Forum Discussion
Calculating full-time equivalent (FTE)
Your simple exceptions aren't that simple. In the first case you can use the MIN() function, for example
hourstocount = MIN(8,hoursworked)
to cap at 8 (ignoring the fact that you are penalizing the worker for doing more work). For the second scenario use ALL() or ALLSELECTED() or ALLEXCEPT() to modify the filter context inside the computation.
Hi jsangerman ,
Try to show values as "Percent of column total". Or create a measure like below.
Measure = DIVIDE ( SUM ( 'Table'[Efforts (hours)] ), CALCULATE ( SUM ( 'Table'[Efforts (hours)] ), ALLEXCEPT ( 'Table', 'Table'[Date], 'Table'[User] ) ) )Best regards
Icey
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Users enter time they spend on each project each day. So, the raw data looks like this:
User Project | 1/1/21 | 1/2/21 | 1/3/21 | 1/4/21 | 1/5/21 |
Mike Wallace Project A Project B Project C |
3 hr 4 hr 1 hr |
1 hr 0 hr 7 hr |
3 hr 0 hr 5 hr |
0 hr 2 hr 6 hr |
3 hr 3 hr 2 hr |
The FTE could calculate this like this
User Project | 1/1/21 | 1/2/21 | 1/3/21 | 1/4/21 | 1/5/21 |
Mike Wallace Project A Project B Project C |
0.375 hr 0.5 hr 0.125 hr |
0.125 hr 0 hr 0.875 hr |
0.375 hr 0 hr 0.625 hr |
0 hr 0.25 hr 0.75 hr |
0.375 hr 0.375 hr 0.25 hr |
Projects run for full months, but the monthly data is really a summary shown in a visualization.
Hi jsangerman ,
Try to show values as "Percent of column total". Or create a measure like below.
Measure =
DIVIDE (
SUM ( 'Table'[Efforts (hours)] ),
CALCULATE (
SUM ( 'Table'[Efforts (hours)] ),
ALLEXCEPT ( 'Table', 'Table'[Date], 'Table'[User] )
)
)
Best regards
Icey
If this post helps, then consider Accepting it as the solution to help other members find it faster.