Gmontano Posted October 7 Share Posted October 7 (edited) Apologies if the answer is implied or present in existing documentation. In my PVSyst report under 'P50-P90 Evaluation' section - I have the following data: Weather Data: Year-to-year Variability: 4.9% Global Variability (Weather Data + System): Variability (Quadratic Sum): 5.2% Annual Production Probability: Variability: 3.11 GWh P50: 60.13 GWh P99: 52.88 GWh For simplicity - assume no degradation. From this report the first year the P99 value for my project is 52.88GWh. Though - what would be the "year-2 P99"? This is an interesting question. Let X1 and X2 represent the production levels in the first and second years respectively. If X1 is 52.88GWh, then what is X2 such that I am 99% confident that I will produce at least X1 + X2 after 2 years. I can assume that X2 ~ N(Mean, Variance) = N(60.13, 3.11^{2}), and assuming it is independent from X1, infer the distribution of (X1 + X2) ~ N(2*60.13, 2*3.11^{2}). While the maths makes sense - I am unsure the present assumptions, and therefore, have small confidence in the numbers I am calculating for the intended purpose. I appreciate your help - thank you very much, Gus Edited October 7 by Gmontano Link to comment Share on other sites More sharing options...

dtarin Posted October 9 Share Posted October 9 For a single year, σ = 3.11. For two consecutive years, σ = σ / √2 Var( X̄ ) = ( 1/n )^2 * Var( X₁ + X₂ + ... + Xₙ ) = ( 1/n )^2 * ( nσ² ) = σ² / n Link to comment Share on other sites More sharing options...

André Mermoud Posted October 11 Share Posted October 11 The variance of the meteo data multi-year distribution is supposed to be the same for any future year. Therefore the ratio P99/P50 remains the same whatever the year. You can simply calculate P99 (YearN) = P50(YearN) * P99/P50 (Year0) Link to comment Share on other sites More sharing options...

dtarin Posted October 14 Share Posted October 14 For any single year that is the case. For multiple years, the uncertainty due to annual variability decreases, so this parameter needs to be separated out and used with the equation in my previous post, and then added to the other uncertainty quantities in full. My example was not quite correct in this regard. Link to comment Share on other sites More sharing options...

dtarin Posted October 14 Share Posted October 14 1 hour ago, dtarin said: For any single year that is the case. For multiple years, the uncertainty due to annual variability decreases, so this parameter needs to be separated out and used with the equation in my previous post, and then added to combined with the other uncertainty quantities in full. My example was not quite correct in this regard. Link to comment Share on other sites More sharing options...

Gmontano Posted October 23 Author Share Posted October 23 Thank you, Andre. On 10/11/2024 at 1:43 PM, André Mermoud said: The variance of the meteo data multi-year distribution is supposed to be the same for any future year. Therefore the ratio P99/P50 remains the same whatever the year. You can simply calculate P99 (YearN) = P50(YearN) * P99/P50 (Year0) I'm not sure if this answers my original question. Using your last notation - I'm not looking for P99 (YearN), I'm looking for a production level in month N, call X_{N}, such that X_{1} + ... + X_{N} is a P99 number,or, X_{N} | P(X_{1} + ... X_{N}) >= 99% Said differently - I am looking for a 99% confidence interval on a random variable that is the sum of production over N years. My original post was for the 2-year case. Link to comment Share on other sites More sharing options...

André Mermoud Posted October 23 Share Posted October 23 This is indeed an interesting question. But sorry, PVsyst doesn't provide this calculation. You should do that by yourself, using the suggestions mentione above. Link to comment Share on other sites More sharing options...

Gmontano Posted October 24 Author Share Posted October 24 On 10/8/2024 at 10:58 PM, dtarin said: For a single year, σ = 3.11. For two consecutive years, σ = σ / √2 Var( X̄ ) = ( 1/n )^2 * Var( X₁ + X₂ + ... + Xₙ ) = ( 1/n )^2 * ( nσ² ) = σ² / n Hi @dtarin - thank you for your original response. It is insightful, though, I'm trying to understand how it's applicable. Please allow me to lay out the mathematical context of your response below. Let X represent the production level of solar in some year with an expected value of E[X] and variance of Var[X]. Then X₁ is an observation of X. Given n observations one may estimate E[X] through the simple average formula given by X̄ = (1/n)( X₁ + X₂ + ... + Xₙ ). X̄ itself is a random variable, as it differs depending the the observations themselves. The usual statistics of X̄ are as follows: E[X̄] = E[(1/n)( X₁ + X₂ + ... + Xₙ )] = (1/n)E[( X₁ + X₂ + ... + Xₙ )] = (1/n)(E[X₁] + E[X₂] + ... + E[Xₙ]) = (1/n)*nE[X] = E[X]. As such, X̄ is an nonbiased estimator for E[X]; said differently the simple average of observations of production is a good estimate the expected production level. Var[X̄] = E[X̄ - E[X̄]] = E[X̄ - E[X]]. This is measuring the expected difference between the estimator of E[X] and E[X] itself. This, can be estimated by σ² / n. Naturally - as the number of observations increase, or approach infinity, the closer the estimator for E[X] moves to E[X] and therefore the standard error approaches 0. This makes sense as the more observations of the population we have, the closer our statistics will move towards the expected population averages! Going across years - we aren't changing n, i.e. the number of empirical observations that drive X̄. We are taking established estimators for E[X] (P50) and Var[X] making inferences through time. I've written this to recall my statistics, and confirm that I don't think this change in variance applies here. Intuitively - the distribution of production between years 1 and 10 shouldn't change (in the absence of physical/mechanical changes such as degradation). The variance of production in year 10, shouldn't be (1/sqrt(10)) times lower than year 1. The same stochasticity of solar irradiation applies. Link to comment Share on other sites More sharing options...

dtarin Posted October 25 Share Posted October 25 When you multiply a random variable by a constant, the mean gets multiplied by the same constant and the variance gets multiplied by that constant squared. Xi are iid random variables. Link to comment Share on other sites More sharing options...

Gmontano Posted October 28 Author Share Posted October 28 I acknowledge this, though it doesn't answer my question. The variance of the estimator of the expected value of a random variable, is not the variance of the random variable, which should be the point of this conversation. Let's close this. It's clear to me that the answer requires an elaborate discussion. Link to comment Share on other sites More sharing options...

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