Question 1 OR (b)
Three departments: equilibrium prices, value added and feasibility of a new output plan
12 marks
- Time to write
- About 24 minutes
Question
Question 1 OR (b) · 12 marksA manufacturing unit has three departments that produce output for inter-departmental consumption and for final consumers. The inter-departmental flow of the products of its three departments is as follows:
| Producer | Inter-Department Consumption (tons) | Final Demand (tons) | ||
|---|---|---|---|---|
| Department X | Department Y | Department Z | ||
| Department X | 120 | 60 | 0 | 60 |
| Department Y | 0 | 30 | 50 | 50 |
| Department Z | 50 | 80 | 90 | 70 |
Labour required by departments X, Y and Z is 70, 0 and 150 labour days respectively and the wage rate is ₹350 per labour day.
Using matrices, you are required to:
Compute the equilibrium prices for the three departments from the above table.
Compute the total value added.
If the domestic demand changes by 60, 0 and 30 tons respectively for departments X, Y and Z, what change in the gross output of each department is needed to meet the demand?
If 200 labour days are available in all, is the new gross output feasible?
Find the change in 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
- Compute…What…?Are…?5 questions in all: answer every one.
- ComputeFormula 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 Equilibrium prices 3 marks 5 min
- 3 Total value added 1 mark 2 min
- 4 Change in gross output 2 marks 3 min
- 5 Labour feasibility 2 marks 3 min
- 6 Change in value added 1 mark 2 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.
This is a Leontief open model with three producing departments and labour as the only primary input.
Gross outputs and technology matrix. Gross output of a department = its inter-departmental sales + final demand: X = 120 + 60 + 0 + 60 = 240 (X), 0 + 30 + 50 + 50 = 130 (Y), 50 + 80 + 90 + 70 = 290 (Z).
| Producer | Dept X | Dept Y | Dept Z | Final demand | Gross output |
|---|---|---|---|---|---|
| Department X | 120 | 60 | 0 | 60 | 240 |
| Department Y | 0 | 30 | 50 | 50 | 130 |
| Department Z | 50 | 80 | 90 | 70 | 290 |
| Labour (days) | 70 | 0 | 150 |
Dividing each column by the gross output of the using department gives aᵢⱼ = xᵢⱼ / Xⱼ:
A = [[120/240, 60/130, 0/290], [0/240, 30/130, 50/290], [50/240, 80/130, 90/290]] = [[1/2, 6/13, 0], [0, 3/13, 5/29], [5/24, 8/13, 9/29]]
Labour per unit of output: l = [70/240, 0/130, 150/290] = [7/24, 0, 15/29] labour days.
I − A = [[1/2, −6/13, 0], [0, 10/13, −5/29], [−5/24, −8/13, 20/29]], with |I − A| = 295/1508 ≈ 0.1956 > 0 and leading minors 1/2 and 5/13 positive, so the system is viable (Hawkins-Simon).
(I − A)⁻¹ = (adj (I − A)) / |I − A| = [[128/59, 96/59, 24/59], [65/354, 104/59, 26/59], [145/177, 609/295, 116/59]] ≈ [[2.1695, 1.6271, 0.4068], [0.1836, 1.7627, 0.4407], [0.8192, 2.0644, 1.9661]].
Value added per ton is the wage cost per ton: v = 350 × l = [350 × 7/24, 0, 350 × 15/29] = [102.0833, 0, 181.0345]. Prices satisfy (I − A)ᵀP = v:
(1/2)P₁ − (5/24)P₃ = 102.0833 −(6/13)P₁ + (10/13)P₂ − (8/13)P₃ = 0 −(5/29)P₂ + (20/29)P₃ = 181.0345
Solving, P = [(I − A)ᵀ]⁻¹v:
P₁ = ₹369.77, P₂ = ₹539.83, P₃ = ₹397.46 per ton.
Check (column X): 0.5 × 369.77 − (5/24) × 397.46 = 184.89 − 82.80 = 102.08 ✓.
Value added is the wage bill: 350 × (70 + 0 + 150) = ₹77,000. Check: value of final demand at equilibrium prices = 60 × 369.77 + 50 × 539.83 + 70 × 397.46 = ₹77,000 (exactly, before rounding).
(iii) Change in gross output for the change in demand ΔD = [60, 0, 30]ᵀ
ΔX = (I − A)⁻¹ ΔD:
ΔX₁ = (128 × 60 + 24 × 30)/59 = 8400/59 = 142.37 tons ΔX₂ = (65/354) × 60 + (26/59) × 30 = 1430/59 = 24.24 tons ΔX₃ = (145/177) × 60 + (116/59) × 30 = 6380/59 = 108.14 tons
| Department | Present gross output | Change | New gross output |
|---|---|---|---|
| X | 240.00 | 142.37 | 382.37 |
| Y | 130.00 | 24.24 | 154.24 |
| Z | 290.00 | 108.14 | 398.14 |
(iv) Is the new gross output feasible with 200 labour days?
Labour needed = (7/24)(382.37) + 0 × (154.24) + (15/29)(398.14) = 111.53 + 0 + 205.93 = 317.46 labour days. Only 200 labour days are available, so the new gross output is not feasible (short by 117.46 days). (Even the present output needs 220 days.)
(v) Change in total value added
ΔVA = 350 × (labour for ΔX) = 350 × [(7/24)(142.37) + (15/29)(108.14)] = 350 × 97.46 = ₹34,110.17 (equivalently, ΔD valued at equilibrium prices: 60 × 369.77 + 30 × 397.46 = ₹34,110.17).
Where the 12 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Technology matrix and (I − A)⁻¹ | 3 |
| 2 | Equilibrium prices | 3 |
| 3 | Total value added | 1 |
| 4 | Change in gross output | 2 |
| 5 | Labour feasibility | 2 |
| 6 | Change in value added | 1 |
| Total | 12 |
Report a problemopens an email to admin@getkunji.com with the question ID1402-Q1bORalready filled in.