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Informatik 7
Rechnernetze und
Kommunikationssysteme
Using Optimization in Smart Grid Applications
Dr.-Ing. Abdalkarim Awad
04.11.2015
Why Optimization
Smart Grid allows better coordination between the different entities
Optimization can be used to find the best strategy, size of different components,…
We are not going to focus on the optimization algorithms rather than on how to use them
There are a lot of tools that can be used
We are going to use some tools to solve the problems.
Dr.-Ing. Abdalkarim Awad 1
Several tools
ADMB
CONDOR
Joptimizer
NLOPT
Dr.-Ing. Abdalkarim Awad 2
R
Nlopt
+
3
What is Optimization?
Optimization is the mathematical discipline which is concerned with finding the maxima and minima of functions, possibly subject to constraints.
Dr.-Ing. Abdalkarim Awad
What do we optimize?
A real function of n variables
with or without constraints
),,,( 21 nxxxf
Dr.-Ing. Abdalkarim Awad
Unconstrained optimization
22 2),(min yxyxf
Dr.-Ing. Abdalkarim Awad
Optimization with constraints
2
2),(min
1,52
2),(min
0
2),(min
22
22
22
or
or
yx
yxyxf
yx
yxyxf
x
yxyxf
Dr.-Ing. Abdalkarim Awad
Nloptr Package
We are going to use a package called nloptr to solve non-linear optimization problems.
Dr.-Ing. Abdalkarim Awad 8
Example1
GA GB
Max=120 Mvar
LB=300LA=300
Euro/hr 0.00149PG11.83PG616.9)(PGC
Euro/hr 0.00334PG11.69PG 399.8)(PGC
2
BBBB
2
AAAA
AB
Dr.-Ing. Abdalkarim Awad 9
Example1
Total Hourly Cost :
Bus A Bus B
300.0 MWMW
199.6 MWMW 400.4 MWMW
300.0 MWMW
8459 $/hr
Area Lambda : 13.02
AGC ON AGC ON
Dr.-Ing. Abdalkarim Awad 10
Generator cost curves
PGA+PGB=600
Euro/hr 0.00149PG11.83PG616.9)(PGC
Euro/hr 0.00334PG11.69PG 399.8)(PGC
2
BBBB
2
AAAA
Dr.-Ing. Abdalkarim Awad 11
Generator cost curves
How much each generator should produce to cover the 600MW demand
The Problem can be written as:
MINIMIZE (CA(PGA)+CB(PGB))
Subject to
PGA+PGB=600
PGA>=0
PGB>=0
Dr.-Ing. Abdalkarim Awad 12
Code
Dr.-Ing. Abdalkarim Awad 13
Code
Dr.-Ing. Abdalkarim Awad 14
Run
Dr.-Ing. Abdalkarim Awad 15
Example2
Add constraints to generator 2
0<PGB<250
What is the total costs now?
Dr.-Ing. Abdalkarim Awad 16
Example3
Max=50 MvarZ=0.1j
GA GBLB=300 MW
LA=300 MW
AB
Write R script to find P1 and P2 that minimizes the operation
costs and takes into considerations line limitations
Dr.-Ing. Abdalkarim Awad 17
Including power Flow
2
1
1010
1010
3
3
PB
PA
N
k
kiiki BP1
))((
P1=(B12)(θ1-θ2)
P1=(B12)θ1-B12(θ2)
P2=B21(θ2- θ1)
P2=-B21(θ1)+(B21)(θ2)
Z=0.1j y=-10jB=-10
Dr.-Ing. Abdalkarim Awad 18
2
1
1010
1010
3
3
PB
PA
PB-3=-(10)θ2
PB-3+(10)(θ2)=0
Solve with respect
to bus #1. i.e θ1=0
(Slack bus)
Dr.-Ing. Abdalkarim Awad 19
P12=B12*(θ1-θ2)
|P12|=|10(θ1-θ2)|<0.5
-0.5<10*θ2<0.5
-0.05<θ2<0.05
Example 4
20
P1 P2
BUS2
BUS3
500 MW
Dr.- Ing. Abdalkarim Awad
100MW
L1=50
Generator cost curves
C1(P1) = 400 + 9 x P1 + 0.0015 x (P1)2
100 MW ≤ P1 ≤ 600 MW
C3(P2) = 100 + 8 x P2 + 0.0048 x (P2)2
50 MW ≤ P2 ≤ 400 MW
Minimize (C1(P1)+C3(P2))
Dr.- Ing. Abdalkarim Awad 21
Optimization Problem
Minimize (C1(P1)+C2(P2))
Subject to:
100 MW ≤ P1 ≤ 600 MW
50 MW ≤ P2 ≤ 400 MW
Dr.- Ing. Abdalkarim Awad 22
Solution
9 + 2*0.0015 x (P1)=λ
8+ 2*0.0048 x (P2) =λ
P1+P2=500+50-100=450
P1=263.5
P2=186.5
Minimum Cost = 4634.60
Dr.- Ing. Abdalkarim Awad 23
Do they satisfy all constraints?
24
P1 P2
BUS2
BUS3
BUS1
y12=-10j
y13=-10jy23=-10j
L3=500= 5 pu
P13<200=2 pu
SB=100 MVA
100MW= 1 pu
P13=221.2MW
P12=42.3
P23=178.8
P1=263.5MW 186.5MWL1=50= 0.5 pu
Including power Flow
Dr.- Ing. Abdalkarim Awad 25
3
2
1
32213231
23232121
13121312
3
2
1
BBBB
BBBB
BBBB
P
P
P
N
k
kiiki BP1
))((
P1=(B12)(θ1-θ2)+B13(θ1-θ3
P1=(B12+B13)θ1-B12(θ2)-B13(θ3)
P2=-B21(θ1)+(B21+B23)(θ2)-B23(θ3)
P3=-B31(θ1)-B32(θ2)+(B31+B32)(θ3)
Compare to Y matrix!
Dr.- Ing. Abdalkarim Awad 26
3
2
1
201010
102010
101020
51
5.02
1
P
P
singular P1 P2
BUS2
BUS3
BUS1
y12=-10j
y13=-10jy23=-10j
L3=500= 5 pu
P13<200=2 pu
SB=100 MVA
100MW= 1 pu
L1=50= 0.5 pu
Dr.- Ing. Abdalkarim Awad 27
3
2
2010
1020
4
5.02
P
P13<200 B13(θ1-θ3)<2 (pu) θ3<0.20
(B13=-10)
Solve with respect
to bus #1. i.e θ1=0
(Slack bus)
3
2
1
20100
102010
01020
4
5.02
1
P
P
Optimization Problem
Minimize (C1(P1)+C2(P2)) (in pu)
Subject to:
1 pu ≤ P1 ≤ 6 pu
0.5 pu ≤ P2 ≤ 4 pu
P1+P2=4
Dr.- Ing. Abdalkarim Awad 28
3
2
2010
1020
4
5.02
P
Θ3<0.20
Solution (using R)
Dr.- Ing. Abdalkarim Awad 29
2.0
0
0
3
2
1
P1=200 MW
P2=250 MW
P13=-B13(θ1-θ3)=-10*(0-0.2)=2 pu=200 MW
Costs=4659.95>4634.60 Euro
30
P1 P2
BUS2
BUS3
BUS1-10j
-10j -10j
5
P13<1.7
1
0.5 pu
Optimization Problem
Minimize (C1(P1)+C2(P2)) (in pu)
Subject to:
1 pu ≤ P1 ≤ 6 pu
0.5 pu ≤ P2 ≤ 4 pu
Dr.- Ing. Abdalkarim Awad 31
P13<170 B13(θ1-θ3)<1.70 θ3<0.17
3
2
2010
1020
4
2
P
Solution (using R)
The total generation cost becomes: 4783.03> 4634.60
Here, we had to sacrifice “cost” for “implementation”.
Dr.- Ing. Abdalkarim Awad 32
P1=110
P2=340
θ2=-0.06
θ3=0.17