Im trying to get the global minimum of a non linear function with linear constraints. I share you the code:
rho = 1.0
g = 9.8
C = 1.0 #roughness coefficient
J = rho*g/(float(1e6)) #constant of function
capacity = [0.3875, 0.607, 0.374] #Supply data
demand = [0.768, 0.315, 0.38,] #Demand data
m= len(capacity)
n = len(demand)
x0 = np.array(m*n*[0.01]) #Initial point for algorithm
# In[59]:
#READ L AND d FROM ARCGIS !!!
L = (314376.57, 277097.9663, 253756.9869 ,265786.5632, 316712.6028, 232857.1468, 112063.9914, 135762.94, 131152.8206)
h= (75, 75, 75, 75, 75, 75, 1320,75,75)
K = np.zeros(m*n)
N=np.zeros(m*n)
# In[60]:
#Each path has its own L,d, therefore, own constant
# In[61]:
#np.seterr(all='ignore')
def diameter(x):
d = 1.484
if x >= 0.276:
d = 1.484
elif x >= 0.212:
d = 1.299
elif x >= 0.148:
d = 1.086
elif x >= 0.0975:
d = 0.881
elif x >= 0.079:
d = 0.793
elif x >= 0.062:
d = 0.705
elif x >= 0.049:
d = 0.626
elif x >= 0.038:
d = 0.555
elif x >= 0.030:
d = 0.494
elif x >= 0.024:
d = 0.441
elif x >= 0.0198:
d = 0.397
elif x >= 0.01565:
d = 0.353
elif x >= 0.0123:
d = 0.313
elif x >= 0.0097:
d = 0.278
elif x >= 0.0077:
d = 0.247
elif x >= 0.0061:
d = 0.22
elif x >= 0.0049:
d = 0.198
elif x >= 0.00389:
d = 0.176
elif x >= 0.0032:
d = 0.159
elif x >= 0.0025:
d = 0.141
elif x >= 0:
d = 0.123
return d
#Definition of Objetive function
def objective(x):
sum_obj = 0.0
for i in range(len(L)):
K = 10.674*L[i]/(C**1.852*diameter(x[i])**4.871)
N[i] = K*x[i]**2.852*J+x[i]*J*h[i]
sum_obj = sum_obj + N[i]
return sum_obj
print(str(objective(x0)))
#Definition of the constraints
GB=[]
for i in range(m):
GB.append(n*i*[0]+n*[1]+(m-1-i)*n*[0])
P=[]
for i in range(n):
P.append(m*(i*[0]+[1]+(n-1-i)*[0]))
DU=np.array(GB+P)
lb = np.array(m*[0] + demand) # Supply
ub = np.array(capacity + n*[np.inf]) # Demand
# In[62]:
b = (0, 1)
bnds = []
for i in range(m*n):
bnds.append(b)
cons = LinearConstraint(DU,lb,ub)
solution = differential_evolution(objective,x0,cons,bnds)
x = solution.x
# show initial objective
print('Initial SSE Objective: ' + str(objective(x0)))
# show final objective
print('Final SSE Objective: ' + str(objective(x)))
# print solution
print('Solution Supply to Demand:')
print('Q = ' + str(((np.around(x,6)).reshape(10,18))))
I dont know why, but when I run appear the following:
"if strategy in self._binomial:
TypeError: unhashable type: 'list'
Anyone have gotten the same mistake ? Its my first time trying to solve optimization problem, so I need a little bit of help. Any advice is welcome !!
Thanks a lot !
You're not calling differential_evolution correctly. For your example you should call it as:
differential_evolution(objective, bnds, contraints=(cons,))
(I'm not sure if there are additional problems)
Related
I write a formula to solve nqueen problem. It finds one of the solution but I want find all solution how I generalize to all solution for this formula:
from z3 import *
import time
def queens(n,all=0):
start = time.time()
sol = Solver()
# q = [Int("q_%s" % (i)) for i in range(n) ] # n=100: ???s
# q = IntVector("q", n) # this is much faster # n=100: 28.1s
q = Array("q", IntSort(), BitVecSort(8)) # n=100: ??s
# Domains
sol.add([And(q[i]>=0, q[i] <= n-1) for i in range(n)])
# Constraints
for i in range(n):
for j in range(i):
sol.add(q[i] != q[j], q[i]+i != q[j]+j, q[i]-i != q[j]-j)
if sol.check() == sat:
mod = sol.model()
ss = [mod.evaluate(q[i]) for i in range(n)]
print(ss)
# Show all solutions
if all==1:
num_solutions = 0
while sol.check() == sat:
m = sol.model()
ss = [mod.evaluate(q[i]) for i in range(n)]
sol.add( Or([q[i] != ss[i] for i in range(n)]) )
print("q=",ss)
num_solutions = num_solutions + 1
print("num_solutions:", num_solutions)
else:
print("failed to solve")
end = time.time()
value = end - start
print("Time: ", value)
for n in [8,10,12,20,50,100,200]:
print("Testing ", n)
queens(n,0)
For N=4 I try to show 2 solution
For N=8 I try to show all 92 solution
You got most of it correct, though there are issues with how you coded the find-all-solutions part. There's a solution for N-queens that comes with the z3 tutorial here: https://ericpony.github.io/z3py-tutorial/guide-examples.htm
You can turn it into an "find-all-solutions" versions like this:
from z3 import *
def queens(n):
Q = [Int('Q_%i' % (i + 1)) for i in range(n)]
val_c = [And(1 <= Q[i], Q[i] <= n) for i in range(n)]
col_c = [Distinct(Q)]
diag_c = [If(i == j, True, And(Q[i] - Q[j] != i - j, Q[i] - Q[j] != j - i)) for i in range(n) for j in range(i)]
sol = Solver()
sol.add(val_c + col_c + diag_c)
num_solutions = 0
while sol.check() == sat:
mod = sol.model()
ss = [mod.evaluate(Q[i]) for i in range(n)]
print(ss)
num_solutions += 1
sol.add(Or([Q[i] != ss[i] for i in range(n)]))
print("num_solutions:", num_solutions)
queens(4)
queens(8)
This'll print 2 solutions for N=4 and 92 for N=8.
I have a problem where I want to limit the range of a real variable between the maximum and minimum value of another set of real variables.
s = Solver()
y = Real('y')
Z = RealVector('z', 10)
s.add(And(y >= min(Z), y <= max(Z)))
Is there a way to do this in z3py?
You can use Axel's solution; though that one requires you to create an extra variable and also asserts more constraints than needed. Moreover, it doesn't let you use min and max as simple functions. It might be easier to just program this in a functional way, like this:
# Return minimum of a vector; error if empty
def min(vs):
m = vs[0]
for v in vs[1:]:
m = If(v < m, v, m)
return m
# Return maximum of a vector; error if empty
def max(vs):
m = vs[0]
for v in vs[1:]:
m = If(v > m, v, m)
return m
Another difference is that in the functional style we throw an error if the vector is empty. In the other style, the result will essentially be unconstrained. (i.e., min/max can take any value.) You should consider which semantics is right for your application, in case the vector you're passing might be empty. (At the least, you should change it so it prints out a nicer error message. Currently it'll throw an IndexError: list index out of range error if given an empty vector.)
Now you can say:
s = Solver()
y = Real('y')
Z = RealVector('z', 10)
s.add(And(y >= min(Z), y <= max(Z)))
print (s.check())
print (s.model())
This prints:
sat
[z__7 = -1,
z__0 = -7/2,
z__4 = -5/2,
z__5 = -2,
z__3 = -9/2,
z__2 = -4,
z__8 = -1/2,
y = 0,
z__9 = 0,
z__6 = -3/2,
z__1 = -3]
You could benefit from Hakan Kjellerstrand's collection of useful z3py definitions:
from z3 import *
# Functions written by Hakan Kjellerstrand
# http://hakank.org/z3/
# The following can be used by importing http://www.hakank.org/z3/z3_utils_hakank.py
# v is the maximum value of x
def maximum(sol, v, x):
sol.add(Or([v == x[i] for i in range(len(x))])) # v is an element in x)
for i in range(len(x)):
sol.add(v >= x[i]) # and it's the greatest
# v is the minimum value of x
def minimum(sol, v, x):
sol.add(Or([v == x[i] for i in range(len(x))])) # v is an element in x)
for i in range(len(x)):
sol.add(v <= x[i]) # and it's the smallest
s = Solver()
y = Real('y')
zMin = Real('zMin')
zMax = Real('zMax')
Z = RealVector('z', 10)
maximum(s, zMin, Z)
minimum(s, zMax, Z)
s.add(And(y >= zMin, y <= zMax))
print(s.check())
print(s.model())
local delay = math.random(25, 50)
[string "LuaVM"]:5: attempt to index a nil value (global 'math')
I can't use math.random anymore is there any way to fix this ?
If math library is missed you can insert the following code block at the beginning of your script.
It will not fix the whole math library, but only some of the most frequently used functions (including math.random).
It will also fix the following errors:
bad argument #1 to 'Sleep' (number has no integer representation)
attempt to call a nil value (field 'getn')
do
local state_8, state_45, cached_bits, cached_bits_qty = 2, 0, 0, 0
local prev_width, prev_bits_in_factor, prev_k = 0
for c in GetDate():gmatch"." do
state_45 = state_45 % 65537 * 23456 + c:byte()
end
local function get_53_random_bits()
local value53 = 0
for shift = 26, 27 do
local p = 2^shift
state_45 = (state_45 * 233 + 7161722017421) % 35184372088832
repeat state_8 = state_8 * 76 % 257 until state_8 ~= 1
local r = state_8 % 32
local n = state_45 / 2^(13 - (state_8 - r) / 32)
n = (n - n%1) % 2^32 / 2^r
value53 = value53 * p + ((n%1 * 2^32) + (n - n%1)) % p
end
return value53
end
for j = 1, 10 do get_53_random_bits() end
local function get_random_bits(number_of_bits)
local pwr_number_of_bits = 2^number_of_bits
local result
if number_of_bits <= cached_bits_qty then
result = cached_bits % pwr_number_of_bits
cached_bits = (cached_bits - result) / pwr_number_of_bits
else
local new_bits = get_53_random_bits()
result = new_bits % pwr_number_of_bits
cached_bits = (new_bits - result) / pwr_number_of_bits * 2^cached_bits_qty + cached_bits
cached_bits_qty = 53 + cached_bits_qty
end
cached_bits_qty = cached_bits_qty - number_of_bits
return result
end
table = table or {}
table.getn = table.getn or function(x) return #x end
math = math or {}
math.huge = math.huge or 1/0
math.abs = math.abs or function(x) return x < 0 and -x or x end
math.floor = math.floor or function(x) return x - x%1 end
math.ceil = math.ceil or function(x) return x + (-x)%1 end
math.min = math.min or function(x, y) return x < y and x or y end
math.max = math.max or function(x, y) return x > y and x or y end
math.sqrt = math.sqrt or function(x) return x^0.5 end
math.pow = math.pow or function(x, y) return x^y end
math.frexp = math.frexp or
function(x)
local e = 0
if x == 0 then
return x, e
end
local sign = x < 0 and -1 or 1
x = x * sign
while x >= 1 do
x = x / 2
e = e + 1
end
while x < 0.5 do
x = x * 2
e = e - 1
end
return x * sign, e
end
math.exp = math.exp or
function(x)
local e, t, k, p = 0, 1, 1
repeat e, t, k, p = e + t, t * x / k, k + 1, e
until e == p
return e
end
math.log = math.log or
function(x)
assert(x > 0)
local a, b, c, d, e, f = x < 1 and x or 1/x, 0, 0, 1, 1
repeat
repeat
c, d, e, f = c + d, b * d / e, e + 1, c
until c == f
b, c, d, e, f = b + 1 - a * c, 0, 1, 1, b
until b <= f
return a == x and -f or f
end
math.log10 = math.log10 or
function(x)
return math.log(x) / 2.3025850929940459
end
math.random = math.random or
function(m, n)
if m then
if not n then
m, n = 1, m
end
local k = n - m + 1
if k < 1 or k > 2^53 then
error("Invalid arguments for function 'random()'", 2)
end
local width, bits_in_factor, modk
if k == prev_k then
width, bits_in_factor = prev_width, prev_bits_in_factor
else
local pwr_prev_width = 2^prev_width
if k > pwr_prev_width / 2 and k <= pwr_prev_width then
width = prev_width
else
width = 53
local width_low = -1
repeat
local w = (width_low + width) / 2
w = w - w%1
if k <= 2^w then
width = w
else
width_low = w
end
until width - width_low == 1
prev_width = width
end
bits_in_factor = 0
local bits_in_factor_high = width + 1
while bits_in_factor_high - bits_in_factor > 1 do
local bits_in_new_factor = (bits_in_factor + bits_in_factor_high) / 2
bits_in_new_factor = bits_in_new_factor - bits_in_new_factor%1
if k % 2^bits_in_new_factor == 0 then
bits_in_factor = bits_in_new_factor
else
bits_in_factor_high = bits_in_new_factor
end
end
prev_k, prev_bits_in_factor = k, bits_in_factor
end
local factor, saved_bits, saved_bits_qty, pwr_saved_bits_qty = 2^bits_in_factor, 0, 0, 2^0
k = k / factor
width = width - bits_in_factor
local pwr_width = 2^width
local gap = pwr_width - k
repeat
modk = get_random_bits(width - saved_bits_qty) * pwr_saved_bits_qty + saved_bits
local modk_in_range = modk < k
if not modk_in_range then
local interval = gap
saved_bits = modk - k
saved_bits_qty = width - 1
pwr_saved_bits_qty = pwr_width / 2
repeat
saved_bits_qty = saved_bits_qty - 1
pwr_saved_bits_qty = pwr_saved_bits_qty / 2
if pwr_saved_bits_qty <= interval then
if saved_bits < pwr_saved_bits_qty then
interval = nil
else
interval = interval - pwr_saved_bits_qty
saved_bits = saved_bits - pwr_saved_bits_qty
end
end
until not interval
end
until modk_in_range
return m + modk * factor + get_random_bits(bits_in_factor)
else
return get_53_random_bits() / 2^53
end
end
local orig_Sleep = Sleep
function Sleep(x)
return orig_Sleep(x - x%1)
end
end
Using the code below to to understand how Scipy optmization/minimization works. The results are not matching what I am expecting.
"""
Minimize: f = 2*x[0]*x[1] + 2*x[0] - x[0]**2 - 2*x[1]**2
Subject to: -2*x[0] + 2*x[1] <= -2
2*x[0] - 4*x[1] <= 0
x[0]**3 -x[1] == 0
where: 0 <= x[0] <= inf
1 <= x[1] <= inf
"""
import numpy as np
from scipy.optimize import minimize
def objective(x):
return 2.0*x[0]*x[1] + 2.0*x[0] - x[0]**2 - 2.0*x[1]**2
def constraint1(x):
return +2.0*x[0] - 2.0*x[1] - 2.0
def constraint2(x):
return -2.0*x[0] + 4.0*x[1]
def constraint3(x):
sum_eq = x[0]**3.0 -x[1]
return sum_eq
# initial guesses
n = 2
x0 = np.zeros(n)
x0[0] = 10.0
x0[1] = 100.0
# show initial objective
print('Initial SSE Objective: ' + str(objective(x0)))
# optimize
#b = (1.0,None)
bnds = ((0.0,1000.0), (1.0,1000.0))
con1 = {'type': 'ineq', 'fun': constraint1}
con2 = {'type': 'ineq', 'fun': constraint2}
con3 = {'type': 'eq', 'fun': constraint3}
cons = ([con1, con2, con3])
solution = minimize(objective,
x0,
method='SLSQP',
bounds=bnds,
constraints=cons)
x = solution.x
print(solution)
# show final objective
print('Final SSE Objective: ' + str(objective(x)))
# print solution
print('Solution')
print('x1 = ' + str(x[0]))
print('x2 = ' + str(x[1]))
print('\n')
print('x', x)
print('constraint1', constraint1(x))
print('constraint2', constraint2(x))
print('constraint3', constraint3(x))
When I run, this is what Python throws on its output console:
Initial SSE Objective: -18080.0
fun: 2.0
jac: array([ 0.00000000e+00, -2.98023224e-08])
message: 'Optimization terminated successfully.'
nfev: 122
nit: 17
njev: 13
status: 0
success: True
x: array([2., 1.])
Final SSE Objective: 2.0
Solution
x1 = 2.0000000000010196
x2 = 1.0000000000012386
x [2. 1.]
constraint1 -4.3787196091216174e-13
constraint2 2.915001573455811e-12
constraint3 7.000000000010997
Despite the optimizer says the result was successful, the constraint3 is not respected because the result should be zero. What am I missing?
Your problem is incompatible. You can eliminate the 3rd constraint (which makes your problem simpler in the first place - only a scalar optimization), after this it is a bit more clear to see what is the problem. From constraint 3 and the lower bound on the original x1 follows, that x0 is not feasible from 0 to 1, so the lower bound in the 1D problem should be 1. It is easy to see that constraint 2 will be always positive, when x0 is larger than 1, therefore it will never be satisfied.
When I run your original problem for me it stops with positive directional derivative (and for the rewritten problem with 'Inequality constraints incompatible').
Which SciPy are you using? For me it is 1.4.1.
On the picture below you can see the objective and the remaining constraints for the 1D problem (horizontal axis is the original x0 variable)
"""
Minimize: f = 2*x[0]*x1 + 2*x[0] - x[0]**2 - 2*x1**2
Subject to: -2*x[0] + 2*x[1] <= -2
2*x[0] - 4*x[1] <= 0
x[0]**3 -x[1] == 0
where: 0 <= x[0] <= inf
1 <= x[1] <= inf
"""
import numpy as np
from scipy.optimize import minimize
def objective(x):
return 2*x**4 + 2*x - x**2 - 2*x**6
def constraint1(x):
return x - x**3 - 1
def constraint2(x):
return 2 * x**3 - x
#
# def constraint3(x):
# sum_eq = x[0]**3.0 -x[1]
# return sum_eq
# initial guesses
n = 1
x0 = np.zeros(n)
x0[0] = 2.
# x0[1] = 100.0
# show initial objective
print('Initial SSE Objective: ' + str(objective(x0)))
# optimize
#b = (1.0,None)
bnds = ((1.0,1000.0),)
con1 = {'type': 'ineq', 'fun': constraint1}
con2 = {'type': 'ineq', 'fun': constraint2}
# con3 = {'type': 'eq', 'fun': constraint3}
cons = [
# con1,
con2,
# con3,
]
solution = minimize(objective,
x0,
method='SLSQP',
bounds=bnds,
constraints=cons)
x = solution.x
print(solution)
# show final objective
print('Final SSE Objective: ' + str(objective(x)))
# print solution
print('Solution')
print('x1 = ' + str(x[0]))
# print('x2 = ' + str(x[1]))
print('\n')
print('x', x)
print('constraint1', constraint1(x))
print('constraint2', constraint2(x))
# print('constraint3', constraint3(x))
x_a = np.linspace(1, 2, 200)
f = objective(x_a)
c1 = constraint1(x_a)
c2 = constraint2(x_a)
import matplotlib.pyplot as plt
plt.figure()
plt.plot(x_a, f, label="f")
plt.plot(x_a, c1, label="c1")
plt.plot(x_a, c2, label="c2")
plt.legend()
plt.show()
Certain problem in transport Phenomena is solved using the following code:
T_max, T_0, S, R, k, I, k_e, L, R, E, a = Reals('T_max T_0 S R k I k_e L R E a')
k = a*k_e*T_0
I = k_e*E/L
S = (I**2)/k_e
eq = T_0 + S* R**2/(4*k)
print eq
equations = [
T_max == eq,
]
print "Temperature equations:"
print equations
problem = [
R == 2, L == 5000,
T_0 == 20 + 273,
T_max == 30 + 273, k_e == 1,
a == 2.23*10**(-8), E > 0
]
print "Problem:"
print problem
print "Solution:"
solve(equations + problem)
using this code online we obtain
This output gives the correct answer but there are two issues in the code: a) the expresion named "eq" is not fully simplified and then it is necessary to give an arbitrary value for k_e . My question is: How to simplify the expression "eq" in such way that k_e be eliminated from "eq"?
Other example: To determine the radius of a tube
Code:
def inte(n,a,b):
return (b**(n+1))/(n+1)-(a**(n+1))/(n+1)
P_0, P_1, L, R, mu, q, C = Reals('P_0 P_1 L R mu q C')
k = (P_0 - P_1)/(2*mu*L)
equations = [0 == -k*inte(1,0,R) +C,
q == 2*3.1416*(-(k/2)*inte(3,0,R) + C*inte(1,0,R))]
print "Fluid equations:"
print equations
problem = [
L == 50.02/100, mu == (4.03*10**(-5)),
P_0 == 4.829*10**5, P_1==0,
q == 2.997*10**(-3), R >0
]
print "Problem:"
print problem
print "Solution:"
solve(equations + problem)
Output:
Fluid equations:
[-((P_0 - P_1)/(2·mu·L))·(R2/2 - 0) + C = 0, q =
3927/625·
(-(((P_0 - P_1)/(2·mu·L))/2)·(R4/4 - 0) + C·(R2/2 - 0))]
Problem:
[L = 2501/5000, mu = 403/10000000, P_0 = 482900, P_1 = 0, q = 2997/1000000, R > 0]
Solution:
[R = 0.0007512843?,
q = 2997/1000000,
P_1 = 0,
P_0 = 482900,
mu = 403/10000000,
L = 2501/5000,
C = 3380.3149444289?]