7.2 Application to economics: Leontief Model

[Pages:7]7.2 Application to economics: Leontief Model

Wassily Leontief won the Nobel prize in economics in 1973. The Leontief model is a model for the economics of a whole country or region.

In the model there are n industries producing n different products such that the input equals the output or, in other words, consumption equals production. One distinguishes two models: open model: some production consumed internally by industries, rest consumed

by external bodies. Problem: Find production level if external demand is given. closed model: entire production consumed by industries. Problem: Find relative price of each product.

The open Leontief Model

Let the n industries denoted by S1, S2, . . ., Sn. The exchange of products can be described by an

input-output graph

Here, aij denotes the number of units produced by industry Si necessary to produce one unit by industry Sj and bi is the number of externally demanded units of industry Si. Example: Primitive model of the economy of Kansas in the 19th century.

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The following equations are satisfied:

Production of

Total output = Internal consumption + External Demand

farming industry (in tons): horse industry:

x = 0.05 x + 0.5 y y = 0.01 x

+ 8000 + 2000

.

(in 1000km horse rides)

In general, let x1, x2, . . ., xn, be the total output of industry S1, S2, . . ., Sn, respectively. Then

x1 x2

= =

a11 x1 + a12 x1 + ? ? ? + a1n x1 + b1 a21 x1 + a22 x1 + ? ? ? + a2n x2 + b2

xn

... =

an1 x1 + an2 x1 + ? ? ? + ann xn + bn ,

since aij xj is the number of units produced by industry Si and consumed by indus-

try Sj. The total consumption equals the total production for the product of each

industry Si.

Let

a11 . . . a1n

b1

x1

A = ...

... , B = ... , X = ...

an1 . . . ann

bn

xn .

A is called the input-output matrix, B the external demand vector and X the

production level vector. The above system of linear equations is equivalent to the

matrix equation

X = AX + B.

0 In the open Leontief model, A and B = ... are given and the problem is to

0

determine X from this matrix equation.

We can transform this equation as follows:

InX - AX = B (In - A)X = B

X = (In - A)-1B

if the inverse of the matrix In - A exists. ((In - A)-1 is then called the Leontief inverse.) For a given realistic economy, a solution obviously must exist.

For our example we have:

A=

0.05 0.5 0.1 0

, B=

8, 000 2, 000

, X=

x y

.

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We obtain therefore the solution

X = (I2 - A)-1B

=

10 01

-

0.05 0.5 0.1 0

-1

=

0.95 -0.5 -1 8, 000

-0.1 1

2, 000

8, 000 2, 000

=

1 9

10 5 1 9.5

8, 000 2, 000

=

10, 000 3, 000

,

i.e., x = 10, 000 tons wheat and y = 3 Million km horse ride.

If the external demand changes, ex. B =

7, 300 2, 500

, we get

x y

= (I2 - A)-1B

=

1 9

10 5 1 9.5

7, 300 2, 500

=

9, 500 3, 450

,

i.e., one doesn't need to recompute (I2 - A)-1.

One difficulty withthex1model: How to determibn1e the matrix A from a given economy? Typically, X = ... is known, B = ... is known and (aijxj)i,j=1, ... n

xn

bn

is known. One takes therefore the matrix (aijxj)i,j=1, ... n and divides the j-th column

by xj for j = 1, . . ., n to get A.

Example: An economy has the two industries R and S. The current consumption is given by the table

consumption R S external

Industry R production 50 50 20

Industry S production 60 40 100

Assume the new external demand is 100 units of R and 100 units of S. Determine

the new production levels.

Solution: The total production is 120 units for R and 200 units for S. We obtain

X=

120 200

,B =

20 100

,A=

50 50 120 200 60 40 120 200

, and B =

100 100

. The

solution is

X

= (I2 - A)-1B

=

1 41

96 30 60 70

100 100

=

307.3 317.0

.

The new production levels are 307.3 and 317.0 for R and S, respectively.

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The closed Leontief Model

The closed Leontief model can be described by the matrix equation X = AX,

i.e., there is no external demand. The matrix0 In - A is usually not invertible. (Otherwise, the only solution would be X = ... .)

0 The input-output graph looks now as follows:

There is only internal consumption. Example: Extended model of the economy of Kansas in the 19th century including labor.

The corresponding matrix equation is:

x

0.05 0.5

y = 0.1 0

z

0.4 0.1

0.5

x

0.1 y

1331 1800

z

.

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If X is a solution, also t ? X for every t > 0 is a solution. (Usually, one gets a one parameter family of solutions.) If x = 0, we can assume x = 1, 000 by choosing the appropriate parameter t. One obtains then the solution

x = 1, 000,

y

=

2900 11

263.63,

z

=

18000 11

1636.36.

For this computation, it is important to use rational numbers (i.e., fractions) as matrix entries since otherwise the approximation to the matrix In - A usually will be invertible and only the trivial uninteresting solution x = 0, y = 0, and z = 0 will exist. This is also the reason, why the entry a33 has large numerator and denominator.

In a closed economy, the absolute units of output are less interesting. More

important is the relative consumption of a product.

We can normalize thereforeth1e matrix Asu1ch that the sum of every row is 1.

This is a matrix A, such that ... = A ? ... . The recipe is: Divide the i-th

1

1

1

row of A by the i-th component of A ? ... (that is the sum of the i-th row).

1

For our example, we have

21

A

?

1 1

=

20 1 5

,

1

2231

1800

leading to the matrix

1

10

10

A =

21 1 2

21

0

21 1 2

,

720 180 1331

2231 2231 2231

1

1

A ? 1 = 1 .

1

1

The entries of the matrix A = (ai,j)i, j=1, ..., n have the following meaning: aij is the relative consumption of the product of industry Si by industry Sj.

Market prices

The consumption of products is regulated by prices. All income of an industry is used for buying other (or the own) products, i.e., income equals expenditure.

Let P = (p1, . . . , pn) the price vector; pi is the relative price of the product of industry Si. We can draw the flow of money into the input-output graph, the money flows in exchange for the products:

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One has

p1 p2

= =

a11 p1 + a21 p2 + ? ? ? + an1 pn a12 p1 + a22 p2 + ? ? ? + an2 p2

pn

... =

a1n p1 + an2 pn + ? ? ? + ann pn ,

since aij pi is the amount paid by industry Sj for products produced by industry Si. The total income of industry Sj equals the total price Sj has to pay to all other industries.

Again, one can write this as a matrix equation:

P A = P.

This equation can be transformed in the following way

P ? In = P ? A P ? (In - A) = (0, . . . , 0).

1 0

The matrix In-A is (similar as In-A) not invertible, since (In-A) ... = ... .

1

0

0

One can show that this implies that there is also a solution P = ... . Since with

0

P also t ? P for t > 0 is a solution, only the relative price between the different

products has a well-defined meaning.

Example

(continued):

Assume

p1

= $1, 000.

One

gets p2

=

$

40000 63

$ 634.92

and

p3

=

$

11155500 567

$ 1967.37.

We can compare these relative prices with the

production levels measured by the original units and obtain the following relative

prices per

unit:

p1/x

=

1000 1000

=

1 for one ton of wheat,

p2/y

634.92 263

2.4 for

1000km

horse

ride,

and

p3/z

1967.37 1636.36

1.2

for

one

man-year.

Since the above matrix equation for P is not of the usual form which we have

studied so far, we make a final modification. We define

A = (ai,j )i, j=1, ..., n,

where ai,j = aj,i.

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This gives us (just by switching the ro^le of rows and columns) the price equation

P = AP ,

where ai,j is now the relative consumppti1onof industry Sj by industry Si, so that the sum of each column is 1, and P = ... is the price column vector.

pn

In the textbook, our matrix A is again denoted by A and our P is denoted by X. The price equation is therefore X = A ? X. However, one has to keep in mind that this matrix A is different from the input-output matrix A we used in the open Leontief model!

Example: Let

1 1 1

A =

2 1 4

3 1 3

4 1 4

.

111

432

Compute all wages, given that the wages for the 3rd product is $ 30, 000. x

Solution: Let X = y be the different wages with z = 30, 000. We have to z

solve

X = AX 0

(I3 - A)X = 0

0

1

2

-

3 4

-

2 3

2

3

-

3 4

-

3 4

x y

0

= 0 .

-

3 4

-

2 3

1 2

30, 000

0

This system of linear equations for x and y has the solution x = 30, 000 and y = 22, 500. The wages for the first and second product are therefore $, 30, 000 and $ 22, 500, respectively.

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