A network flow-based algorithm for bus driver rerostering

Abstract

Bus driver rostering generates the work plan for a pool of drivers during a planning period of predefined length. This plan, called the roster, must consider the balance between the pressure of costs, the provision of a service of high quality, labour agreements, and the goodwill of the workers. The bus driver rerostering problem occurs during real-time operational planning, when unexpected events—such as non-planned absences of drivers—disrupt the roster. To reconstruct a roster which is originally built in a context of days off schedules for drivers, we propose a reactive methodology based on a multicommodity flow assignment mixed integer linear programming model. The objective is to minimise the number of depot drivers who are assigned to drive and the number of postponed days off, as well as the dissimilarity between the reconstructed and the original roster and the balancing of the workload. The proposed algorithm enables the disrupted roster to be reconstructed at the expense of a relatively small number of changes in drivers’ work and rest periods, while, at the same time, controlling the dimension of the multicommodity flow network. Computational experience based on real-life based instances revealed that the algorithm has the ability to produce reconstructed rosters with few changes to the drivers’ original work assignment, in a short CPU time.

Appendix

Example of the random generation of absences according to Scenario C1 for the real roster P6 instance.

• (1) Randomly generate r according to a uniform distribution U[0,1]. If $r > 0.5$ then NAD = 1 else NAD = 2.

  $${\text{Instance}}\;P6{:}\;{\text{NAD}} = 2.$$

• (2) Defining the first absent driver

  Driver $i \in \overline{M}$ is selected according to a discrete uniform distribution, taking values on {1,…, $\left| {\overline{M}} \right|$}

  $${\text{Instance}}\;P6{:}\;i = 47$$

  • (i) Defining the first day of the absence period of driver i:

    Let $d_{i,1}$ be the first day of absence for driver i, $d_{i,1}$ is randomly generated according to a discrete uniform distribution, taking values on {8,…,14};

    $${\text{Instance}}\;P6{:}\;d_{i,1} = 10$$

  • (ii) Defining the number of consecutive absent days for driver i:

    Let X be a random variable which represents the number of consecutive absent days, and let $dc_{i}$ be the number of consecutive days that driver i will be absent. Number $dc_{i}$ is randomly generated according to the following distribution:

    x i	1	2	3	4	7	14
    P [X = xi]	0.2	0.2	0.2	0.2	0.1	0.1

    Instance P6: $dc_{i} = 4$. This means that driver 47 will be absent on days 10, 11, 12, and 13.

    If NAD = 1 stop.

• (3) Defining the second absent driver

  Driver $j \in \overline{M}$ is selected according to a discrete uniform distribution, taking values on {1,…, $\left| {\overline{M}} \right|$}\{i}.

  $${\text{Instance}}\;P6{:}\; j = 20$$

  • (i) Defining the first day of the absence period for driver j

    Let $d_{j,1}$ be the first day of absence for driver j, $d_{j,1}$ is randomly generated according to a discrete uniform distribution, taking values on {8,…,49};

    $${\text{Instance}}\;P6{:}\;d_{j,1} = 16$$

  • (ii) Defining the number of consecutive absent days for driver j

    Let X be a random variable that represents the number of consecutive absent days, and let $dc_{j}$ be the number of consecutive days driver j will be absent. The probability distribution of X depends on the value of $d_{j,1} { }$. Therefore, $dc_{j}$ is randomly generated according to the following distribution:

  • If $\left| H \right| - d_{j,1} + 1 \ge 14$:

    xj	1	2	3	4	7	14
    P [X = xj]	0.2	0.2	0.2	0.2	0.1	0.1

    Instance P6: $\left| H \right| - d_{j,1} + 1 = 49 - 16 + 1 \ge 14$ and we obtain $dc_{j}$ = 3. This means that driver 20 will be absent on days 16, 17, and 18.

  • Elseif $\left| H \right| - d_{j,1} + 1 \ge 7$:

    xj	1	2	3	4	5	7
    P [X = xj]	0.2	0.2	0.2	0.2	0.1	0.1

  • Elseif $\left| H \right| - d_{j,1} + 1 < 7$:

x j	1	…	$\left| H \right| - d_{j,1}$
P [X = xj]	1/($\left| H \right| - d_{j,1}$)	1/($\left| H \right| - d_{j,1}$)	1/($\left| H \right| - d_{j,1}$)