Iterating an equation through a range of values to satisfy a condition

I have two equations, A and B, that utilize two variables, X and Y.

I aim to find the minimum value Y that makes it so that equation B is less than equation A. X must also be a value between 11 and 15. I also need to store the values of X and Y throughout each iteration. I am unsure how to use nested while loops to accomplish this, as well as how to store the values. I have tried other variations of this, the condition is always overriden. Thank you for the help!

X = 11.1; % Initiate the loop 
while 11 < X < 15 % Needs to be a range of values from 11 to 15 
    diff = 1; % Initiate the loop, not sure about placement 
while diff > 0 % Implies equation B < A
Y = 1; % Random initial value for the equations to run  
    % By manually typing in numbers until it works, 816 is desired value 
V_i = 1; % Random initial value for the equations to run  
A = (abs((sqrt((2*Y)/(1+Y))-1))+abs(sqrt(2/Y)*((sqrt((1)/(1+(Y/X))))-sqrt(1/(1+Y))))+abs(sqrt(1/X)*(sqrt((2*Y)/(X+Y))-1)))*V_i;
B = ((1-(1/X))*sqrt((2*X)/(1+X))+sqrt(1/X)-1)*V_i; 
diff = A-B; % To calculate condition that must be met 
Y = Y+1; % Y should increase by some amount (ex: 1) each time until condition is met 
end 
    X = X+1
end 
% Not sure how to store X and Y values through each iteration

Answer 1

[KALYAN ACHARJYA]

Sufficint Hints:

X = 11:increment spacing:15; % Initiate the loop. check the increment spacing 
Y = 1; 
V_i = 1; % Random initial value for the equations to run
% Any A & B initial value
while B>A
    A= 
    B= 
    Y=Y+1; 
end 
Y+1 % This Y+1 minimum vlaue B<A (Iteration number too)
% Be careful on typical values & conditions, 
% so that no loop runs for infinite times

If you wish to store the A & B Value too, use the array A(i) & B(i)

[Mary Jane]

Thank you, I am still confused on how to then initiate the while loop if there is no B or A value specified before hand. Whenever I try and give an arbitrary value for A and B to initiate the while loop, it only iterates one more time and does not continue despite the condition still being true. For Y it only does the first step and does not continue the iterations, but for everything else, it will continue the entire loop.

Answer 2

[Torsten]

count = 0;
for X = 11:15 % Needs to be a range of values from 11 to 15
    count = count + 1;
    X_array(count) = X;
    diff = 1; % Initiate the loop, not sure about placement 
    Y = 0; % Random initial value for the equations to run
    while diff > 0 % Implies equation B < A
        Y = Y+1; % Y should increase by some amount (ex: 1) each time until condition is met
        % By manually typing in numbers until it works, 816 is desired value 
        V_i = 1; % Random initial value for the equations to run  
        A = (abs((sqrt((2*Y)/(1+Y))-1))+abs(sqrt(2/Y)*((sqrt((1)/(1+(Y/X))))-sqrt(1/(1+Y))))+abs(sqrt(1/X)*(sqrt((2*Y)/(X+Y))-1)))*V_i
        B = ((1-(1/X))*sqrt((2*X)/(1+X))+sqrt(1/X)-1)*V_i 
        diff = A-B % To calculate condition that must be met 
    end 
    Y_array(count) = Y;
end 

A = 0.5324

B = 0.5324

diff = 0

A = 0.5342

B = 0.5342

diff = 0

A = 0.5353

B = 0.5353

diff = 0

A = 0.5359

B = 0.5359

diff = 0

A = 0.5362

B = 0.5362

diff = 0

X_array

X_array = 1×5

11 12 13 14 15

Y_array

Y_array = 1×5

1 1 1 1 1

[Mary Jane]

Thanks, what is the -10 inside the parenthesis for?

When I do individual iterations, my Y value decreases as my X value increases. But in this instance, I get 2 for each iteration

[Torsten]

Thanks, what is the -10 inside the parenthesis for?

X goes from 11 to 15, but in order to start the array index for X_array and Y_array with 1, you must subtract 10.

Maybe it's easier to use a counter ( see above).

When I do individual iterations, my Y value decreases as my X value increases. But in this instance, I get 2 for each iteration

Maybe you copied A and/or B wrong for use in this code.

As you can see, A and B are equal already for Y = 1 in all cases - thus the while loop is exited here.