ECE 476
POWER SYSTEM ANALYSIS
Lecture 17
Economic Dispatch and
Optimal Power Flow
Professor Tom Overbye
Department of Electrical and
Computer Engineering
Announcements


Homework 8 is 11.19, 11.21, 11.26, 11.27, due on
Thursday
Homework 9 is 7.1, 7.17, 7.20, 7.24, 7.27
–


you do not need to turn it in, but should do it before the
exam
Second exam is Tuesday Nov 13 in class
Design Project 2 from the book (page 345 to 348)
was due on Nov 15, but I have given you an
extension to Nov 29. The Nov 29 date is firm!
1
Power System Economic Operation


Power system loads are cyclical. Therefore the
installed generation capacity is usually much greater
than the current load. This allows options on how
to meet the current load
Generation costs can vary widely, with different
technologies balancing
–
–
–
the capital costs necessary to build the generator
the costs to actually produce electric power
for example, nuclear and some hydro have high capital
costs and low operating costs. Natural gas generators
have low capital costs, and higher operating costs
2
Thermal versus Hydro Generation


The two main types of generating units are thermal
and hydro
For hydro the fuel (water) is free but there may be
many constraints on operation
–
–
–


fixed amounts of water available
reservoir levels must be managed and coordinated
downstream flow rates for fish and navigation
Hydro optimization is typically longer term (many
months or years)
In 476 we will concentrate on thermal units, looking
at short-term optimization
3
Generator types

Traditionally utilities have had three broad groups
of generators
–
–
–
baseload units: large coal/nuclear; always on at max.
midload units: smaller coal that cycle on/off daily
peaker units: combustion turbines used only for several
hours during periods of high demand
4
Generator Cost Curves

Generator costs are typically represented by up to
four different curves
–
–
–
–

input/output (I/O) curve
fuel-cost curve
heat-rate curve
incremental cost curve
For reference
-
1 Btu (British thermal unit) = 1054 J
1 MBtu = 1x106 Btu
1 MBtu = 0.29 MWh
5
Heat-rate Curve
Plots the average number of MBtu/hr of fuel input
needed per MW of output.
Heat-rate curve is the I/O curve scaled by MW
Best for most efficient units are
around 9.0
6
Incremental (Marginal) cost Curve


Plots the incremental $/MWh as a function of MW.
Found by differentiating the cost curve
7
Economic Dispatch: Formulation

The goal of economic dispatch is to determine the
generation dispatch that minimizes the
instantaneous operating cost, subject to the
constraint that total generation = total load + losses
m
Minimize
CT
 Ci ( PGi )
i 1
Such that
m
 PGi  PD  PLosses
i=1
Initially we'll
ignore generator
limits and the
losses
8
Unconstrained Minimization



This is a minimization problem with a single
inequality constraint
For an unconstrained minimization a necessary (but
not sufficient) condition for a minimum is the
gradient of the function must be zero, f (x)  0
The gradient generalizes the first derivative for
multi-variable problems:
f (x)
 f (x) f (x)
f (x) 
 x , x , , x 
 1
2
n 
9
Minimization with Equality Constraint


When the minimization is constrained with an
equality constraint we can solve the problem using
the method of Lagrange Multipliers
Key idea is to modify a constrained minimization
problem to be an unconstrained problem
That is, for the general problem
minimize f (x) s.t. g(x)  0
We define the Lagrangian L(x,λ )  f (x)  λ T g(x)
Then a necessary condition for a minimum is the
L x (x,λ )  0
and
L λ (x,λ )  0
10
Economic Dispatch Lagrangian
For the economic dispatch we have a minimization
constrained with a single equality constraint
L(PG ,  ) 
m
m
 Ci ( PGi )   ( PD   PGi )
i 1
(no losses)
i 1
The necessary conditions for a minimum are
L(PG ,  )
dCi ( PGi )

   0 (for i  1 to m)
PGi
dPGi
m
PD   PGi  0
i 1
11
Minimization with Equality Constraint


When the minimization is constrained with an
equality constraint we can solve the problem using
the method of Lagrange Multipliers
Key idea is to modify a constrained minimization
problem to be an unconstrained problem
That is, for the general problem
minimize f (x) s.t. g(x)  0
We define the Lagrangian L(x,λ )  f (x)  λ T g(x)
Then a necessary condition for a minimum is the
L x (x,λ )  0
and
L λ (x,λ )  0
12
Economic Dispatch Lagrangian
For the economic dispatch we have a minimization
constrained with a single equality constraint
L(PG ,  ) 
m
m
 Ci ( PGi )   ( PD   PGi )
i 1
(no losses)
i 1
The necessary conditions for a minimum are
L(PG ,  )
dCi ( PGi )

   0 (for i  1 to m)
PGi
dPGi
m
PD   PGi  0
i 1
13
Economic Dispatch Example
What is economic dispatch for a two generator
system PD  PG1  PG 2  500 MW and
C1 ( PG1 )  1000 20 PG1  0.01PG21
$ / hr
C2 ( PG 2 )  400 15 PG 2  0.03PG22
$ / hr
Using the Largrange multiplier method we know
dC1 ( PG1 )

dPG1
 20  0.02 PG1  
0
dC2 ( PG 2 )

dPG 2
 15  0.06 PG 2  
0
500  PG1  PG 2  0
14
Economic Dispatch Example, cont’d
We therefore need to solve three linear equations
20  0.02 PG1  
0
15  0.06 PG 2  
0
500  PG1  PG 2  0
0
1  PG1   20 
0.02
 0
0.06 1  PG 2    15 


 

1 0      500 
 1
 PG1 
 312.5 MW 
 P    187.5 MW 
 G2 


  
 26.2 $/MWh 
15
Lambda-Iteration Solution Method


The direct solution only works well if the
incremental cost curves are linear and no generators
are at their limits
A more general method is known as the lambdaiteration
–
–
the method requires that there be a unique mapping
between a value of lambda and each generator’s MW
output
the method then starts with values of lambda below and
above the optimal value, and then iteratively brackets the
optimal value
16
Lambda-Iteration Algorithm
Pick  L and  H such that
m
m
L
P
(

 Gi )  PD  0
H
P
(

 Gi )  PD  0
i=1
i=1
While  H   L   Do
 M  ( H   L ) / 2
m
If
M
H
M
P
(

)

P

0
Then



 Gi
D
i=1
Else  L   M
End While
17
Lambda-Iteration: Graphical View
In the graph shown below for each value of lambda
there is a unique PGi for each generator. This
relationship is the PGi() function.
18
Lambda-Iteration Example
Consider a three generator system with
IC1 ( PG1 )  15  0.02 PG1
  $/MWh
IC2 ( PG 2 )  20  0.01PG 2

$/MWh
IC3 ( PG 3 )  18  0.025 PG 3

$/MWh
and with constraint PG1  PG 2  PG 3  1000 MW
Rewriting as a function of  , PGi ( ), we have
PG1 ( ) 
  15
0.02
  18
PG3 ( ) 
0.025
PG2 ( ) 
  20
0.01
19
Lambda-Iteration Example, cont’d
Pick  L so
m
L
P
(

 Gi )  1000  0 and
i=1
m
H
P
(

 Gi )  1000  0
i=1
Try  L  20 then
m
 PGi (20)  1000

i 1
  15   20   18
0.02

0.01
Try  H  30 then

0.025
m
 1000  670 MW
 PGi (30)  1000
 1230 MW
i 1
20
Lambda-Iteration Example, cont’d
Pick convergence tolerance   0.05 $/MWh
Then iterate since  H   L  0.05
 M  ( H   L ) / 2  25
m
Then since
H
P
(25)

1000

280
we
set

 25
 Gi
i 1
Since 25  20  0.05
 M  (25  20) / 2  22.5
m
L
P
(22.5)

1000


195
we
set

 22.5
 Gi
i 1
21
Lambda-Iteration Example, cont’d
Continue iterating until  H   L  0.05
The solution value of  ,  , is 23.53 $/MWh
*
Once  * is known we can calculate the PGi
23.53  15
PG1 (23.5) 
 426 MW
0.02
23.53  20
PG2 (23.5) 
 353 MW
0.01
23.53  18
PG3 (23.5) 
 221 MW
0.025
22
Lambda-Iteration Solution Method


The direct solution only works well if the
incremental cost curves are linear and no generators
are at their limits
A more general method is known as the lambdaiteration
–
–
the method requires that there be a unique mapping
between a value of lambda and each generator’s MW
output
the method then starts with values of lambda below and
above the optimal value, and then iteratively brackets the
optimal value
23
Generator MW Limits



Generators have limits on the minimum and
maximum amount of power they can produce
Often times the minimum limit is not zero. This
represents a limit on the generator’s operation with
the desired fuel type
Because of varying system economics usually many
generators in a system are operated at their
maximum MW limits.
24
Lambda-Iteration with Gen Limits
In the lambda-iteration method the limits are taken
into account when calculating PGi ( ) :
if PGi ( )  PGi ,max then PGi ( )  PGi ,max
if PGi ( )  PGi ,min then PGi ( )  PGi ,min
25
Lambda-Iteration Gen Limit Example
In the previous three generator example assume
the same cost characteristics but also with limits
0  PG1  300 MW
100  PG2  500 MW
200  PG3  600 MW
With limits we get
m
 PGi (20)  1000
i 1
 PG1 (20)  PG 2 (20)  PG 3 (20)  1000
 250  100  200  450 MW (compared to -670MW)
m
 PGi (30)  1000
i 1
 300  500  480  1000  280 MW
26
Lambda-Iteration Limit Example,cont’d
Again we continue iterating until the convergence
condition is satisfied. With limits the final solution
of  , is 24.43 $/MWh (compared to 23.53 $/MWh
without limits). The presence of limits will always
cause  to either increase or remain the same.
Final solution is
PG1 (24.43)  300 MW
PG2 (24.43)  443 MW
PG3 (24.43)  257 MW
27
Lambda-Iteration Limit Example,cont’d
Again we continue iterating until the convergence
condition is satisfied. With limits the final solution
of  , is 24.43 $/MWh (compared to 23.53 $/MWh
without limits). The presence of limits will always
cause  to either increase or remain the same.
Final solution is
PG1 (24.43)  300 MW
PG2 (24.43)  443 MW
PG3 (24.43)  257 MW
28
Lambda-Iteration Limit Example,cont’d
Again we continue iterating until the convergence
condition is satisfied. With limits the final solution
of  , is 24.43 $/MWh (compared to 23.53 $/MWh
without limits). The presence of limits will always
cause  to either increase or remain the same.
Final solution is
PG1 (24.43)  300 MW
PG2 (24.43)  443 MW
PG3 (24.43)  257 MW
29
Thirty Bus ED Example
Case is economically dispatched without considering
the incremental impact of the system losses
30
Back of Envelope Values

Often times incremental costs can be approximated
by a constant value:
–
–
–
–
$/MWhr = fuelcost * heatrate + variable O&M
Typical heatrate for a coal plant is 10, modern
combustion turbine is 10, combined cycle plant is 7 to 8,
older combustion turbine 15.
Fuel costs ($/MBtu) are about 1 to 1.5 for coal, 12 for
natural gas, 0.5 for nuclear, probably 13 to 15 for fuel oil.
Hydro costs tend to be quite low, but are fuel (water)
constrained
31
Aside: Cost of Electricity Generation
All values are in 2003 dollars. Nuclear costs do not include
decommissioning costs, which are < 0.1 cents/kWh
Source: California Energy Commission:
http://www.energy.ca.gov/electricity/comparative_costs-v1.html
32
Natural Gas Prices Over the Years
(adjusted for inflation)
Source: US FERC,
http://www.ferc.gov/market-oversight/mkt-gas/overview/2007/ngas-ovr-hh-pr.pdf
33
Professor Chapman’s Solar House

Our own Professor Chapman recently installed a
2870 W solar system for his new house
–
Details at http://www.patrickchapman.com/solar.htm
34
Inclusion of Transmission Losses



The losses on the transmission system are a function
of the generation dispatch. In general, using
generators closer to the load results in lower losses
This impact on losses should be included when
doing the economic dispatch
Losses can be included by slightly rewriting the
Lagrangian:
L(PG ,  ) 
m
m
i 1
i 1
 Ci ( PGi )   ( PD  PL ( PG )   PGi )
35
Impact of Transmission Losses
This small change then impacts the necessary
conditions for an optimal economic dispatch
L(PG ,  ) 
m
m
i 1
i 1
 Ci ( PGi )   ( PD  PL ( PG )   PGi )
The necessary conditions for a minimum are now
L(PG ,  )
PGi
dCi ( PGi )
PL ( PG )

  (1 
)0
dPGi
PGi
m
PD  PL ( PG )   PGi  0
i 1
36
Impact of Transmission Losses
Solving each equation for  we get
dCi ( PGi )
PL ( PG )
  (1 
0
dPGi
PGi
dCi ( PGi )
1
 
 PL ( PG )  dPGi
1  P


Gi 
Define the penalty factor Li for the i th generator
1
Li 
 PL ( PG ) 
1  P


Gi 
The penalty factor
at the slack bus is
always unity!
37
Impact of Transmission Losses
The condition for optimal dispatch with losses is then
L1IC1 ( PG1 )  L2 IC2 ( PG 2 )  Lm ICm ( PGm )  
1
Since Li 
if increasing PGi increases
 PL ( PG ) 
1  P


Gi 
PL ( PG )
the losses then
 0  Li  1.0
PGi
This makes generator i appear to be more expensive
(i.e., it is penalized). Likewise Li  1.0 makes a generator
appear less expensive.
38
Calculation of Penalty Factors
Unfortunately, the analytic calculation of Li is
somewhat involved. The problem is a small change
in the generation at PGi impacts the flows and hence
the losses throughout the entire system. However,
using a power flow you can approximate this function
by making a small change to PGi and then seeing how
the losses change:
PL ( PG ) PL ( PG )

PGi
PGi
1
Li 
PL ( PG )
1
PGi
39
Two Bus Penalty Factor Example
PL ( PG )
 0.0387
PG 2
L2  0.9627
PL ( PG ) 0.37 MW

 0.037
PGi
10MW
L2  0.9643
40
Thirty Bus ED Example
Because of the penalty factors the generator incremental
costs are no longer identical.
41
Area Supply Curve
The area supply curve shows the cost to produce the
next MW of electricity, assuming area is economically
dispatched
10.00
7.50
Supply
curve for
thirty bus
system
5.00
2.50
0.00
0
100
200
Total Area Generation (MW)
300
400
42
Eastern US Supply Curve
100.0
The y-axis units are $/MWh
75.0
50.0
25.0
0.0
0
150000
300000
Total Area Generation (MW)
450000
600000
43
Economic Dispatch - Summary


Economic dispatch determines the best way to
minimize the current generator operating costs
The lambda-iteration method is a good approach for
solving the economic dispatch problem
–
–


generator limits are easily handled
penalty factors are used to consider the impact of losses
Economic dispatch is not concerned with
determining which units to turn on/off (this is the
unit commitment problem)
Economic dispatch ignores the transmission system
limitations
44