveyor-based DMMs. The Genetic Algorithm, which is popular for 2D systems, was adapted to deal with the 3D system. The Lowest-Waiting-Time-First heuristic was proposed based on the optimality condition of the scheduling problem and was demonstrated to outperform the other three algorithms in terms of solution quality and computational time. Further numerical results revealed insights for improving 3D AS/RS productivity. When the number of retrieval tasks is small (e.g., when a short planning horizon is adopted for high responsiveness), having more shuttles can improve the system performance. When there are many tasks to schedule, for example, in a situation with a long planning horizon, using a crane with higher speed rather than adding more shuttles can improve system efficiency more.

Appendices

Appendices

1.1 Appendix 1

In that study, only lanes with retrieval tasks are considered for the scheduling problem. However, in practice, a system may have more lanes than the SKU types and one type of goods is possibly stored in more than one lane. The lane-to-task assignment, which selects lanes to fulfill demand, can be modeled based on the MIP (1-21) proposed for the scheduling problem in this paper. In addition to the notations defined before, set $A$ is defined as the set of SKU types, and $N_{a}$, $a \in A$, is the set of lanes used for storing SKU $a$. The demand on SKU type $a$ is $D_{a}$. The lane-to-task assignment problem is formulated as follows based on the scheduling problem (1-21) by having parameter $Q_{i}$ be a decision variable and adding an additional constraint set (22) to meet the demand.

$$ \begin{array}{*{20}l} {\text{Min}} \hfill & {(1)} \hfill & {} \hfill \\ {{\text{s}} . {\text{t}} .} \hfill & {\mathop \sum \limits_{{i \in N_{a} }} Q_{i} = D_{a} ,} \hfill & {a \in A,} \hfill \\ {} \hfill & {\left( {2 - 21} \right)} \hfill & {} \hfill \\ {} \hfill & {Q_{i} \ge 0.} \hfill & {} \hfill \\ \end{array} $$

(22)

1.2 Appendix 2

1.2.1 Parameters

$N$: set of lanes with shuttles or having retrieval tasks, $i$ and $j$ are their indices;

$N^{0}$: set of lanes having a shuttle at the beginning, $N^{0} \subseteq N$ and $\left| {N^{0} } \right| =$ number of shuttles;

$N^{1}$: set of lanes without a shuttle at the beginning, $N^{0} \cup N^{1} = N$

$Q_{i}$: number of retrieval tasks from lane $i$;

$Q^{\prime}$: largest number of tasks at a lane over all lanes, $Q^{\prime} = \mathop {\hbox{max} }\limits_{i \in N} Q_{i} + 1$;

$M$: number of total SC tasks, $M = \left| {N^{1} } \right| + \mathop \sum \limits_{i \in N} Q_{i}$;

$p_{iq}^{1}$: time for a shuttle finishing the $q$th task in lane $i$, $p_{{i,Q_{i} + 1}}^{1} = 0$;

$p_{ij}^{2}$: time for the crane traveling from lane $i$ to lane $j$;

$p_{ij}^{3}$: time for the crane traveling from lane $i$ to the picking position, unloading a unit, and then traveling to lane $j$ for the next task;

$p_{i}^{4}$: time for the crane traveling from lane $i$ to the picking position and unloading a unit;

$P_{{i_{0} ,i}}$: time for the crane traveling from original position to lane $i$; and

$T$: planning horizon.

1.2.2 Variables

$y_{mi}$ = 1 if the $m$th SC task is for retrieving a load or a shuttle from lane $i$; = 0, otherwise;

$t_{m}$: starting time for the crane to handle the $m$th task;

$z_{mq}$ = 1 if the $q$th task at the lane is the $m$th SC task; = 0, otherwise;

$x_{ij}$ = 1 if lane $j$ is immediately handled by lane $i$ with the same shuttle; = 0, otherwise;

$\theta_{m}$ = 1 if task $m$ is for moving a shuttle; = 0, otherwise;

$r_{iq}$: moment when the $q$th task in lane $i$ is ready for the crane to pick up, including the shuttle moving task; $r_{i0} = 0$

$l_{i}$ and $\mu_{i}$: an artificial variable to avoid sub-tours and make sure that $m$th SC task associated with a lane which has a shuttle.

1.3 Appendix 3

Theorem 1

The 3D AS/RS retrieval task scheduling problem is NP-hard.

Proof

According to Han et al. (1987), the task scheduling problem in 2D AS/RS operating in the DC mode with multiple open locations is equivalent to the traveling salesman problem (TSP) in its simplified version in which there is only one open location. Since the TSP is NP-hard (Cormen et al. 2009), the task scheduling problem in 2D AS/RS is NP-hard. For the 3D AS/RS scheduling problem considered in our study, if the shuttle speed is assumed to be infinite, the crane can pick up an SKU/shuttle instantly when arrives at a target lane. In that case, after moving a shuttle to a lane, the crane will pick up an SKU from the lane immediately instead of going to serve another lane. If we consider all locations of shuttles as open locations, the crane operates in the DC mode when transferring shuttles: starts from the I/O point to an open location, picks up the shuttle there and transfers it to lane $j$, hands over the shuttle and picks up an SKU form lane $j$, and travels back to I/O point. It is obvious that the problem considered by Han et al. (1987) is a special case of our problem with infinite shuttle speed. As this special case is NP-hard, our scheduling problem for the 3D AS/RS is also NP-hard.

Theorem 2

(Optimality Condition) Consider the $m{\rm th}$ and $\left( {m + 1} \right){\rm th}$ tasks in an optimal task schedule. If the $m{\rm th}$ task is for retrieving the $\left( {q_{1} } \right){\rm th}$ SKU from lane $j$, the $\left( {m + 1} \right){\rm th}$ task is for retrieving the $\left( {q_{1} } \right){\rm th}$ SKU from lane $k$, and the $\left( {m - 1} \right){\rm th}$ task is associated with lane $i,$ we should always have $p_{m} \ge s_{m}$ where $p_{m} = max\left\{ {r_{{k,q_{2} }} - t_{m - 1} - p_{ik}^{3} ,0} \right\}$ and $s_{m} = \hbox{max} \left\{ {r_{{j,q_{1} }} - t_{m - 1} - p_{ij}^{3} ,0} \right\}$ if the $\left( {m - 1} \right){\rm th}$ task is a retrieval task, otherwise; $p_{m} = \hbox{max} \{ r_{{k,q_{2} }} - t_{m - 1} - p_{i,l}^{2} - p_{l,k}^{2} ,0\}$ and $s_{m} = \hbox{max} \left\{ {r_{{j,q_{1} }} - t_{m - 1} - p_{i,l}^{2} - p_{l,j}^{2} ,0} \right\}$ if the $\left( {m - 1} \right){\rm th}$ task is to reallocate a shuttle from lane $i$ to $l$.

Proof

Suppose the $\left( {m - 1} \right){\rm th}$ task is for retrieving an SKU, $p_{m} = { \hbox{max} }\left\{ {r_{{k,q_{2} }} - t_{m - 1} - p_{ik}^{3} ,0} \right\}$ and $s_{m} = \hbox{max} \left\{ {r_{{j,q_{1} }} - t_{m - 1} - p_{ij}^{3} ,0} \right\}$. In addition, Let’s define $w_{m}$ and $w_{m + 1}$ as the crane’s waiting time of the $m{\rm th}$ and $m + 1{\rm th}$ task. Clearly, $w_{m} = s_{m}$ and $w_{m + 1} = { \hbox{max} }\left\{ {r_{{k,q_{2} }} - t_{m} - p_{jk}^{3} ,0} \right\}$. To prove the optimality, we have to show that the total waiting time of these two tasks will increase if we switch the sequence. Let’s define $w_{m}^{'}$ and $w_{m + 1}^{'}$ as the crane’s waiting time of the new $m{\rm th}$ and $m + 1{\rm th}$ task after switching the sequence. It is clear that when $s_{m} = w_{m} = 0$, the solution is optimal, since we can always have $w_{m} = w_{m + 1}^{'} = 0$ and $w_{m}^{'} \ge w_{m + 1}$. When $s_{m} > 0$, two cases can be considered.

Case 1 $p_{m} = 0, r_{{k,q_{2} }} \le t_{m - 1} + p_{ik}^{3}$

In that case, we can have

$$ w_{m}^{'} = \hbox{max} \left\{ {p_{m} ,0} \right\} = 0, $$

and

$$ w_{m}^{'} + w_{m + 1}^{'} = w_{m + 1}^{'} = \hbox{max} \left\{ {r_{{j,q_{1} }} - t_{m - 1} - p_{ik}^{3} - p_{jk}^{3} ,0} \right\} < r_{{j,q_{1} }} - t_{m - 1} - p_{ij}^{3} . $$

Clearly, when $s_{m} > 0$, we must have $p_{m} > 0$ to guarantee the optimality.

Case 2 $p_{m} > 0$

When $p_{m} > 0$, $w_{m}^{'} = r_{{k,q_{2} }} - t_{m - 1} - p_{ik}^{3} > 0$, and $w_{m + 1}^{'} = \hbox{max} \left\{ {r_{{j,q_{1} }} - r_{{k,q_{k} }} - p_{jk}^{3} ,0} \right\}$. If $w_{m + 1}^{'} > 0$, which is equivalent to $r_{{k,q_{2} }} - r_{{j,q_{1} }} < - p_{jk}^{3}$, $w_{m + 1} = \hbox{max} \left\{ {r_{{k,q_{2} }} - r_{{j,q_{1} }} - p_{jk}^{3} ,0} \right\} = 0$. Therefore, after switching, the total waiting time of these two tasks will be

$$ w_{m}^{'} + w_{m + 1}^{'} = r_{{j,q_{1} }} - t_{m - 1} - p_{ik}^{3} - p_{jk}^{3} < r_{{j,q_{1} }} - t_{m - 1} - p_{ij}^{3} = w_{m} + w_{m + 1} = w_{m} . $$

To guarantee the optimality of the given solution, $r_{{k,q_{2} }} - r_{{j,q_{1} }} \ge - p_{jk}^{3}$ and $w_{m + 1}^{'} = 0$. In that case, if $w_{m + 1} = 0$ and $r_{{k,q_{2} }} - r_{{j,q_{1} }} \le p_{jk}^{3}$,

$$ \begin{aligned} & w_{m}^{'} + w_{m + 1}^{'} = w_{m}^{'} = r_{{k,q_{2} }} - t_{m - 1} - p_{ik}^{3} ,{\text{and}} \\ & w_{m} + w_{m + 1} = w_{m} = r_{{j,q_{1} }} - t_{m - 1} - p_{ij}^{3} . \\ \end{aligned} $$

To guarantee the optimally,

$$ r_{{k,q_{k} }} - t_{m - 1} - p_{ik}^{3} \ge r_{{j,q_{1} }} - t_{m - 1} - p_{ij}^{3} , $$

which is equivalent to

$$ p_{jk}^{3} \ge r_{{k,q_{2} }} - r_{{j,q_{1} }} \ge p_{ik}^{3} - p_{ij}^{3} . $$

And if $w_{m + 1} > 0$,

$$ w_{m}^{'} + w_{m + 1}^{'} = w_{m}^{'} = r_{{k,q_{2} }} - t_{m - 1} - p_{ik}^{3} > r_{{k,q_{2} }} - t_{m - 1} - p_{ij}^{3} - p_{jk}^{3} = w_{m} + w_{m + 1} . $$

According to the discussion, if $s_{m} = 0$, $p_{m}$ can take any value. However, when $s_{m} > 0$, we need $r_{{k,q_{2} }} - r_{{j,q_{1} }} \ge p_{ik}^{3} - p_{ij}^{3}$, which is equivalent to $p_{m} \ge s_{m} .$ The same logic can be applied for the scenario when $m - 1{\rm th}$ task is for reallocating a shuttle.