I'm trying to get the week number of a given quarter based on the date.
I currently have this formula
=1+(WEEKNUM(EDATE(Y4,-1)))-(WEEKNUM(DATE(YEAR(EDATE(Y4,-1)),
LOOKUP(MONTH(EDATE(Y4,-1)),{1,4,7,10}),1)))
But for January, it should be giving me 1 but it's giving me 10. Any suggestions?
How do you expect this to work at the start and end of the quarter? Default WEEKNUM function starts week 1 on the 1st of January every year and week 2 starts on the next Sunday after 1st January.
Assuming your quarter week numbers should work the same way, i.e. week 1 starts on the 1st of Jan/Apr/Jul/Oct and week 2 starts on the next Sunday then that's actually equivalent to counting Sundays since 6 days back into the previous quarter.
You can do that using NETWORKDAYS.INTL function, i.e. with this formula:
=NETWORKDAYS.INTL(EOMONTH(Y4,MOD(1-MONTH(Y4),-3)-1)-5,Y4,"1111110")
format result as number with no decimal places
NETWORKDAYS.INTL function is available in Excel 2010 and later versions - for older versions of Excel you can get the same results with this formula:
=INT((13-WEEKDAY(Y4)+Y4-EOMONTH(Y4,MOD(1-MONTH(Y4),-3)-1))/7)
(Expanded from comment)
when you choose a date in January, it's going back to December. 12 in your lookup array gives 10 as the result. Perhaps instead of EDATE, you should use EOMONTH(Y4,-1)+1, so you look at the 1st of the current month for your calculation
=1+(WEEKNUM(EOMONTH(Y4,-1)+1))-(WEEKNUM(DATE(YEAR(EOMONTH(Y4,-1)+1), LOOKUP(MONTH(EOMONTH(Y4,-1)+1),{1,4,7,10}),1)))
This is fairly interesting, since it changes with the year, and changes with what day of the week is the "start" of the week. So if a quarter starts on Saturday, and the week starts on a Saturday, the entire week is week 1. However, if it starts on a Sunday, week 1 is only one day long, and week 2 starts on Sunday.
The first question we have is, what day is it?
=DayCheck
Additionally, I'm going to call the start of each quarter the following:
Q1Start = Date(Year(DayCheck),1,1)
Q2Start = Date(Year(DayCheck),4,1)
Q3Start = Date(Year(DayCheck),7,1)
Q4Start = Date(Year(DayCheck),10,1)
The next question is, what's the first day of the week? We have some control over this with the Weekday function. For the sake of keeping it simple, Sunday is the start of the week.
Ok, that's our day. Next, what quarter is it?
`Quarter=ROUNDDOWN(MONTH(O16)/4,0)+1`
This gives us 1 for Q1, 2 for Q2, etc.
What day of the week is it now?
=WEEKDAY(DayCheck,1)
Ok, and now, what week are we on?
=WEEKNUM(DayCheck,1)
I'm going to put it together in a not very elegant fashion. I'm sure there's a better way out there.
=(Quarter=1)*((Weeknum(DayCheck)-WeekNum(Q1Start)+1)+(Quarter=2)*((Weeknum(DayCheck)-WeekNum(Q2Start)+1)+(Quarter=3)*((Weeknum(DayCheck)-WeekNum(Q3Start)+1)+(Quarter=4)*((Weeknum(DayCheck)-WeekNum(Q4Start)+1)
Try this:
=CHOOSE((MOD(WEEKNUM(Y4),13)=0)+1,WEEKNUM(Y4)-(ROUNDDOWN(WEEKNUM(Y4)/13,0)*13),13)
This will get the week number of a given date within a quarter.
I used this in one of my applications so you might be able to use it too. HTH.
Note: If you use 1st day other than Sunday, then adjust the WEEKNUM formula.
Can try this as I got this as combination of 2 formula
=WEEKNUM(A1,1)-(INT((MONTH(A1)-1)/3)*13)
second part - INT((MONTH(A1)-1)/3) gives us the quarter number of previous quarter which then multiplied with 13 weeks/quarter gives us how many weeks have passed in all previous quarter before current quarter.
First part - "WEEKNUM(A1,1)" gives us the week number of current week in the year.
so by deducting all the previous weeks in previous quarters from current week number of year, we get the current week number in current quarter.
Related
I have a formula to calculate the first Friday of the Year and it works okay.
I understand how the formula gets the answer however, I can't seem to find the reasoning behind the formula. The formula is:
=DATE(YEAR(TODAY()),1,8)-WEEKDAY(DATE(YEAR(TODAY()),1,2))
This gives us 2020/01/08 - 5 = 2020/01/03 which is the First Friday of the Year. But why does the formula choose 8 and 2 as the dates?
Can someone please explain the reason.
The first, obvious, part of the formula is that DATE(YEAR(TODAY()),1,8) gives you the eighth day of the month which can't be the first Friday since one of the 7 days before it would must be a Friday.
My guess is that the second part has a little bit of kludgery going on. WEEKDAY(DATE(YEAR(TODAY()),1,1)) would give you the number of days from the first day of the month to the previous Sunday. But if you're looking for the previous Saturday you'd need WEEKDAY(DATE(YEAR(TODAY()),1,1)) + 1 or WEEKDAY(DATE(YEAR(TODAY()),1,2)).
Finally, the day of the week WEEKDAY(DATE(YEAR(TODAY()),1,2)) before the 1st of the month is the same day of the week before the 8th of the month.
Working on a formula that will take a date and translate it to the format FYxxPxxWx.
For example. Input the date of 03/22/20 and the formula will give you FY20P06W4 which is correct.
However if you input 02/02/20 the formula will give you FY20P05W2. The correct output would be FY20P05W1. This issue also rears its head with the date 09/29/19. It gives you FY20P12W5. The correct output would be FY20P1W1.
Something else weird happens when you put in the date 04/5/20 you get FY21P07W2 when it should be FY20P07W2.
The formula is
=CONCATENATE("FY",RIGHT(YEAR(DATE(YEAR(D5),MONTH(D5)+(10-1),1)),2),"P",TEXT(CHOOSE(MONTH(D5),4,5,6,7,8,9,10,11,12,1,2,3),"0#"),"W",WEEKNUM(D5,1)-WEEKNUM(DATE(YEAR(D5),MONTH(D5),1),1)+1)
I think this issue is caused by the strange weeks where the the month ends and another begins throwing off the formula.
I do have a formula that calculates the years fiscal year start date
=(DATE(YEAR(TODAY())-1,10,1)-(WEEKDAY(DATE(YEAR(TODAY())-1,10,1),1)))+1
This outputs 09/29/19 as the start date of the Fiscal year as its the same week as 10/1/19 which is the first month of the fiscal year. IF that makes sense.
The separate formulas are
For FY and grabs only last two digits of year
RIGHT(YEAR(DATE(YEAR(D5),MONTH(D5)+(10-1),1)),2)
For Period (gives me a two digit Period
TEXT(CHOOSE(MONTH(D5),4,5,6,7,8,9,10,11,12,1,2,3),"0#")
For Week
WEEKNUM(D5,1)-WEEKNUM(DATE(YEAR(D5),MONTH(D5),1),1)+1)
I believe I have a solution for you. Discussion to follow, but here's the full formula:
=CONCAT("FY",RIGHT(YEAR(D5+91+WEEKDAY(DATE(YEAR(D5),10,1))),2),"P",TEXT(IF(MONTH(D5+(7-WEEKDAY(D5)))<>MONTH(D5),IF(MONTH(D5)=9,1,CHOOSE(MONTH(D5),5,6,7,8,9,10,11,12,1,2,3,4)),CHOOSE(MONTH(D5),4,5,6,7,8,9,10,11,12,1,2,3)),"0#"),"W",ROUNDUP(((D5-IF(MONTH(D5+(7-WEEKDAY(D5)))<>MONTH(D5),DATE(YEAR(D5),MONTH(D5)+1,1)-WEEKDAY(DATE(YEAR(D5),MONTH(D5)+1,1))+1,DATE(YEAR(D5),MONTH(D5),1)-WEEKDAY(DATE(YEAR(D5),MONTH(D5),1))+1))/7)+0.01,0))
One issue is that this still calculates 2/2/2020 the way you said was incorrect. When I verify it against a calendar, though, it seems that FY20P05W02 should be correct. If the week that includes the first of the month begins a new pay period, that would mean 2/1/2020, falling on a Saturday, would be the last day of fiscal week 1. That would make 2/2/2020 the first day of fiscal week 2.
To calculate fiscal year, I used RIGHT(YEAR(D5+91+WEEKDAY(DATE(YEAR(D5),10,1))),2). Since you can count on there always being 91 days from the beginning of October to the end of December, it helps with this calculation. In your formula, you had MONTH(D5)+(10-1), which would push you 9 months out past the month in D5. This explains why your result for 4/5/2020 was off by a year.
Fiscal period was a bit trickier, requiring a couple nested IF statements. I used IF(MONTH(D5+(7-WEEKDAY(D5)))<>MONTH(D5) first to account for days at the end of the month that would fall into the next fiscal period, then IF(MONTH(D5)=9 to account for the few days at the end of September that might fall into the next fiscal year. Days at the end of September would default to 1, days at the end of a month that are included in the next fiscal period use the first CHOOSE function (they need the next month's number), and everything else gets the CHOOSE function as you wrote it.
The fiscal week took a bit more, but in the end I evaluated the beginning of the current fiscal month and subtracted it from the date in D5, then divided by 7 and added 0.01 so that even numbers would round up correctly.
I tested this out over a few years of dates and it seemed to be functioning correctly, but let me know if you have questions or issues.
One thing to consider when using WEEKNUM is that you'll have a week that is counted twice at the beginning of the year unless you use option 21 or ISOWEEKNUM. These give the same result as each other, and ensure that only one week number is assigned to any given day, no matter the year.
Last time I posted a quite vague story about a date difference challenge which I haven't solved yet. I will try to elaborate since I have tried everything in my power and the problem still isn't fixed.
I currently have three columns.
Column 1 (F)
the date a car starts its repairs (format DayOfWeek-DD-MM-YYYY)
Column 2 (G)
the number of days in which the car is repaired (service level agreement [SLA]; the standard is 10 days)
Column 3 (H)
the output, which is the date the car should be finished. So the number of days after the startdate*
*Th thing which makes this case difficult is that only weekdays are included.
So, for example:
If a car starts repairs on Monday 1st of August, the finish date is Tuesday the 14th of August.
I tried to solve this with the following formula:
=IF(WEEKDAY(F218)=2;(F218+11);
IF(WEEKDAY(F218)=3;F218+12;
IF(WEEKDAY(F218)=4;F218+13;
IF(WEEKDAY(F218)=5;F218+14;
IF(WEEKDAY(F218)=6;F218+15)))))
In other words:
If startdate = Monday then startdate + 11,
if startdate = Tuesday then startdate + 12, etc.
This works, but I have 300+ rows and dragging this function down doesn't change the cell references.
I know about the NETWORKDAYS and WEEKDAY functions, but I encounter problems with any Monday where only 1 weekend passes and other days where 2 weekends pass.
First of all, I am assuming that your first day - whatever day that may be - is considered day one (1). So in my scenario, if a SLA states 2 days to complete a repair and the start date is a Monday, I'm assuming the repair should be completed by Tuesday.
My assumption is based off this comment by #RonRosenfeld:
...although you might have to subtract 1 from the number of days
With all that being said, try this formula in your cell instead:
NOTE: You may need to change things like commas and semi-colons to adjust for your region.
=WORKDAY($F2,$G2-1)+LOOKUP(WEEKDAY(WORKDAY($F2,$G2-1),16),{1;2;3},{2;1;0})
What it does:
WORKDAY($F2,$G2-1)
First we want to find out exactly what day the repairs should be completed by if weekend days (Saturday and Sunday) were included. This part of the formula will simply give us a place to start.
$F2 is your repair start date
$G2 is the number of days a repair is supposed to take (you may need to add a column for this, because, as you stated, the SLA may change and you need the formula to be easily adjusted)
WEEKDAY(WORKDAY($F2,$G2-1),16)
The WORKDAY function from above is wrapped inside a WEEKDAY function. This WEEKDAY function is written to account for each day of a week to be assigned numbers. The [return_type] parameter of 16 tells Excel to label them as "Numbers 1 (Saturday) through 7 (Friday)". We chose 16 so that our LOOKUP function is easier to write. This part of the formula only returns a one-digit number, which in turn will be used to figure out what day of the week we actually want when excluding weekends.
LOOKUP(WEEKDAY(WORKDAY($F2,$G2-1),16),{1;2;3},{2;1;0})
We finish the formula by adding the result from a LOOKUP function using the first form of the function: LOOKUP(lookup_value,lookup_vector,[result_vector])
We found our lookup_value in the previous bullet point using the WEEKDAY function. Now we want Excel to use the lookup_vector - {1;2;3} in our formula - to find the correct value to add to the first part of our formula (which is found using the [result_vector] - {2;1;0} in our formula).
The lookup_vector only has three values: 1, 2, and 3.
1 signals Saturday
2 signals Sunday
3 signals all other days
Think of the lookup_vector and [result_vector] as forming a matrix/table from which our value is found:
1 2
2 1
3 0
If our number of repair days pushes us to:
a Saturday (1), the formula adds 2.
a Sunday (2), the formula adds 1.
any weekday, the formula adds 0 (since weekdays are acceptable).
Hopefully all of this makes sense. Best of luck to you!
This is a follow up to an earlier question.
#ForwardED I am trying to convert your original single static formula into a single dynamic formula.
Unfortunately my employer's filters will let me up certain things to a hyperlink, but will not let me download or view from the same site. I am also trying to come up with a formula for floating dates.
Below is a copy of the expanded explanation I gave on the original question. I am not sure if you missed it or not. It deals with holidays that have a set date like Christmas, 25th of December of every year. However if it fall on a Saturday the time of work is the Friday and if it is on a Sunday the day off is the Monday.
Dealing with a holiday falling on a Saturday or Sunday
Again we need to refer to some cell in your spreadsheet with the year so I will again use Q10 as the example and we will assume a date of 2014/10/24.
=IF(WEEKDAY(DATE(YEAR(Q10),12,25))=7,DATE(YEAR(Q10),12,24),IF(WEEKDAY(DATE(YEAR(Q10),12,25))=1,DATE(YEAR(Q10),12,26),DATE(YEAR(Q10),12,25)))
The formula checks first if the weekday is a Saturday. We do this using a function that will return the day of the week See step 2) from the original question. It is this part from the equation above:
WEEKDAY(DATE(YEAR(Q10),12,25))
It will return a single integer 1 through 7 corresponding to the day of the week the date function results in, in this case. If its a 1 we known its Sunday, if its 7 we know its Saturday. So the check for Saturday is:
WEEKDAY(DATE(YEAR(Q10),12,25))=7
If WEEKDAY()=7 is true then we provides the date of the day before which is really just subtracting 1 from the date we were looking at. We use this part of the formula to calculate that:
DATE(YEAR(Q10),12,24)
notice how I changed the day from 25 to 24. An alternate way would be to recycle our date and make the computer do one more calculation using this formula:
DATE(YEAR(Q10),12,25)-1
or
DATE(YEAR(Q10),12,25-1)
That all sits in the TRUE portion of the if statement. so if the date does not fall on Saturday then we wind up in the FLASE portion of the IF statement. Here we check with a second IF for the date falling on a Sunday. we use the same theory and process as we did for the Saturday check.
IF(WEEKDAY(DATE(YEAR(Q10),12,25))=1,DATE(YEAR(Q10),12,26),DATE(YEAR(Q10),12,25))
Placing an IF statement inside an IF statement is commonly referred to as "nesting". This whole IF statement happens in the FALSE portion of the previous IF that checked to see if it was Saturday. This time we checked for Sunday:
WEEKDAY(DATE(YEAR(Q10),12,25))=1
When this is true, then we need to increase the date by 1 day instead of decreasing it like was done for Saturday:
DATE(YEAR(Q10),12,26)
or
DATE(YEAR(Q10),12,25)+1
or
DATE(YEAR(Q10),12,25+1)
So that was the true portion of the Sunday check. Logically speaking the only way to get to the FALSE portion of this nested IF statement is to fail the Saturday check and then fail the Sunday check. Which means you do not need to go through and check if is the WEEKDAY comes out as 2, 3, 4, 5 or 6! Its one of those by the process of eliminating Sunday and Saturday (1 and 7). And if the date falls on Monday-Friday we dont need to change the date and can leave it just as is:
DATE(YEAR(Q10),12,25)
And I realized I did not explain how the date function works, though I think I tried to in one of the previous questions...regardless! DATE(arg1,arg2,arg3) requires three different arguments as integers or other functions that return integers.
arg1 is the year so 2014, 1995, 1965 are all acceptable integers. Also we could use YEAR(Q10), where the cell Q10 holds the date of 2014/10/24. In this case YEAR(Q10) would return 2014.
arg2 is the month and needs to be an integer in the range of 1 to 12. Again you can always use a formula that returns an integer in that range as well such as MONTH(Q10) which from our previous value of Q10 would return 10.
arg3 is day and similar to the above it needs to be an integer. A formula such as DAY(Q10) would return a value of 24.
What this means is if we know what day a holiday is on we can force it to a date by supplying a set month and day, and letting the year be determined by a formula that supplies the year you are interested in. So if you look at the last formula you can see we fixed the month at 12 and the day at 25. They year will be determine from the year of the date supplied in cell Q10.
I am trying to write a formula that will subtract my current value from my last data point. I run a production report and over the weekends we do not produce any product. I would like the value typically from a Friday be used for calculating the change in production for Monday.
I have an input page for all the data and then a calculation page for the reports.
(Input page)
Thursday 1000
Friday 5000
Saturday "blank"
Sunday "blank"
Monday 2000
Ideally the output page would look something like this:
(output)
Friday 4000
Saturday "blank"
Sunday "blank"
Monday -3000
Having the last inputted data (being the 5000 value from Friday) subtracted from the Monday value of 2000.
Please try:
=IF(LEFT(A2,1)="S","",Sheet4!B3-INDEX(Sheet4!B:B,MATCH(1E+100,Sheet4!B$1:B2)))
with adjustment of cell references to suit.
My idea was:
=IF(B2=0,"Nonproductive",IF(B1<>0,B2-B1,B2-OFFSET(B2,-(COUNTIF(OFFSET(C2,-3,0,3),"nonproductive")+1),0)))
Imagine you've got your days in column A, and production in column B.
Copy and paste this from cell C2 onwards.
What this does, is first test if there was any production on a given day, if there was 0 production, or in your above example, "blank", then it will give the value "nonproductive". If there was production on that day, it will test if the day prior had any production, if it did, it will simply deduct yesterdays from today. If the day prior had no production, it will count the number of times that "nonproductive" has occurred in the 3 days prior and add that to the number of days in the past it looks.
On a long weekend, it would count 3 nonproductives, so it would compare Monday to Thursday.
Monday - (1 (default offset for the day prior) + 3 (due to nonproductives)
Monday - 4 = Thursday.
Notes:
This will not work for the first 2 rows because the offset will be trying to find cells that don't exist. Even if
The only problems arise when a break greater than 3 days happens, or you have a day off, then a day on, then a day off, then a day on, such as a Tuesday bank holiday, this is because there are 2 "nonproductive" days in the 3 days prior, but we only want it to increase the offset by 1. This could be avoided through continuing a chain of "if" condition checks, but it doesn't sound as though that is required.
This will work for a 2 or 3 day weekend, and some mid-week holidays