Example 2. A Model of General Economic Supply-Demand We present a linear model d

Question

Example 2. A Model of General Economic Supply-Demand We present a linear model due to W. Leontief, a Nobel Prize-winning economist. The model seeks to balance supply and demand throughout a whole economy. For each industry, there will be one supply equation. In practical applications, Leontief economic models can have hundreds or thousands of specific industries. We consider an example with four industries -demand The left-hand side of each equation is the supply, the amount produced by the ith industry. Call this quantity x; it is meas dollars. On the right-hand side, we have the demand for the product of the it h industry. There are two parts to the demand. The first part is demand for the output by other industries (to create other products requires some of this product as input). The second part is consumer demand for the product For a concrete instance, let us consider an economy of four gen- eral industries: energy, construction, transportation, and steel. Suppose that the supply-demand equations are

Explanation / Answer

Iteration 1 :

x1 = 100, x2 = 50, x3 = 100, x4 = 0

x1 = .4(100) + .2(50) + .2(100) + .2(0) = 40 + 10 + 20 = 70

x2 = .3(100) + .3(50) + .2(100) + .1(0) = 30 + 15 + 20 = 65

x3 = .1(100) + .1(50) + 0 + .2(0) = 10 + 5 = 15

x4 = 0 + .1(50) + .1(100) + 0 = 5 + 10 = 15

Iteration 2 :

x1 = 70, x2 = 65, x3 = 15, x4 = 15

x1 = .4(70) + .2(65) + .2(15) + .2(15) = 28 + 13 + 3 + 3 = 47

x2 = .3(70) + .3(65) + .2(15) + .1(15) = 21 + 19.5 + 3 + 1.5 = 45

x3 = .1(70) + .1(65) + 0 + .2(15) = 7 + 6.5 + 3 =16.5

x4 = 0 + .1(65) + .1(15) + 0 = 6.5 + 1.5 = 8

Iteration 3 :

x1 = 47, x2 = 45, x3 = 16.5, x4 = 8

x1 = .4(47) + .2(45) + .2(16.5) + .2(8) = 18.8 + 9 + 3.3 + 1.6 = 32.7

x2 = .3(47) + .3(45) + .2(16.5) + .1(8) = 14.1 + 13.5 + 3.3 + 0.8 = 31.7

x3 = .1(47) + .1(45) + 0 + .2(8) = 4.7 + 4.5 + 1.6 =10.8

x4 = 0 + .1(45) + .1(16.5) + 0 = 4.5 + 1.65 = 6.15

Iteration 4 :

x1 = 32.7, x2 = 31.7, x3 = 10.8, x4 = 6.15

x1 = .4(32.7) + .2(31.7) + .2(10.8) + .2(6.15) = 13.08 + 6.34 + .216 + 1.23 = 20.866

x2 = .3(32.7) + .3(31.7) + .2(10.8) + .1(6.15) = 9.81 + 9.51 + 2.16 + 0.615 = 22.095

x3 = .1(32.7) + .1(31.7) + 0 + .2(6.15) = 3.27 + 3.17 + 1.23 = 7.67

x4 = 0 + .1(31.7) + .1(10.8) + 0 = 3.17 + 1.08 = 4.25

Iteration 5 :

x1 = 20.866, x2 = 22.095, x3 = 7.67, x4 = 4.25

x1 = .4(20.866) + .2(22.095) + .2(7.67) + .2(4.25) = 8.3464 + 4.419 + 1.534 + 0.85 = 15.1494

x2 = .3(20.866) + .3(22.095) + .2(7.67) + .1(4.25) = 6.2598 + 6.6285 + 1.534 + 0.425 = 14.8473

x3 = .1(20.866) + .1(22.095) + 0 + .2(4.25) = 2.0866 + 2.2095 + 0.85 = 5.1461

x4 = 0 + .1(22.095) + .1(7.67) + 0 = 2.2095 + 0.767 = 2.9765

These are the final values of x1, x2, x3 and x4.

Cheers!

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