Leontieff - Yousef's Notes

#Rules

There exist equilibrium prices that can be assigned to the total outputs of the various sectors in such a way that the income of each sector exactly balances its expenses. A sector looks down a column to see where its output goes and it looks across a row to see what it needs as inputs.

#Example 1

A simple economy with 3 sectors: Coal, Electric, and Steel.

Coal receives 40% of Electric output & 60% Steel.
Electric receives 60% Coal, 10% Electric & 20% Steel.
Steel receives 40% Coal, 50% Electric & 20% Steel.

#Exchange Table

Output			Purchased by
Coal	Electric	Steel
0.0	0.4	0.6	Coal
0.6	0.1	0.2	Electric
0.4	0.5	0.2	Steel

#Equations

$$ \begin{align*} p_C &= 0p_C + 0.4p_E + 0.6p_S \\ p_E &= 0.6p_C + 0.1p_E + 0.2p_S \\ p_S &= 0.4p_C + 0.5p_E + 0.2p_S \end{align*} $$ Move unknowns to the left and numbers to the right $$ \begin{cases} p_C - 0.4p_E - 0.6p_S = 0 \\ - 0.6p_C + 0.9p_E - 0.2p_S = 0 \\ - 0.4p_C - 0.5p_E + 0.8p_S = 0 \end{cases} $$ Matrix form (Reduced Row Echelon Form) $$ \begin{bmatrix} 10 & -4 & -6 & 0 \\ - 6 & 9 & -2 & 0 \\ - 4 & -5 & 8 & 0 \end{bmatrix} \rightarrow \text{RREF} \rightarrow \begin{bmatrix} 1 & 0 & -0.94 & 0 \\ 0 & 1 & -0.85 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$

Move to equations to solve

Only 2 equations because the last row is all 0’s:

$$ \begin{align*} p_C - 0.94p_S &= 0 \\ p_E - 0.85p_S &= 0 \end{align*} $$ Solutions dependent on one variable: $$ \begin{align*} p_C &= 0.94p_S \\ p_E &= 0.85p_S \\ p_S &= p_S \end{align*} $$ For $( p_S = 1 )$, the other prices are proportional: $$ \begin{align*} p_C &= 0.94 \\ p_E &= 0.85 \\ p_S &= 1 \end{align*} $$

#Example 2 (Video)

put video here