Mathe IIILecture 2Mathe IIILecture 2
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Tutorien für Mathematik IIIIm WS 05/06
Tutor: ChongDae KIM
Mo. 11:00 Uhr - 12.30Uhr HS N. Mo. 12.30 Uhr - 14.00Uhr HS N.
Di. 9.30 Uhr - 11.00Uhr HS N.Di.13.30 Uhr - 15.00Uhr HS N.
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The Cobweb Model
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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
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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
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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*
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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:
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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
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αN + 2βγp* =
2βδ + N
t
t 0
Np = p* + - p - p*
2δβ
N- < -1 N,
2δβ> 2δβ
t
t
p p*
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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
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(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
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(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
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(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
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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
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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
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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
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Ad - Kan
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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
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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
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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
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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
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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
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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
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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:
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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
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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 )
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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
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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
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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.
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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 , ............