示例#1
0
def build_dqm(W, C, n, p, a, verbose=True):
    """Builds discrete quadratic model representing the optimization problem.

    Args:
        - W: Numpy matrix. Represents passenger demand. Normalized with total demand equal to 1.
        - C: Numpy matrix. Represents airline leg cost.
        - n: Int. Number of cities in play.
        - p: Int. Number of hubs airports allowed.
        - a: Float in [0.0, 1.0]. Discount allowed for hub-hub legs.
        - verbose: Print to command-line for user.

    Returns:
        - dqm: DiscreteQuadraticModel representing the optimization problem.
    """

    if verbose:
        print("\nBuilding DQM...\n")

    # Initialize DQM object.
    dqm = DiscreteQuadraticModel()
    for i in range(n):
        dqm.add_variable(n, label=i)

    # Objective: Minimize cost.
    for i in range(n):
        for j in range(n):
            for k in range(n):
                dqm.set_linear_case(
                    i, k,
                    dqm.get_linear_case(i, k) + C[i][k] * W[i][j])
                dqm.set_linear_case(
                    j, k,
                    dqm.get_linear_case(j, k) + C[j][k] * W[i][j])
                for m in range(n):
                    if i != j:
                        dqm.set_quadratic_case(i, k, j, m,
                                               a * C[k][m] * W[i][j])

    # Constraint: Every leg must connect to a hub.
    gamma1 = 150
    for i in range(n):
        for j in range(n):
            if i != j:
                dqm.set_linear_case(i, j,
                                    dqm.get_linear_case(i, j) + 1 * gamma1)
                dqm.set_quadratic_case(
                    i, j, j, j,
                    dqm.get_quadratic_case(i, j, j, j) - 1 * gamma1)

    # Constraint: Exactly p hubs required.
    gamma2 = 250
    for i in range(n):
        dqm.set_linear_case(i, i,
                            dqm.get_linear_case(i, i) + (1 - 2 * p) * gamma2)
        for j in range(i + 1, n):
            dqm.set_quadratic_case(
                i, i, j, j,
                dqm.get_quadratic_case(i, i, j, j) + 2 * gamma2)

    return dqm
示例#2
0
文件: gc_solver.py 项目: adadima/QUBO
def solve(adj_nodes, adj_edges, even_dist=False):

    dqm = DiscreteQuadraticModel()

    for i in adj_nodes:
        dqm.add_variable(NURSES, label=i)

    # classic graph coloring constraint that no two adjacent nodes have the same color
    for i0, i1 in adj_edges:
        dqm.set_quadratic(i0, i1, {(c, c): LAGRANGE for c in range(NURSES)})

    shifts_per_nurse = DAYS * SHIFTS // NURSES

    if even_dist:
        # we should ensure that nurses get assigned a roughly equal amount of work
        for i in range(NURSES):
            for index, j in enumerate(adj_nodes):

                dqm.set_linear_case(
                    j, i,
                    dqm.get_linear_case(j, i) - LAGRANGE *
                    (2 * shifts_per_nurse + 1))

                for k_index in range(index + 1, len(adj_nodes)):
                    k = adj_nodes[k_index]
                    dqm.set_quadratic_case(
                        j, i, k, i,
                        LAGRANGE * (dqm.get_quadratic_case(j, i, k, i) + 2))

    # some nurses may hate each other, so we should do out best to not put them in the same shift!
    for d in range(DAYS):
        for s in range(SHIFTS):
            for l1 in range(NURSES_PER_SHIFT):
                for l2 in range(l1 + 1, NURSES_PER_SHIFT):

                    j = f"l{l1}_d{d}_s{s}"
                    k = f"l{l2}_d{d}_s{s}"

                    for conflict in CONFLICTS:

                        for n1 in conflict:
                            for n2 in conflict:

                                if n1 == n2:
                                    continue

                                dqm.set_quadratic_case(
                                    j, n1, k, n2,
                                    LAGRANGE2 *
                                    (dqm.get_quadratic_case(j, n1, k, n2) +
                                     10))

    sampler = LeapHybridDQMSampler(token=API_TOKEN)
    sampleset = sampler.sample_dqm(dqm, time_limit=10)
    sample = sampleset.first.sample
    energy = sampleset.first.energy

    return sample, energy
示例#3
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文件: scheduler.py 项目: RodTol/Tesi
preferences = preferences[rows, cols]

# Initialize the DQM object
dqm = DiscreteQuadraticModel()

# Build the DQM starting by adding variables
for name in range(num_employees):
    dqm.add_variable(num_shifts, label=name)

# Distribute employees equally across shifts according to preferences
num_per_shift = int(num_employees/num_shifts)
gamma = 1/(num_employees*num_shifts)

for i in range(num_shifts):
    for j in range(num_employees):
        dqm.set_linear_case(j, i, dqm.get_linear_case(j, i) - gamma*(2*num_per_shift+1))
        for k in range(j+1, num_employees):
            dqm.set_quadratic_case(j, i, k, i, gamma*(dqm.get_quadratic_case(j, i, k, i) + 2))

# Initialize the DQM solver
sampler = LeapHybridDQMSampler()

# Solve the problem using the DQM solver
sampleset = sampler.sample_dqm(dqm, label='Example - Employee Scheduling')

# Get the first solution, and print it
# Mi da la soluzione migliore, quella in cui sono caduto più volte e me la salvo
sample = sampleset.first.sample
energy = sampleset.first.energy
print("\nSchedule score:", energy)