I want to find unknown value for nonlinear equation.

I want to find t1 value by search method with mesh grid plot between t1 and TC. Anyone help me to do this problem.

function tcfun()
clc
close all
t1=0:0.5:3;
TC=f(t1);
r0=[0.20;10];
alfa=0.2;
while abs (f(r0(1))) > 1e-2
    r0 = r0 - alfa.*(f(r0(1)))./fprime(r0(1));
end
hold on
r0=[0.1]
alfa=0.2;
for index = 1:100
    r0 = r0 - alfa.*inv(fdb1prime(r0(1)).*fprime(r0(1)));
end
r0
f(r0(1))
plot(r0(1),f(r0(1)),'rs','Markersize',20)
    function TC = f(t1)
        A0 =500;
        D0=100;
        c1=5;
        c2=10;
        c3=10;
        c4=8;
        a=30;
        a2=40;
        m=0.5;
        b=5;
        b2=7;
        mu1=4;
        mu2=8;
        T=12;
        k1=0.01;
        k0=0.03;
        TC=(1./T).*(A0+c1.*((b-a).*t1-(b.*t1^2./2)+(a+b.*(1+m)).*log((1+m-t1)./(1+m)))+c2.*((b.*t1^2./2)+(a+b.*(1+m)).*(t1+(1+m).*log(1+m-t1)))+c3.*((a./k1)-(b.*(1-k1.*T)./k1^2).*(mu1.*(log((1+k1.*(mu1-T))./(1+k1.*(t1-T)))-1)+t1)+((1-k1.*T)./k1).*(log((1+k1.*(mu1-T))./(1+k1.*(t1-T))))+(b.*k0./2.*k1).*(mu1-t1).^2+((a./k1)-(b.*k0.*(1-k1.*T)./k1^2).*(log((1+k1.*(mu1-T))./(1+k1.*(t1-T)))).*(mu2-mu1)+(k0.*D0./k1).*(mu2.*(log((1+k1.*(mu2-T))./(1+k1.*(t1-T)))-1)-mu1.*(log((1+k1.*(mu1-T))./(1+k1.*(t1-T)))-1)+((1-k1.*T)./k1).*(log((1+k1.*(mu2-T))./(1+k1.*(mu1-T))))+((a./k1)-(b.*k0.*(1-k1.*T)./k1^2).*(log((1+k1.*(mu1-T))./(1+k1.*(t1-T)))).*(T-mu2)+(b.*k0./k1).*(mu1-t1).*(T-mu2)+(k0.*D0./k1).*(log((1+k1.*(mu2-T))./(1+k1.*(mu1-T)))).*(T-mu2)-(k0.*b2./2.*k1).*(mu2-T).^2+((1./k1)-T)+(T-(1./k1)).*(T.*(log(1+k1(t1-T))-1)-mu2.*(log((1+k1.*(mu2-T))./(1+k1.*(t1-T)))-1))+(T-(1./k1)).*log(1+k1.*(mu2-T))))+c4.*(a.*(mu1-t1)+(b2./2).*(mu1.^2-t1.^2)-k0.*(a./k1)-(b.*(1-k1.*T)./k1^2).*(log((1+k1.*(mu1-T))./(1+k1.*(t1-T))))+(b./k1).*(mu1-t1))+D0.*(mu2-mu1)-(D0.*k0./k1).*(log((1+k1.*(mu2-T))./(1+k1.*(mu1-T))))+(1./2.*b2).*((a2-b2.*mu2).^2-(a2-b2.*T).^2)+k0.*a2.*log(1+k1.*(mu2-T))-(b2./k1).*(T-mu2)+(b2./k1).*(T-(1./k1)).*log(1+k1.*(mu2-T)))));
        function TCprime = fprime(t1)
            dfdt1=(1./T).*(c1.*((b-a)-b.*t1-((a+b.*(1+m))./(1+m-t1)))+c2.*(b.*t1-((a+b.*(1+m).*t1)./(1+m-t1)))+c3.*((a./k1)-(b.*(1-k1.*T)./k1^2).*(((-mu1.*k1)./(1+k1.*(t1-T)))+1)-((1-k1.*T)./(1+k1.*(t1-T)))+((b.*k0.*(t1-mu1))./k1)-((a-b.*k0.*(1-k1.*T))./k1).*((mu2-mu1)./(1+k1.*(t1-T)))+(k0.*D0./k1).*((k1.*(mu1-mu2)./(1+k1.*(t1-T)))-(a-((b.*k0.*(1-k1.*T))./k1)).*((T-k2)./(1+k1.*(t1-T)))+(b.*k0.*(mu2-T)./k1)-(1-k1.*T).*((T+mu2)./(1+k1.*(t1-T)))))+c4.*(-a2+b2.*t1+k0.*((a-b.*(1-k1.*T))./k1).*(1./(1+k1.*(t1-T)))+(b./k1)));
        end
    end
end
mesh(t1,TC)

[Walter Roberson]

The definition for fprime() needs to be moved out of the definition for f() . When it is inside the definition for f(), only f() is able to call it, but your script needs to be able to call fprime()

Also your script appears to use fdb1prime() as a function but nothing defines that.

[M.Rameswari Sudha]

How to correct it? I couldn't understand.

[Torsten]

Anything wrong with the solution given under

https://de.mathworks.com/matlabcentral/answers/1669894-i-want-to-solve-the-unknown-variable-t1?s_tid=srchtitle

?

[M.Rameswari Sudha]

I got the answer. But I need feasible solution. so , I want to apply optimization by using any one of search method. I didnt get the answer. I also want to show the answer by mesh grid to minimum value for TC. https://de.mathworks.com/matlabcentral/answers/1669894-i-want-to-solve-the-unknown-variable-t1?s_tid=srchtitle by using this method I got the anwer instead of perfect answer. I thank to help to get the solution.

[Walter Roberson]

@M.Rameswari Sudha

function tcfun()
A0 =500;
D0=100;
c1=5;
c2=10;
c3=10;
c4=8;
a=30;
a2=40;
m=0.5;
b=5;
b2=7;
mu1=4;
mu2=8;
T=12;
k1=0.01;
k0=0.03;


t1=0:0.5:3;
TC=f(t1);
r0=[0.20;10];
alfa=0.2;
while abs (f(r0(1))) > 1e-2
    r0 = r0 - alfa.*(f(r0(1)))./fprime(r0(1));
end
hold on
r0=[0.1]
alfa=0.2;
for index = 1:100
    r0 = r0 - alfa.*inv(fdb1prime(r0(1)).*fprime(r0(1)));
end
r0
f(r0(1))
plot(r0(1),f(r0(1)),'rs','Markersize',20)

mesh(t1,TC)

    function TC = f(t1)
        TC=(1./T).*(A0+c1.*((b-a).*t1-(b.*t1^2./2)+(a+b.*(1+m)).*log((1+m-t1)./(1+m)))+c2.*((b.*t1^2./2)+(a+b.*(1+m)).*(t1+(1+m).*log(1+m-t1)))+c3.*((a./k1)-(b.*(1-k1.*T)./k1^2).*(mu1.*(log((1+k1.*(mu1-T))./(1+k1.*(t1-T)))-1)+t1)+((1-k1.*T)./k1).*(log((1+k1.*(mu1-T))./(1+k1.*(t1-T))))+(b.*k0./2.*k1).*(mu1-t1).^2+((a./k1)-(b.*k0.*(1-k1.*T)./k1^2).*(log((1+k1.*(mu1-T))./(1+k1.*(t1-T)))).*(mu2-mu1)+(k0.*D0./k1).*(mu2.*(log((1+k1.*(mu2-T))./(1+k1.*(t1-T)))-1)-mu1.*(log((1+k1.*(mu1-T))./(1+k1.*(t1-T)))-1)+((1-k1.*T)./k1).*(log((1+k1.*(mu2-T))./(1+k1.*(mu1-T))))+((a./k1)-(b.*k0.*(1-k1.*T)./k1^2).*(log((1+k1.*(mu1-T))./(1+k1.*(t1-T)))).*(T-mu2)+(b.*k0./k1).*(mu1-t1).*(T-mu2)+(k0.*D0./k1).*(log((1+k1.*(mu2-T))./(1+k1.*(mu1-T)))).*(T-mu2)-(k0.*b2./2.*k1).*(mu2-T).^2+((1./k1)-T)+(T-(1./k1)).*(T.*(log(1+k1(t1-T))-1)-mu2.*(log((1+k1.*(mu2-T))./(1+k1.*(t1-T)))-1))+(T-(1./k1)).*log(1+k1.*(mu2-T))))+c4.*(a.*(mu1-t1)+(b2./2).*(mu1.^2-t1.^2)-k0.*(a./k1)-(b.*(1-k1.*T)./k1^2).*(log((1+k1.*(mu1-T))./(1+k1.*(t1-T))))+(b./k1).*(mu1-t1))+D0.*(mu2-mu1)-(D0.*k0./k1).*(log((1+k1.*(mu2-T))./(1+k1.*(mu1-T))))+(1./2.*b2).*((a2-b2.*mu2).^2-(a2-b2.*T).^2)+k0.*a2.*log(1+k1.*(mu2-T))-(b2./k1).*(T-mu2)+(b2./k1).*(T-(1./k1)).*log(1+k1.*(mu2-T)))));
    end
    function TCprime = fprime(t1)
            dfdt1=(1./T).*(c1.*((b-a)-b.*t1-((a+b.*(1+m))./(1+m-t1)))+c2.*(b.*t1-((a+b.*(1+m).*t1)./(1+m-t1)))+c3.*((a./k1)-(b.*(1-k1.*T)./k1^2).*(((-mu1.*k1)./(1+k1.*(t1-T)))+1)-((1-k1.*T)./(1+k1.*(t1-T)))+((b.*k0.*(t1-mu1))./k1)-((a-b.*k0.*(1-k1.*T))./k1).*((mu2-mu1)./(1+k1.*(t1-T)))+(k0.*D0./k1).*((k1.*(mu1-mu2)./(1+k1.*(t1-T)))-(a-((b.*k0.*(1-k1.*T))./k1)).*((T-k2)./(1+k1.*(t1-T)))+(b.*k0.*(mu2-T)./k1)-(1-k1.*T).*((T+mu2)./(1+k1.*(t1-T)))))+c4.*(-a2+b2.*t1+k0.*((a-b.*(1-k1.*T))./k1).*(1./(1+k1.*(t1-T)))+(b./k1)));
    end
end

and you also need to define fdb1prime

You should probably not be using inv(): you should probably be using the \ operator

[Torsten]

@M.Rameswari Sudha

I got the answer. But I need feasible solution. so , I want to apply optimization by using any one of search method.

But how can you be sure your method gives a feasible solution ?

As shown by the symbolic computation, there are three solutions for t1. Your Newton method will converge to any of them depending on the starting guess you choose for t1.

[M.Rameswari Sudha]

@Torsten

Yes, You are correct. I misunderstood the method of finding the solution. Thank you for providing the right suggestion. Thank you.