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railtons's avatar
railtons
Helper I
2 years ago
Solved

Questions about creating a column with projected values according to the last remaining work to be d

Hello team, I'm just getting started with Power BI and I have a question that seems simple, but I can't figure it out.

I have a table with the following columns:

1 - Estimated remaining work: where I record my last estimated effort;
2 - Capacity: My team's capacity to perform in each iteration;
3 - Projected Estimated remaining work: Based on my last estimated effort, according to my capabilities, in which iteration this estimate would reach zero.

 

 


My question is:

I would like to create a column in DAX that exactly represents column 3.

 

 

 

Thanks

 


Railton

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  • Anonymous's avatar
    Anonymous
    2 years ago

    Hi railtons ,

    Here some steps that I want to share, you can check them if they suitable for your requirement.

    Here is my test data:

    1.Open power query and add a index column

    2.Create a calculate column

    Projected Estimated remaining work = 
VAR _a = 
CALCULATE(
    SUM('Table'[Capacity]),
    FILTER('Table','Table'[Index] <= EARLIER ( 'Table'[Index] ) )
)
RETURN
CALCULATE(
    MAX('Table'[Estimated remaing work]),
    FILTER(
        'Table',
        'Table'[Index] = 1
    )
)-_a

    Final output

     

    Best regards,

    Albert He

     

    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly

2 Replies

  • Anonymous's avatar
    Anonymous
    Not applicable
    2 years ago

    Hi railtons ,

    Here some steps that I want to share, you can check them if they suitable for your requirement.

    Here is my test data:

    1.Open power query and add a index column

    2.Create a calculate column

    Projected Estimated remaining work = 
VAR _a = 
CALCULATE(
    SUM('Table'[Capacity]),
    FILTER('Table','Table'[Index] <= EARLIER ( 'Table'[Index] ) )
)
RETURN
CALCULATE(
    MAX('Table'[Estimated remaing work]),
    FILTER(
        'Table',
        'Table'[Index] = 1
    )
)-_a

    Final output

     

    Best regards,

    Albert He

     

    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly