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Antonio_Gomez
Resolver I
Resolver I

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 know that a line chart have the option of forecast, but I need the numbers in a table.

 

Please help!!!!!!!

 

 

 

 

 

 

2 ACCEPTED SOLUTIONS

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:

vchenwuzmsft_0-1661845678621.png

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.

View solution in original post

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.

View solution in original post

4 REPLIES 4
amitchandak
Super User
Super User

@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:

vchenwuzmsft_0-1661845678621.png

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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