Thank you
I think I'm getting the hang of it.
I'll study your explanation carefully tomorrow morning.
Thanks Again for the xtra effort. It is appreciated.
John F.

" wrote:

John F wrote...
....
I don't understand the =Lookup statement.
Could I trouble you to explain it to me?
....
" wrote:
....
=LOOKUP(B27-0.000001,{-1E+300;0;24;30;36;42;48;54;60;66},{0;24;30;36;42;4 8;54;60;66;72})

Basically, LOOKUP with 3 arguments assumes its 2nd argument is sorted
in ascending order, locates the largest value in it less than or equal
to its 1st argument, and returns the corresponding value in the list
given as its 3rd argument. For example,

LOOKUP(2.5,{1;2;3},{10;100;1000})

would return 100 because the 2 in the 2nd argument is the largest value
less than or equal to 2.5, and 100 in the 3rd argument corresponds to 2
in the 2nd argument.

What you seem to want to do is a lookup into intervals. However, you
want to treat exact matches against your interval boundary points as
matching the *upper* end of your intervals rather than the lower end,
which is what LOOKUP does. By subtracting a small value from the 1st
argument, LOOKUP can be made to do what you want.

For example, in the topmost formula above, if B27 (sorry for not using
W58) were 30, then the 1st argument would evaluate to 29.999999. This
would match the 24 in the 2nd argument, and the corresponding value in
the 3rd argument would be 30. Note that the n_th entry in the 2nd
argument equals the (n-1)_th entry in the 3rd argument with the 1st
entry in the 2nd argument an extremely small number and the last entry
in the 3rd argument the highest allowed return value.