excel formula lookup with % min variable - excel-formula

Question

I am working with a lookup formula where a amount is typed into a column and returns a fee value based on which fee scale is selected. The variable that i'm having trouble with is that the higher range ie. 500.00-1,000.00 the fee is a % or minimum fee whichever is higher. I cannot get the lookup to add this variable into the equation.
=IFERROR(IFS($B1="scale1",LOOKUP($A1,'scale1'!,$D$2:$D$10,'scale1'!,$F$2:$F$F10)),"")

Answer 1

UPPER LIMIT LOOKUP
If you're looking up a value that could be in between two different values on a scale, use INDEX(MATCH()) instead of LOOKUP() because it allows you to set a Match Type parameter, allowing you to select the lower limit of your lookup results (750.00 will return 500.00 in your example). Use Match Type '1' if your Lookup Array is arranged in ascending order, '-1' if descending, and '0' if you want only exact matches. If you want to select the upper limit (750.00 to return 1,000.00 in your example), simply allow the match to select the lower limit, but bump it up by 1 interval with a +1 at the end:
=INDEX(scale1!$F$2:$F$10,MATCH($A1,scale1!$D$2:$D$10,1)+1)
WHICHEVER IS HIGHER
To apply a "whichever is higher" rule in any formula, simply utilize the MAX() function to contain all the possible options which you'd like to compare and retrieve the highest value from. Let's pretend your fee is either a minimum of $70 or 15%, whichever is higher, and the input in $A1 was 550. The formula would look something like this:
=MAX(70,550*15%)
=MAX(70,82)
=82
These two concepts together, your formula might look something like:
=IFERROR(IFS($B1="scale1",MAX(INDEX(scale1!$F$2:$F$10,MATCH($A1,scale1!$D$2:$D$10,1)+1)*10%,70)),"")