Input-Output Model for Three Farm Cooperatives
Three cooperatives-grain, dairy and equipment-trade inputs with one another. Let x = (x₁, x₂, x₃)ᵀ be their total outputs in suitable units, and let A be the input matrix where aᵢⱼ is the amount of sector i needed to produce one unit of sector j. The cooperatives estimate A = [[0.1, 0.2, 0.1], [0.2, 0.1, 0.3], [0.1, 0.3, 0.1]]. External final demand is d = (68, 34, 20)ᵀ. The board wants an output plan x that exactly meets this demand after all inter-industry inputs are supplied, so it must solve (I − A)x = d. A consultant proposes raising dairy's final demand by 10 units and asks how much each sector's output must change. The board also wants to know whether the system has a unique solution and why the row sums of A matter for the credibility of the plan.
1. Write down the Leontief system and solve for the output vector x that exactly meets final demand d = (68, 34, 20)ᵀ. Show enough elimination steps to justify the answer.
Read the worked answer
The system is (I − A)x = d:
0.9x₁ − 0.2x₂ − 0.1x₃ = 68 −0.2x₁ + 0.9x₂ − 0.3x₃ = 34 −0.1x₁ − 0.3x₂ + 0.9x₃ = 20
Multiplying by 10 gives:
9x₁ − 2x₂ − x₃ = 680 −2x₁ + 9x₂ − 3x₃ = 340 −x₁ − 3x₂ + 9x₃ = 200
From the first equation, x₃ = 9x₁ − 2x₂ − 680. Substituting into the second and third equations and simplifying gives 29x₁ − 15x₂ = 1700 and 80x₁ − 21x₂ = 4900. Solving these simultaneously gives x₁ = 100 and x₂ = 80. Then x₃ = 9(100) − 2(80) − 680 = 60.
So x = (100, 80, 60)ᵀ. Checking: 0.9(100) − 0.2(80) − 0.1(60) = 68, and the other equations check similarly.
2. Dairy's final demand rises by 10 units, with all other final demands unchanged. Find the change in each sector's output and comment on the size of the dairy change.
Read the worked answer
We need Δx satisfying (I − A)Δx = (0, 10, 0)ᵀ. The equations are:
0.9a − 0.2b − 0.1c = 0 −0.2a + 0.9b − 0.3c = 10 −0.1a − 0.3b + 0.9c = 0
From the first equation, c = 9a − 2b. Substituting into the second and third equations gives −29a + 15b = 100 and 80a − 21b = 0. Solving gives a = 525/148 ≈ 3.55, b = 500/37 ≈ 13.51, and c = 725/148 ≈ 4.90.
Thus the output changes are approximately (3.55, 13.51, 4.90). Dairy's own output rises by about 13.5 units, more than the 10-unit increase in final demand, because extra dairy output requires grain and equipment inputs, which in turn require further inputs. The positive changes in all sectors are consistent with a productive input-output system.
3. Compute det(I − A). Explain what a zero determinant would mean for the board's output plan, and why the fact that all row sums of A are less than 1 is relevant.
Read the worked answer
For I − A = [[0.9, −0.2, −0.1], [−0.2, 0.9, −0.3], [−0.1, −0.3, 0.9]],
det(I − A) = 0.9(0.81 − 0.09) − (−0.2)(−0.18 − 0.03) + (−0.1)(0.06 + 0.09) = 0.9(0.72) + 0.2(−0.21) − 0.1(0.15) = 0.648 − 0.042 − 0.015 = 0.591.
Since the determinant is nonzero, I − A is invertible and the system has a unique solution for any demand vector. If det(I − A) were zero, the system would either have no solution or infinitely many solutions, so the board could not determine a unique output plan from the demand data alone.
The row sums of A are 0.4, 0.6 and 0.5, all less than 1. This means each sector uses less than one unit of total input per unit of output, which is consistent with the system being productive and with I − A being invertible in this nonnegative setting. It supports, but does not by itself replace, the determinant check.