Why have negative value in result

Hello,

In the following code, I try to exclude the case when there is any element of the vector "kseq" in my code. So, I expect the "kseq" achived should have all positive element.

But I dont know why in the result, I still got negative value for some elements of that vector "kseq".

If anyone can see what I did wrong with my code, plz kindly let me know. Thank you!

alpha  = 1/3;              
beta   = 0.9999;            
sigma = 0.001;              
delta  = 0.1;           
  
% Horizon
T=200;
kseq=zeros(T+1,1);
cseq=zeros(T+1,1);
kseq(1)=((alpha*beta)/(1-beta+delta*beta))^(1/(1-alpha));
cguessl=1;
cguessh=6;
iter=0;
tol=0.01;
gapk=1;
while gapk > 0.01 && iter < 1000
    cguess=0.5*(cguessl+cguessh);
    cseq(1)=cguess;
   for i = 2:T+1
     
        kseq(i) = (kseq(i-1))^alpha+(1-delta)*kseq(i-1)-cseq(i-1);
       
        if kseq(i)<0; flag=1; break; end;                      
        cseq(i)=(beta*(1+alpha*(kseq(i))^(alpha-1)-delta))^sigma*cseq(i-1);
   end
  
   if flag==0 & kseq(T+1) >0
       cguessl=cguess;       
   else
       cguessh=cguess;
   end
     
   gapk = abs(kseq(T+1));
   iter=iter+1;
end

Answer 1

[madhan ravi]

kseq(i) = (kseq(i-1))^alpha+(1-delta)*kseq(i-1)-cseq(i-1);
                                                ^^^^^^^^^---- if greater than the other values the result would be negative

[heidi pham]

Hi,

Could you kindly say a bit clearer? I do not understand fully.

In the code, I have if kseq(i)<0; flag=1; break; end; so I should have negative values?

Thanks

Answer 2

[John D'Errico]

LEARN TO USE THE DEBUGGER!

If I step through your code, when i == 4, I get to this point:

i
i =
     4
K>> [kseq(1:i),cseq(1:i)]
ans =
       6.0767          3.5
       3.7938       3.5001
        1.474       3.5006
            0            0

That is just before kseq(4) is assigned. What will it be?

(kseq(i-1))^alpha+(1-delta)*kseq(i-1)-cseq(i-1)
ans =
       -1.036

USE THE DEBUGGER! It was put there for you to use. What you are doing and why, I have no clue, because it is impossible to understand uncommented code, to know if it should be written differently.

[per isakson]

See: Debug a MATLAB Program and Examine Values While Debugging