Call Now: (800) 537-1660
 Home    Why?     Free Solver   Testimonials    FAQs    Product    Contact

May 20th

May 20th

# MATH 115 - FINITE MATHEMATICS

Leontief Model

The problem section is located on the last page of this activity. Turn in your answers, written in complete
sentences on the sheet, next class.

Introduction. Suppose we consider a (very) simplified model of the economy, in which we only distinguish
three different sectors : Resources (growing crops, mining coal, and other activities that produce important
materials), Manufacturing (turning crops into meals and iron and coal into cars), and Services (such as
teaching Finite Mathematics). Of course, doing each of these things requires the others. Manufacturing
requires both resources to provide the raw materials and energy and the services of skilled craftsmen and
designers for example. Indeed, each of these sectors also consumes some of its own outputs. The robots that
manufacture cars must themselves be manufactured. Suppose we have the fol lowing table to describe the
situation (all numbers are made up).

 Matrix A To Produce each unit of Resources Manufacturing Services Requires Resources 0.3 0.5 0.2 this many Manufacturing 0.2 0.4 0.4 units of Services 0.4 0.2 0.3

The right-hand column tells us that to produce one unit of services requires 0.2 units of resources, 0.4 units
of manufacturing, and 0.3 units of services. Suppose now that we need to produce one additional unit of
services (as more students clamor to have an opportunity to take Finite Mathematics say). How many more
resources will be needed? As noted, the table says that producing 1 unit of services requires 0.2 units of
resources, 0.4 units of manufacturing, and 0.3 units of services. But the middle column tells us that each unit
of manufacturing requires 0.5 units of resources (and 0.4 units of manufacturing, and 0.2 units of services),
while the left column tells us that producing a unit of resources consumes 0.3 units of resources. So that
extra 0.4 units of manufacturing will then require 0.5*0.4 = 0.2 additional units of resources, the extra 0.3
units of services will also require 0.3 * 0:2 = 0.06 additional units of resources, and the extra 0.2 units of
resources will itself require an additional 0.2 *0.3 = 0:06 units of resources, which adds up to another 0.32
units of resources. And now we can continue and ask how many resources and services and manufacturing
will be needed for this additional 0.32 units of resources. Indeed, it doesn't seem we will ever stop. Therefore,
it would be best not to start. Rather than trying to answer how increasing demand for services will affect
resources by multiplying and adding individual terms, it is best to treat this problem using matrices.

Suppose the economy currently produces

 Production Resources 2485 units Manufacturing 2475 units Services 2270 units

We will represent this by a production vector P, and also represent the data in the table of input-output
values above as a matrix A. Note the word \vector" is often used to denote a matrix with just one column
(or just one row).

Much of this production is actually consumed by the production process of course. For example, the economy
will have consumed 0:3*2485 = 74.5 units of resources in the production of resources, 0.5*2475 = 1237.5
units of resources in the production of manufacturing, and 0.2* 2270 = 454 units of resources in the

production services, for a total internal consumption of 2437 units of resources, with just 2485 - 2437 = 48
units of resources left for the satisfaction of external demand. We could similarly compute how many units
of manufacturing and services were consumed internally by the economy in the production process. Looking
at the computations, I hope you recognize that this is another example of a matrix multiplication.
Internal Consumption = AP =

and the amount left over to satisfy external demand is then

External Demand = P- AP =

Now suppose external demand changes from 48 resources, 80 manufacturing, 100 services to a new vector
D. To find the production required to meet this demand, we can solve the matrix equation P - AP = D.
This will require using some of our tricks about matrix ope rations (including one we didn't mention before,
that the distributive law applies to matrices).

Note that we can't factor P - AP as (1 - A) P since 1 - A isn't defined, as the 1 is a scalar number while
A is a 3* 3 matrix and hence we use the identity matrix I in place of the 1. Of course, this means that to
solve our problem we will want to invert a 3 * 3 matrix and so we will use a spreadsheet.

A. Open the Leontief Studio spreadsheet (posted on the class web site). The is data already filled from
the input-output table and a 3 * 3 identity matrix, as well as labels for the other terms we will
compute.

B. First we will compute I - A. We compute the difference of two matrices element by element, which
is easy to do on a spreadsheet. Go to cell C13 and enter the formula =C8-C3. Click and drag to
copy the formula first to the range C13..E13, and then click and drag again to copy this range down
to rows 14 and 15 as well. The range C13..E15 should now contain the matrix I - A.

C. To invert this matrix, we will have to use an array formula. Select the entire range C19..E21. Enter
the formula =minverse(C13..E15) and press Ctrl-Shift-Enter to enter it as an array formula (so it
will fill the whole range selected). This will compute the inverse matrix. The range from C19..E21
should now have the values

D. Now enter the external demands in the labeled cells. Put 48 (the demand for resources) in cell G19.
Put 80 (the demand for manufacturing) in cell G20. Put 100 (the demand for services) in cell G21.

E. Now we have to multiply the demand vector D (entered in cells G19..G21) by the inverse matrix
(I - A)-1 to compute the required production vector. Click and drag to select the range I19..I21.
Enter the formula =mmult(C19..E21,G19..G21). Press Ctrl- Shift-Enter to enter this as an array
formula and fill in all the values in the selected range. This will compute the matrix product, as in
the last part of Activity 4. You should see that the production vector needed to meet this demand
is exactly what we started with, 2485 resources, 2475 manufacturing, and 2270 services.

1. We are now in a position to answer the question from the start of the activity. Increase the demand
for services by 1 unit to 101.

(a) How must the production of resources, manufacturing, and services increase to meet the in-
creased demand for services?

(b) Now increase the demand for services by another unit to 102. How must the production of
resources, manufacturing, and services increase to meet the increased demand for services?
Here, we are just looking for the increase from the production needed to go from 101 to 102
(not the total increase from 100 to 102).

2. How do the answers in 1(a) and 1(b) compare? You should try some other values and persuade
yourself the pattern you observe will always hold.

3. Compare your answers in problems 1(a) and 1(b) to the values shown in the inverse matrix. You
should observe a pattern. Using this pattern, you should be able to quickly answer the following
questions.
(a) How must the production of resources, manufacturing, and services increase to meet an increased
demand for 1 more unit of manufacturing?

(b) How must the production of resources, manufacturing, and services increase to meet an increased
demand for 1 more unit of resources?

(c) Next, we can compute the determinant of the IA matrix with the formula =mdeterm(c13..e15).
What is the determinant of the I - A matrix?

4. Suppose the amount of services required to produce 1 unit of services increases from 0.3 to 0.4.
(a) Under these circumstances, what happens to the values in the inverse matrix?

(b) What is the determinant of the I-A matrix after the increase from 0.3 to 0.4? Warning: one of
the answers produced by the spreadsheet is not strictly accurate (think round-o® error). Values
of the input-output table that lead to problems similar to this are called \non-productive."

Comments. The model we have just developed is called the (open) Leontief model of an economy. Wassily
Leontief introduced input-output analysis and applied matrix algebra (which had been developed by math-
ematicians a century earlier) to the study of the U.S. economy in the 1940s. Of course, he used rather more
than 3 sectors for his work, and you may contemplate the fun of inverting a 42*42 matrix in the days before
computers, working by hand assisted only by mechanical 4- function calculators (not to mention the fun of
determining the actual values for all the entries in a 42 * 42 input-output table). His work is now widely
used to track how changes in one sector will affect other sectors. In 1973, Leontief received the Nobel prize
in economics.
The notion of how changes in one area can propagate into many other areas is a central theme of problems
where matrices are used. In todays lab, changing the demand for services will also change the production of
every other sector. The apparently odd rules for matrix multiplication give us a tool to deal with problems
where different values are entangled and changing one will cause further changes down the line . Recognizing
such follow-on effects is very important to understanding economic development and many other situations
(and can lead to some profitable investment opportunities as well).

 Prev Next

Home    Why Algebra Buster?    Guarantee    Testimonials    Ordering    FAQ    About Us
What's new?    Resources    Animated demo    Algebra lessons    Bibliography of     textbooks