how can i avoid Nan in matlab expression and return 0

The below three expression gives three different values,

and can someone advice me how can i avoid Nan values for 0 input and return 0 value for it

(sqrt((-220).^2)./-220+1)/2.*(10-5) + 5
(sqrt((220).^2)./220+1)/2.*(10-5) + 5
(sqrt((0).^2)./0+1)/2.*(10-5) + 5

>> (sqrt((-220).^2)./-220+1)/2.*(10-5) + 5
(sqrt((220).^2)./220+1)/2.*(10-5) + 5
(sqrt((0).^2)./0+1)/2.*(10-5) + 5

ans =

     5


ans =

    10


ans =

   NaN

[Dyuman Joshi]

Is it necessary to use this particular expression only?

[NN]

yes,

[Walter Roberson]

@Dyuman Joshi

NN is trying to use Problem Based Optimization for something it was never designed to do, so the straight-forward methods are not available, and instead it is necessary to find some mathematical "cheat" that the Optimization Toolbox will permit.

[NN]

:-(

Yes, but how logical comparison can be performed inside optimisation problem ? I am worried if there is no solution for this

Answer 1

[Matt J]

Replace

sqrt(x.^2)./x

with

sign(x)

[NN]

>> (sign(0).*(10-5) + 5)

ans =

5

It avoids Nan,

But i would like to perform the same expression ,

if X is positive it must give 10,

If X is negative it must give 5

If X is zero, it must give 0.

Please advice

Thanks very much for the support

[Matt J]

fun=@(x) 5*(x<0)+10*(x>0);

fun(-2.7)

ans = 5

fun(0)

ans = 0

fun(2.7)

ans = 10

[NN]

Thank you very much for the support .

i tried this, it works, but the expression i have to use for optimisation variable, and it shows error for sign(X)

[Matt J]

But I removed the sign(X)...

[Walter Roberson]

Optimization variables cannot do logical multiplication. They can only use comparisons in the context of constraints.

Answer 2

[Stephen23]

"if X is positive it must give 10, If X is negative it must give 5, If X is zero, it must give 0."

X = randi([-3,3],1,9) % random data

X = 1×9

0 -3 0 1 2 -2 3 -3 0

V = [5,0,10];
Z = V(2+sign(X))

Z = 1×9

0 5 0 10 10 5 10 5 0

[NN]

Thank you very much, i used this expression in an optimisation problem and i got this error

An error occurred while running the simulation and the simulation was terminated

Caused by:

• sign

Answer 3

[Image Analyst]

Why can't you just simply assign it to a variable and check if that variable is nan, and if it is, assign it to zero?

result = (sqrt((0).^2)./0+1)/2.*(10-5) + 5;
if isnan(result)
    result = 0;
end

[Walter Roberson]

Because the context is Problem Based Optimization, which does not support if or isnan() and only permits comparisons as part of constraints.

Answer 4

[Walter Roberson]

The following is not exactly right:

M = optimvar('M', 1, 5)
costa = 5
costb = 10
delta = eps(realmin)
part1a = (M - sqrt(M.^2))/2
part1b = part1a./(part1a - delta)
part2a = (M + sqrt(M.^2))/2
part2b = part2a./(part2a + delta)
cost = part1b * costa + part2b * costb

Mathematically it is wrong at exactly two points, M = -eps(realmin) and M = +eps(realmin) . For those two points, the output should be costa and costb respectively, but instead the formula mathematically gives costa/2 and costb/2 at those two points instead.

In practice, though, for values sufficiently close to +/- realmin, numeric evaluation might return costa+costb and for values sufficiently clost to eps(realmin) numeric evaluation might return 0 instead of costa or costb .

The exact result in a range close to +/- realmin is going to depend on the exact order of evaluation, which is not something that we have control over; optimization could potentially re-arrange the evaluation.

This code has been constructed so that it should never return NaN.

How important is it for your purposes that the values must be correct near +/- realmin, given that it is designed to return 0 for 0 exactly?

[Walter Roberson]

I found a different and much more simple way.

cost = fcn2optimexpr(@(PV) (PV < r) .* Costa + (PV > r) .* Costb, PbattV);

[NN]

Thank you very much, i will try and update

[NN]

I got this,

An error occurred while running the simulation and the simulation was terminated

Caused by:

• SOLVE requires an initial point structure to solve a nonlinear problem.

[Walter Roberson]

Example. This needs to be upgraded to have Costb passed in as well, but your scripts and the Simulink model currently only pass in a single cost vector.

function [Pgrid,Pbatt,Ebatt] = battSolarOptimize(N,dt,Ppv,Pload,Einit,Cost,FinalWeight,batteryMinMax)

% Minimize the cost of power from the grid while meeting load with power 
% from PV, battery and grid 

prob = optimproblem;

% Decision variables
PgridV = optimvar('PgridV',N);
PbattV = optimvar('PbattV',N,'LowerBound',batteryMinMax.Pmin,'UpperBound',batteryMinMax.Pmax);
EbattV = optimvar('EbattV',N,'LowerBound',batteryMinMax.Emin,'UpperBound',batteryMinMax.Emax);

% Minimize cost of electricity from the grid
prob.ObjectiveSense = 'minimize';

%{
prob.Objective = dt*Cost'*PgridV - FinalWeight*EbattV(N);
%}

Costa = Cost(:,1);
Costb = Cost(:,end) + randn(size(Costa))/20;

r = 200;
%Cost = (PbattV<=r).*Costa+ (~PbattV>=r).*Costb;


Cost = fcn2optimexpr(@(PV) (PV < r) .* Costa + (PV > r) .* Costb, PbattV);

%P1 = dt*Cost'*PbattV;
%P2 = dt*Cost'*PbattV;
P = dt*Cost'*PbattV;

prob.Objective = dt*Cost'*PgridV - FinalWeight*EbattV(N);

% Power input/output to battery
prob.Constraints.energyBalance = optimconstr(N);
prob.Constraints.energyBalance(1) = EbattV(1) == Einit;
prob.Constraints.energyBalance(2:N) = EbattV(2:N) == EbattV(1:N-1) - PbattV(1:N-1)*dt;

% Satisfy power load with power from PV, grid and battery
prob.Constraints.loadBalance = Ppv + PgridV + PbattV == Pload;

x0.PgridV = ones(1,N);
x0.PbattV = batteryMinMax.Pmin * ones(1,N);
x0.EbattV = batteryMinMax.Emin * ones(1,N);

% Solve the linear program
options = optimoptions(prob.optimoptions, 'Display', 'iter', 'MaxFunctionEvaluations', 1e7, 'MaxIterations', 1024 );
[values,~,exitflag] = solve(prob, x0, 'Options', options);

% Parse optmization results
if exitflag <= 0
    fprintf('warning: ran into iteration or function evaluation limit!\n');
end
    Pgrid = values.PgridV;
    Pbatt = values.PbattV;
    Ebatt = values.EbattV;

Answer 5

[Matt J]

Here's a way you can do it by adding some additional binary variables and linear constraints. It requires that x be bounded to the interval [-1,1]. You can easily introduce a variable z=A*x+B if you need a variable that spans a different range.

x=optimvar('x','LowerBound',-1,'UpperBound',+1);

b1=optimvar('b1','LowerBound',0,'UpperBound',1,'type','integer'); %Binary variables
b2=optimvar('b2','LowerBound',0,'UpperBound',1,'type','integer');

%%EDITED
con1(1)= b1>=x+eps(0);      % b1 "ON" when x>=0
con2(1)= b1<=x+1;           % b1 "OFF when x<0

con1(2)= b2>=x;              % b2 "ON" when x>0
con2(2)= b2<=x+1-eps(1)      % b2 "OFF when x<=0


y = 5*(1 + (b1+b2)/2 -3*(b1-b2)/2);
    % x>0 ==> b1=b2=1    ==> y=10
    % x<0 ==> b1=b2=0    ==> y=5
    % x=0 ==> b1=1, b2=0 ==> y=0

[Walter Roberson]

Multiple relations per OptimizationConstraint not supported.

Create a new OptimizationConstraint to capture the constraint with the different relation.

con1(1)= b1>=x+eps(0);      % b1 "ON" when x>=0
con2(1)= b1<=x+1;           % b1 "OFF when x<0
con1(2)= b2>=x;              % b2 "ON" when x>0
con2(2)= b2<=x+1-eps(1)      % b2 "OFF when x<=0

[Matt J]

Hmmm. I wonder why...

[Walter Roberson]

Each constraint is structed by constraint type (same for all elements in the constraint), and two pieces of data indicating what the constraint is operating on.

I do not know why they choose to implement that way..