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colonetti authored Feb 29, 2024
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# Under construction


$$
A = \begin{bmatrix}
1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 &-1\\
\end{bmatrix}
$$
## Original system

$$
Y = diag(\begin{bmatrix}
10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 \\
\end{bmatrix})
$$
Consider the following 14-node system.

$$
B = A ^ \top \cdot (Y \cdot A)
$$
<img src="https://drive.google.com/uc?id=1xeFtii1CcXhDphXqDH5UOAmYSSBtGnJD"
alt="original system"
style="width: 50%" />

$$
PTDF = (Y \cdot A) \cdot \hat{B}
$$
The system has 18 branches, 5 generators (represented in green), and 11 nodes with load (represented by the red arrows).
Assume for simplicity that all branches have reactances of 0.1 p.u.. Moreover, throughout this example, node 2 is used as the reference node for the voltage angles.
The first order of businness is to compute the Power Transfer Distribution Factors (PTDF) matrix for this system.

For this system, the PTDF matrix (rounded) is

$$
PTDF = \begin{bmatrix}
Expand All @@ -63,11 +40,11 @@ $$

The above steps are taken in https://github.com/colonetti/ward_UC_pscc2024/blob/e71b4caff9d4ca70743ca9296af9262a6ec3dfd7/pre_processing/build_ptdf.py#L148.

$$
\begin{align}
\sum_{g \in \mathcal{G}\_{b}}p_{g} - \sum_{l \in \mathcal{L}^{-}}f_{l} + \sum_{l \in \mathcal{L}^{+}}f_{l} + s_{b} = \mathrm{\mathbf{D}}_{b} & \qquad \forall b \in \mathcal {B}\\
\end{align}
$$
Now, assume that the individual loads are 1 p.u., then adding up to a total system load of 11 p.u..
Moreover, assume that only 4 of all 18 branches can possibly reach their limits under any feasible operation of this system for this load profile --- these are the branches 4, 13, 14 and 16, represented in red in the image.
For the possibly binding branches, assume lower and upper limits for the flows of 10 p.u.. For all other branches, the limits -inf and +inf.

For this example, the power flow equations under the B-theta formulation are the following.

$$
\begin{align}
Expand All @@ -88,6 +65,20 @@ $$
\end{align}
$$

Where $p$ are generation outputs, $f$ are flows in the branches, and $s$ are slack variables that account for shortage of generation.

For this formulation, the flow expressions and their respective limits are as follows.

$$
\begin{align}
f_{l} - 10 \cdot \left( \theta_{\text{from}(l)} - \theta_{\text{to}(l)} \right)= 0 & \qquad \forall l \in \mathcal {L}\\
{-}10 \leq f_{l} \leq 10 & \qquad \forall l \in \\{4, 13, 14\\}\\
{-}inf \leq f_{l} \leq inf & \qquad \forall l \in \mathcal {L} \setminus \\{4, 13, 14\\}\\
\end{align}
$$

On the other hand, for the PTDF formulation, we first have the flow expressions and limits of the possibly binding branches, as below.

$$
\begin{align}
{-}10 \leq -0.103 \cdot p_1 -0.23 \cdot (p_3 + s_3 - 1) -0.461 \cdot (s_4 - 1 ) -0.206 \cdot (s_5 - 1) -0.261 \cdot (p_4 + s_6 - 1) -0.442 \cdot p_5 -0.424 \cdot (s_9 - 1) -0.37 \cdot (s_{10} - 1) -0.315 \cdot (s_{11} - 1) -0.261 \cdot (s_{12} - 1) -0.261 \cdot (s_{13} - 1) -0.261 \cdot (s_{14} - 1) \leq 10 & \qquad \text{(l = 4)}\\
Expand All @@ -97,39 +88,47 @@ $$
\end{align}
$$

And the global power balance for the system.

$$
\begin{align}
f_{l} - 10 \cdot \left( \theta_{\text{from}(l)} - \theta_{\text{to}(l)} \right)= 0 & \qquad \forall l \in \mathcal {L}\\
{-}10 \leq f_{l} \leq 10 & \qquad \forall l \in \\{4, 13, 14\\}\\
{-}inf \leq f_{l} \leq inf & \qquad \forall l \in \mathcal {L} \setminus \\{4, 13, 14\\}\\
p_1 + p_2 + p_3 + p_4 + p_5 + s_2 + s_3 + s_4 + s_5 + s_6 + s_9 + s_{10} + s_{11} + s_{12} + s_{13} + s_{14} == 11\\
\end{align}
$$

After the first iteration:
$$\mathcal{B} = \mathcal{B} \setminus \\{8, 14\\}$$
$$\mathcal{L} = \mathcal{L} \setminus \\{13, 18\\}$$
The above formulations are used when the network is not reduced. That is, for the full network, or original system.
As the network is reduced, nodes, branches and slack variables are eliminated from the model. Thus reflecting in changes both in the B-theta and the PTDF formulations

Below we show the step-by-step transformation that these formulations undergo as the network is reduced.

### First iteration

At the first iteration, nodes 8 and 14 are removed and their respective injections are reassigned.
The load previously allocated to node 14 goes to node 13, and the generator then connected to node 8 is now placed at node 7.

<img src="https://drive.google.com/uc?id=1xhHf9JmMk1Fjzs-Ar2V0nPQ_Psa8oapB"
alt="system after removing nodes 8 and 14"
style="width: 50%" />

In the B-theta formulation, in addition to removing the constraints associated with nodes 8 and 14 and branches 13 and 18, the power balance constraints of nodes 7 and 14 now become

$$
\begin{align}
\sum_{g \in \mathcal{G}\_{b}}p_{g} - \sum_{l \in \mathcal{L}^{-}}f_{l} + \sum_{l \in \mathcal{L}^{+}}f_{l} + s_{b} = \mathrm{\mathbf{D}}\_{b}& \qquad \forall b \in \mathcal{B} \setminus \\{7, 13\\}\\
p_{5} + f_{8} - f_{14} = 0 & \\
f_{17} + s_{13} = 1 + 1 & \\
\end{align}
$$

In our strategy, because the limits of branch 13 are possibly binding but it has been removed, its limits are now enforced on the injections connected to the node being deleted.
For branch 13, this simply means enforcing its limits on the generation of G5.

$$
\begin{align}
p_{5} \leq 10 & \\
\end{align}
$$

$$
\begin{align}
f_{l} - 10 \cdot \left( \theta_{\text{from}(l)} - \theta_{\text{to}(l)} \right)= 0 & \qquad \forall l \in \mathcal {L}\\
{-}10 \leq f_{l} \leq 10 & \qquad \forall l \in \\{4, 14\\}\\
{-}inf \leq f_{l} \leq inf & \qquad \forall l \in \mathcal {L} \setminus \\{4, 14\\}\\
\end{align}
$$
For the PTDF formulation, the changes in the flow expressions are due to the deletion of the slack variables, as seen below.

$$
\begin{align}
Expand All @@ -139,53 +138,82 @@ $$
\end{align}
$$

After the second iteration:
$$\mathcal{B} = \mathcal{B} \setminus \\{13\\}$$
$$\mathcal{L} = \mathcal{L} \setminus \\{17\\}$$

After the third iteration:
$$\mathcal{B} = \mathcal{B} \setminus \\{12\\}$$
$$\mathcal{L} = \mathcal{L} \setminus \\{12\\}$$
Naturally, the slack variables associated with the deleted nodes 8 and 14 also need to be removed from the global balance equation.

$$
\begin{align}
\sum_{g \in \mathcal{G}\_{b}}p_{g} - \sum_{l \in \mathcal{L}^{-}}f_{l} + \sum_{l \in \mathcal{L}^{+}}f_{l} + s_{b} = \mathrm{\mathbf{D}}\_{b}& \qquad \forall b \in \mathcal{B} \setminus \\{6, 7\\}\\
p_{4} + f_{10} - f_{11} + s_{6} = 4 & \\
p_{5} + f_{8} - f_{14} = 0 & \\
p_1 + p_2 + p_3 + p_4 + p_5 + s_2 + s_3 + s_4 + s_5 + s_6 + s_9 + s_{10} + s_{11} + s_{12} + s_{13} == 11\\
\end{align}
$$

$$
\begin{align}
p_{5} \leq 10 & \\
\end{align}
$$
### Second iteration

$$
\begin{align}
f_{l} - 10 \cdot \left( \theta_{\text{from}(l)} - \theta_{\text{to}(l)} \right)= 0 & \qquad \forall l \in \mathcal {L}\\
{-}10 \leq f_{l} \leq 10 & \qquad \forall l \in \\{4, 14\\}\\
{-}inf \leq f_{l} \leq inf & \qquad \forall l \in \mathcal {L} \setminus \\{4, 14\\}\\
\end{align}
$$
Now node 13 is removed and the load connected to it is transfered to node 12.

<img src="https://drive.google.com/uc?id=1xiSH66igfQHCcyNyEzn6_MeAqEqGYKHN"
alt="system after removing node 13"
style="width: 50%" />

The implications of removing node 13 to the formulation are similar to what happened in the previous iteration.

### Third iteration

The next step is to remove node 12.

<img src="https://drive.google.com/uc?id=1xiyhlPZYVeFgEhG5Za-HZXoDiN3Ei0Pm"
alt="system after removing node 12"
style="width: 50%" />

After removing the end-of-line nodes 8, 14, 13 and 12, the system`s PTDFs for the remaining monitored branches 4, 14 and 16 have not changed w.r.t. the remaining nodes. As seen below

$$
PTDF = \begin{bmatrix}
0 & -0.615 & -0.23 & -0.103 & -0.13 & -0.221 & -0.212 & -0.185 & -0.158\\
0 & -0.23 & -0.461 & -0.206 & -0.261 & -0.442 & -0.424 & -0.37 & -0.315\\
0 & 0.385 & -0.23 & -0.103 & -0.13 & -0.221 & -0.212 & -0.185 & -0.158\\
0 & 0.127 & 0.255 & -0.255 & -0.145 & 0.218 & 0.182 & 0.073 & -0.036\\
0 & 0.009 & 0.018 & -0.018 & -0.082 & -0.627 & -0.273 & -0.209 & -0.145\\
0 & 0.018 & 0.036 & -0.036 & -0.164 & -0.255 & -0.545 & -0.418 & -0.291\\
0 & -0.027 & -0.055 & 0.055 & -0.755 & -0.118 & -0.182 & -0.373 & -0.564\\
0 & -0.027 & -0.055 & 0.055 & 0.245 & -0.118 & -0.182 & -0.373 & -0.564\\
0 & 0.009 & 0.018 & -0.018 & -0.082 & 0.373 & -0.273 & -0.209 & -0.145\\
0 & 0.027 & 0.055 & -0.055 & -0.245 & 0.118 & 0.182 & -0.627 & -0.436\\
0 & 0.027 & 0.055 & -0.055 & -0.245 & 0.118 & 0.182 & 0.373 & -0.436\\
0 & -0.155 & -0.309 & -0.691 & -0.609 & -0.336 & -0.364 & -0.445 & -0.527\\
0.615&0&0.052&0.103&0.23&0.203&0.112&0.121&0.148&0.176\\
0.385&0&-0.052&-0.103&-0.23&-0.203&-0.112&-0.121&-0.148&-0.176\\
-0.052&0&-0.615&-0.23&-0.103&-0.13&-0.221&-0.212&-0.185&-0.158\\
-0.103&0&-0.23&-0.461&-0.206&-0.261&-0.442&-0.424&-0.37&-0.315\\
-0.23&0&-0.103&-0.206&-0.461&-0.406&-0.224&-0.242&-0.297&-0.352\\
-0.052&0&0.385&-0.23&-0.103&-0.13&-0.221&-0.212&-0.185&-0.158\\
-0.127&0&0.127&0.255&-0.255&-0.145&0.218&0.182&0.073&-0.036\\
-0.009&0&0.009&0.018&-0.018&-0.082&-0.627&-0.273&-0.209&-0.145\\
-0.018&0&0.018&0.036&-0.036&-0.164&-0.255&-0.545&-0.418&-0.291\\
0.027&0&-0.027&-0.055&0.055&-0.755&-0.118&-0.182&-0.373&-0.564\\
0.027&0&-0.027&-0.055&0.055&0.245&-0.118&-0.182&-0.373&-0.564\\
-0.009&0&0.009&0.018&-0.018&-0.082&0.373&-0.273&-0.209&-0.145\\
-0.027&0&0.027&0.055&-0.055&-0.245&0.118&0.182&-0.627&-0.436\\
-0.027&0&0.027&0.055&-0.055&-0.245&0.118&0.182&0.373&-0.436\\
\end{bmatrix}
$$

### Fourth iteration

Now nodes 1 and 6 are removed. Different from the previous nodes, they are not connected to a single branch. Thus, the reassignment of their injections is not as straightforward.
Moreover, after removing them, new connections will be created.

<img src="https://drive.google.com/uc?id=1xj7OeO--F4eKBktQBJHkzb7onh7IoK1x"
alt="system after removing node 12"
style="width: 50%" />

After removing the end-of-line nodes 8, 14, 13 and 12, the system`s PTDFs for the remaining monitored branches 4, 14 and 16 have not changed w.r.t. the remaining nodes. As seen below

$$
PTDF = \begin{bmatrix}
0.615&0&0.052&0.103&0.23&0.203&0.112&0.121&0.148&0.176\\
0.385&0&-0.052&-0.103&-0.23&-0.203&-0.112&-0.121&-0.148&-0.176\\
-0.052&0&-0.615&-0.23&-0.103&-0.13&-0.221&-0.212&-0.185&-0.158\\
-0.103&0&-0.23&-0.461&-0.206&-0.261&-0.442&-0.424&-0.37&-0.315\\
-0.23&0&-0.103&-0.206&-0.461&-0.406&-0.224&-0.242&-0.297&-0.352\\
-0.052&0&0.385&-0.23&-0.103&-0.13&-0.221&-0.212&-0.185&-0.158\\
-0.127&0&0.127&0.255&-0.255&-0.145&0.218&0.182&0.073&-0.036\\
-0.009&0&0.009&0.018&-0.018&-0.082&-0.627&-0.273&-0.209&-0.145\\
-0.018&0&0.018&0.036&-0.036&-0.164&-0.255&-0.545&-0.418&-0.291\\
0.027&0&-0.027&-0.055&0.055&-0.755&-0.118&-0.182&-0.373&-0.564\\
0.027&0&-0.027&-0.055&0.055&0.245&-0.118&-0.182&-0.373&-0.564\\
-0.009&0&0.009&0.018&-0.018&-0.082&0.373&-0.273&-0.209&-0.145\\
-0.027&0&0.027&0.055&-0.055&-0.245&0.118&0.182&-0.627&-0.436\\
-0.027&0&0.027&0.055&-0.055&-0.245&0.118&0.182&0.373&-0.436\\
\end{bmatrix}
$$


# Ward Reduction in Unit-Commitment Problems
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