I have noticed that when using Pyomo + Ipopt, some optimization dae problems that converge to an optimal solution, when expanded in complexity (e.g. larger distance in a car example) and consequetly in the number of finite elements to keep accuracy, the solver displays:
EXIT: Solved To Acceptable Level.
instead of the previous "Optimal solution found".
As an example of stated above, I will use a modified code of "ampl car sample" from Pyomo repository.
# Ampl Car Example
#
# Shows how to convert a minimize final time optimal control problem
# to a format pyomo.dae can handle by removing the time scaling from
# the ContinuousSet.
#
# min tf
# dxdt = v
# dvdt = a-R*v^2
# x(0)=0; x(tf)=L
# v(0)=0; v(tf)=0
# -3<=a<=1
from pyomo.environ import *
from pyomo.dae import *
m = ConcreteModel()
m.R = Param(initialize=0.001) # Friction factor
m.L = Param(initialize=1000000.0) # Final position
m.tau = ContinuousSet(bounds=(0,1)) # Unscaled time
m.time = Var(m.tau) # Scaled time
m.tf = Var()
m.x = Var(m.tau,bounds=(0,m.L+50))
m.v = Var(m.tau,bounds=(0,None))
m.a = Var(m.tau, bounds=(-3.0,1.0),initialize=0)
m.dtime = DerivativeVar(m.time)
m.dx = DerivativeVar(m.x)
m.dv = DerivativeVar(m.v)
m.obj = Objective(expr=m.tf)
def _ode1(m,i):
if i == 0 :
return Constraint.Skip
return m.dx[i] == m.tf * m.v[i]
m.ode1 = Constraint(m.tau, rule=_ode1)
def _ode2(m,i):
if i == 0 :
return Constraint.Skip
return m.dv[i] == m.tf*(m.a[i] - m.R*m.v[i]**2)
m.ode2 = Constraint(m.tau, rule=_ode2)
def _ode3(m,i):
if i == 0:
return Constraint.Skip
return m.dtime[i] == m.tf
m.ode3 = Constraint(m.tau, rule=_ode3)
def _init(m):
yield m.x[0] == 0
yield m.x[1] == m.L
yield m.v[0] == 0
yield m.v[1] == 0
yield m.time[0] == 0
m.initcon = ConstraintList(rule=_init)
discretizer = TransformationFactory('dae.finite_difference')
discretizer.apply_to(m,nfe=5000,scheme='BACKWARD')
solver = SolverFactory('ipopt')
solver.solve(m,tee=True)
print("final time = %6.2f" %(value(m.tf)))
x = []
v = []
a = []
time=[]
for i in m.tau:
time.append(value(m.time[i]))
x.append(value(m.x[i]))
v.append(value(m.v[i]))
a.append(value(m.a[i]))
import matplotlib.pyplot as plt
plt.subplot(131)
plt.plot(time,x,label='x')
plt.title('location')
plt.xlabel('time')
plt.subplot(132)
plt.plot(time,v,label='v')
plt.xlabel('time')
plt.title('velocity')
plt.subplot(133)
plt.plot(time,a,label='a')
plt.xlabel('time')
plt.title('acceleration')
plt.show()
NOTE: The original source code can be colsulted here to compare with mine modified: https://github.com/Pyomo/pyomo/blob/main/examples/dae/car_example.py
Is there anything I can do about this? May I lower the ipopt tolerance so it keeps finding for an optimal solution?
You can disable the heuristic that makes Ipopt stop with an "acceptable" solution by setting option acceptable_iter to 0. See https://coin-or.github.io/Ipopt/OPTIONS.html#OPT_Termination for all options that determine termination of Ipopt.
Related
I am trying to validate the findings of a paper by testing it on the same model architecture as well as the same dataset reported by the paper. I have been using the imagenet script provided in the official pytorch repository's examples section to do the same.
class AverageMeter(object):
"""Computes and stores the average and current value
Imported from https://github.com/pytorch/examples/blob/master/imagenet/main.py#L247-L262
"""
def init(self):
self.reset()
def reset(self):
self.val = 0
self.avg = 0
self.sum = 0
self.count = 0
def update(self, val, n=1):
self.val = val
self.sum += val * n
self.count += n
self.avg = self.sum / self.count
def accuracy(output, target, topk=(1,)):
"""Computes the precision#k for the specified values of k"""
maxk = max(topk)
batchsize = target.size(0)
, pred = output.topk(maxk, 1, True, True)
pred = pred.t()
correct = pred.eq(target.view(1, -1).expand_as(pred))
res = []
for k in topk:
correct_k = correct[:k].reshape(-1).float().sum(0)
res.append(correctk.mul(100.0 / batch_size))
return res
top1 = AverageMeter()
top5 = AverageMeter()
# switch to evaluate mode
model.eval()
with torch.no_grad():
for batch_idx, (inputs, targets) in enumerate(test_loader):
# measure data loading time
print(f"Processing {batch_idx+1}/{len(test_loader)}")
inputs, targets = inputs.cuda(), targets.cuda()
inputs, targets = torch.autograd.Variable(inputs, volatile=True), torch.autograd.Variable(targets)
# compute output
outputs = model(inputs)
# measure accuracy and record loss
prec1, prec5 = accuracy(outputs.data, targets.data, topk=(1, 5))
print(prec1,prec5)
top1.update(prec1.item(), inputs.size(0))
top5.update(prec5.item(), inputs.size(0))
print(top1)
print(top5)
However the top 5 error which I am getting by using this script is not matching with the one in the paper. Can anyone tell me what is wrong in this particular snippet?
I wanna get all integer solutions in a limited time, is it possible?
This is a linear, integer constraint satisfaction problem, which can be solved efficiently by OR Tools' CP-SAT. I've modified their example to solve your problem in Python:
from ortools.sat.python import cp_model
class VarArraySolutionPrinter(cp_model.CpSolverSolutionCallback):
"""Print intermediate solutions."""
def __init__(self, variables):
cp_model.CpSolverSolutionCallback.__init__(self)
self.__variables = variables
self.__solution_count = 0
def on_solution_callback(self):
self.__solution_count += 1
for v in self.__variables:
print('%s=%i' % (v, self.Value(v)), end=' ')
print()
def solution_count(self):
return self.__solution_count
def SearchForAllSolutionsSampleSat():
"""Showcases calling the solver to search for all solutions."""
# Creates the model.
model = cp_model.CpModel()
p = [1, 2, 3, 4]
ceq = 30
cgeq = 2
N = len(p)
# Creates the variables
x = [model.NewIntVar(0, 100, f'x{i}') for i in range(N)]
# Create the constraints.
model.Add(sum([xi*pi for xi, pi in zip(x, p)]) == ceq)
model.Add(sum(x) >= cgeq)
# Create a solver and solve.
solver = cp_model.CpSolver()
solution_printer = VarArraySolutionPrinter(x)
status = solver.SearchForAllSolutions(model, solution_printer)
print('Status = %s' % solver.StatusName(status))
print('Number of solutions found: %i' % solution_printer.solution_count())
SearchForAllSolutionsSampleSat()
I was trying to plot a loss curve, but is always abnormal (just like a circle, I really don't know how to describe it in English properly), I had found many topics about question like this and just can't solve, my tensorflow version is 0.10.0.
import tensorflow as tf
from tensorflow.core.util.event_pb2 import SessionLog
import os
# initialize variables/model parameters
# define the training loop operations
def inputs():
# read/generate input training data X and expected outputs Y
weight_age = [[84,46],[73,20],[65,52],[70,30],[76,57],[69,25],[63,28],[72,36],[79,57],[75,44],[27,24]
,[89,31],[65,52],[57,23],[59,60],[69,48],[60,34],[79,51],[75,50],[82,34],[59,46],[67,23],
[85,37],[55,40],[63,30]]
blodd_fat_content = [354,190,405,263,451,302,288,385,402,365,209,290,346,
254,395,434,220,374,308,220,311,181,274,303,244]
return tf.to_float(weight_age), tf.to_float(blodd_fat_content)
def inference(X):
# compute inference model over data X and return the result
return tf.matmul(X, W) + b
def loss(X, Y):
# compute loss over training data X and expected outputs Y
Y_predicted = inference(X)
return tf.reduce_sum(tf.squared_difference(Y, Y_predicted))
def train(total_loss):
# train / adjust model parameters according to computed total loss
learning_rate = 1e-7
return tf.train.GradientDescentOptimizer(learning_rate).minimize(total_loss)
def evaluate(sess, X, Y):
# evaluate the resulting trained model
print (sess.run(inference([[80., 25.]])))
print (sess.run(inference([[60., 25.]])))
g1 = tf.Graph()
with tf.Session(graph=g1) as sess:
W = tf.Variable(tf.zeros([2,1]), name="weights")
b = tf.Variable(0., name="bias")
tf.initialize_all_variables().run()
X, Y = inputs()
print (sess.run(W))
total_loss = loss(X, Y)
train_op = train(total_loss)
tf.scalar_summary("loss", total_loss)
summaries = tf.merge_all_summaries()
coord = tf.train.Coordinator()
threads = tf.train.start_queue_runners(sess=sess, coord=coord)
summary_writer = tf.train.SummaryWriter('linear', g1)
summary_writer.add_session_log(session_log= SessionLog(status=SessionLog.START), global_step=1)
# actual training loop
training_steps = 100
tolerance = 100
total_loss_last = 0
initial_step = 0
# Create a saver.
saver = tf.train.Saver()
# verify if we don't have a checkpoint saved already
ckpt = tf.train.get_checkpoint_state(os.path.dirname('my_model'))
if ckpt and ckpt.model_checkpoint_path:
# Restores from checkpoint
saver.restore(sess, ckpt.model_checkpoint_path)
initial_step = int(ckpt.model_checkpoint_path.rsplit('-', 1)[1])
# summary_writer.add_session_log(SessionLog(status=SessionLog.START), global_step=initial_step)
for step in range(initial_step, training_steps):
sess.run([train_op])
if step%20 == 0:
saver.save(sess, 'my-model', global_step=step)
gap = abs(sess.run(total_loss) - total_loss_last)
total_loss_last = sess.run(total_loss)
summary_writer.add_summary(sess.run(summaries), step)
# for debugging and learning purposes, see how the loss gets decremented thru training steps
if step % 10 == 0:
print ("loss: ", sess.run([total_loss]))
print("step: ", step)
if gap < tolerance:
break
# evaluation...
evaluate(sess, X, Y)
coord.request_stop()
coord.join(threads)
saver.save(sess, 'my-model', global_step=training_steps)
summary_writer.flush()
sess.close()
I am trying migrating the ampl car problem that comes in the Ipopt source code tarball as example. I am having got problems with the end condition (reach a place with zero speed at final iteration) and with the cost function (minimize final time).
Can someone help me revise the following model?
# min tf
# dx/dt = 0
# dv/dt = a - R*v^2
# x(0) = 0; x(tf) = 100
# v(0) = 0; v(tf) = 0
# -3 <= a <= 1 (a is the control variable)
#!Python3.5
from pyomo.environ import *
from pyomo.dae import *
N = 20;
T = 10;
L = 100;
m = ConcreteModel()
# Parameters
m.R = Param(initialize=0.001)
# Variables
def x_init(m, i):
return i*L/N
m.t = ContinuousSet(bounds=(0,1000))
m.x = Var(m.t, bounds=(0,None), initialize=x_init)
m.v = Var(m.t, bounds=(0,None), initialize=L/T)
m.a = Var(m.t, bounds=(-3.0,1.0), initialize=0)
# Derivatives
m.dxdt = DerivativeVar(m.x, wrt=m.t)
m.dvdt = DerivativeVar(m.v, wrt=m.t)
# Objetives
m.obj = Objective(expr=m.t[N])
# DAE
def _ode1(m, i):
if i==0:
return Constraint.Skip
return m.dxdt[i] == m.v[i]
m.ode1 = Constraint(m.t, rule=_ode1)
def _ode2(m, i):
if i==0:
return Constraint.Skip
return m.dvdt[i] == m.a[i] - m.R*m.v[i]**2
m.ode2 = Constraint(m.t, rule=_ode2)
# Constraints
def _init(m):
yield m.x[0] == 0
yield m.v[0] == 0
yield ConstraintList.End
m.init = ConstraintList(rule=_init)
'''
def _end(m, i):
if i==N:
return m.x[i] == L amd m.v[i] == 0
return Constraint.Skip
m.end = ConstraintList(rule=_end)
'''
# Discretize
discretizer = TransformationFactory('dae.finite_difference')
discretizer.apply_to(m, nfe=N, wrt=m.t, scheme='BACKWARD')
# Solve
solver = SolverFactory('ipopt', executable='C:\\EXTERNOS\\COIN-OR\\win32-msvc12\\bin\\ipopt')
results = solver.solve(m, tee=True)
Currently, a ContinuousSet in Pyomo has to be bounded. This means that in order to solve a minimum time optimal control problem using this tool, the problem must be reformulated to remove the time scaling from the ContinuousSet. In addition, you have to introduce an extra variable to represent the final time. I've added an example to the Pyomo github repository showing how this can be done for your problem.
After having seen the nice implementation of the "ampl car example" in Pyomo repository, I would like to keep extending the problem with new features and constraints, but I have found the next problems during development. Is someone able of fix them?
1) Added new constraint "electric car": Now the acceleration is limited by adherence until a determined speed and then constant power model is used. I am not able of implement this constraint as i would think. It is commented in the, but Pyomo complains about that a constraint is related to a variable. (now Umax depends of the car speed).
2) Added new comfort acceleration and jerk constraints. It seems they are working right, but should be nice if a Pyomo guru supervise them and tell me if they are really implemented in the correct way.
3) About last one, in order of reducing verbosity. Is there any way of combine accelerationL and accelerationU in a unique constraint? Same for jerkL and jerkU.
4) The last feature is a speed limit constraint divided in two steps. Again, I am not able of getting it works, so it is commented in code. Does anybody dare to fix it?
# Ampl Car Example (Extended)
#
# Shows how to convert a minimize final time optimal control problem
# to a format pyomo.dae can handle by removing the time scaling from
# the ContinuousSet.
#
# min tf
# dx/dt = v
# dv/dt = u - R*v^2
# x(0)=0; x(tf)=L
# v(0)=0; v(tf)=0
# -3 <= u <= 1 (engine constraint)
#
# {v <= 7m/s ===> u < 1
# u <= { (electric car constraint)
# {v > 7m/s ===> u < 1*7/v
#
# -1.5 <= dv/dt <= 0.8 (comfort constraint -> smooth driving)
# -0.5 <= d2v/dt2 <= 0.5 (comfort constraint -> jerk)
# v <= Vmax (40 kmh[0-500m] + 25 kmh(500-1000m])
from pyomo.environ import *
from pyomo.dae import *
m = ConcreteModel()
m.R = Param(initialize=0.001) # Friction factor
m.L = Param(initialize=1000.0) # Final position
m.T = Param(initialize=50.0) # Estimated time
m.aU = Param(initialize=0.8) # Acceleration upper bound
m.aL = Param(initialize=-1.5) # Acceleration lower bound
m.jU = Param(initialize=0.5) # Jerk upper bound
m.jL = Param(initialize=-0.5) # Jerk lower bound
m.NFE = Param(initialize=100) # Number of finite elements
'''
def _initX(m, i):
return m.x[i] == i*m.L/m.NFE
def _initV(m):
return m.v[i] == m.L/50
'''
m.tf = Var()
m.tau = ContinuousSet(bounds=(0,1)) # Unscaled time
m.t = Var(m.tau) # Scaled time
m.x = Var(m.tau, bounds=(0,m.L))
m.v = Var(m.tau, bounds=(0,None))
m.u = Var(m.tau, bounds=(-3,1), initialize=0)
m.dt = DerivativeVar(m.t)
m.dx = DerivativeVar(m.x)
m.dv = DerivativeVar(m.v)
m.da = DerivativeVar(m.v, wrt=(m.tau, m.tau))
m.obj = Objective(expr=m.tf)
def _ode1(m, i):
if i==0:
return Constraint.Skip
return m.dt[i] == m.tf
m.ode1 = Constraint(m.tau, rule=_ode1)
def _ode2(m, i):
if i==0:
return Constraint.Skip
return m.dx[i] == m.tf * m.v[i]
m.ode2 = Constraint(m.tau, rule=_ode2)
def _ode3(m, i):
if i==0:
return Constraint.Skip
return m.dv[i] == m.tf*(m.u[i] - m.R*m.v[i]**2)
m.ode3 = Constraint(m.tau, rule=_ode3)
def _accelerationL(m, i):
if i==0:
return Constraint.Skip
return m.dv[i] >= m.aL*m.tf
m.accelerationL = Constraint(m.tau, rule=_accelerationL)
def _accelerationU(m, i):
if i==0:
return Constraint.Skip
return m.dv[i] <= m.aU*m.tf
m.accelerationU = Constraint(m.tau, rule=_accelerationU)
def _jerkL(m, i):
if i==0:
return Constraint.Skip
return m.da[i] >= m.jL*m.tf**2
m.jerkL = Constraint(m.tau, rule=_jerkL)
def _jerkU(m, i):
if i==0:
return Constraint.Skip
return m.da[i] <= m.jU*m.tf**2
m.jerkU = Constraint(m.tau, rule=_jerkU)
'''
def _electric(m, i):
if i==0:
return Constraint.Skip
elif value(m.v[i])<=7:
return m.a[i] <= 1
else:
return m.v[i] <= 1*7/m.v[i]
m.electric = Constraint(m.tau, rule=_electric)
'''
'''
def _speed(m, i):
if i==0:
return Constraint.Skip
elif value(m.x[i])<=500:
return m.v[i] <= 40/3.6
else:
return m.v[i] <= 25/3.6
m.speed = Constraint(m.tau, rule=_speed)
'''
def _initial(m):
yield m.x[0] == 0
yield m.x[1] == m.L
yield m.v[0] == 0
yield m.v[1] == 0
yield m.t[0] == 0
m.initial = ConstraintList(rule=_initial)
discretizer = TransformationFactory('dae.finite_difference')
discretizer.apply_to(m, nfe=value(m.NFE), wrt=m.tau, scheme='BACKWARD')
#discretizer = TransformationFactory('dae.collocation')
#discretizer.apply_to(m, nfe=value(m.NFE), ncp=4, wrt=m.tau, scheme='LAGRANGE-RADAU')
solver = SolverFactory('ipopt')
solver.solve(m,tee=True)
print("final time = %6.2f" %(value(m.tf)))
t = []
x = []
v = []
a = []
u = []
for i in m.tau:
t.append(value(m.t[i]))
x.append(value(m.x[i]))
v.append(3.6*value(m.v[i]))
a.append(10*value(m.u[i] - m.R*m.v[i]**2))
u.append(10*value(m.u[i]))
import matplotlib.pyplot as plt
plt.plot(x, v, label='v (km/h)')
plt.plot(x, a, label='a (dm/s2)')
plt.plot(x, u, label='u (dm/s2)')
plt.xlabel('distance')
plt.grid('on')
plt.legend()
plt.show()
Thanks a lot in advance,
Pablo
(1) You should not think of Pyomo constraint rules as callbacks that are used by the solver. You should think of them more as a function to generate a container of constraint objects that gets called once for each index when the model is constructed. Meaning it is invalid to use a variable in an if statement unless you are really only using its initial value to define the constraint expression. There are ways to express what I think you are trying to do, but they involve introducing binary variables into the problem, in which case you can no longer use Ipopt.
(2) Can't really provide any help. Syntax looks fine.
(3) Pyomo allows you to return double-sided inequality expressions (e.g., L <= f(x) <= U) from constraint rules, but they can not involve variable expressions in the L and U locations. It doesn't look like the constraints you are referring to can be combined into this form.
(4) See (1)