Ch02. Matrix Algebra

2.6 The Leontief Input-Output Model

Linear algebra played an essential role in the Nobel prize-winning work of Wassily Leontief.

• Suppose a nation’s economy is divided into $n$ sectors that produce goods or services, and let $\textbf{x}$ be a production vector in $\mathbb{R}^n$ that lists the output of each sector for one year.
• Also, suppose another part of the economy (called the open sector) does not produce goods or services but only consumes them, and let $\textbf{d}$ be a final demand vector (or bill of final demands) that lists the values of the goods and services demanded from the various sectors by the nonproductive part of the economy.
• As the various sectors produce goods to meet consumer demand, the producers themselves create additional intermediate demand for goods they need as inputs for their own production.
• Leontief asked if there is a production level $\textbf{x}$ such that the amounts produced (or “supplied”) will exactly balance the total demand for that production, so that

$\left\{ \begin{matrix} \text{amount produced}\\ \textbf{x} \end{matrix}\right\} = \left\{ \begin{matrix} \text{intermediate} \\ \text{demand} \end{matrix} \right\} + \left\{ \begin{matrix} \text{final demand}\\ \textbf{d} \end{matrix} \right\} \tag{1}$

• The basic assumption of Leontief’s input-output model is that for each sector, there is a unit consumption vector in that lists the inputs needed per unit of output of the sector.
• As a simple example, suppose the economy consists of three sectors($n=3$)—manufacturing, agriculture, and services—with unit consumption vectors $\textbf{c}_1, \textbf{c}_2,$ and $\textbf{c}_3$ as shown in the table that follows.

Inputs Consumed per Unit of Output

$\textbf{c}_1 = \begin{bmatrix} .5 \\ .2 \\ .1 \end{bmatrix} \\ \textbf{c}_2 = \begin{bmatrix} .4 \\ .3 \\ .1 \end{bmatrix} \\ \textbf{c}_3 = \begin{bmatrix} .2 \\ .1 \\ .3 \end{bmatrix} \\ $

Example 1

What amounts will be consumed by the manufacturing sector if it decides to produce 100 units?

Solution

Compute

$100\textbf{c}_1 = 100 \begin{bmatrix} .5 \\ .2 \\ .1 \end{bmatrix} = \begin{bmatrix} 50 \\ 20 \\ 10 \end{bmatrix}$

• To produce 100 units, manufacturing will order (i.e., “demand”) and consume 50 units from other parts of the manufacturing sector, 20 units from agriculture, and 10 units from services.
• If manufacturing decides to produce $x_1$ units of output, then $x_1\textbf{c}_1$ represents the intermediate demands of manufacturing, because the amounts in $x_1\textbf{c}_1$ will be consumed in the process of creating the $x_1$ units of output.
• Likewise, if $x_2$ and $x_3$ denote the planed outputs of the agriculture and services sectors, $x_2\textbf{c}_2$ and $x_3\textbf{c}_3$ list their corresponding intermediate demands.
• The total intermediate demand from all three sectors is given by $\begin{aligned} \left\{ \begin{matrix}\text{intermediate} \text{demand} \end{matrix}\right\} &= x_1\textbf{c}_1+x_2\textbf{c}_2+x_3\textbf{c}_3 \\ &= C\textbf{x} \end{aligned} \tag{2}$
  • where $C$ is the consumer matrix $\begin{bmatrix} \textbf{c}_1 & \textbf{c}_2 & \textbf{c}_3 \end{bmatrix}$, namely, $C=\begin{bmatrix} .5 & .4 & .2 \\ .2 & .3 & .1 \\ .1 & .1 & .3 \end{bmatrix} \tag{3}$
• Equation (1) and (2) yield Leontief's Model. $\textbf{x} = C\textbf{x} + \textbf{d} \tag{4}$
  • $\textbf{x}$ is the amount produced.
  • $C\textbf{x}$ is the intermediate demand.
  • $\textbf{d}$ is the final demand.
• Equation (4) may also be written as $I\textbf{x}-C\textbf{x}=\textbf{d}$, or $(I-C)\textbf{x}=\textbf{d} \tag{5}$

Example 2

Consider the economy whose consumption matrix is given by (3). Suppose the final demand is 50 units for manufacturing, 30 units for agriculture, and 20 units for services. Find the production level $\textbf{x}$ that will satisfy this demand.

Solution

The coefficient matrix in (5) is

$\begin{aligned} I-C &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} .5 & .4 & .2 \\ .2 & .3 & .1 \\ .1 & .1 & .3 \end{bmatrix} \\ &= \begin{bmatrix} .5 & -.4 & -.2 \\ -.2 & .7 & -.1 \\ -.1 & -.1 & .7 \end{bmatrix} \end{aligned}$

• To solve (5), row reduce the augmented matrix

$\begin{bmatrix} .5 & -.4 & -.2 & 50\\ -.2 & .7 & -.1 & 30\\ -.1 & -.1 & .7 & 20 \end{bmatrix} \sim \begin{bmatrix} 5 & -4 & -2 & 500\\ -2 & 7 & -1 & 300\\ -1 & -1 & 7 & 200 \end{bmatrix} \sim \cdots \sim \begin{bmatrix} 1 & 0 & 0 & 226\\ 0 & 1 & 0 & 119\\ 0 & 0 & 1 & 78 \end{bmatrix}$

• The last column is rounded to the nearest whole unit. Manufacturing must produce approximately 226 units, agriculture 119 units, and services only 78 units.

Theorem 11 :

Let $C$ be the consumption matrix for an economy, and let $\textbf{d}$ be the final demand. If $C$ and $\textbf{d}$ have nonnegative entries and if each column sum of $C$ is less than 1, then $(I-C)^{-1}$ exists and the production vector $\textbf{x} = (I-C)^{-1} \textbf{d}$ has nonnegative entries and is the unique solution of $\textbf{x} = C\textbf{x} + \textbf{d}$

• In the theorem, the term column sum denotes the sum of the entries in a column of a matrix.
• Under ordinary circumstances, the column sums of a consumption matrix are less than 1 because a sector should require less than one unit’s worth of inputs to produce one unit of output.

A Formula for $(I-C)^{-1}$

• Imagine that the demand represented by $\textbf{d}$ is presented to the various industries at the beginning of the year, and the industries respond by setting their production levels at $\textbf{x} = \textbf{d}$, which will exactly meet the final demand. As the industries prepare to produce $\textbf{d}$, they send out orders for their raw materials and other inputs. This creates an intermediate demand of $C\textbf{d}$ for inputs.
• To meet the additional demand of $C\textbf{d}$, the industries will need as additional inputs the amounts in $C(C\textbf{d})= C^2\textbf{d}$. Of course, this creates a second round of intermediate demand, and when the industries decide to produce even more to meet this new demand, they create a third round of demand, namely, $C(C^2\textbf{d}) = C^3\textbf{d}$. And so it goes.
• Theoretically, this process could continue indefinitely, although in real life it would not take place in such a rigid sequence of events. We can diagram this hypothetical situation as follows:

• The production level $\textbf{x}$ that will meet all of this demand is

$\begin{aligned} \textbf{x} &= \textbf{d} + C\textbf{d} + C^2\textbf{d} + C^3\textbf{d} + \cdots \\ &= (I+C+C^2+C^3+\cdots)\textbf{d} \end{aligned} \tag{6}$

• To make sense of equation (6), consider the following algebraic identity:

$(I-C)(I+C+C^2+\cdots+C^m)=I-C^{m+1} \tag{7}$

It can be shown that if the column sums in $C$ are all strictly less than 1, then $I-C$ is invertible, $C^m$ approaches the zero matrix as $m$ gets arbitrarily large, and $I-C^{m+1} \rightarrow I$.(This fact is analogous to the fact that if a positive number $t$ is less than 1, then $t^m \rightarrow 0$ as $m$ increases.) Using equation (7), write

$(I-C)^{-1} \approx I + C + C^2+C^3+\cdots+C^m \tag{8}$

when the column sums of $C$ are less than 1.

• The approximation in (8) means that the right side can be made as close to $(I-C)^{-1}$ as desired by taking $m$ sufficiently large.
• In actual input–output models, powers of the consumption matrix approach the zero matrix rather quickly. So (8) really provides a practical way to compute $(I-C)^{-1}$.
• Likewise, for any $\textbf{d}$, the vectors $C^m\textbf{d}$ approach the zero vector quickly, and (6) is a practical way to solve $(I-C)\textbf{x}= \textbf{d}$. If the entries in $C$ and $\textbf{d}$ are nonnegative, then (6) shows that the entries in $\textbf{x}$ are nonnegative, too.