Introduction to Optimization - PowerPoint Presentation

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Introduction to Optimization - PPT Presentation

Part 1 Daniel Kirschen 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 ID: 179789

washington kirschen 2011 university kirschen washington university 2011 cost point function optimality objective conditions minimum stationary lagrange output gradient

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Slide1

Introduction to Optimization(Part 1)

Daniel KirschenSlide2

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 unitsIgnore the limits of the network

2Slide3

Thermal generating units

Consider

the running costs only

Input / Output curveFuel vs. electric powerFuel consumption measured by its energy content

Upper and lower limit on output of the generating unit

(Input)

Electric Power

Fuel

(Output)

Output

min

max

Input

J/h

MWSlide4

Cost Curve

Multiply fuel input by fuel cost

No-load cost

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

Output

min

max

Cost

No-load cost

4Slide5

Incremental Cost CurveIncremental cost curve

Derivative of the cost curve

MWhCost of the next MWh

∆F

∆P

Cost

[$/

Incremental Cost

[$/

MWh]

MWSlide6

Mathematical formulation

Objective function

Constraints

Load / Generation balance:Unit Constraints:

This is an

optimization

problemSlide7

Introduction to OptimizationSlide8

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

A. M. Wellington (1847-1895)

8Slide9

ObjectiveMost 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 transformerminimize cost of supplying power

minimize losses in a power systemmaximize profit from a bidding strategy

9Slide10

Decision VariablesThe 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 tapsParameters of bids for selling electrical energy

10Slide11

Optimization ProblemWhat value should the decision variables take so that

minimum or maximum?

11Slide12

f(x)

f(x

)

f(x) is maximum for x = x* Slide13

Minimization and Maximization

f(x)

f(x

)

If x = x*

maximizes

f(x) then it

minimizes

- f(x)

-f(x)

-f(x

)Slide14

Minimization and Maximization

maximizing

f(x)

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

Minimization and maximization problems are thus interchangeable

14Slide15

f(x)

f(x

)Slide16

f(x)

*Slide17

f(x)

For what values of

is ?

In other words, for what values of

is the necessary condition for optimality satisfied?Slide18

Example

A, B, C, D are stationary points

A and D are maxima

B is a minimumC is an inflexion point

f(x)

18Slide19

f(x)

The objective function is

concave

around a maximumSlide20

How can we distinguish minima and maxima?

f(x)

The objective function is

convex

around a minimum

20Slide21

f(x)

The objective function is flat around an inflexion pointSlide22

Necessary and Sufficient Conditions of Optimality

Necessary condition:

Sufficient condition:

For a maximum:For a minimum:

22Slide23

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

23Slide24

Feasible SetThe 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)

24Slide25

Feasible Set

f(x)

MAX

MIN

Feasible Set

The values of the objective function outside

the feasible set do not matter

25Slide26

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

f(x)

MAX

MIN

Do not satisfy

the

ptimality

conditions

26Slide27

Two-Dimensional Case

f(x

)

f(x

) is minimum for x

, x

27Slide28

Necessary Conditions for Optimality

f(x

)

28Slide29

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

29Slide30

Sufficient Conditions for Optimality

f(x

)

minimum

maximum

30Slide31

Sufficient Conditions for Optimality

f(x

)

Saddle point

31Slide32

Sufficient Conditions for Optimality

Calculate the Hessian matrix at the stationary point:

32Slide33

Sufficient Conditions for OptimalityCalculate 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-definiteThe stationary point is a maximumIf some or the eigenvalues are positive and other are negative:The stationary point is a saddle point

33Slide34

Contours

f(x

)

34Slide35

Contours

Minimum or maximum

A contour is the locus of all the point that give the same value

to the objective function

35Slide36

Example 1

is a stationary

point

36Slide37

Example 1

Sufficient conditions for optimality:

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

The stationary point

is a minimum

37Slide38

C=1

C=4

C=9

Minimum: C=0Slide39

Example 2

is a stationary

point

39Slide40

Example 2

Sufficient conditions for optimality:

The stationary point

is a saddle point

40Slide41

C=1

C=4

C=9

C=1

C=4

C=9

C=-1

C=-4

C=-9

C=0

C=-9

C=-4

This stationary point is a saddle pointSlide42

Optimization with ConstraintsSlide43

Optimization with Equality Constraints

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

Objective function

Equality constraintsSlide44

Number of ConstraintsN decision variables

equality constraints

If M > N, the problems is over-constrainedThere is usually no solutionIf M = N, the problem is determinedThere may be a solution

If M < N, the problem is under-constrainedThere is usually room for optimization

44Slide45

Example 1

Minimum

45Slide46

Example 2: Economic Dispatch

Cost of running unit 1

Cost of running unit 2

Total cost

Optimization

problem:

46Slide47

Solution by substitution

Unconstrained

minimization

47Slide48

Solution by substitution

Difficult

Usually impossible

when constraints are non-linearProvides little or no insight into solution

Solution using Lagrange multipliers

48Slide49

Gradient

49Slide50

Properties of the GradientEach 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

50Slide51

Example 3

DSlide52

Example 4

52Slide53

Lagrange multipliers

53Slide54

Lagrange multipliers

54Slide55

Lagrange multipliers

55Slide56

Lagrange multipliers

The

solution must be on the constraint

To reduce the value of

, we must move

a direction opposite to the gradient

?Slide57

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

CSlide58

Lagrange multipliers

At 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:

58Slide59

Lagrangian function

To 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:

59Slide60

Example

60Slide61

Example

61Slide62

Example

Minimum

62Slide63

Important Note!

If the constraint is of the form:

It must be included in the Lagrangian as follows:

And not as follows:

63Slide64

Application to Economic Dispatch

Equal incremental cost

solution

64Slide65

Equal incremental cost solution

Cost curves:

Incremental

cost curves:

65Slide66

Interpretation of this solution

If < 0, reduce

If > 0, increase

66Slide67

Physical interpretation

The incremental cost is the cost of

one additional MW for one hour.

This cost depends on the output of

the generator.

67Slide68

Physical interpretation

68Slide69

Physical interpretation

It pays to increase the output of unit 2 and decrease the

output of unit 1 until we have:

The Lagrange multiplier

is thus the cost of one more MW

at the optimal solution.

This is a very important result with many applications in

economics

69Slide70

Generalization

Lagrangian:

One Lagrange multiplier for each constraint

n + m variables:

, …,

and

, …,

70Slide71

Optimality conditions

n equations

m equations

n + m

equations in

n + m variables

Introduction to Optimization - PowerPoint Presentation

Introduction to Optimization - PPT Presentation

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