Question 1 OR (b)

Two-sector transaction matrix: technology matrix, viability and the new transaction matrix

12 marks

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

Question

Question 1 OR (b) · 12 marks

The transaction matrix for a two-sector economy is given below:

Purchase Sector	Sales Sector		Final Demand	Gross Output
	I	II
I	6	2	16	24
II	4	6	2	12
Primary Input	14	4

Determine the technology matrix, and test the Simon-Hawkins conditions of viability of the system.
Rewrite the transaction matrix afresh when the final demand for the output of sector I rises to 26 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

24 minfor 12 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 2 marks 3 min
2 Hawkins-Simon test 2 marks 3 min
3 (I − A)⁻¹ and new gross outputs 4 marks 7 min
4 New transaction matrix with checks 4 marks 7 min
Check: arithmetic, units (₹, %), final answer underlined 2 min

1234

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 with one primary input.

(i) Technology matrix and viability

	Sector I	Sector II
Sector I	6/24 = 1/4	2/12 = 1/6
Sector II	4/24 = 1/6	6/12 = 1/2
Primary input (per unit)	14/24 = 7/12	4/12 = 1/3

A = [[1/4, 1/6], [1/6, 1/2]]; I − A = [[3/4, −1/6], [−1/6, 1/2]].

Hawkins-Simon (Simon-Hawkins) conditions: diagonal elements 3/4 and 1/2 are positive, and |I − A| = 3/8 − 1/36 = 25/72 ≈ 0.347 > 0. The system is viable.

(Each column of A plus the primary-input coefficient sums to 1, since the table is in value terms: 1/4 + 1/6 + 7/12 = 1 and 1/6 + 1/2 + 1/3 = 1.)

(ii) New transaction matrix when final demand for sector I rises to 26 (sector II stays 2)

(I − A)⁻¹ = (72/25) [[1/2, 1/6], [1/6, 3/4]] = [[1.44, 0.48], [0.48, 2.16]].

X₁ = 1.44 × 26 + 0.48 × 2 = 37.44 + 0.96 = 38.4 X₂ = 0.48 × 26 + 2.16 × 2 = 12.48 + 4.32 = 16.8

New flows xᵢⱼ = aᵢⱼ × Xⱼ:

Purchase sector	Sector I	Sector II	Final demand	Gross output
I	9.6	2.8	26.0	38.4
II	6.4	8.4	2.0	16.8
Primary input	22.4	5.6
Total input	38.4	16.8

(For example, x₁₁ = (1/4)(38.4) = 9.6 and x₂₁ = (1/6)(38.4) = 6.4; primary input in sector I = (7/12)(38.4) = 22.4.)

Check: the rows sum to gross output (9.6 + 2.8 + 26 = 38.4; 6.4 + 8.4 + 2 = 16.8) and so do the columns (9.6 + 6.4 + 22.4 = 38.4; 2.8 + 8.4 + 5.6 = 16.8). Total primary input 28 = total final demand 28.

Where the 12 marks sit

#	Component	Marks
1	Technology matrix	2
2	Hawkins-Simon test	2
3	(I − A)⁻¹ and new gross outputs	4
4	New transaction matrix with checks	4
	Total	12

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