Suppose an economy has four sectors: Mining, Lumber, Energy and Transportation t

Question

Suppose an economy has four sectors: Mining, Lumber, Energy and Transportation that produce an annual output in billions of dollars represented by the variables p_M, rho_L, rho_E and rho_r Mining sells 10% of its output to Lumber, 60% to Energy and retains the rest. Lumber sells 15% of its output to Mining, 50% to Energy, 20% to Transportation and retains the rest. Energy sells 20% of its output to Mining, 15% to Lumber, 20% to Transportation and retains the rest. Transportation sells 20% of its output to Mining, 10% to Lumber, 50% to Energy and retains the rest. Construct an exchange table and a digraph for this economy and use them to set up a system of equations for a set of equilibrium prices for this economy. What are the values of PM. rho_L, rho_E and rho_T if each sector's income exactly balances its expenses? Select one: a. P_M = $136.88/unit P_L = $83.69/unit P_E = $316.31/unit P_T = $100/units b. P_M = $152.22/unit P_L = $89.29/unit P_E = $311.35/unit P_T = $100/unit c. P_M = $139.38/unit P_L = $73.69/unit P_E = $243.36/unit P_T = $100/unit d. P_M = $128.68/unit P_L = $85.62/unit P_E = $325.11/unit P_T = $100/unit

Explanation / Answer

So first we have to setup a table with the info provided for part A:

( The columns represent the source and how much each sector recieved from it. )

Mining | 30 15 20 20

Lumber | 10 15 15 10

Energy | 60 50 45 50

Trans | 0 20 20 20

Write the equations for the matrix to find equalibrium prices:

Pm = 0.3Pm + 0.15Pl + 0.2Pe + 0.2Pe

Pl = 0.1Pm + 0.15Pl + 0.15Pe + 0.5Pt

Pe = 0.6Pm + 0.5Pl + 0.45Pe + 0.5Pt

Pt = 0Pm + 0.2Pl + 0.2Pe + 0.5Pt

(1 - 0.3)Pm -0.15Pl - 0.2Pe - 0.2Pt = 0

-0.1Pm + (1 - 0.15)Pl - 0.15Pe - 0.1Pt = 0

-0.6Pm - 0.5Pl + ( 1 - 0.45)Pe - 0.5Pt = 0

0Pm - 0.2Pl - 0.2Pe + (1 - 0.2)Pt = 0

So our matrix is now ( you can use a program like Octave/MatLab to input this matrix)

m1 =

0.70 -0.15 -0.20 -0.20 0.00

-0.10 0.85 -0.15 -0.10 0.00

-0.60 -0.50 0.55 -0.50 0.00

0.00 -0.20 -0.20 0.80 0.00

Next we find the row reduced echelon form ( rref(m1) in octave )

Solution =

1.00 0.00 0.00 -1.37 0.00

0.00 1.00 0.00 -0.84 0.00

0.00 0.00 1.00 -3.16 0.00

0.00 0.00 0.00 0.00 0.00

Each row corresponds to the unknown variables:

Pm = 1.37Pt

Pl = 0.84Pt

Pe = 3.16Pt

Pt = free variable, we set it to any single one non-negative value.

Easiest of all is to let Pt = 100

Then:

Pm = 137

Pl = 84

Pe = 316

Pt = 100

So, A option

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