I want to implement the following optimization problem from the following paper Randomized Gossip Algorithms, Page 10 Eq 53
The screenshot of the optimization problem.
1- In this problem, W, P, and P_{ij} are n-by-n matrices. I would appreciate if you help me with implementing the following constraint in CVX.
W=\frac{1}{n}\sum_{i,j=1}^{n}P_{i,j}W_{i,j}
2- Also, in this problem, E is a set of neighbors of a nod i. Constraint P_{ij}=0 if {i,j}\not\in{E} means that P_{ij} is zero if node i and j are not neighbors.
Does anyone can help with how to implement this neighborhood relationship?
For $n=3$, neighbors.xlsx can look like:
screenshot of neighbors.xlsx
This means node 1 is neighbor with node 2, node 2 is neighbor with node 1 and 3, and node 3 is neighbor with node2.
I have the written the following piece of code for that in Matlab.
cvx_begin sdp
agt = struct([]);
neighbors = readcell('neighbors.xlsx');
N = 2;
for i = 1:N
agt(i).neighbors = neighbors{i};
end
variable s
variable P(N,N) symmetric
variable W_ij(N,N) symmetric
expression W
minimize (s)
subject to
P(:) >= 0;
j = 1;
for i = 1:N
D =[i,j];
if ~ismember(D,agt(i).neighbors)
P(i,j)== 0;
end
j = j+1;
end
for i = 1:N
for j = 1:N
W = P(i,j).*W_ij;
end
end
W = (1/N).*W;
W-(1/N)*ones(N,1)*ones(1,N) - s*eye(N) == semidefinite(N);
cvx_end
It does not work, and I get the following error. Any help is greatly appreciated.
Error using .* (line 262)
Disciplined convex programming error:
Invalid quadratic form(s): not a square.
Error in lambda (line 35)
W = P(i,j).*W_ij;
Related
I am trying to implement a simplex algorithm following the rules I was given at my optimization course. The problem is
min c'*x s.t.
Ax = b
x >= 0
All vectors are assumes to be columns, ' denotes the transpose. The algorithm should also return the solution to dual LP. The rules to follow are:
Here, A_J denotes columns from A with indices in J and x_J, x_K denotes elements of vector x with indices in J or K respectively. Vector a_s is column s of matrix A.
Now I do not understand how this algorithm takes care of condition x >= 0, but I decided to give it a try and follow it step by step. I used Matlab for this and got the following code.
X = zeros(n, 1);
Y = zeros(m, 1);
% i. Choose starting basis J and K = {1,2,...,n} \ J
J = [4 5 6] % for our problem
K = setdiff(1:n, J)
% this while is for goto
while 1
% ii. Solve system A_J*\bar{x}_J = b.
xbar = A(:,J) \ b
% iii. Calculate value of criterion function with respect to current x_J.
fval = c(J)' * xbar
% iv. Calculate dual solution y from A_J^T*y = c_J.
y = A(:,J)' \ c(J)
% v. Calculate \bar{c}^T = c_K^T - u^T A_K. If \bar{c}^T >= 0, we have
% found the optimal solution. If not, select the smallest s \in K, such
% that c_s < 0. Variable x_s enters basis.
cbar = c(K)' - c(J)' * inv(A(:,J)) * A(:,K)
cbar = cbar'
tmp = findnegative(cbar)
if tmp == -1 % we have found the optimal solution since cbar >= 0
X(J) = xbar;
Y = y;
FVAL = fval;
return
end
s = findnegative(c, K) %x_s enters basis
% vi. Solve system A_J*\bar{a} = a_s. If \bar{a} <= 0, then the problem is
% unbounded.
abar = A(:,J) \ A(:,s)
if findpositive(abar) == -1 % we failed to find positive number
disp('The problem is unbounded.')
return;
end
% vii. Calculate v = \bar{x}_J / \bar{a} and find the smallest rho \in J,
% such that v_rho > 0. Variable x_rho exits basis.
v = xbar ./ abar
rho = J(findpositive(v))
% viii. Update J and K and goto ii.
J = setdiff(J, rho)
J = union(J, s)
K = setdiff(K, s)
K = union(K, rho)
end
Functions findpositive(x) and findnegative(x, S) return the first index of positive or negative value in x. S is the set of indices, over which we look at. If S is omitted, whole vector is checked. Semicolons are omitted for debugging purposes.
The problem I tested this code on is
c = [-3 -1 -3 zeros(1,3)];
A = [2 1 1; 1 2 3; 2 2 1];
A = [A eye(3)];
b = [2; 5; 6];
The reason for zeros(1,3) and eye(3) is that the problem is inequalities and we need slack variables. I have set starting basis to [4 5 6] because the notes say that starting basis should be set to slack variables.
Now, what happens during execution is that on first run of while, variable with index 1 enters basis (in Matlab, indices go from 1 on) and 4 exits it and that is reasonable. On the second run, 2 enters the basis (since it is the smallest index such that c(idx) < 0 and 1 leaves it. But now on the next iteration, 1 enters basis again and I understand why it enters, because it is the smallest index, such that c(idx) < 0. But here the looping starts. I assume that should not have happened, but following the rules I cannot see how to prevent this.
I guess that there has to be something wrong with my interpretation of the notes but I just cannot see where I am wrong. I also remember that when we solved LP on the paper, we were updating our subjective function on each go, since when a variable entered basis, we removed it from the subjective function and expressed that variable in subj. function with the expression from one of the equalities, but I assume that is different algorithm.
Any remarks or help will be highly appreciated.
The problem has been solved. Turned out that the point 7 in the notes was wrong. Instead, point 7 should be
So, I'm trying to do the Gauss-Seidel method in Matlab and I found a code that does this but when I apply it to my matrices I get the Subscripted assignment dimension mismatch. error. I will show you my code in order to get a better idea.
%size of the matrix
n = 10;
%my matrices are empty in the beginning because my professor wants to run the algorithm for n = 100
and n = 1000. A's diagonal values are 3 and every other value is -1. b has the constants and the
first and last value will be 2,while every other value will be 1.
A = [];
b = [];
%assign the values to my matrices
for i=1:n
for j=1:n
if i == j
A(i,j) = 3;
else
A(i,j) = -1;
end
end
end
for i=2:n-1
b(i) = 1;
end
%here is the Gauss-Seidel algorithm
idx = 0;
while max(error) > 0.5 * 10^(-4)
idx = idx + 1;
Z = X;
for i = 1:n
j = 1:n; % define an array of the coefficients' elements
j(i) = []; % eliminate the unknow's coefficient from the remaining coefficients
Xtemp = X; % copy the unknows to a new variable
Xtemp(i) = []; % eliminate the unknown under question from the set of values
X(i) = (b(i) - sum(A(i,j) * Xtemp)) / A(i,i);
end
Xsolution(:,idx) = X;
error = abs(X - Z);
end
GaussSeidelTable = [1:idx;Xsolution]'
MaTrIx = [A X b]
I get the error for the Xsolution(:,idx) = X; part. I don't know what else to do. The code posted online works though, and the only difference is that the matrices are hardcoded in the m-file and A is a 5x5 matrix while b is a 5x1 matrix.
I am unable to run your code because some variables are not initialised, at least error and X. I assume the Problem is caused because Xsolution is already initialised from a previous run with a different size. Insert a Xsolution=[] to fix this.
Besides removing the error I have some suggestions to improve your code:
Use Functions, there are no "left over" variables from a previous run, causing errors like you got here.
Don't use the variable name error or i. error is a build-in function to throw errors and i is the imaginary unit. Both can cause hard to debug errors.
Initialise A with A=-1*ones(n,n);A(eye(size(A))==1)=3;, it's faster not to use a for loop in this case. To initialise b you can simply write b(1)=0;b(2:n-1)=1;
Use preallocation
the first time you run the code, Xsolution(:,idx) = X will create a Xsolution with the size of X.
the second time you run it, the existing Xsolution does not fit the size of new X.
this is another reason why you always want to allocate the array before using it.
I am having difficulty in finding roots of a nonlinear equation. I have tried Matlab and Maple both, and both give me the same error which is
Error, (in RootFinding:-NextZero) can only handle isolated zeros
The equation goes like
-100 + 0.1335600000e-5*H + (1/20)*H*arcsinh(2003.40/H)
The variable is H in the equation.
How do I find the roots (or the approximate roots) of this equation?
Matlab Code:
The function file:
function hor_force = horizontal(XY, XZ, Lo, EAo, qc, VA)
syms H
equation = (-1*ZZ) + (H/qc)*(cosh((qc/H)*(XZ- XB))) - H/qc + ZB;
hor_force = `solve(equation);`
The main file:
EAo = 7.5*10^7;
Lo = 100.17;
VA = 2002;
XY = 0;
ZY = 0;
XB = 50;
ZB = -2;
XZ = 100;
ZZ = 0;
ql = 40;
Error which Matlab shows:
Error using sym/solve (line 22)
Error using maplemex
Error, (in RootFinding:-NextZero) can only handle isolated zeros
Error in horizontal (line 8)
hor_force = solve(equation);
Error in main (line 34)
h = horizontal(XY, XZ, Lo, EAo, ql, VA)
http://postimg.org/image/gm93z3b7z/
You don't need the symbolic toolbox for this:
First, create an anonymous function that can take vectors at input (use .* and ./:
equation = #(H) ((-1*ZZ) + (H./qc).*(cosh((qc./H).*(XZ- XB))) - H./qc + ZB);
Second, create a vector that you afterwards insert into the equation to find approximately when the sign of the function changes. In the end, use fzero with x0 as the second input parameter.
H = linspace(1,1e6,1e4);
x0 = H(find(diff(sign(equation(H))))); %// Approximation of when the line crosses zero
x = fzero(equation, x0) %// Use fzero to find the crossing point, using the initial guess x0
x =
2.5013e+04
equation(x)
ans =
0
To verify:
You might want to check out this question for more information about how to find roots of non-polynomials.
In Maple, using the expression from your question,
restart:
ee := -100 + 0.1335600000e-5*H + (1/20)*H*arcsinh(2003.40/H):
Student:-Calculus1:-Roots(ee, -1e6..1e6);
[ 5 ]
[-1.240222868 10 , -21763.54830, 18502.23816]
#plot(ee, H=-1e6..1e6, view=-1..1);
One of the tasks for my assignment is to add an appropriate noise function to the following equation:
x = A*(1+a1*E)*sin(w*(1+a2*E)*t+y)+ a3*E
We must then plot the noise function vs. time as well as the original function with the noise added. I have asked the professor if a random number generator between -1 and 1 will suffice and he has agreed. I have the following code so far:
t = 0:0.1:6.5;
A = 2;
a1 = 2;
a2 = 4;
a3 = 3;
w = 1;
y = 2;
for i=1:length(t)
E(i) = random('unif', -1, 1, 1, 1);
x(i) = A*(1+a1*E(i))*sin(w*(1+a2*E(i))*t+y)+ a3*E(i);
i=i+1;
end
plot(t,E)
figure
stem(t,x)
I keep getting the following error In an assignment A(I) = B, the number of elements in B and I must
be the same.
Error in Try1 (line 58)
x(i) = A*(1+a1*E(i))*sin(w*(1+a2*E(i))*t+y)+ a3*E(i);
I don't understand the error because the E is just one number.
Any help appreciated!! Thanks!!
You're correct that E is just one number, but you're multiplying by t, which is not just one number - it's an array. I think you meant to multiply by t(i).
here------v
x(i) = A*(1+a1*E(i))*sin(w*(1+a2*E(i))*t(i)+y)+ a3*E(i);
Hello guys I am writing program to compute determinant(this part i already did) and Inverse matrix with GEPP. Here problem arises since i have completely no idea how to inverse Matrix using GEPP, i know how to inverse using Gauss Elimination ([A|I]=>[I|B]). I have searched through internet but still no clue, could you please explain me?
Here is my matlab code (maybe someone will find it useful), as of now it solves AX=b and computes determinant:
function [det1,X ] = gauss_czesciowy( A, b )
%GEPP
perm=0;
n = length(b);
if n~=m
error('vector has wrong size');
end
for j = 1:n
p=j;
% choice of main element
for i = j:n
if abs(A(i,j)) >= abs(A(p,j))
p = i;
end
end
if A(p,j) == 0
error('Matrix A is singular');
end
%rows permutation
t = A(p,:);
A(p,:) = A(j,:);
A(j,:) = t;
t = b(p);
b(p) = b(j);
b(j) = t;
if~(p==i)
perm=perm+1;
end
% reduction
for i = j+1:n
t = (A(i,j)/A(j,j));
A(i,:) = A(i,:)-A(j,:)*t;
b(i) = b(i)-b(j)*t;
end
end
%determinant
mn=1;
for i=1:n
mn=mn*A(i,i);
end
det1=mn*(-1)^perm;
% solution
X = zeros(1,n);
X(n) = b(n)/A(n,n);
if (det1~=0)
for i = 1:n
s = sum( A(i, (i+1):n) .* X((i+1):n) );
X(i) = (b(i) - s) / A(i,i);
end
end
end
Here is the algorithm for Guassian elimination with partial pivoting. Basically you do Gaussian elimination as usual, but at each step you exchange rows to pick the largest-valued pivot available.
To get the inverse, you have to keep track of how you are switching rows and create a permutation matrix P. The permutation matrix is just the identity matrix of the same size as your A-matrix, but with the same row switches performed. Then you have:
[A] --> GEPP --> [B] and [P]
[A]^(-1) = [B]*[P]
I would try this on a couple of matrices just to be sure.
EDIT: Rather than empirically testing this, let's reason it out. Basically what you are doing when you switch rows in A is you are multiplying it by your permutation matrix P. You could just do this before you started GE and end up with the same result, which would be:
[P*A|I] --> GE --> [I|B] or
(P*A)^(-1) = B
Due to the properties of the inverse operation, this can be rewritten:
A^(-1) * P^(-1) = B
And you can multiply both sides by P on the right to get:
A^(-1) * P^(-1)*P = B*P
A^(-1) * I = B*P
A^(-1) = B*P