Line flows by G-S after one iteration

The line values $0.01 + j0.05$ pu etc. have the size of series impedances (as admittances they would be far too small), so they are treated as series impedances. Bus 1 is the slack bus ($1.02\angle 0^\circ$); buses 2 and 3 are load buses: $P_2 = -0.8$, $Q_2 = -0.4$, $P_3 = -1.0$, $Q_3 = -0.6$ pu. Flat start $V_2^{(0)} = V_3^{(0)} = 1\angle 0^\circ$.

Y-bus

$$\begin{aligned} y_{12} &= \frac{1}{0.010 + j0.050} = 3.8462 - j19.2308 \\ y_{23} &= \frac{1}{0.007 + j0.037} = 4.9365 - j26.0931 \\ y_{31} &= \frac{1}{0.012 + j0.064} = 2.8302 - j15.0943 \end{aligned}$$ $$Y_{bus} = \begin{bmatrix} 6.6763 - j34.3251 & -3.8462 + j19.2308 & -2.8302 + j15.0943 \\ -3.8462 + j19.2308 & 8.7827 - j45.3239 & -4.9365 + j26.0931 \\ -2.8302 + j15.0943 & -4.9365 + j26.0931 & 7.7667 - j41.1874 \end{bmatrix}\ \text{pu}$$

Gauss-Seidel: first iteration

Iteration 1

$$\begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{-0.8000 + j0.4000}{1.0000} = -0.8000 + j0.4000 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -8.8596 + j45.7085 \\ V_2^{(1)} &= \frac{(-0.8000 + j0.4000) - (-8.8596 + j45.7085)}{8.7827 - j45.3239} = 0.9967 - j0.0153 \\ &= 0.9968\angle -0.8802^\circ\ \text{pu} \end{aligned}$$ $$\begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.0000 + j0.6000}{1.0000} = -1.0000 + j0.6000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -7.4074 + j41.4786 \\ V_3^{(1)} &= \frac{(-1.0000 + j0.6000) - (-7.4074 + j41.4786)}{7.7667 - j41.1874} = 0.9868 - j0.0305 \\ &= 0.9872\angle -1.7706^\circ\ \text{pu} \end{aligned}$$

Voltages after one iteration: $V_1 = 1.02\angle 0^\circ$, $V_2 = 0.9968\angle -0.8802^\circ$, $V_3 = 0.9872\angle -1.7706^\circ$ pu.

Line flows

$S_{ij} = V_iI_{ij}^{*}$ with $I_{ij} = (V_i - V_j)y_{ij}$ (no shunt admittance). For line 1-2:

$$\begin{aligned} I_{12} &= (V_1 - V_2)y_{12} = (1.0200 - (0.9967 - j0.0153))(3.8462 - j19.2308) = 0.3841 - j0.3893 \\ S_{12} &= V_1I_{12}^{*} = 0.3918 + j0.3971\ \text{pu} \\ S_{21} &= V_2(-I_{12})^{*} = -0.3888 - j0.3821\ \text{pu} \end{aligned}$$

All lines:

Total loss $= 0.0104 + j0.0543$ pu. Slack bus injection $S_1 = S_{12} + S_{13} = 0.9574 + j0.8210$ pu.

Answer: $S_{12} = 0.3918 + j0.3971$, $S_{23}$ and $S_{31}$ as in the table; total line loss $= 0.0104$ pu real and 0.0543 pu reactive (with first-iteration voltages).