Scheduling quay cranes and yard trucks for unloading operations in container ports

Abstract

This paper studies an integrated optimization problem on quay crane and yard truck scheduling in container terminals. A mixed-integer programming model is formulated. For the model, we show the integrated scheduling problem is strongly NP-hard and investigate some properties that can considerably reduce the computational complexity. For solving the proposed model within a reasonable time, a particle swarm optimization based solution method is developed. Numerical experiments are conducted to compare the proposed method with the CPLEX solver and the genetic algorithm. The results validate the effectiveness of the proposed model and the efficiency of the proposed solution method.

Appendices

Appendix 1: Proof of Proposition 1

Proposition 1

Finding an optimal solution for the model is strongly NP-hard.

Proof

We can prove the proposition by reducing the problem in polynomial time to the three-partition problem (Garey and Johnson 1979; Liu and Tang 2008), which is well-known to be strongly NP-hard. Given 3h items, $H=\{ {1,2,\ldots ,3h} \}$, each item $j\in H$ has a positive integer size $a_j$ satisfying $a/4<a_j <a/2$, and $\mathop \sum \nolimits _{j=1}^{3h} a_j =ha$, for some integer a. The three-partition problem asks whether there are h disjoint subsets $H_1$, $H_{2,} \ldots ,H_h$ of H such that $|H_i |=3$ and $\mathop \sum \nolimits _{j\in H_i } a_j =a$, $i=1,2,\ldots ,h.$

Given any instance of a three-partition problem, consider the following instance that we construct for our problem: number of tasks: $N=3h$, number of QCs: $K=h$, processing time: $t_j =P_{j1} +P_{j2} =a_j$, $j=1,2,\ldots 3h$. We will show that there exists a solution to the three-partition problem if and only if there is a feasible solution to our integrated QC and YT scheduling problem. (i) The “only-if” direction: given a solution to the three-partition problem, $H_1$, $H_{2,} \ldots ,H_h$, we can simply let tasks in set $S_j$ correspond to the elements of $H_j , 1<j<h$, and construct a schedule to our integrated QC and YT scheduling problem. (ii) The “if” direction: suppose there exists a schedule for the constructed instance of our integrated QC and YT scheduling problem. Since the total processing time of all tasks is $\mathop \sum \nolimits _{j=1}^{3h} a_j =ha$, a total of h QCs are fully utilized and the completion time of each QC is a. So we know the schedule contains exactly h sets, the total processing time of tasks in each set $S_j$ is a. A partition for the set H is obtained by mapping the elements corresponding to the tasks in set $S_j$ to the elements in the sub-set $H_j$, $1<j<h$. Then $\mathop \sum \nolimits _{j\in H_i } a_j =a$, and $|H_i |=3$ because $a/4<a_j <a/2$.

It can be thus seen that our integrated QC and YT scheduling problem has a feasible solution if and only if there exists a solution to the three-partition problem. The reduction of the integrated QC and YT scheduling problem to the three-partition problem can be done in polynomial time. Therefore, finding an optimal solution for the model is strongly NP-hard. $\square$

Appendix 2: Proof of Proposition 2

Proposition 2

A formula used to calculate a proper value of M about the integrated QC and YT scheduling problem is as follows:

$$\begin{aligned} M^{\textit{UB}}= & {} f_q +g\times \left| {l_0^q -L_1 } \right| +P_{11}+\mathop \sum \limits _{i\in \varOmega \backslash \left\{ 1 \right\} } max\left\{ {P_{i-1,2} ,g\times \left| {L_i -L_{i-1} } \right| +P_{i1} } \right\} +P_{\left| \varOmega \right| 2} \end{aligned}$$

Proof

Suppose that ($\hbox {I}$) there is only one QC (the first QC in the set Q) and one YT; ($\hbox {II}$) the QC moves from one side to the next; ($\hbox {III}$) the containers are labeled such that container 1 can be unloaded first, container 2 can be unloaded second, etc. Then we can calculate the time required to complete all of the tasks, i.e., $max_{i\in \varOmega } C_{i2}$ as the formula (29).

In formula (29), $q=1$. The first term is the ready time of the QC. The second term is the moving time of the QC from the initial position to the first task. The third term is the QC time for the first task. In the fourth term, the YT can receive task i from the QC if (I) the YT has completed task $i-1$ (the component $P_{i-1,2}$), and (II) the QC has moved to the bay where task i is located and has unloaded task i (the component $g\times | {L_i -L_{i-1} } |+P_{i1}$). The fifth term is the transportation time of the last task. Through this formula a valid value for M can be equal to the above time. We can see that for different-scale problems the corresponding data M is not the same. When the problem size is increasing, the value of M is increasing. Compared with the traditional data M that usually is a single and sufficiently large value, the value of M that we calculated by the proposed formula is more reasonable. In our model there are seven constraints that use the value of M, including Constraints (9), (10), (12)–(15), (22). A proper value of M can improve the computational efficiency. Through the computational experiments we can also confirmed the validity of the proposed proposition. $\square$

Appendix 3: Proof of Proposition 3

Proposition 3

To reduce the influence of symmetry, we propose the following method as follows: $\mathop \sum \nolimits _{j\in \varOmega ^{F}} j \cdot Y_{0j}^k \le \mathop \sum \nolimits _{j\in \varOmega ^{F}} j \cdot Y_{0j}^{k+1}$, $k\in T/\{ {| T |} \}$.

Proof

We prove in the following that the above constraint removes some feasible solutions and optimal solutions, but at least one optimal solution is not removed. We prove it in two steps.

(i) Evidently, the problem has optimal solutions. We let $\{ {Y_{ij}^{k^{*}} } \},({i,j\in \varOmega ^{F},k\in T} )$ be an optimal solution that does not satisfy Constraint (30).

Then at least $\exists \bar{K} \in T$, such that $\mathop \sum \nolimits _{j\in \varOmega ^{F}} j \cdot Y_{0j}^{\bar{k} } \ne 0$. Through the solution $\{ {Y_{ij}^{k^{*}} } \}$, $( {i,j\in \varOmega ^{F},k\in T} )$, we will construct a new optimal solution satisfying Constraint (30) below.

Firstly we define a parameter $\pi _k =\mathop \sum \nolimits _{j\in \varOmega ^{F}} j\in Y_{0j}^{k^{*}}$, $k\in T \cdot \pi _{k}$ denotes the ID of the first task that YT k transports. Note that if YT k is not used, then $\pi _k =0$. And then define a function $\{ {\theta ( i )} \}$:

$$\begin{aligned} \left\{ {\theta (i)} \right\} = \left\{ {1,2,\ldots | T |} \right\} ,i=1,2\ldots | T |,\quad \hbox {and}\quad \pi _{\theta \left( 1 \right) } \le \pi _{\theta ( 2)} \le \pi _{\theta (3)} \ldots \le \pi _{\theta ( {| T |} )}. \end{aligned}$$

That is, we sort the first task’s ID in ascending order and $\theta (i)$ denotes the ID of the YT whose first task is the ith smallest.

Let us give an example to illustrate. If YT 1 transports tasks 3, 5, and 9, YT 2 transports task 8, and YT 3 transports tasks 1, 2, 4, 6 and 7, then $\pi _1 =3,\pi _2 =8,\pi _3 =1$ and $\theta ( 1 )=3,\theta (2)=1,\theta ( 3)=2.$ Thus, we have $\pi _{\theta ( 1)} =1\le \pi _{\theta ( 2 )} =3\le \pi _{\theta ( 3 )} =8$. Now a new solution

$$\begin{aligned} \left\{ {\bar{Y} _{ij}^k } \right\} =\left\{ {Y_{ij}^{\theta ( k )^{*}} } \right\} ,\quad \left( {i,j\in \varOmega ^{F},k\in T} \right) \end{aligned}$$

is defined. We can observe the new solution $\{ {\bar{Y} _{ij}^k } \}$ is also a feasible solution with the same objective function value as $\{ {Y_{ij}^{\theta ( k )^{*}} } \}$, and satisfy Constraint (30). Therefore, there exists at least one optimal solution that satisfies Constraint (30).

(ii) Now we would prove Proposition 3 could remove a part of the optimal solutions. Suppose $\{ {\bar{Y} _{ij}^k } \}$ is an optimal solution and satisfies Constraint (30). Define $\bar{k}$:

$$\begin{aligned} \bar{k}= argmin\left\{ {\mathop \sum \limits _j \bar{Y} _{0j}^k >0} \right\} , \quad k\in T,\quad j\in \varOmega ^{F} \end{aligned}$$

If $\bar{k}<|T|$, we exchange $\bar{k}$ and $\bar{k}+1$, and then a new optimal solution to the original model that does not satisfy Constraint (30) is generated. If $\bar{k}=|T|$, we exchange $\bar{k}$ and $\bar{k}-1$, then a new optimal solution that does not satisfy Constraint (30) is generated. In sum, Proposition 3 removes some optimal solutions to the original model.

Proposition 3 can remove some feasible solutions and a part of optimal solutions but do not remove all the optimal solutions. In this way we can reduce the influence of symmetry. $\square$

Appendix 4: The sub-procedure Adjust(m)in in the PSO based method

Appendix 5: The pseudo code of PSO algorithm

Appendix 6: The pseudo code of GA algorithm