2.6 The Leontief input-output model

Big idea

Sectors of an economy buy from each other. Making steel consumes electricity, making electricity consumes steel, and both consume transport. So the output you need is not just the output people want to buy; it also includes everything consumed along the way, including the output consumed while producing the output consumed while producing it.

That regress is what a linear model handles cleanly. Fix the amount of each sector’s output consumed per unit produced by each other sector, and the internal consumption becomes a matrix times the production vector. Total production has to cover internal consumption plus outside demand, which is one vector equation: $\mathbf{x} = C\mathbf{x} + \mathbf{d}$.

Rearranged, that is $(I - C)\mathbf{x} = \mathbf{d}$, an ordinary square system. When $I - C$ is invertible the answer is $\mathbf{x} = (I - C)^{-1}\mathbf{d}$, and here the inverse earns its keep. A planner faces many demand vectors and one fixed technology, so the matrix is computed once and reused, and its columns answer the question a planner really faces: how much more of everything does one more unit of this product cost.

Decoder

Three vectors carry the whole model and they are easy to confuse. The production vector $\mathbf{x}$ lists what each sector makes in total. The final demand vector $\mathbf{d}$ lists what is consumed by everyone outside the sectors, which is households, government, exports. The intermediate demand $C\mathbf{x}$ lists what the sectors consume from each other. Total output splits into those last two parts and nothing else.

The consumption matrix is read down its columns. Column $j$ says what sector $j$ consumes, per unit of its own output, from each sector in turn. So entry $c_{ij}$ is input from sector $i$ per unit of output of sector $j$, which is the transpose of the order most people guess. A column that sums to more than $1$ describes a sector that consumes more value than it produces.

Units are money, not tonnes. Every quantity is the value of a flow over a period, which is what lets outputs of different sectors be added at all.

Definitions and results

The setup. An economy has $n$ sectors. The production vector $\mathbf{x}$ in $\mathbf{R}^n$ lists the output of each sector, the final demand vector $\mathbf{d}$ lists the output claimed from outside the sectors, and the unit consumption columns are collected into the consumption matrix $C$. All entries of $C$, $\mathbf{x}$ and $\mathbf{d}$ are nonnegative.

Intermediate demand. The amount consumed internally when production is $\mathbf{x}$ is $C\mathbf{x}$. This is where the matrix-vector product does the work: entry $i$ of $C\mathbf{x}$ sums the demands placed on sector $i$ by every sector, each weighted by that sector’s output.

Production equation. Production has to cover both claims:

$$ \mathbf{x} = C\mathbf{x} + \mathbf{d} $$

which rearranges to $(I - C)\mathbf{x} = \mathbf{d}$. The matrix $I - C$ is called the Leontief matrix.

Solution. If $I - C$ is invertible, then for every demand there is exactly one production vector, and it is $\mathbf{x} = (I - C)^{-1}\mathbf{d}$. Never write it as $\mathbf{d}/(I - C)$, and never drop the $I$; $1 - C$ is not defined for a matrix.

When the model behaves. If $C$ has nonnegative entries and every column sum is less than $1$, then $I - C$ is invertible, the entries of $(I - C)^{-1}$ are nonnegative, and the production vector for any nonnegative demand is nonnegative. The column sum condition says each sector produces more value than it consumes, which is the economic meaning of a productive economy.

Power series. Under the same condition,

$$ (I - C)^{-1} = I + C + C^2 + C^3 + \cdots $$

with the powers of $C$ shrinking toward the zero matrix. Read the terms as a story: $\mathbf{d}$ is the output demanded directly, $C\mathbf{d}$ is what producing that consumes, $C^2\mathbf{d}$ is what producing that consumes, and so on. The regress converges because each round is smaller than the last. A few terms give a usable estimate, which is how the inverse is approximated when the economy has hundreds of sectors.

The marginal reading. Column $j$ of $(I - C)^{-1}$ is the extra production needed from each sector when final demand for sector $j$ rises by one unit, with all other demands unchanged. That follows from the solution formula, because adding $1$ to entry $j$ of $\mathbf{d}$ adds column $j$ of the inverse to $\mathbf{x}$.

Worked examples

A two-sector economy

Let sector $1$ be manufacturing and sector $2$ be services, with

$$ C = \begin{bmatrix} 0.4 & 0.2 \\ 0.5 & 0.5 \end{bmatrix}, \qquad \mathbf{d} = \begin{bmatrix} 20 \\ 10 \end{bmatrix} $$

Read column $1$: producing one unit of manufacturing consumes $0.4$ units of manufacturing and $0.5$ units of services. The column sums are $0.9$ and $0.7$, both under $1$, so the model is productive.

Form the Leontief matrix and invert it. Its determinant is $(0.6)(0.5) - (-0.2)(-0.5) = 0.30 - 0.10 = 0.20$:

$$ I - C = \begin{bmatrix} 0.6 & -0.2 \\ -0.5 & 0.5 \end{bmatrix}, \qquad (I - C)^{-1} = \frac{1}{0.2}\begin{bmatrix} 0.5 & 0.2 \\ 0.5 & 0.6 \end{bmatrix} = \begin{bmatrix} 2.5 & 1 \\ 2.5 & 3 \end{bmatrix} $$

Then

$$ \mathbf{x} = \begin{bmatrix} 2.5 & 1 \\ 2.5 & 3 \end{bmatrix}\begin{bmatrix} 20 \\ 10 \end{bmatrix} = \begin{bmatrix} 50 + 10 \\ 50 + 30 \end{bmatrix} = \begin{bmatrix} 60 \\ 80 \end{bmatrix} $$

Check by going back to the production equation. Intermediate demand is $C\mathbf{x} = (0.4(60) + 0.2(80), \; 0.5(60) + 0.5(80)) = (40, 70)$. Subtract it from production: $(60 - 40, \; 80 - 70) = (20, 10)$, which is $\mathbf{d}$.

Now the marginal reading. Column $1$ of the inverse is $(2.5, 2.5)$, so one extra unit of manufacturing demand needs $2.5$ extra units of manufacturing and $2.5$ of services. Test it: with $\mathbf{d} = (21, 10)$ the formula gives $(2.5(21) + 10, \; 2.5(21) + 30) = (62.5, 82.5)$, which is the old answer plus that column.

Chasing the regress term by term

Keep the same $C$ and $\mathbf{d}$ and use the power series instead of the inverse. Each round is the previous round multiplied by $C$:

$$ \mathbf{d} = (20, 10), \quad C\mathbf{d} = (10, 15), \quad C^2\mathbf{d} = (7, 12.5), \quad C^3\mathbf{d} = (5.3, 9.75) $$

Check the second of these: $0.4(20) + 0.2(10) = 10$ and $0.5(20) + 0.5(10) = 15$.

Running totals are $(20, 10)$, then $(30, 25)$, then $(37, 37.5)$, then $(42.3, 47.25)$. They climb toward $(60, 80)$ and never pass it. Convergence is slow here because the sectors consume a lot of each other, so four terms are nowhere near enough for an exact answer; they do show the shape of the limit, and they show that the direct demand of $(20, 10)$ badly understates the production required.

A three-sector check

Let

$$ C = \begin{bmatrix} 0.2 & 0.3 & 0.1 \\ 0.1 & 0.2 & 0.3 \\ 0.2 & 0.1 & 0.2 \end{bmatrix}, \qquad \mathbf{d} = \begin{bmatrix} 40 \\ 40 \\ 50 \end{bmatrix} $$

Column sums are $0.5$, $0.6$ and $0.6$, so the economy is productive and a unique nonnegative production vector exists. Solving $(I - C)\mathbf{x} = \mathbf{d}$ by row reduction is easier after multiplying both sides by $10$ to clear the decimals, which turns the coefficient matrix into the integer matrix with rows $(8, -3, -1)$, $(-1, 8, -3)$, $(-2, -1, 8)$ and the right side into $(400, 400, 500)$.

The answer is $\mathbf{x} = (100, 100, 100)$. Verify it in the integer system: $8(100) - 3(100) - 1(100) = 400$, then $-1(100) + 8(100) - 3(100) = 400$, then $-2(100) - 1(100) + 8(100) = 500$. All three hold.

Read it back in the original terms: intermediate demand is $C\mathbf{x} = (60, 60, 50)$, and adding the final demand $(40, 40, 50)$ gives $(100, 100, 100)$, the production vector again. More than half of what this economy makes is consumed making it.

Practice

Compute intermediate demand, which is one matrix-vector product per production vector.

Solve the production equation with an inverse, then check the answer in the original equation.

Quiz

Five items on the production equation and on solving it.