Question 1 (b)
Electricity and water: Leontief open model, viability, transactions, prices and value added
12 marks
- Time to write
- About 24 minutes
Question
Question 1 (b) · 12 marksA hydro-electricity power plant in a small village produces electricity and water for irrigation, each of which is also an intermediate input in the other's production. Generating a unit of electricity needs 0.30 units of electricity and 0.25 units of water. Similarly, a unit of water needs 0.40 units of water and 0.20 units of electricity. Producing a unit of electricity and a unit of water needs 0.20 and 0.10 units of labour respectively, and 0.10 and 0.05 units of capital respectively. The wage rate and the interest rate are ₹600 per man-day and 8% respectively. For final consumption the village needs 2000 kW (kilowatt) of electricity and 3600 cubic feet of water.
With the help of matrix algebra, you are required to:
Write down the balancing equations.
Test the Hawkins-Simon viability conditions of the system.
Construct the input-output transaction table of the gross output needed to meet the above final demand.
Determine the equilibrium prices of the two commodities.
Compute the total value added.
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
- Write…Determine…Compute…3 questions in all: answer every one.
- 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 Balancing equations 2 marks 3 min
- 2 Hawkins-Simon conditions 2 marks 3 min
- 3 Gross outputs and transaction table 3 marks 5 min
- 4 Equilibrium prices 3 marks 5 min
- 5 Total value added 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.
The plant is a two-sector Leontief open model (input-output, open model only, as the syllabus sets it). Let X₁ be the gross output of electricity (kW) and X₂ that of water (cubic feet).
Technology (input-coefficient) matrix. Column j shows the inputs used per unit of output j:
| Input ↓ / used per unit of → | Electricity | Water |
|---|---|---|
| Electricity | 0.30 | 0.20 |
| Water | 0.25 | 0.40 |
| Labour (man-days) | 0.20 | 0.10 |
| Capital (units) | 0.10 | 0.05 |
A = [[0.30, 0.20], [0.25, 0.40]], final demand D = [2000, 3600]ᵀ.
Gross output = intermediate use + final demand:
X₁ = 0.30X₁ + 0.20X₂ + 2000 X₂ = 0.25X₁ + 0.40X₂ + 3600
i.e. 0.70X₁ − 0.20X₂ = 2000 and −0.25X₁ + 0.60X₂ = 3600, or (I − A)X = D with I − A = [[0.70, −0.20], [−0.25, 0.60]].
- Every diagonal element of I − A is positive: 0.70 > 0 and 0.60 > 0.
- |I − A| = 0.70 × 0.60 − (−0.20)(−0.25) = 0.42 − 0.05 = 0.37 > 0.
Both conditions hold, so the system is viable: it can meet any non-negative final demand with non-negative gross outputs.
Gross output. X = (I − A)⁻¹D, where (I − A)⁻¹ = (1/0.37) [[0.60, 0.20], [0.25, 0.70]] (adjoint over determinant):
X₁ = (0.60 × 2000 + 0.20 × 3600)/0.37 = 1920/0.37 = 5,189.19 kW X₂ = (0.25 × 2000 + 0.70 × 3600)/0.37 = 3020/0.37 = 8,162.16 cubic feet
(iii) Input-output transaction table (gross output to meet the final demand)
| Producing sector ↓ / using sector → | Electricity | Water | Final demand | Gross output |
|---|---|---|---|---|
| Electricity (kW) | 1,556.76 | 1,632.43 | 2,000.00 | 5,189.19 |
| Water (cu. ft.) | 1,297.30 | 3,264.86 | 3,600.00 | 8,162.16 |
| Labour (man-days) | 1,037.84 | 816.22 | ||
| Capital (units) | 518.92 | 408.11 |
(Each entry is the coefficient times the gross output of the using sector, e.g. 0.30 × 5,189.19 = 1,556.76.) Total labour needed is 1,854.05 man-days and total capital 927.03 units.
The price of each good covers the cost of the produced inputs it uses plus its value added, wages at ₹600 per man-day and interest at 8% on the capital it uses (read as interest at 8% on each unit of capital):
Value added per unit of electricity: v₁ = 600 × 0.20 + 0.08 × 0.10 = 120 + 0.008 = ₹120.008 Value added per unit of water: v₂ = 600 × 0.10 + 0.08 × 0.05 = 60 + 0.004 = ₹60.004
Price equations (I − A)ᵀP = V:
0.70P₁ − 0.25P₂ = 120.008 −0.20P₁ + 0.60P₂ = 60.004
P₁ = (0.60 × 120.008 + 0.25 × 60.004)/0.37 = 87.0058/0.37 = ₹235.15 per kW of electricity P₂ = (0.20 × 120.008 + 0.70 × 60.004)/0.37 = 66.0044/0.37 = ₹178.39 per cubic foot of water
Check: 0.70(235.15) − 0.25(178.39) = 120.01 and −0.20(235.15) + 0.60(178.39) = 60.00 (to rounding).
| Item | Electricity (₹) | Water (₹) | Total (₹) |
|---|---|---|---|
| Wages (₹600 × man-days) | 6,22,702.70 | 4,89,729.73 | 11,12,432.43 |
| Interest (8% × capital units) | 41.51 | 32.65 | 74.16 |
| Value added | 11,12,506.59 |
By sector, value added is ₹6,22,744.22 in electricity and ₹4,89,762.38 in water (each rounded to the paisa; the total is taken before rounding).
Total value added = ₹11,12,506.59. As a check, it equals the value of the final demand at the equilibrium prices: 2000 × 235.1508 + 3600 × 178.3903 = ₹11,12,506.59.
Reading. The question gives capital in units and an interest rate, so interest is taken as 8% of each unit of capital (₹0.08 a unit). If capital were instead treated as having no separate cost, the prices would be ₹235.14 and ₹178.38 and the value added ₹11,12,432.43 (wages only); the interest term changes the result by less than ₹75.
Where the 12 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Balancing equations | 2 |
| 2 | Hawkins-Simon conditions | 2 |
| 3 | Gross outputs and transaction table | 3 |
| 4 | Equilibrium prices | 3 |
| 5 | Total value added | 2 |
| Total | 12 |
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