Lecture 10: Last updated 01/06/2010
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Chapter 17
General Equilibrium (Linear Models)
Assume: 2 goods, 2 factors in the economy.
y1,y2: final goods,
x1,x2: factors of production,
p1,p2: prices of final goods (assumed to be fixed by the world market è small country (economy) assumption)
xij = amount of factor i used in industry j.
max
s.t.
for simplification, we can assume that the functions f1 & f2 exhibit constant returns to scale.
for all
let
Input – Output table
I – O coefficients (Technological Coefficients)
L / L1 / L2K / K1 / K2
Y1 / Y2
so; the constant returns to scale enables us to completely describe the isoquants by the knowledge of one isoquant (unit – isoquant).
Then the model can be written as
Max
s.t.
With fixed technological coefficients , the model becomes a “linear programming model”.
Max
s.t. & &
(here; and
Fixed – Coefficients Production Function
Linear Activity Analysis
Suppose the firm is faced with 3 production coefficient possibilities.
Assumption: the firm can choose any one or more activities simultaneously.
Suppose: 1 unit of output can be produced by ½ of A and ½ of B.
½ of A è ½ of L and 1 of K
½ of B è 3/2 of L and ½ of K
Total è 2 of Labor and 1.5 of Capital è A new combination activity D.
Using the connected line, we get another point E on the same activity line of C but using less of both inputs.
C becomes irrelevant in a cost minimizing problem.
Production is done where slope of isoquant = slope of isocost
· If , tangency will occur at “A” à relatively more capital intensive activity.
· If , tangency will occur at “B” à labor intensive activity will be used.
· If , infinitely many number of solutions along
Algebraic Formulation (Kuhn Tucker can be used) (Minimization Problem)
Min
s.t. &
if A is used (1,2) then,
if B is used (3,1) then,
if c is used (2,3) then,
Problem becomes
Min
s.t
Foc.
If
If
If
At the cost minimizing point, the process with minimum MC will be used.
Numerical Example
A firm produces 2 goods, y1 (food) and y2 (clothing). 3 inputs are land (H), Labor (L) and capital (K). inputs are combined in fixed proportions to produce 1 unit of output.
1 unit of food requires 3 ac/land, 2 w/h, 1 m/h
1 unit of clothing requires 2 ac/land, 2 w/h, 2 m/h
I – O Matrix
Also
F C (food is more land intensive)
Maximize Output
Max
s. t. “ A linear programming model”
Slope of the obj function : 40/30 = 1.333
Comparing the slopes, we see that production will be at point C. it is the intersection of Land and Labor and capital is non – binding.
Therefore, simultaneous solutions of (1) & (2) yields
Derivation of Input Prices
Suppose land is 55 (1 increased). The previous solution would yield;
ð value of output increased by $10.
VMPLand = 10 = Price of Land in competitive market.
“Shadow Price” or “Imputed rent”
Suppose change in Land: ΔH
similarly by increasing L by ΔL
shadow price of labor.
Finally doing the same for K, we will get
The Rybczynski Theorem
Effects of a change in factor endowment on the output.
Theorem: if the endowment of some resources increases, the industry which uses that resources more intensively will increase its output, value of the other will decrease.
(recall in the example y1 (food) was made land intensive and y2 (clothing) labor intensive).
First industry is land intensive.
Proof of the Theorem
When land increases, output of land intensive industry will increase.
Matrix Format
Cramer’s Rule
for land intensive industry
Rybczynski: (proof)
2. Similarly
For land intensive industry denominator > 0
ECON 603
Chapter 17 continued
Summary
Max
s.t.
With technological coefficients
Problem è Linear programming model
Max
s.t.
Rybz. Theorem
by Cramer’s Rule
Suppose industry 1 is more land – intensive.
Then,
è
Then, for a 1 unit increase in Land (H).
output of y1 which is land intensive good increases.
Also,
in addition for where
then,
è
So, when some factor endowment changes, the output that is intensive in that factor, will change in greater absolute proportion than the parameter.
The Stolper Samuelson Theorem
Effect of an increase in the price of output on the factor prices.
Theorem. If the price of (say) labor intensive good increases, the price of labor will increase and in greater proportion. The price of the other factor falls, (but not necessarily in greater proportion).
Proof: Consider the unit factor cost of y1 and y2. remember we found as a shadow prices of factors.
Here the coefficient matrix is the transpose of technical coefficients matrix of the primal problem.
So the algebra of the relations between the factor and output prices is virtually identical to the relations between physical output and resource endowment.
Solving the system with we get
When
The Dual Problem
Original problem è revenue was maximized constraints were statements of limited resource endowments.
(Primal problem)à shadow factor prices derived.
Recall (envelope theorem)
Lagrance multipliers in primal problem are shadow factor prices.
These are the non-positive profit conditions.
The remaining foc. are the inequality constraints:
For a competitive economy we expect revenue maximization to imply (and to be implied by) “minimization of total value of recourses subject to the constraint that profits are non-positive”. So, Rearranging Lagrange function we get;
A minimization problem w.r.t. and objective function , the total value of recourses.
and,
Primal Problem Dual Problem
Max min
s.t. s.t
obviously, when these two problems are solved, the values of objective functions are identical.
Using the numerical example.
actually, by Euler’s Theorem
The Fundamental Theorem of Linear Programming
General Linear Programming Problem
Primal Dual
Max Min
s.t. s.t.
Theorem: suppose there exists an such that (x0 is the feasible solution) and such that (U0 is the feasible solution of minimum problem).
Then both problems possess an optimal solution (i.e a finate solution) and these two solution values are identical.
Meaning, suppose is the vector that maximizes , such that .
The max value of is
Similarly, let be the vector for which is minimum and . Then .
Simplex Algorithm for Solving Linear Programming Models
Max
s.t.
let be the slack variables, then problem becomes:
Max
s.t. (1)
(2)
Choose as basis. Solve in terms of
(2) & (3) è
(3) è
(1) è
set . This corresponds to point B in the previous diagram.
This is a solution but is it optimal? Does it maximize the objective function?
(if y5 increases, max value increases so that is not optimal because by increasing y5 we can increase z*).
Instead of choosing y5=0, we must increase it but how much?
PAY ATTENTION
For y4=0.
for let y3 be out of the system.
New basis; and write in terms of
(6) and put into (4)
(5)
for
is it optimal?
it is definitely optimum.
Deriving Shadow Prices
Let factor endowments be H, L, K where y3 and y4 are non basis variables.