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Antonio_Gomez's avatar
Antonio_Gomez
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4 years ago
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Forecast

Hello everyone!   Here I am again needing help 😕   I have data from sensors, they measure the volume inside of tanks. I would like to forecast the volume for 1, 2, 3 and 4 weeks ahead.   I kno...
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Antonio_Gomez's avatar
Antonio_Gomez
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4 years ago

Hi amitchandak 

 

I just realized that I used an incorrect term. I mean forecast, not predict. (And I edited the original post to avoid confusion)

 

For example, the product in the tank have a constant rate of use of 1 Lt/week.

 

So, if the remaining volume in the tank is 30 Lt, the forecast would show that in 1, 2, 3 and 4 weeks the volume will be 29, 28, 27 and 26 Lt.

 

There no need of use AI nor ML, things that I don't domine well 😞

 

.

v-chenwuz-msft's avatar
v-chenwuz-msft
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4 years ago

Hi Antonio_Gomez ,

 

First I think you should calculate the [constant rate of use], then get the [remaining volume in the tank], and finally use a formula like the one below to get the forecast you want.

 

Measure = 
SUM(FactTable[remaining volume in the tank])-SELECTEDVALUE('Week'[Week])*SUM(FactTable[constant rate of use])

 

Result:

Pbix in the end you can refer.

Best Regards

Community Support Team _ chenwu zhu

 

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

  • Antonio_Gomez's avatar
    Antonio_Gomez
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    4 years ago

    Hi v-chenwuz-msft amitchandak 

     

    I decided to do this:

     

    Starting from the equation of the line Y = aX + b and the historical values I calculated for several days differents approximations. For example:

     

    With today's value and yesterday's values I could construct one equation

    Today volume = 890

    Yesterday volume = 910

    Today (X) = 0

    Yesterday (X) = -1

     

    a = (Y1 - Y2) / (X1 - X2)

     

    So,

     

    a = (890 - 910) / (0 - (-1))

     

    b = Y - aX (isolated from the original equation)

    And if I have the "a" from the last result; Y = today's volume and X = Today (X) = 0; I could have "b" and construct one equation.

     

    Now, with this equation: Y = -20X + 890; I could predict for the day I wanted.

     

    If I want to know the volume in 7 days, just put Y = (-20 * 7) + 890 = 750

     

    I did this for several days in the past and always with today's reference

    Y1 = Same as above

    X1 = Same as above

    Y2 = 915

    X2 = -2 (the day before yesterday)

     

    Y1 = Same as above

    X1 = Same as above

    Y2 = 930

    X2 = -3

     

    So, for each equation created taking into account the days in the past (in this example are 3; X = -1, X = -2 and X = -3), I forecast 7 days in the future Y = a(7) + b and average the three results to have the more approximated value.