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Exercise E · Q3
Q.

Solve the following problem using Leontief input-output model.

Sector 1Sector 2Total
Sector 1122040
Sector 2152030

If the system is viable then discuss the situation for new demand 8 and 8 from sector 1 and sector 2 respectively.

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The technology matrix fails the Hawkins–Simon test (det⁡(I−A)<0\det(I-A)<0), so the economy cannot meet any positive final demand; solving anyway yields negative outputs.

aij=xijXja_{ij}=\dfrac{x_{ij}}{X_j}; viable iff every leading principal minor of (I−A)(I-A) is positive. If not, X=(I−A)−1DX=(I-A)^{-1}D has negative components — economically infeasible.

Totals X1=40,  X2=30X_1=40,\;X_2=30:

From \ ToS1S2Total
S1122040
S2152030
  1. Coefficients: a11=1240=310,  a21=1540=38,  a12=2030=23,  a22=2030=23.a_{11}=\dfrac{12}{40}=\dfrac{3}{10},\;a_{21}=\dfrac{15}{40}=\dfrac38,\;a_{12}=\dfrac{20}{30}=\dfrac23,\;a_{22}=\dfrac{20}{30}=\dfrac23.

A=[3/102/33/82/3].A=\begin{bmatrix}3/10&2/3\\3/8&2/3\end{bmatrix}.

  1. I−A=[7/10−2/3−3/81/3].I-A=\begin{bmatrix}7/10&-2/3\\-3/8&1/3\end{bmatrix}.
  2. det⁡(I−A)=710⋅13−23⋅38=730−14=14−1560=−160<0.\det(I-A)=\dfrac{7}{10}\cdot\dfrac13-\dfrac23\cdot\dfrac38=\dfrac{7}{30}-\dfrac14=\dfrac{14-15}{60}=-\dfrac{1}{60}<0.
  3. Since det⁡(I−A)<0\det(I-A)<0, the Hawkins–Simon conditions fail ⇒\Rightarrow the system is not viable. …

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