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This is a perennial question for retailers, for which there are a number of solutions in existence:
How can you calculate the "forward cover" of a product knowing its current inventory and armed with forward sales estimates.
eg.
current inventory 100 units (cell A1)
weekly forward sales estimates: 25, 30, 10, 40, 90... (in range
A2:AX)
Here the answer would be 3.875 weeks (3 full weeks plus 0.875 of week 4)
I have a UDF to do this already.
I also have some slightly complicated array functions to do this, eg.
=MATCH(TRUE,SUBTOTAL(9,OFFSET(A2:A13,,,ROW(A2:A13)-ROW(A2)+1))>A1,0)-1+(A1-SUM(A2:INDEX(A2:A13,MATCH(TRUE,SUBTOTAL(9,OFFSET(A2:A13,,,ROW(A2:A13)-ROW(A2)+1))>A1,0)-1)))/INDEX(A2:A13,MATCH(TRUE,SUBTOTAL(9,OFFSET(A2:A13,,,ROW(A2:A13)-ROW(A2)+1))>A1,0)-1+1)
I was wondering if there is a neater way with these 'new-fangled' array functions which have been available for the last few years in later versions of Excel?
Here is another possible solution, although it requires the LET() function which is only available to newer version of excel (2021, 365 and later I believe).
The solution would be the following formula:
=LET(
sales,A2:A50,
inventory,A1,
cum_sum,MMULT(SEQUENCE(1,ROWS(sales),1,0),(ROW(sales)<=TRANSPOSE(ROW(sales)))*sales),
week_full,MATCH(TRUE,inventory<cum_sum,0) - 1,
week_frac,(inventory - INDEX(cum_sum,week_full)) / INDEX(sales,week_full + 1),
week_full + week_frac
)
Explanation
Given inventory and the forward looking sales estimates, the formula calculates the running total (i.e. cumulated sum) of the sales estimates as shown in the table here below
Inv and Sales
Cumulated Sum
Inv > Cum_Sum
Week
100
25
25
0
1
30
55
0
2
10
65
0
3
40
105
1
4
90
195
1
5
...
...
1
6
The formula goes on to get the number of full weeks of 'forward cover' by finding the the value for the cumulated sum that exceeds the inventory minus one (here 4 - 1 = 3).
Lastly, for the value of the week fraction covered in the last week, the formula calculates inventory minus sum of sales estimates of all previous weeks divided by sales estimate of final week of cover (i.e. (100 - 65) / 40 = 0.875).
Edit
After simplifying the formula you used with the LET() function, I noticed it's doing exactly the same calculation with the only difference of how the cumulated sum is being calculated. Here's your formula using LET():
=LET(
sales,A2:A50,
inventory,A1,
cum_sum,SUBTOTAL(9,OFFSET(sales,,,SEQUENCE(ROWS(sales)))),
week_full,MATCH(TRUE,cum_sum>inventory,0)-1,
week_frac,(inventory - INDEX(cum_sum,week_full)) / INDEX(sales,week_full+1),
week_full + week_frac
)
=LET(inv,A1,
sales,A2:A6,
cs,SCAN(0,sales,LAMBDA(x,y,x+y)),
m,XMATCH(A1,cs,1)-1,
m+(inv-
IF(m=0,
0,
INDEX(cs,m)))
/INDEX(sales,m+1))
SCAN() is perfect for creating a cumulative sum.
It can be referenced inside XMATCH because of the use of LET.
Here m returns the number of full weeks and the final calculation is the number of full weeks + (inv- cumulative sum up to the full week)/sales of the following week.
I need a formula to calculate between two date and time excluding lunch time, holidays, weekend and between 09:00 and 18:00 work hours.
For example, 25/07/2022 12:00 and 29/07/2022 10:00 and answer has to be 1 day, 06:00
Thanks in advance.
I had a formula but it didn't work when hours bigger than 24 hours.
I don't know how you got to 1 day and 6 hours, but here is a customizable way to filter your time difference calculation:
=LET(
start,E3,
end,E4,
holidays,$B$3:$B$5,
array,SEQUENCE(INT(end)-INT(start)+1,24,INT(start),TIME(1,0,0)),
crit_1,array>=start,
crit_2,array<=end,
crit_3,WEEKDAY(array,2)<6,
crit_4,HOUR(array)>=9,
crit_5,HOUR(array)<=18,
crit_6,HOUR(array)<>13,
crit_7,ISERROR(MATCH(DATE(YEAR(array),MONTH(array),DAY(array)),holidays,0)),
result,SUM(crit_1*crit_2*crit_3*crit_4*crit_5*crit_6*crit_7),
result
)
Limitation
This solution only works on an hourly level, i.e. the start and end dates and times will only be considered on a full hour basis. When providing times like 12:45 as input, the 15 minute increment won't be accounted for.
Explanation
The 4th item in the LET() function SEQUENCE(INT(end)-INT(start)+1,24,INT(start),TIME(1,0,0)) creates an array that contains all hours within the start and end date of the range:
(transposed for illustrative purposes)
then, based on that array, the different 'crit_n' statements are the individual criteria you mentioned. For example, crit_1,array>=start means that only the dates and times after the start date and time will be counted, or crit_6,HOUR(array)<>13 is the lunch break (assuming the 13th hour is lunch time), ...
All of the individual crit_n's are then arrays of the same size containing TRUE and FALSE elements.
At the end of the LET() function, by multiplying all the individual crit_n arrays, the product returns a single array that will then only contain those hours where all individual criteria statements are TRUE:
So then the SUM() function is simply returning the total number of hours that fit all criteria.
Example
I assumed lunch hour to be hour 13, and I assumed the 28th to be a holiday within the given range. With those assumptions and the other criteria you already specified above, I'm getting the following result:
Which looks like this when going into the formula bar:
In cell G2, you can put the following formula:
=LET(from,A2:A4,to,B2:B4,holidays,C2:C2,startHr,E1,endHr,E2, lunchS, E3, lunchE, E4,
CALC, LAMBDA(date,isFrom, LET(noWkDay, NETWORKDAYS(date,date,holidays)=0,
IF(noWkDay, 0, LET(d, INT(date), start, d + startHr, end, d + endHr,
noOverlap, IF(isFrom, date > end, date < start), lunchDur, lunchE-lunchS,
ls, d + lunchS, le, d + lunchE,
isInner, IF(isFrom, date > start, date < end),
diff, IF(isFrom, end-date-1 - IF(date < ls, lunchDur, 0),
date-start-1 - IF(date > le, lunchDur, 0)),
IF(noOverlap, -1, IF(isInner, diff, 0)))))),
MAP(from,to,LAMBDA(ff,tt, LET(wkdays, NETWORKDAYS(ff,tt,holidays),
duration, wkdays + CALC(ff, TRUE) + CALC(tt, FALSE),
days, INT(duration), time, duration - TRUNC(duration),
TEXT(days, "d") &" days "& TEXT(time, "hh:mm") &" hrs "
)))
)
and here is the output:
Explanation
Used LET function for easy reading and composition. The main idea is first to calculate the number of working days excluding holidays from column value to to column value. We use for that NETWORKDAYS function. Once we have this value for each row, we need to adjust it considering the first day and last day of the interval, in case we cannot count as a full day and instead considering hours. For inner days (not start/end of the interval) it is counted as an entire day.
We use MAP function to do the calculation over all values of from and to names. For each corresponding value (ff, tt) we calculate the working days (wkdays). Once we have this value, we use the user LAMBDA function CALC to adjust it. The function has a second input argument isFrom to consider both scenarios, i.e. adjustment at the beginning of the interval (isFrom = TRUE) or to the end of the interval (isFrom=FALSE). The first input argument is the given date.
In case the input date of CALC is a non working day, we don't need to make any adjustment. We check it with the name noWkDay. If that is not the case, then we need we need to determine if there is no overlap (noOverlap):
IF(isFrom, date > end, date < start)
where start, end names correspond to the same date as date, but with different hours corresponding to start Hr and end Hr (E1:E2). For example for the first row, there is no overlap, because the end date doesn't have hour information, i.e. (12:00 AM), in such case the corresponding date should not be taken into account and CALC returns -1, i.e. one day needs to be subtracted.
In case we have overlap, then we need to consider the case the working hours are lower than the maximum working hours (from 9:00 to 18:00). It is identified with the name isInner. If that is the case, we calculate the actual hours. We need to subtract 1 because it is going to be one less full working day and instead to consider the corresponding hours (that should be less than 9hrs, which is the maximum workday duration). The calculation is carried under the name diff:
IF(isFrom, end-date-1 - IF(date < ls, lunchDur, 0),
date-start-1 - IF(date > le, lunchDur, 0))
If the actual start is before the start of the lunch time (ls), then we need to subtract lunch duration (lunchDur). Similarly if the actual end is is after lunch time, we need to discount it too.
Finally, we use CALC to calculate the interval duration:
wkdays + CALC(ff, TRUE) + CALC(tt, FALSE)
Once we have this information, it is just to put in the specified format indicating days and hours.
Now let's review some of the sample input data and results:
The interval starts on Monday 7/25 and ends on Friday 7/29, therefore we have 5 working days, but 7/26 is a holiday, so the maximum number of working days will be 4 days.
For the interval [7/25, 7/29] starts and ends on midnight (12:00 AM), therefore the last day of the interval should not be considered, so actual working days will be 3.
Interval [7/25 10:00, 7/29 17:00]. For the start of the interval we cannot count one day, instead 8hrs and for the end of the interval, the same situation 8hrs, so instead of 4days we are goin to have 2days plus 16hrs, but we need to subtract in both cases the lunch duration (1hr) so the final result will be 2 days 14hrs.
I am trying to do a time subtraction in excel of 30 minutes and I am running into a speed bump. So the table I have are as follows.
Table "Schedule"
Column 1 is day of the week (Mon-Sun) (formated as general, as this is plain text)
Column 2 is start time of the shift (formated as h:mm AM/PM)
Column 3 is end time of the shift (formated as h:mm AM/PM)
Column 4 is duration of the shift (start to end) (formated by formula (TEXT(col3-col2,"h:mm")) )
Column 5 is paid hours (if the total hours is over 6.5 then subtract 0.5 hours for an unpaid lunch) (formula IF(col5>"6:30",col5-"0:30",D5) )
The issue is any time allotment over 10 hours start to end (where column 4, the duration hits 10 hours) no lunch is subtracted at all.
So...
Start 9:00 AM, End 6:59 PM, Hours Total 9:59, Hours Paid 9:29
But...
Start 9:00 AM, End 7:00 PM, Hours Total 10:00, Hours Paid 10:00
and that should obviously not happen. I can't find anything on google so I figured the excel gurus here may have some advice.
Thanks!
If your time columns are stores using excel's dedicated time format, this should be straightforward. Mixed data types are likely your problem.
First, be sure your time columns (columns 2 and 3) are set using the time function, i.e.,
=time(hours,minutes,seconds)
Then, you should be able to add and subtract easily.
Column 4: = column 3 - column 2
... then subtract 30 minutes also using the time() function:
Column 5: = if(column 4 > time(6,30,0),column 4 -time(0,30,0),column 4)
Excel stores time values from 0 to 1. So 24 hours=1, 12 hours=.5 etc. That means 6.5 hours=0.270833333 and .5 hours=0.020833333. As a result you can just do a simple if statement.
=IF(D2>0.270833333,D2-0.020833333,D2)
To turn it into a time format, is to just use excel's time formating options.
I have a date in column H10 and need to add 45 days to this date in the next Column I
If there are not dates Column I must be blank
If the 45th day falls on a weekend the calculation must move to the next workday which is Monday
You need to combine two fundamental functions.
First, DATE + INT = DATE. For example, if H10 = 1/8/2015 and H11 = H10 + 10 then H11 will show 1/18/2015.
In your case, you want to use H10 + 45.
Second, you can use the Weekday(date,mode) function to determine the day of the week. Personally, for your purpose, you could use weekday(h10 + 45, 2) which would give a 1-5 for MTWRF, and a 6-7 for a weekend day. So something like
=if(weekday(h10+45,2) < 6, "weekday", "weekend")
=if(weekday(h10+45,2) = 1, "Monday!!", "not monday...")
But we aren't done yet - you need to make sure your day actually ends up on a weekday. So we can do something like this - when determining a weekday, we can use that to determine how much we need to add. If we end up with a 6 (Saturday) we want to add 2 days to push it to a Monday. In the case of a 7, we want to add 1 day to push it to a Monday. Thus, we can simply take the 8 - weekday(h10+45) to add when it's a weekday. So our add value becomes
// determine day type weekday weekend, so add the offset
= if(weekday(h10+45) < 5, h10+45, h10 + 45 + (8 - weekday(h10+45))
You also have a requirement about being blank, so you'll want to wrap whatever you use with
=if(isblank(h10),"", /* your real function here */)
You can combine the functions for IF(), WEEKDAY() and WORKDAY() to calculate your finish date and ensure that it does not fall on a weekend.
I've used
WEEKDAY(WORKDAY(H10+45),16)
to have Saturday and Sunday be represented as days 1&2 respectively.
IF(WEEKDAY(WORKDAY(H10,45),16)=1,WORKDAY(H10,45)+2,IF(WEEKDAY(WORKDAY(H10,45),16)=2,WORKDAY(H10,46),H10))
I am stuff on a problem regarding setting up a dynamic shift rota at work.
What it has to do is there are 3 people per team and 3 jobs to do (which they do as equals as it can be per week) and when someone is off lets call it job 3 is just not done that day.
I have been able to make it to set true or false if there in but having trouble with assigning a different job per day with priority on job numbers 1 and 2.
edit: I thought I would explain it better what I am looking for
alt text http://img688.imageshack.us/img688/3032/spreadsheet.jpg
This is the spreadsheet i have at the top is the rota 1 = in 0 = day off
I have only been working on Group 3 to get it working, In the grey next to the days at the bottom are the jobs in which that person would be doing if it was a full staff day.
What i need to do is if there are only 2 people in then they do job number 1 and 2 alternativly .
i currently have this formula it works for some combo but not all (this is formula for H33)
=IF(F9 > 0, IF(OR(F9=0,F10=0,F11 = 0),IF(OR(I33 = 1, I33 = 2),I33,I33-1),I33), 0)
Any advise woudl be great
Use this formula for John (column C row 4), and the copy it down to the other 2 people in the same day
=IF(B4="in",COUNTIF(B4:B$4,"in"),"")
for next day you should change the B$4 to B$8, etc
EDIT / I did not undestand your rotation pattern
Now, refering to the excel image you posted, for Group 1, Monday
You need two auxiliar columns
1. In P11-> =D3*E23 // To eliminate zeroes
2. Copy down P11 till P13
3. In Q11-> =IF(P11=0,"",RANK(P11,P$11:P$13,-1)) // To get 1, 2 or 3 values
4. Copy down Q11 till Q13
5. In D23 -> =IFERROR(Q11-MIN(Q$11:Q$13)+1,"") // Offset Factor 1,1-2,1-3
4. Copy down D23 till D25
It was a tricky one