Mathe IIILecture 2Mathe IIILecture 2

2

Mathematik IIIChong-Dae Kim

Donnerstag 9.15 – 10.45 Uhr – HS A Donnerstag 10.45 – 12.15 Uhr – HS A Donnerstag 12.15 – 13.45 Uhr – HS A Freitag 13.00 – 14.30 Uhr – HS G

3

The Cobweb Model

4

The Cobweb Model

• Cost of raising q pigs::

• Demand for pigs: D q = γ - δp

• N (profit maximizing) farms producing pigs

Each farm, takes p as given and maximizes:

2= pq - αq - βq π q = pq - C q

α, β,γ,δ > 0t = 0,1,2,3, .....• It takes one period to raise a pig

2C q = αq + βq

5

2= pq - αq - βq π q = pq - C q

π' q = p - α - 2βq

p - αq =

2β

= 0

p - αS = N

2βSupply:

Farmers decide at t-1 on the production at t

t -1t

p - αS = N

2β t= γ - δpt= D

6

t -1t

p - αN = γ - δp

2β

t t -1

N αN + 2βγp = - p +

2δβ 2βδ

The Stationary State p* :

t t -1

N αN + 2βγp = - p +

2δβ 2βδ

p*p*αN + 2βγ

p* =2βδ + N

p* -αN = γ - δp*

2β S p* = D p*

7

t t -1

N αN + 2βγp = - p +

2δβ 2βδ

αN + 2βγp* =

2βδ + N

t

t 0

Np = p* + - p - p*

2δβ

N- < 1 N,

2δβ< 2δβ

tt

p p*

Graphic illustration:

8

p

q S,D

S = N p - α /2β

D = γ - δp

p*

S *

0p

1S

1p 1 0S = S p

1 1D p = S

2S

2 1S = S p

2p

2 2D p = S

3S

3 2S = S p

3p

3 3D p = S

etc. etc

4S5S 6S

9

αN + 2βγp* =

2βδ + N

t

t 0

Np = p* + - p - p*

2δβ

N- < -1 N,

2δβ> 2δβ

t

t

p p*

10

p

q S,D

S = N p - α /2β

D = γ - δp

N- < -1 N,

2δβ> 2δβ S = N p - α /2β

1S3S5S = 0 2S 4S

11a reminder

t

tt 0

1+ r - 1x = + (1+ r) x

r

t t -1x = 1+ (1+ r)xSavings Equation:

Its solution:

12a 2nd reminder

First order equation with constant coefficient

Its solution:

t t -1 tx = ax + b

t

t t - jt 0 j

j=1

x = a x + a b

13

(t = 1, 2, 3, ......)t t -1 t tw = (1+ r)w + (y - c )

t

t t -kt 0 t t

k=1

w = (1+ r) w + (1+ r) (y - c )

0w0 0w + rw 0= (1+ r)w

0(1+ r)w 0 0(1+ r)w r(1+ r)w2

0= (1+ r) wEtc.

After 1 period

14

(t = 1, 2, 3, ......)t t -1 t tw = (1+ r)w + (y - c )

t

t t -kt 0 k k

k=1

w = (1+ r) w + (1+ r) (y - c )

0w t0(1+ r) w

After t periods

is the Present Value of t

0(1+ r) w0w

is the Present Value of -t(1+ r) w w

15

(t = 1, 2, 3, ......)t t -1 t tw = (1+ r)w + (y - c )

t

t t -kt 0 k k

k=1

w = (1+ r) w + (1+ r) (y - c )

t

-t -kt 0 k k

k=1

(1+ r) w = w + (1+ r) (y - c )

-t(1+ r)X

16

t

-t -kt 0 k k

k=1

(1+ r) w = w + (1+ r) (y - c )

:if then0 tw = w = 0

t t

-k -kk k

k=1 k=1

(1+ r) c = (1+ r) y

The present values of the streams of consumption and income are equal

The present values of the streams of consumption and income are equal

17

t t -1b = (1+ r)b - z

0 Tb = B, b = 0

Mortgage Repayments

Outstanding BalanceAt time t

Repayment per periodz

b* = (1+ r)b* -z b* =rt

t 0

z zb = (1+ r) (b - )+

r r

tt

z zb = (1+ r) (B - )+

r r

18

Mortgage Repayments

tt

z zb = (1+ r) (B - )+

r rfor t t = T : b = 0

T z z0 = (1+ r) (B - )+

r r

T

-T -t

t=1

zB = 1 - (1+ r) = z (1+ r)

r

19

T

-T -t

t=1

zB = 1 - (1+ r) = z (1+ r)

r

The loan equals the present value of T payments of z

-T

-T -T

rB (1+ r) rBz = rB +

1 - (1+ r) 1 - (1+ r)> rB

20

tt

z zb = (1+ r) (B - )+

r r

T z z0 = (1+ r) (B - )+

r r

We found that:

and:

-Tz zB - = - (1+ r)

r r

t -T

t

zb = 1 - (1+ r)

r

21

t -T

t

zb = 1 - (1+ r)

r

For t-1 this becomes:

t -1-T

t -1

zb = 1 - (1+ r)

r

t -1-Tt -1rb = z 1 - (1+ r)

t -1-Tt -1z - rb = z(1+ r)

Interest on last period’s principal

Payment towards the principal

22

t -1-Tt -1z - rb = z(1+ r)

Payment towards the principal

For t=1:

For t=T:

Interest: rBT

z

(1+ r)principal repayment:

Interest: rz

1+ rz

1+ rprincipal repayment:

small

largesmall

large

23

Linear Equations with a Variable Coefficient

t t t -1 tx = a x + b

1 1 0 1x = a x + b

2 2 1 2x = a x + b =

2 1 0 1 2= a a x + b + b =

1 0 1a x + b

1 0 1a x + b

2 1 0 2 1 2= a a x + a b + b

24

Linear Equations with a Variable Coefficient

t t t -1 tx = a x + b

2 1 0 2 1 2= a a x + a b + b2x =

3 3 2 1 0 3 2 1 3 2 3x = a a a x + a a b + a b + b

4 4 3 2 1 0 4 3 2 1

4 3 2 4 3 4

x = a a a a x + a a a b +

+ a a b + a b + b

t t t

t s 0 s 1 s t -1 ts=1 s=2 s=t

x = a x + a b + .....+ a b + b

25

t t t

t s 0 s 1 s t -1 ts=1 s=2 s=t

x = a x + a b + .....+ a b + b

t tt

t s 0 s kk=1s=1 s=k+1

x = a x + a b

t t t -1 tx = a x + b

example

The solution to:

is:

or:

26

t t t -1 t tw = 1+ r w + y - c

t tt

t s 0 s kk=1s=1 s=k+1

x = a x + a b

The general solution

becomes:

t tt

t s 0 s k kk=1s=1 s=k+1

w = 1+ r w + 1+ r y - c

27

t tt

t s 0 s k kk=1s=1 s=k+1

w = 1+ r w + 1+ r y - c

t

-1

t ss=1

define the discount factor D = 1+ r

then:

t

sts=k+1

t t 0 k ktk=1

ss=1

1+ rD w = w + y - c

1+ r

( the present value, at t = 0, of $1 at t )

28

t

sts=k+1

t t 0 k ktk=1

ss=1

1+ rD w = w + y - c

1+ r

t

ss=k+1

t

ss=1

1+ r

1+ r

...

... ...

t

sk+1 ts=k+1

t1 k k+1 t

ss=1

1+ r1+ r 1+ r

1+ r 1+ r 1+ r 1+ r1+ r

k1 k

1= = D

1+ r ... 1+ r

29

t

sts=k+1

t t 0 k ktk=1

ss=1

1+ rD w = w + y - c

1+ r

t

t t 0 k k kk=1

D w = w + D y - c

or:

t

0 kt k k

k=1t t

w Dw = + y - c

D D

the present value at period 0

the present value at period t t

ks

s=k+1t

D= 1+ r

D

30

Second Order Equations

t+2 t t+1 (t = 0, 1, ........)x = f t, x , x

0 1

2 0 1

given x , x :

x = f 2, x , x 0 1f 2, x , x

1 0 1f 2, x= 3, x , , xf

3 1 2x = f 3, x , x2x

etc. etc.

31

Existence and Uniqueness :

The equation

t+2 t t+1 (t = 0, 1, ........)x = f t, x , x

has a unique solution for any given 0 1x , x

0 1 2 3x , x , x , x , ............