# Low-Rate Non-Intrusive Load Disaggregation with Graph Shift Quadratic Form Constraint
## 2 Methodology  
### Problem Statement  
As described previously, load disaggregation can be interpreted as a single channel source separation problem, whose main objective is to extract individual source signals from the aggregated signal taken from a single point power meter. The aggregated signal is considered to be a linear mixture of the individual sources. That is,  
\[P(i)=\sum_{k=1}^{N_{p}}p_{k}(i)+e(i) \tag{1}\]  
where \(P(i)\) is the observed mixture signal at time \(t_{i}\), \(p_{k}(i)\) is the individual active power consumption of appliance \(k\) at time \(t_{i}\), and \(e(i)\) is the error component including random noise and some unknown appliances in the house. Additionally, \(N_{p}\) denotes the number of all known appliances in the house.  
In this paper, we use the active power variation as a signature to recover the original component signals from the mixture signal. Thus, we can rewrite the Equation (1) as,  
\[\Delta P(i)=\sum_{k=1}^{N_{p}}\Delta p_{k}(i)+\Delta e(i) \tag{2}\]  
where \(\Delta P(i)=P(i+1)-P(i)\), \(\Delta p_{k}(i)=p_{k}(i+1)-p_{k}(i)\), \(\Delta e(i)=e(i+1)-e(i)\).  
An effective method to deal with the source separation problem is MF, where source signal \(\Delta p_{k}\) at time \(t_{i}\) is modeled by \(\Delta p_{k}(i)=\sum_{l=1}^{l}p_{k}^{l}a_{k}^{l}(i)\), where \(b_{k}^{l}\) denotes the bases collecting the main signatures of source \(\Delta p_{k}\) and \(a_{k}^{l}\) is the corresponding activation. In this paper, we adopt the difference between mean power value of appliance \(k\)'s consecutive states, \(\Delta\overline{p}_{k}\), as a base. According to the above description, source signal can be expressed as,  
\[\Delta p_{k}(i)=\Delta\overline{p}_{k}a_{k}(i) \tag{3}\]  
where \(\Delta\overline{p}_{k}\) represents the base of device \(k\), and \(a_{k}\) is the activation coefficient.  
By performing MF, we need to calculate the activation coefficient for each appliance. Assuming that \(\Delta p_{k}(i),i=1,2,\cdots,n\) is available for all appliances during training, we can formulate the corresponding optimization problem before imposing any constraints as,  
\[\mathbf{A}_{1:N_{p}} =\text{argmin}\sum_{i=n+1}^{N}f(\mathbf{\alpha}(i)) \tag{4}\] \[=\text{argmin}\sum_{i=n+1}^{N}\left\|\Delta P(i)-\mathbf{b}\mathbf{ \alpha}(i)\right\|_{2}^{2}=\text{argmin}\sum_{i=n+1}^{N}\left\|\Delta P(i)- \left[\Delta\overline{p}_{1},\cdots,\Delta\overline{p}_{N_{p}}\right]\left[ \begin{array}{c}a_{1}(i)\\ a_{2}(i)\\ \vdots\\ a_{N_{p}}(i)\\ \end{array}\right]\right\|_{2}^{2}\]In Equation (4), \(\mathbf{A}_{1:N_{p}}=[\mathbf{\alpha}(n+1),\cdots,\mathbf{\alpha}(N)]\) is an activation matrix and each row of \(\mathbf{A}_{1:N_{p}}\), i.e., \(\mathbf{A}_{k}=[a_{k}(n+1),\cdots,a_{k}(N)]\) denotes the activation vector of appliance \(k\). \(f(\mathbf{\alpha}(i))\) is the objective function. \(\mathbf{b}\) is the base vector and \(\mathbf{\alpha}(i)\) is the activation vector of all appliances at time \(t_{i}\). Note that \(N\) is the total number of samples of interest and \(n+1<i\leq N\) denotes the testing phase.  
After activation coefficient matrix \(\mathbf{A}_{1:N_{p}}\) is acquired, we can calculate the power variation of each appliance using Equation (3).