Proof of Sufficient Conditions ReviewReview

WITHOUT DELTA CONNECTIONS

3.6 Proof of Sufficient Conditions ReviewReview

The existing works [15, 67] prove that the optimal solution of SDP relaxation is of rank 1 in single phase networks. A crucial step in their proof uses the strong duality to show that the product of the primal optimal solution𝑾^∗ and the dual matrix 𝑨^∗ is a zero matrix, and hence the rank of𝑾^∗cannot exceed the dimension of 𝑨^∗’s null space. Under certain conditions [15, 67] prove that 𝑨^∗’s null space is of dimension at most 1. Hence the optimal primal solution𝑾^∗ must be of rank at most 1.

This argument however breaks down in a multiphase network for the following two reasons. First, although the underlying graph for𝑚phase network is still a tree, each

3For example, if Re(𝑠

𝜙

𝑗)is minimized in the objective function, then the lower bound of Re(𝑠

𝜙 𝑗) should not be active in the constraints.

bus now has 𝑚 different phases and might have𝑚 unbalanced voltages in general.

If we extend each phase to a separate vertex in the new graph and connect every phase pair between every two neighboring buses, then the𝑚 phase network will be transformed into an (𝑚 𝑛)-node meshed network with multiple cycles [7, 22, 44].

Hence the theory for single-phase radial network is not applicable. Second, in an𝑚 phase network, it is unknown wether the null space of 𝑨^∗at the optimal point is still of dimension 1. It is therefore not clear how to prove rank(𝑾^∗) = 1 via analyzing the dimension of null(𝑨^∗).

In the following argument, we use a similar proof framework to that in [15], but the proof will be based on the eigenvectors of𝑾^∗instead of the dimension of null(𝑨^∗).

From now on, we suppose A1, A2, A3, and A5 hold.

Preliminaries

Our strategy is to prove the exactness of the perturbed OPF problem and then use Lemma 12 to show (3.4) is also exact. It is important to make sure that all the non-active constraints will remain non-active in the perturbation neighborhood.

Lemma 13. For any nonzero𝑪₁, there exists a positive sequence𝜀↓0such that for each 𝜀in the sequence, one can collect (𝜇

𝜙

𝑗(𝜀), 𝜇^𝜙

𝑗

(𝜀), 𝜂

𝜙 𝑗(𝜀), 𝜂^𝜙

𝑗

(𝜀)) from at least one of its KKT multiplier tuples satisfying

𝑓_𝑝(𝑗 , 𝜙)=0 =⇒ 𝜇^𝜙

𝑗(𝜀) =𝜇^𝜙

𝑗

(𝜀) =0 (3.14a)

𝑓_𝑝(𝑗 , 𝜙) ≠0 =⇒ 𝑓_𝑝(𝑗 , 𝜙) · (𝜇

𝜙

𝑗(𝜀) −𝜇^𝜙

𝑗

(𝜀)) ≥0 (3.14b) 𝑓_𝑞(𝑗 , 𝜙)=0 =⇒ 𝜂

𝜙

𝑗(𝜀) =𝜂^𝜙

𝑗

(𝜀) =0 (3.14c)

𝑓_𝑞(𝑗 , 𝜙) ≠0 =⇒ 𝑓_𝑞(𝑗 , 𝜙) · (𝜂

𝜙

𝑗(𝜀) −𝜂^𝜙

𝑗

(𝜀)) ≥0. (3.14d)

Proof. First consider any positive sequence {𝜀_𝑙}^∞

𝑙=1such that lim𝑙→∞𝜀_𝑙 =0. Sup- pose the optimal solution to (3.5) under𝜀_𝑙is𝑾𝑙(if there are multiple solutions then select one of them). As (3.5b) prescribes a compact set, using a similar argument as in the proof of Lemma 12 we know there must be a subsequence of {𝜀_𝑙}^∞

𝑙=1, denoted by{𝜀_𝑧

𝑡}^∞

𝑡=1, such that𝑾𝑧_𝑡 converges to𝑾^∗in the max norm. The difference

∥𝑾𝑧_𝑡−𝑾^∗∥_∞can be arbitrarily small for sufficiently large𝑡. When𝑡is large enough, the non-active constraints in (3.5b) under𝑾^∗will remain non-active under𝑾𝑧_𝑡, and

the corresponding KKT multipliers will remain 0. As a result, 𝑓_𝑝(𝑗 , 𝜙) =0 =⇒ 𝑝^𝜙

𝑗

< tr(𝚽^𝜙_𝑗𝑾^∗) < 𝑝

𝜙 𝑗

=⇒ 𝑝^𝜙

𝑗

< tr(𝚽^𝜙_𝑗𝑾𝑧_𝑡) < 𝑝

𝜙

𝑗 =⇒ 𝜇

𝜙 𝑗(𝜀_𝑧

𝑡) =𝜇^𝜙

𝑗

(𝜀_𝑧

𝑡) =0, 𝑓_𝑝(𝑗 , 𝜙) =+1 =⇒ 𝑝^𝜙

𝑗

< tr(𝚽^𝜙_𝑗𝑾^∗)

=⇒ 𝑝^𝜙

𝑗

< tr(𝚽^𝜙_𝑗𝑾𝑧_𝑡) =⇒ 𝜇^𝜙

𝑗

(𝜀_𝑧

𝑡) =0

=⇒ 𝑓_𝑝(𝑗 , 𝜙) · (𝜇

𝜙 𝑗(𝜀_𝑧

𝑡) −𝜇^𝜙

𝑗

(𝜀_𝑧

𝑡)) ≥ 0, 𝑓_𝑝(𝑗 , 𝜙) =−1 =⇒ tr(𝚽^𝜙_𝑗𝑾^∗) < 𝑝

𝜙 𝑗

=⇒ tr(𝚽^𝜙

𝑗𝑾𝑧𝑡) < 𝑝

𝜙

𝑗 =⇒ 𝜇

𝜙 𝑗(𝜀_𝑧

𝑡) =0

=⇒ 𝑓_𝑝(𝑗 , 𝜙) · (𝜇

𝜙 𝑗(𝜀_𝑧

𝑡) −𝜇^𝜙

𝑗

(𝜀_𝑧

𝑡)) ≥0

all hold. A similar argument can also be applied to prove (3.14c) and (3.14d).

Properties of Dual Matrix 𝑨^∗(𝜀)

In order to apply Lemma 12, we construct𝑪₁ ∈C^{𝑚 𝑛×𝑚 𝑛}in the following manner:

[𝑪₁]𝑗 𝑗 =0∈C^𝑚×𝑚, for 𝑗 ∈ V [𝑪₁]𝑗 𝑘 =0∈C^𝑚×𝑚, for (𝑗 , 𝑘) ∉E.

When (𝑗 , 𝑘) ∈ E, we assume 𝑗 < 𝑘. If neither 𝑗 nor 𝑘 is in S_o ∪ S_c, then we construct[𝑪₁]𝑗 𝑘 =𝒀𝑗 𝑘.

If 𝑗 ∈ S_o∪ S_c, then A3 guarantees 𝑘 ∉S_o∪ S_c. ∀𝜙 ∈ M, we set [𝑪₁]^𝜙,:

𝑗 𝑘 to𝒀^𝜙,_{𝑗 𝑘}^:if 𝑐

𝜙 𝑗 ,r𝑒 =𝑐

𝜙

𝑗 ,i𝑚 = 𝑓_𝑝(𝑗 , 𝜙) = 𝑓_𝑞(𝑗 , 𝜙) =0, and to (𝑓_𝑝(𝑗 , 𝜙) + 𝑓_𝑞(𝑗 , 𝜙)𝒊)𝒀^𝜙,_{𝑗 𝑘}^:otherwise.

If 𝑘 ∈ S_o ∪ S_c, then A3 guarantees 𝑗 ∉ S_o ∪ S_c. ∀𝜙 ∈ M, we similarly set [𝑪₁]^:,𝜙

𝑗 𝑘 to (𝒀^𝜙,:

𝑘 𝑗)^H if 𝑐

𝜙 𝑘 ,r𝑒 = 𝑐

𝜙

𝑘 ,i𝑚 = 𝑓_𝑝(𝑘 , 𝜙) = 𝑓_𝑞(𝑘 , 𝜙) = 0, and to (𝑓_𝑝(𝑘 , 𝜙) − 𝑓_𝑞(𝑘 , 𝜙)𝒊) (𝒀^𝜙,:

𝑘 𝑗)^Hotherwise.

Finally, we set[𝑪₁]𝑘 𝑗 := [𝑪₁]^H

𝑗 𝑘 for all 𝑗 < 𝑘 to make𝑪₁Hermitian.

The next theorem provides a key intermediate result to prove Theorem 10. Suppose under such𝑪₁, the sequence guaranteed by Lemma 13 is{𝜀_𝑙}^∞

𝑙=1.

Theorem 11. Under A1, A2, A3, and A5, for each 𝜀_𝑙, the dual matrix 𝑨^∗(𝜀_𝑙) is G-invertible. 4

4If the KKT multiplier tuple at𝜀_𝑙is non-unique, then𝑨^∗(𝜀_𝑙)is evaluated at the multiplier tuple in Lemma 13 satisfying (3.14).

Proof. The value of𝑨^∗(𝜀_𝑙)is the same as the right hand side of (3.11) when all dual variables take values at their corresponding KKT multipliers (with respect to 𝜀_𝑙).

If not otherwise specified, all the (𝜇

𝜙 𝑗, 𝜇^𝜙

𝑗

, 𝜂

𝜙 𝑗, 𝜂^𝜙

𝑗

) in this proof refer to the tuple in Lemma 13 with respect to𝜀_𝑙. Since for all𝑎≠ 𝑏, [𝑬^𝜙_𝑗]𝑎 𝑏 and[Π(𝜅)]𝑎 𝑏 are always zero matrices, it is sufficient to show

𝑸:=∑︁

𝑗 ,𝜙

(𝜇

𝜙 𝑗 −𝜇^𝜙

𝑗

)𝚽^𝜙

𝑗 + (𝜂

𝜙 𝑗 −𝜂^𝜙

𝑗

)𝚿^𝜙

𝑗

+𝑪₀+𝜀_𝑙𝑪₁ satisfies the two conditions in Definition 15.5

For𝑎 ≠ 𝑏 and (𝑎, 𝑏) ∉ E, recall that 𝑪₀ is the linear combination of𝚽^𝜙

𝑗 and 𝚿^𝜙

𝑗. When (𝑎, 𝑏) ∉ E, 𝒀𝑎 𝑏 is a zero matrix and so are all [𝚽^𝜙

𝑗]𝑎 𝑏 and [𝚿^𝜙

𝑗]𝑎 𝑏. The construction of 𝑪₁ also guarantees [𝑪₁]𝑎 𝑏 is all zero. Hence [𝑸]𝑎 𝑏 is all zero as well.

Now assume𝑎 < 𝑏. If(𝑎, 𝑏) ∈ E, we have [𝑸]𝑎 𝑏

=∑︁

𝜙

(𝜇

𝜙 𝑎− 𝜇^𝜙

𝑎

+𝑐

𝜙

𝑎,r𝑒) [𝚽^𝜙𝑎]𝑎 𝑏+ (𝜂

𝜙 𝑎−𝜂^𝜙

𝑎

+𝑐

𝜙

𝑎,i𝑚) [𝚿^𝜙𝑎]𝑎 𝑏

+∑︁

𝜙

(𝜇

𝜙 𝑏− 𝜇^𝜙

𝑏

+𝑐

𝜙

𝑏,r𝑒) [𝚽^𝜙

𝑏]𝑎 𝑏+ (𝜂

𝜙 𝑏−𝜂^𝜙

𝑏

+𝑐

𝜙

𝑏,i𝑚) [𝚿^𝜙

𝑏]𝑎 𝑏

+𝜀_𝑙[𝑪₁]𝑎 𝑏. (3.15)

If neither𝑎nor𝑏is inS_o∪ S_c, then by definition, for all𝜙 ∈ M there must be 𝑐

𝜙 𝑎,r𝑒 =𝑐

𝜙

𝑎,i𝑚 = 𝑓_𝑝(𝑎, 𝜙) = 𝑓_𝑞(𝑎, 𝜙) =0, (3.16a) 𝑐

𝜙 𝑏,r𝑒 =𝑐

𝜙

𝑏,i𝑚 = 𝑓_𝑝(𝑏, 𝜙) = 𝑓_𝑞(𝑏, 𝜙) =0. (3.16b) Equation (3.15) and Lemma 13 imply[𝑸]𝑎 𝑏 =𝜀_𝑙[𝑪₁]𝑎 𝑏. By construction,[𝑪₁]𝑎 𝑏 = 𝒀𝑎 𝑏 is invertible, and so is [𝑸]𝑎 𝑏.

If 𝑎 ∈ S_o ∪ S_c, then A3 guarantees 𝑏 ∉ S_o ∪ S_c. Thus (3.16b) holds for all 𝜙 ∈ M. For a given 𝜙 ∈ M, if (3.16a) holds, then by construction, we have [𝑸]^𝜙,:

𝑎 𝑏 =𝜀_𝑙[𝑪₁]^𝜙,:

𝑎 𝑏 =𝜀_𝑙𝒀^𝜙,:

𝑎 𝑏. If (3.16a) does not hold for the given𝜙, then we have [𝑸]^𝜙,:

𝑎 𝑏 =(𝜇

𝜙 𝑎− 𝜇^𝜙

𝑎

+𝑐

𝜙

𝑎,r𝑒+2𝜀_𝑙𝑓_𝑝(𝑎, 𝜙))𝒀^𝜙,:

𝑎 𝑏

2 + (𝜂^𝜙

𝑎−𝜂^𝜙

𝑎

+𝑐^𝜙

𝑎,i𝑚 +2𝜀_𝑙𝑓_𝑞(𝑎, 𝜙))𝒀_{𝑎 𝑏}^𝜙,: 2 𝒊.

5The matrix𝑸 itself might not be G-invertible as 𝑸 might not be positive semidefinite, but 𝑨^∗≥0 always holds.

Note that Condition A5 and Lemma 13 imply both {𝜇

𝜙 𝑎 − 𝜇^𝜙

𝑎

, 𝑓_𝑝(𝑎, 𝜙), 𝑐

𝜙

𝑎,r𝑒} and {𝜂^𝜙

𝑎−𝜂^𝜙

𝑎

, 𝑓_𝑞(𝑎, 𝜙), 𝑐^𝜙

𝑎,i𝑚}are sign-semidefinite sets, respectively. When (3.16a) does not hold, at least one of {𝑐

𝜙 𝑎,r𝑒, 𝑐

𝜙 𝑎,i𝑚

, 𝑓_𝑝(𝑎, 𝜙), 𝑓_𝑞(𝑎, 𝜙)} is non-zero. As a result, there exists some non-zero𝜎

𝜙,:

𝑎 𝑏 ∈Csuch that [𝑸]^𝜙,_{𝑎 𝑏}^: =𝜎

𝜙,:

𝑎 𝑏𝒀_{𝑎 𝑏}^𝜙,:. In short, in the case 𝑎 ∈ S_o∪ S_c, [𝑸]^𝜙,_{𝑎 𝑏}^: is always a non-zero multiple of𝒀_{𝑎 𝑏}^𝜙,:. The invertibility of𝒀𝑎 𝑏

indicates all the𝒀_{𝑎 𝑏}^𝜙,:are independent for𝜙∈ M, so[𝑸]𝑎 𝑏is also invertible.

If 𝑏 ∈ S_o ∪ S_c, then A3 guarantees 𝑎 ∉ S_o ∪ S_c. Then (3.16a) holds for all 𝜙 ∈ M. For a given 𝜙 ∈ M, if (3.16b) holds, then by construction, we have [𝑸]^:,𝜙

𝑎 𝑏 =𝜀_𝑙[𝑪₁]^:,𝜙

𝑎 𝑏 =𝜀_𝑙(𝒀^𝜙,:

𝑏 𝑎)^H. If (3.16b) does not hold, then similar to the previous case, there exists some non-zero 𝜎^:

,𝜙

𝑎 𝑏 ∈ C such that [𝑸]^:_{𝑎 𝑏}^,𝜙 = 𝜎^:

,𝜙

𝑎 𝑏(𝒀_{𝑏 𝑎}^𝜙,:)^H. Hence [𝑸]^:_{𝑎 𝑏}^,𝜙is always a non-zero multiple of(𝒀_{𝑏 𝑎}^𝜙,:)^H. The invertibility of𝒀𝑏 𝑎indicates all the𝒀_{𝑏 𝑎}^𝜙,:are independent for𝜙 ∈ M, so[𝑸]𝑎 𝑏 is also invertible.

Proof of Theorem 10

Theorem 11 and Corollary 9 imply that (3.5) is exact under conditions A1, A2, A3, and A5 for any𝜀. By Lemma 12, Theorem 10 is proved.

3.7 Discussion and Example