How to change all elements of a vector V to achieve the condition sum(V) equal with a upper and lower boundaries?

I have a vector V, for example, V=[0.1 0.002 0.5 0.2 0.1 0.003 0.4].

The boundaries of all elements of the matrix V should always be between 0.01 and 0.8.

I want to create a function to change the elements of the vector V where the sum of V becomes equal to one.

I am looking to create a function in Matlab V= Editor(V, lp, up );

• where lp: is the lower boundary, in my example 0.01
• and up: is the upper boundary, in my example 0.8

[Rik]

Is there a reason why you still want to provide V as an input?

There is a function on the file exchange that allows you to generate a random vector with a fixed sum. Did you try it?

[DGM]

What exactly does "change the elements" mean? You could "change" V to meet the constraints in many different ways. You could scale or shift the values. You could just truncate instead.

... or do the range constraints only apply to the input?

[Rahim Rahim]

I want to change all the variable of the vector !

@Rik I know that, but I want to change the variable of V and the V after the modification shold alwaus respects the up and lower bounderies

[Rahim Rahim]

@DGM change means that reducing or incresing the value of each element

Answer 1

[DGM]

Well with requirements like that, I guess this is fair game.

% input
V = [0.1 0.002 0.5 0.2 0.1 0.003 0.4];

% parameters
limits = [0.01 0.8];
outsum = 1;

% constraints cannot be met generally
if outsum/numel(V) < limits(1)
    error('too many elements')
elseif outsum/numel(V) > limits(2)
    error('too few elements')
end

% i'm going to be super lazy 
sumv = sum(V);
while abs(sumv-outsum) > 1E-9 % or some arbitrary tolerance
    scale = outsum/sumv;
    V = min(max(V*scale,limits(1)),limits(2));
    sumv = sum(V);
end

V

V = 1×7

0.0754 0.0100 0.3769 0.1508 0.0754 0.0100 0.3015

[Rahim Rahim]

The code does not work

the is an open loop

[DGM]

If you had changed outsum to something other than 1, the loop wouldn't exit

while abs(sumv-1) > 1E-9 % presumes that outsum is always 1

Which can be fixed like so.

while abs(sumv-outsum) > 1E-9 % this fixes that bug

If the issue is something else, let me know what inputs/parameters are that break it.

If it's an unacceptable answer, don't accept it.

[Rahim Rahim]

@DGM Thank you so much

But I want everytime the function generat a different matrix ... ( many solution )

and thank you so much

[DGM]

Nothing about the question suggested that the output was random. If the output is random, what's the purpose of V? Does the output depend on V? If so, how?

Answer 2

[Matt J]

V=[0.1 0.002 0.5 0.2 0.1 0.003 0.4];
lp=0.01;
up=0.8;

n=numel(V);

V(:)=(1-n*lp)/n+lp

V = 1×7

0.1429 0.1429 0.1429 0.1429 0.1429 0.1429 0.1429

sum(V)

ans = 1.0000

V>=lp

ans = 1×7 logical array
   1   1   1   1   1   1   1

V<=up

ans = 1×7 logical array
   1   1   1   1   1   1   1

Answer 3

[Walter Roberson]

if the requirement is that you apply a linear transform to the elements of V producing a new vector W such that sum(W) == 1 and the elements respect the upper and lower bounds, then the problem cannot generally be solved.

A simple proof is that V might consist of more than 100 elements. With the minimum being 0.01 then the sum would have to be more than 100*0.01 == 1

[Walter Roberson]

That said

syms a
w = (v-0.01)*a + 0.01;
sola = solve(sum(w)==1,a);
w = simplify(w, a, sola)

sola can be expressed analytically so it can be computed without the symbolic toolbox

[Walter Roberson]

each entry w(k) is a*(v(k)-0.01)+0.01 which is a*v(k) - a*0.01 + 0.01

Let N = length(v). Then sum(w) is a*sum(v) - N*a*0.01 + N*0.01 = a*(sum(v) - N*0.01) + N*0.01. The unknown is a and the sum is 1 so a = (1-N*0.01) / (sum(v) - N*0.01)

Note that if the original values in v could be negative or less than 0.01 then the denominator could be 0 which would be a problem. If the denominator is negative because entries in v are less than 0.01 then I do not promise at the moment that the transformed values are within the required limits.