Subjects Linear Algebra

Water Sector Pricing

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Water Sector Pricing


**Problem statement:** Three sectors A (Agriculture), U (Urban), I (Industry) consume and supply water interdependently with given usage coefficients. Let $p_1, p_2, p_3$ be prices per unit output for A, U, I respectively. **1. Constructing the system of linear equations:** Each sector's income equals its total water expenditure, so for sector $i$, income $p_i$ equals the sum of water consumed from each sector times that sector's price. From the table, consumption matrix $C$ is: $$C=\begin{bmatrix}0.4 & 0.2 & 0.3 \\ 0.2 & 0.6 & 0.4 \\ 0.4 & 0.2 & 0.3 \end{bmatrix}$$ Equations: 1. $p_1 = 0.4 p_1 + 0.2 p_2 + 0.3 p_3$ 2. $p_2 = 0.2 p_1 + 0.6 p_2 + 0.4 p_3$ 3. $p_3 = 0.4 p_1 + 0.2 p_2 + 0.3 p_3$ Rearranged: $$ \begin{cases} p_1 - 0.4 p_1 - 0.2 p_2 - 0.3 p_3 = 0 \\ p_2 - 0.2 p_1 - 0.6 p_2 - 0.4 p_3 = 0 \\ p_3 - 0.4 p_1 - 0.2 p_2 - 0.3 p_3 = 0 \end{cases} $$ Simplified: $$ \begin{cases} 0.6 p_1 - 0.2 p_2 - 0.3 p_3 = 0 \\ -0.2 p_1 + 0.4 p_2 - 0.4 p_3 = 0 \\ -0.4 p_1 - 0.2 p_2 + 0.7 p_3 = 0 \end{cases} $$ **2. Matrix form and solution:** Matrix form: $Mp=0$ with $$ M = \begin{bmatrix} 0.6 & -0.2 & -0.3 \\ -0.2 & 0.4 & -0.4 \\ -0.4 & -0.2 & 0.7 \end{bmatrix}, \quad p = \begin{bmatrix}p_1 \\ p_2 \\ p_3 \end{bmatrix} $$ Since trivial $p=0$ is trivial solution, we want nontrivial prices $p$ which satisfy $\det(M)=0$, indicative of eigenvector corresponding to eigenvalue 0. Using Gaussian elimination or matrix methods, solving system for ratios: From 1st eq: $$0.6 p_1 = 0.2 p_2 + 0.3 p_3 \Rightarrow p_1 = \frac{0.2}{0.6} p_2 + \frac{0.3}{0.6} p_3 = \frac{1}{3} p_2 + \frac{1}{2} p_3$$ Substitute $p_1$ into 2nd eq: $$-0.2 \left(\frac{1}{3} p_2 + \frac{1}{2} p_3\right) + 0.4 p_2 - 0.4 p_3 = 0$$ $$-\frac{0.2}{3} p_2 - 0.1 p_3 + 0.4 p_2 - 0.4 p_3=0$$ $$\left(0.4 - \frac{0.2}{3}\right) p_2 + (-0.1 - 0.4) p_3=0$$ $$\frac{1.2 - 0.2}{3} p_2 -0.5 p_3 = 0$$ $$\frac{1}{3} p_2 = 0.5 p_3 \Rightarrow p_2 = 1.5 p_3$$ Substitute $p_2=1.5 p_3$ into $p_1$: $$p_1 = \frac{1}{3} \times 1.5 p_3 + \frac{1}{2} p_3 = 0.5 p_3 + 0.5 p_3 = p_3$$ Check 3rd eq: $$-0.4 p_1 - 0.2 p_2 + 0.7 p_3= -0.4 p_3 - 0.2 (1.5 p_3) + 0.7 p_3 = (-0.4 -0.3 +0.7) p_3 = 0$$ Valid. Prices up to scale: $$p_1 : p_2 : p_3 = 1 : 1.5 : 1$$ **3. Economic interpretation:** These relative prices represent a consistent sustainable pricing scheme where each sector's income balances its water usage expenditure. The price for Urban ($p_2$) is higher reflecting its greater water interdependence. Such pricing can guide water resource allocation ensuring equitable and balanced sectoral demand and supply.