1. **Stating the problem:** We have a Markov chain with transition matrix $$P = \begin{bmatrix}0.825 & 0.175 & 0.000 \\ 0.060 & 0.919 & 0.021 \\ 0.049 & 0.000 & 0.951 \end{bmatrix}$$ representing transitions between energy sources O, G, and E. Initial distribution vector (1995) is $$\mathbf{v}_0 = \begin{bmatrix}0.25 & 0.60 & 0.15\end{bmatrix}$$. We want to find the distribution after 1, 2, 3 years and the long-run equilibrium vector. 2. **Formula and rules:** The distribution after $n$ years is $$\mathbf{v}_n = \mathbf{v}_0 P^n$$. The long-run equilibrium vector $$\mathbf{v}^*$$ satisfies $$\mathbf{v}^* = \mathbf{v}^* P$$ and $$\sum_i v_i^* = 1$$. 3. **Calculations:** - After 1 year: $$\mathbf{v}_1 = \mathbf{v}_0 P = \begin{bmatrix}0.25 & 0.60 & 0.15\end{bmatrix} \begin{bmatrix}0.825 & 0.175 & 0.000 \\ 0.060 & 0.919 & 0.021 \\ 0.049 & 0.000 & 0.951 \end{bmatrix}$$ Calculate each component: $$v_{1,O} = 0.25 \times 0.825 + 0.60 \times 0.060 + 0.15 \times 0.049 = 0.20625 + 0.036 + 0.00735 = 0.2496$$ $$v_{1,G} = 0.25 \times 0.175 + 0.60 \times 0.919 + 0.15 \times 0.000 = 0.04375 + 0.5514 + 0 = 0.59515$$ $$v_{1,E} = 0.25 \times 0.000 + 0.60 \times 0.021 + 0.15 \times 0.951 = 0 + 0.0126 + 0.14265 = 0.15525$$ So, $$\mathbf{v}_1 = \begin{bmatrix}0.2496 & 0.59515 & 0.15525\end{bmatrix}$$ - After 2 years: $$\mathbf{v}_2 = \mathbf{v}_1 P$$ Calculate each component: $$v_{2,O} = 0.2496 \times 0.825 + 0.59515 \times 0.060 + 0.15525 \times 0.049 = 0.20592 + 0.035709 + 0.007611 = 0.24924$$ $$v_{2,G} = 0.2496 \times 0.175 + 0.59515 \times 0.919 + 0.15525 \times 0.000 = 0.04368 + 0.54658 + 0 = 0.59026$$ $$v_{2,E} = 0.2496 \times 0.000 + 0.59515 \times 0.021 + 0.15525 \times 0.951 = 0 + 0.012498 + 0.14768 = 0.16018$$ So, $$\mathbf{v}_2 = \begin{bmatrix}0.24924 & 0.59026 & 0.16018\end{bmatrix}$$ - After 3 years: $$\mathbf{v}_3 = \mathbf{v}_2 P$$ Calculate each component: $$v_{3,O} = 0.24924 \times 0.825 + 0.59026 \times 0.060 + 0.16018 \times 0.049 = 0.20559 + 0.03542 + 0.00785 = 0.24886$$ $$v_{3,G} = 0.24924 \times 0.175 + 0.59026 \times 0.919 + 0.16018 \times 0.000 = 0.04362 + 0.54244 + 0 = 0.58606$$ $$v_{3,E} = 0.24924 \times 0.000 + 0.59026 \times 0.021 + 0.16018 \times 0.951 = 0 + 0.01240 + 0.15238 = 0.16478$$ So, $$\mathbf{v}_3 = \begin{bmatrix}0.24886 & 0.58606 & 0.16478\end{bmatrix}$$ 4. **Long-run equilibrium vector:** Solve $$\mathbf{v}^* = \mathbf{v}^* P$$ with $$v_O^* + v_G^* + v_E^* = 1$$. Set $$\mathbf{v}^* = \begin{bmatrix}x & y & z\end{bmatrix}$$. From $$\mathbf{v}^* = \mathbf{v}^* P$$: $$x = 0.825x + 0.060y + 0.049z$$ $$y = 0.175x + 0.919y + 0.000z$$ $$z = 0.000x + 0.021y + 0.951z$$ Rewrite: $$x - 0.825x - 0.060y - 0.049z = 0 \Rightarrow 0.175x - 0.060y - 0.049z = 0$$ $$y - 0.175x - 0.919y = 0 \Rightarrow -0.175x + 0.081y = 0$$ $$z - 0.021y - 0.951z = 0 \Rightarrow -0.021y + 0.049z = 0$$ From second equation: $$0.081y = 0.175x \Rightarrow y = \frac{0.175}{0.081} x \approx 2.1605 x$$ From third equation: $$0.049z = 0.021y \Rightarrow z = \frac{0.021}{0.049} y \approx 0.4286 y$$ Substitute $$y$$ and $$z$$ in first equation: $$0.175x - 0.060(2.1605x) - 0.049(0.4286 \times 2.1605 x) = 0$$ Calculate: $$0.175x - 0.12963x - 0.0454x = 0$$ $$0.175x - 0.175x = 0$$ This confirms consistency. Use normalization: $$x + y + z = 1$$ Substitute $$y$$ and $$z$$: $$x + 2.1605x + 0.4286 \times 2.1605 x = 1$$ Calculate: $$x + 2.1605x + 0.925 x = 1$$ $$4.0855 x = 1 \Rightarrow x = \frac{1}{4.0855} \approx 0.2447$$ Then: $$y = 2.1605 \times 0.2447 \approx 0.5285$$ $$z = 0.4286 \times 0.5285 \approx 0.2268$$ 5. **Final answers:** - After 1 year: $$\mathbf{v}_1 = \begin{bmatrix}24.96\% & 59.52\% & 15.53\%\end{bmatrix}$$ - After 2 years: $$\mathbf{v}_2 = \begin{bmatrix}24.92\% & 59.03\% & 16.02\%\end{bmatrix}$$ - After 3 years: $$\mathbf{v}_3 = \begin{bmatrix}24.89\% & 58.61\% & 16.48\%\end{bmatrix}$$ - Long-run equilibrium: $$\mathbf{v}^* = \begin{bmatrix}24.47\% & 52.85\% & 22.68\%\end{bmatrix}$$ This means in the long run, about 24.47% use energy O, 52.85% use G, and 22.68% use E.