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I am trying to solve some equations using a while loop. I have a known and fixed output. Workflow is as follows:

P = 100; %Desired output
x = 1;   %Initial guess

while abs(something) > 1e-6
    x1 = (25 * x)/2
    x2 = 10x - x1
    x3 = 20x - (x - x1 - x2)*2
    x4 = (x - x1 - x2 -x3)*12
    x5 = (x - x1 - x2 -x3) * 10 + x4    

    P1 = 2005x3 
    P2 = 1500x5

    Pnew = P1 + P2

end

I was hoping to calculate x1, x2, ..., x5 and stop looping when the condition P = Pnew is reached. Due to my modest knowledge of MATLAB any help would be appreciated. Many thanks in advance. APPENDIX: Perhaps i didn't explained well. My goal was to stop looping when the condition P = Pnew is achieved. I have a known value P and initial value of x. Pnew should be generated through a given series of equations. When the condition is met x, x1, x2, ..., x5 will have their values. So x is not only the initial value for calculating other unknowns (x1, x2, ..., x5). I tried to modify it but got NaN Pnew, P1, error and inf for P2, x, ...

P = 100; %Desired output
x = 1;   %Initial guess
Pnew = P + 1

while abs(P - Pnew) > 1e-6
    x1 = (25 * x)/2;
    x2 = 10*x - x1;
    x3 = 20*x - (x - x1 - x2)*2;
    x4 = (x - x1 - x2 -x3)*12;
    x5 = (x - x1 - x2 -x3) * 10 + x4    

    P1 = 2005 * x3; 
    P2 = 150 * x5;

    Pnew = P1 - P2;
    error = abs(P - Pnew);
    x_new = x - .001 * error;
    x = x_new;

end
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  • You should note what error you're getting or what the problem is. I don't know what something is. Please clarify if my answer is the issue you're getting. If not, please clarify what your exact issue is. Commented Apr 21, 2015 at 6:53

2 Answers 2

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Your code will error out if you don't use * to multiply. I have fixed the code below for you. The something you are looking for is P - Pnew. This is because you are trying to find the difference and making sure it is over 1e-6 to continue the loop.

Since you need Pnew, you should also do an initial calculation for it. You should note that your calculation doesn't actually make Pnew converge to P.

P = 100; %Desired output
x = 1;   %Initial guess

x1 = ( 25 * x ) / 2
x2 = 10 * x - x1
x3 = 20 * x - ( x - x1 - x2 ) * 2
x4 = ( x - x1 - x2 -x3 ) * 12
x5 = ( x - x1 - x2 -x3 ) * 10 + x4    

P1 = 2005 * x3 
P2 = 1500 * x5

Pnew = P1 + P2

while abs( P - Pnew ) > 1e-6
    x1 = ( 25 * x ) / 2
    x2 = 10 * x - x1
    x3 = 20 * x - ( x - x1 - x2 ) * 2
    x4 = ( x - x1 - x2 -x3 ) * 12
    x5 = ( x - x1 - x2 -x3 ) * 10 + x4    

    P1 = 2005 * x3 
    P2 = 1500 * x5

    Pnew = P1 + P2

end

You should probably also put ; to terminate the line to suppress the output of the calculation.

A do-while loop here would work, but Matlab doesn't have one. You can use encapsulation to hide the code duplication or use a for loop as follows as well.

P = 100; %Desired output
x = 1;   %Initial guess

for i = 1:Inf

    x1 = ( 25 * x ) / 2
    x2 = 10 * x - x1
    x3 = 20 * x - ( x - x1 - x2 ) * 2
    x4 = ( x - x1 - x2 -x3 ) * 12
    x5 = ( x - x1 - x2 -x3 ) * 10 + x4    

    P1 = 2005 * x3 
    P2 = 1500 * x5

    Pnew = P1 + P2

    if abs( P - Pnew ) < 1e-6
        break;
    end
end

To set a predefined iteration termination in case you have an infinite loop, you can replace i = 1:Inf with i = 1:iterMax where iterMax is the number of max iterations.

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2 Comments

Thank You so much for a prompt answer. I will definitely try both solution as soon as I am done with lectures today.
@PeeledPotatoes no worries, please note that I don't think your calculation converges, but please accept the answer if it's correct (and for the promptness!) or let me know if there's anything wrong with it :)
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Usually in numeric calculations one does not test for equality as the computers cannot distinguish all real numbers. Your condition is actually satisfying for something == P - Pnew

In code it would look like this

P = 100; % Desired output
x = 1;   % Initial guess
Pnew = P + 1; % Something far off, just to get through the first check
MAX_ITERATION_COUNT = 10000; % a reasonable upper bound to your loop
loopCount = 0;    

while abs(P - Pnew) > 1e-6 && loopCount < MAX_ITERATION_COUNT
    loopCount = loopCount + 1;
    x1 = (25 * x) / 2;
    x2 = 10 * x - x1;
    x3 = 20 * x - (x - x1 - x2) * 2;
    x4 = (x - x1 - x2 - x3) * 12;
    x5 = (x - x1 - x2 - x3) * 10 + x4;

    P1 = 2005 * x3;
    P2 = 1500 * x5;

    Pnew = P1 + P2;
end

This should work for your problem.

EDIT: I added a loop counter in case you're loop does not finish regularly.

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7 Comments

Thank You so much for a prompt answer. I tried to do something similar like this but somehow I forgot to write the third line of code from above. Will definitely try it as soon as I am done with lectures today.
@wiseveri this is pretty much the same answer as mine except you're initializing the variable first. I believe my for loop is better because you can easily set a maximum iteration termination in the case that there's an infinite loop, which I think @PeeledPotatoes has.
@krisdestruction that's true. It's just bad style to introduce variables you do not need. Modelling do while in matlab a while true with a condition to break is cleaner, if you want to have a maximum iteration count a while loop is still easier to read. But you're right, our solutions are basically the same.
@wiseveri I agree, that works too. Sometimes I forget to put the break though which sometimes ends in an infinite loop :P Anyways there are multiple solutions, both are valid
It should be && not and
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