3 Month Prior Sales in set-analysis/Scripting

[abhaytrip]

Hi folks,

I'm stuck in a problem where I need to present the 3 Month prior sales (including the 3rd Month) in a straight table , but im unable to do so.

here im attaching the image where you can find the sample example of the problem stated above.

any sort of help is appreciate.

Thanks,

Abhay

[chinnuchinni]

try this:

=Sum({<Month-Year={">=$(=AddMonths(Max(Month-Year),-3)) <$(=Max(Month-Year))"}>}Sales)

[abhaytrip]

tried this , but it doesn't seem to work.

I used this formula.

=num(Sum({<[Month-Year]={">=$(=AddMonths(Max([Month-Year]),-3)) <$(=Max([Month-Year]))"}>}Sales)/1000,'$###.##k')

[sunny_talwar]

May be try this

Above(Sum(Sales), 3)

Missing Manual - Above() and Below()

[abhaytrip]

hey this works if i dont make any selection from the month or year.

but as you see if i select a particular year i wont be able to see data for JAN FEB MAR as according to the formula there's nothing above it.

Any other alternative would also work .

Thanks alot.

[devyanshu_gupta]

Hi,

I prefer using a Month Number in the calendar for these type of queries.

MonthNum=Year*12 + Month.

Now using this you can calculate the value in Set

Try  Sum({<MonthNum={Max(MonthNum)-3},Month=,Year=>}Sales)

Regards,

Devyanshu

[abhaytrip]

I did try using this expression, It din seem to work out.

Sum({<MonthNumber={'Max(MonthNumber)-3'} ,Month=,Year=>}Sales)

[rodjager]

Hi Abhay,

I have done this before where I used a resident load to bring the data back in as a comparative figure.  It gives more flexibility for doing period comparisons and as far as I know it's more efficient than a complex expression with set analysis.

I have attached a simple model of how this could work.

Hope this helps.

Rod

[abhaytrip · Accepted solution]

=If(Sum(Sales) > 0, Above(sum({<[Month-Year]>}Sales),3))

This seems to be working as expected.

[abhaytrip]

This works too Rod , Thanks alot