I am trying to calculate the difference between days but receiving the wrong number of total days for that year. For the dates 2016-01-01 and 2016-12-31 I receive 365 days but not 366. I used the following function where I receive the same output:
=DAYS("2016-12-31","2016-1-1")
=DATEDIF("2016-1-1","2016-12-31","d")
="2016-12-31"-"2016-1-1"
I am not sure what I am doing wrong? Soemhow the leap year's are not properly accounted for.
I think you are misinterpreting the output from the DATEDIF function. Consider the following function call:
=DATEDIF("2016-12-30", "2016-12-31", "d")
This returns 1, which is the number of days between the two dates. Similarly, the following call returns 365:
=DATEDIF("2016-01-01", "2016-12-31", "d")
A value of 365 means that there are 365 days between the first day of the year (day #1, January 1), and the 366 day of the year (day #366, December 31).
As a sanity check, run the following:
=DATEDIF("2015-01-01", "2015-12-31", "d")
This returns 364 days, because 2015 was not a leap year and had no February 29 in its calendar.
Note that the end date must always be greater than the start date, otherwise you will get the #NUM! error. The call you made in your question will not even work.
Related
I have the Year, Week-of-Year and Day-of-the-Week as follows:
Year = 2022 (A2) ; Week Year = 35 (B2); Week Day = 4 or Thursday (C2)
and I would like to estimate the Date as dd.mm.yyyy, which is highlighted in yellow as it shows in the EXCEL picture.
I tried many formulas, but I am sure there might be an easy one.
I think you are counting the weeks starting from zero because for 9/1/2022 (YYYY/MM/DD format) the corresponding week is 36 as per the result of function WEEKNUM(DATE(2022,9,1)). In order to use the logic to multiply the number of weeks by 7. You need to use as a reference the first day of the year, if it was a Sunday, if not then go back to the previous Sunday, so you can count the entire week. Bottom line use as a reference date, the Sunday of the first week of the year, not the first day of the year (YYYY/1/1)
Here is the approach we use in cell E2:
=LET(y, A2:A6, wk, B2:B6, wDay, C2:C6, fDay, DATE(y,1,1), seq, SEQUENCE(7),
fDay - IF(WEEKDAY(fDay)=1,0, WEEKDAY(fDay,2)) + 7*wk
+ XLOOKUP(wDay, TEXT(seq,"dddd"), seq-1))
We use the LET function to avoid repeating the same calculation. The following expression finds the previous Sunday if the first day of the year (fDay) was not a Sunday:
fDay - IF(WEEKDAY(fDay)=1,0, WEEKDAY(fDay,2))
The XLOOKUP function is used to get the numeric representation of the weekday and use the TEXT function to generate the weekdays in a long format. Since we count the entire week, if the weekday is a Sunday (column C in my screenshot), then we don't need to add any day to our reference date, that is why we use seq-1.
Here is the output for several sample data. Assuming the week count starts with zero, if not the formula needs to be adjusted as also the input data.
Notice that the year 2021 started on a Friday, so if we want to find a day for the first week (0) before Friday it will return a date from the previous year. Like in the case of Monday. If you want an error message instead, then the formula can be modified as follow:
=LET(y, A2:A6, wk, B2:B6, wDay, C2:C6, fDay, DATE(y,1,1), seq, SEQUENCE(7),
result, fDay - IF(WEEKDAY(fDay)=1,0, WEEKDAY(fDay,2)) + 7*wk
+ XLOOKUP(wDay, TEXT(seq,"dddd"), seq-1),
IF(YEAR(result) <> y, "ERROR: Date from previous year", result))
I found the solution:
Year = 2022 (A2) ; Week Year = 35 (B2); Week Day = 4 or Thursday (C2)
=DATE (A2,1,3)-WEEKDAY(DATE(A2,1,3)) + 7 * B2 + C2 - 6
I found this solution, but you need to do further testing if it really works.
I calculate month from week: =+MONTH(DATE(YEAR(A2);1;1)+B2*7-1)
I calculate week day number from week day name: =MATCH(D2;{"Monday";"Tuesday";"Wednesday";"Thursday";"Friday";"Saturday";"Sunday"};0)
And then make date using: =DATE(A2;C2;E2)
I have an excel sheet with about 50,000 records where I need to find the number of minutes between two date timestamps but I need to exclude any minutes that occurred during the times we are not working.
Our schedule is M-F 8:30am-5:30pm, Saturdays 8:30am-1:30pm
We don't work Sundays or holidays.
As an example
Cell B2: [7/3/2020 2:16:21 PM]
Cell C2: [7/6/2020 9:20:23 AM]
The manually calculated answer for this one should be about 244 minutes. Task started Friday afternoon, Saturday was a holiday, don't work Sundays, task completed at 9:20am on Monday.
Usually, I come here and start writing a question and by the time I've understood my own problem well enough to post a question I have figured it out on my own but not this time! Help!
Update:
#ForwardEd shared this...
=((I2-H2)
-MAX(0,(NETWORKDAYS.INTL(H2,I2,"0000011",$M$2:$M$12)-1+(WEEKDAY(I2,1)=7)))*TIME(15,0,0)
-MAX(0,(NETWORKDAYS.INTL(H2,I2,"1111101",$M$2:$M$12)-(WEEKDAY(I2,1)=7)))*TIME(19,0,0)
-NETWORKDAYS.INTL(H2,I2,"1111110",$M$2:$M$12)-(NETWORKDAYS.INTL(H2,I2,"0000000")
-NETWORKDAYS.INTL(H2,I2,"0000000",$M$2:$M$12)))*24*60
Where H:H is the Start Date Timestamp and I:I is the Response Date Timestamp and M2:M12 contains my holiday list.
It worked beautifully until I ran into an example like this:
H2 - 07/26/2020 7:48:45 PM
I2 - 07/27/2020 8:57:58 AM
The net result was -650.78333. It looks like anything that starts one one day and ends on the next is coming back as negative.
We want to measure the average response time in minutes for the applications that require manual underwriting. These start timestamps are times that loan applications were received online so they could come in any time of day. The stop times are timestamps that represent the system recorded response time. i.e. the timestamp where an underwriter first did something with the loan application. If a loan application was received at 7pm and was not auto-decisioned then a manual underwriter will need to do something with it the next day when we start working.
If that application came in at 7pm on Wed and is decisioned by an underwriter at 8:46am on Tuursday, we would want to document 16 minutes for that application - not 826 counting the hours between 7pm and 8:30am.
What you want to look at is NETWORKDAYS.INTL. Use this in conjunction with the custom settings to determine the number of Saturdays, Sundays and for the number of days in between your start and end time. You know you have X amount of time per day that is non working time, and Y amount per Saturday.
Then you formula in essence becomes
(End time - start time) - X * No. Weekdays - Y * No. Saturdays - No. Sundays - No. Holidays
Now there will be some tricks in there in order to count your days. but that is the gist of what it boils down to in a formula.
The formulas that are doing the brunt of the work are:
WORKDAY
NETWORKDAYS.INTL
TIME
I avoided the use of an if statement by using a boolean operation that excel will resolve from TRUE/FALSE to 1/0 when sent through a math operator. Side note: I read somewhere that this is also faster than an IF statement, but have no way of proving it and really does not matter on a small number of calculations.
WORKDAY
This formula will return the day of the week for a given date, and a set day of the week to be 1. It will be need in this solution to determine if the end date is a Saturday which has a value of 7 in default setup up as well when option 1 is picked. The format for the formula is:
WORKDAY(Excel Serial date, day 1 of the week)
For this solution
WEEKDAY(B3,1)
NETWORKDAYS.INTL
This formula will be used to count the number of specific days a start and an end date. It can exclude a custom weekend or count a custom week. If it is supplied with a list of dates that are holidays they can be excluded as well. The basic format of the formula is:
NETWORKDAYS.INTL(Start Date, End Date, Custom week choice or workweek pattern, range of holiday dates)
When entering the formula it will give you a list of predefined options for the weekend choices. It will not talk about the pattern.
The pattern is a string 7 digits long consisting of 1 or 0. 0's represents the days you want to count and 1's are days you want to ignore. An important part of the pattern is that the first entry is MONDAY. "1010111" would count only Tuesdays and Thursdays.
TIME
Excel stores date as an integer. 1 represents 1st of January 1900, 2 the 2nd of January 1900 and so on. Time is stored as a decimal or if you prefer the percentage/fraction of a day or 24 hour period. So rather than figuring out the math to determine what percentage of a day X number of hours is, it is simpler to let excel calculate it for us and make the number a little more understandable to someone who may be deciphering the formula later. The basic format of the formula is:
TIME(Hours, Minutes, Seconds)
So as stated earlier, 6 key components need to be determined:
X - Amount of non working time after a weekday
Y - Amount of non working time after a Saturday
Number of weekdays
Number of Saturdays
Number of Sundays
Number of holidays
1) Determine Weekday Non-Working Hours
Based on the supplied information that work day stops at 1730 and starts as 0830. There are a couple of ways of doing the math. Subtract the working hours from 24 hours or count the non work hours at the end of the day and add them to the non work hours at the start of the day.
24 - (17.5 - 8.5) = 15
or
(24 - 17.5) + (8.5 - 0) = 15
For this example 15 will be hard coded into the final formula
2) Determine Saturday Non-Working Hours
Similar to above. Note that we are ignoring Sunday as it is a designated non working day which we already know is 24 hours or 1 day. We are just interested in the time between end of shift Saturday and start of the next normal working Monday. So it really gets calculated the same with just with difference end of shift time.
24 - (13.5 - 8.5) = 19
or
(24- 13.5) - (8.5 - 0) = 19
For this example 19 will be hard coded into the final formula
3) Determine Number of Weekdays
Based on the description earlier of of NETWORKDAYS.INT and working with the assumption that holidays are stored in the range F2:F2, and using a pattern of "0000011" the number of weekdays the formula will be as follows:
=NETWORKDAYS.INTL(B2,B3,"0000011",F2)
For this example the formula is place in cell F6
4) Determine Number of Saturdays
Similar 3) adjust the pattern to only select Saturdays by using "1111101"
=NETWORKDAYS.INTL(B2,B3,"1111101",F2)
For this example the formula is place in cell F7
5) Determine Number of Sundays
Similar 4) adjust the pattern to only select Saturdays by using "1111110"
=NETWORKDAYS.INTL(B2,B3,"1111110",F2)
For this example the formula is place in cell F8
6) Determine Number of Holidays
To get the number of holidays there is not a direct way of doing it. Instead take the difference between all days counted without holidays being factored in and all days counted with holidays counted in.
=NETWORKDAYS.INTL(B2,B3,"0000000")-NETWORKDAYS.INTL(B2,B3,"0000000",$F$2:F2)
For this example the formula is place in cell F9
Now at this point I would love to say just substitute all of the above into the generic formula, but there are a couple of special cases that need to be taken care of. You may have also noted I have not used the WEEKDAY formula yet.
So in order to count the number of days to which X is going to apply, its really the number of days minus 1. The minus 1 is because you want to cont the intervals between days, not the number of days themselves. This gets a little bit more trickier when the end day is a Saturday because there is still an interval there but Saturday is not counted as a weekday. So the True count for number of weekday intervals is:
=MAX(0,(F6-1+(WEEKDAY(B3,1)=7)))
I originally had the MAX(0, calc) in there to prevent the posibility of the day count being negative. After arriving at this final format it may not be needed and you might get away with the following but its untested:
=F6-1+(WEEKDAY(B3,1)=7)
This same concept needs to be applied to your Saturday count. If you job ends on Saturday you do not need to subtract the non working hours after the last Saturday. You formula will look like:
=MAX(0,(F7-(WEEKDAY(B3,1)=7)))
and again further testing is required to make sure MAX can be removed, but if it can then the formula would look like:
=F7-(WEEKDAY(B3,1)=7)
So now with the understanding how dates and times are stored, determine the time difference between start end end time and subtract all the non working hours.
=(B3-B2)-MAX(0,(F6-1+(WEEKDAY(B3,1)=7)))*TIME(15,0,0)-MAX(0,(F7-(WEEKDAY(B3,1)=7)))*TIME(19,0,0)-F8-F9
Now you will not want to use helper cells, so you can take each of the individual formula from F6 to F9 and wind up with:
=(B3-B2)-MAX(0,(NETWORKDAYS.INTL(B2,B3,"0000011",F2)-1+(WEEKDAY(B3,1)=7)))*TIME(15,0,0)-MAX(0,(NETWORKDAYS.INTL(B2,B3,"1111101",F2)-(WEEKDAY(B3,1)=7)))*TIME(19,0,0)-NETWORKDAYS.INTL(B2,B3,"1111110",F2)-(NETWORKDAYS.INTL(B2,B3,"0000000")-NETWORKDAYS.INTL(B2,B3,"0000000",$F$2:F2))
The formula looks unruly, but is easier to understand when broken down into its parts.
Now the last step is to get the answer to display in minutes. There are two choices.
You can leave it as it is in an excel serial date format and change the formatting of to a custom format of [m]. The [ ] will force it into minutes and prevent spill over to hours. It will also round to the nearest minute.
You can convert the results to minutes by multiplying by 24*60 and the value will be in minutes and decimal of minutes.
Note that:
A11 has Time formatting applied
A12 has General formatting applied
A14 has custom formatting of [m] applied
It should be something like this:
Create a calendar table with the workinghours for each days in the year you have data in
Date | StartTime | End time
1/1/2020 1/1/2020 8:30:00 PM 1/1/2020 5:30:00 PM
...
7/3/2020 7/3/2020 8:30:00 PM 7/6/2020 5:30:00 PM
...
12/31/2020
Then paste this code in a module
Function CalcDays(dStart As Date, dEnd As Date, daysCalendar As Range)
Dim Cell As Range
Dim MinDaysCalendar As Date, MaxDaysCalendar As Date
Dim aWSF As WorksheetFunction
Set aWSF = Application.WorksheetFunction
'check the minimum en the maximum date in the calendar
With aWSF
MinDaysCalendar = .Min(daysCalendar)
MaxDaysCalendar = .Max(daysCalendar)
End With
'if the date you check is not in the calendar, exit the function
If dStart < MinDaysCalendar Or dStart > MaxDaysCalendar Then
MsgBox "Date not in calendar"
Exit Function
End If
If dEnd < MinDaysCalendar Or dEnd > MaxDaysCalendar Then
MsgBox "date not in calendar"
Exit Function
End If
'sum the time of all the dates between the start and the end
'pick min and max in order to start and stop at the right time per day
Dim tempTime As Integer
With daysCalendar
For i = 1 To .Rows.Count
If .Cells(i, 2).Value >= CLng(dStart) And .Cells(i, 3).Value <= CLng(dEnd) Then
daytime = aWSF.Max(.Cells(i, 2).Value, dStart) - aWSF.Min(.Cells(i, 3).Value, dEnd)
End If
tempTime = tempTime + daytime
Next i
End With
'return the total time
CalcDays = tempTime
End Function
You can call the function by typing =calcdays in a cell and then give the startDay, endDay and calendar column as parameters.
There might still be some flaws in this code but I think we can manage those.
I have this table in Google Sheets (or excel). The year is the two last digit of my code.
Code Duration Months
1 AC-26482-17 60
2 AC-26482-18 30
3
I would like to return the date in this format (If no data, just leave blanks).
Code Duration Months Start Expiration
1 AC-26482-17 60 01/01/2017 01/01/2022
2 AC-26482-18 30 01/01/2018 01/07/2020
3
Is there a way to achieve this?
You mean you want to add the duration in months to the start date? If so, your sample has the wrong expiration date. 30 months added to 1/1/2018 is not June 1st, but July 1st.
The formula in Excel is
=EDATE(C2,B2)
If you also want to calculate the start date from the last two characters of the code, given all dates are in this millennium, then you can use this for the start date:
=DATE(RIGHT(A2,2)+2000,1,1)
edit: To handle blank cells, you can check with IsBlank()
=if(isblank(a2),"",DATE(RIGHT(A2,2)+2000,1,1))
use:
=ARRAYFORMULA(IF(A2:A="",,DATE(IF(RIGHT(A2:A, 2)*1>=40,,20)&RIGHT(A2:A, 2), 1, 1)))
=ARRAYFORMULA(IF(A2:A="",,DATE(YEAR(C2:C), MONTH(C2:C)+B2:B, 1)))
The Excel/Google-Sheets/LibreOffice function DAYS360() returns the number of days between two dates based on a 360-day year. 0 (default) is used for the US-based method and here are some examples
A = 30 Apr 2016, B = 29 Feb 2016, DAYS360(A, B) = -61
A = 29 Feb 2016, B = 30 Apr 2016, DAYS360(A, B) = 60
This seems ok according to the rules here
But the Excel/Google-Sheets/LibreOffice function YEARFRAC() returns the number of years, including fractional years, between two dates using a specified day count convention. Even here 0 (default) uses US method, (US (NASD) 30/360) which I presumed will also be equal to the value of number of days calculated by DAYS360 * the number of seconds in a day/number of seconds in 360 days. The values in the sheets are as follows
A = 30 Apr 2016, B = 29 Feb 2016, YEARFRAC(A, B) = 0.1666666667
A = 29 Feb 2016, B = 30 Apr 2016, YEARFRAC(A, B) = 0.1666666667
Since it can be seen that the absolute value of the DAYS360 is different by one, the YEARFRAC value is same and assumes 60 days according to the presumption made above, so are the US-based convention mentioned here is the same as mentioned for DAYS360.
If not, what are the exact rules for this one, or is there some other problem?
NOTE: Tested these values on Google Sheets and Libre Office.
DAYS360 parameter 3:
0 indicates the US method - Under the US method, if start_date is the last day of a month, the day of month of start_date is changed to
30 for the purposes of the calculation. Furthermore if end_date is the
last day of a month and the day of the month of start_date is earlier
than the 30th, end_date is changed to the first day of the month
following end_date, otherwise the day of month of end_date is changed
to 30.
1 or any other value indicates the European method - Under the European method, any start_date or end_date that falls on the 31st of
a month has its day of month changed to 30.
YEARFRAC parameter 3:
0 indicates US (NASD) 30/360 - This assumes 30 day months and 360 day
years as per the National Association of Securities Dealers standard,
and performs specific adjustments to entered dates which fall at the
end of months.
1 indicates Actual/Actual - This calculates based upon the actual
number of days between the specified dates, and the actual number of
days in the intervening years. Used for US Treasury Bonds and Bills,
but also the most relevant for non-financial use.
2 indicates Actual/360 - This calculates based on the actual number of
days between the specified dates, but assumes a 360 day year.
3 indicates Actual/365 - This calculates based on the actual number of
days between the specified dates, but assumes a 365 day year.
4 indicates European 30/360 - Similar to 0, this calculates based on a
30 day month and 360 day year, but adjusts end-of-month dates
according to European financial conventions.
For the calculation I'm trying to perform one year = one season and the data I have is start year and end year.
So 1952 1953 needs to add up to 2 (seasons) but if I use =YEAR(A1)-YEAR(A2) the result is 0 - is there a simple way to include the start year and end year as a value in these calculations?
It looks like you have found the problem and rectified it. For the benefit of others who might have found this thread, this is why it was behaving the ways it was.
Background - Excel (and many other applications) treat dates as 1 for every day past Dec 31, 1899. Today happens to be 42,079. Time is a decimal portion of a day so 42,079.75 would be Mar 16, 2015 06:00 PM.
You had the years as numbers in A1:A2; not as full dates. Using the 1-per-day formula, 1952 is May 5, 1905 and 1953 is May 6, 1905. If you peel out the year of each of those with the YEAR() function, you are subtracting 1905 from 1905; resulting in zero.
The solution would be to either type full dates into A1:A2 and format the cells as yyyy so they display 1952 & 1953 but retain their full date nature e.g. =ABS(YEAR(A1) - YEAR(A2)) + 1 , or use the years as numbers only and discard the YEAR() function altogether, e.g. =ABS(A1 - A2) + 1 to get the spanned (inclusive) number of seasons.