# Applications of Matrices Leontif Input-Output Models by X0kg4Mt

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Applications of Matrices
Leontif Input-Output Models
Models for Simple Economies
   Suppose we consider a simple economy
with a lumber industry and a power industry.
   Suppose further that production of 10 units
of power require 4 units of power and 25
units of power require 5 units of lumber.
   10 units of lumber require 1 unit of lumber
and 25 units of lumber require 5 units of
power.
   If surplus of 30 units of lumber and 70 units
of power are desired, find the gross
production of each industry.
Models for Simple Economies:
Creating a Technology Matrix

   In the previous
problem we have               Outputs
the following
information
Inputs   Power   Lumber
(converting all
numbers to
percentages)      Power     0.4     0.2

Lumber    0.2     0.1
Models for Simple Economies:
Creating a Technology Matrix

   We can convert                    Outputs
this information to   Inputs   Power   Lumber
the following         Power     0.4     0.2
technology matrix
Lumber    0.2     0.1
or Leontif matrix

 .4 .2 
A
 .2 .1
       
Models for Simple Economies:
The gross production matrix

   The gross production matrix for the economy
can be represented by the column matrix
 x1 
X     
 x2 
    
   Where x1 is the gross production of power
and x2 is the gross production of lumber.
Models for Simple Economies:
The technological equation

   X (or IX where I is the identity matrix)
is the amount of production that is
desired.
   AX is the amount that is actually
produced.
   So IX-AX=(I-A)X is the amount of
surpluses, D, (also called final
demands)
Models for Simple Economies:
The technology equation

( I  A) X  D
   Is called the technology equation of the
Leontif equation.
Models for Simple Economies

   Original Question :
If surplus of 30

 70
units of lumber and

D 
70 units of power

 30 
are desired, find
the gross
production of each
industry.
 
   Find X
Models for Simple Economies

   To find X, take the inverse of (I-A) if it
exists

( I  A) X  D
1
X  ( I  A) D
Models for Simple Economies
  1 0   .4 .2    .6  .2 
I  A                    
  0 1    .2 .1     .2 .9 
                         
                            

Finding the inverse of I-A
 .6  .2 1 0  1  .333 1.667 0  1  .333 1.667 0 

  .2 .9 0 1     .2 .9 0 1    0 .833 .333 1 
                                    
                                                 

1  .333 1.667 0  1 0 1.8 0.4 
 0 1 0.4 1.2    0 1 0.4 1.2 
                              
                              
Models for Simple Economies
   Then the inverse of I-A
1   1 .8 .4 
is
 .4 1 .2 
( I  A)           
   To find the amount to                         
produce for the given                 1
amount of surplus, we     X  ( I  A) D
must multiply the
inverse of I-A and D         1.8 .4  70  138
 .4 1.2  30    64 
X                    
Hence the gross

production are :                                
   Lumber : 64 units
   Power : 138 units
Models for Simple Economies:
Another Example
   The economy of a developing nation is based on
agricultural products, steel, and coal. An input of 1
ton of agricultural products requires an input of 0.1
ton of agricultural products, 0.02 ton of steel, and
0.05 ton of coal. An output of 1 ton of steel requires
an input of 0.01 ton of agricultural products, 0.13
tons of steel, and 0.18 tons of coal. An output of 1
ton of coal requires an input of 0.01 ton of
agricultural products, .2 tons of steel, and 0.05 ton
of coal. Find the necessary gross productions to
provide surpluses of 2350 tons of agricultural
products, 4552 tons of steel, and 911 tons of coal.
   What is the technology matrix?
Models for Simple Economies:
Another Example

Output

 0 .1 .01 .01 
Input       Agriculture   Steel   Coal                     
A   .02 .13 0.2 
Agriculture       0.1       0.01    0.01        .05 .18 .05 
              
Steel         0.02       0.13    0.2

Coal          0.05       0.18    0.05
Models for Simple Economics:
Another Example

   What is the surplus matrix?
   Find the technological equation.
   What is (I-A)-1?
   What is the production matrix?
Models for Simple Economics:
Another Example

   What is the surplus
matrix?                    2350 
      
D   4552 
 911 
      
Models for Simple Economics:
Another Example

   Find the
technological
equation.
( I  A) X  D
  1 0 0   0.1 .01 .01          2350
                                   
  0 1 0    .02 .13 .20   X   4552
  0 0 1   .05 .18 .05          991 
                                   
Models for Simple Economies:
Another Example

   What is (I-A)-1?

 1.11 .016 .015
1
                
( I  A)   .041 1.20 .254
 .066 .229 1.10 
                
Models for Simple Economics:
Another Example

   What is the production
matrix?
   Thus in our to have the                     2700
given surplus in this                 1
      
problem 2700 units of
X  ( I  A) D   5800 
 2200
agriculture, 5800 units                          
of steel, and 2200
units of coal must be
produced.
Closed vs Open Leontif
Models

   In both of the previous examples some
amount of surplus in production. That is
 0
 
D   0
 0
 

These models are called Open Leontif Models.
Closed vs Open Leontif
Models
   If a company want to find out the amount of
production need with no surplus, that is

 0
 
D   0
 0
 
Then the model is called a Closed Leontif
Model.
Closed Leontif Models

   What is the technological equation for
a closed leontif model?
Closed Leontif Models

   What is the technological equation for a
closed leontif model?

( I  A) X  0
   Where 0 is actually a column matrix of
all zeros.

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