I have successfully managed to calculate the number of days in a specific month between two dates. However I need to calculate those days as work days within the formula. For example.
| ------------- A -------------------- B --------------- C
| 1 ------- 11/12/2014 ----- 17/03/2015 ----- 01/03/2015
| 2
A - Being the start date
B - Being the End date
C - Being the month to check the number of days range A1:B1 falls within. (In this case March)
=MAX(0,MIN(EOMONTH(C$1,0),$B1)-MAX(C$1,$A1)+1)
This should produce 17, which is the number of regular days. How can I produce the number of working days, and incorporate
=NETWORKDAYS()
Ultimately the answer should be 12.
Thanks in advance for your suggestions.
How about this:
=NETWORKDAYS(MAX(A1,C1),MIN(B1,EOMONTH(C1,0)))
It might be wise to expand it with some error checking like so:
=IF(NETWORKDAYS(MAX(A1,C1),MIN(B1,EOMONTH(C1,0)))<0,0,NETWORKDAYS(MAX(A1,C1),MIN(B1,EOMONTH(C1,0))))
Related
I have a home rental business, I wish to calculate the occupancy rate.
For that, I need to calculate the number of nights in a specific month between 2 dates (check-in and check-out).
Here's an example:
A B C D E F
1 | IN | OUT | NIGHTS | 01/01/18 | 01/02/18 |...
------------------------------------------------------------
2 | 15/01/18 | 25/01/18 | 10 | 10 | 0 |...
------------------------------------------------------------
3 | 28/01/18 | 04/02/18 | 7 | 3 | 3 |...
Here are the formulas for:
C3 = B3-A3
D3 = MAX(0; MIN(EOMONTH(D$1;0); $B3) - MAX(D$1; $A3))
I think I'm not too far but the formula is still incorrect.
As you can see, D2 is correct, but D3 is wrong.
This is the tricky part, when the stays overlaps on 2 months, Jan and Feb in thisi case.
D3 should show 4 nights and not 3.
Anyone can help please? Thanks in advance!
PS: sorry for the formatting but I am not able to add a screenshot..
I was thinking that you just add a day to the end of the month unconditionally - I need to try this with one or two test cases though.
MAX(0; MIN(EOMONTH(D$1;0)+1; $B3) - MAX(D$1; $A3))
Just to spell out the four possible scenarios - given start date d1/m1/y1 (A3), end date d2/m2/y2 (B3) ,first day of current month 01/mm/yy (D1) and last day of current month dd/mm/yy (last day of month in D1)
(1) d1/m1/y1>=01/mm/yy and d2/m2/y2 <= dd/mm/yy -> d2/m2/y2 - d1/m1/y1
(Both days in current month - all nights between the two dates)
(2) d1/m1/y1 < 01/mm/yy and d2/m2/y2 <= dd/mm/yy -> d2/m2/y2 - 01/mm/yy
(Start before current month, end in current month - all nights from 1st of month up to end date)
(3) d1/m1/y1 >= 01/mm/yy and d2/m2/y2 > dd/mm/yy -> 01/(mm+1)/yy - d1/m1/y1
(Start in current month, end after current month - all nights from start date up to 1st of following month)
(4) d1/m1/y1 < 01/mm/yy and d2/m2/y2 > dd/mm/yy -> 01/(mm+1)/yy - 01/mm/yy
(Start before current month, end after current month - all nights in month).
In other words, it counts each night following a day in the range, but not the night preceding a day in the range.
You could conditionally add a day when the stay period stretches across month end. This would make that period inclusive as opposed to a simple subtraction.
=MAX(0, MIN(EOMONTH(D$1, 0), $B4)-MAX(D$1, $A4))+(EOMONTH(D$1, 0)<$B4)
Hi I was wondering how I can calculate the number of days difference based on todays date. I would like to create a table like the following:
>50 >40 >30
Agent A
Agent B
based on this data:
Agent | Work Completed
A | 12/07/2017
B | 01/03/2017
A | 12/06/2017
Result should be:
>50 >40 >30
Agent A 0 1 0
Agent B 1 0 0
I have tried a sumif and sumproduct but as I have a field with todays date (=today()) and then I am minusing the 2, this doesn't work.
Would really appreciate some advice.
Here's an example
It caters for future dates, things within 30, 60, 90, or more days old.
Data creates random list of dates relative to today.
Forumulae
I have roughly the following setup:
Values:
MONTH | DURATION | VALUE |
5 | 3 | 120 |
6 | 1 | 100 |
Expected outcome for totals:
MONTH | TOTAL
5 | 120
6 | 220
7 | 120
What I would like to do, is to be able to sum in another table the total values for each month. The logic would be to SUM every value where the total table's month is equal or higher than that of the values, but lower than the value's month + duration.
Does that make any sense? Is that possible? I'm cracking my head and I can't seem to find a way to solve it.
Thank you very much.
The easiest solution is probably to make another column with the end month. And then use SUMIFS to check if month is >= starting month and <= ending month.
=SUMIFS(<Range of Values>,<Range of starting>,"<="& "Target month",<Range of ending>,">="& <Target month>)
DSUM is the best candidate for this, I believe. See Microsft's documentation and this site for help understanding function, and what it is doing. I have made very complex calculations possible in Excel by using the "database" methods (DSUM, DCOUNT, etc).
I’m trying to compare a measure as of today through the same day and month for the prior 4 years (e.g. through June 6 of 2016, 2015, 2014, etc.).
For each year, I decided to count the number of days since the beginning of the year, and sum my values through that number of days for each year.
To identify whether a date should be included in the year to date comparison, I used the formula where my date is in cell A1:
=IF((A1-DATE(YEAR(A1),1,1)+1)<=(TODAY()-DATE(YEAR(TODAY()),1,1)+1),1,0)
I’m looking for a way around the issue of the extra day added to leap years. In other words, after February 28th, the day count will always be off by one in a leap year, and trying to use Februrary 29th in a non-leap year will return an error.
I’d like to adjust this formula, but I’m open to using a different function & formula if it gets me the right results.
you can check any information about February, 29th. If an error occurs, you know its no leap year. Catch that error with =IFERROR(;).
Assuming a table structure like this:
A:Date | B:Value
----------------------
01/01/2016 | 0
01/01/2015 | 1
01/01/2014 | 2
01/01/2013 | 3
01/01/2012 | 4
Formula
To - for example - calculate the average of the previous four (excluding the current) years on January 1st (today is 01/01/2016):
=SUMPRODUCT(
(MONTH(A:A)=MONTH(compare))*
(DAY(A:A)=DAY(compare))*
(YEAR(A:A)>YEAR(compare)-5)*
(YEAR(A:A)<YEAR(compare))*
(B:B)
) / (
SUMPRODUCT(
(MONTH(A:A)=MONTH(compare))*
(DAY(A:A)=DAY(compare))*
(YEAR(A:A)>YEAR(compare)-5)*
(YEAR(A:A)<YEAR(compare))*
1
)
)
Result
For the above example, the result is 2.5
Explanation
To select only those rows representing the same month and day:
(MONTH(A:A)=MONTH(compare))*(DAY(A:A)=DAY(compare))
To select only those values from the previous 4 years (excluding the current):
(YEAR(A:A)>YEAR(compare)-5)*(YEAR(A:A)<YEAR(compare))*
The actual values we are interested in:
(B:B)
Divide by 4 for the average over the last four years. This assumes there is no missing data which might be an issue. You could use another SUMPRODUCT (replace B:B with 1) to count the number of resulting rows and divide by that number to handles this case. This seems to be rather slow, but it works.
Note
For performance reason you should not use A:A (a full column) in the formula, just use the actual range you need, which will likely be much faster.
I'm working on an excel sheet which has a simple structure like this.
Date | Order
-----------------------
08-15-2014 | 84
08-16-2014 | 50
08-17-2014 | 68
08-17-2014 | 78
08-18-2014 | 23
Here, in a separate column I'd like to calculate the sum of Orders by week days(Sunday, Monday and so on). Meaning, I want to see how many Orders received on Sunday, Monday etc.
So far I have come up with this non working formula.
=SUM(LOOKUP(1, WEEKDAY(date_column), order_column))
This was example for summing all order received on Sunday. And, yes Date column is in DateTime Data type so that shouldn't be a problem.
Thanks.
Consider adding a new column after the dates, representing the day values. So in column B you might have something like =TEXT(WEEKDAY(A2),"ddd").
Then if you wanted to sum up order numbers by day, you could use an equation like =SUMIF(B1:B5,"Sun",C1:C5) where column B has the day values and column C has the order numbers.
To show an example (with headers, etc):