Don't think so. It looks like 'MFE < CumGn' always equates to False.
Is there a typo? 'MFE < CumGn' should be 'bMFE < CumGn'? What is the
value of CumGn?
If you replace 'MFE < CumGn' with 'True' bMFE rises above 2

Following on from what Pascal said, I'm also not happy that your False2
case - which is the default case - is equivalent to

bMFE = bMFE

Is this really what you want?

However, I think Pascal is on the right track. Do you have "Option
Explicit" at the top of your code? If not, you should. Otherwise, a typo
like MFE instead of bMFE will not cause a compile error.

I'm embarassed to say that after the umpteenth time of looking through
this code  and thinking about it, the solution jumped out at me.
Amending the code to read:

bMFE = IIf(TT = 2, TT, IIf(TT > 2 And MFE <= CumGn, TT, bMFE))

fixed it. The complete sequence is shown below.

MFE = IIf(TT = 2, CumGn, IIf(TT > 2 And MFE < CumGn, CumGn, MFE))
bMFE = IIf(TT = 2, TT, IIf(TT > 2 And MFE <= CumGn, TT, bMFE))

MFE and CumGn are double precision calculated values that are compared
and the higher chosen (first line). Then, an identifying number, or
the "bar" number (which actually is the TT)of the winner is chosen in
the second line, and that becomes the new bMFE, an integer. There is
no MFE or CumGn at TT = 1, or bar 1; at bar 2 the CumGn is calculated
and it also becomes the first MFE. From bar 3 on the new CumGn is
calculated and then compared to the MFE from the step before it, etc.
The MFE increases only when the CumGn increases; it never decreases,
and the bar number, the bMFE, will then change. I had overlooked the
instance when the CumGn and MFE are the same.