How can I solve this?

2 views (last 30 days)
CHENG WEI LI
CHENG WEI LI on 23 Oct 2022
Answered: Torsten on 24 Oct 2022
clc
clear
syms PL c k
eqn = [PL * exp(-c*exp(-k*0)) == 179323 ; PL * exp(-c*exp(-k*10)) == 203302 ; PL * exp(-c*exp(-k*20)) == 226542]
S = solve(eqn , [PL;c;k] )
S =
struct with fields:
PL: [0×1 sym]
c: [0×1 sym]
k: [0×1 sym]
  2 Comments
Muhammad Usman
Muhammad Usman on 23 Oct 2022
the set of equations are non linear in nature that's why you can't use solve to compute the solution
Walter Roberson
Walter Roberson on 23 Oct 2022
The equations have no solution over reals.

Sign in to comment.

Answers (2)

Muhammad Usman
Muhammad Usman on 23 Oct 2022
% Solve the system of equations starting at the point [0,0,0].
% PL = x(1); c = x(2); k = x(3);
% Initial guess is [0,0,0], you can change it accordingily
fun = @root2d;
x0 = [0,0,0];
x = fsolve(fun,x0)
Warning: Trust-region-dogleg algorithm of FSOLVE cannot handle non-square systems; using Levenberg-Marquardt algorithm instead.
No solution found. fsolve stopped because the relative size of the current step is less than the value of the step size tolerance, but the vector of function values is not near zero as measured by the value of the function tolerance.
x = 1×3
1.0e+05 * 2.0192 -0.0000 0
function F = root2d(x)
F(1) = x(1) * exp(-x(2)*exp(-x(3)*0)) - 179323;
F(2) = x(1) * exp(-x(2)*exp(-x(3)*10)) - 203302;
F(2) = x(1) * exp(-x(2)*exp(-x(3)*20)) - 226542;
end
  2 Comments
CHENG WEI LI
CHENG WEI LI on 23 Oct 2022
The function has correct answer.
why this answer is wrong
Alex Sha
Alex Sha on 24 Oct 2022
There are actually numerical solutions like below:
x1: 446505.431672107
x2: 0.912262916225993
x3: 0.0148006249649759

Sign in to comment.


Torsten
Torsten on 24 Oct 2022
syms PL c k
eqn1 = PL * exp(-c*exp(-k*0)) == 179323;
eqn2 = PL * exp(-c*exp(-k*10)) == 203302;
eqn3 = PL * exp(-c*exp(-k*20)) == 226542;
sPL = solve([eqn1 eqn2],[c,k]);
Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.
PLsol = solve(subs(eqn3,[c,k],[sPL.c sPL.k]),PL);
sck = solve([subs(eqn1,PL,PLsol),subs(eqn2,PL,PLsol)],[c,k]);
csol = sck.c;
ksol = sck.k;
vpa(PLsol)
ans = 
446505.43167210849515476976968181
vpa(csol)
ans = 
0.91226291622599614437018856917074
vpa(ksol)
ans = 
0.014800624964975891465693800088887

Categories

Find more on Symbolic Math Toolbox in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!