Question 1 (b)

Two-sector input-output: technology matrix, viability, new outputs, labour, value added and prices

12 marks

Time to write: About 24 minutes
Topic: Leontief input-output model

Question

Question 1 (b) · 12 marks

The inter-industry flow of products of two industries is as follows:

Production Sector	Consumption Sector		Domestic Demand	Total Output
	X	Y
X	40	180	180	400
Y	60	30	210	300
Labour (hours)	100	60

Determine the technology matrix, and test the Simon-Hawkins conditions of viability of the system.
If the domestic demand changes to 288 and 144 units respectively, what should each sector's gross output be to meet the demands?
If the total labour available is 200 hours, is the solution feasible?
Find the value added by each sector.
Find the equilibrium prices, if the wage rate is ₹300.

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

24 minfor 12 marks

Read from the question

Determine…What…?Is…?5 asks, each with its own marks below. Answer all 5.
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 Hawkins-Simon test 3 marks 5 min
2 New gross outputs 3 marks 5 min
3 Labour feasibility 2 marks 3 min
4 Value added by each sector 2 marks 3 min
5 Equilibrium prices 2 marks 3 min
Check: arithmetic, units (₹, %), final answer underlined 3 min

12345

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 of two sectors with labour as the primary input.

(i) Technology matrix and viability

aᵢⱼ = xᵢⱼ / Xⱼ (input of sector i per unit of output of sector j):

	X	Y
X	40/400 = 0.10	180/300 = 0.60
Y	60/400 = 0.15	30/300 = 0.10
Labour (hours per unit)	100/400 = 0.25	60/300 = 0.20

A = [[0.10, 0.60], [0.15, 0.10]], I − A = [[0.90, −0.60], [−0.15, 0.90]].

Hawkins-Simon (Simon-Hawkins) conditions: the diagonal elements 0.90 and 0.90 are positive, and |I − A| = 0.81 − 0.09 = 0.72 > 0. The system is viable.

(ii) Gross outputs for domestic demand of 288 and 144

(I − A)⁻¹ = (1/0.72) [[0.90, 0.60], [0.15, 0.90]] = [[1.25, 0.8333], [0.2083, 1.25]] (adjoint over determinant).

X₁ = (0.90 × 288 + 0.60 × 144)/0.72 = 345.6/0.72 = 480 X₂ = (0.15 × 288 + 0.90 × 144)/0.72 = 172.8/0.72 = 240

Check: 0.10(480) + 0.60(240) + 288 = 48 + 144 + 288 = 480; 0.15(480) + 0.10(240) + 144 = 72 + 24 + 144 = 240.

(iii) Labour feasibility

Labour needed = 0.25 × 480 + 0.20 × 240 = 120 + 48 = 168 hours ≤ 200 hours available, so the solution is feasible (32 hours to spare).

(iv) Value added by each sector (from the given transaction table: gross output less inputs bought from the two sectors)

Sector	Inputs from X	Inputs from Y	Value added	Gross output
X	40	60	300	400
Y	180	30	90	300

Value added: sector X 300, sector Y 90 (total 390, which equals the total domestic demand 180 + 210).

The other reading (at the new outputs of part (ii)): X: 480 − 0.10(480) − 0.15(480) = 360; Y: 240 − 0.60(240) − 0.10(240) = 72 (total 432 = 288 + 144).

(v) Equilibrium prices at ₹300 an hour

Wage cost per unit: v = [300 × 0.25, 300 × 0.20] = [75, 60]. Prices satisfy (I − A)ᵀP = v:

0.90P₁ − 0.15P₂ = 75 −0.60P₁ + 0.90P₂ = 60

P₁ = (0.90 × 75 + 0.15 × 60)/0.72 = 76.5/0.72 = ₹106.25; P₂ = (0.60 × 75 + 0.90 × 60)/0.72 = 99/0.72 = ₹137.50.

Check: 0.90(106.25) − 0.15(137.5) = 95.625 − 20.625 = 75; −0.60(106.25) + 0.90(137.5) = −63.75 + 123.75 = 60.

Where the 12 marks sit

#	Component	Marks
1	Technology matrix and Hawkins-Simon test	3
2	New gross outputs	3
3	Labour feasibility	2
4	Value added by each sector	2
5	Equilibrium prices	2
	Total	12

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