The information in this article applies to:
SYMPTOMS
If you use the EOMONTH function to return the last day of a month, the
function may return the first day of the next month instead.
CAUSEThis problem occurs when the date returned by the EOMONTH function meets the following conditions:
-and- WORKAROUNDBecause the EOMONTH function should always return a date at the end of the month, you can modify the formulas so that they subtract one from the date returned by the EOMONTH function if the date is not at the end of the month. This forces the function to return the correct date, for example: A1: 1/1/2100The formula in cell A2 returns 3/1/2100, an incorrect date. However, if you change the formula A1: 1/1/2100the formula returns the correct date (2/28/2100) even though the date falls in February of a century year that is not evenly divisible by 400. This is how the formula works: If the day of the end of the month is 1, recalculate the end of the month, subtract one, and return the date. Otherwise, recalculate the end of the month and return the date. STATUSMicrosoft has confirmed this to be a problem in the Microsoft products listed at the beginning of this article. This problem no longer occurs in Microsoft Excel 2000. MORE INFORMATIONThe EOMONTH function, which is included in the Analysis ToolPak, allows you to return the date of the last day in a month. For example, if you want to determine the last day of the month six months from today's date, use the following formula: =EOMONTH(TODAY(),6)If today is 1/13/98, the function returns 7/31/98: the last day of the month six months from today. The problem described in this article occurs only when the EOMONTH function returns a date in February of a century year that is not a leap year, for example: A1: 1/1/2000 B1: =EOMONTH(A1,1)The formulas in cells B1 and B5 return the correct results (2/29/2000 and 2/29/2400) because the dates are in century years that are evenly divisible by 400. The formulas in cells B2, B3, and B4 return incorrect results (3/1/2100, 3/1/2200, 3/1/2300) because the dates are in century years that are not evenly divisible by 400. For more information about leap years, please see the following article in the Microsoft Knowledge Base: Q118923 Method to Determine Whether a Year Is a Leap Year Additional query words: XL97 y2k year2000
Keywords : kb2000 xladdins |
Last Reviewed: November 5, 1999 © 2000 Microsoft Corporation. All rights reserved. Terms of Use. |