# Trying to find the date every 5 years from a start date. I have (AdjustYear([Start Date], 5)). How do I find the 5 year increment from the Start Date but after Today?

• 0
• Question
• Updated 3 years ago
• Answered
• 40 Points

Posted 3 years ago

• 0

Xavier Fan, Champion

• 630 Points
Can you give an example?  Say [Start Date] is 2/2/2016, and today is 2/19/2016.  What is the 5-year increment you're looking for?
• 40 Points
Start date is 1990. 5 year date will be 1995. Every 5 years you get a bonus. When is the next bonus year from today 2/19/2016. Does that make sense?

Xavier Fan, Champion

• 630 Points
Does the Day and Month of [Start Date] matter?  Or do you care about just the Year?

Say the [Start Date] is 2/1/1990.  Then the next bonus year after today would be 2/1/2020.

Are you looking to get back "2/1/2020" as the result?  Or just "2020"?
• 40 Points
month, day and year. Thanks.

Xavier Fan, Champion

• 630 Points
Try this:

Create a Formula Date field called [Bonus Date], with the following formula:

var Number BonusYearIncrement = 5;

var Number YearsofEmployment = ToDays(Today() - [Start Date]) / 365;

var Number YearsToNextBonus = Ceil(\$YearsofEmployment, \$BonusYearIncrement);

(AdjustYear([Start Date], \$YearsToNextBonus))
• 40 Points
Thanks Very Much!

Xavier Fan, Champion

• 630 Points
You're welcome!