Transcript Optimization - University of Washington

Introduction to Optimization
(Part 2)
Daniel Kirschen
Optimization with inequality constraints
Minimise
f ( x1 , x2 ,.. , xn )
Objective function
subject to:
w 1 ( x1 , x2 ,.. , xn ) = 0
Equality constraints
w m ( x1 , x2 ,.. , xn ) = 0
and:
g1 ( x1 , x2 ,.. , xn ) £ 0
Inequality constraints
gp ( x1 , x2 ,.. , xn ) £ 0
© 2011 D. Kirschen and University of Washington
2
Example: Economic Dispatch
x1
x2
G1
G2
x 1min £ x 1 £ x 1max
L
x 2min £ x 2 £ x 2max
Minimise C = a1 + b1 x 12 + a 2 + b 2 x 22
Subject to:
x1 + x 2 = L
Equality constraints
x 1 - x 1max £ 0
x 1min - x 1 £ 0
x2 - x
max
2
£0
x 2min - x 2 £ 0
© 2011 D. Kirschen and University of Washington
Inequality constraints
3
Example: Economic Dispatch
Minimise C = a1 + b1 x 12 + a 2 + b 2 x 22
x2
x1 + x 2 = L
Family of ellipses
Ellipse tangent to equality constraint at A
Inequality constraints are satisfied
x2max
A
x2min
x1
min
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x1max
x1
4
Example: Economic Dispatch
What is the solution for a larger load?
Ellipse tangent to equality constraint at B
Inequality constraints are NOT satisfied!
x2
x 1 + x 2 = L'
x1 + x 2 = L
x2max
A
B
x2min
x1min
© 2011 D. Kirschen and University of Washington
x1max
x1
5
Example: Economic Dispatch
C is the solution because it is the point on
the equality constraint that satisfies the
inequality constraints at minimum cost
x2
x 1 + x 2 = L'
x1 + x 2 = L
x2max
C
A
B
x2min
x1min
© 2011 D. Kirschen and University of Washington
x1max
x1
6
Binding Inequality Constraints
• A binding inequality constraint is an inequality constraint that is
satisfied exactly
• Example:
– If we must have x1 ≤ x1max
– And at the solution we have x1 = x1max
– Then the constraint x1 ≤ x1max is said to be binding
• ALL of the inequality constraints must be satisfied
• Only a FEW will be binding (or active) at any given time
• But we don’t know ahead of time which inequality constraints
will be binding!
• All equality constraints are always binding
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Solution using Lagrange multipliers
Minimise f ( x1 , x2 ,.. , xn )
subject to:
w i ( x1 , x2 ,.. , xn ) = 0
g j ( x1 , x2 ,.. , xn ) £ 0
i = 1,...m
j = 1,...p
Lagrangian function:
( x1,.. , xn , l1,..., lm , m1,..., m p ) = f ( x1,… , xn )
m
+ å liw i ( x1 ,.. , xn )
i=1
p
+ å m j g j ( x1 ,.. , xn )
j=1
© 2011 D. Kirschen and University of Washington
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Optimality Conditions
(Known as the Karush Kuhn Tucker or KKT conditions)
m
p
i=1
j =1
( x ,l ,m ) = f ( x ) + å l i w i ( x ) + å m j g j ( x )
¶g j ( x )
¶ ( x, l, m ) ¶f ( x ) m
¶w k ( x ) p
º
+ å lk
+ å mj
=0
¶xi
¶xi
¶xi
¶xi
k=1
j=1
i = 1,...n
¶ ( x, l, m )
º w k ( x) = 0
¶lk
k = 1,...m
¶ ( x, l, m )
º gj ( x) £ 0
¶m j
j = 1,... p
m j gj ( x) = 0
j = 1,... p
mj ³ 0
Complementary Slackness Conditions
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j = 1,... p
9
Complementary slackness conditions
m j g j ( x) = 0
mj ³0
Two possibilities for each constraint j:
m j = 0 then the constraint g j ( x ) is non - binding Þ g j ( x ) < 0
OR
g j ( x ) = 0 then the constraint g j ( x ) is binding Þ m j > 0
© 2011 D. Kirschen and University of Washington
10
Solving the problem
• The KKT conditions are not quite enough to
solve the problem
• The complementary slackness conditions:
– Tell us that an inequality constraint is either
binding or non-binding
– DON’T tell us which constraints are binding and
non-binding
• The binding constraints have to be identified
through trial and error
© 2011 D. Kirschen and University of Washington
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Example
Minimise
f ( x 1 , x 2 ) = 0.25 x + x
2
1
2
2
Subject to:
w ( x1 ,x 2 ) º 5 - x1 - x 2 = 0
g ( x 1 , x 2 ) º x 1 + 0.2 x 2 - 3 £ 0
© 2011 D. Kirschen and University of Washington
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Example
x2
g ( x 1 , x 2 ) º x 1 + 0.2 x 2 - 3 £ 0
f ( x 1 , x 2 ) = 0.25 x 12 + x 22
x1
© 2011 D. Kirschen and University of Washington
w ( x1 , x 2 ) º 5 - x1 - x 2 = 0
13
Example
( x 1 , x 2 , l , m ) = f ( x 1 , x 2 ) + lw ( x 1 , x 2 ) + mg ( x 1 , x 2 )
= 0.25 x 12 + x 22 + l ( 5 - x 1 - x 2 ) + m ( x 1 + 0.2 x 2 - 3 )
¶
º 0.5x1 - l + m = 0
¶x1
¶
º 2x2 - l + 0.2 m = 0
¶x2
¶
º 5 - x1 - x2 = 0
¶l
¶
º x1 + 0.2x2 - 3 £ 0
¶m
m g(x) º m ( x1 + 0.2x2 - 3) = 0 and m ³ 0
© 2011 D. Kirschen and University of Washington
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Example
KKT conditions do not tell us if inequality constraint is binding
Must use a trial and error approach
Trial 1: Assume inequality constraint is not binding μ = 0
From solution of example without inequality constraint, we
know that the solution is then:
x1 = 4; x2 = 1
Inserting these values in the inequality constraint, we get:
x1 + 0.2x2 - 3 = 1.2 ³ 0
This solution is thus not an acceptable solution
© 2011 D. Kirschen and University of Washington
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Example
Trial 2: Assume that the inequality constraint is binding
¶
º 5 - x1 - x 2 = 0
¶l
¶
º x 1 + 0.2 x 2 - 3 = 0
¶m
¶
º 0.5 x 1 - l + m = 0
¶x 1
¶
º 2 x 2 - l + 0.2 m = 0
¶x 2
x 1 = 2.5
x 2 = 2.5
l = 5.9375
m = 4.6875
All KKT conditions are satisfied. This solution is acceptable
© 2011 D. Kirschen and University of Washington
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Example: graphical solution
x2
g ( x 1 , x 2 ) º x 1 + 0.2 x 2 - 3 £ 0
Solution of problem
with inequality constraint
f ( x 1 , x 2 ) = 0.25 x + x
2
1
Solution of problem
without constraints
© 2011 D. Kirschen and University of Washington
2
2
Solution of problem
without inequality
constraint
x1
w ( x1 , x 2 ) º 5 - x1 - x 2 = 0
17
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 1min £ x 1 £ x 1max
x 2min £ x 2 £ x 2max
© 2011 D. Kirschen and University of Washington
g1 ( x 1 , x 2 ) º x 1 - x 1max £ 0
g 2 ( x 1 , x 2 ) º x 1min - x 1 £ 0
g 3 ( x 1 , x 2 ) º x 2 - x 2max £ 0
g 4 ( x 1 , x 2 ) º x 2min - x 2 £ 0
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Application to Economic Dispatch
( x 1 , x 2 ,l ,m 1 ,m 2 ,m 3 ,m 4 ) = C1 ( x 1 ) + C 2 ( x 2 ) + l ( L - x 1 - x 2 )
+ m 1 ( x 1 - x 1max ) + m 2 ( x 1min - x 1 )
+ m 3 ( x 2 - x 2max ) + m 4 ( x 2min - x 2
)
KKT Conditions:
dC 1
¶
º
- l + m1 - m 2 = 0
¶x 1 dx 1
dC 2
¶
º
-l +m3 -m4 = 0
¶x 2 dx 2
¶
º L - x1 - x 2 = 0
¶l
© 2011 D. Kirschen and University of Washington
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Application to Economic Dispatch
KKT Conditions (continued):
¶
º x 1 - x 1max £ 0
¶m 1
m 1 ( x 1 - x 1max ) = 0 ; m 1 ³ 0
¶
º x 1min - x 1 £ 0
¶m 2
m 2 ( x 1min - x 1 ) = 0 ; m 2 ³ 0
¶
º x 2 - x 2max £ 0
¶m 3
m 3 ( x 2 - x 2max ) = 0 ; m 3 ³ 0
¶
º x 2min - x 2 £ 0
¶m 4
m 4 ( x 2min - x 2 ) = 0 ; m 4 ³ 0
© 2011 D. Kirschen and University of Washington
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“Solving” these equations
Trial #1: No generator is at a limit
No inequality constraint is binding and all μ’s are equal to zero
dC 1
¶
º
-l =0
¶x 1 dx 1
dC 2
¶
º
-l =0
¶x 2 dx 2
dC 1
dx 1
=
dC 2
dx 2
=l
¶
º L - x1 - x 2 = 0
¶l
All generators operate at the same incremental cost
© 2011 D. Kirschen and University of Washington
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Solving the KKT Equations
Trial #2: Generator 1 is at its upper limit; Other limits not binding
x1 - x1max = 0 Þ m1 ³ 0; m2 = m3 = m4 = 0
dC 1
¶
º
- l + m1 = 0
¶x 1 dx 1
dC 1
dC 2
¶
º
-l =0
¶x 2 dx 2
dC 2
dx 1
dx 2
= l - m1 £ l
=l
All generators do NOT operate at the same incremental cost!
The incremental cost of unit 1 is lower…. Why?
If that was possible, more power would be produced by unit 1
© 2011 D. Kirschen and University of Washington
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Solving the KKT Equations
Trial #3: Generator 1 is at its lower limit; other limits not binding
x1min - x1 = 0 Þ m2 ³ 0; m1 = m3 = m4 = 0
dC 1
¶
º
-l -m2 = 0
¶x 1 dx 1
dC 1
dC 2
¶
º
-l =0
¶x 2 dx 2
dC 2
dx 1
dx 2
= l + m2 ³ l
=l
Again, all generators do NOT operate at the same incremental cost!
The incremental cost of unit 1 is higher… Why?
If that was possible, less power would be produced by unit 1
© 2011 D. Kirschen and University of Washington
23
Physical interpretation of the Lagrangian
Minimise f (x) subject to: g ( x ) = K or K - g ( x ) = 0
Where g ( x ) is either an equality constraint or a
binding inequality constraint
Lagrangian:
( x, l ) = f (x) + l éë K - g ( x ) ùû
The optimum values of the decision variables and the
Lagrange multipliers are given by the optimality conditions:
¶
¶f
¶g
º
-l
=0
¶x ¶x
¶x
¶
º K - g( x) = 0
¶l
© 2011 D. Kirschen and University of Washington
x ,l
*
*
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What happens when I move the constraint?
Equality constraint:
g(x) = K + DK
Inequality constraint:
g(x) ³ K + DK
Keep DK sufficiently small that an optimal solution still exists
and involves the same binding inequality constraints
The optimal value of the decision variables and of the Lagrange
multipliers change as DK changes:
x * = x(K )
l * = l (K )
© 2011 D. Kirschen and University of Washington
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What happens when I move the constraint?
Lagrangian:
( x, l ) = f (x) + l éë K - g ( x ) ùû
d
df dx
dg dx dl
=
+l-l
+
K - g(x)]
[
dK dx dK
dx dK dK
Re-arranging the terms, we get
d
dg ù dx d l
é df
=l+ê -l ú
+
K - g(x)]
[
dK
dx û dK dK
ë dx
At the optimum:
=0
=0
d
dK
© 2011 D. Kirschen and University of Washington
=l
*
opt
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Physical interpretation of the Lagrangian
At the optimum:
(
)
( )
x* , l * = f (x * ) + l * éë K - g x* ùû
Complementary slackness 
=0
The value of the Lagrangian function is thus equal to the
value of the cost or objective function at the optimum
d
dK
=l
opt
*
The value of the Lagrange multipliers at the
optimal solution thus tell us by how much the
the objective changes when the constraints change
The Lagrange multipliers are the marginal cost of the constraints
© 2011 D. Kirschen and University of Washington
27
Physical Interpretation of Lagrange Multipliers
• The Lagrange multipliers are thus often called
shadow costs
• Non-binding inequality constraints have a zero
marginal cost
• Adding binding constraints always increases
the cost of a solution
• Tightening a constraint increases the cost
• Relaxing a constraint decreases the cost
© 2011 D. Kirschen and University of Washington
28
Physical Interpretation of Lagrange Multipliers
• Note that these observations are valid only for
sufficiently small changes in the constraints
• If the constraints change substantially, the set
of constraints that is binding at the optimum
may change
• The constraints that are no longer binding
would then have a zero marginal cost
• The newly binding constraints would then
have a non-zero marginal cost
© 2011 D. Kirschen and University of Washington
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Example: Economic Dispatch
Using Lagrange Multipliers
Example: Economic Dispatch
A load L = 800 MW is supplied by three generating units.
The cost characteristics of these generating units are
given by the following expressions:
C1 = 100 + 8 P1 + 0.1 P12 [$/h] 230 ≤ P1 ≤ 400 MW
C2 = 200 + 7 P2 + 0.06 P22 [$/h] 100 ≤ P2 ≤ 500 MW
C3 = 250 + 9 P3 + 0.07 P32 [$/h] 100 ≤ P3 ≤ 260 MW
Calculate the optimal economic dispatch.
© 2011 D. Kirschen & University of Washington
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Example: Form the Lagrangian
= 100 + 8P1 + 0.1P
2
1
+200 + 7P2 + 0.06P
2
2
+250 + 9P3 + 0.07P
2
3
+ l (L - P1 - P2 - P3 )
+ m1 (P1 - 400) + m2 (230 - P1 )
+ m3 (P2 - 500) + m4 (100 - P2 )
+ m5 (P3 - 260) + m6 (100 - P3 )
© 2011 D. Kirschen & University of Washington
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Example: Optimality Conditions
Set the partial derivatives of the Lagrangian with respect to the
decision variables and λ to zero:
¶
º 8 + 0.2P1 - l + m1 - m2 = 0
¶P1
¶
º 7 + 0.12P2 - l + m 3 - m 4 = 0
¶P2
¶
º 9 + 0.14P3 - l + m5 - m6 = 0
¶P3
¶
º L - P1 - P2 - P3 = 0
¶l
© 2011 D. Kirschen & University of Washington
(1)
(2)
(3)
(4)
33
Example: Optimality Conditions
Set the partial derivatives of the Lagrangian with respect to the
μ’s to zero:
¶
º P1 - 400 £ 0
¶ m1
¶
º 230 - P1 £ 0
¶ m2
¶
º P2 - 500 £ 0
¶ m3
© 2011 D. Kirschen & University of Washington
34
Example: Optimality Conditions
Set the partial derivatives of the Lagrangian with respect to the
μ’s to zero:
¶
º 100 - P2 £ 0
¶m4
¶
º P3 - 260 £ 0
¶ m5
¶
º 100 - P3 £ 0
¶ m6
© 2011 D. Kirschen & University of Washington
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Example: Complementary slackness
m1.(P1 - 400) = 0; m1 ³ 0
m2 .(230 - P1 ) = 0; m2 ³ 0
m3 .(P2 - 500) = 0; m3 ³ 0
m4 .(100 - P2 ) = 0; m4 ³ 0
m5 .(P3 - 260) = 0; m5 ³ 0
m6 .(100 - P3 ) = 0; m6 ³ 0
© 2011 D. Kirschen & University of Washington
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Example: First Trial (1)
Assume that none of the inequality constraints is binding
This means that all the μ‘s are zero
Set the μ’s equal to zero in the optimality conditions.
Equations (1) to (3) become:
¶
º 8 + 0.2P1 - l = 0
¶P1
¶
º 7 + 0.12P2 - l = 0
¶P2
¶
º 9 + 0.14P3 - l = 0
¶P3
© 2011 D. Kirschen & University of Washington
Þ P1 =
Þ P2 =
Þ P3 =
l-8
0.2
l-7
0.12
l-9
0.14
37
Example: First Trial (2)
Inserting these expressions in Equation (4), we get:
L - 5(l - 8) - 8.33(l - 7) - 7.14(l - 9) = 0
l = 46.96 $/MWh
With L = 800MW, we get:
Replacing λ in the equations on the previous slide, we have:
P1 = 194.8 MW
<P
min
1
= 230 MW
Not a valid solution
 trial fails
P2 = 333.0 MW
P3 = 271.2 MW
© 2011 D. Kirschen & University of Washington
>P
max
3
= 260 MW
38
Example: Second Trial (1)
We must try another combination of binding constraints.
6 inequality constraints  26 = 64 possible combinations
Realistic system  ridiculously large number of combinations
Must use information from previous trials
Let us try:
P1 = P1min = 230 MW
P3 = P
max
3
= 260 MW
Þ m2 ¹ 0
Þ m5 ¹ 0
m1 = m3 = m4 = m6 = 0
© 2011 D. Kirschen & University of Washington
39
Example: Second Trial (2)
L - P1 - P2 - P3 = 0
P1 = P1min = 230 MW
P3 = P
max
3
Þ P2 = 310 MW
= 260 MW
However, we must check that the μ‘s meet the
complementary slackness conditions
© 2011 D. Kirschen & University of Washington
40
Example: Second Trial (3)
m1 = m3 = m4 = m6 = 0
¶
º 8 + 0.2P1 - l - m2 = 0
¶P1
Þ m2 = 9.8 $/MWh
¶
º 7 + 0.12P2 - l = 0
¶P2
Þ l = 44.2 $/MWh
¶
º 9 + 0.14P3 - l + m5 = 0 Þ m5 = -1.2 $/MWh
¶P3
£ 0!
© 2011 D. Kirschen & University of Washington
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Example: Second Trial (4)
The constraint
P3 £ P
max
3
should not be binding!
What is happening?
First trial:
P1
P1
P
max
1
P
min
1
P2
P2max
P2min
P3
P
min
3
© 2011 D. Kirschen & University of Washington
P
max
3
P2
P3
42
Example: Second Trial (5)
In the second trial, we “pushed” P1 towards its lower limit and P3
towards its upper limit. However, there was no need to push P3
because increasing P1 would naturally reduce P3 below its limit.
Second trial:
P1
P1
P
max
1
P
min
1
P2
P2max
P2min
P3
P
min
3
© 2011 D. Kirschen & University of Washington
P
max
3
P2
P3
43
Example: Third Trial (1)
P1 = P1min = 230 MW Þ m2 ¹ 0
m1 = m3 = m4 = m5 = m6 = 0
¶
º 7 + 0.12P2 - l = 0
¶P2
¶
º 9 + 0.14P3 - l = 0
¶P3
¶
º L - P1 - P2 - P3 = 0
¶l
© 2011 D. Kirschen & University of Washington
l = 44.75 $/MWh
P2 = 315 MW
P3 = 255 MW
44
Example: Third Trial (2)
¶
º 8 + 0.2P1 - l - m2 = 0 Þ m2 = 9.25 $/MWh
¶P1
Solution that satisfies all the optimality conditions:
P1 = 230 MW
P2 = 315 MW
P3 = 255 MW
l = 44.75 $/MWh
 Cost of an extra MW of load
m2 = 9.25 $/MWh
 Value of reducing P1min by 1MW
© 2011 D. Kirschen & University of Washington
45
Example: Optimal Solution
Third trial:
P1
P1
P
max
1
P
min
1
P2
P2max
P2min
P3
P
min
3
© 2011 D. Kirschen & University of Washington
P
max
3
P2
P3
46
Practical Economic Dispatch
Equal Incremental Cost Dispatch
dC A
dC B
dC C
dPA
dPB
dPC
l
PA
PB
PC
+
PA + PB + PC
© 2011 D. Kirschen and University of Washington
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Lambda search algorithm
1. Choose a starting value for l
2. Calculate PA ,PB ,PC such that
dC A (PA ) dC B (PB ) dCC (PC )
=
=
=l
dPA
dPB
dPC
3. If one of these values exceeds its lower or upper limit, fix it at that limit
4. Calculate PTOTAL = PA + PB + PC
5. If PTOTAL < L, then increase l
Else If PTOTAL > L, then decrease l
Else If PTOTAL » L, then exit
6. Go To Step 2
© 2011 D. Kirschen and University of Washington
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Linear Cost Curves
CA
CB
PA
PB
PAMAX
PBMAX
dC A
dC B
dPA
dPB
l
PA
PAMAX
© 2011 D. Kirschen and University of Washington
PB
PBMAX
50
Linear Cost Curves
CA
CB
PA
PB
PAMAX
PBMAX
dC A
dC B
dPA
dPB
l
PA
PAMAX
© 2011 D. Kirschen and University of Washington
PB
PBMAX
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Linear Cost Curves
CA
CB
PA
PB
PAMAX
PBMAX
dC A
dC B
dPA
dPB
l
PA
PAMAX
© 2011 D. Kirschen and University of Washington
PB
PBMAX
52
Linear Cost Curves
• The output of the unit with the lowest
incremental cost (in this case unit A) is
increased first (it is “loaded” first)
• The output of the other units is not increased
until unit A is loaded up to its maximum
• The output of the unit with second lowest
incremental cost is then increased (i.e. the
“second cheapest” unit)
• And so on until the generation meets the load
© 2011 D. Kirschen and University of Washington
53
Linear cost curves with minimum generation
CA
CB
PA
PAMIN
PB
PAMAX
PBMIN
dC A
dC B
dPA
dPB
PBMAX
PA
PAMIN
© 2011 D. Kirschen and University of Washington
PAMAX
PB
PBMIN
PBMAX
54
Linear cost curves with minimum generation
• All generating units are first loaded up to their
minimum generation
• The generating units are then loaded in order
of incremental cost
© 2011 D. Kirschen and University of Washington
55
Piecewise Linear Cost Curves
CA
CB
PA
PB
dC A
dC B
dPA
dPB
PA
© 2011 D. Kirschen and University of Washington
PB
56
Economic Dispatch with Piecewise Linear Cost Curves
CA
CB
PA
PB
dC A
dC B
dPA
dPB
l
PA
© 2011 D. Kirschen and University of Washington
PB
57
Economic Dispatch with Piecewise Linear Cost Curves
CA
CB
PA
PB
dC A
dC B
dPA
dPB
l
PA
© 2011 D. Kirschen and University of Washington
PB
58
Economic Dispatch with Piecewise Linear Cost Curves
CA
CB
PA
PB
dC A
dC B
dPA
dPB
l
PA
© 2011 D. Kirschen and University of Washington
PB
59
Economic Dispatch with Piecewise Linear Cost Curves
CA
CB
PA
PB
dC A
dC B
dPA
dPB
l
PA
© 2011 D. Kirschen and University of Washington
PB
60
Economic Dispatch with Piecewise Linear Cost Curves
CA
CB
PA
PB
dC A
dC B
dPA
dPB
l
PA
© 2011 D. Kirschen and University of Washington
PB
61
Economic Dispatch with Piecewise Linear Cost Curves
• All generators except one are at breakpoints of their cost
curve
• The marginal generator is between breakpoints to balance the
load and generation
• Not well-suited for lambda-search algorithm
• Very fast table lookup algorithm:
– Rank all the segments of the piecewise linear cost
curves in order of incremental cost
– First dispatch all the units at their minimum
generation
– Then go down the table until the generation
matches the load
© 2011 D. Kirschen and University of Washington
62
Example
dC A
dPA
0.7
0.5
0.3
30
80
120 150
PA
dC B
Unit
PSegment
Ptotal Lambda
A&B min
20+30 = 50
50
B
70-20=50
100
0.1
A
80-30=50
150
0.3
A
120-80=40
190
0.5
B
110-70=40
230
0.6
A
150-120=30
260
0.7
B
140-110=30
290
0.8
dPB
If Pload = 210MW
0.8
0.6
0.1
20
70
© 2011 D. Kirschen and University of Washington
110 140
PB
The optimal economic dispatch is:
PA = 120MW
PB = 90 MW
λ = 0.6
63