def test_options(self):
from cvxopt import glpk, solvers
c,G,h,A,b = self._prob_data
glpk.options = {'msg_lev' : 'GLP_MSG_OFF'}
sol1 = glpk.lp(c,G,h)
self.assertTrue(sol1[0]=='optimal')
sol2 = glpk.lp(c,G,h,A,b)
self.assertTrue(sol2[0]=='optimal')
sol3 = glpk.lp(c,G,h,options={'msg_lev' : 'GLP_MSG_ON'})
self.assertTrue(sol3[0]=='optimal')
sol4 = glpk.lp(c,G,h,A,b,options={'msg_lev' : 'GLP_MSG_ERR'})
self.assertTrue(sol4[0]=='optimal')
sol5 = solvers.lp(c,G,h,solver='glpk',options={'glpk':{'msg_lev' : 'GLP_MSG_ON'}})
self.assertTrue(sol5['status']=='optimal')
sol1 = glpk.ilp(c,G,h,None,None,set(),set([0,1]))
self.assertTrue(sol1[0]=='optimal')
sol2 = glpk.ilp(c,G,h,A,b,set([0,1]),set())
self.assertTrue(sol2[0]=='optimal')
sol3 = glpk.ilp(c,G,h,None,None,set(),set([0,1]),options={'msg_lev' : 'GLP_MSG_ALL'})
self.assertTrue(sol3[0]=='optimal')
sol4 = glpk.ilp(c,G,h,A,b,set(),set([0]),options={'msg_lev' : 'GLP_MSG_ALL'})
self.assertTrue(sol4[0]=='optimal')
solvers.options['glpk'] = {'msg_lev' : 'GLP_MSG_ON'}
sol5 = solvers.lp(c,G,h,solver='glpk')
self.assertTrue(sol5['status']=='optimal')
def test_ilp(self):
from cvxopt import glpk, matrix
c,G,h,A,b = self._prob_data
sol1 = glpk.ilp(c, G, h, A, b, set([0]), set())
self.assertTrue(sol1[0]=='optimal')
sol2 = glpk.ilp(c, G, h, A, b, set([0]), set())
self.assertTrue(sol2[0]=='optimal')
sol3 = glpk.ilp(c, G, h, None, None, set([0, 1]), set())
self.assertTrue(sol3[0]=='optimal')
sol4 = glpk.ilp(c, G, h, None, None, set(), set([1]))
self.assertTrue(sol4[0]=='optimal')
sol5 = glpk.ilp(c, G, h, A, matrix(-1.0), set(), set([0,1]))
self.assertTrue(sol5[0]=='LP relaxation is primal infeasible')
def lst_l1(A, y, integer=False, xmax=1000):
"""
Returns the solution of least L1-norm problem
minimize || x ||_1.
subject to A'x == y
x can be float or integer.
Return None if no optimal/feasible solution found.
This problem can be converted into a linear programming problem
by setting v = |u|, x = [u' v']',
minimize [0]' [u]
[1] [v]
subject to [ I -I] [ 0 ]
[-I -I] [u] = [ 0 ]
[ 0 -I] [v] [ 0 ]
[ I 0] [xmax]
[A]' [u] = [y]
[0] [v]
"""
m, n = A.size
c = matrix(0.0, (2*n,1))
c[n:] = 1.0
# inequality constraint
I = spmatrix(1.0, range(n), range(n))
O = matrix(0.0, (n,n))
G = sparse(matrix([[I, -I, O, I], [-I, -I, -I, O]]))
h = matrix(0.0, (4*n,1))
h[3*n:] = xmax
# equality constraint
Al = sparse(matrix([[A], [matrix(0.0, (m,n))]]))
bl = y
# solve the linear programming problem
if integer:
(status, x) = glpk.ilp(c, G, h, Al, bl)[0:2]
else:
(status, x) = glpk.ilp(c, G, h, Al, bl)[0:2]
return x
def solveIP(team, initialCutoff, finalCutoff):
global reignsByYear, reignDict, indexToReign
setReignsByYear(team, initialCutoff)
c,G,h,A,b,I,B= genConstraintMatrix(finalCutoff)
#sol=solvers.lp(c,A,b)
(status,x)= ilp(c,G,h,A,b,I,B)
#print "Num reigns used", sum(list(x))
#x has value for each reign: 1 if reign should be included, 0 otherwise
reigns= sorted([(indexToReign[i].player,indexToReign[i].start,indexToReign[i].end) for i,val in enumerate(x) if val==1],key=lambda tup: tup[1])
playerStats= []
for player,start,end in reigns:
entry= {"player": player, "stats": []}
for year in range(start,end+1):
seasonLine= getStatLine(team,player,year)
entry["stats"].append(seasonLine)
playerStats.append(entry)
playerStats= consolidateReigns(playerStats)
reignsByYear= {}
reignDict= {}
indexToReign= {}
with open(os.getcwd()+'/jsons/'+team+'.json', 'w') as outfile:
json.dump(playerStats, outfile)
return playerStats
def CVXOPT_LP_Solver(p, solverName):
#os.close(1); os.close(2)
if solverName == 'native_CVXOPT_LP_Solver': solverName = None
cvxopt_solvers.options['maxiters'] = p.maxIter
cvxopt_solvers.options['feastol'] = p.contol
cvxopt_solvers.options['abstol'] = p.ftol
if p.iprint <= 0:
cvxopt_solvers.options['show_progress'] = False
cvxopt_solvers.options['LPX_K_MSGLEV'] = 0
cvxopt_solvers.options['MSK_IPAR_LOG'] = 0
xBounds2Matrix(p)
#WholeRepr2LinConst(p)
# CVXOPT have some problems with x0 so currently I decided to avoid using the one
#if p.x0.size>0 and p.x0.flatten()[0] != None and all(isfinite(p.x0)):
# sol= cvxopt_solvers.solvers.lp(Matrix(p.f), Matrix(p.A), Matrix(p.b), Matrix(p.Aeq), Matrix(p.beq), solverName)
#else:
if (len(p.intVars)>0 or len(p.binVars)>0) and solverName == 'glpk':
from cvxopt.glpk import ilp
c = Matrix(p.f)
A, b = Matrix(p.Aeq), Matrix(p.beq)
G, h = Matrix(p.A), Matrix(p.b)
if A is None:
A = matrix(0.0, (0, p.n))
b = matrix(0.0,(0,1))
if G is None:
G = matrix(0.0, (0, p.n))
h = matrix(0.0,(0,1))
(status, x) = ilp(c, G, h, A, b, set(p.intVars), B=set(p.binVars))
if status == 'optimal': p.istop = SOLVED_WITH_UNIMPLEMENTED_OR_UNKNOWN_REASON
elif status == 'maxiters exceeded': p.istop = IS_MAX_ITER_REACHED
elif status == 'time limit exceeded': p.istop = IS_MAX_TIME_REACHED
elif status == 'unknown': p.istop = UNDEFINED
else: p.istop = FAILED_WITH_UNIMPLEMENTED_OR_UNKNOWN_REASON
if x is None:
p.xf = nan*ones(p.n)
else:
p.xf = array(x).flatten()#w/o flatten it yields incorrect result in ff!
p.ff = sum(p.dotmult(p.f, p.xf))
p.msg = status
else:
# if len(p.b) != 0:
# s0 = matrix(p.b - dot(p.A, p.x0))
# else:
# s0 = matrix()
# primalstart = {'x': matrix(p.x0), 's': s0}
# sol = cvxopt_solvers.lp(Matrix(p.f), Matrix(p.A), Matrix(p.b), Matrix(p.Aeq), Matrix(p.beq), solverName, primalstart)
sol = cvxopt_solvers.lp(Matrix(p.f), Matrix(p.A), Matrix(p.b), Matrix(p.Aeq), Matrix(p.beq), solverName)
p.msg = sol['status']
if p.msg == 'optimal' : p.istop = SOLVED_WITH_UNIMPLEMENTED_OR_UNKNOWN_REASON
else: p.istop = -100
if sol['x'] is not None:
p.xf = asarray(sol['x']).flatten()
# ! don't involve p.ff - it can be different because of goal and additional constant from FuncDesigner
p.duals = concatenate((asarray(sol['y']).flatten(), asarray(sol['z']).flatten()))
else:
p.ff = nan
p.xf = nan*ones(p.n)
def optimization(distanceMatrix, numAutos, numCust):
distanceMatrix = numpy.array(distanceMatrix)
c = numpy.ones((numAutos * numCust, 1))
count = 0
for i in range(numAutos):
for j in range(numCust):
c[count][0] = distanceMatrix[i][j]
count = count + 1
if numAutos < numCust:
G, h, A, b, I, B = allAutoAssignment(numAutos, numCust)
elif numAutos > numCust:
G, h, A, b, I, B = allCustAssignment(numAutos, numCust)
else:
G, h, A, b, I, B = allAutoCustAssignment(numAutos, numCust)
(status, x) = ilp(c, G, h, A, b, I, B)
return x, numAutos, numCust
def nlp_exec_cvx(c, ops, matstrats, trips, G, h, A, b):
"""
@return newassignments, prov_to_remove
prov_to_remove is a list of (op, runid) pairs
"""
from cvxopt.glpk import ilp
from cvxopt import *
from cvxopt import glpk
# solvers.options['show_progress'] = False
# solvers.options['LPX_K_MSGLEV'] = 0
# solvers.options['MessageLevel'] = 3
# glpk.options['LPX_K_MSGLEV'] = 0
I = set()
B = set(range(len(c)))
start = time.time()
status, x = ilp(c, G, h, A, b, I, B)
end = time.time()
nlog.info( "optruntime \t%.5f", (end-start))
nlog.info( "status \t%s", status )
nlog.info( "resources \t%s", np.array((G*x).trans())[0] )
nlog.info( "constraints\t%s", np.array(h.trans())[0] )
nlog.info( "cost \t%s", np.array(x.trans() * c)[0] )
pairsize = len(ops)*len(matstrats)
newassignments = x[:pairsize]
rmassignments = x[pairsize:]
newassignments = numpy.array(newassignments).reshape((len(ops), len(matstrats))).tolist()
strategies = assign_strategies(newassignments, ops, matstrats)
torm = []
for strat, ass in zip(trips[pairsize:], rmassignments):
if strat[2] == Strat.query() and ass:
torm.append((strat[1], strat[0]))
return strategies, torm
[0.0, -1.0, 0.0, -1.0, 0.0, 0.0, 0.0, -1.0, 0.0, 0.0, 0.0],
[0.0, -1.0, 0.0, 0.0, -1.0, 0.0, 0.0, 0.0, -1.0, 0.0, 0.0],
[0.0, 0.0, -1.0, -1.0, 0.0, 0.0, 0.0, 0.0, 0.0, -1.0, 0.0],
[0.0, 0.0, -1.0, 0.0, -1.0, 0.0, 0.0, 0.0, 0.0, 0.0, -1.0],
]
)
h = matrix([-7.0, -2.0, -4.0, -5.0, -8.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0])
o = matrix([4.0, 7.0, 6.0, 8.0, 8.0, 9])
sol = solvers.lp(o, G, h)
print sol["x"]
print sol["primal objective"]
sol = glpk.ilp(o, G, h)
print sol[1]
##print sol[1]
##print sol[1][0]*4 + sol[1][1]*7 + sol[1][2]*6 + sol[1][3]*8 + sol[1][4]*8\
## + sol[1][5]*9
G = matrix(
[
[-1.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, -1.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, -1.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, -1.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, -1.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, -1.0],
]
for j in range(700):
if i == j:
array.append(-1.)
else:
array.append(0.)
constraint2.append(array)
preferences = np.array([preferences])
constraint1 = np.array(constraint1)
demand = np.array(demand)
constraint2 = np.array(constraint2)
print constraint2.shape
c = matrix(preferences.T)
G = matrix(constraint2)
h = matrix(demand)
A = matrix(constraint1)
b = matrix(b.T)
sol = glpk.ilp(c,G,h,A,b,I=set([0,1]))
solution = []
for i in range(10):
array = []
for j in range(70):
k = sol[1][(i)*70+j]
if k == 1:
print "nurse", i, "takes shift", j
array.append(k)
solution.append(array)
I = set()
B = set(range(numAutos * numCust))
return G,h,A,b,I,B
## The following function only prints the results in an easily interpretable way
def print_my_soln(status,solution, num_autos,num_customers):
print "The optimization solution is", status
for i in range(len(solution)):
if solution[i]==1.0:
print "Auto Number",i/num_customers + 1, "is going to the customer number",i%num_customers + 1
# print "Auto Driver at the place", origins[0][i/num_customers], "should go to the customer at the place", destinations[0][i%num_customers]
## This is where the code execution starts
dataRead=jsonDistanceMatrix() ## dataRead is the json sent by the google maps
dataRead_dict = json.loads(dataRead)
distanceMatrix, timeMatrix, numAutos, numCust = parseJsonText(dataRead_dict)
distanceMatrix=numpy.array(timeMatrix)
c=numpy.ones((numAutos*numCust,1))
count=0
for i in range(numAutos):
for j in range(numCust):
c[count][0]=distanceMatrix[i][j]
count=count+1
G, h, A, b, I, B = assignmentOptimization(numAutos, numCust)
(status, x) = ilp(c, G, h, A, b, I, B)
print "Number of Autos is", numAutos
print "Number of Customers is",numCust
print_my_soln(status, x, numAutos, numCust)
#standard is left less or equal to right
#G = matrix([[-1.0, -1.0, 0.0, 1.0], [1.0, -1.0, -1.0, -2.0]])
#h = matrix(json.load(open('h_constraints_right.json')))
#print h
#h = matrix([1.0, -2.0, 0.0, 4.0])
#print glpk.options
#glpk.options['LPX_K_MIPGAP'] = '1.0'
#glpk.options['LPX_K_MIP_GAP'] = 1.0
#glpk.options['LPX_K_GMI_CUTS'] = 'GLP_ON'
#glpk.options['LPX_K_TOLOBJ'] = 0.007
glpk.options['LPX_K_TMLIM'] = max_time
#glpk.options['LPX_K_PRESOL']= 1
#print glpk.options
#quit()
status, result = glpk.ilp(c = c, G=G, h=h, A = A, b = b, B= set(range(0,len(c))))
final_spi = -int(np.sum(np.multiply(np.array(result), np.array(c))))
print '[INFO] final optimal spi value is %s' % (str(final_spi))
with open('output/config.txt','a') as f:
if final_spi >= 0:
f.write(spi_file+'\t1\t'+str(final_spi)+'\n')
else:
f.write(spi_file+'\t-1\t'+str(final_spi)+'\n')
#print result
result = np.array(result, np.int)
result = np.reshape(result, (number_of_company, number_of_reps), 'F')
array = []
for j in range(700):
if i == j:
array.append(-1.)
else:
array.append(0.)
constraint2.append(array)
preferences = np.array([preferences])
constraint1 = np.array(constraint1)
demand = np.array(demand)
constraint2 = np.array(constraint2)
print constraint2.shape
c = matrix(preferences.T)
G = matrix(constraint2)
h = matrix(demand)
A = matrix(constraint1)
b = matrix(b.T)
sol = glpk.ilp(c, G, h, A, b, I=set([0, 1]))
solution = []
for i in range(10):
array = []
for j in range(70):
k = sol[1][(i) * 70 + j]
if k == 1:
print "nurse", i, "takes shift", j
array.append(k)
solution.append(array)
A = np.kron(np.eye(n_data1),np.ones((1,n_data2)))
b = np.ones((n_data1,1))
C = np.tile(np.eye(n_data2),(1,n_data1))
d = np.ones((n_data2,1))
f = dist.ravel()
# <markdowncell>
# ## Implementation using CVXOPT and GLPK
#
# Having setup the constraints above, the remaining steps for performing the optimization using CVXOPT are relatively simple. CVXOPT is capable of many types of optimization, thus the variable `binVars` is used to define the position of all elements in $\matrix{X}$ which are both integer and binary. (For this optimization, that is every element of $\matrix{X}$. Because $\matrix{f}$ is the same length as $\matrix{X}$ and is already defined, its length is used to determine `binVars`.) Also note that CVXOPT's integer programming function, `ilp`, does not accept numpy matrices and instead requires its own `matrix` function that is imported above. Lastly, all the functions use their own variable names, so they are listed as named arguments to eliminate confusion.
# <codecell>
binVars = range(len(f)) # ilp won't accept numpy's arange even when as a set
output_cvxopt = ilp(c=matrix(f),G=matrix(C),h=matrix(d),A=matrix(A),b=matrix(b),\
I=set(binVars),B=set(binVars))
# <markdowncell>
# The variable `output_cvxopt` contains several variables, one of which indicates if the optimization converged and, if it did, another variable containing $\matrix{X}$. Normally, it's good practice to check that the optimization converged, but in this case it's known to work. The transformation matrix, `X_cvxopt`, is reshaped into a two-dimensional matrix so it can easily be used to determine the order of `data2` data points that matches them with those in `data1`.
# <codecell>
X_cvxopt = np.array(output_cvxopt[1]).reshape((n_data1,n_data2))
match_cvxopt = np.nonzero(X_cvxopt)[1]
# <markdowncell>
# ### Sanity check the optimization results
# It is incredibly easy to get dimensions transposed, particularly when matrices have to be flattened and reshaped. Thus, it is critical to check the results. In the above optimization, the peak labels should be identical for each point if the peaks are correctly matched.
if (j >= i): # not found - add this cost
total_cost = total_cost + array[i][2]
numb_plans=numb_plans+1
break
i = i+1
if (i >= nmb): break
#print("Plans:")
numb_of_trips=0
for i in range(nmb):
if (array[i][0]!=-1):
#print(array[i])
numb_of_trips = numb_of_trips +1
#print ("Routes: %d, total cost: %d" % (numb_of_trips, total_cost))
print ("====================")
# now optimal solution
(status,x)=ilp(c,g.T,h,a,b,I,B)
#print(sum(c.T*x))
tot_cost=0
numb_var = 0
for i in range(n*n):
if (x[i]==1 and c[i]<n*n):
#print ("%d %d (%d)" % (int(i/n), i % n, c[i]))
tot_cost = tot_cost + c[i]
numb_var = numb_var +1
#print ("Routes: %d, total cost: %d" % (numb_var, tot_cost))
if (numb_var == numb_of_trips): # it is easier to compare, small difference in numbers of plans in LCM and solver
f.write("%d; %d; %5.0f\n" % (total_cost, tot_cost, 100* (total_cost - tot_cost)/tot_cost) )
f.close()
