I'm trying to multiply (element wise) a vector V of length N by a randomly generated number in the range (a,b), while keeping the sum of the vector equal to a total amount, E. I want to do this in MATLAB, but I'm not sure how. Getting random numbers between a certain range I know how to do:
minrand = 0;
maxrand = 1;
randfac = (maxrand-minrand).*rand(1,N) + minrand;
But yeah, beyond that I'm pretty clueless. I guess the random numbers can't really be generated like this, because if we call the random numbers the vector R, then I want that
R_1*V1 + R_2*V2 .... + R_N*V_N = E. So I guess it's a big equation. Is there any way to solve it, while putting constraints on the max and min values of R?
You can pick pairs of two elements (in all combinations) and add and subtract an equal random number.
% Make up a random vector
N=10;
randfac = 10*rand(1,N);
%OP Answer here: Given randfac with sum E re-randomize it
E = sum(randfac);
minrand = 0;
maxrand = 2;
disp(randfac)
% v = [6.4685 2.9652 6.6567 1.6153 7.3581 0.0237 7.1025
% 3.2381 1.9176 1.3561]
disp(sum(randfac))
% E = 38.7019
r = minrand + (maxrand-minrand)*rand(N*N,1);
k = 1;
for i=1:N
for j=1:N
randfac(i) = randfac(i)-r(k);
randfac(j) = randfac(j)+r(k);
k = k + 1;
end
end
disp(randfac)
% v = [5.4905 0.7051 4.7646 1.3479 9.3722 -1.4222 7.9275
% 7.5777 1.7549 1.1836]
disp(sum(randfac))
% E = 38.7019
Just divide the vector with the sum and multiply with the target E.
randfac = (maxrand-minrand).*rand(1,N) + minrand;
randfac = E*randfac/sum(randfac);
as long as the operator is linear, the result is going to retain it's randomness. Below is some sample code:
minrand = 0;
maxrand = 1;
N = 1000; %size
v = (maxrand-minrand).*rand(1,N) + minrand;
E = 100; %Target sum
A = sum(v);
randfac = (E/A)*v;
disp(sum(randfac))
% 100.0000
First of all with random numbers in the interval of [a b] you can't guarantee that you will have the same summation (same E). For example if [a b]=[1 2] of course the E will increase.
Here is an idea, I don't know how random is this!
For even N I randomize V then divide it in two rows and multiply one of them with random numbers in [a b] but the second column will be multiplied to a vector to hold the summation fixed.
N = 10;
V = randi(100,[1 N]);
E = sum(V);
idx = randperm(N);
Vr = V(idx);
[~,ridx] = sort(idx);
Vr = reshape(Vr,[2 N/2]);
a = 1;
b = 3;
r1 = (b - a).*rand(1,N/2) + a;
r2 = (sum(Vr) - r1.*Vr(1,:))./Vr(2,:);
r = reshape([r1;r2],1,[]);
r = r(ridx);
Enew = sum(V.*r);
The example results are,
V = [12 82 25 51 81 51 31 87 6 74];
r = [2.8018 0.7363 1.9281 0.5451 1.9387 -0.4909 1.3076 0.8904 2.9236 0.8440];
with E = 500 as well as Enew.
I'm simply assigning one random number to a pair (It can be considered as half random!).
Okay, I have found a way to somewhat do this, but it is not elegant and there are probably better solutions. Starting with an initial vector e, for which sum(e) = E, I can randomize its values and end up with an e for which sum(e) is in the range [(1-threshold)E,(1+thresholdE)]. It is computationally expensive, and not pretty.
The idea is to first multiply e by random numbers in a certain range. Then, I will check what the sum is. If it is too big, I will decrease the value of the random numbers smaller than half of the range until the sum is no longer too big. If it is too small, I do the converse, and iterate until the sum is within the desired range.
e = somepredefinedvector
minrand = 0;
maxrand = 2;
randfac = (maxrand-minrand).*rand(1,N) + minrand;
e = randfac.*e;
threshold = 0.001;
while sum(e) < (1-threshold)*E || sum(e) > (1+threshold)*E
if sum(e) > (1+threshold)*E
for j = 1:N
if randfac(j) > (maxrand-minrand)/2
e(j) = e(j)/randfac(j);
randfac(j) = ((maxrand-minrand)/2-minrand).*rand(1,1) + minrand;
e(j) = randfac(j)*e(j);
end
if sum(e) > (1-threshold)*E && sum(e) < (1+threshold)*E
break
end
end
elseif sum(e) < (1-threshold)*E
for j = 1:N
if randfac(j) < (maxrand-minrand)/2
e(j) = e(j)/randfac(j);
randfac(j) = (maxrand-(maxrand-minrand)/2).*rand(1,1) + (maxrand-minrand)/2;
e(j) = randfac(j)*e(j);
end
if sum(e) > (1-threshold)*E && sum(e) < (1+threshold)*E
break
end
end
end
end
Related
The problem is there are c no. of firms bidding on p no. of projects. The winning bidders should collectively have the lowest cost on the client. Each firm can win a maximum of 2 projects.
I have written this code. It works, but takes forever to produce the result, and is very inefficient.
==========================================================================
function FINANCIAL_RESULTS
clear all; clc;
%This Matlab Program aims to select a large number of random combinations,
%filter those with more than two allocations per firm, and select the
%lowest price.
%number of companies
c = 7;
%number of projects
p = 9;
%max number of projects per company
lim = 2;
%upper and lower random limits
a = 1;
b = c;
%Results Matrix: each row represents the bidding price of one firm on all projects
Results = [382200,444050,725200,279250,750800,190200,528150,297700,297700;339040,393420,649520,243960,695760,157960,454550,259700,256980;388032,499002,721216,9999999,773184,204114,512148,293608,300934;385220,453130,737860,287480,9999999,188960,506690,274260,285670;351600,9999999,9999999,276150,722400,9999999,484150,266000,281400;404776,476444,722540,311634,778424,210776,521520,413130,442160;333400,403810,614720,232200,656140,165660,9999999,274180,274180];
Output = zeros(1,p+1);
n=1;
i=1;
for i = 1:10000000
rndm = round(a + (b-a).*rand(1,p));
%random checker with the criteria (max 2 allocations)
Check = tabulate(rndm);
if max(Check(:,2)) > lim
continue
end
Output(n,1:end-1) = rndm;
%Cumulative addition of random results
for k = 1:p
Output(n,end) = Output(n,end) + Results(rndm(k),k);
end
n = n+1;
end
disp(Results);
[Min_pay,Indx] = min(Output(:,end));
disp(Output(Indx,:));
%You know the program is done when Handel plays
load handel
sound(y,Fs);
%Done !
end
Since the first dimension is much greater than the second dimension it would be more efficient to perform loop along the second dimension:
i = 10000000;
rndm = round(a + (b-a) .* rand(i, p));
Check = zeros(size(rndm, 1), 1);
for k = 1:p
Check = max(Check, sum(rndm == k, 2));
end
rndm = rndm(Check <= lim, :);
OutputEnd = zeros(size(rndm, 1), 1);
for k = 1:p
OutputEnd = OutputEnd + Results(rndm(:, k), k);
end
Output = [rndm OutputEnd];
Note that if the compute has a limited memory put the above code inside a loop and concatenate the results of iterations to produce the final result:
n = 10;
Outputc = cell(n, 1);
for j = 1:n
i = 1000000;
....
Outputc{j} = Output;
end
Output = cat(1, Outputc{:});
I'm writing a user-defined function to convert integers to binary. The largest number that could be converted with the function should be a binary number with
16 1 s. If a larger number is entered as d, the function should display an error
message. With my code, I'm trying to add the numbers 0 or 1 to my vector x based on the remainder, then I want to reverse my final vector to display a number in binary. Here's what I have:
function [b] = bina(d)
% Bina is a function that converts integers to binary
x = [];
y = 2;
in = d/2;
if d >=(2^16 -1)
fprintf('This number is too big')
else
while in > 1
if in >= 1
r = rem(in,y);
x = [x r]
end
end
end
end
As you insist on a loop:
x = [];
y = 2;
in = d;
if d >=(2^16 -1)
fprintf('This number is too big')
else
ii = 1;
while in > 0
r = logical(rem(in,y^ii));
x = [r x];
in = in - r*2^(ii-1);
ii = ii+1;
end
end
b = x;
You had the right ideas, but you need to update the variables in your while-loop with every iteration. This is mainly in, where you need to subtract the remainder. And just store the binary remainders in your variable x.
You can check your result with
x = double( dec2bin(d, 16) ) - 48
You could also use a for loop, by pre-calculating the number of iterations with
find( d < 2.^(1:16),1)
and then
if d >=(2^16 -1)
fprintf('This number is too big')
else
for ii = 1:find( d < 2.^(1:16),1)
r = logical(rem(in,y^ii));
x = [r x];
in = in - r*2^(ii-1)
end
end
I'm trying to write my own program to sort vectors in matlab. I know you can use the sort(A) on a vector, but I'm trying to code this on my own. My goal is to also sort it in the fewest amount of swaps which is kept track of by the ctr variable. I find and sort the min and max elements first, and then have a loop that looks at the ii minimum vector value and swaps it accordingly.
This is where I start to run into problems, I'm trying to remove all the ii minimum values from my starting vector but I'm not sure how to use the ~= on a vector. Is there a way do this this with a loop? Thanks!
clc;
a = [8 9 13 3 2 8 74 3 1] %random vector, will be function a once I get this to work
[one, len] = size(a);
[mx, posmx] = max(a);
ctr = 0; %counter set to zero to start
%setting min and max at first and last elements
if a(1,len) ~= mx
b = mx;
c = a(1,len);
a(1,len) = b;
a(1,posmx) = c;
ctr = ctr + 1;
end
[mn, posmn] = min(a);
if a(1,1) ~= mn
b = mn;
c = a(1,1);
a(1,1) = b;
a(1,posmn) = c;
ctr = ctr + 1;
end
ii = 2; %starting at 2 since first element already sorted
mini = [mn];
posmini = [];
while a(1,ii) < mx
[mini(ii), posmini(ii - 1)] = min(a(a~=mini))
if a(1,ii) ~= mini(ii)
b = mini(ii)
c = a(1,ii)
a(1,ii) = b
a(1,ii) = c
ctr = ctr + 1;
end
ii = ii + 1;
end
I have 2 matrices = X in R^(n*m) and W in R^(k*m) where k<<n.
Let x_i be the i-th row of X and w_j be the j-th row of W.
I need to find, for each x_i what is the j that maximizes <w_j,x_i>
I can't see a way around iterating over all the rows in X, but it there a way to find the maximum dot product without iterating every time over all of W?
A naive implementation would be:
n = 100;
m = 50;
k = 10;
X = rand(n,m);
W = rand(k,m);
Y = zeros(n, 1);
for i = 1 : n
max_ind = 1;
max_val = dot(W(1,:), X(i,:));
for j = 2 : k
cur_val = dot(W(j,:),X(i,:));
if cur_val > max_val
max_val = cur_val;
max_ind = j;
end
end
Y(i,:) = max_ind;
end
Dot product is essentially matrix multiplication:
[~, Y] = max(W*X');
bsxfun based approach to speed-up things for you -
[~,Y] = max(sum(bsxfun(#times,X,permute(W,[3 2 1])),2),[],3)
On my system, using your dataset I am getting a 100x+ speedup with this.
One can think of two more "closeby" approaches, but they don't seem to give any huge improvement over the earlier one -
[~,Y] = max(squeeze(sum(bsxfun(#times,X,permute(W,[3 2 1])),2)),[],2)
and
[~,Y] = max(squeeze(sum(bsxfun(#times,X',permute(W,[2 3 1]))))')
I have two matrices of big sizes, which are something similar to the following matrices.
m; with size 1000 by 10
n; with size 1 by 10.
I would like to subtract each element of n from all elements of m to get ten different matrices, each has size of 1000 by 10.
I started as follows
clc;clear;
nrow = 10000;
ncol = 10;
t = length(n)
for i = 1:nrow;
for j = 1:ncol;
for t = 1:length(n);
m1(i,j) = m(i,j)-n(1);
m2(i,j) = m(i,j)-n(2);
m3(i,j) = m(i,j)-n(3);
m4(i,j) = m(i,j)-n(4);
m5(i,j) = m(i,j)-n(5);
m6(i,j) = m(i,j)-n(6);
m7(i,j) = m(i,j)-n(7);
m8(i,j) = m(i,j)-n(8);
m9(i,j) = m(i,j)-n(9);
m10(i,j) = m(i,j)-n(10);
end
end
end
can any one help me how can I do it without writing the ten equations inside the loop? Or can suggest me any convenient way especially when the two matrices has many columns.
Why can't you just do this:
m01 = m - n(1);
...
m10 = m - n(10);
What do you need the loop for?
Even better:
N = length(n);
m2 = cell(N, 1);
for k = 1:N
m2{k} = m - n(k);
end
Here we go loopless:
nrow = 10000;
ncol = 10;
%example data
m = ones(nrow,ncol);
n = 1:ncol;
M = repmat(m,1,1,ncol);
N = permute( repmat(n,nrow,1,ncol) , [1 3 2] );
result = bsxfun(#minus, M, N );
%or just
result = M-N;
Elapsed time is 0.018499 seconds.
or as recommended by Luis Mendo:
M = repmat(m,1,1,ncol);
result = bsxfun(#minus, m, permute(n, [1 3 2]) );
Elapsed time is 0.000094 seconds.
please make sure that your input vectors have the same orientation like in my example, otherwise you could get in trouble. You should be able to obtain that by transposements or you have to modify this line:
permute( repmat(n,nrow,1,ncol) , [1 3 2] )
according to your needs.
You mentioned in a comment that you want to count the negative elements in each of the obtained columns:
A = result; %backup results
A(A > 0) = 0; %set non-negative elements to zero
D = sum( logical(A),3 );
which will return the desired 10000x10 matrix with quantities of negative elements. (Please verify it, I may got a little confused with the dimensions ;))
Create the three dimensional result matrix. Store your results, for example, in third dimension.
clc;clear;
nrow = 10000;
ncol = 10;
N = length(n);
resultMatrix = zeros(nrow, ncol, N);
neg = zeros(ncol, N); % amount of negative values
for j = 1:ncol
for i = 1:nrow
for t = 1:N
resultMatrix(i,j,t) = m(i,j) - n(t);
end
end
for t = 1:N
neg(j,t) = length( find(resultMatrix(:,j,t) < 0) );
end
end