df_iter=4,
legend=solver,
show=False,
contol=contol,
maxTime=maxTime,
maxFunEvals=maxFunEvals,
name="NLP_5",
)
if solver == "algencan":
p.gtol = 1e-2
elif solver == "ralg":
pass
# p.debug = 1
p.debug = 1
r = p.solve(solver)
for fn in ("h", "c"):
if not r.evals.has_key(fn):
r.evals[fn] = 0 # if no c or h are used in problem
results[solver] = (
r.ff,
p.getMaxResidual(r.xf),
r.elapsed["solver_time"],
r.elapsed["solver_cputime"],
r.evals["f"],
r.evals["c"],
r.evals["h"],
)
if PLOT:
def minimize(self, **kwargs):
""" solve the nonlinear problem using OpenOpt
Returns:
obj_value, solution
obj_value -- value of the objective function at the discovered solution
solution -- the solution flux vector (indexed like matrix columns)
"""
if self.iterator is None:
nlp = NLP(self.obj,
self.x0,
df=self.d_obj,
c=self.nlc,
dc=self.d_nlc,
A=self.Aineq,
Aeq=self.Aeq,
b=self.bineq,
beq=self.beq,
lb=self.lb,
ub=self.ub,
**kwargs)
nlp.debug = 1
nlp.plot = False
nlp.checkdf()
if self.nlc is not None:
nlp.checkdc()
r = nlp.solve(self.solver)
else:
self.rlist = []
for x0 in self.iterator:
nlp = NLP(self.obj,
x0,
df=self.d_obj,
c=self.nlc,
dc=self.d_nlc,
A=self.Aineq,
Aeq=self.Aeq,
b=self.bineq,
beq=self.beq,
lb=self.lb,
ub=self.ub,
**kwargs)
r = nlp.solve(self.solver)
if r.istop > 0 and r.ff == r.ff:
self.rlist.append(r)
if self.rlist != []:
r = min(self.rlist, key=lambda x: x.ff)
if r.istop <= 0 or r.ff != r.ff: # check halting condition
self.obj_value = 0.
self.solution = []
self.istop = r.istop
else:
self.obj_value = r.ff
self.solution = r.xf
self.istop = r.istop
return self.obj_value, self.solution
# that are connected to / used in lincher and some other solvers
# optional: check of user-supplied derivatives
p.checkdf()
p.checkdc()
p.checkdh()
# last but not least:
# please don't forget,
# Python indexing starts from ZERO!!
p.plot = 0
p.iprint = 0
p.df_iter = 4
p.maxTime = 4000
p.debug=1
#r = p.solve('algencan')
r = p.solve('ralg')
#r = p.solve('lincher')
"""
typical output:
OpenOpt checks user-supplied gradient df (size: (50,))
according to:
prob.diffInt = 1e-07
prob.check.maxViolation = 1e-05
max(abs(df_user - df_numerical)) = 2.50111104094e-06
(is registered in df number 41)
sum(abs(df_user - df_numerical)) = 4.45203815948e-05
========================
