I have the following problem and I can't figure out if cvxpy can do what I need.
Context: I optimize portfolios. When buying bonds and optimizing the quantity of each bond to buy, it's only possible to buy each bond only in multiples of 1,000 units.
However, the minimum piece required to be bought is most of the time 10,000.
This means we either don't buy a bond at all or if we buy it, the quantity bought has to be either 10,000, 11,000, 12,000 and so on.
Is there a way (it seems it doesn't) to restrict certain values from the possible solutions an integer variable can have?
So let's assume we have an integer variable x that is non negative.
We basically want to buy 1000x but we know that x can be x = {0, 10, 11, 12, ...}
Is it possible to skip values 1.. 9 without adding other variables?
For example:
import numpy as np
import pandas as pd
import cvxpy as cvx
np.random.seed(1)
# np.random.rand(3)
p = pd.DataFrame({'bond_id': ['s1','s2', 's3', 's4', 's5', 's6', 's7','s8','s9', 's10'],
'er': np.random.rand(10),
'units': [10000,2000,3000,4000,27000,4000,0,0,0,0] })
final_units = cvx.Variable( 10, integer=True)
constraints = list()
constraints.append( final_units >= 0)
constraints.append(sum(final_units*1000) <= 50000)
constraints.append(sum(final_units*1000) >= 50000)
constraints.append(final_units <= 15)
obj = cvx.Maximize( final_units # np.array(list(p['er'])) )
prob = cvx.Problem(obj, constraints)
solve_val = prob.solve()
print("\n* solve_val = {}".format(solve_val))
solution_value = prob.value
solution = str(prob.status).lower()
print("\n** SOLUTION 3: {} Value: {} ".format(solution, solution_value))
print("\n* final_units -> \n{}\n".format(final_units.value))
p['FINAL_SOL'] = final_units.value * 1000
print("\n* Final Portfolio: \n{}\n".format(p))
This solution is a very simplified version of the problem I face. The final vector final_units can suggest values like in this example where we have to buy 5,000 units of bond s9, however I can't since the min I can buy is 10,000.
I know I could add an additional integer vector to express an OR condition, but in reality my real problem is way bigger than this, I have thousand of integer variables already. Hence, I wonder if there's a way to exclude values from 1 to 9 without adding additional variables to the problem.
Thank you
No, not with CVXPY. You can model it with an integer variable x[i] plus a binary variable y[i], and using the constraints (in math notation):
y[i] * 10 <= x[i] <= y[i] * 15
This results in x[i] ∈ {0, 10..15}.
Some solvers have a variable type for this: semi-integer variables. Using this you don't need to have an extra binary variable and these 2 constraints. CVXPY does not support this variable type AFAIK.
Related
I'm struggling a bit finding a fast algorithm that's suitable.
I just want to minimize:
norm2(x-s)
st
G.x <= h
x >= 0
sum(x) = R
G is sparse and contains only 1s (and zeros obviously).
In the case of iterative algorithms, it would be nice to get the interim solutions to show to the user.
The context is that s is a vector of current results, and the user is saying "well the sum of these few entries (entries indicated by a few 1.0's in a row in G) should be less than this value (a row in h). So we have to remove quantities from the entries the user specified (indicated by 1.0 entries in G) in a least-squares optimal way, but since we have a global constraint on the total (R) the values removed need to be allocated in a least-squares optimal way amongst the other entries. The entries can't go negative.
All the algorithms I'm looking at are much more general, and as a result are much more complex. Also, they seem quite slow. I don't see this as a complex problem, although mixes of equality and inequality constraints always seem to make things more complex.
This has to be called from Python, so I'm looking at Python libraries like qpsolvers and scipy.optimize. But I suppose Java or C++ libraries could be used and called from Python, which might be good since multithreading is better in Java and C++.
Any thoughts on what library/package/approach to use to best solve this problem?
The size of the problem is about 150,000 rows in s, and a few dozen rows in G.
Thanks!
Your problem is a linear least squares:
minimize_x norm2(x-s)
such that G x <= h
x >= 0
1^T x = R
Thus it fits the bill of the solve_ls function in qpsolvers.
Here is an instance of how I imagine your problem matrices would look like, given what you specified. Since it is sparse we should use SciPy CSC matrices, and regular NumPy arrays for vectors:
import numpy as np
import scipy.sparse as spa
n = 150_000
# minimize 1/2 || x - s ||^2
R = spa.eye(n, format="csc")
s = np.array(range(n), dtype=float)
# such that G * x <= h
G = spa.diags(
diagonals=[
[1.0 if i % 2 == 0 else 0.0 for i in range(n)],
[1.0 if i % 3 == 0 else 0.0 for i in range(n - 1)],
[1.0 if i % 5 == 0 else 0.0 for i in range(n - 1)],
],
offsets=[0, 1, -1],
)
a_dozen_rows = np.linspace(0, n - 1, 12, dtype=int)
G = G[a_dozen_rows]
h = np.ones(12)
# such that sum(x) == 42
A = spa.csc_matrix(np.ones((1, n)))
b = np.array([42.0]).reshape((1,))
# such that x >= 0
lb = np.zeros(n)
Next, we can solve this problem with:
from qpsolvers import solve_ls
x = solve_ls(R, s, G, h, A, b, lb, solver="osqp", verbose=True)
Here I picked CVXOPT but there are other open-source solvers you can install such as ProxQP, OSQP or SCS. You can install a set of open-source solvers by: pip install qpsolvers[open_source_solvers]. After some solvers are installed, you can list those for sparse matrices by:
print(qpsolvers.sparse_solvers)
Finally, here is some code to check that the solution returned by the solver satisfies our constraints:
tol = 1e-6 # tolerance for checks
print(f"- Objective: {0.5 * (x - s).dot(x - s):.1f}")
print(f"- G * x <= h: {(G.dot(x) <= h + tol).all()}")
print(f"- x >= 0: {(x + tol >= 0.0).all()}")
print(f"- sum(x) = {x.sum():.1f}")
I just tried it with OSQP (adding the eps_rel=1e-5 keyword argument when calling solve_ls, otherwise the returned solution would be less accurate than the tol = 1e-6 tolerance) and it found a solution is 737 milliseconds on my (rather old) CPU with:
- Objective: 562494373088866.8
- G * x <= h: True
- x >= 0: True
- sum(x) = 42.0
Hoping this helps. Happy solving!
Suppose there are three variables that take on discrete integer values, say w1 = {1,2,3,4,5,6,7,8,9,10,11,12}, w2 = {1,2,3,4,5,6,7,8,9,10,11,12}, and w3 = {1,2,3,4,5,6,7,8,9,10,11,12}. The task is to pick one value from each set such that the resulting triplet minimizes some (black box, computationally expensive) cost function.
I've tried the surrogate optimization in Matlab but I'm not sure it is appropriate. I've also heard about simulated annealing but found no implementation applied to this instance.
Which algorithm, apart from exhaustive search, can solve this combinatorial optimization problem?
Any help would be much appreciated.
The requirement/benefit of Simulated Annealing (SA), is that the objective surface is somewhat smooth, that is, we can be close to a solution.
For a completely random spiky surface- you might as well do a random search
If it is anything smooth, or even sometimes, it makes sense to try SA.
The idea is that (sometimes) changing only 1 of the 3 values, we have little effect on out blackbox function.
Here is a basic example to do this with Simulated Annealing, using frigidum in Python
import numpy as np
w1 = np.array( [1,2,3,4,5,6,7,8,9,10,11,12] )
w2 = np.array( [1,2,3,4,5,6,7,8,9,10,11,12] )
w3 = np.array( [1,2,3,4,5,6,7,8,9,10,11,12] )
W = np.array([w1,w2,w3])
LENGTH = 12
I define a black-box using the Rastrigin function.
def rastrigin_function_n( x ):
"""
N-dimensional Rastrigin
https://en.wikipedia.org/wiki/Rastrigin_function
x_i is in [-5.12, 5.12]
"""
A = 10
n = x.shape[0]
return A*n + np.sum( x**2- A*np.cos(2*np.pi * x) )
def black_box( x ):
"""
Transform from domain [1,12] to [-5,5]
to be able to push to rastrigin
"""
x = (x - 6.5) * (5/5.5)
return rastrigin_function_n(x)
Simulated Annealing needs to modify state X. Instead of taking/modifying values directly, we keep track of indices. This simplifies creating new proposals as an index is always an integer we can simply add/subtract 1 modulo LENGTH.
def random_start():
"""
returns 3 random indices
"""
return np.random.randint(0, LENGTH, size=3)
def random_small_step(x):
"""
change only 1 index
"""
d = np.array( [1,0,0] )
if np.random.random() < .5:
d = np.array( [-1,0,0] )
np.random.shuffle(d)
return (x+d) % LENGTH
def random_big_step(x):
"""
change 2 indici
"""
d = np.array( [1,-1,0] )
np.random.shuffle(d)
return (x+d) % LENGTH
def obj(x):
"""
We have a triplet of indici,
1. Calculate corresponding values in W = [w1,w2,w3]
2. Push the values in out black-box function
"""
indices = x
values = W[np.array([0,1,2]), indices]
return black_box(values)
And throw a SA Scheme at it
import frigidum
local_opt = frigidum.sa(random_start=random_start,
neighbours=[random_small_step, random_big_step],
objective_function=obj,
T_start=10**4,
T_stop=0.000001,
repeats=10**3,
copy_state=frigidum.annealing.naked)
I am not sure what the minimum for this function should be, but it found a objective with 47.9095 with indicis np.array([9, 2, 2])
Edit:
For frigidum to change the cooling schedule, use alpha=.9. My experience is that all the work of experiment which cooling scheme works best doesn't out-weight simply let it run a little longer. The multiplication you proposed, (sometimes called geometric) is the standard one, also implemented in frigidum. So to implement Tn+1 = 0.9*Tn you need a alpha=.9. Be aware this cooling step is done after N repeats, so if repeats=100, it will first do 100 proposals before lowering the temperature with factor alpha
Simple variations on current state often works best. Since its best practice to set the initial temperature high enough to make most proposals (>90%) accepted, it doesn't matter the steps are small. But if you fear its soo small, try 2 or 3 variations. Frigidum accepts a list of proposal functions, and combinations can enforce each other.
I have no experience with MINLP. But even if, so many times experiments can surprise us. So if time/cost is small to bring another competitor to the table, yes!
Try every possible combination of the three values and see which has the lowest cost.
Let's say I have a list of 5 words:
[this, is, a, short, list]
Furthermore, I can classify some text by counting the occurrences of the words from the list above and representing these counts as a vector:
N = [1,0,2,5,10] # 1x this, 0x is, 2x a, 5x short, 10x list found in the given text
In the same way, I classify many other texts (count the 5 words per text, and represent them as counts - each row represents a different text which we will be comparing to N):
M = [[1,0,2,0,5],
[0,0,0,0,0],
[2,0,0,0,20],
[4,0,8,20,40],
...]
Now, I want to find the top 1 (2, 3 etc) rows from M that are most similar to N. Or on simple words, the most similar texts to my initial text.
The challenge is, just checking the distances between N and each row from M is not enough, since for example row M4 [4,0,8,20,40] is very different by distance from N, but still proportional (by a factor of 4) and therefore very similar. For example, the text in row M4 can be just 4x as long as the text represented by N, so naturally all counts will be 4x as high.
What is the best approach to solve this problem (of finding the most 1,2,3 etc similar texts from M to the text in N)?
Generally speaking, the most widely standard technique of bag of words (i.e. you arrays) for similarity is to check cosine similarity measure. This maps your bag of n (here 5) words to a n-dimensional space and each array is a point (which is essentially also a point vector) in that space. The most similar vectors(/points) would be ones that have the least angle to your text N in that space (this automatically takes care of proportional ones as they would be close in angle). Therefore, here is a code for it (assuming M and N are numpy arrays of the similar shape introduced in the question):
import numpy as np
cos_sim = M[np.argmax(np.dot(N, M.T)/(np.linalg.norm(M)*np.linalg.norm(N)))]
which gives output [ 4 0 8 20 40] for your inputs.
You can normalise your row counts to remove the length effect as you discussed. Row normalisation of M can be done as M / M.sum(axis=1)[:, np.newaxis]. The residual values can then be calculated as the sum of the square difference between N and M per row. The minimum difference (ignoring NaN or inf values obtained if the row sum is 0) is then the most similar.
Here is an example:
import numpy as np
N = np.array([1,0,2,5,10])
M = np.array([[1,0,2,0,5],
[0,0,0,0,0],
[2,0,0,0,20],
[4,0,8,20,40]])
# sqrt of sum of normalised square differences
similarity = np.sqrt(np.sum((M / M.sum(axis=1)[:, np.newaxis] - N / np.sum(N))**2, axis=1))
# remove any Nan values obtained by dividing by 0 by making them larger than one element
similarity[np.isnan(similarity)] = similarity[0]+1
result = M[similarity.argmin()]
result
>>> array([ 4, 0, 8, 20, 40])
You could then use np.argsort(similarity)[:n] to get the n most similar rows.
I'm trying to optimize a configuration X (boolean), such that the total price : base_price + discount, on a configuration is minimized, but the problem formulation gives a Matmul error since x is a cvxpy Variable and thus doesn't conform to the Numpy shape even though it was defined with the correct length.
n = len(Configuration)
x = cp.Variable(n, boolean=True)
problem = cp.Problem(cp.Minimize(base_price + price#(price_rules_A#x <= price_rules_B)), [
config_rules_A#x <= config_rules_B,
config_rules_2A#x == config_rules_2B
])
# where price#(price_rules_A#x <= price_rules_B) is the total discount
# and price, price_rules_A and price_rules_B are numpy arrays
The error i get is
ValueError: matmul: Input operand 1 does not have enough dimensions (has 0, gufunc core with signature (n?,k),(k,m?)->(n?,m?) requires 1)
I expect it to find an optimal config for x ( 0010110...) such that the discount is minimized but it doesn't. Any idea what might be causing this?
Assuming the evaluation of the inequality in the objective function is suppose to work as index to price, you can rewrite the function as
cp.Minimize(base_price + price#(1-(price_rules_B - price_rules_A#x))
Then the elements in price where the inequality is true will be summed.
I am trying to debug an index problem I am having on my CUDA machine
Cuda Machine Info:
{1->{Name->Tesla C2050,Clock Rate->1147000,Compute Capabilities->2.,GPU Overlap->1,Maximum Block Dimensions->{1024,1024,64},Maximum Grid Dimensions->{65535,65535,65535},Maximum Threads Per Block->1024,Maximum Shared Memory Per Block->49152,Total Constant Memory->65536,Warp Size->32,Maximum Pitch->2147483647,Maximum Registers Per Block->32768,Texture Alignment->512,Multiprocessor Count->14,Core Count->448,Execution Timeout->0,Integrated->False,Can Map Host Memory->True,Compute Mode->Default,Texture1D Width->65536,Texture2D Width->65536,Texture2D Height->65535,Texture3D Width->2048,Texture3D Height->2048,Texture3D Depth->2048,Texture2D Array Width->16384,Texture2D Array Height->16384,Texture2D Array Slices->2048,Surface Alignment->512,Concurrent Kernels->True,ECC Enabled->True,Total Memory->2817982462},
All this code does is set the values of a 3D array equal to the index that CUDA is using:
__global __ void cudaMatExp(
float *matrix1, float *matrixStore, int lengthx, int lengthy, int lengthz){
long UniqueBlockIndex = blockIdx.y * gridDim.x + blockIdx.x;
long index = UniqueBlockIndex * blockDim.z * blockDim.y * blockDim.x +
threadIdx.z * blockDim.y * blockDim.x + threadIdx.y * blockDim.x +
threadIdx.x;
if (index < lengthx*lengthy*lengthz) {
matrixStore[index] = index;
}
}
For some reason, once the dimension of my 3D array becomes too large, the indexing stops.
I have tried different block dimensions (blockDim.x by blockDim.y by blockDim.z):
8x8x8 only gives correct indexing up to array dimension 12x12x12
9x9x9 only gives correct indexing up to array dimension 14x14x14
10x10x10 only gives correct indexing up to array dimension 15x15x15
For dimensions larger than these all of the different block sizes eventually start to increase again, but they never reach a value of dim^3-1 (which is the maximum index that the cuda thread should reach)
Here are some plots that illustrate this behavior:
For example: This is plotting on the x axis the dimension of the 3D array (which is xxx), and on the y axis the maximum index number that is processed during the cuda execution. This particular plot is for block dimensions of 10x10x10.
Here is the (Mathematica) code to generate that plot, but when I ran this one, I used block dimensions of 1024x1x1:
CUDAExp = CUDAFunctionLoad[codeexp, "cudaMatExp",
{{"Float", _,"Input"}, {"Float", _,"Output"},
_Integer, _Integer, _Integer},
{1024, 1, 1}]; (*These last three numbers are the block dimensions*)
max = 100; (* the maximum dimension of the 3D array *)
hold = Table[1, {i, 1, max}];
compare = Table[i^3, {i, 1, max}];
Do[
dim = ii;
AA = CUDAMemoryLoad[ConstantArray[1.0, {dim, dim, dim}], Real,
"TargetPrecision" -> "Single"];
BB = CUDAMemoryLoad[ConstantArray[1.0, {dim, dim, dim}], Real,
"TargetPrecision" -> "Single"];
hold[[ii]] = Max[Flatten[
CUDAMemoryGet[CUDAExp[AA, BB, dim, dim, dim][[1]]]]];
, {ii, 1, max}]
ListLinePlot[{compare, Flatten[hold]}, PlotRange -> All]
This is the same plot, but now plotting x^3 to compare to where it should be. Notice that it diverges after the dimension of the array is >32
I test the dimensions of the 3D array and look at how far the indexing goes and compare it with dim^3-1. E.g. for dim=32, the cuda max index is 32767 (which is 32^3 -1), but for dim=33 the cuda output is 33791 when it should be 35936 (33^3 -1). Notice that 33791-32767 = 1024 = blockDim.x
Question:
Is there a way to correctly index an array with dimensions larger than the block dimensions in Mathematica?
Now, I know that some people use __mul24(threadIdx.y,blockDim.x) in their index equation to prevent errors in bit multiplication, but it doesn't seem to help in my case.
Also, I have seen someone mention that you should compile your code with -arch=sm_11 because by default it's compiled for compute capability 1.0. I don't know if this is the case in Mathematica though. I would assume that CUDAFunctionLoad[] knows to compile with 2.0 capability. Any one know?
Any suggestions would be extremely helpful!
So, Mathematica kind of has a hidden way of dealing with grid dimensions, to fix your grid dimension to something that will work, you have to add another number to the end of the function you are calling.
The argument denotes the number of threads to launch (or grid dimension times block dimension).
For example, in my code above:
CUDAExp =
CUDAFunctionLoad[codeexp,
"cudaMatExp", {
{"Float", _, "Input"}, {"Float", _,"Output"},
_Integer, _Integer, _Integer},
{8, 8, 8}, "ShellOutputFunction" -> Print];
(8,8,8) denotes the dimension of the block.
When you call CUDAExp[] in mathematica, you can add an argument that denotes the number of threads to launch:
In this example I finally got it to work with the following:
// AA and BB are 3D arrays of 0 with dimensions dim^3
dim = 64;
CUDAExp[AA, BB, dim, dim, dim, 4089];
Note that when you compile with CUDAFunctionLoad[], it only expects 5 inputs, the first is the array you pass it (of dimensions dim x dim x dim) and the second is where the memory of it is stored. The third, fourth, and fifth are the dimensions.
When you pass it a 6th, mathematica translates that as gridDim.x * blockDim.x, so, since I know I need gridDim.x = 512 in order for every element in the array to be dealt with, I set this number equal to 512 * 8 = 4089.
I hope this is clear and useful to someone in the future that comes across this issue.