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Hedging the Risk of Wind Power Production Using Dispatchable Energy Source

  • Guglielmo D’Amico ORCID logo , Bice Di Basilio EMAIL logo and Filippo Petroni ORCID logo

Abstract

In this paper we advance a nonlinear optimization problem for hedging wind power variability by using a dispatchable energy source (DES) like gas. The model considers several important aspects such as modeling of wind power production, electricity price, nonlinear penalization scheme for energy underproduction and interrelations among the considered variables. Results are given in terms of optimal co-generation policy with DES. The optimal policy is interpreted and analyzed in different penalization scenarios and related to a 48 MW hypothetical wind park. The model is suitable for integration of wind energy especially for isolated grids. Some probabilistic results for special moments of a Log-Normal distribution are obtained; they are necessary for the evolution of the optimal policy.

MSC 2010: 90B25

A Proofs

A.1 Proof of Lemma 1

Let 𝕄>:=𝔼[R>] and 𝕄=:=𝔼[R=]. Then

=𝕄>+𝕄=
=𝔼[χ{We+PgK}χ{We>0}(πeK-πgPg)]+𝔼[χ{We+Pg<K}χ{We>0}[πe(We+Pg)-πgPg
-C(K-(We+Pg))α]+𝔼[χ{We=0}[πePg-πgPg-C(K-Pg)α],

where

(A.1)

𝔼[χ{WeK-Pg}χ{We>0}πeK]=KK-Pg+(0+xf(we,πe)(x,p)dx)dp,
-𝔼[χ{WeK-Pg}χ{We>0}πgPg]=-K-Pg+πgPgfwe(x)dx=-πgPgF¯we(K-Pg),
𝔼[χ{We<K-Pg}χ{We>0}πe(We+Pg)]=0K-Pg0+x(p+Pg)f(we,πe)(x,p)dxdp,
-𝔼[χ{We<K-Pg}χ{We>0}[πgPg+C(K-(We+Pg))α]=-0K-Pgfwe(p)[πgPg+C(K-(p+Pg))α]dp,
𝔼[χ{We=0}[πePg-πgPg-C(K-Pg)α]=aPg𝔼[πe|We=0]-a[πgPg+C(K-Pg)α].

Now we calculate the first-order derivative of with respect to the variable Pg. In order to do this, we need to evaluate the derivatives of all the five addenda in formula (A.1) using Leibnitz’s formula for differentiation under an integral sign. Let us start computing the derivative of the first addendum of (A.1). We have

(A.2)KK-Pg+(0+xf(we,πe)(x,p)𝑑x)𝑑pPg=K𝔼[πe|We=K-Pg]fwe(K-Pg).

The derivatives of the second addendum is

(A.3)-PgπgPgF¯we(K-Pg)=-fwe(K-Pg)πgPg-F¯we(K-Pg)πg.

The derivatives of the third addendum is

(A.4)

Pg0K-Pg0+x(p+Pg)f(we,πe)(x,p)𝑑x𝑑p
=-Kfwe(K-Pg)𝔼[πeWe=K-Pg]+0K-Pgfwe(p)𝔼[πeWe=p]dp.

The derivatives of the fourth addendum is

(A.5)

-Pg0K-Pgfwe(p)[πgPg+C(K-(p+Pg))α]𝑑p
=fwe(K-Pg)[πgPg+C(K-(K-Pg+Pg))α]-0K-Pgfwe(p)[πg-Cα(K-(p+Pg))α-1]𝑑p
=fwe(K-Pg)πgPg-0K-Pgfwe(p)[πg-Cα(K-(p+Pg))α-1dp].

Finally, we report the derivative of the fifth addendum:

(A.6)PgaPg𝔼[πe|We=0]-a[πgPg+C(K-Pg)α]=a𝔼[πe|We=0]-aπg+aCα(K-Pg)α-1.

The summation of formulas (A.2), (A.3), (A.4), (A.5), (A.6) gives

(Pg)Pg=K𝔼[πe|We=K-Pg]fwe(K-Pg)-fwe(K-Pg)πgPg-F¯we(K-Pg)πg
-Kfwe(K-Pg)𝔼[πeWe=K-Pg]+0K-Pgfwe(p)𝔼[πeWe=p]dp+fwe(K-Pg)πgPg
-0K-Pgfwe(p)[πg-Cα(K-(p+Pg))α-1dp]+a𝔼[πe|We=0]-aπg+aCα(K-Pg)α-1,

and through some algebraic manipulations we have

(Pg)Pg=-πg(1+a)+0K-Pgfwe(p)Cα[K-(p+Pg)]α-1𝑑p+aCα(K-Pg)α-1
+0K-Pgfwe(p)𝔼[πe|We=p]dp+a𝔼[πe|We=0].

To prove the concavity of (Pg), we proceed to compute the second-order derivative:

(A.7)

2(Pg)Pg2=-aCα(α-1)(K-Pg)α-2-0K-Pgfwe(p)(1-a)Cα(α-1)(K-(p+Pg))α-2𝑑p
-(1-a)fwe(K-Pg)𝔼[πe|We=K-Pg].

Note that if α1, the right-hand side of (A.7) is negative and the function (Pg) is concave. If 0<α<1, we have concavity only if 2(Pg)Pg2<0, i.e. when

𝔼[πe|We=K-Pg]>1(1-a)fwe(K-Pg){-aCα(α-1)(K-Pg)α-2
-0K-Pgfwe(p)(1-a)Cα(α-1)(K-(p+Pg))α-2dp}.

The lemma is proved.

A.2 Proof of Proposition 2

This proof is based on the following two lemmas (Lemma 3 and Lemma 4).

Lemma 3.

Let ϕ(a) be the probability density function of a standard Normal distribution. Then:

  1. If k is even, we have

    (A.8)-+akϕ2(a)𝑑a=12πiOkk-i2.
  2. If k is odd, we have

    (A.9)-+akϕ2(a)𝑑a=0.

Proof.

We start to prove formula (A.8). If k is even, we have

-+akϕ2(a)𝑑a=-+ak12πe-a2𝑑a=12π-+ak-1(-12)(-2ae-a2)𝑑a.

Using integration by parts, we obtain

-+akϕ2(a)𝑑a=(-14π){lima+ak-1ea2-lima-ak-1ea2-(k-1)-+ak-22π12πe-a2𝑑a}
=(-14π){0-limb+(-b)k-1eb2-2π(k-1)-+ak-2ϕ2(a)𝑑a}.

Since limb+-(bk-1)eb2=0, we have

(A.10)-+akϕ2(a)𝑑a=(-14π){0-2π(k-1)-+ak-2ϕ2(a)𝑑a}=k-12-+ak-2ϕ2(a)𝑑a.

This relation is of recursive type and from [14] we have that for k=2,

-+akϕ2(a)𝑑a=-+a2ϕ2(a)𝑑a=14π.

Then according to relation (A.10) for k=4 we get

-+a4ϕ2(a)𝑑a=3214π,

and in general

-+akϕ2(a)𝑑a=12πiOkk-i2,

where Ok={i:ik with i odd}.

Let us now prove formula (A.9). If k is odd, we have

(A.11)-+akϕ2(a)𝑑a=-+ak12πe-a2𝑑a,

and F(a)=ak12πe-a2 is an odd function. Thus

F(-a)=-ak12πe-a2=-F(a)

and for this reason -+F(a)𝑑a=0. ∎

Lemma 4.

Let ϕ(a) and Φ(a) be the probability density function and the distribution function of a standard Normal distribution, respectively. Then:

  1. If k is even, we have

    (A.12)IkP:=-+akϕ(a)Φ(a)𝑑a=Ik-2P(k-1)𝑎𝑛𝑑I0P=0.5.
  2. If k is odd, we have

    (A.13)IkD:=-+akϕ(a)Φ(a)𝑑a=(k-1)Ik-2D+12πiOkk-i2=12πj=1k+12tk,j,

    and the numbers tk,j are those defined in Proposition 2.

Proof.

Consider formula (A.12):

-+akϕ(a)Φ(a)𝑑a=-+aϕ(a)Φ(a)ak-1𝑑a.

Set u(a)=aϕ(a) and v(a)=Φ(a)ak-1. Then u(a)=-ϕ(a); indeed,

D(-ϕ(a))=D(-12πe-0.5a2)=aϕ(a).

Moreover, v(a)=(k-1)ak-2Φ(a)+ak-1ϕ(a) and using integrations by parts, we obtain

(A.14)-+akϕ(a)Φ(a)𝑑a=[-ϕ(a)ak-1Φ(a)]-+--+-ϕ(a)[(k-1)ak-2Φ(a)+ak-1ϕ(a)]da.

Note that

lima+-ϕ(a)ak-1Φ(a)=0

because

lima+ak-1Φ(a)e0.5a2lima+ak-1e0.5a2=0.

Moreover, for a>0,

ak-1Φ(a)e0.5a20

and

lima+ak-1Φ(a)e0.5a2=0.

Note that changing the variable x=-a we get

lima-aΦ(a)e0.5a2=limx+-xΦ(-x)e0.5x2
=-limx+xΦ(-x)e0.5x2
limx+xe0.5x2=0

and the observing that xΦ(-x)e0.5x20 implies

limx+xΦ(-x)e0.5x2=0.

Consequently, formula (A.14) becomes

(k-1)-+ϕ(a)ak-2Φ(a)𝑑a+-+ϕ(a)2ak-1𝑑a.

Being k even, (k-2) is even and (k-1) is odd so using the result of Lemma 3, we have

-+ϕ(a)akΦ(a)𝑑a=(k-1)-+ak-2ϕ(a)Φ(a)𝑑a.

If k=2, then

-+ϕ(a)(k-1)a2Φ(a)𝑑a=(2-1)-+ϕ(a)Φ(a)𝑑a=12.

If k=4, we have

-+ϕ(a)a4Φ(a)𝑑a=(3)-+ϕ(a)Φ(a)a2𝑑a=32.

In general, if k is even, then

IkP=-+ϕ(a)akΦ(a)𝑑a=12iOki,

or recursively

IkP=Ik-2P(k-1).

Now let us consider formula (A.13):

(A.15)-+akϕ(a)Φ(a)𝑑a=-+aϕ(a)Φ(a)ak-1𝑑a.

By using integrations by parts, integral (A.15) becomes

-+(k-1)ϕ(a)Φ(a)ak-2𝑑a+-+ϕ2(a)ak-1𝑑a.

Being k odd, (k-1) is even and (k-2) is odd so using formula (A.8), we have

(k-1)-+ϕ(a)Φ(a)ak-2𝑑a+12πiOk-1(k-1-i)2.

In general, we have the following recursive relation:

IkD=(k-1)Ik-2D+-+ak-1ϕ2(a)𝑑a
=(k-1)Ik-2D+12πiOk(k-i)2.

The lemma is proved. ∎

Table 4

The triangular array shows coefficients ti,j.

Row Indexti,j
11
3122
53428
7158921248
910516153696384
m(m-2)12(m-1)/2

From a practical point of view it can be interesting to note that coefficients ti,j can be obtained using the triangular array described in Table 4. In the first column of the table called “Row index” there are the rows from 1 to m and in the second column of the table there are coefficients ti,j where i refers to the row and j refers to the column. For example the element t1,1 is the first element of the fist row and t3,2 is the second element of the third row and so on. In particular, the first elements of each row are obtained with the following formula: (m-2)12(m-1)/2 so the element t3,1 comes from (3-2)12, the element t5,1 comes from (5-2)122, etc., until the m-row. Indeed, the first column’s coefficients ti,1 are the result of the multiplication of (m-2) and the element of the previous row with the same position, i.e. t(i-1),1. As regards all of the other coefficients, they are obtained multiplying (m-1) by the element in the north-west corner. For example, applying this rule we have that the coefficient t11,3 comes from (11-1)(6)(5)(3)(1)(2)(2)(2), where (6)(5)(3)(1)(2)(2)(2) is t9,2; t11,4 is the result of the product of (m-1) by t10,3, and so on. Finally, to obtain Tk is necessary adding the rows of this triangular array.

Proof of Proposition 2.

We have

𝔼[πeFπe(πe)]=0+x1x2πσ2e-(log(x)-μ)22σ2Φ(log(x)-μ)σ𝑑x.

By setting t=(log(x)-μ)σ, we obtain

𝔼[πeFπe(πe)]=-+etσeμ12πexp(-t22)Φ(t)𝑑t=eμ-+etσϕ(t)Φ(t)𝑑t,

but etσ=k=0+(tσ)kk! so we have

𝔼[πeFπe(πe)]=eμk=0+(σ)kk!-+tkϕ(t)Φ(t)𝑑t.

Using Lemma 3 and Lemma 4, we have

𝔼[πeFπe(πe)]=eμ{kEσkk!IkP+kOσkk!IkD}.

Acknowledgements

The authors would like to express their gratitude to their friend and colleague Robert Adam Sobolewski for suggesting the possibility to use data from MERRA-2 and for providing related wind speed measures.

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Received: 2020-11-05
Revised: 2020-12-07
Accepted: 2021-01-04
Published Online: 2021-01-12
Published in Print: 2021-06-01

© 2021 Walter de Gruyter GmbH, Berlin/Boston

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