Concluding remarks

of real-time control is small (<15%) over the whole range of 5%–25% of wind penetration levels. Of course, without knowledge of future deferrable loads, the suboptimality of real-time control becomes bigger. However, it still eventually outperforms the optimal static controller at around 6% wind penetration, despite the fact that the optimal static controller is using exact information about deferrable loads.

Appendix

3.A Proof of Lemma 3.5

Whenbt=b andE(a(t)) =λfort= 1, . . . , T, the model (3.17) for Algorithm 2 reduces to

p0(t) = 1 T −t+ 1



 XT τ=t

b(τ) +λ(T−t) +

n(t)X

i=1

Pi(t)



 (3.19)

fort= 1, . . . , T. Then

(T−t+ 1)p0(t) = XT τ=t

b(τ) +λ(T−t) +

n(t)X

i=1

Pi(t),

(T−t+ 2)p0(t−1) = XT τ=t−1

b(τ) +λ(T−t+ 1) +

n(t−1)X

i=1

Pi(t−1)

fort= 2, . . . , T. Subtract the two equations and simplify using the fact that

b(t−1) +

n(t−1)

X

i=1

(Pi(t−1)−Pi(t)) =b(t−1) +

n(t−1)

X

i=1

pi(t−1) =p0(t−1)

and the definition ofa(t) to obtain

p0(t)−p0(t−1) = 1

T−t+ 1(a(t)−λ)

fort= 2, . . . , T. Substituting t= 1 into (3.19), it can be verified that p0(1) =λ+PT

τ=1b(τ)/T+ (a(1)−λ)/T, therefore

p0(t) =λ+ 1 T

XT τ=1

b(τ) + Xt τ=1

1

T−τ+ 1(a(τ)−λ)

fort= 1, . . . , T. The average aggregate load is

u= 1 T

XT t=1

p0(t) =λ+ 1 T

XT τ=1

b(τ) + XT τ=1

(a(τ)−λ)

! .

Hence,

E(p0(t)−u)² = E Xt τ=1

1

T−τ+ 1(a(τ)−λ)− 1 T

XT τ=1

(a(τ)−λ)

!²

= E Xt τ=1

τ−1

T(T−τ+ 1)(a(τ)−λ)− 1 T

XT τ=t+1

(a(τ)−λ)

!²

= s² T²

Xt τ=1

(τ−1)²

(T−τ+ 1)² +T−t

!

for t= 1, . . . , T. The last equality holds because (a(τ)−λ) are independent for allτ and each of them have mean zero and variances². It follows that

E(V) = 1 T

XT t=1

E(d(t)−u)²

= s² T³

XT t=1

Xt τ=1

(τ−1)² (T−τ+ 1)² +

XT t=1

(T−t)

!

= s² T³

XT τ=1

(τ−1)² T−τ+ 1+

XT t=1

(T−t)

!

= s² T³

XT t=1

(T−t)²

t +

XT t=1

(T−t)t t

!

= s² PT

t=21 t

T ≈ s²lnT T .

3.B Proof of Lemma 3.6

In the case where no deferrable arrival after t= 1, i.e.,n(t) =nfort = 1, . . . , T, the model (3.17) for Algorithm 2 reduces to

(T−t+ 1)p0(t) = XT τ=t

bt(τ) + Xn i=1

Pi(t) (3.20)

fort= 1, . . . , T. Substitutetbyt−1 to obtain

(T−t+ 2)p0(t−1) = XT τ=t−1

bt−1(τ) + Xn i=1

Pi(t−1)

fort= 2, . . . , T. Subtract the two equations to obtain

(T−t+ 1)p0(t)−(T−t+ 2)p0(t−1) = XT τ=t

e(t)f(τ−t)−b(t−1)− Xn i=1

pi(t−1)

= e(t)F(T −t)−p0(t−1), which implies

p0(t)−p0(t−1) = 1

T−t+ 1e(t)F(T−t)

fort = 2, . . . , T. Substitutingt= 1 into (3.20) and recalling the definition ofbt in (3.1), it can be verified that

p0(1) = 1 T

Xn i=1

Pi+ XT τ=1

¯b(τ)

! + 1

Te(1)F(T−1).

Therefore,

p0(t) = 1 T

Xn i=1

Pi+ XT τ=1

¯b(τ)

! +

Xt τ=1

1

T−τ+ 1e(τ)F(T−τ) fort= 1, . . . , T. The average aggregate load is

u = 1

T Xn i=1

Pi+ XT t=1

¯b(t)

! + 1

T XT τ=1

e(τ)F(T−τ).

Hence,

E(p0(t)−u)² = E Xt τ=1

1

T−τ+ 1e(τ)F(T−τ)− XT τ=1

1

Te(τ)F(T−τ)

!²

= E Xt τ=1

τ−1

T(T−τ+ 1)e(τ)F(T−τ)− XT τ=t+1

1

Te(τ)F(T−τ)

!²

= σ² T²

Xt τ=1

(τ−1)²

(T−τ+ 1)²F²(T−τ) + XT τ=t+1

F²(T−τ)

!

fort= 1, . . . , T. The last equality holds becausee(τ) are uncorrelated random variables with mean zero and varianceσ². It follows that

E(V) = 1 T

XT t=1

E(p0(t)−u)²

= σ² T³

XT t=1

Xt τ=1

(τ−1)²

(T −τ+ 1)²F²(T−τ) + XT τ=t+1

F²(T−τ)

!

= σ² T³

XT τ=1

F²(T−τ) (τ−1)² T−τ+ 1+ σ²

T³ XT τ=2

(τ−1)F²(T−τ)

= σ² T²

XT τ=1

F²(T−τ) τ−1

T−τ+ 1 = σ² T²

TX−1 t=0

F²(t)T−t−1 t+ 1 .

3.C Proof of Theorem 3.3

Similar to the proof of Lemma 3.5 and 3.6, use the model (3.17) to obtain

p0(t) =λ+ 1 T

XT τ=1

¯b(τ) + Xt τ=1

1

T−τ+ 1(e(τ)F(T−τ) +a(τ)−λ) fort= 1, . . . , T and

u=λ+ 1 T

XT τ=1

¯b(τ) + XT τ=1

1

T (e(τ)F(T−τ) +a(τ)−λ). Hence,

E[p0(t)−u]² = E

" _t X

τ=1

1

T−τ+ 1(e(τ)F(T−τ) +a(τ)−λ)− XT τ=1

1

T (e(τ)F(T−τ) +a(τ)−λ)

#²

= E

" _t X

τ=1

1

T−τ+ 1e(τ)F(T−τ)− XT τ=1

1

Te(τ)F(T−τ)

#²

+E

" _t X

τ=1

1

T−τ+ 1(a(τ)−λ)− XT τ=1

1

T (a(τ)−λ)

#² .

The first term is exactly that in Lemma 3.6, and the second term is exactly that in Lemma 3.5.

Hence, the expected load variance is

E(V) = σ² T²

TX−1 t=0

F²(t)T−t−1 t+ 1 +s²

T XT t=2

1 t.

3.D Proof of Corollary 3.7

If |f(t)| ∼ O(t^−1/2−α) for some α > 0, then |f(t)| ≤ Ct^−1/2−α for some C > 0 and all t ≥ 1.

Without loss of generality, assume that 0< α <1/2 and C ≥(1−2α)/(1 + 2α). Then F(0) = 1 and

|F(t)|=

Xt τ=0

f(τ) ≤1 +

Xt τ=1

Cτ^−1/2−α ≤ 1 +C+ Z t

1

Cτ^−1/2−αdτ ≤ 2C

1−2αt^1/2−α fort= 1, . . . , T. The last inequality holds becauseC≥(1−2α)/(1 + 2α). Therefore it follows from Lemma 3.6 that

E(V) ≤ σ² T

TX−1 s=0

F²(s) 1 s+ 1

≤ σ² T +σ²

T

TX−1 s=1

4C²

(1−2α)²s^1−2α 1 s+ 1

≤ σ² T +σ²

T

4C² (1−2α)²

TX−1 s=1

1 s^2α

≤ σ² T +σ²

T

4C²

(1−2α)²+σ² T

4C² (1−2α)²

Z T−1 1

1 s^2αds

≤ σ²

T + 4σ²C²

(1−2α)²T + 4σ²C² (1−2α)³T^2α. Hence,E(V)→0 asT → ∞.

3.E Proof of Lemma 3.8

The aggregate loaddobtained by the optimal static algorithm is

p0(t) = 1 T

Xn i=1

Pi+ XT τ=1

¯b(τ)

!

−¯b(t) +b(t)

= 1 T

Xn i=1

Pi+ XT τ=1

¯b(τ)

! +

XT τ=1

e(τ)f(t−τ)

fort= 1, . . . , T. Hence,

E(p0(t)−u)² = E XT τ=1

e(τ)

f(t−τ)− 1

TF(T−τ)

!²

= σ² T²

XT τ=1

T²f²(t−τ)−2T f(t−τ)F(T−τ) +F²(T−τ)

fort= 1, . . . , T. It follows that

E(V⁰) = 1 T

XT t=1

E(p0(t)−u)²

= σ² T

XT t=1

XT τ=1

f²(t−τ)−2σ² T²

XT τ=1

F(T −τ) XT t=1

f(t−τ) + σ² T²

XT τ=1

F²(T−τ)

= σ² T

XT t=1

Xt−1 τ=0

f²(τ)−σ² T²

XT τ=1

F²(T−τ)

= σ² T

TX−1 τ=0

(T−τ)f²(τ)− σ² T²

TX−1 τ=0

F²(τ)

= σ² T²

TX−1 t=0

T(T−t)f²(t)−F²(t) .

3.F Proof of Corollary 3.9

Corollary 3.9 follows from Lemma 3.6 and Lemma 3.8 and the definition ofF:

E(V⁰)−E(V) = σ² T

TX−1 t=0

(T−t)f²(t)−σ² T²

TX−1 t=0

F²(t)

1 + T−t−1 t+ 1

= σ² T

XT n=0

(T−n)f²(n)−σ² T

TX−1 t=0

1 t+ 1F²(t)

= σ² T

XT n=0

XT t=n+1

f²(n)−σ² T

TX−1 t=0

1 t+ 1F²(t)

= σ² T

XT t=1

Xt−1 n=0

f²(n)−σ² T

XT t=1

1 t

Xt−1 n=0

f(n)

!²

= σ² T

XT t=1

1 t



t

t−1

X

n=0

f²(n)−

t−1

X

n=0

f(n)

!²



= σ² T

XT t=1

1 2t

t−1X

m=0

Xt−1 n=0

(f(m)−f(n))².

3.G Proof of Corollary 3.10

We have

F(t) =







t+ 1 if 0≤t <∆

∆ if t≥∆

fort= 0, . . . , T. It follows that

E(V) = σ² T²

TX−1 t=0

F²(t)T−t−1 t+ 1

= σ² T²

∆−1X

t=0

(t+ 1)²T−t−1

t+ 1 +σ²∆² T²

T−1X

t=∆

T−t−1 t+ 1

= σ² T²

X∆ t=1

t(T −t) +σ²∆² T²

XT t=∆+1

T t −1

∈ σ²∆² T²

Tln T+ 1

∆ + 1−T+ ∆, TlnT

∆ −T+ ∆ + 1

= ∆σ²ln(T /∆) T /∆

1 +O

1 ln(T /∆)

and

E(V⁰) = σ² T

TX−1 t=0

(T−t)f²(t)− σ² T²

TX−1 t=0

F²(t)

= σ² T

∆−1X

t=0

(T−t)−σ² T²

∆−1X

t=0

(t+ 1)²−σ²∆²

T² (T−∆)

= ∆σ²(1 +O(∆/T)).

Hence, the expected load variance reduction is E(V⁰)

E(V) = T /∆

ln (T /∆)

1 +O 1

ln(T /∆)

.

3.H Proof of Corollary 3.11

We haveF(t) = (1−a^t+1)/(1−a) fort= 0, . . . , T. Note that for any ∆∈ {1, . . . , T},

E(V) = σ² (1−a)²T²

XT t=1

1−a^t2T−t

t = σ²

(1−a)²T²(A+B) where

A:=

X∆ t=1

T−t

t 1−a^t2

, B:=

XT t=∆+1

T−t

t 1−a^t2

by splitting the sum at ∆. We can bound the termsAandB separately, on the one hand,

0≤A≤T X∆ t=1

1

t ≤T(1 + ln ∆).

On the other hand,

B ≥ 1−a^∆+12 X^T

t=∆+1

T−t t

≥ 1−a^∆+12

TlnT + 1

∆ + 1−T+ ∆

= 1−a^∆+12 TlnT

1 +O

ln ∆ lnT

and

B ≤

XT t=∆+1

T−t t

≤ Tln T

∆ −T+ ∆

= TlnT

1 +O ln ∆

lnT

.

Select ∆ = [lnT] and letT → ∞to obtain

TlnT

1 +O

ln lnT lnT

≤A+B ≤TlnT

1 +O

ln lnT lnT

.

ThereforeA+B=TlnT(1 +O(ln lnT /lnT)) and it follows that

E(V) = σ² (1−a)²

lnT T

1 +O

ln lnT lnT

.

We also have

E(V⁰) = σ² T

TX−1 t=0

(T−t)a^2t− σ² (1−a)²T²

XT t=1

1−a^t2

= σ²

1−a²

1−a²(1−a^2T) T(1−a²)

− σ² (1−a)²T

1−a²+ 2a−2a^T+1−2a^T+2+a^2+2T T(1−a²)

= σ²

1−a²

1 +O 1

T

.

Hence, the expected load variance reduction is E(V⁰)

E(V) = 1−a 1 +a

T lnT

1 +O

ln lnT lnT

.

Chapter 4

Optimal Power Flow

Distribution networks are usually multiphase and radial. To facilitate power flow computation and optimization, two semidefinite programming (SDP) relaxations of the optimal power flow problem and a linear approximation of the power flow are proposed. We prove that the first SDP relaxation is exact if and only if the second one is exact. Case studies show that the second SDP relaxation is numerically exact and that the linear approximation obtains voltages within 0.0016 per unit of their true values for the IEEE 13, 34, 37, 123-bus networks and a real-world 2065-bus network.

Literature The optimal power flow (OPF) problem is nonconvex, and approximations and relax- ations have been developed to solve it; see recent surveys in [20, 25, 39, 54, 81, 84]. For convex relax- ations, it is first proposed in [57] to solve OPF as a second-order cone programming for single-phase radial networks and in [10] as a semidefinite programming (SDP) for single-phase mesh networks.

While numerically illustrated in [57] and [10], whether or when the convex relaxations are exact is not studied until [66]; see [73, 74] for a survey and references to a growing literature on convex relaxations of OPF.

Most of these works assume a single-phase network, while distribution networks are typically multiphase and unbalanced [63]. It has been observed in [29, 60] that a multiphase network has an equivalent single-phase circuit model where each bus-phase pair in the multiphase network is identified with a single bus in the equivalent model. Hence methods for single-phase networks can be applied to the equivalent model of a multiphase unbalanced network. This approach is taken in [33] for solving optimal power flow problems. Additionally, [33] develops distributed solutions.

Summary This chapter develops convex relaxations of OPF and a linear approximation of power flow. Solving OPF through convex relaxation offers several advantages. It provides the ability to check if a solution is globally optimal. If it is not, the solution provides a lower bound on the minimum cost and hence a bound on how far any feasible solution is from optimality. Unlike approximations, if a relaxation is infeasible, it certifies that the original OPF is infeasible.

There are three questions on convex relaxations: 1) how to compute convex relaxations efficiently, 2) how to attain numerical stability, and 3) when can a globally optimum of OPF be obtained by solving its convex relaxation?

To address 1), the relaxation BIM-SDP is proposed in Section 4.2 to improve the computational efficiency of a standard SDP relaxation by exploiting the radial network topology. While the standard SDP relaxation declaresO(n²) variables wheren+1 is the number of buses in the network, BIM-SDP only declaresO(n) variables and is therefore more efficient.

To address 2), the relaxation BFM-SDP is proposed in Section 4.3 to improve the numerical stability of BIM-SDP by avoiding ill-conditioned operations. BIM-SDP is ill-conditioned due to subtractions of voltages that are close in value. Using alternative variables, BFM-SDP avoids these subtractions and is therefore numerically more stable.

To partially address 3), we prove in Section 4.5 that BIM-SDP is exact if and only if BFM-SDP is exact, and empirically show in Section 4.5 that BFM-SDP is numerically exact for the IEEE 13, 34, 37, 123-bus networks and a real-world 2065-bus network. Remarkably, BIM-SDP is numerically exact only for the IEEE 13 and 37-bus networks. This highlights the numerical stability of BFM- SDP.

Approximation LPF is proposed in Section 4.4 to estimate voltages and power flows. LPF is accurate when line loss is small compared with power flow and voltages are nearly balanced, i.e., the voltages of different phases have similar magnitudes and differ in angle by∼120^◦. Empirically, it is presented in Section 4.5 that LPF computes voltages within 0.0016 per unit of their true values for the IEEE 13, 34, 37, and 123-bus networks and a real-world 2065-bus network.

4.1 Optimal Power Flow Problem

OPF in multiphase radial networks is applicable for demand response and volt/var control.

4.1.1 A Standard Nonlinear Power Flow Model

A distribution network is composed of buses and lines connecting these buses. It is usually multiphase and radial. There is a substation bus in the network with a fixed voltage. Index the substation bus by 0 and the other buses by 1,2, . . . , n. Let N ={0,1, . . . , n} denote the set of buses and define N⁺ =N \{0}. Each line connects an ordered pair (i, j) of buses where busilies between bus 0 and bus j. Let E denote the set of lines. Use (i, j)∈ E andi →j interchangeably. If i→j or j →i, denotei∼j.

Let a, b, c denote the three phases of the network, let Φi denote the phases of bus i ∈ N, and let Φij denote the phases of line i ∼ j. For each busi ∈ N, let V_i^φ denote its phase φ complex voltage forφ∈Φi and define Vi := [V_i^φ]_φ∈Φi; letI_i^φ denote its phaseφcurrent injection for φ∈Φi

and defineIi := [I_i^φ]φ∈Φi; let s^φ_i denote its phaseφcomplex power injection forφ∈Φi and define si:= [s^φ_i]φ∈Φi. For each linei∼j, letI_ij^φ denote the phaseφcurrent from busito busjforφ∈Φij

and defineIij:= [I_ij^φ]φ∈Φij; letzij denote the phase impedance matrix and defineyij :=z⁻_ij¹.

bus 0 bus i bus j

V0 Vi yij=z_ij¹ Vj

Iij I_i si

Figure 4.1: Summary of notations

Some notations are summarized in Figure 4.1. Further, let superscripts denote projection to specified phases, e.g., if Φi=abc, then

V_i^ab= (V_i^a, V_i^b)^T. Fill nonexisting phase entries by 0, e.g., if Φi =ab, then

V_i^abc= (V_i^a, V_i^b,0)^T.

Let a letter without subscripts denote a vector of the corresponding quantity, e.g.,z= [zij]i∼j and s= [si]i∈N.

Power flows are governed by [63]:

1) Ohm’s law: Iij =yij(V_i^Φij−V_j^Φij) fori∼j.

2) Current balance: Ii=P

j:i∼jI_ij^Φi fori∈ N. 3) Power balance: si= diag(ViI_i^H) fori∈ N.

Eliminate current variablesIi andIij to obtain the followingbus injection model (BIM):

si= X

j:i∼j

diagh

V_i^Φij(V_i^Φij −V_j^Φij)^Hy^H_ijiΦi

, i∈ N. (4.1)

4.1.2 Optimal Power Flow

OPF determines the power injection that minimizes generation cost subject to physical and oper- ational constraints. Generation cost is separable. In particular, let Ci(si) : C^|Φi|7→ Rdenote the generation cost at busi∈ N, and

C(s) =X

i∈N

Ci(si) is the generation cost of the network.

OPF has operational constraints on power injections and voltages besides physical constraints (4.1). First, while the substation power injections0 is unconstrained, a branch bus power injection si can only vary within some externally specified setSⁱ, i.e.,

si∈ Sⁱ, i∈ N⁺. (4.2)

For example, the setsSⁱof two types of devices are illustrated in Figure 4.2. Note thatSⁱis usually not a box, and thatSⁱ can be nonconvex or even disconnected.

0 Si

Re(si) Im(si)

0 Re(si)

Im(si)

A

B Si={A,B}

Figure 4.2: The left figure illustrates the setSⁱ of an inverter, and the right figure illustrates the set Sⁱ of a shunt capacitor. Note that the setSⁱ is usually not a box, and thatSⁱ can be nonconvex or even disconnected.

Second, while the substation voltageV0is fixed and given (denote byV₀^refthat is nonzero compo- nentwise), a branch bus voltage can be regulated within a range, i.e., there exists [V^φ_i, V^φ_i]i∈N⁺,φ∈Φi

such that

V0=V₀^ref; (4.3a)

V^φ_i ≤ |V_i^φ| ≤V^φ_i, i∈ N⁺, φ∈Φi. (4.3b) For example, if voltages must stay within 5% from their nominal values, then 0.95 ≤ |V_i^φ| ≤1.05 per unit.

To summarize, OPF can be formulated as

OPF: min X

i∈N

Ci(si) over s, V

s.t. (4.1)−(4.3).

The following assumptions are made throughout this paper.

1. The network (N,E) is connected.

2. Voltage lower bounds are strictly positive, i.e.,

V^φ_i >0, i∈ N⁺, φ∈Φi.

3. Bus and line phases satisfy

Φi⊇Φij = Φj, i→j.