Ch_3_Eqm_ana_in_Econ

Chapter 3 Equilibrium analysis in Economics

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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


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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)


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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


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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


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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


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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


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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


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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]


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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


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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


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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


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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?


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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

= + +

= +=

=

Ch_3_Eqm_ana_in_Econ

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