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Lecture 14' Inputoutput analysis II'

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Title: Lecture 14' Inputoutput analysis II'


1
Lecture 14. Input-output analysis II.
  • Learning objectives. By the end of this lecture
    you should
  • Know more about input output analysis
  • Understand how to check for the productiveness of
    an input-output system.
  • Introduction Inputs and outputs.
  • In the last lecture we learnt the basics of
    input-output analysis.
  • A key question is whether a given input
    coefficients matrix makes sense, meaning
  • Given the input coefficient matrix and provided
    there are sufficient primary inputs, can any
    pattern of final demands be produced?
  • Example. OK Computers PLC buys PCs from 3
    suppliers. From the first it keeps the keyboard
    and throws everything else away. From the second
    it keeps the monitor, throwing everything else
    away and from the third it throws away the
    monitor and keyboard. So from 3 computers it
    produces 1 new one.

2
2. Example
  • A I-A
  • So (I-A)-1
  • If then x (I-A)-1d
  • In other words, to meet final demand for 1 unit
    of the first good, 1.33 units of that good must
    be produced along with 0.46 units of good 2 and
    0.53 units of good 3.

3
3. Meeting demand
  • Another way to think about this
  • To meet final demand d we require
  • d the final demand
  • Ad the direct intermediate inputs
  • A(Ad) the inputs required for the direct
    intermediate inputs
  • A(A2d) the inputs required
  • Or x d Ad A2d A3d . (IAA2A3)d
  • Question on feasibility
  • Given any final demand vector d, is there an
    input vector x that will produce d?
  • Obviously to be feasible no element of x can be
    negative
  • And no element can be infinite

4
3. Meeting demand
  • Given that,
  • x d Ad A2d A3d . (IAA2A3)d and
  • All the elements of A are non-negative (so that
    all the elements of An must be positive), then
  • Its clear that x wont be negative. But can it
    be infinite?
  • Define S (IAA2A3)
  • Define Sn (IAA2A3An)
  • In other words S is the limit of Sn as n ?8.
  • Note that ASn AA2A3An1
  • So Sn ASn (IAA2A3An) (AA2A3An1)
  • I - An1
  • Or
  • (I-A)Sn I - An1
  • So if (I-A)S I or S (I-A)-1 and x exists.

5
4. Hawkins-Simons conditions
  • The Hawkins-Simons conditions are conditions on A
    which guarantee that given any final demand
    vector, d, there is an input vector, x, which
    will produce d.
  • There are different, but equivalent statements of
    the conditions. We shall state 3 and consider the
    first 2
  • The principal minors of (I-A) are all positive.
  • The dominant eigenvalue is less than 1.
  • (dont know what an eigenvalue is? Given a matrix
    A its a solution, ? to the equation . The
    dominant eigenvalue is the one with the largest
    absolute value)
  • we are going to define this term on the next
    slide
  • If the conditions are satisfied then the
    input-output system is said to be productive.

6
4. Hawkins-Simons conditions
  • This is the condition we derived earlier. Its
    simple, but may be hard to calculate.
  • The principal minors of (I-A) are all positive.
  • Given an nxn matrix, A, a principal matrix is
    found from A by deleting k rows (e.g. rows 2, 5
    and 7) and the same k columns (so columns 2, 5
    and 7). 0 k n-1. k is called the order of the
    principal matrix.
  • The principal minor is the determinant of the
    relevant principal matrix.
  • So for a 3x3 matrix, there are
  • 1 0-order principal matrix (A itself)
  • 3 first order principal matrices
  • 3 second order principal matrices.
  • So you would need to check 7 determinants.

7
4. Hawkins-Simons conditions -example
  • Suppose A Are the conditions satisfied?
  • I-A is
  • The Principal matrices are I-A itself and,
  • The relevant determinants are
  • 0.352, 0.52, 0.59, 0.54, 0.8, 0.7, 0.8, so the
    conditions are satisfied.
  • Note how when we eliminate n rows and columns we
    end up with the diagonal elements of A.

8
4. Hawkins-Simons conditions -exercise
  • Suppose A
  • Find the principal matrices (no need to calculate
    their determinants).
  • Note principal minors will resurface in the next
    topic.

9
4. Hawkins-Simons conditions OK computers example
  • Suppose two goods, computers and software. To
    produce the computer 0.1 unit of software is
    required and to produce 1 unit of software, 0.3
    units of computer and 0.2 units of software are
    needed.
  • A
  • I-A
  • Principal minors are, -2, 0.8 and -1.63.
    Obviously not all positive.
  • Note the basic point here if it takes more than
    one unit of a good to produce that good, then the
    Hawkins-Simons conditions can never be satisfied.

10
5. Employment multipliers
  • Often the number of jobs apparently lost as the
    result of a business closing or the number of
    jobs created as the result of new business
    opening seems far in excess of the number of jobs
    actually with the specific company.
  • Input output analysis can be used to calculate
    the knock on effects of changes in employment.

11
5. Employment multipliers
  • Suppose the only primary input is labour. Recall
    that,
  • Note first that this labour demand is in value
    terms
  • Consider a change in final demand of ?d, then the
    change in the value of labour demand will simply
    be ?d (because of the circular flow of payment
    within the economy and the fact that we assumed
    that there is only one input).
  • It is still useful to break this down into
    sectors using

12
How we calculated l (previous lecture)
  • Recall final inputs receive the final payments
    the requirement for primary inputs is given by 1
    sum of the column entries in the input matrix
  • In the example, 1 of IT goods needs 0.25 input
    of primary inputs.
  • The primary inputs coefficient vector, l, is the
    vector of these values

13
5. Employment multipliers
  • Example. Suppose d1 falls by one unit, what
    happens to the value of employment?

14
5. Employment multipliers
  • Now l1 0.25, so from a 1 unit drop in demand
    we get
  • a 0.25 direct drop in the value of employment in
    sector 1
  • a further 0.25 indirect drop in the value of
    employment in sector 1
  • a 0.27 indirect drop in sector 2
  • a 0.23 indirect drop in sector 3.
  • The employment multiplier is the ratio of the
    total change in employment to the direct drop.
  • The multiplier is therefore
  • I.e. for every one job lost due to the immediate
    effect of the drop in demand, there are 3 jobs
    lost indirectly.

15
Conclusion.
  • Exercise.
  • identify the direct and indirect effects on
    employment if ?d2 2
  • what is the employment multiplier for d2?

16
Conclusion.
  • 7. Summary
  • 4 definitions learnt
  • Hawkins-Simons conditions
  • Principal minors
  • 4 skills you should be able to do
  • Identify the Principal minors of a square matrix
  • Check that the Hawkins Simons are satisfied
  • Calculate employment multipliers.
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