Question 1 OR (a)
Three-sector economy: gross outputs for a final demand and for its increase
12 marks
- Time to write
- About 24 minutes
Question
Question 1 OR (a) · 12 marksThe input-output coefficients aᵢⱼ of a three-sector economy are:
a₁₁ = 0.4, a₂₁ = 0.1, a₃₁ = 0.2, a₁₂ = 0.2, a₂₂ = 0.5, a₃₂ = 0.1, a₁₃ = 0.2,
a₂₃ = 0.1, a₃₃ = 0.5.
Determine the gross output of each of the three sectors when their final demand values are 18,000, 36,000 and 54,000 units respectively.
Find the new output level if the final demand rises by 1200, 2400 and 3600 units.
Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.
Before you write
Read off the question and its marks
Read from the question
- Determine…Find…Two asks in one question: answer both.
- DetermineFormula first, then the working, then the answer with its unit (₹, %, units).
Write it in this order
- Read the data; note exactly what is asked 2 min
- 1 Technology matrix and I − A 2 marks 3 min
- 2 Determinant and inverse 4 marks 7 min
- 3 Gross outputs 3 marks 5 min
- 4 New output levels 3 marks 5 min
- Check: arithmetic, units (₹, %), final answer underlined 2 min
Try it first. About 24 minutes for 12 marks. Write it out, then open the solution.
Solution
Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.
Open Leontief model. The coefficient aᵢⱼ is the input from sector i per unit of output of sector j, so
A = [[a₁₁, a₁₂, a₁₃], [a₂₁, a₂₂, a₂₃], [a₃₁, a₃₂, a₃₃]] = [[0.4, 0.2, 0.2], [0.1, 0.5, 0.1], [0.2, 0.1, 0.5]],
I − A = [[0.6, −0.2, −0.2], [−0.1, 0.5, −0.1], [−0.2, −0.1, 0.5]].
|I − A| = 0.6(0.25 − 0.01) + 0.2(−0.05 − 0.02) − 0.2(0.01 + 0.10) = 0.144 − 0.014 − 0.022 = 0.108 > 0 (the leading minors 0.6 and 0.28 are also positive, so the system is viable).
Cofactors (adj(I − A) is their transpose):
adj(I − A) = [[0.24, 0.12, 0.12], [0.07, 0.26, 0.08], [0.11, 0.10, 0.28]]
(I − A)⁻¹ = adj/0.108 = [[20/9, 10/9, 10/9], [35/54, 65/27, 20/27], [55/54, 25/27, 70/27]] ≈ [[2.2222, 1.1111, 1.1111], [0.6481, 2.4074, 0.7407], [1.0185, 0.9259, 2.5926]].
(i) Gross outputs for final demand 18,000, 36,000 and 54,000
X₁ = (0.24 × 18,000 + 0.12 × 36,000 + 0.12 × 54,000)/0.108 = 15,120/0.108 = 1,40,000.00 X₂ = (0.07 × 18,000 + 0.26 × 36,000 + 0.08 × 54,000)/0.108 = 14,940/0.108 = 1,38,333.33 X₃ = (0.11 × 18,000 + 0.10 × 36,000 + 0.28 × 54,000)/0.108 = 20,700/0.108 = 1,91,666.67
Check (sector 1): its output is used as follows.
| Use of sector 1's output | Working | Units |
|---|---|---|
| By sector 1 | 0.4 × 1,40,000.00 | 56,000.00 |
| By sector 2 | 0.2 × 1,38,333.33 | 27,666.67 |
| By sector 3 | 0.2 × 1,91,666.67 | 38,333.33 |
| Final demand | 18,000.00 | |
| Gross output of sector 1 | 1,40,000.00 |
(ii) New output levels when final demand rises by 1,200, 2,400 and 3,600
ΔX = (I − A)⁻¹ ΔD:
ΔX₁ = (0.24 × 1,200 + 0.12 × 2,400 + 0.12 × 3,600)/0.108 = 1,008/0.108 = 9,333.33 ΔX₂ = (0.07 × 1,200 + 0.26 × 2,400 + 0.08 × 3,600)/0.108 = 996/0.108 = 9,222.22 ΔX₃ = (0.11 × 1,200 + 0.10 × 2,400 + 0.28 × 3,600)/0.108 = 1,380/0.108 = 12,777.78
| Sector | Gross output (i) | Increase | New gross output |
|---|---|---|---|
| 1 | 1,40,000.00 | 9,333.33 | 1,49,333.33 |
| 2 | 1,38,333.33 | 9,222.22 | 1,47,555.56 |
| 3 | 1,91,666.67 | 12,777.78 | 2,04,444.44 |
(Each figure is rounded to two decimals from the exact fractions, so a row may differ by 0.01 in the last place. The rise in final demand is 6.67% of the original in each sector, so every gross output rises by the same 6.67%.)
Where the 12 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Technology matrix and I − A | 2 |
| 2 | Determinant and inverse | 4 |
| 3 | Gross outputs | 3 |
| 4 | New output levels | 3 |
| Total | 12 |
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