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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 know that a line chart have the option of forecast, but I need the numbers in a table.
Please help!!!!!!!
Solved! Go to Solution.
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.
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.
@Antonio_Gomez , You can explore to bring data using R or Python connector and use forecasting by that R/Python
refer
https://www.youtube.com/watch?v=IP76UJ4nZ70
https://community.powerbi.com/t5/Desktop/Forecast-Using-AI-ML/td-p/1184505
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 😞
.
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.
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.
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