introduction to optimization (part 1)

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Introduction to Optimization (Part 1). Daniel Kirschen. L. A. B. C. Economic d ispatch problem. Several generating units serving the load What share of the load should each generating unit produce? Consider the limits of the generating units Ignore the limits of the network. - PowerPoint PPT Presentation

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Introduction to Optimization(Part 1)

Daniel Kirschen

2

Economic dispatch problem

• Several generating units serving the load• What share of the load should each

generating unit produce?• Consider the limits of the generating units• Ignore the limits of the network

A B C L

© 2011 D. Kirschen and University of Washington

3

Characteristics of the generating units

© 2011 D. Kirschen and University of Washington

• Thermal generating units• Consider the running costs only• Input / Output curve

– Fuel vs. electric power• Fuel consumption measured by its energy content• Upper and lower limit on output of the generating unit

B T G

(Input)Electric PowerFuel

(Output)

OutputPmin Pmax

Inpu

t

J/h

MW

4

Cost Curve

• Multiply fuel input by fuel cost• No-load cost

– Cost of keeping the unit running if it could produce zero MW

OutputPmin Pmax

Cost

$/h

MWNo-load cost

© 2011 D. Kirschen and University of Washington

5

Incremental Cost Curve

• Incremental cost curve

• Derivative of the cost curve• In $/MWh• Cost of the next MWh

© 2011 D. Kirschen and University of Washington

∆F∆P

Cost [$/h]

MW

Incremental Cost [$/MWh]

MW

6

Mathematical formulation

• Objective function

• Constraints– Load / Generation balance:

– Unit Constraints:

© 2011 D. Kirschen and University of Washington

A B C L

This is an optimization problem

Introduction to Optimization

8

“An engineer can do with one dollar which any bungler can do with two”

A. M. Wellington (1847-1895)

© 2011 D. Kirschen and University of Washington

9

Objective

• Most engineering activities have an objective:– Achieve the best possible design – Achieve the most economical operating conditions

• This objective is usually quantifiable• Examples:

– minimize cost of building a transformer– minimize cost of supplying power– minimize losses in a power system– maximize profit from a bidding strategy

© 2011 D. Kirschen and University of Washington

10

Decision Variables

• The value of the objective is a function of some decision variables:

• Examples of decision variables:– Dimensions of the transformer– Output of generating units, position of taps– Parameters of bids for selling electrical energy

© 2011 D. Kirschen and University of Washington

11

Optimization Problem

• What value should the decision variables take so that

is minimum or maximum?

© 2011 D. Kirschen and University of Washington

12

Example: function of one variable

© 2011 D. Kirschen and University of Washington

x

f(x)

x*

f(x*)

f(x) is maximum for x = x*

13

Minimization and Maximization

© 2011 D. Kirschen and University of Washington

x

f(x)

x*

f(x*)

If x = x* maximizes f(x) then it minimizes - f(x)

-f(x)-f(x*)

14

Minimization and Maximization

• maximizing f(x) is thus the same thing as minimizing g(x) = -f(x)

• Minimization and maximization problems are thus interchangeable

• Depending on the problem, the optimum is either a maximum or a minimum

© 2011 D. Kirschen and University of Washington

15

Necessary Condition for Optimality

© 2011 D. Kirschen and University of Washington

x

f(x)

x*

f(x*)

16

Necessary Condition for Optimality

© 2011 D. Kirschen and University of Washington

x

f(x)

x*

17

Example

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x

f(x)

For what values of x is ?

In other words, for what values of x is the necessary condition for optimality satisfied?

18

Example

• A, B, C, D are stationary points• A and D are maxima• B is a minimum• C is an inflexion point

x

f(x)

A B C D

© 2011 D. Kirschen and University of Washington

19

How can we distinguish minima and maxima?

© 2011 D. Kirschen and University of Washington

x

f(x)

A B C D

The objective function is concave around a maximum

20

How can we distinguish minima and maxima?

x

f(x)

A B C D

The objective function is convex around a minimum© 2011 D. Kirschen and University of Washington

21

How can we distinguish minima and maxima?

© 2011 D. Kirschen and University of Washington

x

f(x)

A B C D

The objective function is flat around an inflexion point

22

Necessary and Sufficient Conditions of Optimality

• Necessary condition:

• Sufficient condition:– For a maximum:

– For a minimum:

© 2011 D. Kirschen and University of Washington

23

Isn’t all this obvious?

• Can’t we tell all this by looking at the objective function?– Yes, for a simple, one-dimensional case when we

know the shape of the objective function– For complex, multi-dimensional cases (i.e. with

many decision variables) we can’t visualize the shape of the objective function

– We must then rely on mathematical techniques

© 2011 D. Kirschen and University of Washington

24

Feasible Set

• The values that the decision variables can take are usually limited

• Examples:– Physical dimensions of a transformer must be

positive– Active power output of a generator may be limited

to a certain range (e.g. 200 MW to 500 MW)– Reactive power output of a generator may be

limited to a certain range (e.g. -100 MVAr to 150 MVAr)

© 2011 D. Kirschen and University of Washington

25

Feasible Set

x

f(x)

A D xMAXxMIN

Feasible Set

The values of the objective function outside the feasible set do not matter

© 2011 D. Kirschen and University of Washington

26

Interior and Boundary Solutions

• A and D are interior maxima• B and E are interior minima• XMIN is a boundary minimum• XMAX is a boundary maximum

x

f(x)

A D xMAXxMIN B E

Do not satisfy theOptimality conditions!

© 2011 D. Kirschen and University of Washington

27

Two-Dimensional Case

x1

x2

f(x1,x2)

x2*

x1*

f(x1,x2) is minimum for x1*, x2

* © 2011 D. Kirschen and University of Washington

28

Necessary Conditions for Optimality

x1

x2

f(x1,x2)

x2*

x1*

© 2011 D. Kirschen and University of Washington

29

Multi-Dimensional Case

At a maximum or minimum value of

we must have:

A point where these conditions are satisfied is called a stationary point

© 2011 D. Kirschen and University of Washington

30

Sufficient Conditions for Optimality

x1

x2

f(x1,x2) minimum maximum

© 2011 D. Kirschen and University of Washington

31

Sufficient Conditions for Optimality

x1

x2

f(x1,x2)

Saddle point

© 2011 D. Kirschen and University of Washington

32

Sufficient Conditions for Optimality

Calculate the Hessian matrix at the stationary point:

© 2011 D. Kirschen and University of Washington

33

Sufficient Conditions for Optimality

• Calculate the eigenvalues of the Hessian matrix at the stationary point

• If all the eigenvalues are greater or equal to zero:– The matrix is positive semi-definite– The stationary point is a minimum

• If all the eigenvalues are less or equal to zero:– The matrix is negative semi-definite– The stationary point is a maximum

• If some or the eigenvalues are positive and other are negative:– The stationary point is a saddle point

© 2011 D. Kirschen and University of Washington

34

Contours

x1

x2

f(x1,x2)

F1 F2

F2

F1

© 2011 D. Kirschen and University of Washington

35

Contours

x1

x2

Minimum or maximum

A contour is the locus of all the point that give the same valueto the objective function

© 2011 D. Kirschen and University of Washington

36

Example 1

is a stationarypoint

© 2011 D. Kirschen and University of Washington

37

Example 1Sufficient conditions for optimality:

must be positive definite (i.e. all eigenvalues must be positive)

The stationary point is a minimum

© 2011 D. Kirschen and University of Washington

38

Example 1

© 2011 D. Kirschen and University of Washington

x1

x2

C=1C=4

C=9

Minimum: C=0

39

Example 2

is a stationarypoint

© 2011 D. Kirschen and University of Washington

40

Example 2Sufficient conditions for optimality:

The stationary point is a saddle point

© 2011 D. Kirschen and University of Washington

41

Example 2

© 2011 D. Kirschen and University of Washington

x1

x2

C=1

C=4

C=9

C=1

C=4

C=9

C=-1 C=-4 C=-9

C=0

C=0

C=-9 C=-4

This stationary point is a saddle point

Optimization with Constraints

43

Optimization with Equality Constraints

• There are usually restrictions on the values that the decision variables can take

© 2011 D. Kirschen and University of Washington

Objective function

Equality constraints

44

Number of Constraints

• N decision variables• M equality constraints• If M > N, the problems is over-constrained

– There is usually no solution• If M = N, the problem is determined

– There may be a solution• If M < N, the problem is under-constrained

– There is usually room for optimization

© 2011 D. Kirschen and University of Washington

45

Example 1

x1

x2

Minimum

© 2011 D. Kirschen and University of Washington

46

Example 2: Economic Dispatch

LG1 G2

x1 x2

Cost of running unit 1

Cost of running unit 2

Total cost

Optimization problem:

© 2011 D. Kirschen and University of Washington

47

Solution by substitution

Unconstrained minimization

© 2011 D. Kirschen and University of Washington

48

Solution by substitution

• Difficult• Usually impossible when constraints are non-

linear• Provides little or no insight into solution

• Solution using Lagrange multipliers

© 2011 D. Kirschen and University of Washington

49

Gradient

© 2011 D. Kirschen and University of Washington

50

Properties of the Gradient

• Each component of the gradient vector indicates the rate of change of the function in that direction

• The gradient indicates the direction in which a function of several variables increases most rapidly

• The magnitude and direction of the gradient usually depend on the point considered

• At each point, the gradient is perpendicular to the contour of the function

© 2011 D. Kirschen and University of Washington

51

Example 3

x

y

© 2011 D. Kirschen and University of Washington

A

B

C

D

52

Example 4

x

y

© 2011 D. Kirschen and University of Washington

53

Lagrange multipliers

© 2011 D. Kirschen and University of Washington

54

Lagrange multipliers

© 2011 D. Kirschen and University of Washington

55

Lagrange multipliers

© 2011 D. Kirschen and University of Washington

56

Lagrange multipliersThe solution must be on the constraint

© 2011 D. Kirschen and University of Washington

A

B

To reduce the value of f, we must move in a direction opposite to the gradient

?

57

Lagrange multipliers• We stop when the gradient of the function

is perpendicular to the constraint because moving further would increase the value of the function

At the optimum, the gradient of the function is parallel to the gradient of the constraint

© 2011 D. Kirschen and University of Washington

A

B

C

58

Lagrange multipliersAt the optimum, we must have:

Which can be expressed as:

is called the Lagrange multiplier

The constraint must also be satisfied:

In terms of the co-ordinates:

© 2011 D. Kirschen and University of Washington

59

Lagrangian functionTo simplify the writing of the conditions for optimality,it is useful to define the Lagrangian function:

The necessary conditions for optimality are then given by the partial derivatives of the Lagrangian:

© 2011 D. Kirschen and University of Washington

60

Example

© 2011 D. Kirschen and University of Washington

61

Example

© 2011 D. Kirschen and University of Washington

62

Example

x1

x2

Minimum

4

1

© 2011 D. Kirschen and University of Washington

63

Important Note!If the constraint is of the form:

It must be included in the Lagrangian as follows:

And not as follows:

© 2011 D. Kirschen and University of Washington

64

Application to Economic Dispatch

LG1 G2

x1 x2

Equal incremental costsolution

© 2011 D. Kirschen and University of Washington

65

Equal incremental cost solution

x1 x2

Cost curves:

x1 x2

Incrementalcost curves:

© 2011 D. Kirschen and University of Washington

66

Interpretation of this solution

x1 x2

L+

--

If < 0, reduce λIf > 0, increase λ

© 2011 D. Kirschen and University of Washington

67

Physical interpretation

x

x

The incremental cost is the cost ofone additional MW for one hour. This cost depends on the output of the generator.

© 2011 D. Kirschen and University of Washington

68

Physical interpretation

© 2011 D. Kirschen and University of Washington

69

Physical interpretation

It pays to increase the output of unit 2 and decrease the output of unit 1 until we have:

The Lagrange multiplier λ is thus the cost of one more MWat the optimal solution.

This is a very important result with many applications in economics.

© 2011 D. Kirschen and University of Washington

70

Generalization

Lagrangian:

• One Lagrange multiplier for each constraint• n + m variables: x1, …, xn and λ1, …, λm

© 2011 D. Kirschen and University of Washington

71

Optimality conditions

n equations

m equations

n + m equations inn + m variables

© 2011 D. Kirschen and University of Washington

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