San José State University
Department of Economics

applet-magic.com
Thayer Watkins
Silicon Valley
& Tornado Alley
USA

Input-Output Analysis and Related Methods

Introduction
Hawkins-Simon
Condition
Dynamics

Input-output analysis is one of a set of related methods which show how the parts of a system are affected by a change in one part of that system. Input-output analysis specifically shows how industries are linked together through supplying inputs for the output of an economy. Suppose there are only two industries producing Coal and Steel. Coal is required to produce steel and some amount of steel in the form of tools is required to produce coal. Suppose the input requirements per ton output of the two products are:

Industry Coal Steel
Coal 0 3
Steel 0.1 0

Suppose it desired that the Coal industry produce a net output of 200,000 tons of coal and the Steel industry a net output of 50,000 tons. If the Coal industry just produces 200,000 tons and the Steel industry produces 50,000 tons some of outputs are used up in producing the other output. To produce 50,000 tons of Steel requires 3(50,000)=150,000 tons of coal. Likewise the production of 200,000 tons of coal requires (0.1)(200,000)=20,000 tons of steel. The net outputs of coal and steel would then be 200,000-150,000=50,000 tons of coal and 50,000-20,000=30,000 tons of steel. In other words, in order to get net outputs of 50,000 tons of coal and 30,000 tons of steel it is necessary to produce 200,000 tons of coal and 50,000 tons of steel. But we want net outputs of 200,000 tons of coal and 50,000 tons of steel. We would at least have to produce an additional amount to replace the coal and steel used up in producing 200,000 tons of coal and 50,000 tons of steel. Those amounts, as we saw above, are 150,000 tons of coal and 20,000 tons of steel. But in producing these amount we will also use up coal and steel. In fact, we will use up 3(20,000)=60,000 tons of coal and (0.1)(150,000)=15,000 tons of steel. So we must increase production again to cover these amounts.

We can think of these production increases as Rounds of production. In Round 1 we produce the net outputs we are trying to achieve. In Round 2 we produce the outputs that were used up in producing Round 1 outputs. Then in Round 3 we produce the outputs used up in producing Round 2, and so on. The figures are given in the following table.

Round Coal Steel
1 200,000 50,000
2 150,000 20,000
3 60,000 15,000
4 45,000 6,000
5 18,000 4,500
6 13,500 1,800
7 5,400 1,350
8 4,050 540
9 1,620 405

The totals for the first nine rounds are 497,570 tons of coal and 99,595 tons of steel. This approximate how much we must produce to achieve the net outputs we seek. The exact amounts, found by more advanced methods, are 500,000 tons of coal and 100,000 tons of steel. We see that 3(100,000)=300,000 tons of coal are used up in producing the 100,000 tons of steel leaving 500,000-300,000=200,000 tons of coal as net output. Also (0.1)(500,000)=50,000 tons of steel are used up in producing the 500,000 tons of coal leaving us with 100,000-50,000=50,000 tons of steel as net output.

We find the exact figures by solving two algebraic equations. If x1 is the output of coal and x2 is the output of steel, then the conditions that have to be satisfied are:

x1 - 3x2 = 200,000
x2 - 0.1x1 = 50,000.

This is a set of two equations in two unknowns and we can solve it using simple algebra.

A systematic way to solve for any set of net outputs is to find how much coal and steel are needed to produce a net output of one ton of coal. The equations which have to be satisfied are:

x1 - 3x2 = 1
x2 - 0.1x1 = 0.

The solution is x1=1.42857 and x2=0.14286. To get the amounts necessary to produce a net output of 200,000 tons we multiply these figures by 200,000. The outputs required are 285,714 tons of coal and 28,571 tons of steel.

For a net output of one ton of steel the equations to be satisfied are:

x1 - 3x2 = 0
x2 - 0.1x1 = 1.

The solution is x1=4.28571 and x2=1.42857. To get the amounts necessary to produce a net output of 50,000 tons of steel we multiply these figures by 50,000. We get an output of coal of 214,286 tons and 71,429 tons of steel. The total outputs required for 200,000 tons of coal and 50,000 tons of steel are then

285,714 + 214,286 = 500,000 tons of coal
and
28,571 + 71,429 = 100,000 tons of steel.

The outputs of coal and steel required to achieve a net output of one ton of coal may be considered the direct and indirect requirements for one ton of coal. Likewise for the outputs necessary for a net output of one ton of steel. These can be put together in a table to contrast them with the direct requirements of production:

Industry Coal Steel
Coal 0 3
Steel 0.1 0

Direct Requirements

Industry Coal Steel
Coal 1.42857 4.28571
Steel 0.14286 1.42857
Direct and Indirect Requirements

The about table is the one that gives the most information about how the industries are inter-related.


The direct and indirect requirements are usually determined using matrix operations. A matrix is simply a rectangular array of numbers; i.e., a table. A matrix is characterized by the number of rows and columns. An n by m matrix, written n×m, is a matrix with n rows and m columns. A matrix with only one column is also called a column vector. If a matrix has only one row it is usually called a row vector.

The element of matrix A that is in the i-th row and j-th column is denoted as Ai,j. The addition of two matrices A and B with the same number of rows and columns is defined as the matrix C such that Ci,j=Ai,j+Bi,j. Subtraction of matrices is defined in an analogous manner.

The definition of multiplication of matrices that is useful is not the multiplication of corresponding elements. Instead multiplication is defined only for matrices having the proper number of rows and columns; i.e., the number of rows of the second matrix has to be equal to the number of columns of the first matrix. If A is an n×m matrix and B is a m×p matrix then their product AB is defined to be an n×p matrix C such that

Ci,j = ΣAi,kBk,j

where the summation is over k=1 to k=m.

The levels of production of the Coal and Steel industries can be represented as a 2×1 matrix (a column vector) X where

X = | x1 |
  | x2 |

Likewise the levels of net output required to be met, usually called final demands, f1 and f2 can be represented as a column vector (2×1 matrix) F where

F = | f1 |
  | f2 |

The direct requirements per units of output for an economy with two industries can be represent as a 2×2 matrix A; i.e.,

A = | A1,1   A1,2 |
  | A2,1   A2,2 |

For the Coal/Steel economy above

A = |   0   3 |
  | 0.1   0 |

The amounts of production used up in producing x1 and x2 are equal to

A1,1x1 + A1,2x2
A2,1x1 + A2,2x2

This is just the product of the matrix A and the matrix X; i.e., AX. The production levels are given by X so the net productions after the amounts used up in production are deducted are given by

X - AX

We want this to be equal to the required levels F so the equations to be satisfied are, in matrix form,

X - AX = F

At this point it is necessary to note that there are special type of square matrices, called identity matrices and denoted as I, that consist of 1's on the diagonal that runs from the upper left to the lower right and 0's everywhere else. The 2×2 identity matrix is

I = | 1   0 |
  | 0   1 |

The virtue of the identity matrices is that the product of an identity matrix with any other matrix for which the product is defined is just the other matrix. In particular, IX is just X. It turns out that it is often useful to represent a matrix as a product with the identity matrix. The equations to be satisfied by the levels of output are, in matrix form,

X - AX = F
which can be expressed as
IX - AX = F

The advantage of this second representation is that it can be factored; i.e.,

IX - AX = (I-A)X
so the matrix equations to be satisfied is
(I-A)X = F

For the Coal/Steel economy above

I-A = |  1   -3 |
     | -0.1   1 |

The solution to this equation would involve carrying out some algebraic operation so we end up with X being equal to some matrix. Generally the problem we face is that we have a matrix equation BX=C and we want to end up with X=something. Suppose that we could find a special matrix D such that DB=I. Then we could multiply both sides of the matrix equality BX=C to get DBX=DC. Since DB=I this means we would have IX=DC, but IX is the same as X so we have the solution to the equations as X=DC. The special matrix D such that DB=I is called the inverse of B and it is denoted as B-1. So the solution to the matrix equation BX=C is X=B-1C. The only problem in finding a solution to BX=C is finding B-1.

At this point we do not know even if such a matrix exists. It turns out that there is a simple test to determine whether an inverse exists. The test is based upon the determinant of the matrix. If the determinant is not equal to zero then an inverse exists and if the determinant is equal to zero then an inverse does not exist. For a 2×2 matrix B the determinant of B is

det(B) = B1,1B2,2 - B1,2B2,1

For a 2×2 matrix the determination of the inverse is very simple but it is not as simple for higher order matrices. to get the inverse of a 2×2 matrix B we just interchange B1,1 and B2,2 and change the signs of B1,2 and B2,1 and then divide all of the elements of the resulting matrix by det(B); i.e.,

B-1 = | B2,2/det(B)   -B1,2/det(B) |
  | -B2,1/det(B)    B1,1/det(B) |

For the Coal/Steel economy above the (I-A) was

I-A = | 1   -3 |
      | -0.1   1 |

so it determinant is (1)(1)-(-3)(-0.1)=1-0.3=0.7. Thus it does have an inverse and that inverse is

(I-A)-1 = (1/0.7)| 1     3 |
               | 0.1     1 |

or, carrying out the indicated division,

(I-A)-1 = |  1.4286     4.2857 |
  | 0.1429     1.4286 |

When we multiply this matrix times the vector of final demands F, where

F = | 200,000 |
  |  50,000 |

we get the solution

X = | 500,000 |
  | 100,000 |


In conclusion, the direct and indirect requirements per unit of final demands are given by the columns of the inverse of the matrix I-A; i.e., (I-A)-1.

For the Coal/Steel economy the inverse of (I-A)

(I-A)-1 = |  1.4286   4.2857 |
  | 0.1429     1.4286 |

tells us that for each ton of final demand for coal the economy has to produce 1.4286 tons of coal and 0.1429 tons of steel. For each ton of final demand for steel the economy has to produce 4.2857 tons of coal and 1.4286 tons of steel.


Interregional and International Interaction
Using the Methods of Input-Output Analysis

Input-Output Analysis arose to deal with the problem of interindustry demand, but the same method can be used to show how changes in one region affect the economies of regions linked to it. Suppose we have information on how changes in production in Santa Clara and Santa Cruz Counties affect the demand for each other's output. (Santa Clara County is essentially the famed Silicon Valley and Santa Cruz County is a county to the south of it over the Santa Cruz Mountains and on Monterey Bay of the Pacific Ocean.) If production in Santa Clara County increases there will be more income not only for residents of Santa Clara County but also for the residents of Santa Cruz County because some of the jobs in Santa Clara County will go to Santa Cruz County residents. The residents of both counties will decide how much of their income they will spend, where, and for what. Some of that spending will be in the two counties and be for goods and services that are produced locally. Likewise when production in Santa Cruz County increases some of the jobs will go to Santa Clara County residents and some of these will spend their income in Santa Cruz County as well as in Santa Clara County. Suppose we have that information in matrix form:

 County of Production
County of
Residence
Santa Clara Santa Cruz
Santa Clara 0.5 0.1
Santa Cruz 0.2 0.4

This is like a matrix of marginal propensities to consume in macroeconomic theory. In macroeconomic theory if income is spent for products outside of the economy it is considered a leakage. In this regional setting some of these leakages leak back into the economy.

The matrix above corresponds to the matrix A in input-output analysis.

In macroeconomic theory the income multiplier k is equal to:

1/(1-c)

where c is the marginal propensity to consume.

This could be written as:

k = (1-c)-1.

With the regional interaction there is a matrix of multipliers and the matrix is equal to:

(I-A)-1.

For the above matrix the matrix I-A is:

 County of Production
County of
Residence
Santa Clara Santa Cruz
Santa Clara 0.5 -0.1
Santa Cruz -0.2 0.6

The determinant of I-A is (0.5)(0.6)-(-0.1)(-0.2)=0.30-0.02=0.28.

This means the matrix I-A does have an inverse. Remember that for a 2x2 matrix the inverse is found by interchanging the diagonal elements and changing the sign of the off-diagonal elements, then dividing every element by the determinant. This gives:

 County of Production
County of
Residence
Santa Clara Santa Cruz
Santa Clara 2.14286 0.35714
Santa Cruz 0.71429 1.78571

This means that when the demand for Santa Clara County's output increases by $1 the output in Santa Clara County increases by $2.14 and in Santa Cruz County by $0.71. On the other hand, if the demand for Santa Cruz County's output increases by $1 then output in Santa Cruz County increases by $1.79 and in Santa Clara County by $0.36.


The above shows how once the matrix A is known how the inter-relationships between the parts is determined in the form of the inverse of the (I-A) matrix. So once A is determined the rest is merely numerical computation. But the matrix A first has to be established. The derivation of the matrix A involves several economic processes. First, there is the distribution of income (and jobs) to the subregions. This is given in the form of a matrix which will be called the matrix J (for jobs). Suppose J has the following value:

 County of Production
County of
Residence
Santa Clara Santa Cruz
Santa Clara 0.75 0.20
Santa Cruz 0.25 0.80

This says that 75% of the jobs and income go to Santa Clara County residents and 25% go to residents of Santa Cruz County. On the other hand, 20% of the jobs and income in Santa Cruz County go to Santa Clara County residents and the other 80% to Santa Cruz County residents.

But not all of a dollar of production goes for labor income. Let us say that in both counties one third of the revenue goes for labor income. This means that the effect of additional dollars of production would have the following effects on incomes. This is the matrix Y (for income).

 County of Production
County of
Residence
Santa Clara Santa Cruz
Santa Clara 0.250 0.067
Santa Cruz 0.083 0.267

There is also the matrix that tells where people spend their money and how much of it goes for local production. This is the matrix S (for spending).

 County of Residence
County of
Spending
Santa Clara Santa Cruz
Santa Clara 0.80 0.30
Santa Cruz 0.10 0.60

This says that that when Santa Clara residents get another dollar of income 80% is spent in Santa Clara County and another 10% is spent in Santa Cruz County. On the other hand, when Santa Cruz residents get another dollar of income 30% is spent in Santa Clara County and 60% at home in Santa Cruz County. In both cases all of the spending goes for goods or services which are produced in the county of the spending.

Note that the orientation of this table is opposite of the previous tables.

To construct the matrix A we have to follow a dollar of production to its dispersement as income to the two counties and the allocation of the recipients spending between the two counties. This illustration is going to leave out several other important economic processes such how much of labor income goes for taxes, savings and imports. These omissions are to keep the detail to a minimum.

According to matrix J, when a dollar of production is produced in Santa Clara County, $0.25 goes to Santa Clara County residents who spend 80% of it in Santa Clara County and $0.083 goes to Santa Cruz County residents who spend 30% of it in Santa Clara County. Altogether then the $1 of production in Santa Clara County leads to (0.25)(.8)+(0.083)(.3)=0.225 of addition consumer demand in Santa Clara County. This is the element in the first row, first column of the matrix A. The dollar of additional production also leads to increased demand in Santa Cruz County; i.e,. (0.25)(0.1)+(0.083)(.6)=0.075. This is the element in the second row, first column of A.

When an additional dollar of production takes place in Santa Cruz County the additional spending in Santa Clara County is (0.063)(0.8)+(0.267)(.3)=0.131, the element of the A matrix in the first row, second column. The final element is the spending in Santa Cruz County resulting from an additional dollar of production in Santa Cruz County. This is (0.63)(0.1)+(0.267)(0.6)=0.167. Thus the A matrix is

 County of Production
County of
Residence
Santa Clara Santa Cruz
Santa Clara 0.225 0.131
Santa Cruz 0.075 0.167

The (I-A) is then

 County of Production
County of
Residence
Santa Clara Santa Cruz
Santa Clara 0.775 -0.131
Santa Cruz -0.075 0.834

The determinant of this matrix is 0.6365, so there is an inverse. That inverse is

 County of Production
County of
Residence
Santa Clara Santa Cruz
Santa Clara 1.3102 0.2058
Santa Cruz 0.1178 1.2175

This says that when there is an additional dollar of demand in Santa Clara County production goes up by $1.31 in Santa Clara County and about $0.12 in Santa Cruz County. On the other hand, when demand in Santa Cruz County increase by one dollar production in Santa Clara County increases by about $0.21 and about $1.22 in Santa Cruz County. But these are increases in sales and production rather than income.

The increases in income are found by determining the proportion of production going to income (one third in this example) and then distributing it according to the J matrix. The result of this computation for Santa Clara County income due to an increase in Santa Clara production is (1/3)(1.3102)(0.75)+(1/3)(0.1178)(0.20)=0.335. The results of the computation is then

 County of Production
County of
Residence
Santa Clara Santa Cruz
Santa Clara 0.335 0.133
Santa Cruz 0.141 0.342

(To be continued.)

For the application of this methodology to prices see Input-Output and Prices. Also see Linear Programming.


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