Economics at Zero Marginal Cost: Leontief Input-Output Model

2. A large energy company produces electricity, natural gas and oil. The production of a dollar's worth of electricity requires input of $0.30 from electricity, $0.10 from natural gas and $0.20 from oil. The production of a dollar's worth of natural gas requires input of $0.30 from electricity, $0.10 from natural gas and $0.20 from oil. Production of a dollar's worth of oil requires input of $0.10 from each sector. Find the output for each sector that is needed to satisfy a final demand of $25 billion for electricity, $15 billion for natural gas and $20 billion for oil.

Solution:

Given:

Let E -> Electricity N -> Natural Gas and O -> Oil

By question,

$0.30 of E, $0.10 of N and $0.20 of O is used for producing $1 of E

$0.30 of E, $0.10 of N and $0.20 of O is used for producing $1 of N

$0.10 of E, $0.10 of N and $0.10 of O is used for producing $1 of O

The final demand for electricity (E), natural gas (N) and Oil (O) are $25 billion, $15 billion and $20 billion respectively.

To find: Output levels for the three sectors.

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+0.30.N+0.10.O+25

N=0.10.E+0.10.N+0.10.O+15

O=0.20.E+0.20.N+0.30.O+20

Rearranging the above equation in matrix form, we have

$\begin{bmatrix} E\\N\\O \end{bmatrix}= \begin{bmatrix} 0.30\ \ \ \ \ 0.30 \ \ \ \ \ 0.10\\ 0.10\ \ \ \ \ 0.10\ \ \ \ \ 0.10\\ 0.20\ \ \ \ \ 0.20\ \ \ \ \ 0.10\\ \end{bmatrix}\begin{bmatrix} E\\N\\O \end{bmatrix}+ \begin{bmatrix} 25\\15\\20 \end{bmatrix}$

$\\ Let,\ Output\ levels,\ X=\begin{bmatrix} E\\N\\O \end{bmatrix}\\$

$\\ Input\ co-efficient\ matrix,\ A= \begin{bmatrix} 0.30\ \ \ \ \ 0.30 \ \ \ \ \ 0.10\\ 0.10\ \ \ \ \ 0.10\ \ \ \ \ 0.10\\ 0.20\ \ \ \ \ 0.20\ \ \ \ \ 0.10\\ \end{bmatrix}$

$\\ and\ Final\ demand\ vector,\ F= \begin{bmatrix} 25\\15\\20 \end{bmatrix}\\$

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),

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

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

$\\I-A=\begin{bmatrix} 0.70\ \ \ \ \ \ -0.30 \ \ \ \ \ -0.10\\ -0.10\ \ \ \ \ \ \ \ 0.90 \ \ \ \ \ -0.10\\ -0.20\ \ \ \ -0.20 \ \ \ \ \ \ \ \ \ 0.90\\ \end{bmatrix}\$

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

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

$\\\left |I-A \right |=0.70\begin{vmatrix} 0.90\ \ \ \ -0.10\\ -0.20\ \ \ \ \0.90 \end{vmatrix}-(-0.30) \begin{vmatrix} -0.10\ \ -0.10\\ -0.20\ \ \ \ \0.90 \end{vmatrix}+(-0.10) \begin{vmatrix} -0.10\ \ \ \ \ \ 0.90\\ -0.20\ \ \ -0.20 \end{vmatrix}\\\ \\\left |I-A \right |=0.70(0.81-0.02)+0.30(-0.09-0.02)-0.10(0.02-(-0.18))\\ \left |I-A \right |=0.70(0.79)+0.30(-0.11)-0.10(0.02-(0.02+0.18))\\ \left |I-A \right |=0.553-0.033-0.10(0.20))\\ \left |I-A \right |=0.553-0.033-0.02\\ \left |I-A \right |=0.5$

$\\Co-factor\ of\ a_{11}, i.e.,\ 0.70=\begin{vmatrix} 0.90\ \ \ -0.10\\ -0.20\ \ \ \ \ 0.90\\ \end{vmatrix}=0.81-0.02=0.79\\ Co-factor\ of\ a_{12}, i.e.,\ -0.30=-\begin{vmatrix} -0.10\ \ \ -0.10\\ -0.20\ \ \ \ \ \ 0.90\\ \end{vmatrix}=-(-0.09-0.02)=-(-0.11)=0.11\\ Co-factor\ of\ a_{13},i.e.,\ -0.10=\begin{vmatrix} -0.10\ \ \ \0.90\\ -0.20\ -0.20\\ \end{vmatrix}=0.2-(-0.18)=0.02+0.18=0.20\\$

$\\Co-factor\ of\ a_{21}, i.e.,\ -0.10=-\begin{vmatrix} -0.30\ \ \ -0.10\\ -0.20\ \ \ \ \ 0.90\\ \end{vmatrix}=-(-0.27-0.02)=-(-0.29)=0.29\\ Co-factor\ of\ a_{22}, i.e.,\ 0.90=\begin{vmatrix} 0.70\ \ \ -0.10\\ -0.20\ \ \ \ \ \ 0.90\\ \end{vmatrix}=0.63-0.02=0.61\\ Co-factor\ of\ a_{23},i.e.,\ -0.10=-\begin{vmatrix} 0.70\ \ \ \ -0.30\\ -0.20\ \ -0.20\\ \end{vmatrix}=-(--0.14-0.06)=-(-0.20)=0.20\\$

$\\Co-factor\ of\ a_{31},i.e.,\ -0.20=\begin{vmatrix} -0.30\ \ \ \ -0.10\\ 0.90\ \ \ \ \ \ -0.10\\ \end{vmatrix}=0.03-(-0.09)=0.03+0.09=0.12\\ Co-factor\ of\ a_{32},i.e.,\ -0.20=-\begin{vmatrix} 0.70\ \ \ \ -0.10\\ -0.10\ \ -0.10\\ \end{vmatrix}=-(-0.07-0.01)=-(-0.08)=0.08\\ Co-factor\ of\ a_{33},i.e.,\ 0.90=\begin{vmatrix} 0.70\ \ \ \ -0.30\\ -0.10\ \ \ \ \ 0.90\\ \end{vmatrix}=0.63-0.03=0.60\\$

$Co-factor\ matrix\ (I-A)=\begin{bmatrix} 0.79\ \ \ \ \ 0.11\ \ \ \ \ 0.20\\ 0.29\ \ \ \ \ 0.61\ \ \ \ \ 0.20\\ 0.12\ \ \ \ \ 0.08\ \ \ \ \ 0.60\\ \end{bmatrix}$

$Adjoint\ matrix,\ Adj. (I-A)=\begin{bmatrix} 0.79\ \ \ \ \ 0.29\ \ \ \ \ 0.12\\ 0.11\ \ \ \ \ 0.61\ \ \ \ \ 0.08\\ 0.20\ \ \ \ \ 0.20\ \ \ \ \ 0.60\\ \end{bmatrix}\$

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

$(I-A)^{-1}=\frac{1}{0.5}\begin{bmatrix} 0.79\ \ \ \ \ 0.29\ \ \ \ \ 0.12\\ 0.11\ \ \ \ \ 0.61\ \ \ \ \ 0.08\\ 0.20\ \ \ \ \ 0.20\ \ \ \ \ 0.60\\ \end{bmatrix}$

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

$X^*=\frac{1}{0.5}\begin{bmatrix} 0.79\ \ \ \ \ 0.29\ \ \ \ \ 0.12\\ 0.11\ \ \ \ \ 0.61\ \ \ \ \ 0.08\\ 0.20\ \ \ \ \ 0.20\ \ \ \ \ 0.60\\ \end{bmatrix}\begin{bmatrix} 25\\15\\20 \end{bmatrix}\$

$X^*=\frac{1}{0.5}\begin{bmatrix} 0.79\times25 \ \ \ \ \ 0.29\times15 \ \ \ \ \ 0.12\times20\\ 0.11\times25\ \ \ \ \ 0.61\times15 \ \ \ \ \ 0.08\times20\\ 0.20\times125\ \ \ \ \ 0.20\times15\ \ \ \ 0.60\times20\\ \end{bmatrix}$