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I need a formula to: If 'B8' 80% then 'C3' * .63% and / that by 12, if 'B8'
is 85.01%, * 'C3' by .86% and / that by 12, etc. Please help with example.
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lena
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Hi,

Try this =if(and(b8<=0.8501,b80.80),C3*0.0063/12,C3*0.0086/12,0)

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Ashish Mathur
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"lena" wrote in message
...
I need a formula to: If 'B8' 80% then 'C3' * .63% and / that by 12, if
'B8'
is 85.01%, * 'C3' by .86% and / that by 12, etc. Please help with
example.
--
lena


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Your IF conditions are not consistent (or at least I couldn't understand).
Please verify them and post again.
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R. Khoshravan
Please click "Yes" if it is helpful.

"lena" wrote:

I need a formula to: If 'B8' 80% then 'C3' * .63% and / that by 12, if 'B8'
is 85.01%,


What is the condition for this part: * 'C3' by .86% and / that by 12, etc.
Please help with example.
--
lena

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On Feb 5, 3:00*pm, lena wrote:
I need a formula to: *If 'B8' 80% then 'C3' * .63% and / that by 12, if 'B8'
is 85.01%, * 'C3' by .86% and / that by 12, etc. *Please help with example.


Well, for what you specify, you could do simply:

=if(B8 85.01%, C3*0.86%/12, if(B8 80%, C3*0.63%/12, ""))

Note that I added a 3rd result: "" (null string). The point is:
what do you want if B8<=80%? Replace "" with whatever you want.

Moreover, if all results can be express as "C3 times some percent,
then divided by 12", you can simplify the expression in any of a
number of ways. One way:

=C3 * if(B8 85.01%, 0.86%, if(B8 80%, 0.63%, 50%)) / 12

where I substituted "" with 50% arbitrarily.

One last comment, if I may: comparing with numbers like 85.01% --
even 80% -- is dubious because such numbers are rarely stored
internally exactly as they are displayed in a cell. I suspect you
would be happier if you write:

=C3 * if(round(B8,4) 85.01%, 0.86%, if(round(b8,2)80%, 0.63%,
50%)) / 12

and perhaps you want the second ROUND to be to 4 dp, too.
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lena wrote:
I need a formula to: If 'B8' 80% then 'C3' * .63% and / that by 12, if 'B8'
is 85.01%, * 'C3' by .86% and / that by 12, etc. Please help with example.



The "etc." in your post suggests more variables. If there are many, a standard
IF() formula may not work. Post your entire requirement to get a proper solution.


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On Feb 6, 6:05*am, Glenn wrote:
lena wrote:
I need a formula to: *If 'B8' 80% then 'C3' * .63% and / that by 12, if 'B8'
is 85.01%, * 'C3' by .86% and / that by 12, etc. *Please help with example.


The "etc." in your post suggests more variables. *If there are many, a standard
IF() formula may not work. *Post your entire requirement to get a proper solution.


Good eye! I missed that.
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