Relationship Between Power Loss and Network Topology in Power Systems

Javad Lavaei and Steven H. Low

Abstract— This paper is concerned with studying how theminimum power loss in a power system is related to its networktopology. The existing algorithms in the literature all exploitnonlinear, heuristic, or local search algorithms to find theminimum power loss, which make them blind to the networktopology. Given certain constraints on power level, bus voltages,etc., a linear-matrix-inequality (LMI) optimization problemis derived, which provides a lower bound on the minimumactive loss in the network. The proposed LMI problem has theproperty that its objective function depends on the loads andits matrix inequality constraint is related to the topology ofthe power system. This property makes it possible to addressmany important power problems, such as the optimal networkreconfiguration and the optimal placement/sizing of distributedgeneration units in power systems. Moreover, a condition isprovided under which the solution of the given LMI problemis guaranteed to be exactly equal to the minimum power loss.As justified mathematically and verified on IEEE test systems,this condition is expected to hold widely in practice, implyingthat a practical power loss minimization problem is likely tobe solvable using a convex algorithm.

I. INTRODUCTION

The optimal power flow (OPF) problem deals with findingan optimal operating point of a power system that satisfiescertain constraints on power and voltage parameters and, inaddition, minimizes an appropriate cost function, such asgeneration cost or transmission loss [1]. The OPF problemhas been extensively studied for several decades and numer-ous gradient and nonlinear algorithms have been proposedto find a near-optimal solution of this highly nonconvexproblem [2], [3], [4], [5]. Since the power losses occurring inthe transmission lines of a low-voltage distribution networkare significant in comparison to the total amount of the powerto be delivered, the power loss minimization problem, asa particular type of the OPF problem, has attracted muchattention in the literature since 1962.

Power loss is minimized in a distribution network bymeans of various techniques, some of which are: (i) ex-ploitation of capacitor banks, (ii) network reconfiguration and(iii) installation of distributed generation units. Method (i)relies on the fact that capacitor banks compensate for thereactive powers at the load buses, which could accordinglyreduce the active power loss. The paper [6] employs agenetic algorithm to find the optimal places of the capacitors.In contrast, Method (ii) only modifies the configuration ofthe radial distribution network to diminish the power loss.Finding an optimal configuration leads to an OPF problem

Javad Lavaei is with the Department of Control and Dynamical Systems,California Institute of Technology (email: lavaei@cds.caltech.edu).

Steven H. Low is with Computer Science and Electrical EngineeringDepartments, California Institute of Technology (email: slow@caltech.edu).

mixed with a graph combinatorial problem, which turnsout to be extremely hard to solve. The papers [7], [8], [9]propose heuristic, meta-heuristic and fuzzy algorithms to findan optimal configuration. Motivated by the availability ofrenewable resources such as hydro, wind, solar, biomass andocean energy, distributed generation (DG) units, e.g. micro-turbine, fuel cell, mini-hydro, battery storage, are becomingan important part of power systems. Method (iii) minimizesthe power loss by integrating DGs in a distribution network.The problem of finding the optimal locations (placement)and values (sizing) of the DGs have been recently studied inmany papers [10], [11], [12], [13].

To address the loss minimization problem using eachof Methods (i), (ii) or (iii), it is crucial to understandthe relationship between the power loss and the networktopology. To this end, the present paper proposes a linear-matrix-inequality (LMI) optimization problem (a special typeof convex optimization) that yields a nonnegative lowerbound on the power loss [14]. This LMI problem has theremarkable property that its feasibility region only dependson the network topology, and the load profiles (as well asvoltage magnitudes) just appear in the objective function.This decomposition property of the proposed LMI problemopens up the possibility of designing convex, numericallyefficient algorithms to achieve an optimal placement ofcapacitor backs, optimal network reconfiguration or evenoptimal placement/sizing of DGs. Interestingly, the LMIproblem given here attains the exact value of the loss (asopposed to a lower bound on it) for all IEEE test systemswith 14, 30, 57, 118 and 300 buses. We connect thisobservation to the recent results on rank minimization andjustify it rigorously [16], [17]. In fact, we show that a powersystem has certain physical properties, which can make ageneral OPF problem efficiently solvable.

The paper is organized as follows. The problem is for-mulated in Section II. The results are developed for a radialnetwork in Sections III and IV, and then extended to a generalnetwork in Section V. Simulation results are provided inSection VI. Finally, some concluding remarks are drawn inSection VII.

Notations: We introduce the following notations:

• i : The imaginary unit.• R: The set of real numbers.• Re{·} and Im{·}: The operators returning the real and

imaginary parts of a complex matrix.• ∗ : The conjugate transpose operator.• T : The transpose operator.• ≽ : The matrix inequality sign in the positive semidef-

49th IEEE Conference on Decision and ControlDecember 15-17, 2010Hilton Atlanta Hotel, Atlanta, GA, USA

978-1-4244-7746-3/10/$26.00 ©2010 IEEE 4004

inite sense [14].• ∥·∥∗: The nuclear-norm operator (taking the sum of the

singular values of a matrix).

II. PROBLEM FORMULATION

Consider a power system with m load buses and n −mgenerator buses. With no loss of generality, assume that thepower network consists of only transmission lines (and notransformers), where

• The resistance of every transmission line is strictlypositive.

• The power network is a strongly connected graph.Figure 1 illustrates a transmission system with three gener-ator buses and three load buses, in which

• Each generator is represented by a circle.• Two types of loads are connected to each of buses

1, 2 and 3: (i) constant-impedance loads shown byshunt admittances (rectangles), (ii) constant-power loadsshown by triangles.

For every k ∈ {1, 2, ..., n}, let yk denote the admittanceof the constant-impedance load connected to bus k and Vk

represent the complex voltage of this bus. Each transmissionline in the network can be described by its equivalent Πmodel. To be more precise, a transmission line connectingarbitrary buses i, j ∈ {1, 2, ..., n} (shown in Figure 2(a)) canbe replaced by its Π model given in Figure 2(b), where zijand yij denote the series impedance and shunt admittance,respectively. Define the admittance matrix Y of the networkas an n×n complex-valued matrix whose (i, j) entry, i, j ∈{1, 2, ..., n}, is given as:

(i, j) entry of Y =

{− 1

zijif i = j

yi +∑

j∈N (i)

(yij

2 + 1zij

)if i = j

where N (i) is the set of those buses that are connectedto bus i. Note that the matrix Y captures the topology ofthe power network and plays the role of the Laplacian ofthe weighted graph associated with the network. Stack thevoltages V1, V2, ..., Vn in a column vector, denoted by V .Define the current vector I as Y V and denote its ith elementwith Ii for every i ∈ {1, 2, ..., n}. It is easy to show that thecomplex power injected at a generator bus j ∈ {m+1, ..., n}by its generator is equal to VjI

∗j and that the complex power

consumed at a load bus k ∈ {1, 2, ...,m} by its constant-power load is equal to −VkI

∗k .

Given some desired constant-power loads at the loadbuses, there may exist several ways to provision the gen-erators’ powers to supply such loads. Hence, it is importantto find a set of generators’ powers which not only minimizesthe active/reactive power loss in the network, but also meetsother practical constraints on voltages, powers, etc. This aimhas been long studied in the context of optimal power flowproblem, which is notorious for its high non-convexity. Notethat active and reactive power losses occur in a power systemdue to the fact that transmission lines in the network haveresistive, inductive and capacitive components. The objectiveof this paper is to study how the power loss is related

Fig. 1. An example of a power network.

Bus i Bus j

Transmission Line

(a)

zij

2

ijy

Bus jBus i 2

ijy

(b)

Fig. 2. (a): A transmission line connecting bus i to bus j; (b): EquivalentΠ model of the transmission line connecting bus i to bus j.

to the network topology. Indeed, finding a convex formularelating the power loss to the network configuration hasdirect applications in two important problems that have beenstudied in the literature using heuristic algorithms:

• Optimal reconfiguration in distribution networks: Thelast stage of power delivery is often accomplished by aradial distributed network, such as the one depicted inFigure 3(c), in which all load buses are connected to afeeder (slack bus 1 in the figure) through a tree-shapednetwork. This network is reconfigurable to some extentin practice; for example, bus 4 in Figure 3(c) can bedisconnected from bus 1 and get connected to bus 2.The goal is to find a configuration whose associatedpower loss is minimum.

• Optimal placement/sizing of the distributed generationin distribution networks: This problem is concernedwith finding the optimal location and power values ofdistributed generators, mainly based on renewable re-sources, whose addition to a radial distribution networkminimizes the loss.

III. POWER LOSS AND NETWORK TOPOLOGY

To simplify the presentation, assume for now that• Bus n is the only generator bus, with a prescribed

magnitude of voltage equal to V0.• The load bus k is required to deliver the pre-specified

power Pk +Qki, for every k ∈ {1, 2, ..., n− 1}.Define Ploss as the minimum of the active power loss thatmust be incurred in the network so that the above require-ments are met. In other words, Ploss is equal to the minimum

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of the total loss

Re{VnI∗n} −

n−1∑k=1

Pk, (1)

subject to the following power, voltage and network con-straints:

VkI∗k = −Pk −Qki, k = 1, 2, ..., n− 1, (2a)

|Vn| = V0, (2b)I = Y V. (2c)

The objective is to study how Ploss is related to the topologyof the network. The results being developed here will begeneralized in Section V to the multi-generator case withmore voltage and power constraints.

Let e1, e2, ..., en denote the standard basis vectors in Rn.For every k = 1, 2, ..., n, define

Yk := eke∗kY,

Yk :=1

2

[Re{Yk + Y T

k } Im{Y Tk − Yk}

Im{Yk − Y Tk } Re{Yk + Y T

k }

],

Yk :=−1

2

[Im{Yk + Y T

k } Re{Yk − Y Tk }

Re{Y Tk − Yk} Im{Yk + Y T

k }

],

Y :=

[Re{Y } 0

0 Re{Y }

],

Define also M ∈ R2n×2n as a diagonal matrix whose entriesare all equal to zero, except for its (n, n) and (2n, 2n) entriesthat are equal to 1.

Optimization 1: Given the variables µ ∈ R and

λ :=[λ1 · · · λn−1

]T ∈ Rn−1,

λ :=[λ1 · · · λn−1

]T ∈ Rn−1,

maximize the linear function

f(λ, λ, µ) :=n−1∑k=1

(λk − 1)Pk +n−1∑k=1

λkQk − µV 20 , (3)

subject to the linear matrix inequality

Φ(λ, λ, µ) : = Yn +n−1∑k=1

λkYk +n−1∑k=1

λkYk + µM ≽ 0.

Denote the optimal value of the function (3) with Pmin.Theorem 1: The quantity Pmin is a nonnegative lower

bound on the minimum loss Ploss.Proof: Recall that Ploss can be obtained using the optimiza-

tion problem specified in (1) and (2). To solve this problem,define the Lagrangian

L(λ, λ, µ, V ) : =

n−1∑k=1

λk

(Re{VkI

∗k}+ Pk

),

+n−1∑k=1

λk

(Im{VkI

∗k}+Qk

)+ µ

(VnV

∗n − V 2

0

)+ Re{VnI

∗n} −

n−1∑k=1

Pk,

where I = Y V . The inequality

maxλ,λ,µ

minV

L(λ, λ, µ, V ) ≤ Ploss

holds by the weak duality theorem. In order to prove thetheorem, it suffices to show that the left side of the aboveinequality is nonnegative and equal to Pmin. To this end,notice that

Re{VkI∗k} = Re{V ∗eke

∗kI} = Re{V ∗YkV }

= WT

[Re{Yk} −Im{Yk}Im{Yk} Re{Yk}

]W

=1

2WT

[Re{Yk + Y T

k } Im{Y Tk − Yk}

Im{Yk − Y Tk } Re{Yk + Y T

k }

]W

= WTYkW,

for every k ∈ {1, 2, ..., n}, where

W :=[

Re{V }T Im{V }T]T

.

Similarly,Im{VkI

∗k} = WT YkW.

Thus, one can write

L(λ, λ, µ, V ) = f(λ, λ, µ) +WTΦ(λ, λ, µ)W.

By fixing λ, λ, µ, observe that the minimum of the quadraticfunction WTΦ(λ, λ, µ)W with respect to the complex-valued variable V (or the real-valued variable W ), is either0 or −∞, depending on whether the symmetric matrixΦ(λ, λ, µ) is positive semidefinite. This implies that theunconstrained optimization maxλ,λ,µ minV L(λ, λ, µ, V ) istantamount to the maximization of f(λ, λ, µ) subject to theconstraint Φ(λ, λ, µ) ≽ 0, which is identical to Optimiza-tion 1. Hence, maxλ,λ,µ minV L(λ, λ, µ, V ) = Pmin. Now,it remains to show that Pmin ≥ 0. For this purpose, define0n−1 and 1n−1 as the vector of zeros and the vector of onesin Rn−1, respectively. The proof is completed by noting that(1n−1,0n−1, 0) is a feasible point of Optimization 1, wheref(1n−1,0n−1, 0) = 0 (note that the real part of Y is positivesemidefinite). �

Theorem 1 states that solving Optimization 1 leads to asensible (nonnegative) lower bound on the active power loss.Two important properties of this optimization problem are asfollows:

• Optimization 1 is a semidefinite program and, there-fore, its globally optimal solution can be found effi-ciently [14].

• In the case when the optimal value Pmin is equal to +∞,the associated power flow problem must be infeasible(because a lower bound on the power loss is obtained tobe infinity). As a result, Optimization 1 is a sanity testfor checking the feasibility of a power flow problem.

Remark 1: The objective function f(λ, λ, µ) in Optimiza-tion 1 depends on the load demands as well as the voltageset point. In contrast, the constraint Φ(λ, λ, µ) ≽ 0 inthis optimization problem is contingent upon the physicaltopology of the distribution network (reflected by Y ). Hence,

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Optimization 1 decomposes the network topology from theload profile. This significant property opens up the possibilityof systematically addressing many important problems, suchas (near) optimal network reconfiguration design.

Remark 2: Since Theorem 1 is based on applying the La-grangian theory to a nonlinear optimization problem, strongduality may not hold. As a result, Pmin might not be equalto Ploss. In the case when there exists no duality gap, theoptimal bus voltages can be recovered from Optimization 1.More precisely, the vector

[Re{V opt}T Im{V opt}T

]Tis an eigenvector of the singular matrix Φ(λopt, λ

opt, µopt)

associated with its zero eigenvalue, where (λopt, λopt, µopt)

is a maximizer of Optimization 1.Remark 3: Recall that

∑n−1k=1(λ

optk − 1)Pk +∑n−1

k=1 λoptk Qk − µoptV 2

0 is either equal to or possiblya lower bound on Ploss. Now, assume that the goal isto identify a load bus incurring the most power loss sothat a small generator is installed to compensate for itsactive power consumption. The above lower bound suggestschoosing a bus k whose Lagrange multiplier λopt

k has thehighest value (because any deduction in Pk is amplified bythe factor λopt

k −1). In other words, the Lagrange multipliersλopt1 , ..., λopt

n−1 roughly determine the contributions ofdifferent buses on the power loss. The validity of thisinterpretation will be later verified in simulations. This ideacan be generalized to tackle the problem of sizing/placementof distributed generation.

Although Pmin is only a lower bound on Ploss, we haveverified that the equality Ploss = Pmin holds for all IEEEtest systems with 14, 30, 57, 118 and 300 buses as well asnumerous other power systems. This observation implies thatpower systems have some hidden structures, which couldbridge the duality gap. In what follows, we first derive acondition under which Ploss = Pmin holds and then explainwhy this condition is likely to be satisfied in practice.

IV. DUALITY GAP

The proof of Theorem 1 yields that Ploss can be obtainedby minimizing

WTYnW −n−1∑k=1

Pk (4)

over the real-valued variable W ∈ R2n subject to

WTYkW = −Pk, k = 1, 2, ..., n− 1, (5a)

WT YkW = −Qk, k = 1, 2, ..., n− 1, (5b)

WTMW = V 20 . (5c)

If X is defined as WWT , the above optimization will betantamount to the minimization of

trace{YnX} −n−1∑k=1

Pk (6)

subject to

trace{YkX} = −Pk, k = 1, 2, ..., n− 1, (7a)trace{YkX} = −Qk, k = 1, 2, ..., n− 1, (7b)

trace{MX} = V 20 , (7c)

X ≽ 0, (7d)rank(X) = 1. (7e)

The non-convexity of this optimization problem arises fromthe constraint rank(X) = 1.

Lemma 1: The dual of the nonconvex problem of thepower loss minimization, i.e. Optimization 1, is the sameas the dual of the convex optimization derived from (6) and(7) by removing the rank constraint. Moreover, X in thelatter optimization plays the role of the Lagrange multiplierfor the matrix constraint of Optimization 1.

Proof: The proof is omitted due to space restrictions andmay be found in [15]. �

For simplicity in the proof and with no loss of generality,assume that Ploss is finite and that the feasibility region ofOptimization 1 has a non-empty interior. These assumptionsare made to ensure that strong duality holds between Opti-mization 1 and its dual.

Theorem 2: Assume that Φ(λopt, λopt, µopt) has rank

greater than or equal to 2n − 2. The lower bound Pmin

obtained from Optimization 1 is the same as the minimumpower loss Ploss.

Sketch of Proof: Consider the problem of minimizing

trace{YX} −n−1∑k=1

Pk (8)

subject to

trace{YkX} = −Pk, k = 1, 2, ..., n− 1, (9a)trace{YkX} = −Qk, k = 1, 2, ..., n− 1, (9b)

trace{MX} = V 20 , (9c)

X ≽ 0. (9d)

Following the discussion made before Lemma 1, it sufficesto show that the above optimization problem has a rank-onesolution X . To this end, let Xopt be an arbitrary minimizer ofthis problem. It follows from the complementary slacknessconditions and Lemma 1 that

Φ(λopt, λopt, µopt) ≽ 0, Xopt ≽ 0,

trace{Φ(λopt, λ

opt, µopt)Xopt

}= 0.

(10)

Since Φ(λopt, λopt, µopt) has rank at least 2n − 2, it can be

concluded that Xopt has rank at most 2. If the rank of Xopt

is 1, the proof is complete. Thus, assume that Xopt has rank 2.Let

[UT1 UT

2

]Tbe a nonzero vector in the null space of

Φ(λopt, λopt, µopt), for some real vectors U1, U2 ∈ Rn. It is

easy to verify that[UT2 −UT

1

]Tis also in the null space

of Φ(λopt, λopt, µopt). The relations given in (10) yield that

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there must be two positive scalars α and β such that

Xopt = α

[U1

U2

] [UT1 UT

2

]+ β

[U2

−U1

] [UT2 −UT

1

].

Now, it can be shown that the rank-one matrix

(α+ β)

[U1

U2

] [UT1 UT

2

]is also a minimizer of the optimization problem specified by(8) and (9). This completes the proof. �

The relation rank{Φ(1n−1,0n−1, 0)} ≥ 2n − 2 holdsdue to the power network being strongly connected. Thismeans that the rank of Φ(λ, λ, µ) is generically at least2n − 2. Therefore, Theorem 2 requires that the conditionrank{Φ(λ, λ, µ)} ≥ 2n−2 that holds generically be satisfiedat the optimal point.

A. Zero Duality Gap for Resistive Networks

The active power loss Ploss is a consequence of theresistive part of the network. In other words, the minimumpower loss is zero for a purely inductive/capacitive powernetwork, provided the power flow problem is feasible. In thissubsection, we investigate an abstract, nonetheless important,case to gain insight into the zero-duality-gap conditiongiven in Theorem 2. The general case will be discussed inSection V.

Throughout this subsection, assume that the power systemis resistive (i.e. Im{Y } = 0) and that there is no impedance-to-ground (i.e. y1 = · · · = yn = 0). A practical powernetwork is normally associated with the well-known propertythat its required minimum loss reduces if its load demanddecreases (due to a reduction in the line currents). In otherwords, it is likely that replacing the constraints given in (9a)with

trace{YkX} ≤ −Pk, k = 1, 2, ..., n− 1 (11a)

again leads to the same solution Pmin. Under this cir-cumstance, the optimal Lagrange multipliers λopt

1 , ..., λoptn−1

cannot be negative. This property will be rigorously provedin the sequel for a resistive network.

Lemma 2: The Lagrange multipliers λopt1 , ..., λopt

n−1 are allnonnegative.

Proof: Since the network is resistive, the term Im{Y } andits associated multiplier λ can be removed from Optimiza-tion 1. Hence, one can write:

Φ(λopt, λopt, µopt) = Yn +

n−1∑k=1

λoptk Yk + µoptM

= Dopt(Y

2+

µoptM

2

)+

(Y

2+

µoptM

2

)Dopt,

(12)

where Dopt ∈ R2n×2n is a diagonal matrix with the diagonalentries λopt

1 , ..., λoptn−1, 1, λ

opt1 , ..., λopt

n−1, 1. Define 1n as thevector of ones in Rn. Since there is no impedance-to-groundin the network, the relation Y 1n = 0 holds. Now, it follows

from the positive semi-definiteness of Φ(λopt, λopt, µopt) and

(12) that

0 ≤[1Tn 1T

n

]Φ(λopt, λ

opt, µopt)

[1Tn 1T

n

]T= 2µopt,

which implies that µopt is nonnegative. As the first case,assume that µopt > 0, which makes the matrix Y

2 + µoptM2

positive definite. Using this property, it can be shown that

Dopt =

∫ ∞

0

e−(

Y2 +µoptM

2

)tΦe

−(

Y2 +µoptM

2

)tdt, (13)

where Φ stands for Φ(λopt, λopt, µopt). The above relation

simply shows that Dopt is nonnegative and so are its entriesλopt1 , ..., λopt

n−1. As the second case, let µopt be zero. The onlyextra part in the proof for this case is the necessity to showthat the right side of (13) does not become infinity. Thisfollows from the fact that Y has two orthogonal eigenvectors[1Tn 1T

n

]Tand

[1Tn −1T

n

]Tcorresponding to its zero

eigenvalue, which are both in the null space of Φ. �Lemma 2 states that the Lagrange multipliers

λopt1 , ..., λopt

n−1 are all nonnegative. The next theoremshows that the duality gap is zero under a slightly strongercondition of the strict positivity of these multipliers.

Theorem 3: Assume that the Lagrange multipliersλopt1 , ..., λopt

n−1 are strictly positive. Then, the zero-duality-gap condition given in Theorem 2 holds and thereforePloss = Pmin.

Proof: The key to the proof is the following two importantproperties:

• The matrix Y is irreducible (due to the power networkbeing strongly connected).

• The off-diagonal entries of Y are all non-positive (dueto the definition of Y and the positivity of everytransmission line’s resistance).

On applying the above properties to (12) and using the posi-tivity of λopt

1 , ..., λoptn−1, one can deduce that Φ(λopt, λ

opt, µopt)

is expressible as

Φ(λopt, λopt, µopt) =

[T 00 T

],

where T is an n×n irreducible matrix with non-positive off-diagonal entries. Now, it follows from the Perron-Frobeniustheorem that the multiplicity of the smallest eigenvalue ofΦ(λopt, λ

opt, µopt) is at most 2. Since Φ(λopt, λ

opt, µopt) is

positive semidefinite, this result implies that the rank ofΦ(λopt, λ

opt, µopt) is at least 2n − 2. This completes the

proof. �As can be inferred from Theorem 3, the main reason

behind the convexification of the nonconvex problem of lossminimization is the physical property of the network, i.e. thepositivity of resistance, which has made the matrix Y havea nice “sign” structure.

The above result can be extended to a general powernetwork. Indeed, we provided two different approaches in[15] to justify that the relation Ploss = Pmin is expected towidely hold in practice due to the physical properties of apower network. The details are omitted here for brevity.

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B. Connection to the Rank Minimization Literature

Combining the constraint (5a) with the objective function(4) leads to the conclusion that Ploss is equal to the mini-mum of

WTYW (14)

subject to

WTYkW = −Pk, k = 1, 2, ..., n− 1, (15a)

WT YkW = −Qk, k = 1, 2, ..., n− 1, (15b)

WTMW = V 20 . (15c)

Assume, for simplicity, that det{Y} = 0. Since Y is positivedefinite, its square root exists. Let Y

12 denote the unique,

symmetric, positive definite matrix whose square is equal toY. Define X as

X := Y12WWTY

12 .

Notice that

WTYW = trace{WTYW} = trace{X}

It follows from (14) and (15) that Ploss is equal to theminimum of the function

∥X∥∗subject to

trace{Y− 12YkY

− 12 X} = −Pk, k = 1, 2, ..., n− 1, (16a)

trace{Y− 12 YkY

− 12 X} = −Qk, k = 1, 2, ..., n− 1, (16b)

trace{Y− 12MY− 1

2 X} = V 20 , (16c)

X ≻ 0, (16d)

rank(X) = 1. (16e)

The rank constraint (16e) makes this optimization problemnonconvex, in general. On the other hand, it is shown in [16]and [17] that a good heuristic for solving a rank-constraintfeasibility problem is to replace the constraint rank(X) = 1with an objective function min ∥X∥∗. Interestingly, the aboveoptimization problem automatically minimizes the norm ofX . In other words, minimizing the loss also minimizes thenuclear norm of the variable X , which in turn would reducethe rank of X . This important property of a loss minimizationproblem could be another reason behind the satisfaction ofthe relation Ploss = Pmin widely in practice (note that Pmin

is obtained from the dual of the above optimization problemafter removing its rank constraint).

V. GENERALIZATION

Consider a power network with n buses, labeled as 1, ..., n,where all buses are possibly directly connected to loads, butonly the first m buses are directly connected to generators.For k ∈ {1, ..., n} and l ∈ {1, ...,m}, define the followingquantities:

• P dk and Qd

k: Known active and reactive loads at bus k,respectively.

• P gl and Qg

l : Unknown active and reactive powers gen-erated at bus l, respectively.

• Vk: Unknown complex voltage at bus k.Instead of the loss minimization problem, we equivalentlyconsider the problem of the total generation minimization,which aims to minimize

m∑l=1

PGl

over the parameters V1, ..., Vn, PG1 , ..., PGm , QG1 , ..., QGm ,subject to the constraints

Pminl ≤ PGl

≤ Pmaxl , l = 1, 2, ...,m,

Qminl ≤ QGl

≤ Qmaxl , l = 1, 2, ...,m,

V mink ≤ |Vk| ≤ V max

k , k = 1, 2, ..., n,

VlI∗l = (PGl

− PDl) + (QGl

−QDl)i, l = 1, 2, ...,m,

VkI∗k = −PDk

−QDki, k = m+ 1, ..., n.

In this problem, the first three inequalities limit thepower and voltage parameters by the given boundsPminl , Pmax

l , Qminl , Qmax

l , V mink , V max

k , whereas the last twoequations express the constraints imposed by the network.There could be more constraints, e.g. line flow limits, whichcan be easily incorporated. Note that the loss minimizationproblem is generally hard and NP-complete in the worst-case[15].

Extend the definition of Pmink , Pmax

k , Qmink , Qmax

k to k ∈{m + 1, ..., n}, with Pmin

k = Pmaxk = Qmin

k = Qmaxk = 0

if k ∈ {m + 1, ..., n}. For every l = 1, 2, ..., n, define alsoMl ∈ R2n×2n as a diagonal matrix whose entries are allequal to zero, except for its (l, l) and (n + l, n + l) entriesthat are equal to 1. The following optimization is the generalform of Optimization 1.

Optimization 2: Maximize the function

f(λ, λ,µ) :=n∑

k=1

{λmink Pmin

k − λmaxk Pmax

k + λkPDk

+ λmink Qmin

k − λmaxk Qmax

k + λkQDk+ µmin

k

(V mink

)2− µmax

k (V maxk )

2

}over the nonnegative scalar variables λmin

k , λmaxk ,

λmink , λmax

k , µmink , µmax

k , k = 1, 2, ..., n, subject to

Φ(λ, λ,µ) :=n∑

k=1

(λkYk + λkYk + µkMk

)≽ 0,

where

λk :=

{−λmin

k + λmaxk + 1 if k = 1, ...,m

−λmink + λmax

k otherwise ,

λk := −λmink + λmax

k ,

µk := −µmink + µmax

k ,

for every k ∈ {1, 2, ..., n}.Note that λ, λ and µ in Optimization 2 are the vectors

associated with the sets {λmink , λmax

k }nk=1, {λmink , λmax

k }nk=1

and {µmink , µmax

k }nk=1, respectively.Theorem 4: Assume that there exists a solution

(λopt, λopt,µopt) to Optimization 2 such that Φ(λopt, λ

opt,

4009

µopt) has rank greater than or equal to 2n − 2. Theoptimal objective value of Optimization 2 is identical tothe minimum value of the total generation. Moreover, theoptimal vector of bus voltages satisfies the relation

V opt = (ζ1 + ζ2i)(U1 + U2i)

for some real numbers ζ1, ζ2 and a vector[UT1 UT

2

]Tin

the null space of Φ(λopt, λopt,µopt).

Theorem 4 can be proved in line with the proof ofTheorem 2. As we have studied in [15], the rank conditiongiven in Theorem 4 is likely to hold widely in practice dueto the two properties associated with power systems (seethe proof of Theorem 3): (i) the positivity of the Lagrangemultipliers λopt

1 , ...,λoptn , (ii) the particular sign structures of

the real and imaginary parts of Y .

VI. SIMULATION RESULTS

Our method can be exploited to verify that the dualitygap is zero for all IEEE systems with 14, 30, 57, 118and 300 buses, where the goal is to minimize either thetotal generation cost or the power loss. The details of thisinteresting observation can be found in [15]. Since thesesystems are so large that the specific values of the optimalsolutions cannot be provided here, some smaller exampleswill be analyzed in the sequel instead.

Consider the three systems depicted in Figure 3, whichare referred to as Systems 1, 2 and 3. In each of thesesystems, bus 1 is the slack bus (feeder) whose voltage isset to a specific value and other buses are the load buses.The detailed specifications of these systems (together withthe load demands) are provided in Table I in per unit forthe voltage rating 400kV and the power rating 100MVA.Optimization 1 is solved for each system and the correspond-ing Lagrange multipliers are summarized in Table II. Asexpected from Lemma 2, it can be seen that λopt

2 , λopt3 , λopt

4 areall nonnegative. Interestingly, the zero-duality-gap conditiongiven in Theorem 2 is satisfied for all these systems. Therecovered optimal voltages and power losses are brought inTable III.

We repeated this example several hundred times by ran-domly choosing the parameters of these systems over a widerange of values. In all these trials, Optimization 1 alwaysfound Ploss or detected the infeasibility of the correspondingpower flow problem. We also validated our simulations usingthe toolbox PSAT [18]. Another advantage of Optimization 1is in proving that a specific power flow problem is infeasible.For instance, if the voltage magnitude at the slack busof System 1 is changed from 1.05 to 1, Optimization 1becomes unbounded. Hence, the power loss in this casebecomes infinity, which is an indication of the fact that thecorresponding power flow problem is infeasible.

Optimization 1 also provides valuable insights into theconnection between the power loss and the network topol-ogy. For example, consider System 3 and assume that anyarbitrary amount of reactive power can be injected at everyload bus. It follows from a variant of Optimization 1 thatthe active power loss can be reduced from 0.3877 to 0.3542

TABLE IPARAMETERS OF THE SYSTEMS GIVEN IN FIGURE 3.

Parameters System 1 System 2 System 3z12 0.05 + 0.25i 0.1 + 0.5i 0.10 + 0.1iz13 0.04 + 0.40i None Nonez23 0.02 + 0.10i 0.02 + 0.20i 0.01 + 0.1iz14 None None 0.01 + 0.2iy12 0.06i 0.02i 0.06iy13 0.05i None Noney23 0.02i 0.02i 0.02iy14 None None 0.02iV0 1.05 1.4 1

P2 +Q2i 0.95 + 0.4i 0.7 + 0.02i 0.9 + 0.02iP3 +Q3i 0.9 + 0.6i 0.65 + 0.02i 0.6 + 0.02iP4 +Q4i None None 0.9 + 0.02i

TABLE IILAGRANGE MULTIPLIERS OBTAINED BY SOLVING OPTIMIZATION 1 FOR

THE SYSTEMS GIVEN IN FIGURE 3.

Lagrange Multipliers System 1 System 2 System 3λ

opt2 1.3809 1.4028 1.7176

λopt3 1.4155 1.4917 1.7900

λopt4 None None 1.0207

λopt2 0.4391 0.2508 0.1764

λopt3 0.4955 0.2633 0.1858

λopt4 None None 0.0061

µopt 0.0005 0.0001 0.0005

(per unit) under this circumstance. Since this improvementis not significant, it is desired to further reduce the activepower loss by installing a small active source (generator)at one of buses 2, 3 or 4 with the capacity of 0.3. Forthis purpose, notice that the Lagrange multipliers associatedwith the power loss 0.3542 are λopt

2 = 1.5874, λopt3 =

1.6153, and λopt4 = 1.0186. These numbers indicate that

the order of buses in causing the power loss is 3, 2, 4 (seeRemark 3). Hence, this optimization suggests that the smallgenerator be installed at bus 3. Indeed, it can be verified thatthe power losses associated with installing the generator atbuses 3, 2, 4 are 0.2043, 0.2092, 0.3496, respectively. Hence,the Lagrange multipliers correctly identified the bus whosepower compensation had the greatest effect on power loss.

VII. CONCLUSIONS

Given a power system with certain constraints on its phys-ical variables, this paper proposes a linear-matrix-inequality

TABLE IIIPARAMETERS RECOVERED FROM OPTIMIZATION 1 FOR THE SYSTEMS

GIVEN IN FIGURE 3.

Recovered System 1 System 2 System 3Parameters

Vopt1 1.05∠0◦ 1.4∠0◦ 1∠0◦

Vopt2 0.71∠−20.11◦ 1.10∠−25.73◦ 0.78∠−10.58◦

Vopt3 0.68∠−21.94◦ 1.08∠−31.96◦ 0.76∠−16.31◦

Vopt4 None None 0.95∠−10.82◦

Ploss 0.2193 0.1588 0.3877

4010

(a) (b) (c)

Fig. 3. Figures (a), (b) and (c) depict Systems 1, 2 and 3 studied in Section VI, respectively.

(LMI) optimization problem to find a lower bound on theminimum amount of active (reactive) power loss incurredin the network. This LMI problem has the useful propertythat its objective function depends on the load profile andits matrix inequality constraint is contingent only upon thetopology of the network. This implies that the networktopology determines the shape of the feasibility region withinwhich the Lagrange multipliers that minimize the power lossmust lie. Direct applications of this result are in optimalnetwork reconfiguration and optimal placement/sizing ofdistributed generation units in distribution networks. It isalso observed in many examples, including IEEE benchmarksystems with 14, 30, 57, 118 and 300 buses, that the proposedLMI optimization problem always generates the exact valueof the power loss, as opposed to a lower bound on it. Asjustified mathematically, this result is expected to hold widelyin practice.

ACKNOWLEDGMENTS

The authors are thankful to Prof. John C. Doyle for severalfruitful discussions on this work. This research was supportedby ONR MURI N00014-08-1-0747 “Scalable, Data-driven,and Provably-correct Analysis of Networks,” ARO MURIW911NF-08-1-0233 “Tools for the Analysis and Design ofComplex Multi-Scale Networks,” and the Army’s W911NF-09-D-0001 Institute for Collaborative Biotechnology.

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