Project 1 - Gauss-Seidel Circuit Solver

Annotated Circuit Diagrams

Circuit 1

KCL and KVL Equations:

$$\begin{array}{ll}\text{Junction A: }{I}_1-I_2-I_3-I_4=0& \text{Loop 1: }{I}_1+3I_2=6\\ \text{Junction B: }{I}_4-I_5-I_6=0& \text{Loop 2: }6I_3-3I_2=0\\ \text{Junction C: }{I}_6-I_7-I_8=0& \text{Loop 3: }3I_4+12I_5-6I_3=0\\ & \text{Loop 4: }2I_6+3I_7-12I_5=0\\ & \text{Loop 5: }6I_8-3I_7=0\end{array}$$

In matrix form, and then with partial pivoting to get rid of diagonal $0$s:

$$\begin{array}{r}\left[\begin{array}{cccccccc}1& -1& -1& -1& 0& 0& 0& 0\\ 0& 0& 0& 1& -1& -1& 0& 0\\ 0& 0& 0& 0& 0& 1& -1& -1\\ 1& 3& 0& 0& 0& 0& 0& 0\\ 0& -3& 6& 0& 0& 0& 0& 0\\ 0& 0& -6& 3& 12& 0& 0& 0\\ 0& 0& 0& 0& -12& 2& 3& 0\\ 0& 0& 0& 0& 0& 0& -3& 6\end{array}\right]\left\{\begin{array}{c}{I}_1\\ I_2\\ I_3\\ I_4\\ I_5\\ I_6\\ I_7\\ I_8\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\\ 0\\ 6\\ 0\\ 0\\ 0\\ 0\end{array}\right\}\end{array}⟹\left[\begin{array}{cccccccc}1& -1& -1& -1& 0& 0& 0& 0\\ 1& 3& 0& 0& 0& 0& 0& 0\\ 0& -3& 6& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& -1& -1& 0& 0\\ 0& 0& -6& 3& 12& 0& 0& 0\\ 0& 0& 0& 0& -12& 2& 3& 0\\ 0& 0& 0& 0& 0& 1& -1& -1\\ 0& 0& 0& 0& 0& 0& -3& 6\end{array}\right]\left\{\begin{array}{c}{I}_1\\ I_2\\ I_3\\ I_4\\ I_5\\ I_6\\ I_7\\ I_8\end{array}\right\}=\left\{\begin{array}{c}0\\ 6\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right\}$$

Using the initial relaxation parameter of $1$, the solution diverges. Higher relaxation parameter values ($1\le \omega \le 2$) also caused divergence. For values of $\omega$ below $1$:

$$\begin{array}{llllllllll}\text{Relaxation:}& 0.9& 0.8& 0.7& 0.6& 0.5& 0.4& 0.3& 0.2& 0.1\\ \text{Conv. Iterations:}& \text{N/A}& \text{N/A}& 20& 26& 34& 46& 65& 98& 206\end{array}$$

Thus, the solution converge for values below approximately $\omega \approx 0.8$; the least iterations achieved for all tested values occurred for $\omega =0.7$. The Gauss-Seidel and MATLAB solutions are:

$$\begin{array}{lllllllll}\text{Current (mA)}& I_1& I_2& I_3& I_4& I_5& I_6& I_7& I_8\\ \text{Gauss-Seidel:}& 2.3999997& 1.1999969& 0.5999979& 0.5999984& 0.1499989& 0.4499974& 0.2999967& 0.1499996\\ \text{MATLAB:}& 2.4& 1.2& 0.6& 0.6& 0.15& 0.45& 0.3& 0.15\end{array}$$

Substituting into original equations to verify correctness:

$$\begin{array}{ll}\text{Junction A: }2.4-1.2-0.6-0.6=0& \text{Loop 1: }2.4+3(1.2)=6\\ \text{Junction B: }0.6-0.15-0.45=0& \text{Loop 2: }6(0.6)-3(1.2)=0\\ \text{Junction C: }0.45-0.3-0.15=0& \text{Loop 3: }3(0.6)+12(0.15)-6(0.6)=0\\ & \text{Loop 4: }2(0.45)+3(0.3)-12(0.15)=0\\ & \text{Loop 5: }6(1.5)-3(0.3)=0\end{array}$$

Note that I kept resistor coefficients as $1\text{k}=1$ instead of $1\text{k}=1000$, so the resulting currents are in $\text{mA}$.

Circuit 2

KCL and KVL Equations:

$$\begin{array}{ll}\text{Junction A: }{I}_1-I_2-I_3=0& \text{Loop 1: }5I_3+105I_5=125\\ \text{Junction B: }{I}_3-I_4-I_5=0& \text{Loop 2: }5I_3+10I_4+15I_6=125\\ \text{Junction C: }{I}_2+I_4-I_6=0& \text{Loop 3: }25I_2+15I_6=125\end{array}$$

In matrix form, and then with partial pivoting to get rid of diagonal $0$s:

$$\left[\begin{array}{cccccc}1& -1& -1& 0& 0& 0\\ 0& 0& 1& -1& -1& 0\\ 0& 1& 0& 1& 0& -1\\ 0& 0& 5& 0& 105& 0\\ 0& 0& 5& 10& 0& 15\\ 0& 25& 0& 0& 0& 15\end{array}\right]\left\{\begin{array}{c}{I}_1\\ I_2\\ I_3\\ I_4\\ I_5\\ I_6\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\\ 0\\ 125\\ 125\\ 125\end{array}\right\}⟹\left[\begin{array}{cccccc}1& -1& -1& 0& 0& 0\\ 0& 1& 0& 1& 0& -1\\ 0& 0& 1& -1& -1& 0\\ 0& 0& 5& 10& 0& 15\\ 0& 0& 5& 0& 105& 0\\ 0& 25& 0& 0& 0& 15\end{array}\right]\left\{\begin{array}{c}{I}_1\\ I_2\\ I_3\\ I_4\\ I_5\\ I_6\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\\ 0\\ 125\\ 125\\ 125\end{array}\right\}$$

Using the initial relaxation parameter of $1$, the solution diverges. Higher relaxation parameter values ($1\le \omega \le 2$) also caused divergence. For values of $\omega$ below $1$:

$$\begin{array}{llllllllll}\text{Relaxation:}& 0.9& 0.8& 0.7& 0.6& 0.5& 0.4& 0.3& 0.2& 0.1\\ \text{Conv. Iterations:}& \text{N/A}& \text{N/A}& 2261& 116& 85& 82& 93& 128& 231\end{array}$$

Thus, the solution converge for values below approximately $\omega \approx 0.8$; the least iterations achieved for all tested values occurred for $\omega =0.4$. The Gauss-Seidel and MATLAB solutions are:

$$\begin{array}{lllllll}\text{Current (A)}& I_1& I_2& I_3& I_4& I_5& I_6\\ \text{Gauss-Seidel:}& 5.9999989& 1.9999967& 4.0000020& 3.0000017& 0.9999999& 5.0000024\\ \text{MATLAB:}& 6& 2& 4& 3& 1& 5\end{array}$$

Substituting into the original system of equations to verify correctness:

$$\begin{array}{ll}\text{Junction A: }6-2-4=0& \text{Loop 1: }5(4)+105(1)=125\\ \text{Junction B: }4-3-1=0& \text{Loop 2: }5(4)+10(3)+15(5)=125\\ \text{Junction C: }2+3-5=0& \text{Loop 3: }25(2)+15(5)=125\end{array}$$