I am new to CVXPY, I learn by running examples and I found this post: How to construct a SOCP problem and solve using cvxpy and cvxpylayers #ben
The author provided the codes (Below I've corrected the line that caused an error in OP's EDIT 2):
import cvxpy as cp
import numpy as np
N = 5
Q_sqrt = cp.Parameter((N, N))
Q = cp.Parameter((N, N))
x = cp.Variable(N)
z = cp.Variable(N)
p = cp.Variable()
t = cp.Variable()
objective = cp.Minimize(p - t)
constraint_soc = [z == Q # x, x.value * z >= t ** 2, z >= 0, x >= 0] # <-- my question
constraint_other = [cp.quad_over_lin(Q_sqrt # x, p) <= N * p, cp.sum(x) == 1, p >= 0, t >= 0]
constraint_all = constraint_other + constraint_soc
matrix = np.random.random((N, N))
a_matrix = matrix.T # matrix
Q.value = a_matrix
Q_sqrt.value = np.sqrt(a_matrix)
prob = cp.Problem(objective, constraint_all)
prob.solve(verbose=True)
print("status:", prob.status)
print("optimal value", prob.value)
My question is here: x.value * z >= t ** 2
why only x.value while z not?
Actually I tried x * z, x.value * z.value, they both throw out errors, only the one in the original post works, which is x.value * z.
I googled and found this page and this, looks most relevant, but still failed to find an answer.
But both x and z are variables and defined as such
x = cp.Variable(N)
z = cp.Variable(N)
why only x needs a .value after?? Maybe it's a trivial question to experienced users, but I really don't get it. Thank you.
Update: two follow-up questions
Thanks to #MichalAdamaszek the first question above is clear: x.value didn't make sense here. A suggestion is using constraint_soc = [z == Q # x] + [ z[i]>=cp.quad_over_lin(t, x[i]) for i in range(N) ]
I have two following questions:
is it better to write the second of the soc constraint as : [ x[i]>=cp.quad_over_lin(t, z[i]) for i in range(N) ]? because in the question we only know that z[i] > 0 for sure. Or it doesn't matter at all in cvxpy? I tired both, both returned a similar optimal value.
I noticed that for the two constraints:
$x^TQx <= Np^2 $ and $ x_i z_i >= t^2 $
the quadratic terms were always intentionally split into two linear terms:
cp.quad_over_lin(Q_sqrt # x, p) <= N * p and z[i]>=cp.quad_over_lin(t, x[i]) respectively.
Is it because that in cvxpy, it is not allowed to have quadratic terms in (in)equality constraints? Is there an documentation to those basic rules?
Thank you.
Related
I'm setting up a linear programming optimization model using CPLEX and am wondering if it's possible to accomplish a modification of the cost function dependent upon which binary decision variables are 'active' in an arbitrary solution. This is mostly a question about how to formulate the LP model (if it's even possible), but responses in the context of CPLEX are welcome or even preferred.
Say I have an LP problem in canonical form:
minimize cTx
s.t. Ax <= b
With cost function:
c = [c_1, c_2,...,c_100]
All variables are binary. I have this basic setup modeled and running effectively in CPLEX.
Now say I have a subset of variables:
efficiency_set = [x_1, x_2,...,x_5]
With the condition:
if any x_n in efficiency_set == 1
then c_n for all other x_n in the set = 0.9 * c_n
Essentially there is a dependency where if any x_n in the efficiency set is 'active', it becomes 10% less expensive for other variables in the set to appear in the solution.
I thought that CPLEX indicator constraints were what I was looking for, but after reading through documentation, I don't think I can enforce an on-the-fly change to cost function with them (I could be wrong). So I feel like it needs to be done through formulation of the LP, but I can't reason how to accomplish it. Any ideas?. Thanks.
In CPLEX you have many APIs, let me answer you with the easiest one OPL.
Your canonical form can be written
int n=3;
int m=4;
range N=1..n;
range M=1..m;
float A[N][M]=[[1,4,9,6],[8,5,0,8],[2,9,0,2]];
float B[M]=[3,1,3,0];
float C[N]=[1,1,1];
dvar boolean x[N];
minimize sum(i in N) C[i]*x[i];
subject to
{
forall(j in M) sum(i in N) A[i,j]*x[i]>=B[j];
}
and then you can you write logical constraints:
int n=3;
int m=4;
range N=1..n;
range M=1..m;
float A[N][M]=[[1,4,9,6],[8,5,0,8],[2,9,0,2]];
float B[M]=[3,1,3,0];
float C[N]=[1,1,1];
{int} efficiencySet={1,2};
dvar boolean activeEfficiencySet;
dvar boolean x[N];
minimize sum(i in N) C[i]*x[i]*(1-0.1*activeEfficiencySet*(i not in efficiencySet));
subject to
{
forall(j in M) sum(i in N) A[i,j]*x[i]>=B[j];
activeEfficiencySet==(1<=sum(i in efficiencySet) x[i]);
}
Using Alex's data, I have written the program in docplex (cplex python API)
from docplex.mp.model import Model
n = 3
m = 4
A = {}
A[0, 0] = 1
A[0, 1] = 4
A[0, 2] = 9
A[0, 3] = 6
A[1, 0] = 8
A[1, 1] = 5
A[1, 2] = 0
A[1, 3] = 8
A[2, 0] = 2
A[2, 1] = 9
A[2, 2] = 0
A[2, 3] = 2
B = {}
B[0] = 3
B[1] = 1
B[2] = 3
B[3] = 0
C = {}
C[0] = 1
C[1] = 1
C[2] = 1
efficiencySet = [0, 1]
mdl = Model(name="")
activeEfficiencySet = mdl.binary_var()
x = mdl.binary_var_dict(range(n), name="x")
# constraint 1:
for j in range(m):
mdl.add_constraint(mdl.sum(A[i, j] * x[i] for i in range(n)) >= B[j])
# constraint 2:
mdl.add(activeEfficiencySet == (mdl.sum(x) >= 1))
# objective function:
# expr = mdl.linear_expr()
lst = []
for i in range(n):
if i not in efficiencySet:
lst.append((C[i] * x[i] * (1 - 0.1 * activeEfficiencySet)))
else:
lst.append(C[i] * x[i])
mdl.minimize(mdl.sum(lst))
mdl.solve()
for i in range(n):
print(str(x[i]) + " : " + str(x[i].solution_value))
activeEfficiencySet.solution_value
Using Z3Py, once a model has been checked for an optimization problem, is there a way to convert ArithRef expressions into values?
Such as
y = If(x > 5, 0, 0.5 * x)
Once values have been found for x, can I get the evaluated value for y, without having to calculate again based on the given values for x?
Many thanks.
You need to evaluate, but it can be done by the model for you automatically:
from z3 import *
x = Real('x')
y = If(x > 5, 0, 0.5 * x)
s = Solver()
r = s.check()
if r == sat:
m = s.model();
print("x =", m.eval(x, model_completion=True))
print("y =", m.eval(y, model_completion=True))
else:
print("Solver said:", r)
This prints:
x = 0
y = 0
Note that we used the parameter model_completion=True since there are no constraints to force x (and consequently y) to any value in this model. If you have sufficient constraints added, you wouldn't need that parameter. (Of course, having it does not hurt.)
Suppose x,y,z are int variables and A is a matrix, I want to express a constraint like:
z == A[x][y]
However this leads to an error:
TypeError: object cannot be interpreted as an index
What would be the correct way to do this?
=======================
A specific example:
I want to select 2 items with the best combination score,
where the score is given by the value of each item and a bonus on the selection pair.
For example,
for 3 items: a, b, c with related value [1,2,1], and the bonus on pairs (a,b) = 2, (a,c)=5, (b,c) = 3, the best selection is (a,c), because it has the highest score: 1 + 1 + 5 = 7.
My question is how to represent the constraint of selection bonus.
Suppose CHOICE[0] and CHOICE[1] are the selection variables and B is the bonus variable.
The ideal constraint should be:
B = bonus[CHOICE[0]][CHOICE[1]]
but it results in TypeError: object cannot be interpreted as an index
I know another way is to use a nested for to instantiate first the CHOICE, then represent B, but this is really inefficient for large quantity of data.
Could any expert suggest me a better solution please?
If someone wants to play a toy example, here's the code:
from z3 import *
items = [0,1,2]
value = [1,2,1]
bonus = [[1,2,5],
[2,1,3],
[5,3,1]]
choices = [0,1]
# selection score
SCORE = [ Int('SCORE_%s' % i) for i in choices ]
# bonus
B = Int('B')
# final score
metric = Int('metric')
# selection variable
CHOICE = [ Int('CHOICE_%s' % i) for i in choices ]
# variable domain
domain_choice = [ And(0 <= CHOICE[i], CHOICE[i] < len(items)) for i in choices ]
# selection implication
constraint_sel = []
for c in choices:
for i in items:
constraint_sel += [Implies(CHOICE[c] == i, SCORE[c] == value[i])]
# choice not the same
constraint_neq = [CHOICE[0] != CHOICE[1]]
# bonus constraint. uncomment it to see the issue
# constraint_b = [B == bonus[val(CHOICE[0])][val(CHOICE[1])]]
# metric definition
constraint_sumscore = [metric == sum([SCORE[i] for i in choices ]) + B]
constraints = constraint_sumscore + constraint_sel + domain_choice + constraint_neq + constraint_b
opt = Optimize()
opt.add(constraints)
opt.maximize(metric)
s = []
if opt.check() == sat:
m = opt.model()
print [ m.evaluate(CHOICE[i]) for i in choices ]
print m.evaluate(metric)
else:
print "failed to solve"
Turns out the best way to deal with this problem is to actually not use arrays at all, but simply create integer variables. With this method, the 317x317 item problem originally posted actually gets solved in about 40 seconds on my relatively old computer:
[ 0.01s] Data loaded
[ 2.06s] Variables defined
[37.90s] Constraints added
[38.95s] Solved:
c0 = 19
c1 = 99
maxVal = 27
Note that the actual "solution" is found in about a second! But adding all the required constraints takes the bulk of the 40 seconds spent. Here's the encoding:
from z3 import *
import sys
import json
import sys
import time
start = time.time()
def tprint(s):
global start
now = time.time()
etime = now - start
print "[%ss] %s" % ('{0:5.2f}'.format(etime), s)
# load data
with open('data.json') as data_file:
dic = json.load(data_file)
tprint("Data loaded")
items = dic['items']
valueVals = dic['value']
bonusVals = dic['bonusVals']
vals = [[Int("val_%d_%d" % (i, j)) for j in items if j > i] for i in items]
tprint("Variables defined")
opt = Optimize()
for i in items:
for j in items:
if j > i:
opt.add(vals[i][j-i-1] == valueVals[i] + valueVals[j] + bonusVals[i][j])
c0, c1 = Ints('c0 c1')
maxVal = Int('maxVal')
opt.add(Or([Or([And(c0 == i, c1 == j, maxVal == vals[i][j-i-1]) for j in items if j > i]) for i in items]))
tprint("Constraints added")
opt.maximize(maxVal)
r = opt.check ()
if r == unsat or r == unknown:
raise Z3Exception("Failed")
tprint("Solved:")
m = opt.model()
print " c0 = %s" % m[c0]
print " c1 = %s" % m[c1]
print " maxVal = %s" % m[maxVal]
I think this is as fast as it'll get with Z3 for this problem. Of course, if you want to maximize multiple metrics, then you can probably structure the code so that you can reuse most of the constraints, thus amortizing the cost of constructing the model just once, and incrementally optimizing afterwards for optimal performance.
I'm trying to find the minimal value of the Parabola y=(x+2)**2-3, apparently, the answer should be y==-3, when x ==-2.
But z3 gives the answer [x = 0, y = 1], which doesn't meet the ForAll assertion.
Am I assuming wrongly with something?
Here is the python code:
from z3 import *
x, y, z = Reals('x y z')
print(Tactic('qe').apply(And(y == (x + 2) ** 2 - 3,
ForAll([z], y <= (z + 2) ** 2 - 3))))
solve(y == x * x + 4 * x +1,
ForAll([z], y <= z * z + 4 * z +1))
And the result:
[[y == (x + 2)**2 - 3, True]]
[x = 0, y = 1]
The result shows that 'qe' tactic eliminated that ForAll assertion into True, although it's NOT always true.
Is that the reason that solver gives a wrong answer?
What should I code to find the minimal (or maximal) value of such an expression?
BTW, the Z3 version is 4.3.2 for Mac.
I refered
How does Z3 handle non-linear integer arithmetic?
and found a partial solution, using 'qfnra-nlsat' and 'smt' tactics.
from z3 import *
x, y, z = Reals('x y z')
s1 = Then('qfnra-nlsat','smt').solver()
print s1.check(And(y == (x + 2) ** 2 - 3,
ForAll([z], y <= (z + 2) ** 2 - 3)))
print s1.model()
s2 = Then('qe', 'qfnra-nlsat','smt').solver()
print s2.check(And(y == (x + 2) ** 2 - 3,
ForAll([z], y <= (z + 2) ** 2 - 3)))
print s2.model()
And the result:
sat
[x = -2, y = -3]
sat
[x = 0, y = 1]
Still the 'qe' tactic and the default solver seem buggy. They don't give the correct result.
Further comments and discussions are needed.
I've written some code below to check if two line segments intersect and if they do to tell me where. As input I have the (x,y) coordinates of both ends of each line. It appeared to be working correctly but now in the scenario where line A (532.87,787.79)(486.34,769.85) and line B (490.89,764.018)(478.98,783.129) it says they intersect at (770.136, 487.08) when the lines don't intersect at all.
Has anyone any idea what is incorrect in the below code?
double dy[2], dx[2], m[2], b[2];
double xint, yint, xi, yi;
WsqT_Location_Message *location_msg_ptr = OPC_NIL;
FIN (intersect (<args>));
dy[0] = y2 - y1;
dx[0] = x2 - x1;
dy[1] = y4 - y3;
dx[1] = x4 - x3;
m[0] = dy[0] / dx[0];
m[1] = dy[1] / dx[1];
b[0] = y1 - m[0] * x1;
b[1] = y3 - m[1] * x3;
if (m[0] != m[1])
{
//slopes not equal, compute intercept
xint = (b[0] - b[1]) / (m[1] - m[0]);
yint = m[1] * xint + b[1];
//is intercept in both line segments?
if ((xint <= max(x1, x2)) && (xint >= min(x1, x2)) &&
(yint <= max(y1, y2)) && (yint >= min(y1, y2)) &&
(xint <= max(x3, x4)) && (xint >= min(x3, x4)) &&
(yint <= max(y3, y4)) && (yint >= min(y3, y4)))
{
if (xi && yi)
{
xi = xint;
yi = yint;
location_msg_ptr = (WsqT_Location_Message*)op_prg_mem_alloc(sizeof(WsqT_Location_Message));
location_msg_ptr->current_latitude = xi;
location_msg_ptr->current_longitude = yi;
}
FRET(location_msg_ptr);
}
}
FRET(location_msg_ptr);
}
There is an absolutely great and simple theory about lines and their intersections that is based on adding an extra dimensions to your points and lines. In this theory a line can be created from two points with one line of code and the point of line intersection can be calculated with one line of code. Moreover, points at the Infinity and lines at the Infinity can be represented with real numbers.
You probably heard about homogeneous representation when a point [x, y] is represented as [x, y, 1] and the line ax+by+c=0 is represented as [a, b, c]?
The transitioning to Cartesian coordinates for a general homogeneous representation of a point [x, y, w] is [x/w, y/w]. This little trick makes all the difference including representation of lines at infinity (e.g. [1, 0, 0]) and making line representation look similar to point one. This introduces a GREAT symmetry into formulas for numerous line/point manipulation and is an
absolute MUST to use in programming. For example,
It is very easy to find line intersections through vector product
p = l1xl2
A line can be created from two points is a similar way:
l=p1xp2
In the code of OpenCV it it just:
line = p1.cross(p2);
p = line1.cross(line2);
Note that there are no marginal cases (such as division by zero or parallel lines) to be concerned with here. My point is, I suggest to rewrite your code to take advantage of this elegant theory about lines and points.
Finally, if you don't use openCV, you can use a 3D point class and create your own cross product function similar to this one:
template<typename _Tp> inline Point3_<_Tp> Point3_<_Tp>::cross(const Point3_<_Tp>& pt) const
{
return Point3_<_Tp>(y*pt.z - z*pt.y, z*pt.x - x*pt.z, x*pt.y - y*pt.x);
}