ANSWERS TO SELECTED QUESTIONS


4-10

Given the interindustry transactions matrix (X) and the gross outputs vector (z), calculate values for the final- demand (y), final-payments (v), and input (q) vectors.

IMAGE imgs/chap0912.gif

Answer:

IMAGE imgs/chap0913.gif

4-11

5-6

EMULTj = e -1(I - A) -1

5-Exercise

Consider the following interindustry transactions matrix (X) and total outputs vector (z) for a two- sector economy.

IMAGE imgs/chap0914.gif

a. What are the two elements in the final demand (fd) vector?

IMAGE imgs/chap0915.gif

b. What are the two elements in the total inputs vector (q) and in the final-payments (fp) vector? IMAGE imgs/chap0916.gif

What does the complete transactions table look like now?

c. What is the algebraic representation of the A matrix, where aij = xij/qj?

A = X-1

or

IMAGE imgs/chap0917.gif

Calculate the numeric values for this A matrix and for (I-A):

IMAGE imgs/chap0918.gif

d. Using your hand-held calculator or a spreadsheet, what is the inverse of this Leontief matrix?

IMAGE imgs/chap0919.gif

e. With this A matrix and the following new final-demand vector (now y), calculate the first 3 rounds of spending which would occur in seeking the new outputs which would satisfy the new final-demand.

IMAGE imgs/chap0920.gif

What is the equation for this 'rounds of spending' solution to the model?

It is the matrix equation for the sum of an infinite series:

T = A0y + A1y + A2y + A3y + ... + Any,

Where y is the vector to be traced and A0 turns out to be the identity matrix.

Round 0:

IMAGE imgs/chap0921.gif

Round 1:

IMAGE imgs/chap0922.gif

Round 2:

IMAGE imgs/chap0923.gif

f. How would the solution using the inverse compare with that using rounds of spending as above?

It would be identical. In this case, however, with such high coefficients, convergence would take over 20 rounds.

8-3

3. In Illustration 8-1, provide the details in the "manipulation to solution" from

g = D(Bg + e + x - (Bg + e)) to

g = (I - D(I - )B)-1D((I - )e + x):

g = D(Bg + e + x - (Bg + e))

Expand, regroup, and manipulate to solution in terms of g:

g = DBg + D(e + x) - D (Bg + e))

g = DBg + De + Dx - DBg - De

g = DBg - DBg +De - De + Dx


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