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.
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 |
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 SWe 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 = 0Using rref in MATLAB we get
1 0 -0.94 0 0 1 -0.85 0 0 0 0 0Thus 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.
x = Cx + d
or we can solve for x, via (I-C)x = d or x = (I-C)^{-1}d.
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 |
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 + 2000After writing this as a linear system and using rref, we have (rounded)
1 0 0 9829 0 1 0 12874 0 0 1 8360Thus we should produce 9829 Coal, 12874 Electricity and 8360 Steel to meet all the needs.
2. From 11.9 #5 (open production)
For your solution, show the appropriate linear system and its solution.