Leontief Workshop: Computing Output for Final Demand

From Sector Requirements to an Output Equation

An input-output model records how sectors rely on one another to produce goods or services. A sector’s output may be used by other sectors as an intermediate input, or it may meet final demand, such as household purchases or exports. The production plan must cover both uses. The central calculation is to find the gross output required in each sector, taking account of inputs that themselves need to be produced.

Consider \(n\) sectors. Let \(x_j\) be the gross output of sector \(j\), and let \(d_i\) be the final demand for the output of sector \(i\). The technical coefficient \(a_{ij}\) measures the amount of sector \(i\)'s output needed to produce one unit of sector \(j\)'s output. Thus the first index identifies the supplying sector and the second identifies the producing sector. Put these coefficients in \(\mathbf{A}=[a_{ij}]\), and collect outputs and final demands in columns \(\mathbf{x}\) and \(\mathbf{d}\). The total intermediate use of sector \(i\)'s output is \(\sum_{j=1}^{n}a_{ij}x_j\).

The equation is a balance, not a claim that gross output equals final demand. For each sector, gross output must cover intermediate sales to all sectors plus final demand for that sector’s product. Rearranging gives \((\mathbf{I}_n-\mathbf{A})\mathbf{x}=\mathbf{d}\). This form is a linear system in the unknown gross outputs. When the system has a unique solution, that solution is the production plan matching the stated coefficients and demand.

The model assumes the coefficients stay fixed over the calculation: producing one unit of a sector’s output always requires the same amounts of each input. It also treats the entries and outputs as measured in compatible units. If the entries represent physical quantities, the units must be consistent; if they represent monetary amounts, they must use a consistent valuation. The calculation does not itself determine prices or choose an economically optimal plan.

Worked Solution for Two Sectors

A Sufficient Condition for a Unique Nonnegative Plan

Row reduction answers a particular numerical question. A useful general guarantee is available when all technical coefficients are nonnegative and each column sum is less than \(1\). The sum of column \(j\) is the total amount of intermediate inputs required per unit of output in sector \(j\). The condition is sufficient for a unique solution and, when final demand is nonnegative, for that solution to be nonnegative. It is not necessary: a model can have a unique solution even if this particular condition fails.

Proof. Define \(q=\max_j\sum_i a_{ij}\), so \(0\le q<1\). For a matrix \(\mathbf{M}\), write \(\|\mathbf{M}\|_1=\max_j\sum_i|m_{ij}|\). For any two compatible square matrices \(\mathbf{M},\mathbf{N}\), the triangle inequality gives \(\sum_i|(\mathbf{M}\mathbf{N})_{ij}|\le\sum_k|n_{kj}|\sum_i|m_{ik}|\le\|\mathbf{M}\|_1\sum_k|n_{kj}|\). Taking the maximum over \(j\) shows \(\|\mathbf{M}\mathbf{N}\|_1\le\|\mathbf{M}\|_1\|\mathbf{N}\|_1\). Since \(\|\mathbf{A}\|_1=q\), repeated application gives \(\|\mathbf{A}^k\|_1\le q^k\) for each positive integer \(k\).

Now form \(\mathbf{S}_N=\mathbf{I}_n+\mathbf{A}+\cdots+\mathbf{A}^N\). For \(N>M\), the norm of \(\mathbf{S}_N-\mathbf{S}_M\) is at most \(\sum_{k=M+1}^{N}q^k\), which tends to \(0\) as \(M\) increases because \(q<1\). Thus the entries of \(\mathbf{S}_N\) converge to a matrix \(\mathbf{S}\). Direct multiplication and cancellation of consecutive powers give \((\mathbf{I}_n-\mathbf{A})\mathbf{S}_N=\mathbf{S}_N(\mathbf{I}_n-\mathbf{A})=\mathbf{I}_n-\mathbf{A}^{N+1}\). The bound \(\|\mathbf{A}^{N+1}\|_1\le q^{N+1}\) shows that \(\mathbf{A}^{N+1}\) tends to the zero matrix. Taking entrywise limits in the displayed identities gives \((\mathbf{I}_n-\mathbf{A})\mathbf{S}=\mathbf{S}(\mathbf{I}_n-\mathbf{A})=\mathbf{I}_n\). Hence \(\mathbf{I}_n-\mathbf{A}\) has inverse \(\mathbf{S}\), and \(\mathbf{x}=\mathbf{S}\mathbf{d}\) is the unique solution. Each power of \(\mathbf{A}\) has nonnegative entries, as does \(\mathbf{I}_n\), so each \(\mathbf{S}_N\) and its limit \(\mathbf{S}\) have nonnegative entries. Since \(\mathbf{d}\) is nonnegative, \(\mathbf{x}=\mathbf{S}\mathbf{d}\) is nonnegative. \(\square\)

For the two-sector example, both column sums are \(\frac15+\frac1{10}=\frac3{10}<1\). The theorem’s hypotheses hold, which guarantees uniqueness and nonnegative output for any nonnegative final-demand vector with those same technical coefficients. The row reduction above still matters: it gives the actual numerical plan for the specified demand.

A Three-Sector Calculation

Interpreting a Change in Final Demand

With fixed technical coefficients, a change in final demand changes the right-hand side of the same linear system. The required change in gross output must cover the additional final demand and the intermediate inputs needed to produce that change. This is why the production increase need not equal the final-demand increase in each sector.

Common Pitfall: Confusing Gross Output with Final Demand

A frequent mistake is to set \(\mathbf{x}=\mathbf{d}\), as though final demand were the whole output requirement. In the two-sector example, that would give the proposed plan \(\begin{bmatrix}72&54\end{bmatrix}^{\top}\). But its intermediate use is

The proposed vector fails the sector-balance equation, so it is not a valid production plan. The correction is to solve \((\mathbf{I}_2-\mathbf{A})\mathbf{x}=\mathbf{d}\), then substitute the result into the original balance equation. A final check should compare gross output with intermediate use plus final demand separately for every sector; checking only one sector can miss an indexing or arithmetic error.