EE 553
Security-constrained unit
commitment (SCUC)
J. McCalley
1
Unit Commitment (UC)
The problem of unit commitment (UC) is to decide
which units to interconnect over the next T hours,
where T is commonly as few as 2 but more
commonly it is 24 or 48 hours, or even 1 week.
In the day-ahead market, it is always 24 hours.
In reliability-assessment-commitment (RAC), it can be &often is less.
The problem becomes security constrained (SCUC)
when constraints are imposed to ensure line flows
do not exceed chosen limits following a contingency.
2
SCUC Objective Function (no demand bidding)
min
 z
Fit   g itCit   yit Sit   xit H it
t
i
t
t
t


 
i 
i 
i
it
Fixed (no -load) Costs
Production Costs
Startup Costs
Shutdown Costs
• Decision variables are zit, git, yit, xit
• zit, yit, xit are discrete, git is continuous
git is the MW produced by generator i in period t,
zit is 1 if generator i is dispatched during t, 0 otherwise,
yit is 1 if generator i starts at beginning of period t, 0 otherwise,
xit is 1 if generator i shuts at beginning of period t, 0 otherwise,
Fit is no-load cost ($/period) of operating generator i in period t,
Cit is prod. cost ($/MW/period) of operating gen i in period t;
Sit is startup cost ($) of starting gen i in period t.
Hit is shutdown cost ($) of shutting gen i in period t.
3
SCUC Problem (no demand bidding)
 z
Fit   g itCit   yit Sit   xit H it
t
i
t
t
t


 
i 
i 
i
min
it
Fixed (no -load) Costs
Production Costs
Subject to
power balance
 git  Dt   dit
i
reserve
Startup Costs
 rit  SDt
Shutdown Costs
 t,
(2) Power balance at each period t.
 t,
(3)
i
i
min generation
max generation
max spinning reserve
ramp rate pos limit
ramp rate neg limit
start if off-then-on
shut if on-then-off
normal line flow limit
git  zit MINi
 i, t ,
git  rit  zit MAX i
 i, t ,
rit  zit MAXSPi
 i, t ,
git  git 1  MxInci
 i, t ,
git  git 1  MxDeci
 i, t ,
zit  zit 1  yit
 i, t ,
zit  zit 1  xit
 i, t ,
 aki ( git  dit )  MxFlowk
 k , t,
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
 k , j, t ,
(12)
Max gen,
min gen,
reserves.
i
security line flow limits
 aki( j ) ( git  dit )  MxFlowk( j )
i
Dt is the total demand in period t,
SDt is the spinning reserve required in period t,
MxInci is max ramprate (MW/period) for increasing
gen i output
MxDeci is max ramprate (MW/period) for
decreasing gen i output
aij is linearized coefficient relating bus i injection to
line k flow
MxFlowk is the maximum MW flow on line k
aki( j )is linearized coefficient relating bus
i
injection to line k flow under contingency j,
MxFlowk( j )
is the maximum MW flow on line k
under contingency j
MAXSPi is maximum spinning reserve for4 unit i
Two Categories of Reserves
• “Regulating”: To handle the moment-to-moment
variation in load (or net load where variable
generation has significant presence).
• “Contingency”: To compensate for unexpected
imbalances usually caused by a gen trip.
Contingency reserves must be available within 10
mins following a request. There are 2 types of
contingency reserves:
 Spinning reserve: reserve from units that are connected
 Supplementary reserve: reserve from units not connected
The markets separate regulating (up and down) from contingency
reserves, but it makes the problem more complicated. In this initial
view of the problems, we represent only contingency reserves.
5
Contingency Reserves
min generation
max generation
max spinning reserve
git  zit MINi
 i, t ,
git  rit  zit MAX i
 i, t ,
rit  zit MAXSPi
 i, t ,
(4)
(5)
(6)
Note last constraint: MAXSPi.
The amount of reserve a unit can offer is not unconstrained. If no
value of MAXSPi is entered, then the software should default to
MAXSPi=MAXi-MINi
6
SCUC Problem (no demand bidding)
 z
Fit   g itCit   yit Sit   xit H it
t
i
t
t
t


 
i 
i 
i
min
it
Fixed (no -load) Costs
Production Costs
Subject to
power balance
 git  Dt   dit
i
reserve
Startup Costs
 rit  SDt
Shutdown Costs
 t,
(2)
 t,
(3)
i
i
min generation
max generation
max spinning reserve
ramp rate pos limit
ramp rate neg limit
start if off-then-on
shut if on-then-off
normal line flow limit
git  zit MINi
 i, t ,
git  rit  zit MAX i
 i, t ,
rit  zit MAXSPi
 i, t ,
git  git 1  MxInci
 i, t ,
git  git 1  MxDeci
 i, t ,
zit  zit 1  yit
 i, t ,
zit  zit 1  xit
 i, t ,
 aki ( git  dit )  MxFlowk
 k , t,
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
 k , j, t ,
(12)
Max increase
and max
decrease.
This reflects
ramp rates.
i
security line flow limits
 aki( j ) ( git  dit )  MxFlowk( j )
i
Dt is the total demand in period t,
SDt is the spinning reserve required in period t,
MxInci is max ramprate (MW/period) for increasing
gen i output
MxDeci is max ramprate (MW/period) for
decreasing gen i output
aij is linearized coefficient relating bus i injection to
line k flow
MxFlowk is the maximum MW flow on line k
aki( j )is linearized coefficient relating bus
i
injection to line k flow under contingency j,
MxFlowk( j )
is the maximum MW flow on line k
under contingency j
MAXSPi is maximum spinning reserve for7 unit i
Max Inc and Max Dec (Ramp Rates)
ramp rate pos limit
ramp rate neg limit
git  git 1  MxInci
 i, t ,
git  git 1  MxDeci
 i, t ,
(7)
(8)
%/min
MxInci=RampRateUpi*ΔT
Coal
1-5
Nuclear 1-5
MW/Min
Min
MxDeci=RampRateDowni*ΔT
NGCC
5-10
CT
20
Diesel
40
A unit may be able to ramp down faster than it can ramp up.
Wind is an extreme case (it may not be able to ramp up at all!)
So RampRateUpi and RampRateDowni may differ.
ΔT is amount of time from one period t to the next t+1
8
Max Inc and Max Dec (Ramp Rates)
ramp rate pos limit
ramp rate neg limit
git  git 1  MxInci
 i, t ,
git  git 1  MxDeci
 i, t ,
(7)
(8)
%/min
MxInci=RampRateUpi*ΔT
Coal
1-5
Nuclear 1-5
MW/Min
Min
MxDeci=RampRateDowni*ΔT
NGCC
5-10
CT
20
Diesel
40
Note that ramp rate constraints (RRC) enforce the ramping capability of each
given unit. They are not ramp rate requirements (RRR), which would place a
requirement on the amount of reserve at any given time to be comprised of
units having a certain level of ramping capability. RRC (which depend on 2
consecutive time periods) require, under a given load forecast, the solution
remains feasible. RRR (which depend on only one time period) attempt to
ensure a feasible solution will exist. We are only considering RRC here. We
are NOT considering RRR here. The following paper addresses this further:
V. Krishnan, E. Ibanez, T. Das, Y. Gu, and J. McCalley, “Modeling Operational Effects of Variable Generation within National
9
Long-term Infrastructure Planning Software,” to appear in IEEE Transactions on Power Systems.
SCUC Problem (no demand bidding)
 z
Fit   g itCit   yit Sit   xit H it
t
i
t
t
t


 
i 
i 
i
min
it
Fixed (no -load) Costs
Production Costs
Subject to
power balance
 git  Dt   dit
i
reserve
Startup Costs
 rit  SDt
Shutdown Costs
 t,
(2)
 t,
(3)
i
i
min generation
max generation
max spinning reserve
ramp rate pos limit
ramp rate neg limit
start if off-then-on
shut if on-then-off
normal line flow limit
git  zit MINi
 i, t ,
git  rit  zit MAX i
 i, t ,
rit  zit MAXSPi
 i, t ,
git  git 1  MxInci
 i, t ,
git  git 1  MxDeci
 i, t ,
zit  zit 1  yit
 i, t ,
zit  zit 1  xit
 i, t ,
 aki ( git  dit )  MxFlowk
 k , t,
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
 k , j, t ,
(12)
Start
constraint
Shut
constraint
i
security line flow limits
 aki( j ) ( git  dit )  MxFlowk( j )
i
Dt is the total demand in period t,
SDt is the spinning reserve required in period t,
MxInci is max ramprate (MW/period) for increasing
gen i output
MxDeci is max ramprate (MW/period) for
decreasing gen i output
aij is linearized coefficient relating bus i injection to
line k flow
MxFlowk is the maximum MW flow on line k
aki( j )is linearized coefficient relating bus
i
injection to line k flow under contingency j,
MxFlowk( j )
is the maximum MW flow on line k
under contingency j
10 unit i
MAXSPi is maximum spinning reserve for
Start and Shut Constraints
start if off-then-on
shut if on-then-off
zit  zit 1  yit
 i, t ,
zit  zit 1  xit
 i, t ,
(9)
(10)
Starting constraints. For example, z12≤z11+y12,
or more generally, zkt≤zk,t-1+ykt, which says
Status of unit k in time t ≤status of unit k in time t-1 + start flag in time t
Constraints associated with shutting. For example, z12≥z11-x12,
or more generally, zkt≥zk,t-1-xkt, which says
Status of unit k in time t ≥status of unit k in time t-1 - shut flag in time t
These constraints are very important. You can understand them
better by performing the following exercise for unit 1:
• List all possible combinations of z11, z12, y12, x12
• For each combination, compute z11+y12 and z11-x12
• Eliminate combinations not satisfying above constraints
• Of the “feasible” combinations, identify which ones make no
economic sense, i.e., which ones will give you nothing but cost
you money (e.g., “on, start, on” or “off, shut, off”). Because we
are minimizing costs, these will never be chosen!
11
SCUC Problem (no demand bidding)
 z
Fit   g itCit   yit Sit   xit H it
t
i
t
t
t


 
i 
i 
i
min
it
Fixed (no -load) Costs
Production Costs
Subject to
power balance
 git  Dt   dit
i
reserve
Startup Costs
 rit  SDt
Shutdown Costs
 t,
(2)
 t,
(3)
i
i
min generation
max generation
max spinning reserve
ramp rate pos limit
ramp rate neg limit
start if off-then-on
shut if on-then-off
normal line flow limit
git  zit MINi
 i, t ,
git  rit  zit MAX i
 i, t ,
rit  zit MAXSPi
 i, t ,
git  git 1  MxInci
 i, t ,
git  git 1  MxDeci
 i, t ,
zit  zit 1  yit
 i, t ,
zit  zit 1  xit
 i, t ,
 aki ( git  dit )  MxFlowk
i
security line flow limits
 aki( j ) ( git  dit )  MxFlowk( j )
i
Dt is the total demand in period t,
SDt is the spinning reserve required in period t,
MxInci is max ramprate (MW/period) for increasing
gen i output
MxDeci is max ramprate (MW/period) for
decreasing gen i output
aij is linearized coefficient relating bus i injection to
line k flow
 k , t,
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
Transmission
normal
constraint
Transmisison
security
 k , j, t ,
(12)
constraint,
MxFlowk is the maximum MW flow on line k
aki( j )is linearized coefficient relating bus
i
injection to line k flow under contingency j,
MxFlowk( j )
is the maximum MW flow on line k
under contingency j
12 unit i
MAXSPi is maximum spinning reserve for
Transmission normal, security constraints
normal line flow limit
 aki ( git  dit )  MxFlowk
 k , t,
(11)
 k , j, t ,
(12)
i
security line flow limits
 aki( j ) ( git  dit )  MxFlowk( j )
i
Note: Should have absolute value signs around the both
summations on the left to account for bi-directional flow.
The addition of eq. (11) alone provides that this problem is a
transmission-constrained unit commitment problem.
The addition of eqs. (11) and (12) together provides that this
problem is a security-constrained unit commitment problem.
aki is linearized coefficient relating bus i injection to line k flow
aki( j )
is linearized coefficient relating bus i injection to line k flow under
contingency j,
We have learned how to compute the aij (which are just
generation shift factors). The aki(j) are developed on the next slide.13
Modeling transmission security constraints
al,i: gen shift factor for cct l, given a change in injection at bus i, when all ccts are in service.
dl,k line outage distribution factor for cct l, given outage of circuit k.
Pi: real power injection at bus i.
fk: flow on circuit k.
Consider taking the following actions in the network:
(1) Increase generation at all buses i=1,…,N from 0 to Pi.
(2) Remove circuit k.
Derive total flow on cct , fl, in terms of parameters defined above, following actions (1), (2).
N
f l(1)   ali Pi
i 1
N
f l(2)  d lk f k(1)  d lk  aki Pi
f
(2)
l
 f l  f l
(1)
( 2)
i 1
al(ik )  ali  dlk aki
Effective generation
  ali Pi  d lk  aki Pi   ali  d lk aki Pi shift factor: linearized
coefficient relating bus
i 1
i 1
i 1
i injection to line l flow
under contingency k.
N
N
N
14
Key Concept
 z
Fit   g itCit   yit Sit   xit H it
t
i
t
t
t


 
i 
i 
i
min
it
Fixed (no -load) Costs
Production Costs
Subject to
power balance
 git  Dt   dit
i
reserve
Startup Costs
 rit  SDt
Problem is a
function of t.
Shutdown Costs
 t,
(2)
 t,
(3)
i
i
min generation
max generation
max spinning reserve
ramp rate pos limit
ramp rate neg limit
start if off-then-on
shut if on-then-off
normal line flow limit
git  zit MINi
 i, t ,
git  rit  zit MAX i
 i, t ,
rit  zit MAXSPi
 i, t ,
git  git 1  MxInci
 i, t ,
git  git 1  MxDeci
 i, t ,
zit  zit 1  yit
 i, t ,
zit  zit 1  xit
 i, t ,
 aki ( git  dit )  MxFlowk
 k , t,
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
 k , j, t ,
(12)
These are intertemporal
constraints!
i
security line flow limits
 aki( j ) ( git  dit )  MxFlowk( j )
i
In addition to being MIP, the UC problem has two important features.
Dynamic: It obtains decisions for a sequence of time periods.
Inter-temporal constraints: What happens in one time period affects
what happens in another time period. So we may not solve each time
period independent of solutions in other time periods.
15
Branch & Bound is Best 
Note the financial implications of changing to MIP.
Source: M. Rothleder, presentation to the Harvard Energy Policy Group, Dec 7, 2007,
available at http://www.hks.harvard.edu/hepg/Rothleder_Mark.pdf.
16
Some good industry quotes in notes
“The most popular algorithms for the solutions of the unit
commitment problems are Priority-List schemes [4], Dynamic
Programming [5], and Mixed Integer Linear Programming [6]. Among
these approaches the MILP technique has achieved significant
progress in the recent years [7]. The MILP methodology has been
applied to the SCUC formulation to solve this MOW problem. Recent
developments in the implementation of MILP-based algorithms and
careful attention to the specific problem formulation have made it
possible to meet accuracy and performance requirements for solving
such large scale problems in a practical competitive energy market
environment.”
Q. Zhou, D. Lamb, R. Frowd, E. Ledesma, A. Papalexopoulos, “Minimizing Market
Operation Costs Using A Security-Constrained Unit Commitment Approach,” 2005 IEEE/PES
Transmission and Distribution Conference & Exhibition: Asia and Pacific Dalian, China.
17
Some good industry quotes in notes
“The LR algorithm was adequate for the original market size, but as
the market size increased, PJM desired an approach that had more
flexibility in modeling transmission constraints. In addition, PJM has
seen an increasing need to model Combined-cycle plant operation
more accurately. While these enhancements present a challenge to
the LR formulation, the use of a MIP formulation provides much
more flexibility. For these reasons, PJM began discussion with its
software vendors, in late 2002, concerning the need to develop a
production grade MIP-based approach for large-scale unit
commitment problems….”
D. Streiffert, R. Philbrick, and A. Ott, “A Mixed Integer Programming Solution for Market
Clearing and Reliability Analysis,” Power Engineering Society General Meeting, 2005. IEEE
12-16 June 2005 , pp. 2724 - 2731 Vol. 3..
18
Some good industry quotes in notes
“The Unit Commitment problem is a large-scale non-linear mixed integer programming
problem. Integer variables are required for modeling: 1) Generator hourly On/Off-line
status, 2) generator Startups/Shutdowns, 3) conditional startup costs (hot, intermediate &
cold). Due to the large number of integer variables in this problem, it has long been viewed
as an intractable optimization problem. Most existing solution methods make use of
simplifying assumptions to reduce the dimensionality of the problem and the number of
combinations that need to be evaluated. Examples include priority-based methods,
decomposition schemes (LR) and stochastic (genetic) methods. While many of these
schemes have worked well in the past, there is an increasing need to solve larger (RTO-size)
problems with more complex (e.g. security) constraints, to a greater degree of accuracy.
Over the last several years, the number of units being scheduled by RTOs has increased
dramatically. PJM started with about 500 units a few years ago, and is now clearing over
1100 each day. MISO cases will be larger still….”
“The classical MIP implementation utilizes a Branch and Bound scheme. This method
attempts to perform an implicit enumeration of all combinations of integer variables to
locate the optimal solution. In theory, the MIP is the only method that can make this claim.
It can, in fact, solve non-convex problems with multiple local minima. Since the MIP
methods utilize multiple Linear Programming (LP) executions, they have benefited from
recent advances in both computer hardware and software [6]…”
D. Streiffert, R. Philbrick, and A. Ott, “A Mixed Integer Programming Solution for Market Clearing and Reliability
19
Analysis,” Power Engineering Society General Meeting, 2005. IEEE 12-16 June 2005 , pp. 2724 - 2731 Vol. 3..
Illustration – Problem data
g2
g1
1
2
y12 =-j10
y14 =-j10
y13 =-j10
Pd2
y23 =-j10
y34 =-j10
4
3
g4
Pd3
We illustrate using an example that utilizes the
same system we have been using in our previous
notes, where we had 3 generator buses in a 4
bus network supplying load at 2 different buses,
but this time we model each generator with the
ability to submit 3 offers.
We ignore reserve constraints in this illustration.
Three offers per gen: (gk1t,Pk1t), (gk2t, Pk2t), (gk3t, Pk3t)
Ci
$/hr
gk3t
Unit,
k
gk2t
gk1t
Pi1
Pi,min
Pi2
Pi3
1
2
4
Fixed Startup Shutdown
costs Costs Costs ($)
($/hr)
($)
50
100
20
50
100
20
50
100
20
Production Costs ($/pu-hr)
gk1t
gk2t
gk4t
1246
1129
1183
1307
1211
1254
1358
1282
1320
Pi (MW) 
Notice that for each unit, the offers increase with generation, i.e., gk1t<gk2t<gk3t. This
prevents use of a higher generation level before a lower generation level. It also says that
our offer function is convex.
20
Illustration – Problem data
Constraints on the offers
0  g11t  0.50
0  g  0.60
   12t  

0  g13t  0.40
 
  

0  g 21t  0.35
0   g 22t   0.60, t
 
  

0  g 23t  0.20
0  g  0.45
   41t  

0  g 42t  0.50
0  g  0.40
   43t  

21
Illustration – Problem data
Load, Dt (pu)
1.50
1.40
1.30
1.40
1.70
2.00
2.40
2.80
3.20
3.30
3.30
3.20
3.20
3.30
3.35
3.40
3.30
3.30
3.20
2.80
2.30
2.00
1.70
1.60
Load curve
One Day Load variation
400
Load (MW)
Hour, t
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
350
300
250
200
Series1
150
100
50
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Time (hr)
Notice between the hours of
t=20 and t=21 that the load
drops 0.5 pu. We must have
the reserves available to
handle such a drop!
22
Illustration 1: CPLEX code for 4 hours
23
Illustration 1: CPLEX code for 4 hours
24
Illustration 1: CPLEX result
Note that all y- and xvariables are 0, therefore
there is no starting up or
shutting down.
0  g11t  0.50
0  g  0.60
   12t  

0  g13t  0.40
 
  

g
0
0
.
35
   21t  

0   g 22t   0.60, t
 
  

g
0
0
.
20
   23t  

0  g  0.45
   41t  

g
0  42t  0.50
0  g  0.40
   43t  

25
Illustration 1: CPLEX result
Why did we obtain such a simple solution with
unit 1 down, units 2 and 4 up for all four hours?
Most
expensive unit
This is a result of the fact
Unit, Fixed Startup Shutdown
Production Costs ($/pu-hr)
k
costs Costs Costs ($)
g
g
g
that the initial solution of
($/hr)
($)
initialu1: z11=0
1
50
100
20
1246
1307
1358
2
50
100
20
1129
1211
1282
initialu2: z21=1
4
50
100
20
1183
1254
1320
initialu4: z41=1
was the best one for the initial loading condition, and since the
loading condition hardly changed during the first four hours, there
was no reason to change any of the units.
k1t
k4t
One Day Load variation
400
Load (MW)
To test this, let’s try a different
initial condition:
initialu1: z11=1
initialu2: z21=0
initialu4: z41=1
k2t
350
300
250
200
Series1
150
100
50
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Time (hr)
26
Illustration 2: CPLEX result
Previous solution was
7020.70. Why was this
one more expensive?
One Day Load variation
Load (MW)
400
350
300
250
200
Series1
150
100
50
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Time (hr)
Because of
• U1 shut-down cost
• U2 startup cost
• Difference in cost
between running U1
and U2 during hr 1
27
Illustration 2: CPLEX result
Because we initialized the solution with more
expensive units, to get back to the less
expensive solution, the program forces U2 to
start up (y22=1) and U1 to shut down (x12=1) at
the beginning of period 2. The additional cost
of starting U2 ($100) and shutting U1 ($20) was
less than the savings associated with running
the more efficient unit (U2) over the remaining
3 hours of the simulation, and so the program
ordered starting U2 and shutting U1.
Let’s test our understanding by increasing
startup costs of U2 from $100 to $10,000.
The objective function value in this case is
$7281.25 (higher than the last solution).
28
The decision variables are….
Illustration 3: CPLEX result
We observe U1 was on-line the entire
four hours, i.e, there is no switching,
something we expected since the startup cost of U2 was high.
29
Illustration 4: 24 hours
We refrain from showing the data in this case because it is extensive,
having 426 variables:
72 z-variables
69 y-variables
69 x-variables
216 g-variables
The solution is initialized at
initialu1: z11=0
initialu2: z21=1
initialu4: z41=1
which is the most economic solution for the hour 1 loading level.
30
Illustration 4: 24 hours
The output can be analyzed by using
“display solution variables -”
and then either reading the z-variables or reading the y-variables and
x-variables that are listed (and therefore 1). The x and y variables
indicate changes in the unit commitment.
In studying the load curve, what kind of changes do you expect?
One Day Load variation
Load (MW)
400
350
300
250
200
150
100
50
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Time (hr)
The result, objective value=$77667.3, shows that the only x and y variables that are non-zero
are y1,8 and x1,21. This means that the changes in the unit commitment occur only for unit 1
and only at hours 8 and 21. A pictorial representation of the unit commitment through the 24
hour period is shown on the next slide.
31
Illustration 4: 24 hours
One Day Load variation
400
Load (MW)
350
300
250
200
150
100
50
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Time (hr)
Unit 1
1.2
1
Unit 1 starts at hr 8, shuts at hr 21.
Up or down
0.8
0.6
Unit 1
0.4
0.2
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
Hour
Unit 2
1.2
1
Up or down
0.8
0.6
Unit 2
Unit 2 is always up.
0.4
0.2
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
Hour
Unit 3
1.2
1
Up or down
0.8
0.6
Unit 3
Unit 4 is always up.
0.4
0.2
32
0
1
2
3
4
5
6
7
8
9
10
11
12
13
Hour
14
15
16
17
18
19
20
21
22
23
24
Illustration 5: 24 hours
One Day Load variation
400
350
300
350
300
250
200
250
load
Load (MW)
One day load variation
400
150
100
200
150
100
50
0
50
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Time (hr)
Previous load curve
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
hour
New load curve
Observe initial & final load values are lower in the new load curve.
What effect will this have on the UC?
33
Illustration 5: 24 hours
One day load variation
400
350
300
load
250
200
150
100
50
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
hour
Unit 1
1.2
1
0.8
Unit 1 starts at hr 8, shuts at hr 19.
0.6
0.4
0.2
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
Unit 2
1.2
1
0.8
Unit 2 is always up.
0.6
0.4
0.2
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
Unit 3
1.2
1
0.8
Unit 4 is always up, until hour 24.
0.6
0.4
0.2
34
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
Illustration 6: 24 hours
One day load variation
400
350
300
load
250
200
150
100
50
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
hour
New load curve
Use new load curve but reduce startup costs to $10 and shutdown
costs to $2. All other data remains as before.
What effect do you think this will have on the UC?
The result, with objective function value of $66,867.95, shows that
the only x and y variables that are non-0 are
y1,8, y1,12, y4,5, x1,11, x1,20, x4,2, x4,24.
35
Illustration 6: 24 hours
One day load variation
400
350
300
load
250
200
150
100
50
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
hour
Unit 1
1.2
1
0.8
Unit 1 starts at hr 8, shuts at hr 11,
starts at hr 12, shuts at hr 20.
0.6
0.4
0.2
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
Unit 2
1.2
1
0.8
0.6
Unit 2 is always up.
0.4
0.2
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
Unit 3
1.2
1
0.8
0.6
0.4
0.2
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
Unit 4 shuts at hr 2, starts at hr 5,
and shuts at hr 23.
36
An observation
What do low start up and shut down costs do to the UC solution?
They tend to make UC change more.
What do high start up and shut down costs do to the UC solution?
They tend to make UC change less.
Recall:
Inter-temporal constraints: What happens in one time period
affects what happens in another time period. So we may not solve
each time period independent of solutions in other time periods.
Start-up and shut-down costs make the inter-temporal
constraints influential. In our problem, if these costs were zero,
then the solution we obtain would be the same one we would get
37
if we solved each hour independently.
Another observation
Do we obtain the generation dispatch from the UC solution?
Yes, these are the gkit variables.
Why, then, do we need the SCED?
The SCED provides the LMPs, the SCUC does not.
After solving any MIP, ask CPLEX for the dual variables using:
display solution duals –
CPLEX will tell you “Not available for mixed integer problems.”
SCED is solved using LP, which provides dual variables.
SCUC is solved using B&B, which cannot provide dual variables.
38
Co-optimization: what?
Electricity markets: simultaneous clearing of two or more
commodity markets within one optimization problem.
Planning: simultaneous identification of two or more
classes of infrastructure decisions within one optimization
problem (G&T, G&T&natural gas pipeline) .
Cooptimization optimizes two (or more) objectives which depend
on different but related decisions: min f1(x(y))+f2(y(x))
Choose copper sheet transmission (y) then optimize
generation f1(x) then optimize transmission f2(y).
Multiobjective optimization optimizes two (or more) objectives
which depend on the same decisions: min f1(x)+f2(x)
39
How to choose operating condition x to achieve
good risk-cost f1-f2 trade-off of risk?
Co-optimization (SC-SCUC)
Co-optimization, in the context of electricity markets, refers to the
simultaneous clearing of two or more commodity markets within
the same optimization problem.
Most ISOs clear 3 commodity markets within their co-optimization:
- Energy
Operating
- Regulating reserve
reserve
- Contingency reserve
K. Wissman, “Competitive Electricity Markets and the Special
Role of Ancillary Services, slides presented at the Licensing/
Competition and Tariff/Pricing Comm Meeting, Feb 4-5, 2008.
In the following, I only include contingency reserves.
Inclusion of regulating reserves is a little more complicated because we
need both ramp-up and ramp-down capability.
Contingency reserves require only ramp-up capability since there are
40
usually no single loads as large as the largest single generator.
SCUC Problem (no demand bidding)
 z
Fit   g iktCikt   yit Sit   xit H it   rit Rit
t
i
t
k
t
t
t
i


 
i 
 
i 
i 


min
it
Fixed (no -load) Costs
Subject to
power balance
Production Costs
 git  Dt   dit
i
reserve
Startup Costs
 rit  SDt
Shutdown Costs
 t,
(2)
 t,
(3)
Reserve Costs
i
i
min generation
max generation
max spinning reserve
ramp rate pos limit
ramp rate neg limit
start if off-then-on
shut if on-then-off
normal line flow limit
git  zit MINi
 i, t ,
git  rit  zit MAX i
 i, t ,
rit  zit MAXSPi
 i, t ,
git  git 1  MxInci
 i, t ,
git  git 1  MxDeci
 i, t ,
zit  zit 1  yit
 i, t ,
zit  zit 1  xit
 i, t ,
 aki ( git  dit )  MxFlowk
 k , t,
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
 k , j, t ,
(12)
i
security line flow limits
 aki( j ) ( git  dit )  MxFlowk( j )
i
How does this formulation differ from the one on slide 4?
 Here, we allow offers to be made on reserves and so
have included the last term in the objective function.
41
SCUC Problem (w/demand+reserve bidding)
min
 z F
  g iktCikt   diktU ikt   yit Sit   xit H it   rit Rit   witWit
t
i
t
k
t
k
t
t
t
i
t
i


 
i
 
i 
 
i 
i 

 


it it
Fixed (no -load) Costs
Production Costs
Subject to
power balance
 git  Dt   dit
i
reserve
DemandValu e
 rit  SDt
Startup Costs
Shutdown Costs
 t,
(2)
 t,
(3)
Reserve Costs
Reserve Value
i
i
min generation
max generation
max spinning reserve
ramp rate pos limit
ramp rate neg limit
start if off-then-on
shut if on-then-off
normal line flow limit
git  zit MINi
 i, t ,
git  rit  zit MAX i
 i, t ,
rit  zit MAXSPi
 i, t ,
git  git 1  MxInci
 i, t ,
git  git 1  MxDeci
 i, t ,
zit  zit 1  yit
 i, t ,
zit  zit 1  xit
 i, t ,
 aki ( git  dit )  MxFlowk
 k , t,
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
 k , j, t ,
(12)
This is
interesting….
i
security line flow limits
 aki( j ) ( git  dit )  MxFlowk( j )
i
How does this formulation differ from the one on slide 39?
 Here, we allow offers and bids to be made on energy & reserves
and so have included the terms corresponding to demand value and
reserve value.
42
SCUC Problem (w/demand+reserve bidding)
“The Midwest ISO market-wide OR demand curve is utilized to ensure that
energy and OR are priced to reflect scarcity conditions when OR becomes
scarce. The market-wide OR demand curve price is determined in terms of
the Value of Lost Load (VoLL, currently set to $3,500/MW) and the
estimated conditional probability of loss of load given that a single forced
resource outage of 100 MW or greater will occur at the cleared marketwide OR level for which the price is being determined.”
Ref: Xingwang Ma, Haili Song, Mingguo Hong, Jie Wan, Yonghong Chen, Eugene Zak, “The Security-constrained Commitment and Dispatch For
Midwest ISO Day-ahead Co-optimized Energy and Ancillary Service Market,” Proc. of the 2009 IEEE PES General Meeting.
K. Wissman, “Competitive Electricity Markets and the Special Role of Ancillary Services, slides presented at the Licensing/
Competition and Tariff/Pricing Comm Meeting, Feb 4-5, 2008.
43