How to solve for multiple variables using multiple equations?
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TAHMEED TASNIM
am 18 Aug. 2020
Kommentiert: Alex Sha
am 25 Aug. 2020
I am trying to find the positive values for N1 N2 N3 N4 N5 N6 using the following equations. Running solver gives values for all as 0 by 1. I was suggested to use Newton's iteration but I have no Idea how to do that as I am new to the software.
syms N1 N2 N3 N4 N5 N6;
N7 = 3.135E+00*N1*N2;
N8 = 4.100E-01*N2*N6;
N9 = 1.381E+01*N1*N4;
N10 = 7.610E+01*N2^2*N3;
N11 = 1.720E+05*N2*N3^2*N5^3;
N12 = 4.034E+04*N3*N5^2;
N13 = 2.615E+04*N2*N3^2*N5^2;
N14 = 3.826E+44*N1*N3*N5^2;
N15 = 1.458E+03*N2*N3*N5;
N16 = 2.435E+04*N1*N3^2*N5;
N17 = 1.186E+08*N3^2*N5^3;
N18 = 7.511E+02*N3*N5;
N19 = 6.874E+02*N3*N5;
eqn1 = N1+N2+N3+N4+N5+N6+N7+N8+N9+N10+N11+N12+N13+N14+N15+N16+N17+N18+N19 == 1;
eqn2 = 0.1962*(N2+N7+N8+2*N10+N11+N13+N15)-0.6575*(N1+N7+N9+N14+N16) == 0;
eqn3 = 0.0557*(N3+N10+2*N11+N12+2*N13+N14+N15+2*N16+2*N17+N18+N19)-0.4698*(N4+N9) == 0;
eqn4 = 0.0099*(N5+3*N11+2*N12+2*N13+2*N14+N15+N16+3*N17+N18+N19)-0.3525*(N6+N8) == 0;
eqn5 = 0.0557*(N1+N7+N9+N14+N16)-0.1962*(N4+N9) == 0;
eqn6 = 0.0099*(N2+N7+N8+2*N10+N11+N13+N15)-0.6575*(N6+N8) == 0;
eqns = [eqn1 eqn2 eqn3 eqn4 eqn5 eqn6];
S = solve(eqns,[N1 N2 N3 N4 N5 N6]);
4 Kommentare
Alex Sha
am 25 Aug. 2020
Hi, the results obtained above are from another math software named 1stOpt, the code looks like below. Matlab should be able to solve this problem, however, it is more possible that some difficulties will be encountered, for example, guess of initial start values. In 1stOpt, guess of start values are no longer needed anymore.
ParameterDomain = [0,];
ConstStr
N7 = 3.135E+00*N1*N2,
N8 = 4.100E-01*N2*N6,
N9 = 1.381E+01*N1*N4,
N10 = 7.610E+01*N2^2*N3,
N11 = 1.720E+05*N2*N3^2*N5^3,
N12 = 4.034E+04*N3*N5^2,
N13 = 2.615E+04*N2*N3^2*N5^2,
N14 = 3.826E+44*N1*N3*N5^2, //N14 = 3.826E+04*N1*N3*N5^2,
N15 = 1.458E+03*N2*N3*N5,
N16 = 2.435E+04*N1*N3^2*N5,
N17 = 1.186E+08*N3^2*N5^3,
N18 = 7.511E+02*N3*N5,
N19 = 6.874E+02*N3*N5;
Function
N1+N2+N3+N4+N5+N6+N7+N8+N9+N10+N11+N12+N13+N14+N15+N16+N17+N18+N19 = 1;
0.1962*(N2+N7+N8+2*N10+N11+N13+N15)-0.6575*(N1+N7+N9+N14+N16) = 0;
0.0557*(N3+N10+2*N11+N12+2*N13+N14+N15+2*N16+2*N17+N18+N19)-0.4698*(N4+N9) = 0;
0.0099*(N5+3*N11+2*N12+2*N13+2*N14+N15+N16+3*N17+N18+N19)-0.3525*(N6+N8) = 0;
0.0557*(N1+N7+N9+N14+N16)-0.1962*(N4+N9) = 0;
0.0099*(N2+N7+N8+2*N10+N11+N13+N15)-0.6575*(N6+N8) = 0;
Results:
n1: 0.010296651426512
n2: 0.23988230158704
n3: 0.0756391318899229
n4: 0.0675904371072629
n5: 9.05432955846619E-22
n6: 0.0124929422697438
feval:
2.22044604925031E-16
1.0547118733939E-15
2.77555756156289E-17
2.34187669256869E-17
-6.24500451351651E-17
6.24500451351651E-17
Akzeptierte Antwort
Alan Stevens
am 18 Aug. 2020
Here's a Newton-Raphson routine for the problem:
% Newton-Raphson Simultaneous Equations
% Six variables N1 to N6 need to be found from the six constraints f1 to f6.
% Variables N7 to N19 are found from N1 to N6.
%
% The aim is to find values of the N that make each of the f's equal to
% zero (within some tolerance).
% Set tolerances etc.
f = ones(6,1); % large initial constraint values
tol = 10^-5; % constraint tolerance
its = 0; % iteration counter
flag = true; % error testing flag
% Initial guesses for N1 to N6
N = [-0.5; 0.5; 0.005; 0.01; 0.2; 0.02];
% Newton-Raphson iteration loop
while flag && its < 60
its = its+1;
fold = f;
Nold = N;
N1 = N(1); N2 = N(2); N3 = N(3); N4 = N(4); N5 = N(5); N6 = N(6);
k7 = 3.135;
k8 = 0.41;
k9 = 13.81;
k10 = 76.1;
k11 = 1.720E+05;
k12 = 4.034E+04;
k13 = 2.615E+04;
k14 = 3.826E+04;
k15 = 1.458E+03;
k16 = 2.435E+04;
k17 = 1.186E+08;
k18 = 7.511E+02;
k19 = 6.874E+02;
N7 = k7*N1*N2; dN7dN1 = k7*N2; dN7dN2 = k7*N1;
N8 = k8*N2*N6; dN8dN2 = k8*N6; dN8dN6 = k8*N2;
N9 = k9*N1*N4; dN9dN1 = k9*N4; dN9dN4 = k9*N1;
N10 = k10*N2^2*N3; dN10dN2 = 2*k10*N2*N3; dN10dN3 = k10*N2^2;
N11 = k11*N2*N3^2*N5^3; dN11dN2 = k11*N3^2*N5^3; dN11dN3 = 2*k11*N2*N3*N5^3; dN11dN5 = 3*k11*N2*N3^2*N5^2;
N12 = k12*N3*N5^2; dN12dN3 = k12*N5^2; dN12dN5 = 2*k12*N3*N5;
N13 = k13*N2*N3^2*N5^2; dN13dN2 = k13*N3^2*N5^2; dN13dN3 = 2*k13*N2*N3*N5^2; dN13dN5 = 2*k13*N2*N3^2*N5;
N14 = k14*N1*N3*N5^2; dN14dN1 = k14*N3*N5^2; dN14dN3 = k14*N1*N5^2; dN14dN5 = 2*k14*N1*N3*N5;
N15 = k15*N2*N3*N5; dN15dN2 = k15*N3*N5; dN15dN3 = k15*N2*N5; dN15dN5 = k15*N2*N3;
N16 = k16*N1*N3^2*N5; dN16dN1 = k16*N3^2*N5; dN16dN3 = 2*k16*N1*N3*N5; dN16dN5 = k16*N1*N3^2;
N17 = k17*N3^2*N5^3; dN17dN3 = 2*k17*N3*N5^3; dN17dN5 = 3*k17*N3^2*N5^2;
N18 = k18*N3*N5; dN18dN3 = k18*N5; dN18dN5 = k18*N3;
N19 = k19*N3*N5; dN19dN3 = k19*N5; dN19dN5 = k19*N3;
% Jacobian values
% First row
J11 = 1 + dN7dN1 + dN9dN1 + dN14dN1 + dN16dN1;
J12 = 1 + dN7dN2 + dN8dN2 + dN10dN2 + dN11dN2 +dN13dN2 + dN15dN2;
J13 = 1 + dN10dN3 + dN11dN3 + dN12dN3 + dN13dN3 + dN14dN3 + dN15dN3 + dN16dN3 + dN17dN3 + dN18dN3 + dN19dN3;
J14 = 1 + dN9dN4;
J15 = 1 + dN11dN5 + dN12dN5 + dN13dN5 + dN14dN5 + dN15dN5 + dN16dN5 + dN17dN5 + dN18dN5 + dN19dN5;
J16 = 1 + dN8dN6;
% Second row
J21 = 0.1962*dN7dN1 -0.6575*(1 + dN7dN1 + dN9dN1 + dN14dN1 + dN16dN1);
J22 = 0.1962*(1 + dN7dN2 + dN8dN2 + 2*dN10dN2 + dN11dN2 + dN13dN2 + dN15dN2) -0.6575*dN7dN2;
J23 = 0.1962*(2*dN10dN3 + dN11dN3 +dN13dN3 + dN15dN3) - 0.6575*(dN14dN3 + dN16dN3);
J24 = -0.6575*dN9dN4;
J25 = 0.1962*(dN11dN5 + dN13dN5 + dN15dN5) - 0.6575*(dN14dN5 + dN16dN5);
J26 = 0.1962*dN8dN6;
% Third row
J31 = 0.0557*(dN14dN1 + 2*dN16dN1) - 0.4698*dN9dN1;
J32 = 0.0557*(dN10dN2 + 2*dN11dN2 + 2*dN13dN2 + dN15dN2);
J33 = 0.0557*(1 + dN10dN3 + 2*dN11dN3 + dN12dN3 + 2*dN13dN3 + dN14dN3 +dN15dN3 +2*dN16dN3 + 2*dN17dN3 + dN18dN3 + dN19dN3);
J34 = -0.4698*(1+dN9dN4);
J35 = 0.0557*(2*dN11dN5 + dN12dN5 + 2*dN13dN5 + dN14dN5 + dN15dN5 + 2*dN16dN5 + 2*dN17dN5 + dN18dN5 + dN19dN5);
J36 = 0;
% Fourth row
J41 = 0.0099*(2*dN14dN1 + dN16dN1);
J42 = 0.0099*(3*dN11dN2 + 2*dN13dN2 + dN15dN2) - 0.3525*dN8dN2;
J43 = 0.0099*(3*dN11dN3 + 2*dN12dN3 +2*dN13dN3 + 2*dN14dN3 + dN15dN3 + dN16dN3 +3*dN17dN3 + dN18dN3 + dN19dN3);
J44 = 0;
J45 = 0.0099*(1 + 3*dN11dN5 + 2*dN12dN5 +2*dN14dN5 + dN16dN5 + 3*dN16dN5 + dN18dN5 + dN19dN5);
J46 = -0.3525*(1 + dN8dN6);
% Fifth row
J51 = 0.0557*(1 + dN7dN1 + dN9dN1 + dN14dN1 + dN16dN1) - 0.1962*dN9dN1;
J52 = 0.0557*dN7dN2;
J53 = 0.0557*(dN14dN3 + dN16dN3);
J54 = - 0.1962*(1 + dN9dN4);
J55 = 0.0557*(dN14dN5 + dN16dN5);
J56 = 0;
% Sixth row
J61 = 0.0099*dN7dN1;
J62 = 0.0099*(1 + dN7dN2 + dN8dN2 + 2*dN10dN2 + dN11dN2 + dN13dN2 + dN15dN2) - 0.6575*dN8dN2;
J63 = 0.0099*(2*dN10dN3 + dN11dN3 + dN13dN3 + dN15dN3);
J64 = 0;
J65 = 0.0099*(dN11dN5 + dN13dN5 + dN15dN5);
J66 = 0.0099*dN8dN6 - 0.6575*(1 + dN8dN6);
% Jacobian matrix
J = [J11 J12 J13 J14 J15 J16;
J21 J22 J23 J24 J25 J26;
J31 J32 J33 J34 J35 J36;
J41 J42 J43 J44 J45 J46;
J51 J52 J53 J54 J55 J56;
J61 J62 J63 J64 J65 J66];
% Function values
f1 = N1+N2+N3+N4+N5+N6+N7+N8+N9+N10+N11+N12+N13+N14+N15+N16+N17+N18+N19-1;
f2 = 0.1962*(N2+N7+N8+2*N10+N11+N13+N15)-0.6575*(N1+N7+N9+N14+N16);
f3 = .0557*(N3+N10+2*N11+N12+2*N13+N14+N15+2*N16+2*N17+N18+N19)-0.4698*(N4+N9);
f4 = 0.0099*(N5+3*N11+2*N12+2*N13+2*N14+N15+N16+3*N17+N18+N19)-0.3525*(N6+N8);
f5 = 0.0557*(N1+N7+N9+N14+N16)-0.1962*(N4+N9);
f6 = 0.0099*(N2+N7+N8+2*N10+N11+N13+N15)-0.6575*(N6+N8);
% Function vector
f = [f1; f2; f3; f4; f5; f6];
N = Nold - J\f;
errmax = max(abs(f-fold));
if errmax>tol
flag = true;
else
flag = false;
end
end
% Display resulting values of N1 to N6 and constraint values.
% N7 to N19 not displayed here, but appear in the workspace.
str = ' N f';
disp(str)
disp([N f])
You should check it carefully!
5 Kommentare
Alan Stevens
am 24 Aug. 2020
I can't see a legitimate way to do this I'm afraid. It's possible I've not specified all the values of N and their derivatives correctly - you need to check them carefully. An alternative is to use MATLAB's fsolve if you have the Optimisation toolbox (I haven't).
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