Precision Error in logical evaluation?

In debugging some code I found the following results confusing. Anyone know how to further investigate this? Or do I simply need to avoid dividing a number by itself? I thought MatLab was "smart" enough to use 1 for the fraction of CF/CF. BTW I'm using R2014a student version on a mid 2009 Mac book pro with OSX 10.9.2

Here is the actual code that causes problems only some of the time as this is inside a function that is called 1000's of times but this error only happens a handful of them.

PL=fzero(@(y)(...
        (s.State(6).H-s.State(4).H)*M_Dot/eta_pump...
        -y*CVTModel_NV(RPM_Pump/RPM_Exp,y/CF)),[0.2*CF CF])/CF;

This is what I did at the command prompt in debugger mode at the line.

CVTModel_NV(RPM_Pump/RPM_Exp,0.2*CF/CF)
ans =
   NaN
K>> CVTModel_NV(RPM_Pump/RPM_Exp,0.2)
ans =
   0.828663580586898
K>> 0.2*CF/CF
ans =
   0.200000000000000
K>> CVTModel_NV(RPM_Pump/RPM_Exp,ans)
ans =
   NaN
K>> CF
CF =
     2.669084425186238e+03
K>> CVTModel_NV(RPM_Pump/RPM_Exp,0.200000000000000000000)
ans =
   0.828663580586898

At this point I commented this line out of the function CVTModel_NV:

Eff(SR>2|SR<0.5|PL<0.2|PL>1)=NaN; % to limit results to applicable area

And then got the following results:

K>> CVTModel_NV(RPM_Pump/RPM_Exp,0.2*CF/CF)
ans =
   0.828663580586898

Then further testing reviled the problem.

K>> CF/CF
ans =
     1

K>> 0.2*CF/CF<0.2
ans =
     1
K>> 0.2*CF/CF==0.2
ans =
     0

Answer 1

[Nathan]

Adjusting the code to avoid the division repaired the problem. Goes to show that programing in MatLab while quite user friendly, it still requires some training in numerics....

PL=fzero(@(y)(...
        (s.State(6).H-s.State(4).H)*M_Dot/eta_pump...
        -y*CF*CVTModel_NV(RPM_Pump/RPM_Exp,y)),[0.2 1]);

[Star Strider]

Vectorizing them may also have solved it. If it only occurred a handful of times, those times could have been when the elements were vectors rather than scalars, and required element-wise operations.

You don’t tell us what the error was, so we’ll never know.