# HP Forums

Full Version: Solve Systems of Non-Linear Equations
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
The program solves systems of non-linear equations. Equations are used in a form where the result should be zero. Nelder Mead Optimization is used to find a minimum of the sum of squared results. If the minimum is zero, the solution is found.

Example (taken from a YouTube video):
A quadrilateral is given with the following characteristics:
A, B, C, and D are located on a circle line. The length of the sides is as follows:
AB: 4, BC: 5, CD: 6, DA: 3.
What is the radius R of the circle?

The center of the circle is set to (0,0); A is set to (-R, 0). Then we get the following system of equations of 7 equations with 7 unknowns ( (X1,Y1) coordinates of A, (X2,Y2) of B etc.):
1) (-R-X2)^2+Y2^2=16
2) (X2-X3)^2+(Y2-Y3)^2=25
3) (X3-X4)^2+(Y3-Y4)^2=36
4) (X4+R)^2+Y4^2=9
5) X2^2+Y2^2=R^2
6) X3^2+Y3^2=R^2
7) X4^2+Y4^2=R^2

Application:
Parameters R, X2, X3, X4, Y2, Y3, Y4 are stored in List L1. From a rough drawing, we get the following initial estimates: R=3, X2=1; X3=3; X4=-3; Y2=-1; Y3=4; Y4=3.
We set: {3,1,3,-3,-1,4,3) ▶ L1
Further, the function FUNES is needed with the equations. Equations are changed, so that the results are 0. Results are stored in List L0.

Code:
```EXPORT FUNES() BEGIN LOCAL Q1,I,S1; // PUT HERE IN EQUATIONS f(L1(1),...L1(n))=0 (-L1(1)-L1(2))^2+L1(5)^2-15▶L0(1); (L1(2)-L1(3))^2+(L1(5)-L1(6))^2-25▶L0(2); (L1(3)-L1(4))^2+(L1(6)-L1(7))^2-36▶L0(3); (L1(4)+L1(1))^2+L1(7)^2-9▶L0(4); L1(2)^2+L1(5)^2-L1(1)^2▶L0(5); L1(3)^2+L1(6)^2-L1(1)^2▶L0(6); L1(4)^2+L1(7)^2-L1(1)^2▶L0(7); // End of system of equations SIZE(L1)▶S1; L0(1)^2▶Q1; FOR I FROM 2 TO S1 DO Q1+L0(I)^2▶Q1; END; √Q1▶Q1; RETURN Q1; END;```

Start the program NLES. You are asked for the following parameters:
E0: Termination criterion. Optimization ends if the square root of the sum of squared results of the equations is less than E0, the step length of the Nelder Mead Procedure is less than E0/10, or the maximum number of iterations is reached. We put in 0.0001.
ITM: Maximum number of iterations. We put in 500.

The interim result of the minimum is shown every 25 iterations. After a few seconds, you get the following information on the screen:

--- DONE -----------
Iter: 478
Eps: 8,4281055292E-6
Fmin: 1.70417577211E-4
Parameter: L1
Parameter L2

Interpretation:
Optimization stopped after 478 iterations because of the limit of step length of the Nelder Mead procedure, not because zero was reached (Fmin>1E-4).

So it might make sense to start a new one with a lower step length.
We take S0=0.25, E0=0.0001, ITM=400. After 46 iterations we get Fmin=1.385E-4, so our criterion is nearly reached and we stop.

Parameters are in list L1.
L1(1)=3.2704; this is the estimate for the radius of the circle.
Inspection of list L0 shows that the biggest deviation in the equations is 8.4E-5.

Code:
```EXPORT NLES(S0,E0,ITM) BEGIN // Function with Equations: FUNES() // S0 Starting value of simplex size // E0 Termination criterion of simplex // ITM Maximum number of iterations // Parameters of function: L1 // Function values: L0 LOCAL I1,I2,I3,I4,IP,J1,J2,D0; LOCAL fmin,fmax,fdiff; LOCAL imin,imax,flag,ZAK; Size(L1)▶D0; 1▶ZAK; 0▶J1; 0▶I4; MAKEMAT(0,D0,D0+1)▶M1; MAKELIST(0.,J1,1,D0)▶L2; MAKELIST(0.,J1,1,D0+1)▶L3; MAKELIST(0.,J1,1,D0)▶L0; For I1 FROM 1 TO D0 DO //Simplex J1+I1▶J1; 1▶J2; FOR I2 FROM I1+1 TO D0+1 DO √(J1)/I1/J2*S0▶M1(I1,I2); I1+1▶J2; END; FOR I2 FROM 1 TO D0+1 DO M1(I1,I2)+L1(I1)▶M1(I1,I2); END; END; 0▶flag; REPEAT // Main loop If flag=0 THEN  FOR I1 FROM 1 TO D0+1 DO FOR I2 FROM 1 TO D0 DO M1(I2,I1)▶L1(I2); END; FUNES()▶L3(I1); END; END; 0▶flag; 1ᴇ99▶fmin; −1ᴇ99▶fmax;  FOR I1 FROM 1 TO D0+1 DO IF L3(I1)<fmin Then I1▶imin; L3(I1)▶fmin; END; IF L3(I1)>fmax Then I1▶imax; L3(I1)▶fmax; END; END; FOR I1 FROM 1 TO D0 DO // Reflection center 0▶L2(I1); FOR I2 FROM 1 TO D0+1 DO L2(I1)+M1(I1,I2)▶L2(I1); END; (L2(I1)-M1(I1,imax))/D0▶L2(I1); 2.*L2(I1)-M1(I1,imax)▶L1(I1); // New point END; FUNES()▶Y; IF Y < fmin THEN // New point is min  FOR I1 FROM 1 TO D0 DO 3.*L2(I1)-2.*M1(I1,imax)▶L1(I1); END; Y▶fmin; FUNES()▶Y; IF Y > fmin THEN FOR I1 FROM 1 TO D0 DO L1(I1)-L2(I1)+M1(I1,imax)▶L1(I1); END; ELSE Y▶fmin; END; END; IF Y > fmax THEN // New point is max FOR I1 FROM 1 TO D0 DO 1.5*L3(I1)-0.5*M1(I1,imax)▶L1(I1); END; FUNES()▶Y; If Y > fmax Then FOR I1 FROM 1 TO D0 DO FOR I2 FROM 1 TO D0+1 DO (M1(I1,I2)+M1(I1,imin))*0.5▶M1(I1,I2); END; END; 1▶flag;  END;  END; IF Y < fmax THEN // Updating simplex FOR I1 FROM 1 TO D0 DO L1(I1)▶M1(I1,imax); END; Y▶L3(imax); END; 0▶ZAK; // Termination criterion FOR I1 FROM 1 TO D0 DO FOR I2 FROM 1 TO D0+1 DO ZAK+(M1(I1,I2)-M1(I1,imin))^2▶ZAK; END; END; √(ZAK/D0)▶ZAK; I4+1▶I4; PRINT(I4); PRINT(fmin); (I4>=ITM)+(ZAK<E0)+(fmin<E0)▶IP; UNTIL IP>0; FOR I1 FROM 1 TO D0 DO // Final work M1(I1,imin)▶L1(I1); END; fmin▶Y; fmax-fmin▶fdiff; Print("--- DONE -----------"); Print("Iter: "+I4); Print("Eps: "+ZAK); Print("Fmin: "+fmin); Print("Paramater: L1"); Print("Fct. val: L0"); END;```
Reference URL's
• HP Forums: https://www.hpmuseum.org/forum/index.php
• :