Question 1 (a)
Two industries: new gross outputs, labour days, value added and viability
12 marks
- Time to write
- About 24 minutes
Question
Question 1 (a) · 12 marksThe input-output relationship (in units) of two industries is shown below:
| Industry | I | II | Final Demand | Gross Output |
|---|---|---|---|---|
| I | 24 | 48 | 48 | 120 |
| II | 36 | 60 | 48 | 144 |
| Labour Days | 240 | 360 |
Using matrix notation, determine:
Gross output needed to meet the new final demand of 132 units for industry I and 264 units for industry II respectively.
Total requirement of labour days.
Total value added, when labour cost is ₹240 per labour day.
Determine whether the system is viable under the Hawkins - Simon conditions of viability?
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
- 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)⁻¹ 3 marks 5 min
- 2 New gross outputs 3 marks 5 min
- 3 Total labour days 2 marks 3 min
- 4 Total value added 2 marks 3 min
- 5 Hawkins-Simon conditions 2 marks 3 min
- Check: arithmetic, units (₹, %), final answer underlined 3 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 input-output model with labour as the primary input.
Technology matrix. aᵢⱼ = xᵢⱼ / Xⱼ:
| Industry I | Industry II | |
|---|---|---|
| Industry I | 24/120 = 1/5 | 48/144 = 1/3 |
| Industry II | 36/120 = 3/10 | 60/144 = 5/12 |
| Labour days per unit | 240/120 = 2 | 360/144 = 2.5 |
A = [[1/5, 1/3], [3/10, 5/12]], I − A = [[4/5, −1/3], [−3/10, 7/12]], |I − A| = 7/15 − 1/10 = 11/30.
(i) Gross output for final demand of 132 and 264
(I − A)⁻¹ = (30/11) [[7/12, 1/3], [3/10, 4/5]] = [[35/22, 10/11], [9/11, 24/11]] (adjoint over determinant).
X₁ = (35/22)(132) + (10/11)(264) = 210 + 240 = 450 X₂ = (9/11)(132) + (24/11)(264) = 108 + 576 = 684
Check: (1/5)(450) + (1/3)(684) + 132 = 90 + 228 + 132 = 450; (3/10)(450) + (5/12)(684) + 264 = 135 + 285 + 264 = 684.
| Industry | To industry I | To industry II | Final demand | Gross output |
|---|---|---|---|---|
| I | 90 | 228 | 132 | 450 |
| II | 135 | 285 | 264 | 684 |
| Labour days | 900 | 1,710 |
(ii) Total labour days = 2 × 450 + 2.5 × 684 = 900 + 1,710 = 2,610 labour days.
(iii) Total value added at ₹240 a labour day (labour is the only primary input): 240 × 2,610 = ₹6,26,400.
The diagonal elements of I − A, 4/5 and 7/12, are positive, and |I − A| = 11/30 ≈ 0.367 > 0. Both conditions hold, so the system is viable.
Where the 12 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Technology matrix and (I − A)⁻¹ | 3 |
| 2 | New gross outputs | 3 |
| 3 | Total labour days | 2 |
| 4 | Total value added | 2 |
| 5 | Hawkins-Simon conditions | 2 |
| Total | 12 |
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