EE/Econ 458 SCUC

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EE/Econ 458SCUC

• J. McCalley

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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.
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SCUC Objective Function (no demand bidding)

• 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.
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SCUC Problem (no demand bidding)
Subject to
Power balance at each period t.
Max gen, min gen, reserves.
MxFlowk is the maximum MW flow on line k is
linearized coefficient relating bus i
injection to line k flow under contingency j,
is the maximum MW flow on line k
under contingency j MAXSPi is maximum
spinning reserve for unit 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
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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 these, but there is little
insight to be gained by doing so, and it makes
the problem more complicated. We represent only
contingency reserves.
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Contingency Reserves
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 MAXSPiMAXi-MINi
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SCUC Problem (no demand bidding)
Subject to
Max increase and max decrease. This reflects ramp
rates.
MxFlowk is the maximum MW flow on line k is
linearized coefficient relating bus i
injection to line k flow under contingency j,
is the maximum MW flow on line k
under contingency j MAXSPi is maximum
spinning reserve for unit 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
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Max Inc and Max Dec (Ramp Rates)
/min
Coal 1-5
Nuclear 1-5
NGCC 5-10
CT 20
Diesel 40
MxInciRampRateUpi?T
MW/Min
Min
MxDeciRampRateDowni?T
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.
?T is amount of time from one period t to the
next t1
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SCUC Problem (no demand bidding)
Subject to
Start constraint
Shut constraint
MxFlowk is the maximum MW flow on line k is
linearized coefficient relating bus i
injection to line k flow under contingency j,
is the maximum MW flow on line k
under contingency j MAXSPi is maximum
spinning reserve for unit 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
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Start and Shut Constraints
Starting constraints. For example, z12z11y12,
or more generally, zktzk,t-1ykt, which
says Status of unit k in time t status of unit k
in time t-1start flag in time t
Constraints associated with shutting. For
example, z12z11-x12, or more generally,
zktzk,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 z11y12 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!

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SCUC Problem (no demand bidding)
Subject to
Transmission normal constraint
Transmisison security constraint,
MxFlowk is the maximum MW flow on line k is
linearized coefficient relating bus i
injection to line k flow under contingency j,
is the maximum MW flow on line k
under contingency j MAXSPi is maximum
spinning reserve for unit 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
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Transmission normal, security constraints
?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.
aij is linearized coefficient relating bus i
injection to line k flow
is linearized coefficient relating bus i
injection to line k flow under
contingency j,
We will not have time to learn how to compute
these coefficients. But they can be obtained
using DC-Power flow based approach.
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Key Concept
Problem is a function of t.
Subject to
These are inter-temporal constraints!
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.
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Branch Bound is Best ?
Note the financial implications of changing to
MIP.
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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.
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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..
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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 Analysis, Power Engineering
Society General Meeting, 2005. IEEE 12-16 June
2005 , pp. 2724 - 2731 Vol. 3..
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Illustration Problem data
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)
Notice that for each unit, the offers increase
with generation, i.e., gk1tltgk2tltgk3t. This
prevents use of a higher generation level before
a lower generation level. It also says that our
offer function is convex.
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Illustration Problem data
Constraints on the offers
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Illustration Problem data
Load curve
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Illustration 1 CPLEX code for 4 hours
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Illustration 1 CPLEX code for 4 hours
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Illustration 1 CPLEX result
Note that all y- and x-variables are 0, therefore
there is no starting up or shutting down.
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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?
This is a result of the fact that the initial
solution of initialu1 z110 initialu2
z211 initialu4 z411 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.
To test this, lets try a different initial
condition initialu1 z111 initialu2 z210
initialu4 z411
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Illustration 2 CPLEX result
Previous solution was 7020.70. Why was this one
more expensive?
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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 (y221) and U1 to shut down (x121) 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.
Lets 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). The decision
variables are.
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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 start-up cost of U2 was high.
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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 z110
initialu2 z211 initialu4 z411 which is the
most economic solution for the hour 1 loading
level.
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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?
The result, objective value77667.3, 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.
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Illustration 4 24 hours
Unit 1 starts at hr 8, shuts at hr 21.
Unit 2 is always up.
Unit 4 is always up.
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Illustration 5 24 hours
Previous load curve
New load curve
Observe initial final load values are lower in
the new load curve. What effect will this have on
the UC?
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Illustration 5 24 hours
Unit 1 starts at hr 8, shuts at hr 19.
Unit 2 is always up.
Unit 4 is always up, until hour 24.
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Illustration 6 24 hours
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.
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Illustration 6 24 hours
Unit 1 starts at hr 8, shuts at hr 11, starts at
hr 12, shuts at hr 20.
Unit 2 is always up.
Unit 4 shuts at hr 2, starts at hr 5, and shuts
at hr 23.
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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 if we solved each hour independently.
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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 BB, which cannot
provide dual variables.
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Exam Problem 1
Beginning from the CPLEX file provided, provide
plots of dispatch vs. time for all three units.
Give total cost. Then implement the below ramp
rate constraints for all three units such that
that MxInciMxDeci0.3 pu. Note that this
affects constraints (7) and (8) in our
formulation. Provide plots of dispatch vs. time
for all three units comparing the with ramp rate
constraints and without cases. Give total cost.
I want to see plots for both cases, for all three
units, like I have been showing you.
THIS IS AN EXAM QUESTION AND IS TO BE WORKED
INDIVIDUALLY. BRING TO THE EXAM ON WEDNESDAY.
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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
• Regulating reserve
• Contingency reserve

Operating 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 usually no single loads as large as the
largest single generator.
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SCUC Problem (no demand bidding)
Subject to
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.
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SCUC Problem (w/demandreserve bidding)
Subject to
This is interesting.
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.
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SCUC Problem (w/demandreserve 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
market-wide 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.
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Exam 2 Coverage

• Optimization
• Linear programming
• How SCUC/SCED/RAC are used in the market
• LPOPF (SCED)
• Branch bound
• SCUC

• Suggested priority in studying
• Slides and notes
• Homeworks
• Posted papers
• (All of the above are fair game)

Exam 2 is closed book, closed notes, open
calculator, but no communication devices.
Remember to work and bring to exam problem 1.