1. 2 no lecture on wed february 8th thursday 9 th feb friday 27 th jan friday 10 th feb thursday...

2

No lecture on Wed February 8th

Thursday 9th FebFriday 27th JanFriday 10th Feb

Thursday 14:00 - 17:00 Friday 16:00 – 19:00 HS N

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Non Linear Programming

if is a solution of the problem*x

j j = 1,3 .λ 0,. ..,m

is a stationary point (w.r.t. ) of the Lagrangian :

-

*i

m

i i ii=1

2 x. x

x = f x λ g x -c L

*j j1. j = 1, .g x mc , ..,

if then *j j jg x < c λ = j = 1, .4. 0, ..,m

max s.t.

1 1

m m

x

g x c

g x c

f x ......

Kuhn – Tucker conditions

4

Non Linear Programming

if is a solution of the problem*x

j j jλ g x - c = 0


max s.t.

1 1

m m

x

g x c

g x c

f x ......

Kuhn – Tucker conditions

j j jλ4. j =g x - c = 1, .0, ..,m


5

Nonnegative variables

max . .s t

x

g x, y c

f x, y x 0

y 0

- - 1 2x, y = f x, y λ g x, y - c -x -yL

max . .s t

x

g x, y

-x 0

-y 0

c

f x, y

1

2

x x

y y

f (x, y) - λg (x, y)+ μ = 0x

f (x, y) - λg (x, y)+ μ = 0y

L

L1 2λ, μ , μ 0

1 2λ g x, y - c x = y = 0

6


g x, y c

x 0

y 0

1

2

x x

y y

f (x, y) - λg (x, y)+ μ = 0x

f (x, y) - λg (x, y)+ μ = 0y

L

L

1 2λ, μ , μ 0 1 2λ g x, y - c x = y = 0

1

2

x x

y y

f (x, y) - λg (x, y) = -μ 0

f (x, y) - λg (x, y) = -μ 0

1 x xx > 0 μ = 0 f (x, y) - λg (x, y) = 0

7


g x, y c

x 0

y 0

λ 0 λ g x, y - c 0

x x x x

y y y y

f (x, y) - λg (x, y) x = 0, f (x, y) - λg (x, y) 0

f (x, y) - λg (x, y) y = 0 f (x, y) - λg (x, y) 0

For a general problem:

max s.t.

1 1

1 n

m m

x

g x c

x 0 x 0

g x c

f x ...... ......

8


j j j jλ g x - c 0, λ 0, j = 1, ...m

m

j hj=1 h

jh

g (x)f (x) - λ x = 0, h = 1, ....,n

x

max s.t.

1 1

1 n

m m

x

g x c

x 0 x 0

g x c

f x ...... ......

m

jj=1 h

jh

g (x)f (x) - λ 0, h = 1, ....,n

x

9


Example: Peak Load Pricing

The price of a good (electricity) for time period i is given as pi

The producer chooses how much to produce in each period (xi), and the maximal capacity of his

plant (k).

The total cost of producing(x1,…,xn) is C(x1,…,xn).The cost of capacity k is D(k).

10


Example: Peak Load Pricing

The producer maximizes:

n

1 n i i 1 ni=1

π x , ..., x ,k = p x - C x , ..., x - D k

s.t. i 0 x k i = 1,...,n

n n

1 n i i 1 n i ii=1 i=1

x , ..., x ,k = p x - C x , ..., x - D k λ x - kL

i 1 n ii

ip - C x , ..., x - λ 0x

L

11

Nonnegative variables Example: Peak Load Pricing

n n

1 n i i 1 n i ii=1 i=1

x , ..., x ,k = p x - C x , ..., x - D k λ x - kL

i 1 n ii

ip - C x , ..., x - λ 0x

L

i 1 n i ιip - C x , ..., x - λ x = 0

n

ii=1

-D k + λ 0k

L

n

ii=1

-D k + λ k 0

i i iλ 0, λ x - k 0, i = 1, ...,n

i i i 1 n ιif x 0, p C x , ...x

i i

i i 1 n

if k > x 0, λ = 0

p C x , ...x

price marginal costi.e. equals in off - peak hours.

. ipeak

for peak hours : D k = λ

12

The Maximum Principle

Optimization over time

Stock – state variablesFlow – control variables

A.K. Dixit: Optimization in Economic Theory, Oxford University Press, 1989. Chapter 10

*

*

t = 1,2,3, .....,T

t+1 t t ty - y = Q y ,z ,t

stocks of capital goods

consumption, labor supplyflow variable

production function

13



Stock – state variablesFlow – control variables

t = 1,2,3, .....,T

t+1 t t ty - y = Q y ,z ,t

t+1 t t ty y + Q y ,z ,t t tG y ,z ,t 0

14



t+1 t t ty - y = Q y ,z ,t

t tG y ,z ,t 0 t = 0,1,2, .....,T

T

t tt=0

F y ,z ,t

additively separable utility function

F a,0 + F b,1 + F c,2

The marginal rate of substitution between periods 1,2

F c,2-

F b,1is independent of the quantitiy consumed in period 0

15


. .s t

t+1 t t t

t t

t = 0,1, .....,Ty - y = Q y ,z ,t

G y ,z ,t 0

maxT

t tt =0t t

y ,zF y ,z ,t

{

}

T

t tt=0

t+1 t t t t+1 t t t

F y ,z ,t

+ π y + Q y ,z ,t - y - λ G y ,z ,t

L

16


. .s t

t+1 t t t

t t

t = 0,1, .....,Ty - y = Q y ,z ,t

G y ,z ,t 0

maxT

t tt =0t t

y ,zF y ,z ,t

{

}

T

t tt=0

t+1 t t t t+1 t t t

F y ,z ,t

+ π y + Q y ,z ,t - y - λ G y ,z ,t

L

17

.......

1 0 0 0 1 t t -1 t -1 t -1 t

t+1 t t t t+1

π y + Q y ,z ,0 - y π y + Q y ,z ,t - 1 - y

+ π y + Q y ,z ,t - y + .........

{

}

T

t tt=0

t+1 t t t t+1 t t t

F y ,z ,t

+ π y + Q y ,z ,t - y - λ G y ,z ,t

L

z t t t+1 z t t t z t tt

F y ,z ,t π Q y ,z ,t - λ G y ,z ,t = 0z

Lderivative w.r.t. zt:

derivative w.r.t. yt:

.......

1 0 0 0 1 t t -1 t -1 t -1

t+1 t tt t +1

tπ y + Q y ,z ,0 - y π y + Q y ,z ,t - 1 -

+

y

π + Q ,z ,t - y + .....y y ....

T

t+1 t t t t+1t=0

π y + Q y ,z ,t - y

y t t t+1 y t t t y t tt

t+1 t

F y ,z ,t π Q y ,z ,t - λ G y ,z ,ty

+ π - π = 0

L

18

z t t t+1 z t t t z t tt

F y ,z ,t π Q y ,z ,t - λ G y ,z ,t = 0z

L

y t t t+1 y t t t y t tt

t+1 t

F y ,z ,t π Q y ,z ,t - λ G y ,z ,ty

+ π - π = 0

L

t+1 t y t t t+1 y t t t y t tπ - π = - F y ,z ,t π Q y ,z ,t - λ G y ,z ,t

Define the Hamiltonian:

H y,z,π,t = F y,z,t πQ y,z,t

maximizes s.t. t t t t+1 t tz H y ,z ,π ,t G y ,z ,t 0

Let be the value at maximum*t t+1H y ,π ,t

19

z t t t+1 z t t t z t tF y ,z ,t π Q y ,z ,t - λ G y ,z ,t = 0



The Hamiltonian is maximized at s.t. t t t z G y ,z ,t 0

is the value at maximum.*t t+1H y ,π ,t

The two Lagrange conditions:

The Hamiltonian:

20




Define :

(the Lagrangian at )t t t+1 t t tL = H(y ,z ,π ,t) - λ G y ,z ,t

t


The Hamiltonian:

The Hamiltonian is maximized at s.t. t t t z G y ,z ,t 0

is the value at maximum.*t t+1H y ,π ,t

t+1 t y t t t+1π - π = -L y ,z ,π ,t

From the envelope theorem:

*t+1 t y t t+1π - π = -H y ,π ,t

21

max

s.t. j

x

g x,r = 0, j = 1, ...,m

f x,r

the solution is

and the maximum is

,

*

* *

x r

f r = f x r ,r

Envelope Theorem

* **

j j

x , λ ,rf r=

r r

L

22




Define :


t


The Hamiltonian:

t+1 t y t t t+1π - π = -L y ,z ,π ,t


*t+1 t y t t+1π - π = -H y ,π ,t

23




Define :


t


The Hamiltonian:

t+1 t y t t t+1π - π = -L y ,z ,π ,t


*t+1 t y t t+1π - π = -H y ,π ,t

Similarly from the envelope theorem:*π πH = L = Q

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Define :


t


The Hamiltonian:

*t+1 t y t t+1π - π = -H y ,π ,t

Similarly from the envelope theorem:*π πH = L = Q

*t+1 t π t t+1y - y = H y ,π ,t *

π t t+1= H y ,π ,t t+1 t t ty - y = Q y ,z ,t

25

The Maximum Principle:

. .s t

t+1 t t t

t t

y - y = Q y ,z ,t

G y ,z ,t 0 max

T

t tt =0t t

y ,zF y ,z ,t

. .

For each : maximizes

the Hamiltonian s t t

t t t+1 t t

t z

H y ,z ,π ,t G y ,z

.

,t

1

0

t t t+1 t t t+1 t tH y ,z ,π ,t = F y ,z ,t π Q y ,z ,t

*t+1 t y t t+1 π - π = -H y ,π2. ,t

*t+1 t π t t+1 y -3. y = H y , π ,t

max s.t. *t t+1 t t t+1 t t

tzH y ,π ,t = H y ,z ,π ,t G y ,z ,t 0

26


. .s t

t+1 t t t

t t

y - y = Q y ,z ,t

G y ,z ,t 0 max

T

t tt =0t t

y ,zF y ,z ,t

. .



t t t+1 t t

t z


.

,t

1

0

t t t+1 t t t+1 t tH y ,z ,π ,t = F y ,z ,t π Q y ,z ,t

*t+1 t y t t+1 π - π = -H y ,π2. ,t

*t+1 t π t t+1 y -3. y = H y , π ,t

max s.t. *t t+1 t t t+1 t t

tzH y ,π ,t = H y ,z ,π ,t G y ,z ,t 0 t t t+1 t t t+1 t tH y ,z ,π ,t = F y ,z ,t π Q y ,z ,t

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. .s t

t+1 t t t

t t

y - y = Q y ,z ,t

G y ,z ,t 0 max

T

t tt =0t t

y ,zF y ,z ,t

. .



t t t+1 t t

t z


.

,t

1

0

*t+1 t y t t+1 π - π = -H y ,π2. ,t

*t+1 t π t t+1 y -3. y = H y , π ,t

t t t+1 t t t+1 t tH y ,z ,π ,t = F y ,z ,t π Q y ,z ,t t t t+1 t t t tH y ,z ,π ,t = F y ,z ,t Q yt+1 ,z ,tπ

1. 2 no lecture on wed february 8th thursday 9 th feb friday 27 th jan friday 10 th feb thursday...

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