This is going to be a long question :
I have written a code in MATLAB for updating the weights of MLP with one hidden layer . Here is the code :
function [ weights_1,weights_2 ] = changeWeights( X,y,weights_1,weights_2,alpha )
%CHANGEWEIGHTS updates the weight matrices
% This function changes the weight of the weight matrix
% for a given value of alpha using the back propogation algortihm
m = size(X,1) ; % number of samples in the training set
for i = 1:m
% Performing the feed-forward step
X_i = [1 X(i,1:end)] ; % 1-by-5 input
z2_i = X_i*weights_1' ; % 1-by-4
a2_i = sigmoid(z2_i) ; % 1-by-4
a2_i = [1 a2_i] ; % 1-by-5
z3_i = a2_i*weights_2' ; % 1-by-3
h_i = sigmoid(z3_i) ; % 1-by-3
% Calculating the delta_output_layer
delta_output_layer = ( y(i)' - h_i' )...
.*sigmoidGradient(z3_i') ; % 3-by-1 matrix
% Calculating the delta_hidden_layer
delta_hidden_layer = (weights_2'*delta_output_layer)...
.*sigmoidGradient([1;z2_i']) ; % 5-by-1 matrix
delta_hidden_layer = delta_hidden_layer(2:end) ;
% Updating the weight matrices
weights_2 = weights_2 + alpha*delta_output_layer*a2_i ;
weights_1 = weights_1 + alpha*delta_hidden_layer*X_i ;
end
end
Now I wanted to test it on the fisheriris dataset given in MATLAB which can be accesed by load fisheriris command . I renamed meas to X and changed species to a 150-by-3 matrix where each row depicts the name of species (as for example first row is [1 0 0])
I compute error of the output layer using the following function :
function [ g ] = costFunction( X,y,weights_1,weights_2 )
%COST calculates the error
% This function calculates the error in the
% output of the neural network
% Performing the feed-forward propogation
m = size(X,1) ;
X_temp = [ones([m 1]) X] ; % 150-by-5 matrix
z2 = X_temp*weights_1' ; % 150-by-5-by-5-by-4
a2 = sigmoid(z2) ;
a2 = [ones([m 1]) a2] ; % 150-by-5
z3 = a2*weights_2' ; % 150-by-3
h = sigmoid(z3) ; % 150-by-3
g = 0.5*sum(sum((y-h).^2)) ;
g = g/m ;
end
Now in the course the prof gave an example of toy network with 3 iterations , I tested this on that network and it gives the right values but when I test it on the fisheriris data the cost keeps on increasing . And I am not able to understand where it is going wrong .
Related
I have a stepped shaft as per the attached image. Following information available as an input parameters:
Young's modulus 123e3N/mm^2.
Cross-sectional area 300mm^2 for the length of 400mm
Cross-sectional area 400mm^2 for the length of 250mm
Axial force of 200kN acts axially on the shaft and the location of load is at 200mm from the one end of the shaft on cross-sectional area of 300mm^2
I need help to make do finite element analysis in MATALB.
Please help me in making MATLAB code for this.
%% Clearing workspace
clc
clear
close all
%% Element specifications
ne = 3; % Number of elements
nne = 2; % Number of nodes per element
nn = ne*(nne - 1) + 1; % total number of nodes
ndof = 1; % Number of degress of freedom per node
sg = nn*ndof; % size of global stiffness matrix
se = nne*ndof; % size of elemental stiffness matrix
KG = zeros(sg,sg); % Global stiffness matrix
Ke = zeros(se,se); % Elemental Stiffness MAtrix
Fe = zeros(se,1); % Elemental Force Vector
FG = zeros(sg,1); % Global Force Vector
%% Geometrical parameters
E = 123e3*ones(1,ne); % Young's Modulus in N/mm^2
P = 200e3; % Force in N
F = P;
A = ones(1,ne) ; % Area of cross-section
A(1)=300; % Area of cross-section of 1st element in mm^2
A(2)=300; % Area of cross-section of 2nd element in mm^2
A(3)=400; % Area of cross-section of 3rd element in mm^2
L = ones(1,ne); % Length of elements in mm
L(1)=200; % Length of 1st element in mm
L(2)=200; % Length of 2nd element in mm
L(3)=250; % Length of 3rd element in mm
%% Assembly of Global Stiffness Matrix
for i = 1:ne
Ke = (A(i)*E(i)/L(i))*[1 -1;-1 1]; % Element Stiffness Matrix
for j = 1:se
for k = 1:se
KG(i + j - 1, i + k - 1) = KG(i + j - 1, i + k - 1) + Ke(j,k);
end
end
end
%% Concentrated Load Vector at end
FG(2,1) = F; % Defining location of concentrated load
%% Application of boundary conditions
KGS = KG;
cdof = [1 4]; % specify fixed degree of freedom number
Lcdof = length(cdof);
for a = 1:Lcdof
KGS(cdof(a),:) = 0;
KGS(:,cdof(a)) = 1;
FG(cdof(a),1) = 0;
end
FGL = length(FG);
for b = 1:FGL
if(b > length(FG))
elseif(FG(b)<0)
FG(b) = [];
end
end
%% Solving for displacement
U = linsolve(KGS,FG)
U1=KGS\FG
%% Calculation of Reaction Forces
FR = KG*U1
I'm trying to implement a neural network with sigmoid function
But the following code doesnot work
This is the training part of neural network.
It doesnot update the weights properly
What is wrong in this code?
clc; clear all; close all;
% load train_data1
train_data1=[-0.498800000000000,-0.257500000000000;-0.492800000000000,-0.274300000000000;-0.470300000000000,-0.282600000000000;-0.427400000000000,-0.474000000000000;-0.420400000000000,-0.518000000000000;-0.326300000000000,-1.13230000000000;-0.317300000000000,-0.875300000000000;-0.295000000000000,-1.02770000000000;-0.267600000000000,-0.882800000000000;-0.260500000000000,-0.976500000000000;-0.216100000000000,-0.970400000000000;-0.207000000000000,-0.813800000000000;-0.164000000000000,-0.696600000000000;-0.159900000000000,-0.793300000000000;-0.122000000000000,-0.764400000000000;-0.0729000000000000,-0.435300000000000;-0.00640000000000000,-0.0546000000000000;0.132200000000000,0.710300000000000;0.137100000000000,0.587000000000000;0.160300000000000,0.819200000000000;0.230600000000000,0.989200000000000;0.286800000000000,0.737700000000000;0.334000000000000,0.943500000000000;0.375200000000000,0.688200000000000;0.429700000000000,0.567800000000000];
train_data1 = sortrows(train_data1);
% normalize data to [0,1]
data1=[train_data1];
max1=max(max(data1));
min1=min(min(data1));
train_data2 = (train_data1 - min1) / ( max1 - min1);
x = train_data2(:,1); % train input data
r = train_data2(:,2); % train output data
hidden_neurons = 2;
maxepochcount = 1000;
datacount1 = size(x,1);
% add a bias as an input
bias = ones(datacount1,1);
% x = [x bias];
% read how many inputs
inputcount = size(x,2);
% ---------- data loaded -----------
% ---------- set weights -----------
% set initial random weights
WI = (randn(inputcount,hidden_neurons) - 0.5)/10;
WO = (randn(1,hidden_neurons) - 0.5)/10;
%-----------------------------------
%--- Learning Starts Here! ---------
%-----------------------------------
eta1 = 0.5;
eta2 = eta1/5;
% do a number of epochs
for iter = 1:maxepochcount
% loop through the data
for j = 1:datacount1
% read the current sample
I = x(j,:);
D = r(j,1);
% calculate the error for this sample
H = (sigmoid(I * WI))';
O = H' * WO';
error = D-O;
% adjust weight between hidden & output
delta_i = O.*(1-O).*(D-O); % D actual, O calculated output
% Calculate error for each node in layer_(n-1)
delta_j = H.*(1-H).*(WO.'*delta_i); % H.' is the output of hidden layer
% Adjust weights in matrices sequentially
WO = WO + eta2.*delta_i*(H.') % H.' is the output of hidden layer
WI = WI + eta1.*(delta_j*(I))' % I.' is the inputs
% % adjust weight between hidden & output
% delta_HO = error.*eta2 .* hidden_val;
% WO = WO - delta_HO';
% % adjust the weights between input & hidden
% delta_IH = eta1 .* error .* WO' .* (1 - (H .^ 2)) * I;
% WI = WI - delta_IH';
end
O = sigmoid(WO*sigmoid(x * WI)');
% error(iter) = (sum(error .^ 2)) ^ 0.5;
if rem(iter,100)==0 % Every 100 epochs, show how training is doing
plot(x,O, 'color','red','linewidth',2); hold on;
drawnow;
iter
end
% return
end
only the output values are needed to be scaled to the activation function.
If we use tanh we must scale them to [-1,1], in case of sigmoid [0,1].
The code is working fine but sometimes it needs more epochs.
I want to plot the step response. I know that I can use step function with state space equations, but I try to get same results using plot function. Here is my sample of code:
for i=1:201
u(i) = 1;
x1(i+1) = (-(b/J)*x1(i) + (K/J)*x2(i));
x2(i+1) = (-(K/L)*x1(i) - (R/L)*x2(i) + (1/L)*u(i));
y(i) = x1(i);
end
and this is the state space equations:
A = [-b/J K/J
-K/L -R/L];
B = [0
1/L];
C = [1 0];
D = 0;
If i do:
t = 0:1:200;
plot(t, y)
it is not working and I want to have the same results like the step function below:
sys = ss(A,B,C,D);
step(sys)
You can find my state space equation here.
The reason for the mismatch is that sys is a continuous time model, whereas the computation of y treats it as a discrete-time system.
The following is a way of estimating the step-response of a continuous-time system in the discrete-time domain:
% Given from the problem statement
A = [-b/J K/J
-K/L -R/L];
B = [0
1/L];
C = [1 0];
D = 0;
% this is your continuous-time model
sys = ss(A,B,C,D);
% define the sample rate of the equivalent discrete-time model
Ts = 1/10;
% this needs to be something smaller than the time-constants in your model,
% so that you have enough resolution to represent the continuous-time
% signal.
% convert the system to the equivalent discrete-time model
sysd = c2d(sys,Ts);
% define how long a step response you'd like to compute
T = 7;
% this should be long enough to cover the length of the step response
t = 0:Ts:T; % time-grid for the plot
nSmp = length(t); % total number of samples to be computed
% initializations
y = NaN(1, nSmp); % output vector
u = ones(1, nSmp); % unit step input
X = [0; 0]; % state vector, initialized to 0
% compute the samples of the step-response
% (i prefer to use vectorized form to keep the code concise)
for i=1:nSmp
y(i) = sysd.C * X + sysd.D * u(i);
X = sysd.A * X + sysd.B * u(i);
end
% plot continous-time step response
figure;
step(sys);
% plot simulated discrete-time step response
figure;
plot(t, y, 'r')
xlabel('Time (s)');
ylabel('Amplitude');
title('Simulated Step Response');
I want to plot multiple realizations of a stochastic process in matlab. For a single realization I have the following code:
N = 80;
T = dt*N;
dWt = zeros(1,N);
S= repmat(S0,1,N);
S(1) = S0;
dWt = sqrt(dt) * randn;
for t=2:N
dWt(t) = sqrt(dt)*randn;
dSt = k*(mu-S(t-1))*dt + sigma*S(t-1)*dWt(t);
S(t) = S(t-1)+dSt;
end
plot(handles.pMeasure, [0:dt:T],[S0,S]);
I want to replicate this loop n times and plot the results in one plot.
You could add an additional for loop, but it would be best to vectorize everything and calculate all n instances at once:
k = ...
mu = ...
sigma = ...
S0 = ... % Initial condition
dt = ... % Time step
n = ... % Number of instances
N = 80; % Number of time steps, not counting initial condition
T = dt*N; % Final time
rng(1); % Always seed random number generator
dWt = sigma*sqrt(dt)*randn(n,N); % Calculate Wiener increments
S = zeros(n,N+1); % Allocate
S(:,1) = S0; % Set initial conditions
for t = 2:N+1
S(:,t) = S(:,t-1) + k*(mu-S(:,t-1))*dt + S(:,t-1).*dWt(:,t-1);
end
plot(handles.pMeasure,0:dt:T,S)
There are further ways to optimize this if want or you can also try sde_euler in my SDETools Matlab toolbox:
k = ...
mu = ...
sigma = ...
dt = ... % Time step
n = ... % Number of instances
N = 80; % Number of time steps, not counting initial condition
T = dt*N; % Final time
f = #(t,y)k*(mu-y); % Diffusion function
g = #(t,y)sigma*y; % Drift function
t = 0:dt:T; % Time vector
S0 = zeros(n,1); % Initial conditions
opts = sdeset('RandSeed',1,...
'SDEType','Ito'); % Set random seed, specify Ito SDE
S = sde_euler(f,g,t,S0,opts); % Simulate
plot(t,S)
I've been asked to write down a Matlab program in order to solve LPs using the Revised Simplex Method.
The code I wrote runs without problems with input data although I've realised it doesn't solve the problem properly, as it does not update the inverse of the basis B (the real core idea of the abovementioned method).
The problem is only related to a part of the code, the one in the bottom of the script aiming at:
Computing the new inverse basis B^-1 by performing elementary row operations on [B^-1 u] (pivot row index is l_out). The vector u is transformed into a unit vector with u(l_out) = 1 and u(i) = 0 for other i.
Here's the code I wrote:
%% Implementation of the revised Simplex. Solves a linear
% programming problem of the form
%
% min c'*x
% s.t. Ax = b
% x >= 0
%
% The function input parameters are the following:
% A: The constraint matrix
% b: The rhs vector
% c: The vector of cost coefficients
% C: The indices of the basic variables corresponding to an
% initial basic feasible solution
%
% The function returns:
% x_opt: Decision variable values at the optimal solution
% f_opt: Objective function value at the optimal solution
%
% Usage: [x_opt, f_opt] = S12345X(A,b,c,C)
% NOTE: Replace 12345X with your own student number
% and rename the file accordingly
function [x_opt, f_opt] = SXXXXX(A,b,c,C)
%% Initialization phase
% Initialize the vector of decision variables
x = zeros(length(c),1);
% Create the initial Basis matrix, compute its inverse and
% compute the inital basic feasible solution
B=A(:,C);
invB = inv(B);
x(C) = invB*b;
%% Iteration phase
n_max = 10; % At most n_max iterations
for n = 1:n_max % Main loop
% Compute the vector of reduced costs c_r
c_B = c(C); % Basic variable costs
p = (c_B'*invB)'; % Dual variables
c_r = c' - p'*A; % Vector of reduced costs
% Check if the solution is optimal. If optimal, use
% 'return' to break from the function, e.g.
J = find(c_r < 0); % Find indices with negative reduced costs
if (isempty(J))
f_opt = c'*x;
x_opt = x;
return;
end
% Choose the entering variable
j_in = J(1);
% Compute the vector u (i.e., the reverse of the basic directions)
u = invB*A(:,j_in);
I = find(u > 0);
if (isempty(I))
f_opt = -inf; % Optimal objective function cost = -inf
x_opt = []; % Produce empty vector []
return % Break from the function
end
% Compute the optimal step length theta
theta = min(x(C(I))./u(I));
L = find(x(C)./u == theta); % Find all indices with ratio theta
% Select the exiting variable
l_out = L(1);
% Move to the adjacent solution
x(C) = x(C) - theta*u;
% Value of the entering variable is theta
x(j_in) = theta;
% Update the set of basic indices C
C(l_out) = j_in;
% Compute the new inverse basis B^-1 by performing elementary row
% operations on [B^-1 u] (pivot row index is l_out). The vector u is trans-
% formed into a unit vector with u(l_out) = 1 and u(i) = 0 for
% other i.
M=horzcat(invB,u);
[f g]=size(M);
R(l_out,:)=M(l_out,:)/M(l_out,j_in); % Copy row l_out, normalizing M(l_out,j_in) to 1
u(l_out)=1;
for k = 1:f % For all matrix rows
if (k ~= l_out) % Other then l_out
u(k)=0;
R(k,:)=M(k,:)-M(k,j_in)*R(l_out,:); % Set them equal to the original matrix Minus a multiple of normalized row l_out, making R(k,j_in)=0
end
end
invM=horzcat(u,invB);
% Check if too many iterations are performed (increase n_max to
% allow more iterations)
if(n == n_max)
fprintf('Max number of iterations performed!\n\n');
return
end
end % End for (the main iteration loop)
end % End function
%% Example 3.5 from the book (A small test problem)
% Data in standard form:
% A = [1 2 2 1 0 0;
% 2 1 2 0 1 0;
% 2 2 1 0 0 1];
% b = [20 20 20]';
% c = [-10 -12 -12 0 0 0]';
% C = [4 5 6]; % Indices of the basic variables of
% % the initial basic feasible solution
%
% The optimal solution
% x_opt = [4 4 4 0 0 0]' % Optimal decision variable values
% f_opt = -136 % Optimal objective function cost
Ok, after a lot of hrs spent on the intensive use of printmat and disp to understand what was happening inside the code from a mathematical point of view I realized it was a problem with the index j_in and normalization in case of dividing by zero therefore I managed to solve the issue as follows.
Now it runs perfectly. Cheers.
%% Implementation of the revised Simplex. Solves a linear
% programming problem of the form
%
% min c'*x
% s.t. Ax = b
% x >= 0
%
% The function input parameters are the following:
% A: The constraint matrix
% b: The rhs vector
% c: The vector of cost coefficients
% C: The indices of the basic variables corresponding to an
% initial basic feasible solution
%
% The function returns:
% x_opt: Decision variable values at the optimal solution
% f_opt: Objective function value at the optimal solution
%
% Usage: [x_opt, f_opt] = S12345X(A,b,c,C)
% NOTE: Replace 12345X with your own student number
% and rename the file accordingly
function [x_opt, f_opt] = S472366(A,b,c,C)
%% Initialization phase
% Initialize the vector of decision variables
x = zeros(length(c),1);
% Create the initial Basis matrix, compute its inverse and
% compute the inital basic feasible solution
B=A(:,C);
invB = inv(B);
x(C) = invB*b;
%% Iteration phase
n_max = 10; % At most n_max iterations
for n = 1:n_max % Main loop
% Compute the vector of reduced costs c_r
c_B = c(C); % Basic variable costs
p = (c_B'*invB)'; % Dual variables
c_r = c' - p'*A; % Vector of reduced costs
% Check if the solution is optimal. If optimal, use
% 'return' to break from the function, e.g.
J = find(c_r < 0); % Find indices with negative reduced costs
if (isempty(J))
f_opt = c'*x;
x_opt = x;
return;
end
% Choose the entering variable
j_in = J(1);
% Compute the vector u (i.e., the reverse of the basic directions)
u = invB*A(:,j_in);
I = find(u > 0);
if (isempty(I))
f_opt = -inf; % Optimal objective function cost = -inf
x_opt = []; % Produce empty vector []
return % Break from the function
end
% Compute the optimal step length theta
theta = min(x(C(I))./u(I));
L = find(x(C)./u == theta); % Find all indices with ratio theta
% Select the exiting variable
l_out = L(1);
% Move to the adjacent solution
x(C) = x(C) - theta*u;
% Value of the entering variable is theta
x(j_in) = theta;
% Update the set of basic indices C
C(l_out) = j_in;
% Compute the new inverse basis B^-1 by performing elementary row
% operations on [B^-1 u] (pivot row index is l_out). The vector u is trans-
% formed into a unit vector with u(l_out) = 1 and u(i) = 0 for
% other i.
M=horzcat(u, invB);
[f g]=size(M);
if (theta~=0)
M(l_out,:)=M(l_out,:)/M(l_out,1); % Copy row l_out, normalizing M(l_out,1) to 1
end
for k = 1:f % For all matrix rows
if (k ~= l_out) % Other then l_out
M(k,:)=M(k,:)-M(k,1)*M(l_out,:); % Set them equal to the original matrix Minus a multiple of normalized row l_out, making R(k,j_in)=0
end
end
invB=M(1:3,2:end);
% Check if too many iterations are performed (increase n_max to
% allow more iterations)
if(n == n_max)
fprintf('Max number of iterations performed!\n\n');
return
end
end % End for (the main iteration loop)
end % End function
%% Example 3.5 from the book (A small test problem)
% Data in standard form:
% A = [1 2 2 1 0 0;
% 2 1 2 0 1 0;
% 2 2 1 0 0 1];
% b = [20 20 20]';
% c = [-10 -12 -12 0 0 0]';
% C = [4 5 6]; % Indices of the basic variables of
% % the initial basic feasible solution
%
% The optimal solution
% x_opt = [4 4 4 0 0 0]' % Optimal decision variable values
% f_opt = -136 % Optimal objective function cost