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Optimization of Preventative Maintenance Requests

Optimize the timing of preventative maintenance requests to reduce costs and increase efficient operations

Hosted by

EDF

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Problem

The problem is described by a set of

N

PMRQ (Preventive Maintenance ReQuest). The PMRQ are endowed with the following properties:

cost: $\{ C_1, C_2, ..., C_N\}$
frequency: $\{f_1, f_2, ..., f_N\}$
tolerance: $\{\epsilon_1, \epsilon_2, ..., \epsilon_N\}$
offset: $\{ o_1, o_2, ..., o_N \}$
time horizon: $T$
credit: A dictionary mapping $i \rightarrow \{j_1, j_2 ,..., j_N\}$ where $i,j\in [1,N]$ . The inclusion in the mapping denotes $i$ credits $j$ (i.e. the maintenance of $j$ can be done for free is the maintenance of $i$ is done at the same time)

Solution

A solution is to be given by the binary matrix

S

of size

N \times T

with the entries

S_{it}= \begin{cases} 0, \quad \text{if no maintenance is performed on PMRQ-i at time } t\\ 1, \quad \text{if a maintenance is performed on PMRQ-i at time } t\\ \end{cases}

Template

You can download a solution template by running the following command in the terminal

aqora template edf-pmrq
cd edf-pmrq
aqora install

You can then fill in src/submission/__init__.py with your solution. You can test your solution by running

aqora test

And finally you can upload your solution by running

aqora upload

Scoring

Each solution is validated and scored according to the following script

class Problem:
    id: int
    cost: np.ndarray
    frequency: np.ndarray
    tolerance: np.ndarray
    offset: np.ndarray
    T: int
    credit: dict
    epsilon: np.ndarray

    def __init__(self, cost, frequency, tolerance, offset, credit, T):
        self.id = id
        self.cost = np.array(cost)
        self.frequency = np.array(frequency)
        self.tolerance = np.array(tolerance)
        self.offset = np.array(offset)
        self.T = T
        self.credit = credit
        self.epsilon = self.calc_epsilon(self.cost, credit)

    def assert_valid(self, df: pd.DataFrame):
        # Iterate over each row in the DataFrame
        for row in df.index:
            # Get the column indices where the row equals 1
            ones_indices = df.loc[row][df.loc[row] == 1].index.tolist()

            # this asserts whether the time between first maintenance and the previous one verifies the frequency and tolerance constraint
            assert (
                ones_indices[0] - self.offset[row] - 1
                <= self.frequency[row] + self.tolerance[row]
            )
            # this asserts whether the time between consecutive maintenance verifies the frequency and tolerance constraint
            diffs = np.diff(ones_indices)
            assert np.all(np.abs(diffs - self.frequency[row]) <= self.tolerance[row])
            # this asserts whether the last maintenance and the time horizon do not violate the frequency and tolerance constraint
            assert (
                self.T - ones_indices[-1] <= self.frequency[row] + self.tolerance[row]
            )

    def calc_cost(self, input: np.ndarray) -> float:
        return np.trace(self.epsilon @ input @ input.T)

    @classmethod
    def calc_epsilon(cls, cost: np.ndarray, credit: dict) -> np.ndarray:
        # Define the epsilon matrix
        epsilon = np.zeros((len(cost), len(cost)))

        # Populate the epsilon matrix with cost values. The value of the coefficient epsilon[i,j] is the cost of j which is free is the maintenance of i is done at the same time as j
        for _, key in enumerate(credit):
            for j in credit[key]:
                epsilon[int(key) - 1, j - 1] = cost[j - 1]

        return epsilon