Deterministic#
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import differential_evolution
import scipy as sp
def reliability(beta0_b, d_b, L_b, Thoriz):
# --- Reliability estimation
# beta_bt: reliability curve over time for bridge b
# beta0_b: Initial reliability of bridge b at t=0
# d_b, L_b: when the degradation starts and reaches beta=0 for each bridge
# Thoriz: maximum time studied
nb = len(beta0_b)
beta_bt = np.zeros((nb, Thoriz))
for b in range(nb):
#horziontal branch
timesteps = int(np.floor(d_b[b]))
beta_bt[b, :timesteps] = beta0_b[b]
#degradation till L_b
servicelife = int(np.floor(L_b[b]))
t = np.arange(timesteps + 1, servicelife + 1)
beta_bt[b, timesteps:servicelife] = beta0_b[b] / (L_b[b] - d_b[b]) * (L_b[b] - t)
return beta_bt
def MaintenanceApplication(beta_bt, X_bt, Thoriz):
# --- Given a maintenance strategy given by X_bt, where 1 means that bridge
# b undergoes maintenance at time t, this function gives the updated
# reliability profile
updatedbeta_bt = beta_bt.copy()
InterventionTimes = np.where(np.sum(X_bt, axis=0) > 0)[0]
for t in InterventionTimes:
intervbrid = np.where(X_bt[:, t] == 1)[0]
timecells = Thoriz - t - 1
updatedbeta_bt[intervbrid, t+1:Thoriz] = beta_bt[intervbrid, :timecells]
return updatedbeta_bt
def Frequency2Schedule(y_b, nb, nt, Thoriz):
# Transforms a vector indicating the frequency to a schedule matrix [0/1]
freq_b = y_b.reshape((nt, nb)).T
cumFreq_b = cumulateFreq(freq_b)
X_bt = np.zeros((nb, Thoriz))
for b in range(nb):
idx = cumFreq_b[b, cumFreq_b[b, :] != 0].astype(int)
idx = idx[idx <= Thoriz]
if len(idx) > 0:
X_bt[b, idx - 1] = 1
return X_bt
def cumulateFreq(freq_b):
# creates the cumulative frequency
nb, nt = freq_b.shape
cumFreq_b = np.zeros((nb, nt))
for b in range(nb):
f_b = freq_b[b, freq_b[b, :] != 0]
cumFreq_b[b, :len(f_b)] = np.cumsum(f_b)
return cumFreq_b
def CountNinterv(y_b, nt, nb):
# from vector y_b, it computes the number of interventions per bridge
y_b01 = y_b != 0
freq_b = y_b01.reshape((nt, nb)).T
ninterv_b = np.sum(freq_b, axis=1)
return ninterv_b
Thoriz = 25
# for each bridge b: beta0 d(years) L(years) C(Meuros)
data = np.array([
[1, 5, 10, 0.6],
[1, 6, 11, 0.8],
[1, 7, 15, 1.0],
[1, 10, 20, 0.7]
])
beta0_b, d_b, L_b, C_b = data[:, 0], data[:, 1], data[:, 2], data[:, 3]
# Annual resources (M euro)
R = 1.4
# Initial reliability estimation
nb = len(beta0_b)
beta_bt0 = reliability(beta0_b, d_b, L_b, Thoriz)
plt.figure()
col = ['r', 'b', 'g', 'm']
for b in range(nb):
plt.plot(range(1, Thoriz + 1), beta_bt0[b, :], f'-{col[b]}', linewidth=2)
plt.grid()
plt.xlim([1, Thoriz])
plt.ylim([0, 1])
plt.xlabel('Time, t (years)')
plt.ylabel('Reliability index, β')
plt.title('Reliability Index Over Time')
plt.show()
# Deterministic Optimization
# Variables -> e.g., freq=[10 2 0 0 ; 5, 5 0 0; 0 0 0 0];
maxInter_b = np.ceil(Thoriz / L_b).astype(int) + 1 #max number of interventions expected per bridge
nt = max(maxInter_b) #to dimension the matrix, max interventions experienced by the worst case.
nvars = nb * nt #number of design variables
integers = np.array([True] * nvars) #all variables are integers
# Range of definition: [0,max number of years with no interventions==L_b]
lb = np.zeros(nvars)
ub = np.zeros(nvars)
pos = 0
for b in range(nb):
# This guarantees interventions freq. within the service life (if conducted)
ub[pos:pos + maxInter_b[b]] = L_b[b] - 1
# this forces zeros after the max number of interventions expected
if maxInter_b[b] < nb:
ub[pos + maxInter_b[b]:pos + nb - 1] = 0
pos += nt
def nonlcon(y_b):
# From frequency to a schedule matrix [0/1]
X_bt = Frequency2Schedule(y_b, nb, nt, Thoriz)
# Constraint 1 >>> the last intervention is within Thoriz --> LastInterv<Thoriz --> LastInterv-Thoriz<=1 for all b
lastInterv_b = np.max(cumulateFreq(y_b.reshape((nt, nb)).T), axis=1)
c1 = lastInterv_b - Thoriz - 1
# Constraint 2 >>> C_b*X_bt <= R for all t
c2 = np.sum(C_b[:, None] * X_bt, axis=0) - R
# Constraint 3 >>> beta_bt>0 --> sum(sum(beta_bt))<=0
beta_bt = MaintenanceApplication(beta_bt0, X_bt, Thoriz)
c3 = np.array([np.sum(beta_bt == 0)])
c = np.concatenate([c1, c2, c3])
return c
cons = sp.optimize.NonlinearConstraint(nonlcon, np.array([-np.inf] * (nb + Thoriz + 1)), np.array([0] * (nb + Thoriz + 1)))
def fun(y_b):
# minimize cost trying to go for the largest time between interventions (== max y_b)
ninterv_b = CountNinterv(y_b, nt, nb)
return np.dot(C_b, ninterv_b)
bounds = [[lb[i], ub[i]] for i in range(nvars)]
result = differential_evolution(fun, bounds, constraints=cons, integrality=integers, disp = True)
print(result)
differential_evolution step 1: f(x)= 11.0
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differential_evolution step 3: f(x)= 8.200000000000001
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differential_evolution step 417: f(x)= 5.1
differential_evolution step 418: f(x)= 5.1
differential_evolution step 419: f(x)= 4.5
differential_evolution step 420: f(x)= 4.5
differential_evolution step 421: f(x)= 4.5
differential_evolution step 422: f(x)= 4.5
differential_evolution step 423: f(x)= 4.5
differential_evolution step 424: f(x)= 4.5
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differential_evolution step 480: f(x)= 4.5
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differential_evolution step 484: f(x)= 4.5
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differential_evolution step 488: f(x)= 4.5
differential_evolution step 489: f(x)= 4.5
differential_evolution step 490: f(x)= 4.5
differential_evolution step 491: f(x)= 4.5
differential_evolution step 492: f(x)= 4.5
differential_evolution step 493: f(x)= 4.5
differential_evolution step 494: f(x)= 4.5
differential_evolution step 495: f(x)= 4.5
differential_evolution step 496: f(x)= 4.5
differential_evolution step 497: f(x)= 4.5
differential_evolution step 498: f(x)= 4.5
differential_evolution step 499: f(x)= 4.5
differential_evolution step 500: f(x)= 4.5
differential_evolution step 501: f(x)= 4.5
differential_evolution step 502: f(x)= 4.5
differential_evolution step 503: f(x)= 4.5
differential_evolution step 504: f(x)= 4.5
differential_evolution step 505: f(x)= 4.5
differential_evolution step 506: f(x)= 4.5
differential_evolution step 507: f(x)= 4.5
differential_evolution step 508: f(x)= 4.5
differential_evolution step 509: f(x)= 4.5
differential_evolution step 510: f(x)= 4.5
differential_evolution step 511: f(x)= 4.5
differential_evolution step 512: f(x)= 4.5
differential_evolution step 513: f(x)= 4.5
differential_evolution step 514: f(x)= 4.5
differential_evolution step 515: f(x)= 4.5
differential_evolution step 516: f(x)= 4.5
differential_evolution step 517: f(x)= 4.5
message: Optimization terminated successfully.
success: True
fun: 4.5
x: [ 0.000e+00 8.000e+00 ... 1.400e+01 0.000e+00]
nit: 517
nfev: 17373
population: [[ 0.000e+00 8.000e+00 ... 1.400e+01 0.000e+00]
[ 0.000e+00 7.000e+00 ... 1.400e+01 0.000e+00]
...
[ 0.000e+00 8.000e+00 ... 1.300e+01 0.000e+00]
[ 0.000e+00 9.000e+00 ... 1.400e+01 0.000e+00]]
population_energies: [ 4.500e+00 4.500e+00 ... 4.500e+00 4.500e+00]
constr: [array([ 0.000e+00, 0.000e+00, ..., 0.000e+00,
0.000e+00])]
constr_violation: 0.0
maxcv: 0.0
Yopt = result.x
OptInterv = cumulateFreq(Yopt.reshape((nt, nb)).T)
X_bt = Frequency2Schedule(Yopt, nb, nt, Thoriz)
beta_bt = MaintenanceApplication(beta_bt0, X_bt, Thoriz)
print(f'Min Cost = {result.fun:.2f}')
ninterv = np.sum(CountNinterv(Yopt, nt, nb))
print(f'Number of interventions = {ninterv}')
c = nonlcon(Yopt)
BudgetIssues = np.sum(c[-Thoriz - 1:-1] > 0)
print(f'Number of time intervals with exceeding budget = {BudgetIssues}')
reliabIssues = c[-1]
print(f'Number of time intervals with reliability issues = {reliabIssues}')
plt.figure()
h = []
leg = []
for b in range(nb):
h.append(plt.plot(range(1, Thoriz + 1), beta_bt[b, :], f'-{col[b]}', linewidth=2))
leg.append(f'Bridge {b + 1}')
plt.legend(leg, loc='best')
plt.grid()
plt.xlim([1, Thoriz])
plt.ylim([0, 1])
plt.xlabel('Time, t (years)')
plt.ylabel('Reliability index, β')
plt.title('Reliability Index Over Time')
plt.show()
Min Cost = 4.50
Number of interventions = 6
Number of time intervals with exceeding budget = 0
Number of time intervals with reliability issues = 0.0
