## Lab 4 Leontief Economic Models

#### Preliminaries

This is based on Section 11.9 in the text. I've tried to pull out the essential parts, but you should read through that section before you try the exercises.

In 1949 Leontief used Harvard's Mark II computer to solve as system of 42 equations in 42 unknowns. It took 56 hours to produce a solution. The original problem he wanted to solve had 500 equations and unknowns but the computers of the day weren't large enough to handle it. Leontief's use of linear models in economics lead to him wining the Nobel Prize in 1973.

#### Input-Output or Production Model

The basic idea is to try to find the fair price of goods in various sectors that are exchanged between the other sectors. As an example, suppose we have 3 products: coal, electricity and steel, and each uses a portion of the other output to make their product:
 Output Coal Electric Steel Purchased by: 0 0.4 0.6 Coal 0.6 0.1 0.2 Electric 0.4 0.5 0.2 Steel
Note that each column adds to 1 as we assume all the output is used. Let C be the price of a years output of coal, E for electricity and S for steel. Then from this table, assuming that we want each price to equal the cost (equilibrium), we have
```C = 0.4 E + 0.6 S
E = 0.6 C + 0.1 E + 0.2 S
S = 0.4 C + 0.5 E + 0.2 S
```
We can turn this into a homogeneous linear system:
```     C - 0.4 E - 0.6 S = 0
-0.6 C + 0.9 E - 0.2 S = 0
-0.4 C - 0.5 E + 0.8 S = 0
```
Using rref in MATLAB we get
``` 1 0 -0.94 0
0 1 -0.85 0
0 0    0  0
```
Thus C = 0.94 S and E = 0.85 S. So if we set S = \$100 million, then C = \$94 Million and E = \$85 Million, would be the fair (equilibrium) prices for coal and electricity.

If the table of exchange values is E and the set of prices is p then we are solving p = Ep or (I-E)p = 0.

#### Open Production Model

In this case besides the resources used by the other sectors, there is an outside demand for the products. In this case we are deciding what amount to produce to meet both internal and external demand. We have a table relating how many units of each product it takes to produce 1 unit of the other products. This table forms the consumption matrix C. Typically the columns of C sum to less than 1, as we need some product left over to satisfy external demands. Let the amount produced by each sector be stored in a vector x and let d be a vectors of external demands for the products (in units). Then if we have a balance, we have

x = Cx + d

or we can solve for x, via (I-C)x = d or x = (I-C)-1d.

For example, using the sectors Coal, Electricity & Steel as before, assume we have this consumption table:
 Inputs Consumed Per Unit Output Coal Electric Steel Purchased from: 0 0.4 0.5 Coal 0.5 0.1 0.2 Electric 0.3 0.2 0.1 Steel
(So to make 10 units of Electricity it takes 4 units of Coal, 1 unit of Electricity and 2 units of Steel). Assume we have a outside demand of 500 units of Coal, 5000 units of Electricity and 2000 units of Steel, then we have the system:

```C = 0.4 E + 0.5 S + 500
E = 0.5 C + 0.1 E + 0.2 S + 5000
S = 0.3 C + 0.2 E + 0.1 S + 2000
```
After writing this as a linear system and using rref, we have (rounded)
```1 0 0   9829
0 1 0  12874
0 0 1   8360
```
Thus we should produce 9829 Coal, 12874 Electricity and 8360 Steel to meet all the needs.

#### Exercises

1. From 11.9 #4 (input-output)

2. From 11.9 #5 (open production)

For your solution, show the appropriate linear system and its solution.

Mail: ccollins@math.utk.edu