Introduction to Optimization
(Part 1)
Daniel Kirschen
Economic dispatch problem
A
B
C
L
• 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
© 2011 D. Kirschen and University of Washington
2
Characteristics of the generating units
T
G
J/h
Input
B
Fuel
(Input)
Electric Power
(Output)
MW
Pmin
Pmax
• Thermal generating units
Output
• 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
© 2011 D. Kirschen and University of Washington
3
Cost Curve
• Multiply fuel input by fuel cost
• No-load cost
– Cost of keeping the unit running if it could produce zero MW
Cost
$/h
No-load cost
MW
Pmax
Pmin
© 2011 D. Kirschen and University of Washington
Output
4
Incremental Cost Curve
• Incremental cost curve
Cost [$/h]
D FuelCost vs Power
D Power
• Derivative of the cost curve
• In $/MWh
• Cost of the next MWh
∆F
∆P
MW
Incremental Cost
[$/MWh]
MW
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5
Mathematical formulation
• Objective function
C = CA (PA ) + CB (PB ) + CC (PC )
A
B
C
L
• Constraints
– Load / Generation balance:
L = PA + PB + PC
– Unit Constraints:
PAmin £ PA £ PAmax
PBmin £ PB £ PBmax
PCmin £ PC £ PCmax
© 2011 D. Kirschen and University of Washington
This is an optimization problem
6
Introduction to Optimization
“An engineer can do with one dollar which any
bungler can do with two”
A. M. Wellington (1847-1895)
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8
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
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9
Decision Variables
• The value of the objective is a function of
some decision variables:
F = f ( x1 , x2 , x3 , .. xn )
• Examples of decision variables:
– Dimensions of the transformer
– Output of generating units, position of taps
– Parameters of bids for selling electrical energy
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10
Optimization Problem
• What value should the decision variables take
so that F = f ( x1 , x2 , x3 , .. xn ) is minimum or
maximum?
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11
Example: function of one variable
f(x*)
f(x)
x*
x
f(x) is maximum for x = x*
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12
Minimization and Maximization
f(x*)
f(x)
x*
x
-f(x)
-f(x*)
If x = x* maximizes f(x) then it minimizes - f(x)
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13
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
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14
Necessary Condition for Optimality
f(x*)
f(x)
df
>0
dx
df
<0
dx
x*
x
If x = x * maximises f ( x ) then:
df
f ( x ) < f ( x ) for x < x Þ
> 0 for x < x *
dx
*
*
df
f ( x ) < f ( x ) for x > x Þ
< 0 for x > x *
dx
*
© 2011 D. Kirschen and University of Washington
*
15
Necessary Condition for Optimality
df
=0
dx
f(x)
x*
x
df
If x = x maximises f ( x ) then
= 0 for x = x *
dx
*
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16
Example
f(x)
df
For what values of x is
=0?
dx
x
In other words, for what values of x is the necessary condition for
optimality satisfied?
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17
Example
f(x)
•
•
•
•
A
B
C
A, B, C, D are stationary points
A and D are maxima
B is a minimum
C is an inflexion point
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D
x
18
How can we distinguish minima and maxima?
f(x)
A
B
C
D
For x = A and x = D, we have:
x
d2 f
dx
2
<0
The objective function is concave around a maximum
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19
How can we distinguish minima and maxima?
f(x)
A
B
C
For x = B we have:
D
d2 f
dx
2
x
>0
The objective function is convex around a minimum
© 2011 D. Kirschen and University of Washington
20
How can we distinguish minima and maxima?
f(x)
A
B
C
For x = C , we have:
D
d2 f
dx
2
x
=0
The objective function is flat around an inflexion point
© 2011 D. Kirschen and University of Washington
21
Necessary and Sufficient Conditions of Optimality
• Necessary condition: df = 0
dx
• Sufficient condition:
– For a maximum:
d2 f
<0
2
dx
– For a minimum:
d2 f
>0
2
dx
© 2011 D. Kirschen and University of Washington
22
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
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23
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)
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Feasible Set
f(x)
xMIN
A
D
xMAX
x
Feasible Set
The values of the objective function outside
the feasible set do not matter
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25
Interior and Boundary Solutions
f(x)
xMIN
A
•
•
•
•
© 2011 D. Kirschen and University of Washington
B
A and D are interior maxima
B and E are interior minima
XMIN is a boundary minimum
XMAX is a boundary maximum
D
E
xMAX
x
Do not satisfy the
Optimality conditions!
26
Two-Dimensional Case
f(x1,x2)
x1*
x2
*
x2
© 2011 D. Kirschen and University of Washington
x1
f(x1,x2) is minimum for x1*, x2*
27
Necessary Conditions for Optimality
f(x1,x2)
¶f ( x 1 ,x 2 )
¶x 1
=0
x * ,x *
1 2
¶f ( x 1 ,x 2 )
¶x 2
=0
x * ,x *
1 2
x1*
x2*
x1
x2
© 2011 D. Kirschen and University of Washington
28
Multi-Dimensional Case
At a maximum or minimum value of f ( x1 , x2 , x3 , .. xn )
we must have:
¶f
=0
¶x 1
¶f
=0
¶x 2
¶f
=0
¶x n
A point where these conditions are satisfied is called a stationary point
© 2011 D. Kirschen and University of Washington
29
Sufficient Conditions for Optimality
f(x1,x2)
minimum
maximum
x1
x2
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30
Sufficient Conditions for Optimality
f(x1,x2)
Saddle point
x1
x2
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Sufficient Conditions for Optimality
Calculate the Hessian matrix at the stationary point:
æ ¶2 f
ç ¶x 2
1
ç
ç ¶2 f
ç
ç ¶x 2 ¶x 1
ç
ç ¶2 f
çç
è ¶x n ¶x 1
© 2011 D. Kirschen and University of Washington
¶2 f
¶x 1 ¶x 2
¶2 f
¶x 22
¶2 f
¶x n ¶x 2
¶2 f ö
¶x 1 ¶x n ÷
÷
2
¶ f ÷
÷
¶x 2 ¶x n ÷
÷
¶ 2 f ÷÷
÷
¶x n2 ø
32
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
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33
Contours
f(x1,x2)
F2
F1
x1
F1
F2
x2
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34
Contours
A contour is the locus of all the point that give the same value
to the objective function
x2
© 2011 D. Kirschen and University of Washington
Minimum or maximum
x1
35
Example 1
Minimise C = x 12 + 4 x 22 - 2 x 1 x 2
Necessary conditions for optimality:
¶C
= 2 x1 - 2 x 2 = 0
¶x 1
¶C
= -2 x 1 + 8 x 2 = 0
¶x 2
© 2011 D. Kirschen and University of Washington
ì x1 = 0
í
îx 2 = 0
is a stationary
point
36
Example 1
Sufficient conditions for optimality:
æ ¶ 2C
ç ¶x 2
1
ç
Hessian Matrix:
ç ¶ 2C
ç
è ¶x 2 ¶x 1
¶ 2C ö
¶x 1 ¶x 2 ÷ æ 2 -2 ö
÷ =ç
÷
2
¶ C ÷ è -2 8 ø
÷
¶x 22 ø
must be positive definite (i.e. all eigenvalues must be positive)
l-2
2
2
l-8
= 0 Þ l 2 - 10 l + 12 = 0
10 ± 52
Þl =
³0
2
© 2011 D. Kirschen and University of Washington
The stationary point
is a minimum
37
Example 1
x2
C=9
C=1
C=4
x1
Minimum: C=0
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38
Example 2
Minimize C = -x12 + 3x22 + 2x1 x2
Necessary conditions for optimality:
¶C
= -2x1 + 2x2 = 0
¶x1
ì x1 = 0
í
¶C
îx 2 = 0
= 2x1 + 6x2 = 0
¶x2
© 2011 D. Kirschen and University of Washington
is a stationary
point
39
Example 2
Sufficient conditions for optimality:
æ ¶ 2C
ç ¶x 2
1
ç
Hessian Matrix:
ç ¶ 2C
ç
è ¶x 2 ¶x 1
l+2
-2
-2
l-6
= 0 Þ l2 - 4 l - 8 = 0
4 + 80
Þl =
>0
2
or l =
4-
¶ 2C ö
¶x 1 ¶x 2 ÷ æ -2 2 ö
÷ =ç
÷
2
è
2
6
ø
÷
¶ C
÷
2
¶x 2 ø
80
2
© 2011 D. Kirschen and University of Washington
<0
The stationary point
is a saddle point
40
Example 2
x2
C=0
C=9
C=4
C=1
C=-9
C=-4
C=-4
C=-1
C=-9
x1
C=1
C=4
C=9
This stationary point is a saddle point
© 2011 D. Kirschen and University of Washington
C=0
41
Optimization with Constraints
Optimization with Equality Constraints
• There are usually restrictions on the values
that the decision variables can take
Minimise
f ( x1 , x2 ,.. , xn )
Objective function
subject to:
w 1 ( x1 , x2 ,.. , xn ) = 0
Equality constraints
w m ( x1 , x2 ,.. , xn ) = 0
© 2011 D. Kirschen and University of Washington
43
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
44
Example 1
Minimise f ( x 1 , x 2 ) = 0.25 x + x
2
1
2
2
Subject to w ( x 1 , x 2 ) º 5 - x 1 - x 2 = 0
x2
w ( x1 , x 2 ) º 5 - x1 - x 2 = 0
Minimum
f ( x 1 , x 2 ) = 0.25 x + x
2
1
© 2011 D. Kirschen and University of Washington
2
2
x1
45
Example 2: Economic Dispatch
x1
G1
x2
G2
L
C 1 = a1 + b1 x 12
Cost of running unit 1
C 2 = a 2 + b 2 x 22
Cost of running unit 2
C = C 1 + C 2 = a1 + a 2 + b1 x 12 + b 2 x 22
Total cost
Optimization problem:
Minimise C = a1 + a 2 + b1 x 12 + b 2 x 22
Subject to: x 1 + x 2 = L
© 2011 D. Kirschen and University of Washington
46
Solution by substitution
Minimise C = a1 + a 2 + b1 x 12 + b 2 x 22
Subject to: x 1 + x 2 = L
Þ x 2 = L - x1
Þ C = a1 + a 2 + b1 x 12 + b 2 ( L - x 1 )
2
Unconstrained minimization
dC
= 2 b1 x 1 - 2 b 2 ( L - x 1 ) = 0
dx 1
b2 L
Þ x1 =
b1 + b 2
d 2C
© 2011 D. Kirschen and University of Washington
dx
2
1
b1 L ö
æ
çÞ x2 =
÷
è
b1 + b 2 ø
= 2b1 + 2 b 2 > 0 Þ minimum
47
Solution by substitution
• Difficult
• Usually impossible when constraints are nonlinear
• Provides little or no insight into solution
• Solution using Lagrange multipliers
© 2011 D. Kirschen and University of Washington
48
Gradient
Consider a function f (x1 , x2 ,.. , xn )
æ
ç
ç
ç
The gradient of f is the vector Ñf = ç
ç
ç
ç
ç
çè
© 2011 D. Kirschen and University of Washington
¶f ö
¶x1 ÷
÷
¶f ÷
¶x2 ÷÷
÷
¶f ÷
÷
¶xn ÷ø
49
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
50
Example 3
f ( x , y ) = ax 2 + by 2
æ ¶f ö
ç ¶x ÷ æ 2 ax ö
Ñf = ç ÷ = ç
÷
¶f
ç ÷ è 2 by ø
è ¶y ø
y
B
C
A
x
D
© 2011 D. Kirschen and University of Washington
51
Example 4
f ( x , y ) = ax + by
æ ¶f ö
ç ¶x ÷ æ a ö
Ñf = ç ÷ = ç ÷
¶f
ç ÷ è bø
è ¶y ø
f = f3
f = f2
y
f = f1
Ñf
Ñf
Ñf
x
© 2011 D. Kirschen and University of Washington
52
Lagrange multipliers
Minimise f ( x 1 , x 2 ) = 0.25 x 12 + x 22 subject to w ( x 1 , x 2 ) º 5 - x 1 - x 2 = 0
w ( x1 , x 2 ) = 5 - x1 - x 2
f = 0.25 x 12 + x 22 = 6
f = 0.25 x 12 + x 22 = 5
© 2011 D. Kirschen and University of Washington
53
Lagrange multipliers
Ñf
æ ¶f
ç ¶x 1
Ñf = ç
ç ¶f
ç
è ¶x 2
ö
÷
÷
÷
÷
ø
f ( x1 , x 2 ) = 6
f ( x1 , x 2 ) = 5
Ñf
Ñf
© 2011 D. Kirschen and University of Washington
54
Lagrange multipliers
æ ¶w
ç ¶x 1
Ñw = ç
çç ¶w
è ¶x 2
w ( x1 , x 2 )
ö
÷
÷
÷÷
ø
f ( x1 , x 2 ) = 6
f ( x1 , x 2 ) = 5
Ñw
Ñw
© 2011 D. Kirschen and University of Washington
Ñw
55
Lagrange multipliers
The solution must be on the constraint
To reduce the value of f, we must move
in a direction opposite to the gradient
w ( x1 , x 2 )
f ( x1 , x 2 ) = 6
f ( x1 , x 2 ) = 5
Ñf
A
?
Ñf
B
© 2011 D. Kirschen and University of Washington
56
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
w ( x1 , x 2 )
f ( x1 , x 2 ) = 6
f ( x1 , x 2 ) = 5
Ñf
A
Ñf
Ñw
At the optimum, the gradient of the
function is parallel to the gradient
of the constraint
© 2011 D. Kirschen and University of Washington
C
Ñf
Ñw
B
Ñw
57
Lagrange multipliers
At the optimum, we must have: Ñf Ñw
Which can be expressed as:
Ñf + l Ñw = 0
In terms of the co-ordinates:
¶f
¶w
+l
=0
¶x 1
¶x 1
¶f
¶w
+l
=0
¶x 2
¶x 2
The constraint must also be satisfied: w ( x 1 , x 2 ) = 0
l is called the Lagrange multiplier
© 2011 D. Kirschen and University of Washington
58
Lagrangian function
To simplify the writing of the conditions for optimality,
it is useful to define the Lagrangian function:
( x 1 , x 2 ,l ) = f ( x 1 , x 2 ) + lw ( x 1 , x 2 )
The necessary conditions for optimality are then given
by the partial derivatives of the Lagrangian:
¶
( x 1 , x 2 ,l )
¶x 1
¶
( x 1 , x 2 ,l )
¶x 2
¶
© 2011 D. Kirschen and University of Washington
( x 1 , x 2 ,l )
¶l
¶f
¶w
=
+l
=0
¶x 1
¶x 1
¶f
¶w
=
+l
=0
¶x 2
¶x 2
= w ( x1 ,x 2 ) = 0
59
Example
Minimise f ( x 1 , x 2 ) = 0.25 x 12 + x 22 subject to w ( x 1 , x 2 ) º 5 - x 1 - x 2 = 0
( x 1 , x 2 , l ) = 0.25 x 12 + x 22 + l ( 5 - x 1 - x 2 )
¶
( x 1 , x 2 ,l )
¶x 1
¶
( x 1 , x 2 ,l )
¶x 2
¶
( x 1 , x 2 ,l )
¶l
º 0.5 x 1 - l = 0
º 2x2 -l = 0
º 5 - x1 - x 2 = 0
© 2011 D. Kirschen and University of Washington
60
Example
¶
( x 1 , x 2 ,l )
¶x 1
¶
( x 1 , x 2 ,l )
¶x 2
¶
( x 1 , x 2 ,l )
¶l
º 0.5 x 1 - l = 0
Þ x1 = 2 l
º 2x2 -l = 0
1
Þ x2 = l
2
º 5 - x1 - x 2 = 0
1
Þ 5 - 2l - l = 0
2
Þl=2
Þ x1 = 4
Þ x2 =1
© 2011 D. Kirschen and University of Washington
61
Example
Minimise f ( x 1 , x 2 ) = 0.25 x + x
2
1
2
2
Subject to w ( x 1 , x 2 ) º 5 - x 1 - x 2 = 0
x2
w ( x1 , x 2 ) º 5 - x1 - x 2 = 0
1
f ( x1 , x 2 ) = 5
Minimum
4
© 2011 D. Kirschen and University of Washington
x1
62
Important Note!
If the constraint is of the form:
ax 1 + bx 2 = L
It must be included in the Lagrangian as follows:
= f ( x1 ,.. , xn ) + l ( L - ax1 - bx2 )
And not as follows:
= f ( x1 ,.. , xn ) + l ( ax1 + bx2 )
© 2011 D. Kirschen and University of Washington
63
Application to Economic Dispatch
x1
G1
x2
G2
L
minimise f ( x 1 , x 2 ) = C 1 ( x 1 ) + C 2 ( x 2 )
s.t . w ( x 1 , x 2 ) º L - x 1 - x 2 = 0
( x 1 , x 2 , l ) = C1 ( x 1 ) + C 2 ( x 2 ) + l ( L - x 1 - x 2 )
dC 1
¶
º
-l =0
¶x 1 dx 1
dC 2
¶
º
-l =0
¶x 2 dx 2
¶
º L - x1 - x 2 = 0
¶l
© 2011 D. Kirschen and University of Washington
dC 1
dx 1
=
dC 2
dx 2
=l
Equal incremental cost
solution
64
Equal incremental cost solution
Cost curves:
C1 ( x 1 )
C2 ( x 2 )
x1
Incremental
cost curves:
x2
dC 1
dC 2
dx 1
dx 2
© 2011 D. Kirschen and University of Washington
x1
x2
65
Interpretation of this solution
dC 1
dC 2
dx 1
dx 2
l
x1*
L
x1
x2*
-
-
+
L-x -x
*
1
© 2011 D. Kirschen and University of Washington
x2
*
2
If < 0, reduce λ
If > 0, increase λ
66
Physical interpretation
DC
= lim
dx Dx®0 Dx
dC
C( x )
Dx
DC
x
For Dx sufficiently small:
dC
DC »
´ Dx
dx
If Dx = 1 MW :
dC
DC »
dx
dC(x)
dx
x
© 2011 D. Kirschen and University of Washington
The incremental cost is the cost of
one additional MW for one hour.
This cost depends on the output of
the generator.
67
Physical interpretation
dC 1
dx 1
dC 2
dx 2
: Cost of one more MW from unit 1
: Cost of one more MW from unit 2
Suppose that
dC 1
dx 1
>
dC 2
dx 2
Decrease output of unit 1 by 1MW Þ decrease in cost =
Increase output of unit 2 by 1MW Þ increase in cost =
dC 2 dC 1
Net change in cost =
<0
dx 2 dx 1
© 2011 D. Kirschen and University of Washington
dC 1
dx 1
dC 2
dx 2
68
Physical interpretation
It pays to increase the output of unit 2 and decrease the
output of unit 1 until we have:
dC 1
dx 1
=
dC 2
dx 2
=l
The Lagrange multiplier λ is thus the cost of one more MW
at the optimal solution.
This is a very important result with many applications in
economics.
© 2011 D. Kirschen and University of Washington
69
Generalization
Minimise
f ( x1 , x2 ,.. , xn )
subject to:
w 1 ( x1 , x2 ,.. , xn ) = 0
w m ( x1 , x2 ,.. , xn ) = 0
Lagrangian:
= f ( x1 ,.. , xn ) + l1w 1 ( x1,.. , xn ) +
+ lmw m ( x1 ,.. , xn )
• One Lagrange multiplier for each constraint
• n + m variables: x1, …, xn and λ1, …, λm
© 2011 D. Kirschen and University of Washington
70
Optimality conditions
= f ( x1,.. , xn ) + l1w 1 ( x1,.. , xn ) +
¶w 1
¶f
¶
=
+ l1
+
¶x 1 ¶x 1
¶x 1
+ lmw m ( x1 ,.. , xn )
¶w m
+lm
=0
¶x 1
n equations
¶w 1
¶f
¶
=
+ l1
+
¶x n ¶x n
¶x n
+lm
¶w m
¶x n
=0
¶
= w1 ( x1 , ,x n ) = 0
¶l 1
m equations
¶
= w m ( x1 , ,x n ) = 0
¶l m
© 2011 D. Kirschen and University of Washington
n + m equations in
n + m variables
71