ch_3_eqm_ana_in_econ
TRANSCRIPT
Chapter 3 Equilibrium analysis in Economics
3-1
Chapter 3
Equilibrium Analysis in Economics
--------------------------------------------------------------------- Studies or analysis of Economics
1. Static Analysis ‐ Studies focus only on a particular period of time.
2. Comparative Analysis ‐ Studies focus on the external forces that make equilibrium move to a new
one.
3. Dynamic Analysis ‐ Studies focus on the change of time and how the equilibrium change with
time.
Q0 Q1
P S
DQ
P0
Comparative Analysis
D´
P1 E´
E
Static Analysis
P S
DQ
P*
Q*
E
Chapter 3 Equilibrium analysis in Economics
3-2
Dynamic Analysis
Example of dynamic analysis: Cobb-web Theorem
Variable
Converge
Time
Diverge
Variable
Time
S2 S1
0 3 2 1
P
Q
Price
Time
Pt
P1 P2
P3
D
S3
St (Pt-1)
P2
P
Q2 Q1
P1
Q
Dt (Pt)
St (Pt-1)
P2
P
Q2 Q1
P1
Q
Dt (Pt)
Chapter 3 Equilibrium analysis in Economics
3-3
3.1 Partial Market Equilibrium - A linear model
Identify equilibrium price & equilibrium quantity (endogenous)
Constructing the model
1. Equilibrium condition Conditional equation Qd = Qs
2. Demand equation: a decreasing linear function of P (P ↑, Qd ↓)
Behavioral equation Qd = a – bP (a, b > 0)
3. Supply equation: an increasing linear function of P (P ↑, Qs ↑) Behavioral equation Qs = -c + dP (c, d > 0)
Solution
Qd = a – bP QS = -c + dP
The equilibrium condition Qd = Qs a – bP = -c + dP a + c = bP + dP a + c = (b + d)P
P* = a+cb+d
Q*
P P*
Qd= a - bP The solution values are P* and Q*
Qs= -c + dP
Q
Chapter 3 Equilibrium analysis in Economics
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Substitute P* into Qd Q* = a – bP*
= a – b ( a+cb+d
)
= a(b+d) - b(a+c)b+d
= ab+ad - ba-bcb+d
= ad - bcb+d
Since b+d > 0 ∴ (ad - bc) must be positive in order to have positive Q* ad – bc > 0 ad > bc (Restriction)
• What happen if (b + d) = 0, can an equilibrium solution be found by using P*,Q* why or why not ?
- No, there will be division by zero. The solution is undefined
• What can you conclude regarding the position of D & S curve?
- D & S would be parallel, with no equilibrium
3.2 Partial Market Equilibrium: A nonlinear model
Quadratic function: numerical example
Qd = Qs
Qd = 4 – P2
Qs = 4P – 1
4 – P2 = 4P - 1
P2 + 4P - 5 = 0
The Quadratic formula
ax2 + bx + c = 0 a 0
The roots are
*1x , *
2x = 2-b± b -4ac
2a
Chapter 3 Equilibrium analysis in Economics
3-5
According to the above equation P2 + 4P - 5 = 0
∴ The equilibrium price = * *1 2P ,P =
2-4± 4 -4(1)(-5)2(1)
= -4±62
= -5, 1 The equilibrium quantity = Q* = Qs = 4P*- 1 = 4(1) - 1 = 3
Example:
Qd = Qs
Qd = 8 – P2
Qs = P2 – 2
8 - P2 = P2 – 2
2P2 = 10
(P- 5)(P+ 5) = 0
P* = 5 Q* = 3
P Qd = 4 – P2
Qs = 4P - 1
-5
-25
(-5, -25)
1
3(1, 3)
Q
Chapter 3 Equilibrium analysis in Economics
3-6
3.3 General Market Equilibrium
Last 2 sections: an isolated market. ∵consider only 1 good
In the real world, there are more than one good in the economy
Consider more than 1 good market
General Equilibrium
with n commodities, the equilibrium conditions are
Ei Qdi – Qsi = 0 (i = 1, 2, ….. n)
Two commodity mkt. model
Qd1 = 10 – 2P1 + P2 ……. (1)
Qs1 = -2 + 3P1 ……. (2)
Qd2 = 15 + P1 – P2 ……. (3)
Qs2 = -1 + 2P2 ……. (4)
Qd1 = Qs
1
Qd2 = Qs
2
(1) = (2) 10 – 2P1 + P2 = -2 + 3P1
P2 = -12 + 5P1 …….(5) (3) = (4) 15 + P1 - P2 = -1 + 2P2 P1 = -16 + 3P2 ……. (6)
Substitute (6) into (5)
P2 = -12 + 5 (-16 +3P2) = -12 – 80 + 15P2
14P2 = 92
P2* = 92
14
Chapter 3 Equilibrium analysis in Economics
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Substitute P2
* into (6)
P1* = -16 + 3P2
*
= -16 + 3 ( 9214
)
= 5214
Q1* = -2 + 3P1
*
= -2 + 3( 5214
)
= 647
Q2* = -1 + 2P2
*
= -1 + 2( 9214
)
= 857
Solution of a General Equilibrium System
To guarantee that the model yields a unique solution, the equations should have the following properties.
1. Consistency
‐ The satisfaction of any one equation in the model will not preclude the satisfaction of another
x + y = 8 x + y = 9
2. Functional independence
857
Good 2
P2
Q2
9214
Qd2
Qs2
Q1 Qs
1
P1
647
5214
Good 1
Qd1
Chapter 3 Equilibrium analysis in Economics
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‐ No equation is redundant which means that one can be derived from the other.
2(2x + y) = 2(12) 4x + 2y = 24
(Try exercise 3.4 in the text book) 3.4 Equilibrium in National Income Analysis
Keynesian national income model
Y = C + I + G C = a + bY (a > 0, 0 < b < 1) I = I0 G = G0
Y = C + I + G
= a + bY + I0 + G0 (1 - b)Y = a + I0 + G0
Y* = ( 11-b
) (a + I0 + G0)
Then C* = a + b Y*
= a + b ( 11-b
) (a + I0 + G0)
= 0 0a(1 - b) + ba + bI + bG1 - b
= 0 0a + b(I + G )1 - b
restriction b 1
IS – LM Framework
• Equilibrium in good market – IS • Equilibrium in money market - LM
a. Good market
C = a + bYd ; [a > 0, 0 < b < 1] Yd = Y – T T = T0 + tY ; [0 < t < 1]
Chapter 3 Equilibrium analysis in Economics
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I = I0 – er ; [e > 0] G = G๐ X = X๐ M = M๐ Conditional equation (Equilibrium)
Y = C + I + G + X –M = a + bYd + I0 – er + G0 + X0 – M0 = a + b (Y – T0 - tY) + I0 – er + G0 + X0 – M0 = a + bY – bT0 – btY + I0 + er +G0 + X0 – M0
(1 – b + bt)Y = a – bT0 + I0 + G0 + X0 – M0 – er
Y* = 0 0 0 0 0a – bT + I + G + X – M er-1- b + bt 1 - b + bt
Y* = 0α – 1α r
According to the expression of Y*, National income has a negative relationship with interest
b. Money market
Money demand
1. Transaction Demand 2. Precautionary Demand 3. Speculative Demand Md = N0 + myY – mrr ; my , mr > 0
Money Supply
– Monetary policy (constant) Ms = Mo
Conditional equation (Equilibrium)
Md = Ms
N0 + myY – mrr = M0
IS
r
Y
Chapter 3 Equilibrium analysis in Economics
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r* = 0 y 0
r
N + m Y - Mm
rewrite Y = 0 0
y y
M - N mr+ ( )rm m
Y* = β0 + β1r
According to the expression of Y*, National income has a positive relationship with interest.
c. IS – LM Framework
IS = LM 0α – 1α r = β0 + β1r
(β1 + 1α ) r = 0α - β0
r* = 0 0
1 1
α - ββ + α
From IS Y* = 0α – 1α r*
= 0α – 1α 0 0
1 1
α - ββ + α
= 0 1 1 0
1 1
α β + α ββ + α
Substitute 0α , 1α , 0β , 1β by their expressions
r* = y 0 0 0 0 0 0 0
r y
m [a - bT + I + G + X - M ] - [1 - b + bt](M - N )(1 - b + bt)m + em
LM r
Y
Chapter 3 Equilibrium analysis in Economics
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Y* = r 0 0 0 0 0 0 0
r y
m [a - bT + I + G + X - M ] + e(M - N )(1 - b + bt)m + em
Exercise Chapter 3
1. Let the demand and supply functions be as follow:
(a) Qd = 51 – 3P Qs = -10 + 6P (b) Qd = 30 – 2P Qs = -6 + 5P
find P*and Q*
2. Find the zeros of the following functions graphically:
(a) f(x) = x2 – 8x + 15 (b) g(x) = 2x2 – 4x – 16
3. Find the equilibrium solution for each of the models: Qd = Qs
(a) Qd = 3 – P2 Qs = 6P – 4 (b) Qd = 8 – P2 Qs = P2 – 2
4. The demand and supply functions of a two-commodity market model are as
follows:
Qd1 = 18 – 3P1 + P2 Qd
2 = 12 + P1 – P2
Qs1 = -2 + 4P1 Qs
2 = -2 + 3P2 Find and *
iP and *iQ
5. Given the following model:
Y = C + I0 + G0 C = a + b(Y – T) (a > 0, 0 < b < 1) T = d + tY (d > 0, 0 < t < 1)
LM r
Y
IS
r* = 0 0
1 1
α - ββ + α
Y*= 0 1 1 0
1 1
α β + α ββ + α
Chapter 3 Equilibrium analysis in Economics
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(a) How many endogenous variables are there? (b) Find Y*, T* and C*
6. Let the national-income model be:
Y = C + I0 + G C = a + b(Y – T0) (a > 0, 0 < b < 1) G = gY (0 < g < 1)
(a) Identify the endogenous variables. (b) Give the economic meaning of the parameter g (c) Find the equilibrium national income. (d) What restriction on the parameters is needed for a solution to exist?
Chapter 3 Equilibrium analysis in Economics
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7. Using the following money market information to derive an equation of LM:
1375 0.25 25
2500
d
s
d s
M Y r
M
M M
= + −
=
=
8. Find national-income and aggregate consumption at the equilibrium of the
following model
12
0 0
0
0
25 61614
Y C I G
C YI
G
= + +
= +=
=