# I need a column that is always -5 working days before another date column

Hello,

There is actually a formula to do this in the formula help in the use cases section however when i use that formula it doesn’t work and it tells me I’m creating a circular dependency,

So I have 2 columns, {scheduled delivery date} and {detailing cut off date} the latter needs to be always 5 working days before the scheduled delivery date.
The description of the use case is as follows - Subtract X working days from a date, where X is a column value (Exclude weekends)
and the formula examples is as follows

IF(WEEKDAY({Date}) <= MOD({# Working Days Before},5), DATE(YEAR({Date})-1, MONTH({Date}), DAY({Date}) + 365 - {# Working Days Before} - (2*(ROUNDDOWN(({# Working Days Before}) / 5, 0)+1))), IF(WEEKDAY(DATE(YEAR({Date})-1, MONTH({Date}), DAY({Date}) + 365 - {# Working Days Before} - (2ROUNDDOWN(({# Working Days Before}) / 5, 0)))) = 1, DATE(YEAR({Date})-1, MONTH({Date}), DAY({Date}) -2 + 365 - {# Working Days Before} - (2ROUNDDOWN(({# Working Days Before}) / 5, 0))), IF(WEEKDAY(DATE(YEAR({Date})-1, MONTH({Date}), DAY({Date}) + 365 - {# Working Days Before} - (2ROUNDDOWN(({# Working Days Before}) / 5, 0)))) = 7, DATE(YEAR({Date})-1, MONTH({Date}), DAY({Date}) - 1 + 365 - {# Working Days Before} - (2ROUNDDOWN(({# Working Days Before}) / 5, 0))), DATE(YEAR({Date})-1, MONTH({Date}), DAY({Date}) + 365 - {# Working Days Before} - (2*ROUNDDOWN(({# Working Days Before}) / 5, 0))))))

Could someone let me know where i am supposed to put the 5, {scheduled delivery date}, and the {detailing cut off date}

I tried replacing the {# Working Days Before} column with my {detailing cut off date} column and the {Date}'s with my {scheduled delivery date} column but i got the circular dependency issue.