Online integrated allocation of berths and quay cranes in container terminals with 1-lookahead

Abstract

This paper studies an online over-list model of the integrated allocation of berths and quay cranes (QCs) in container terminals with 1-lookahead ability. The objective is to minimize the maximum completion time of container vessels, i.e., the makespan. We focus on two different types of vessels, three berths and a small number of QCs in the hybrid berth layout, with 1-lookahead ability. We propose a ${{(1 + \sqrt{2} )/2}}$-competitive algorithm for the case with four cranes, a 5/4-competitive algorithm for the case with five cranes and a 4/3-competitive algorithm for the case with six cranes, respectively. All of the algorithms are proved to be optimal.

Appendices

Appendix 1: Lower bound of the case with four QCs

We show the complete calculation process as follow.

If A processes $r_1$ with two QCs on berth $b_1$, we get:

$$\begin{aligned}&{C_{\max }}(\mathcal {I}) \ge {1 / 2} + {\varDelta / 4} \end{aligned}$$

(1)

$$\begin{aligned}&\varDelta \ge 6,~{C^ * }(\mathcal {I}) = {1 / 2} + {\varDelta / 4}\end{aligned}$$

(2)

$$\begin{aligned}&6 \ge \varDelta \ge 3,~ {C^ * }(\mathcal {I})= {\varDelta / 3}\end{aligned}$$

(3)

$$\begin{aligned}&3 \ge \varDelta \ge 2,~ {C^ * }(\mathcal {I})=1 \end{aligned}$$

(4)

If $\mathcal {A}$ processes $r_1$ with three QCs on berth $b_1$, $b_2$, we get:

$$\begin{aligned}&\varDelta \ge 6,~{C_{\max }}(\mathcal {I}) \ge {\varDelta / 3}\end{aligned}$$

(5)

$$\begin{aligned}&6 \ge \varDelta \ge {9 / 2},~{C_{\max }}(\mathcal {I}) \ge 2\end{aligned}$$

(6)

$$\begin{aligned}&{9 / 2} \ge \varDelta \ge 2,~{C_{\max }}(\mathcal {I}) \ge {1 / 2} + {\varDelta / 3}\end{aligned}$$

(7)

$$\begin{aligned}&{C^ * }(\mathcal {I}) = {1 / 2} + {\varDelta / 4} \end{aligned}$$

(8)

If $\mathcal {A}$ processes $r_1$ with two QCs on berth $b_1$, $b_2$, we get:

$$\begin{aligned}&{C_{\max }}(\mathcal {I}) \ge {\varDelta / 2}\end{aligned}$$

(9)

$$\begin{aligned}&\varDelta \ge 6,~{C^ * }(\mathcal {I}) = {1 / 2} + {\varDelta / 4}\end{aligned}$$

(10)

$$\begin{aligned}&6 \ge \varDelta \ge 3,~ {C^ * }(\mathcal {I})= {\varDelta / 3} \end{aligned}$$

(11)

$$\begin{aligned}&3 \ge \varDelta \ge 2,~ {C^ * }(\mathcal {I})=1 \end{aligned}$$

(12)

After processed Eqs. (1–4), we have:

$$\begin{aligned}&\varDelta \ge 6,~\rho \ge 1\end{aligned}$$

(13)

$$\begin{aligned}&6 \ge \varDelta \ge 3,~ \rho \ge \max \left\{ {{{6 + 3\varDelta } \over {4\varDelta }}} \right\} \end{aligned}$$

(14)

$$\begin{aligned}&3 \ge \varDelta \ge 2,~ \rho \ge \max \left\{ {{{2 + \varDelta } \over 4}} \right\} \end{aligned}$$

(15)

After processed Eqs. (5–8), we have:

$$\begin{aligned}&\varDelta \ge 6,~\rho \ge \max \left\{ {{{4\varDelta } \over {6 + 3\varDelta }}} \right\} \ge 1\end{aligned}$$

(16)

$$\begin{aligned}&6 \ge \varDelta \ge {9 / 2},~ \rho \ge \max \left\{ {{8 \over {2 + \varDelta }}} \right\} \end{aligned}$$

(17)

$$\begin{aligned}&{9 / 2} \ge \varDelta \ge 2,~ \rho \ge \max \left\{ {{{6 + 4\varDelta } \over {6 + 3\varDelta }}} \right\} \end{aligned}$$

(18)

After processed Eqs. (9–12), we have:

$$\begin{aligned}&\varDelta \ge 6,~\rho \ge \max \left\{ {{{2\varDelta } \over {2 + \varDelta }}} \right\} \ge 1\end{aligned}$$

(19)

$$\begin{aligned}&6 \ge \varDelta \ge 3,~ \rho \ge {3 \over 2} \end{aligned}$$

(20)

$$\begin{aligned}&3 \ge \varDelta \ge 2,~ \rho \ge {\varDelta \over 2} \end{aligned}$$

(21)

From Eqs. (13–21), we get:

$$\begin{aligned}&\varDelta \ge 6,~ \rho \ge 1\end{aligned}$$

(22)

$$\begin{aligned}&6 > \varDelta \ge {9 \over 2},~ \rho \ge \max \left\{ {{{3\varDelta + 6} \over {4\varDelta }}} \right\} ,~\rho \ge {{13} \over {12}}\end{aligned}$$

(23)

$$\begin{aligned}&{9 \over 2} > \varDelta \ge {{6 + 12\sqrt{2} } \over 7},~ \rho \ge \max \left\{ {{{3\varDelta + 6} \over {4\varDelta }}} \right\} ,~\rho \ge {{1 + \sqrt{2} } \over 2}\end{aligned}$$

(24)

$$\begin{aligned}&{{6 + 12\sqrt{2} } \over 7} > \varDelta \ge 3,~ \rho \ge \max \left\{ {{{4\varDelta + 6} \over {3\varDelta + 6}}} \right\} ,~\rho \ge {{1 + \sqrt{2} } \over 2}\end{aligned}$$

(25)

$$\begin{aligned}&3 > \varDelta \ge {{2 + 2\sqrt{10} } \over 3},~ \rho \ge \max \left\{ {{{4\varDelta + 6} \over {3\varDelta + 6}}} \right\} ,~\rho \ge {6 \over 5} \end{aligned}$$

(26)

$$\begin{aligned}&{{2 + 2\sqrt{10} } \over 3} > \varDelta \ge 2,~ \rho \ge {{4 + \sqrt{10} } \over 6} \end{aligned}$$

(27)

Setting $\varDelta = {{(6 + 12\sqrt{2} )} / 7}$, then $\rho \ge {{(1 + \sqrt{2} )} / 2}$.

Appendix 2: Theorem 2

As in case 2, namely, $2 \le \varDelta \le {{(2 + 2\sqrt{10} )} / 3}$ or $\varDelta \ge {{(6 + 12\sqrt{2} )} / 7}$, the algorithm is the same as that in Zheng et al., we can refer to their proof. For the sake of completeness, we present their proof here:

Proof

Given any request input sequence $\mathcal {I} = \left\{ {{r_1},{r_2},\ldots ,{r_n}} \right\}$. let $\sigma$ be the schedule produced by LIST. Let ${C_\sigma }(\mathcal {I})$ be the makespan of $\sigma$, and $C^*(\mathcal {I})$ that of a schedule produced by an optimal online algorithm OPT. We consider two cases below.

• Case 1 The last request is a large one, i.e., ${r_n} = \varDelta$. For the case with $T_{_a}^ * = [0,0)$, we have $C^*(\mathcal {I}) = {C_\sigma }(\mathcal {I}) = {C_{n,1}} + {\varDelta / 4}$ since neither LIST nor OPT makes any waste of QC utility within $\left[ {0,{C_\sigma }(\mathcal {I})} \right)$. For the other case with $T_{_a}^ * = [{t_1},{t_2}) \ne [0,0)$, let $k \ge 1$ be the number of large requests after time $t_2$. ${C_\sigma }(\mathcal {I})= {t_2} + k{\varDelta / 4}$. If ${t_2} = {1 / 2}$ and $k = 1$, implying $n = 2$, ${r_1} = 1$ and ${r_2} = \varDelta$, then $C^*(\mathcal {I}) \ge \min \left\{ {{1 / 2} + {\varDelta / 4},\max \left\{ {1,{\varDelta / 3}} \right\} } \right\}$ and thus ${{{C_\sigma }(\mathcal {I})} / {{C^ * }(\mathcal {I}) \le {5 / 4}}}$; otherwise if either ${t_2} \ge 1$ or $k \ge 2$, ${C^ * }(\mathcal {I}) \ge {t_2} + k{\varDelta / 4} - {1 / 4}$ and ${{{C_\sigma }(\mathcal {I})} / {{C^ * }(\mathcal {I}) \le {6 / 5}}}$.

• Case 2 The last request is a small request, i.e., ${r_n} = 1$. We claim that $T_{_a}^ * = [0,0)$ in $\sigma$ since otherwise $r_n$ would have been processed in the $T_{_a}^ *$. Divide this case into two subcases by whether ${C_{n,1}} = {C_{n,2}}$.

• Case 2.1 ${C_{n,1}} = {C_{n,2}}$, In this case ${C_\sigma }(\mathcal {I}) = {C_{n,1}} + {1 / 2}$. If either $n = 1$ or $n = 3$ with ${r_1} = {r_2} = {r_3} = 1$, we have ${C_{n,1}} \le {1 / 2}$ and ${C_\sigma }(\mathcal {I}) = {C^ * }(\mathcal {I})$; if $n = 2$ and ${r_1} = \varDelta$, then ${C_{n,1}} = {\varDelta / 4}$ and $C^*(\mathcal {I}) \ge \min \left\{ {{1 / 2} + {\varDelta / 4},\max \left\{ {1,{\varDelta / 3}} \right\} } \right\}$, implying ${{{C_\sigma }(\mathcal {I})} / {{C^ * }(\mathcal {I}) \le {5 / 4}}}$; otherwise if ${C_{n,1}} \ge 1$, then ${C^ * }(\mathcal {I}) \ge {C_{n,1}} + {1 / 4}$, implying a ratio of at most 6 / 5.

• Case 2.2 ${C_{n,1}} < {C_{n,2}}$ (or ${C_{n,2}} < {C_{n,1}}$). Then we have by previous analysis that ${C_{n,2}} = {C_{n,1}} + {1 / 2}$ (or ${C_{n,1}} = {C_{n,2}} + {1 / 2}$), and thus ${C_\sigma }(\mathcal {I}) = {C^ * }(\mathcal {I})$.

The theorem follows. $\square$