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Exercise E · Q4
Q.

Solve the following problem using Leontief input-output model.

Sector 1Sector 2Total
Sector 15730
Sector 261421

If the system is viable then discuss the situation for new demand 8 and 12 from sector 1 and sector 2 respectively.

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Build the technology matrix, confirm viability, then solve X=(I−A)−1DX=(I-A)^{-1}D for the new demand (8,12)(8,12).

aij=xijXja_{ij}=\dfrac{x_{ij}}{X_j}; gross output X=(I−A)−1DX=(I-A)^{-1}D; viable when the leading principal minors of (I−A)(I-A) are positive.

Totals X1=30,  X2=21X_1=30,\;X_2=21:

From \ ToS1S2Total
S15730
S261421
  1. Coefficients: a11=530=16,  a21=630=15,  a12=721=13,  a22=1421=23.a_{11}=\dfrac{5}{30}=\dfrac16,\;a_{21}=\dfrac{6}{30}=\dfrac15,\;a_{12}=\dfrac{7}{21}=\dfrac13,\;a_{22}=\dfrac{14}{21}=\dfrac23.

A=[1/61/31/52/3].A=\begin{bmatrix}1/6&1/3\\1/5&2/3\end{bmatrix}.

  1. I−A=[5/6−1/3−1/51/3].I-A=\begin{bmatrix}5/6&-1/3\\-1/5&1/3\end{bmatrix}.
  2. Viability: 56>0\tfrac56>0 and det⁡(I−A)=56⋅13−13⋅15=518−115=25−690=1990>0⇒\det(I-A)=\dfrac56\cdot\dfrac13-\dfrac13\cdot\dfrac15=\dfrac{5}{18}-\dfrac{1}{15}=\dfrac{25-6}{90}=\dfrac{19}{90}>0\Rightarrow viable.
  3. (I−A)−1=119/90[1/31/31/55/6]=9019[1/31/31/55/6].(I-A)^{-1}=\dfrac{1}{19/90}\begin{bmatrix}1/3&1/3\\1/5&5/6\end{bmatrix}=\dfrac{90}{19}\begin{bmatrix}1/3&1/3\\1/5&5/6\end{bmatrix}.
  4. D=[812]D=\begin{bmatrix}8\\12\end{bmatrix}: …

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