Input-Output Model

 

 

Wassily Leontief converted the balance-sheets system into the mathematical form, that allows much faster solution. His model is also more general and can be used not only for planning calculations but also for analysis of non-planned market economies.

Let us formulate the basic most simple Input-Output model by converting our example of the balance sheet system

 

 

Take the right-hand columns of each balance sheet and transpose them into rows of the following table:

 

 agbgcgdgcons.publ.inv.
ag142.4280.76547.6516.42170.0045.0015.00
bg274.66121.140.00459.7282.005.0067.00
cg0.00353.33265.53262.70530.0072.00176.00
dg254.31272.57348.51410.4623.0089.00244.00

 

Each row represent quantities of one of the four products. Numbers in the first row are all quantities of the product ag. The second row are quantities of the product bg etc. The columns represent use of these products. For example the numbers in the column cg are quantities of the products that are used in production of the good cg. So 547.56 units of ag is needed as input in production of cg; the product bg is not used in production of cg; 265.53 units of cg is used in its own production, and 348.51 units of dg is also need in production of cg. The four times four (blue) table on the left represents production and use of the goods as intermediate products. The three (red) columns on the right-hand side represent the final use of the four products, that is their use for private consumption, public consumption and investment.

The sum of each row must give the total output of the respective goods. We can add it as the last column.

 

 agbgcgdgcons.publ.inv.total
ag142.4280.76547.6516.42170.0045.0015.001017.25
bg274.66121.140.00459.7282.005.0067.001009.52
cg0.00353.33265.53262.70530.0072.00176.001659.56
dg254.31272.57348.51410.4623.0089.00244.001641.85

 

For convenience we can also insert columns with partial sums of the two parts of the table. After the column dg we can insert column of total use of goods as intermediate product, and after column inv we can insert total final product.

 

 ag bg cg dg intermed. cons. publ. inv. final total
ag 142.42 80.76 547.65 16.42 787.25 170.00 45.00 15.00 230.00 1017.25
bg 274.66 121.14 0.00 459.72 855.52 82.00 5.00 67.00 154.00 1009.52
cg 0.00 353.33 265.53 262.70 881.56 530.00 72.00 176.00 778.00 1659.56
dg 254.31 272.57 348.51 410.46 1295.85 23.00 89.00 244.00 356.00 1641.85

 

Now recall how the individual items of the square array of intermediate product were made. We used a table of technical coefficients:

 

 

The meaning of these coefficients is:  quantity of the row product that is needed to produce one unit of the column product. If we multiply these coefficients by the total output of the column good we get quantity of the row product that must be delivered for production of the total output of the column good, and these are the figures from the square (blue) left-hand side of the above table.
For example 0.14x1017.25 = 142.42 or 0.27x1017.25 = 274.66.

Now we translate the whole thing into the system of equations.

We shall use the following notation:
i   ...index of the row;
j   ...index of the column;
Xij...quantity of the good i used in the production of the good j;
xi ...total output of the product i;
yi ...final use of product i;
aij...technical coefficient; quantity of good i needed to produce
       one unit of the good j;

 

The row sums of the Input-Output table can be written as follows:

 

X11+ X12+ X13+X14+ y1  =  x1

X21+ X22+ X23+X24+ y2  =  x2

X31+ X32+ X33+X34+ y3  =  x3

X41+ X42+ X43+X44+ y4  =  x4

 

It follows from the definition of technical coefficients that

 

Xij  =  aijxj 

 

Using this relation in the equations above we get

 

a11x1+ a12x2+ a13x3+a14x4+ y1  =  x1

a21x1+ a22x2+ a23x3+a24x4+ y2  =  x2

a31x1+ a32x2+ a33x3+a34x4+ y3  =  x3

a41x1+ a42x2+ a43x3+a44x4+ y4  =  x4

 

This is a system of four linear equations with 8 variables (4 y's and 4 x's) and 16 parameters aij  The technical coefficients are given, so if certain conditions are met we can choose values of 4 variables (e.g. all y's) and solve equations for the remaining 4 variables (all x's). The solution will have the following form:

 

b11y1+ b12y2+ b13y3+b14y4  =  x1

b21y1+ b22y2+ b23y3+b24y4  =  x2

b31y1+ b32y2+ b33y3+b34y4  =  x3

b41y1+ b42y2+ b43y3+b44y4  =  x4

 

Where coefficients bij are calculated from the coefficients aij.
As the equations are linear it is convenient to use matrix notation of linear algebra.

A matrix is a rectangular array of numbers with let us say n rows and m columns. In our case the matrices are square, with the same number of rows and columns. The matrix having one row only is called a row vector and the matrix having one column is called a column vector. Single numbers are called scalars.

The operations with matrices and vectors are defined in the following way:

Matrix addition (subtraction): add (subtract) pairs of equally situated matrix elements. Obviously only matrices of same dimensions can be added (subtracted).

 

Example:

Let  a =  (a1 a2) and b = (b1 b2) then

 a + b =  (a1+ b1  a2+ b2)

 a - b =  (a1 - b1  a2 - b2)

 

Vector multiplication (division is not defined):
inner or scalar product: the first vector must be a row vector and the second must be a column vector. The result is a sum of the products of equally situated elements.

 

Example:
The vectors a and b are row vectors.
Let us transpose b into column vector and denote it b'.
Then

ab'  =  [a1 a2] b1
b
2
 = a1b1 + a2b2

 

 

The result is a scalar.
Matrix vector multiplication: Let C is a 2x3 matrix of elements cij, e.i. a matrix of 2 rows and 3 columns. Such a matrix can be premultiplied (e.i. multiplied from the left) by a row vector of 2 elements and postmultiplied by a column vector of 3 elements.

 

Example:
 

aC =

[a1 a2]

c11

c12

c13

=

c21

c22

c23

 

= [a1c11+a2c21  

  a1c12+a2c22  

 a1c13+a2c23]

                                                                                 
The result is a row vector of 3 elements.

 

Cb' =     b1= 
c11c12c13b2c11b1+c12b2+c13b3
c21c22c23b3c21b1+c22b2+c23b3
       
   


The result is a column vector of 2 elements.

 

Now we can put the system of input-output equations into matrix form:

 Ax + y = x

Where A is a matrix of technical coefficients, x is a vector of total output and y is a vector of final product.To solve the equations for the vector of total product x we can perform the folowing steps:

 y  =  x - Ax

 y  =  (I - A)x

where I is an "identity matrix", e.i. a diagonal matrix with 1's on the main diagonal and 0's elswhere. Matrix I - A is known as the Leontief matrix.

 (I - A)-1y  =  x

(I - A)-1 is the so called "inverse matrix" to the Leontief matrix (I - A). An inverse matrix is defined so that  (I - A)-1 (I - A) = I. The coefficients bij used in the solution above are elements of the inverse of Leontief matrix.

Let us now return to our numerical example. First our matrix of technical coefficients was

  .14.08.33.01
A=.27.12.00.28
.00.35.16.16
  .25.27.21.25

The Leontief matrix is then

  .86-.08-.33-.01
I - A=-.27.88-.00-.28
-.00-.35.84-.16
  -.25-.27-.21.75

and it inverse

  1.41.49.64.34
(I - A)-1=.701.60.45.70
.46.851.53.65
  .85.97.811.88

If this matrix is postmultiplied by the column vector of final product from the balance-sheets example we should get the well balanced vector of total output.

   1.41.49.64.34230=1017.25
(I - A)-1y=.701.60.45.701541009.52
.46.851.53.657781659.56
   .85.97.811.883561641.85

It is now easy to get the new vector of total output as a result of the change in final product. For example let us increase the second element of y by 33 units.

   1.41.49.64.34230=1033.29
(I - A)-1y=.701.60.45.701871062.17
.46.851.53.657781687.62
   .85.97.811.883561674.01

All four elements of the vector of total output increased.

 

 

 

 

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