From a Demand Change to an Output Change
In a Leontief input-output model, each sector’s production supports both final demand and the intermediate inputs needed by other sectors. As a result, a change in final demand can lead to production changes in several sectors, not just the sector whose final demand changed. The sensitivity calculation is especially useful when a demand forecast is revised: it gives the output adjustment implied by the model without requiring the entire production plan to be solved from scratch.
Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\) be the matrix of direct input coefficients, where \(a_{ij}\) is the amount of output from sector \(i\) needed to produce one unit of output in sector \(j\). Let \(\mathbf{x}\in\mathbb{R}^n\) be the column vector of sector outputs and \(\mathbf{d}\in\mathbb{R}^n\) the column vector of final demands. The Leontief system is \(\mathbf{x}=\mathbf{A}\mathbf{x}+\mathbf{d}\), or equivalently \((\mathbf{I}_n-\mathbf{A})\mathbf{x}=\mathbf{d}\). Throughout, assume \(\mathbf{I}_n-\mathbf{A}\) is invertible, so the production plan for each demand vector is uniquely determined.
The central sensitivity calculation follows from the same system equation. If the technology matrix \(\mathbf{A}\) is fixed, then a demand change alters the right-hand side but not the coefficient matrix. The calculation therefore depends on \(\Delta\mathbf{d}\), not on solving a new unrelated model. This tutorial treats the coefficients as fixed; a change in technology would require a different comparison.
Proof. Write \(\mathbf{x}_{\mathrm{old}}\) and \(\mathbf{x}_{\mathrm{new}}\) for the production plans before and after the demand change. By the Leontief system, \((\mathbf{I}_n-\mathbf{A})\mathbf{x}_{\mathrm{old}}=\mathbf{d}\) and \((\mathbf{I}_n-\mathbf{A})\mathbf{x}_{\mathrm{new}}=\mathbf{d}+\Delta\mathbf{d}\). Subtract the first equation from the second: \((\mathbf{I}_n-\mathbf{A})(\mathbf{x}_{\mathrm{new}}-\mathbf{x}_{\mathrm{old}})=\Delta\mathbf{d}\). By definition, \(\mathbf{x}_{\mathrm{new}}-\mathbf{x}_{\mathrm{old}}=\Delta\mathbf{x}\). Multiplying by the inverse of \(\mathbf{I}_n-\mathbf{A}\) gives \(\Delta\mathbf{x}=(\mathbf{I}_n-\mathbf{A})^{-1}\Delta\mathbf{d}=\mathbf{L}\Delta\mathbf{d}\). \(\square\)
Calculating One Demand Shock
Worked Example: A Demand Increase in Sector 1
Consider two sectors with \(\mathbf{A}=\begin{bmatrix}0&\tfrac14\\\tfrac13&0\end{bmatrix}\). Find the production change when final demand increases by \(11\) units in sector 1 and does not change in sector 2. Start from \(\mathbf{d}=\begin{bmatrix}11\\11\end{bmatrix}\).
First form the coefficient matrix and its inverse:
Since the determinant is nonzero, the matrix is invertible. Its inverse is
The demand change is \(\Delta\mathbf{d}=\begin{bmatrix}11\\0\end{bmatrix}\). Multiply by the Leontief inverse:
Thus sector 1 must increase production by \(12\) units and sector 2 by \(4\) units. To check the response directly, multiply it by the original system matrix:
The original production plan is \(\mathbf{x}_{\mathrm{old}}=\mathbf{L}\mathbf{d}=\begin{bmatrix}15\\16\end{bmatrix}\). Indeed, \(\begin{bmatrix}1&-\tfrac14\\-\tfrac13&1\end{bmatrix}\begin{bmatrix}15\\16\end{bmatrix}=\begin{bmatrix}11\\11\end{bmatrix}\). The new plan is \(\mathbf{x}_{\mathrm{new}}=\mathbf{x}_{\mathrm{old}}+\Delta\mathbf{x}=\begin{bmatrix}27\\20\end{bmatrix}\), and the final check is \(\begin{bmatrix}1&-\tfrac14\\-\tfrac13&1\end{bmatrix}\begin{bmatrix}27\\20\end{bmatrix}=\begin{bmatrix}22\\11\end{bmatrix}=\mathbf{d}+\Delta\mathbf{d}\).
Why Responses Add
The formula \(\Delta\mathbf{x}=\mathbf{L}\Delta\mathbf{d}\) is a linear map from demand changes to output changes. This matters when a demand revision has several parts: one can calculate the response to each part separately, then add those responses. The statement applies to signed changes as well as increases. A negative entry in \(\Delta\mathbf{d}\) represents a demand decrease in that sector.
Proof. By the definition of matrix multiplication, each entry of \(\mathbf{L}\mathbf{y}\) is a sum of entries of \(\mathbf{L}\) multiplied by entries of \(\mathbf{y}\). Distributing multiplication over addition and factoring out scalars in those sums gives \(\mathbf{L}(c_1\Delta\mathbf{d}_1+c_2\Delta\mathbf{d}_2)=c_1\mathbf{L}\Delta\mathbf{d}_1+c_2\mathbf{L}\Delta\mathbf{d}_2\). The demand-change response theorem identifies each product \(\mathbf{L}\Delta\mathbf{d}\) with the corresponding change in production. Thus scaling a demand change scales its response, and adding demand changes adds their responses. \(\square\)
Worked Example: Comparing Equal Shocks Across Three Sectors
For a three-sector model, let \(\mathbf{A}=\begin{bmatrix}0&\tfrac14&0\\0&0&\tfrac12\\0&0&0\end{bmatrix}\). Compare the production changes caused by a one-unit increase in final demand in each sector, considered one at a time.
Here \(\mathbf{I}_3-\mathbf{A}=\begin{bmatrix}1&-\tfrac14&0\\0&1&-\tfrac12\\0&0&1\end{bmatrix}\). Its inverse is \(\mathbf{L}=\begin{bmatrix}1&\tfrac14&\tfrac18\\0&1&\tfrac12\\0&0&1\end{bmatrix}\). Check the inverse by multiplying:
A one-unit increase in sector \(i\)'s final demand is \(\mathbf{e}_i\), the column with a \(1\) in position \(i\) and zeros elsewhere. The responses are the columns of \(\mathbf{L}\):
Each response can be checked by multiplying by \(\mathbf{I}_3-\mathbf{A}\). The three products are respectively \(\begin{bmatrix}1\\0\\0\end{bmatrix}\), \(\begin{bmatrix}\tfrac14-\tfrac14\\1\\0\end{bmatrix}=\begin{bmatrix}0\\1\\0\end{bmatrix}\), and \(\begin{bmatrix}\tfrac18-\tfrac18\\\tfrac12-\tfrac12\\1\end{bmatrix}=\begin{bmatrix}0\\0\\1\end{bmatrix}\), as required.
For a simple total-output comparison, add the components of each response: the totals are \(1\), \(\tfrac54\), and \(\tfrac{13}{8}\), respectively. Under this particular measure, the one-unit increase in sector 3 demand creates the largest total output increase. Other measures, such as the change in one selected sector, may lead to a different comparison; the measure should therefore be stated rather than assumed.
Comparing Absolute and Percentage Changes
The entries of \(\Delta\mathbf{x}\) give absolute output changes, in the same units as the outputs. They do not by themselves say which sector has the largest change relative to its starting level. If the baseline output \(x_i\) is nonzero, the percentage change in sector \(i\) is \(100\,\Delta x_i/x_i\). If \(x_i=0\), this percentage is undefined, so an absolute change or another explicitly chosen comparison is needed.
Worked Example: A Unit Shock and Relative Output Changes
Use the two-sector model and baseline plan from the first example, \(\mathbf{x}_{\mathrm{old}}=\begin{bmatrix}15\\16\end{bmatrix}\), but now increase sector 1 final demand by only one unit: \(\Delta\mathbf{d}=\begin{bmatrix}1\\0\end{bmatrix}\).
Apply the same Leontief inverse:
The absolute changes are larger in sector 1. Relative to the baseline outputs, however, the changes are \(\frac{12/11}{15}=\frac{4}{55}\) in sector 1 and \(\frac{4/11}{16}=\frac{1}{44}\) in sector 2. As percentages these are \(\frac{80}{11}\%\) and \(\frac{25}{11}\%\), respectively. Thus sector 1 also has the larger percentage change in this example, but that conclusion comes from dividing by each sector’s own baseline output, not just comparing \(\Delta x_1\) and \(\Delta x_2\).
To verify the updated plan, \(\mathbf{x}_{\mathrm{new}}=\begin{bmatrix}15+\tfrac{12}{11}\\16+\tfrac{4}{11}\end{bmatrix}=\begin{bmatrix}\tfrac{177}{11}\\\tfrac{180}{11}\end{bmatrix}\). Then \(\begin{bmatrix}1&-\tfrac14\\-\tfrac13&1\end{bmatrix}\begin{bmatrix}\tfrac{177}{11}\\\tfrac{180}{11}\end{bmatrix}=\begin{bmatrix}\tfrac{177}{11}-\tfrac{45}{11}\\-\tfrac{59}{11}+\tfrac{180}{11}\end{bmatrix}=\begin{bmatrix}12\\11\end{bmatrix}\), which is the revised final demand.
Common Pitfall: Treating Demand Change as Output Change
A tempting but incorrect shortcut is to set \(\Delta\mathbf{x}=\Delta\mathbf{d}\), on the grounds that the same demand change should be met by the same output change. In the first two-sector model, this shortcut would give \(\Delta\mathbf{x}=\begin{bmatrix}11\\0\end{bmatrix}\) for a demand increase \(\Delta\mathbf{d}=\begin{bmatrix}11\\0\end{bmatrix}\). But substituting that proposed output change into the system gives
So the proposed change does not satisfy the required equation \((\mathbf{I}_2-\mathbf{A})\Delta\mathbf{x}=\Delta\mathbf{d}\). The correction is to multiply the demand change by \(\mathbf{L}\), giving \(\begin{bmatrix}12\\4\end{bmatrix}\), which does satisfy that equation. The distinction is fundamental: final demand is the part of production left after intermediate requirements have been accounted for, while total output must also support those requirements.
For a practical comparison, first calculate the full response vector. Then specify what “larger effect” means: a larger change in one named sector, a larger sum of increases, or a larger percentage change from baseline. If demand changes can have both positive and negative entries, the sum of changes may conceal offsetting effects, so the componentwise response may be more informative.
Check Your Understanding
Use the fixed-technology response formula and show enough work to verify each conclusion.
- If \(\mathbf{L}=\begin{bmatrix}2&1\\1&3\end{bmatrix}\) and \(\Delta\mathbf{d}=\begin{bmatrix}2\\-1\end{bmatrix}\), what is \(\Delta\mathbf{x}\)?
- Explain why the columns of \(\mathbf{L}\) represent responses to unit increases in the corresponding sectors’ final demands.
- For a baseline output of \(20\) units and a change of \(2\) units, calculate the percentage change. Why is the same calculation unavailable when the baseline output is zero?
- State the condition on \(\mathbf{I}_n-\mathbf{A}\) needed for a unique response formula of the form \(\Delta\mathbf{x}=\mathbf{L}\Delta\mathbf{d}\).
- Describe a concrete case where comparing the sum of absolute output changes gives a different interpretation from comparing percentage changes.
