economic load dispatch using lambda iteration method

hi..can anybody help me to solve this error..i could'nt solve it

the code

%N = iteration count limit

%e = iteration tolerance

%lamda = Lagrange multiplier (Lambda)

%del_lamda = change in lambda

%PD = Power Demand

%Pmin & Pmax = minimum and maximum power limits

n=1; lamda=0; del_lamda=0; e=0.01; P=0;

Psum=0;

m=input('Input total number of thermal unit:')

for k=1:m

disp('plant')

disp(k)

Pmin(k)=input('insert minimum power:')

Pmax(k)=input('insert maximum power:')

end

disp('Input cost coefficients per plant in the form below:')

disp('[alpha1 beta1 gamma1;alpha2 beta2 gamma2;...]')

C=input('Insert Cost Coefficients:')

for k=1:m

P(k)=(lamda-C(k,2))/(2*C(k,3));

if P(k)<Pmin(k)

P(k)=Pmin(k);

elseif P(k)>Pmax(k)

P(k)=Pmax(k);

end

Psum=Psum+P(k);

end

if n>N

disp('Solution non-convergence');

disp('Number of Iterations:')

disp(n-1)

else

Pnet=Psum-PD;

del_lamda=abs(Pnet)/P(k);

if abs(Pnet)<e

disp('final value for lamda:')

disp(lamda)

disp('Power for plants 1 to m:')

disp(P)

disp('number of iterations:')

disp(n)

disp('iteration tolerance:')

disp(e)

elseif Pnet>0

lamda=lamda-del_lamda;

else

lamda=lamda+del_lamda;

end

end

Psum=0;

%n=n+1;

the error

Answer 1

[KALYAN ACHARJYA]

Here

P(k)=(lamda-C(k,2))/(2*C(k,3));

As per the P(k) statements, you have to pass the C vector size minimum of mx3

Here m=Input total number of thermal unit,as here P access the data of C having minimum 3rd cloumns,e.g

C(k,2) or C(k,3)

Inaddition, may be you did not define the 'N'.

Suggestion: Improvements needed to simplify code, Is there necessity of so many disp function?

Happy Learning!

:)

[syazwan sabri]

hi there..i dont understand..can u share me a example

thank you

[syazwan sabri]

because i use 15 thermal unit..so how can i solve it.

[Abdalrahman Alharthy]

3. For problem (2), find Using MATLAB M-file program the minimum cost and optimal dispatch of generation when the total demand power is 800 MW if the transmission line losses is expressed as: 0.00014𝑃1 2 + 0.00013𝑃2 2 + 0.00012𝑃3 2 + 0.00011𝑃4 2 Take initial value of  with 8.0 and the acceptance tolerance 10-5 .