Error using matlab.int​ernal.math​.interp1 Input coordinates must be real.

global vlast1 vlast2 beta delta theta k0 kt At p1 p2
hold off
hold all

vlast1=20*ones(1,40);
vlast2=vlast1;
k0=0.4:0.4:16;
kt11=k0;
kt12=k0;
beta=0.98;
delta=0.1;
theta=0.36;
A1=1.75;
p11=0.9;
p12=1-p11;
p21=0.4;
p22=1-p21;
A2=0.75;
numits=250;
for k=1:numits
    for j=1:40
        kt=k0(j);
        At=A1;
        p1=p11;
        p2=p12;
        %we are using a bounded function minimization routine here
        z=fminbnd(@valfunsto,0.41,15.99);
        v1(j)=-valfunsto(z);
        kt11(j)=z;
        At=A2;
        p1=p21;
        p2=p22;
        z=fminbnd(@valfunsto,0.41,15.99);
        v2(j)=-valfunsto(z);
        kt12(j)=z;
    end
    if k/50==round(k/50)
        plot(k0,v1,k0,v2)
        drawnow
    end
    vlast1=v1;
    vlast2=v2;
end

Error using matlab.internal.math.interp1
Input coordinates must be real.

Error in interp1 (line 188)
VqLite = matlab.internal.math.interp1(X,V,method,method,Xqcol);

Error in solution>valfunsto (line 50)
g1=interp1(k0,vlast1,k,'linear');

Error in fminbnd (line 325)
fu = funfcn(x,varargin{:});

hold off
%optimal plotting prgram for the plans
%plot(k0,kt11,k0,kt12)


function val=valfunsto(k)
global vlast1 vlast2 beta delta theta k0 kt At p1 p2
g1=interp1(k0,vlast1,k,'linear');
g2=interp1(k0,vlast2,k,'linear');
kk=At*kt^theta-k+(1-delta)*kt;
val=log(kk)+beta*(p1*g1+p2*g2);
val=-val;
end

This codes from the textbook The ABCs of RBCs. Once I run them, the error-

Error using matlab.internal.math.interp1

Input coordinates must be real

keeps coming up. Is there anything wrong with the code?

Answer 1

[Torsten]

For some reason, the k suggested by "fminbnd" becomes complex-valued so that interpolation in function "valfunsto" is no longer possible.

Instead try

global vlast1 vlast2 beta delta theta k0 kt At p1 p2
%hold off
%hold all
hold on
vlast1=20*ones(1,40);
vlast2=vlast1;
k0=0.4:0.4:16;
kt11=k0;
kt12=k0;
beta=0.98;
delta=0.1;
theta=0.36;
A1=1.75;
p11=0.9;
p12=1-p11;
p21=0.4;
p22=1-p21;
A2=0.75;
numits=250;
interval = 0.41:0.01:15.99;
for k=1:numits
    for j=1:40
        kt=k0(j);
        At=A1;
        p1=p11;
        p2=p12;
        %we are using a bounded function minimization routine here
        %z=fminbnd(@valfunsto,0.41,15.99);
        [~,idx] = min(arrayfun(@(kk)valfunsto(kk),interval));
        z = interval(idx);
        v1(j)=-valfunsto(z);
        kt11(j)=z;
        At=A2;
        p1=p21;
        p2=p22;
        %z=fminbnd(@valfunsto,0.41,15.99);
        [~,idx] = min(arrayfun(@(kk)valfunsto(kk),interval));
        z = interval(idx);
        v2(j)=-valfunsto(z);
        kt12(j)=z;
    end
    if k/50==round(k/50)
        plot(k0,v1,k0,v2)
        drawnow
    end
    vlast1=v1;
    vlast2=v2;
end
hold off
%optimal plotting prgram for the plans
%plot(k0,kt11,k0,kt12)


function val=valfunsto(k)
global vlast1 vlast2 beta delta theta k0 kt At p1 p2
g1=interp1(k0,vlast1,k,'linear');
g2=interp1(k0,vlast2,k,'linear');
kk=At*kt^theta-k+(1-delta)*kt;
val=log(kk)+beta*(p1*g1+p2*g2);
val=-val;
end

[Torsten]

kk in function "valfunsto" becomes negative if k is big enough. Thus val becomes complex-valued. So my code suggestion from above also doesn't solve your problem.

In order for kk being positive, the search interval for k in "fminbnd" must be such that

At*kt^theta-k+(1-delta)*kt > 0

thus

k < At*kt^theta+(1-delta)*kt

Adopt the 15.99 accordingly.