Economics at Zero Marginal Cost: Leontief Input-Output Model

3. Suppose that a hypothetical economy depends on only two industries, the electric company E and the water company W. Output for each company is measured in dollars. The electric company uses (input) both electricity and water in the production (output) of electricity. The water company uses (input) both electricity and water in the production (output) of water. Suppose that the production of each dollar's worth of electricity requires $0.30 worth of electricity and $0.10 worth of water. The production of each dollar's worth of water requires $0.20 worth of electricity and $0.40 worth of water. If the final demand from all other users of electricity and water is $12 million for electricity and $8 million for water. How much electricity and water should be produced?

Solution:

Given:

By question,

E -> Electric company and W -> Water company

$0.30 of E and $0.10 of W is used for producing $1 of E

$0.20 of E and $0.40 of W is used for producing $1 of W

The final demand for electricity (E) and water (O) are $12 billion and $8 billion respectively.

To find: Output levels for electric company (E) and water company (W).

Representing the above given information in an input-output table, we have

Rewriting the above table in equation form, we have

E=0.30.E+0.20.W+12

W=0.10.E+0.40.W+8

Rearranging the above equation in matrix form, we have

$\begin{bmatrix} E\\W \end{bmatrix}= \begin{bmatrix} 0.30\ \ \ \ 0.20\\ 0.10\ \ \ \ 0.40\\ \end{bmatrix}\begin{bmatrix} E\\W \end{bmatrix}+ \begin{bmatrix} 12\\8 \end{bmatrix}$

$\\ Let,\ Output\ levels,\ X=\begin{bmatrix} E\\W \end{bmatrix},\ Input\ co-efficient\ matrix,\ A= \begin{bmatrix} 0.30\ \ \ \ 0.20\\ 0.10\ \ \ \ 0.40\\ \end{bmatrix}$

$\ and\ Final\ demand\ vectr,\ F= \begin{bmatrix} 12\\8 \end{bmatrix}\$

Rewriting the above in notational form, we have

$X=AX+F\\ or,\ X-AX=F\\ or,\ IX-AX=F\ \ \(\because I.X=X)\\ or,\ (I-A)X=F\\$

Multiplying both sides by $(I-A)^{-1}$

This above equation is known as the Leontief Input-Output model equation where $(I-A)^{-1}$ is the inverse of Leontief matrix (I−A).

Computing the Leontief matrix (I−A),

$I-A=\begin{bmatrix} 1\ \ \ \ \ 0\\ 0\ \ \ \ \ 1\\ \end{bmatrix}-\begin{bmatrix} 0.30\ \ \ \ \ 0.20\\ 0.10\ \ \ \ \ 0.40\\ \end{bmatrix}$

$I-A=\begin{bmatrix} 1-0.30\ \ \ \ \ 0-0.20\\ 0-0.10\ \ \ \ \ 1-0.40\\ \end{bmatrix}$

$I-A=\begin{bmatrix} 0.70\ \ \ \ \ -0.20\\ -0.10\ \ \ \ \ \ \0.60\\ \end{bmatrix}$

Computing the inverse of I−A, i.e., $(I-A)^{-1}$ ,

$(I-A)^{-1}=\frac{1}{\left | I-A \right |}.Adj \left (I-A \right )$

$\\Co-factor\ of\ a_{11}, i.e.,\ 0.70=0.60\\ Co-factor\ of\ a_{12}, i.e.,\ -0.20=-(-0.10)=0.10\\ Co-factor\ of\ a_{21}, i.e.,\ -0.10=-(-0.20)=0.20\\ Co-factor\ of\ a_{22}, i.e.,\ 0.60=0.70\\$

$Co-factor\ matrix\ (I-A)=\begin{bmatrix} 0.60\ \ \ \ \ 0.10\\ 0.20\ \ \ \ \ 0.70\\ \end{bmatrix}$

$Adjoint\ matrix,\ Adj. (I-A)=\begin{bmatrix} 0.60\ \ \ \ \ 0.20\\ 0.10\ \ \ \ \ 0.70 \end{bmatrix}\\$

Now, $(I-A)^{-1}=\frac{1}{\left | I-A \right |}.Adj \left (I-A \right )$

$(I-A)^{-1}=\frac{1}{0.40}\begin{bmatrix} 0.60\ \ \ \ \ 0.20\\ 0.10\ \ \ \ \ \0.70\ \end{bmatrix}\$

Next, $X=(I-A)^{-1}.F$

$X^*=\frac{1}{0.40}\begin{bmatrix} 0.60\ \ \ \ \ 0.20\\ 0.10\ \ \ \ \ 0.70\\ \end{bmatrix}\begin{bmatrix} 12\\ 8 \end{bmatrix}\$

$X^*=\frac{1}{0.40}\begin{bmatrix} 0.60\times12 +0.20\times8 \\ 0.10\times12 +0.70\times8 \\ \end{bmatrix}\$

$X^*=\frac{1}{0.40}\begin{bmatrix} 7.2+1.6 \\ 1.2+5.6 \\ \end{bmatrix}\$

$X^*=\frac{1}{0.40}\begin{bmatrix} 8.8\\ 6.8\ \end{bmatrix}$

$X^*=\begin{bmatrix} E\\W \end{bmatrix}=\begin{bmatrix} 22\\ 17\ \end{bmatrix}$

∴The output levels for electric company (E) and water company (W) are $22 billion and $17 billion respectively.

Economics at Zero Marginal Cost

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Wednesday, June 10, 2020

Leontief Input-Output Model - Problem and Solution 3

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