leontief input-output model example problem

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An application problem involving the Leontief input-output model with thorough solution from an honors course in linear algebra.The text for the course was Linear Algebra and Its Applications, 4e, David Lay.

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Page 1: Leontief Input-Output Model Example Problem

The Leontief Input-Output Model (MAT-208)

Joseph Heavner

December 11, 2015

PROBLEM:

An economy is divided into three sectors – services, raw materials, and manufacturing. For each unitof output services requires 0.04 units from other companies in that sector, 0.03 units of raw materials,and 0.02 units from manufacturing. For each unit of output the raw materials sector uses 0.05 unitsof services, 0.04 units of raw materials, and 0.30 units from manufacturing. For each unit of outputmanufacturing uses 0.02 units of services, 0.04 units of raw materials, and 0.20 units of its own output.Determine the production levels needed to satisfy a final demand of 300 units of services, 500 units ofraw materials, and 600 units for manufacturing.

1. Find the system of equations.

2. Show the augmented matrix.

3. Solve the system. You may use your calculator.

4. Interpret the solution in complete sentences, stating exactly what each value represents.

5. Re-state the solution, assuming that all input and output units are measured in millions of dollars.

SOLUTION:

1. Let s be the production of services, r be the production of raw materials, m be the production ofmanufacturing, S be the amount of material demanded from services, R be the amount of materialdemanded from raw materials, and let M be the amount demanded from materials. The coefficients ofthe system represent the amount of input a particular industry needs from another industry to producta unit of output.

.04s + .05r + .02m = S = 300

.02s + .30r + .20m = M = 500

.03s + .04r + .04m = R = 600

2.

I − C =

1 0 00 1 00 0 1

−.04 .05 .02.02 .30 .20.03 .04 .04

=

.96 −.05 −.02−.02 .70 −.20−.03 −.04 .96

=⇒

.96 −.05 −.02 300−.02 .70 −.20 500−.03 −.04 .96 600

3. .96 −.05 −.02 300

−.02 .70 −.20 500−.03 −.04 .96 600

∼ · · · ∼1 0 0 374.365

0 1 0 917.8220 0 1 674.941

4. The above system says that the production of services is 374.365, the production of manufacturing is917.822, and the production of raw materials is 674.941. This is because the basic variables, i.e. thosewith coefficient 1, or those with value 1 in the matrix, represent the production of a particular industry,and we can just translate the matrix to a system to understand what it all means.

5. The production of services is $374,365,000; the production of manufacturing is $917,822,000; theproduction of raw materials is $674,941,000.

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