Disruption Recovery at Airports: Ground Holding, Curfew Restrictions and an Approximation Algorithm

Abstract

We study disruptions at a major airport. Disruptions could be caused by bad weather, for example. Our study is from the perspective of the airport, the air services provider (such as air traffic control) and the travelling public, rather than from the perspective of a single airline. Disruptions cause flights to be subjected to ground holding, or they cause the flights to violate airport curfew hours. We consider curfew and arrival capacities applicable at a single airport. After proving that the problem is NP-hard, we present a polynomial time approximation algorithm based on the primal–dual schema and show that if the problem is feasible, the algorithm finds a feasible solution that is both within a certain additive bound and within a certain multiplicative factor of the optimal solution. The algorithm returns a solution mix of which flights suffer no delay, which ones to be ground-held and which ones may violate the curfew (and hence pay a curfew penalty). Computational results are positive; our heuristic outperforms the integer programming solver by a wide margin.

Approximation upper bound for a general primal–dual heuristic

The background presented in this section can be found in books on approximation algorithms for NP-hard problems such as [15, 17]. We present this here in “Appendix” for completeness and quick reference.

Suppose the primal problem is a minimization and the dual problem is a minimization.

(A1)

(A2)

(A3)

(A4)

(A5)

The four sets of constraints are marked as (A4) and (A5), two sets in the primal and two sets in the dual. Assume that in the integer version of the primal problem, it is required that $x_j \in \{0, 1\}$ (binary values). Let

$$\begin{aligned} P_H= & {} \hbox { Value of the heuristic solution to the primal problem (PHS). }\\ D_H= & {} \hbox { Value of the heuristic solution to the dual problem (DHS).}\\ P_\mathrm{opt}= & {} \hbox { Value of the primal integer optimal solution (IPOS).}\\ P_{L}= & {} \hbox { Value of the primal optimal solution to the linear relaxation (LPOS)}\\= & {} \hbox { Value of the dual optimal solution.} \end{aligned}$$

Then,

$$\begin{aligned} D_H \leqslant P_L \leqslant P_\mathrm{opt} \leqslant P_H. \end{aligned}$$

(A6)

Hence, ($P_H - D_H$) is an upper bound on ($P_H - P_\mathrm{opt}$).

Feasibility: We maintain feasibility of the primal and dual feasibility conditions at all times during the execution of the primal–dual algorithm. That is, we should obey constraint sets (A4–A5) above.

Complementary slackness (CS):

To satisfy primal CS, for every $i \in [1, m]$, for the $i^{th}$ primal constraint, we require that

$$\begin{aligned} \left( b_i - \sum \nolimits _{j=1}^n a_{ij} x_j \right) y_i = 0; \end{aligned}$$

To satisfy dual CS, for every $j \in [1, n]$, for the $j^{th}$ dual constraint, we require that

$$\begin{aligned} \left( c_j - \sum \nolimits _{i=1}^m a_{ij} y_i \right) x_j = 0 . \end{aligned}$$

Condition 1

In other words, $y_i > 0 \Rightarrow \sum _{j=1}^n a_{ij} x_j = b_i$  and  $x_j > 0 \Rightarrow \sum _{i=1}^m a_{ij} y_i = c_j$.

Suppose we relax Condition 1 a little. Instead of requiring strict equality, suppose we allow “minor” violations, as follows.

Condition 2

$y_i > 0 \Rightarrow \beta _i b_i \geqslant \sum _{j=1}^n a_{ij} x_j \geqslant b_i$  and  $x_j > 0 \Rightarrow \frac{c_j}{\alpha _j} \leqslant \sum _{i=1}^m a_{ij} y_i \leqslant c_j$, where $\alpha _j$ and $\beta _i$ are constants such that $\alpha _j, \beta _i \geqslant 1$.

Note that Condition 2 still requires feasibility of all constraints in the sets (A4–A5).

Note: If $b_i < 0$, then the first part of the condition changes to $y_i > 0 \Rightarrow \frac{b_i}{\beta _i} \geqslant \sum _{j=1}^n a_{ij} x_j \geqslant b_i$.  Similarly, if $c_j < 0$, the other condition changes to $x_j > 0 \Rightarrow \alpha _j c_j \leqslant \sum _{i=1}^m a_{ij} y_i \leqslant c_j$.

Active sets. Now consider primal integer solutions for which $x_j \in \{0, 1\}$. Consider a primal active set ${\mathcal {A}}_P$, which contains all $j \in [1,n]$ such that $x_j = 1$, as well a dual active set ${\mathcal {A}}_D$ of all $i \in [1,m]$ such that $y_i > 0$. Then, ${\mathcal {A}}_P = \{j \in [1,n] ~|~x_j = 1\}$ and ${\mathcal {A}}_D = \{i \in [1,m] ~|~y_i > 0\}$.

From the weak duality theorem which can be found in most books on linear programming, we know that if X is a feasible solution to the primal problem and Y is a feasible solution to the dual problem, then the objective function values $Z_1$ and $Z_2$ are such that $Z_2 \leqslant Z_1$.  In other words, $Z_2 = \sum _{i=1}^m b_i y_i \leqslant Z_1 = \sum _{j=1}^n c_j x_j$.

Suppose a primal–dual heuristic, which maintains active sets ${\mathcal {A}}_P$ and ${\mathcal {A}}_D$, obtains a primal feasible integer solution with a value of $P_H$ and a dual feasible solution with a value of $D_H$. Applying the weak duality theorem to these feasible solutions and recalling that $j \in {\mathcal {A}}_P \Rightarrow x_j = 1$, we get

$$\begin{aligned} D_H = \sum _{i \in {\mathcal {A}}_D} b_i y_i \leqslant P_H = \sum _{j \in {\mathcal {A}}_P} c_j x_j = \sum _{j \in {\mathcal {A}}_P} c_j(1) = \sum _{j \in {\mathcal {A}}_P} c_j . \end{aligned}$$

From Condition 2, we know that $c_j \leqslant \alpha _j \sum _{i=1}^m a_{ij} y_i = \alpha _j \sum _{i \in {\mathcal {A}}_D} a_{ij} y_i$.

$$\begin{aligned} D_H = \sum _{i \in {\mathcal {A}}_D} b_i y_i \leqslant P_H = \sum _{j \in {\mathcal {A}}_P} c_j \leqslant \sum _{j \in {\mathcal {A}}_P} \alpha _j \sum _{i \in {\mathcal {A}}_D} a_{ij} y_i = \sum _{i \in {\mathcal {A}}_D} y_i \sum _{j \in {\mathcal {A}}_P} \alpha _j a_{ij} = D_I. \end{aligned}$$

(A7)

Observe that in the last but one equality above, we have swapped the i summation with the j summation. Continuing with $D_I$:

$$\begin{aligned} D_I \leqslant \sum _{i \in {\mathcal {A}}_D} y_i \alpha _i^{*} \sum _{j \in {\mathcal {A}}_P} a_{ij} = \alpha ^+ \sum _{i \in {\mathcal {A}}_D} y_i \sum _{j \in {\mathcal {A}}_P} a_{ij}, \end{aligned}$$

(A8)

where $\alpha _i^{*}$ and $\alpha ^+$ are suitable values such that $\alpha _{\min } \leqslant \alpha _i^{*}, \alpha ^+ \leqslant \alpha _{\max }$, where $\alpha _{\max } = \max _{1 \leqslant j \leqslant n} \{\alpha _j\}$ and $\alpha _{\min } = \min _{1 \leqslant j \leqslant n} \{\alpha _j\}$. Similarly, let $\beta _{\max } = \max _{1 \leqslant i \leqslant m} \{\beta _i\}$ and $\beta _{\min } = \min _{1 \leqslant i \leqslant m} \{\beta _i\}$.

For instance, if every $a_{ij}$ in the summation $\sum _{j \in {\mathcal {A}}_P} \alpha _j a_{ij}$ is negative (positive), then we choose $\alpha _i^{*} = \min _{j \in {\mathcal {A}}_P} \{\alpha _j\}$ $\left( \alpha _i^{*} = \max _{j \in {\mathcal {A}}_P} \{\alpha _j\} \right)$ when we write $\sum _{j \in {\mathcal {A}}_P} \alpha _j a_{ij} \leqslant \alpha _i^{*} \sum _{j \in {\mathcal {A}}_P} a_{ij}$. Otherwise, we pick a suitable value for $\alpha _i^{*}$ between $\alpha _{\min }$ and $\alpha _{\max }$. Similarly, the values $y_i$ in (A8) determine our choice for $\alpha ^+$.

Combine (A7) and (A8) to get

$$\begin{aligned} D_H = \sum _{i \in {\mathcal {A}}_D} b_i y_i \leqslant P_H = \sum _{j \in {\mathcal {A}}_P} c_j \leqslant \alpha ^+ \sum _{i \in {\mathcal {A}}_D} y_i \sum _{j \in {\mathcal {A}}_P} a_{ij}. \end{aligned}$$

(A9)

Let us now apply the primal active set ${\mathcal {A}}_P$ to Condition 2. The first part of Condition 2 can be rewritten as

$y_i > 0 \Rightarrow \beta _i b_i \geqslant \sum _{j=1}^n a_{ij} x_j = \sum _{j \in {\mathcal {A}}_P} a_{ij} \geqslant b_i$. Applying this to (A9), we get

$$\begin{aligned} D_H= & {} \sum _{i \in {\mathcal {A}}_D} b_i y_i \leqslant P_H = \sum _{j \in {\mathcal {A}}_P} c_j \leqslant \alpha ^+ \sum _{i \in {\mathcal {A}}_D} y_i \left( \sum _{j \in {\mathcal {A}}_P} a_{ij} \right) \leqslant \alpha ^+ \sum _{i \in {\mathcal {A}}_D} y_i \left( \beta _i b_i \right) \nonumber \\\leqslant & {} \alpha ^+ \beta ^+ \sum _{i \in {\mathcal {A}}_D} y_i b_i = \alpha ^+ \beta ^+ \sum _{i \in {\mathcal {A}}_D} y_i b_i = \alpha ^+ \beta ^+ D_H , \end{aligned}$$

(A10)

where the value of $\beta ^+$ depends on the set of values $\{y_ib_i: 1 \leqslant i \leqslant m \}$, and $\beta ^+$ is suitably chosen between $\beta _{\min }$ and $\beta _{\max }$.

Hence, from (A10), and then applying ($D_H \leqslant P_{\mathrm{opt}}$) from (A6), we can state:

$$\begin{aligned} P_H \leqslant \alpha ^+ \beta ^+ D_H \leqslant \alpha ^+ \beta ^+ P_\mathrm{opt}. \end{aligned}$$

That is, the approximation ratio $\frac{P_H}{P_\mathrm{opt}}$ achieved by the heuristic is at most   $\alpha ^+ \beta ^+$.

As with many primal–dual heuristics, if $\alpha _{\max } = 1$ is maintained, the approximation ratio is guaranteed to be at most $\beta ^+$.