An optimal data-splitting algorithm for aircraft sequencing on a single runway

Abstract

During peak-hour busy airports have the challenge of turning aircraft around as quickly as possible, which includes sequencing their landings and take-offs with maximum efficiency, without sacrificing safety. This problem, termed aircraft sequencing problem (ASP) has traditionally been hard to solve optimally in real-time, even for flights over a one-hour planning window. In this article, we present a novel data-splitting algorithm to solve the ASP on a single runway with the objective to minimize the total delay in the system both under segregated and mixed mode of operation. The problem is formulated as a 0–1 mixed integer program, taking into account several realistic constraints, including safety separation standards, wide time-windows, and constrained position shifting. Following divide-and-conquer paradigm, the algorithm divides the given set of flights into several disjoint subsets, each of which is optimized using 0–1 MIP while ensuring the optimality of the entire set. One hour peak-traffic instances of this problem, which is NP-hard in general, are computationally difficult to solve with direct application of the commercial solver, as well as existing state-of-the-art dynamic programming method. Using our data-splitting algorithm, various randomly generated instances of the problem can be solved optimally in near real-time, with time savings of over 90%.

Appendices

Appendix A Proof of variable fixing strategies

Proof of Proposition 1:

Without loss of generality, suppose that $j>i$, i.e., aircraft i is ahead of aircraft j in the FCFS sequence. Given ‘k’, in any sequence, aircraft i can only be placed between $(i-k)$ and $(i+k)$ positions; similarly, the position of aircraft j can range from $(j-k)$ to $(j+k)$. If $j-i \ge 2k$, then $j-k \ge i+k$, which implies that, even in the extreme case, flights i and j maintain their nominal (FCFS) precedence order at optimality.

Proof of Proposition 2:

We will prove this by contradiction. Assume a pair of flights $(r,s)\in \mathcal {F}$ that satisfy the above conditions, let $S^*$ denote an optimal sequence, and suppose that flight s is ahead of flight r at optimality, i.e., $x^*_{sr}=1$. Let $t_{r}^*$, and $t_{s}^*$ be the scheduled time of flights r and s in their optimal sequence. Let $D(S^*)$ denote the optimum (minimum) delay resulting from this optimal sequence $S^*$. Therefore, we get,

$~~~~~~~~~~~~~~~~~~~~D(S^*) = \sum \limits _{i \in \mathcal {F},~i\ne r,s} |t_{i}^*- T_{i}|+ |t_{r}^* - T_{r}|+|t_{s}^* - T_{s}|,~ \text {where}~t_{s}^*<t_{r}^*$.

Moreover, by virtue of optimality, $D(S^*)\le \sum \limits _{i \in \mathcal {F}} |t_{i}- T_{i}|$, for any feasible sequence of flights.

Now, construct a new sequence, say $S_{new}$, where flights r and s swap their optimal positions, while all the remaining flights retain their optimal $S^*$ sequence positions. Now, as r and s belong to the same category, there is no change in the scheduled times for all other flights, except r and s. Evaluating the total delay for this new sequence $S_{new}$, we get

$~~~~~~~~~~~~~~~~~~~~D(S_{new})= \sum \limits _{i \in \mathcal {F},~i\ne r,s} |t_{i}^*- T_{i}|+ |t_{r}^{new} - T_{r}|+|t_{s}^{new} - T_{s}|$.

But $t_{r}^{new}=t_{s}^{*}$ and $t_{s}^{new}=t_{r}^{*}$, which yields,

$~~~~~~D(S_{new})= \sum \limits _{i \in \mathcal {F},i\ne r,s} |t_{i}^*- T_{i}|+ |t_{s}^{*} - T_{r}|+|t_{r}^{*} - T_{s}|$

$~~~~~~~~~~~~~~~~~< \sum \limits _{i \in \mathcal {F},i\ne r,s} |t_{i}^*- T_{i}|+ |t_{s}^{*} - T_{s}|+|t_{r}^{*} - T_{r}|~~~~~~$ ( when $T_r\le t_s^*<t_r^*\le T_s$)

$~~~~~~~~~~~~~~~~~= D(S^*)$, which is a contradiction. Hence, $x_{rs}=1$ preserves the optimality proving the assertion.

Appendix B Time-windows tightening

$E_i$ (resp. $L_i$) values for any flight i can be tightened to give tighter early (resp. late) time restrictions, say $E_i^{new}$ (resp. $L_i^{new}$), where $E_i^{new}\ge E_i$ (resp. $L_i^{new}\le L_i$). In this direction, we formulate the following optimization problems by defining two new sets of aircraft, $\mathcal {F}_i^{+}=\{j:x_{ji}=1\}$ (resp. $\mathcal {F}_i^{-}=\{j:x_{ij}=1\}$), which are known in advance to land before (resp. after) aircraft i (after applying variable fixing strategies).

$$\begin{aligned} \begin{aligned} E_i^{new}\equiv \text {Minimize}~ t_i~~~~~~\\ \text {s.t}~x_{ji}=1 ~~\forall ~ j\in \mathcal {F}_i^{+}\\ t_i\ge t_j+\Delta _{ji}~~ \forall ~ j\in \mathcal {F}_i^{+}\\ \{\text {ASP}:1-6\}~~\forall ~j\in \mathcal {F}_i^{+} \end{aligned} \quad ~~~~~~\quad \begin{aligned} L_i^{new}\equiv \text {Maximize}~ t_i~~~~~~\\ \text {s.t}~x_{ij}=1~~\forall ~ j\in \mathcal {F}_i^{-}\\ t_j\ge t_i+\Delta _{ij}~~ \forall ~ j\in \mathcal {F}_i^{-}\\ \{\text {ASP}:1-6\}~~\forall ~j\in \mathcal {F}_i^{-} \end{aligned} \end{aligned}$$

(B.1)

Even though the above time-windows tightening formulation is found to be very effective—computationally it is expensive. Therefore, we propose a simple heuristics based on a analysis done by Psaraftis (1978) that can tighten the time-windows in an efficient manner. According to Psaraftis (1978), group clustering takes place for arrival traffic in case of the objective of maximizing throughput traffic, if CPS requirements and time-windows restrictions are lifted, and all the flights are ready to land at the same time. In the optimal sequence resulting from group clustering, small arrivals land first, followed by large arrivals, which in turn are followed by heavy arrivals. The whole process can be outlined in the form of the following pseudo-code:

Algorithm 3: Heuristic for generating tighter time-windows
1:	Set $\mathcal {P}\leftarrow \mathcal {F}$.
2:	If $\mathcal {P}= \emptyset$, go to Step 6; else, select any flight i from $\mathcal {P}$, update $\mathcal {P}\leftarrow \mathcal {P}\setminus \{i\}$.
3:	Apply variable fixing strategies to get ${F}_i^{+}$ and ${F}_i^{-}$.
4:	Relax time-windows and CPS requirements, and then using the group clustering method calculate the makespan $z^+$ and $z^-$ for sets $\mathcal {F}_i^{+}\cup \{i\}$ and $\mathcal {F}_i^{-}\cup \{i\}$, respectively.
5:	Calculate: $E^{new}_i= \max \{E_i,E_1+ z^+\}$ and $L^{new}_i=\min \{L_i, L_n- z^-\},~\text {where}~n=|F|$. Go to Step 2.
6:	Update: $E_i=E^{new}_i, L_i=L^{new}_i~ \forall ~i \in \mathcal {F}$. STOP.

Appendix C Combinatorial exposition of search space reduction

Our motivation in developing the DS-ASP is to reduce the search space significantly by decomposing the overall data set into two smaller sets, each of which are then independently optimized. This leads to an appreciable reduction in the computational time required to find an optimal solution. In the following discussion, we present a combinatorial exposition of the extent of search space reduction; we also prove that the number of flight sequences is the lowest when the cardinality of the two smaller data-sets, namely, $\dot{A}$ and B, is the same, both in the absence, and presence, of CPS constraints.

Consider a set of aircraft of size n, which is split into two subsets of sizes m and $n-m$. Let $I_m$ denote the resulting flight sequences (permutations) after performing data-splitting. Here we refer to the total number of possible permutations that result after fixing the last aircraft in the first subset (refer to Sect. 3 for further details). Clearly, when $m=n$, i.e., in the absence of data-splitting, $I_n$ reduces to the total number of permutations in the overall ASP formulation.

Now, consider the following two cases:

Case 1. (Without CPS constraint) In the absence of data splitting, considering all n flights simultaneously, the total number of permutations clearly is n!. After data-splitting and generating the instance pairs, the size of the search space becomes:

$~~~~~~~~~~~~~~~I_m={n\atopwithdelims ()m}m[(m-1)!+(n-m)!],~3\le m\le n-2$,

which is clearly less than n!. For $3\le m\le n-2$ to be satisfied, n being an even number (to allow for the data-set to be split into two equal halves), we require $n\ge 6$ flights. Indeed, for values of $n\le 5$, the ASP is a trivial problem, and an optimal solution can be easily determined through a brute force enumeration strategy.

Now, consider the following conditions on m:

Case 1.1. (Equal Splitting) Substituting $m=\frac{n}{2}$, and with some algebraic manipulations, we get:

$~~~~~~~~~~~~~~~I_{\frac{n}{2}}=(\frac{n}{2}+1)(\frac{n}{2}+1)(\frac{n}{2}+2)(\frac{n}{2}+3)\cdots (n) < n!$

Case 1.2. (Unequal Splitting) If $m< \frac{n}{2}$, then,

$~~~~~~~~~~~~~~~I_{m}={n\atopwithdelims ()m}m[(m-1)!+(n-m)!] = {n\atopwithdelims ()m}[m!+m(n-m)!]> m(\frac{n!}{m!}) = (m)(m+1)(m+2)\cdots (\frac{n}{2})(\frac{n}{2}+1)\cdots (n)> I_{\frac{n}{2}}$.

Case 2. (With CPS constraint) For simplicity of exposition and ease of notation, consider the case of $k=1$. When $k=1$, the number of pairs ($\dot{A},B$) can at most be three, where the last aircraft of $\dot{A}$ in the first pair, last two aircraft of $\dot{A}$ in the second pair, and, the last aircraft of $\dot{A}$ and the first aircraft of B in the third pair, will not change their positions in their corresponding optimal sequences. Based on this observation, the number of arrangements for the three listed pairs are $f_{m-1}$+$f_{n-m}$, $f_{m-2}$+$f_{n-m}$, and $f_{m-1}$+$f_{n-m-1}$, respectively, where $f_i$ denotes the number of permutations of an $i-sized$ sequence of aircraft. From Krafft and Schaefer (2002), the number of permutations, $f_i$, satisfy the relation $f_i=f_{i-1}+f_{i-2}$, i.e., the number of permutations follow the Fibonacci sequence, and therefore, the total number of permutations in our case is given by $f_{m+1}+f_{n-m+2}$.

Now, consider the following conditions on m:

Case 2.1. $m=\frac{n}{2}$. In this case,

$~~~~~~~~~~~~~~~I_{\frac{n}{2}}=f_{\frac{n}{2}+1}+f_{n-\frac{n}{2}+2}= f_{\frac{n}{2}+3}<f_n\equiv I_n.$

Case 2.2. $m<\frac{n}{2}$. In this case,

$~~~~~~~~~~~~~~~I_m=f_{m+1}+f_{n-m+2} > f_{n-m+2}\ge I_{\frac{n}{2}}$, where the inequality is strict when $m<\frac{n}{2}-1$.

Similarly, it can be shown that the search space is greater than $I_{\frac{n}{2}}$ for the case of $m> \frac{n}{2}$. Note that, while we have only illustrated the reduction in the search space for $k=1$, the same can also be verified for $k\ge 2$. For further details, we refer the reader to Krafft and Schaefer (2002).

Appendix D DS-ASP for arrival traffic: enhancements

Consider two instance pairs ($\dot{A}_1$, B) and ($\dot{A}_2$, B), resulting from the same parent pair (A, B). Let $\hat{\mathcal {F}}$ represent the set of flights in set B between positions $(m+1)$ to $(m+k+1)$, and let $(\dot{z}_1,\dot{d}_{\dot{z}_1},\dot{l}_1)$ and $(\dot{z}_2,\dot{d}_{\dot{z}_2},\dot{l}_2)$ denote the completion time, delay, and last flight in the optimal solutions for set $\dot{A}_1$ and $\dot{A}_2$, respectively. Furthermore, suppose that Set B has been solved to optimality in pair ($\dot{A}_1$, B), with flight $f \in \hat{\mathcal {F}}$ placed at the first position in the optimal solution of set B. Now, consider the following two cases:

Case 1. Suppose that flight f is required to be at first position in an optimal solution of set B. Calculate,

$~~~~~~~~~~~~~~~~~~~~~~~~~~s= (\dot{z}_2-\dot{z}_1)+ (\Delta _{\dot{l}_2f}-\Delta _{\dot{l}_1f})$, and $D=\dot{d}_{\dot{z}_2}-\dot{d}_{\dot{z}_1}+rs$,

where s measures the difference in start times for first flight f in set B between pairs ($\dot{A}_1$, B) and ($\dot{A}_2$, B), and rs ($r=|B|$) is the corresponding minimum increase or maximum decrease in delay. Now, if $D\ge 0$, it is not necessary to optimize set B in the second (sibling) pair, as the optimal delay value of the second pair is atleast equal to optimal value of the first pair.

Case 2. If a particular flight f is not required to be at first position in every optimal solution of set B, then we need to determine the minimum difference in start times with respect to all possible flights which can occupy the first position in set B. For this , calculate

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~s= (\dot{z}_2-\dot{z}_1)+\min \limits _{\hat{f}\in \hat{\mathcal {F}}}(\Delta _{\dot{l}_2\hat{f}}-\Delta _{\dot{l}_1\hat{f}})$,

if $s\ge 0$, it can be inferred that the delay for set B in instance ($\dot{A}_2$, B) is atleast the same as the delay for set B in instance ($\dot{A}_1$, B). Moreover, if $\dot{d}_{\dot{z}_2}-\dot{d}_{\dot{z}_1}\ge 0$, we can fathom the second pair since the optimum value of the second pair is greater than or equal to that of the first pair.

We are also applying the above enhancement to mixed traffic in case the separation of flights in the following set is needed only from the last flight in the leading set. We set $r=\bar{e}$ when $s\ge 0$, else $r=|B|$.