EEL 6266 Power System Operation and Control© 2002, 2004 Florida State University EEL 6266 Power...

EEL 6266 Power System Operation and Control

Chapter 5

Unit Commitment

© 2002, 2004 Florida State University EEL 6266 Power System Operation and Control 2

Unit Commitment Solution Methods

� Dynamic programming� chief advantage over enumeration schemes is the reduction in

the dimensionality of the problem� in a strict priority order scheme, there are only N combinations

to try for an N unit system

� a strict priority list would result in a theoretically correct dispatch and commitment only if� the no-load costs are zero

� unit input-output characteristics are linear

� there are no other limits, constraints, or restrictions

� start-up costs are a fixed amount



� Dynamic programming� the following assumptions are made in this implementation of

the DP approach� a stateconsists of an array of units

� with specified units operating and the rest decommitted (off-line)

� a feasible state is one in which the committed units can supply the required load and meets the minimum capacity for each period

� start-up costs are independent of the off-line or down-time� i.e., it is a fixed amount w.r.t. time

� no unit shutting-down costs

� a strict priority order will be used within each interval

� a specified minimum amount of capacity must be operating within each interval



� The forward DP approach� runs forward in time from the initial hour to the final hour

� the problem could run from the final hour back to the initial hour

� the forward approach can handle a unit’s start-up costs that are a function of the time it has been off-line (temperature dependent)� the forward approach can readily account for the system’s history

� initial conditions are easier to specified when going forward

� the minimum cost function for hour K with combination I:

� Fcost(K, I) = least total cost to arrive at state (K, I)

� Pcost(K, I) = production cost for state (K, I)

� Scost(K–1, L: K, I) = transition cost from state (K–1, L) to (K, I)

( ) ( ) ( ) ( )[ ]LKFIKLKSIKPIKFL

,1,:,1,min, costcostcost}{

cost −+−+=



� The forward DP approach� state (K, I) is the I th commitment combination in hour K

� a strategyis the transition or path from one state at a given hour to a state at the next hour� X is defined as the number of states to search each period

� N is defined as the number of strategies to be saved at each step� these variable allow control of the computational effort

� for complete enumeration, the maximum value of X or N is 2N – 1

� for a simple priority-list ordering, the upper bound on X is n, the number of units

� reducing N means that information is discarded about the highest cost schedules at each interval and saving only the lowest Npaths or strategies� there is no assurance that the theoretical optimum will be found



� The forward DP approach� restricted

searchpaths� N = 3

� X = 5

IntervalK – 1

IntervalK

IntervalK + 1

X X X

N

N



� Example� consider a system with 4 units to serve an 8 hour load pattern

Incremental No-load Full-load Min. TimeUnit Pmax Pmin Heat Rate Cost Ave. Cost (h)

(MW) (MW) (Btu / kWh) ($ / h) ($ / mWh) Up Down

1 80 25 10440 213.00 23.54 4 22 250 60 9000 585.62 20.34 5 33 300 75 8730 684.74 19.74 5 44 60 20 11900 252.00 28.00 1 1

Initial Condition Start-up CostsUnit off(-) / on(+) Hot Cold Cold start

(h) ($) ($) (h)

1 -5 150 350 42 8 170 400 53 8 500 1100 54 -6 0 0.02 0

Hour Load (MW)

1 4502 5303 6004 5405 4006 2807 2908 500



� Example� to simplify the generator cost function, a straight line

incremental curve is used� the units in this example have linear F(P) functions:

� the units must operate within their limits

PdP

dFFPF loadno

+= −)(

F(P)

Fno-load

Pmin PmaxP

No-load IncrementalUnit Pmax Pmin Cost Cost

(MW) (MW) ($ / h) ($ / MWh)

1 80 25 213.00 20.882 250 60 585.62 18.003 300 75 684.74 17.464 60 20 252.00 23.80



� Case 1: Strict priority-list ordering� the only states examined each hour

consist of the listed four:� state 5: unit 3, state 12: 3 + 2

state 14: 3 + 2 + 1, state 15: all four

� all possible commitments start from state 12 (initial condition)

� minimum unit up and down times are ignored

� in hour 1:� possible states that meet load demand (450 MW): 12, 14, & 15

State Unit status Capacity

5 0 0 1 0 300 MW12 0 1 1 0 550 MW14 1 1 1 0 630 MW15 1 1 1 1 690 MW

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )[ ] 38.1021102.035036.9861

15,1:12,015,115,1

36.98612080.2330046.1710500.182588.2036.1735

203001052515,1

costcostcost

1321cost

=++=+=

=++++=+++=

SPF

FFFFP Economic Dispatch Eq.

DP State Transition Eq.



� Case 1� in hour 1:

� minimum at state 12 (9208)

� in hour 2:� possible states that meet load demand (530 MW): 12, 14, & 15

K Pcost Scost Fcost

15 9861 350 1021114 9493 350 984312 9208 0 9208

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( )

( ) 20859

102110

98430

9208350

min11301

15,2:,115,215,2

113012080.2330046.1718500.182588.201735

203001852515,2

costcostcost

1321cost

=

+++

+=

+==++++=

+++=

LSPF

FFFFP

DP State Transition Eq.



� Case 1: DP diagram

� total cost: 73,439� priority order list, up-times and down-times neglected

151 1 1 1

690 MW

141 1 1 0

630 MW

120 1 1 0

550 MW

50 0 1 0

300 MW

stateunit statuscapacity

hour 1450 MW

hour 2530 MW

hour 3600 MW

hour 4540 MW

hour 5400 MW

hour 6280 MW

hour 7290 MW

hour 8500 MW

hour 0

986110211

12

94939843

12

92089208

12

1130120860

12

1093320492

12

1064819857

12

1341033618

12

1226532472

12

557457182

12

1148143953

14

1111343585

14

1082843300

14

896452615

12

859352244

12

830851609

12

686958828

12

649058449

12

619257800

12

704364976

5

666564597

5

636663949

5

1076174442

5

1039374074

5

1010873439

5

574862930

5

350 350 350 350 350

350 350 350

0

350 350

0

0 0

0

0 0

0

750

750

400

0

750

750

400



� Case 2:� complete enumeration (2.56 × 109 possibilities)

� fortunately, most are not feasible because they do not supply sufficient capacity

� in this case, the true optimal commitment is found� the only difference in the two trajectories occurs in hour 3

� it is less expensive to turn on the less efficient peaking unit #4 for three hours than to start up the more efficient unit #1 for that same time period

� only minor improvement to the total cost� case 1: 73,439

� case 2: 73,274



� Case 2: DP diagram

151 1 1 1

690 MW

141 1 1 0

630 MW

120 1 1 0

550 MW

50 0 1 0

300 MW

stateunit statuscapacity

hour 1450 MW

hour 2530 MW

hour 3600 MW

hour 4540 MW

hour 0

986110211

12

94939843

12

92089208

12

1130120860

12

1093320492

12

1064819857

12

1341033618

12

1226532472

12

1148143953

14

1111343585

14

1082843135

13

350 350

350 35013

0 1 1 1610 MW

95769576

12

1101620221

12

1245032507

12

1111343503

13

350

350



� Lagrange Relaxation� dual variables and dual optimization

� consider the classical constrained optimization problem� primal problem: minimize f(x1,…,xn), subject to ω(x1,…,xn) = 0

� the Lagrangian function: L(x1,…,xn) = f(x1,…,xn) + λ ω(x1,…,xn)

� define a dual function

� then the “dual problem” is to find

� the solution involves two separate optimization problems� in the case of convex functions this procedure is guaranteed to

solve the problem

( ) ( )λλλ

qq0

max≥

∗ =

( ) ( )λλ ,,min 21, 21

xxLqxx

=



� Example� minimize f(x1,x2) = 0.25x1

2 + x22

subject to ω(x1,x2) = 5 –x1 – x2

� the Lagrangian function:

� the dual function:

� the dual problem: ( ) ( ) ( )2

05max 25

0

=

=−=∂∂→=

≥

∗

λ

λλλ

λλλ

qqq

( ) ( )

( ) λλλ

λλλλ

52

&2,,min

245

2121, 21

+−=

==→=

q

xxxxLqxx

( ) ( )2122

2121 525.0,, xxxxxxL −−++= λλ



� Iterative form of Lagrange relaxation method� the optimization may contain non-linear or non-convex

functions

� iterative process based on incremental improvements of λ is required to solve the problem� select a arbitrary starting λ� solve the dual problem such that q(λ) becomes larger

� update λ using a gradient adjustment:

� find closeness to the solution by comparing the gap between the “primal” function and the dual function� primal function: J* = min L

� relative duality gap: , in practice the gap never reaches zero

( ) αλλ

λλ

+= qtt

d

d

( )*

**

q

qJ −



� Lagrange relaxation for unit commitment� loading constraint

� unit limits

� unit minimum up-time and down time constraints

� the objective function

TtUPPN

i

ti

ti

tload K10

1

=∀=−∑=

TtNiPUPPU iti

tii

ti KK 1 and 1minmin ==∀≤≤

( )[ ] ( )ti

ti

T

t

N

i

titiupstart

tii UPFUSPF ,

1 1, =+∑∑

= =−



� Formation of the Lagrange function� in a similar way to the economic dispatch problem

� unit commitment requires that the minimization of the Lagrange function subject to all the constraints� the cost function and the unit constraints are each separated over

the set of units� what is done with one unit does not affect the cost of running

another unit as far as the cost function, unit limits, and the up-time and down-time constraints are concerned

� the loading constraint is a coupling constraint across all the units

� the Lagrange relaxation procedure solves the unit commitment by temporarily ignoring the coupling constraint

( ) ( ) ∑ ∑= =

−+=T

t

N

i

ti

ti

tload

tti

ti UPPUPFUPL

1 1

,,, λλ



� The dual procedure attempts to reach the constrained optimum by maximizing the Lagrangian with respect to the Lagrange multiplier

� done in two basic steps� Step 1: find a value for each λt which moves q(λ) towards a

larger value

� Step 2: assuming that λt found in Step 1 is fixed, find the minimum of L by adjusting the values of Pt and Ut

� minimizing L

( ) ( ) ( ) ( )λλλλλ

,,minwheremax,

UPLqqqti

ti

t UP==∗

( )[ ]

( )[ ]{ } ∑∑∑

∑∑∑∑∑

== =

= === =

+−+=

−++=

T

t

tload

tN

i

T

t

ti

ti

ttiti

tii

T

t

N

i

ti

ti

tT

t

tload

tT

t

N

i

titi

tii

PUPUSPF

UPPUSPFL

11 1,

1 111 1,

λλ

λλ



� separation of the units from one another; the inside term can now be solved independently for each generating unit

� the minimum of the Lagrangian is found by solving for the minimum for each generating unit over all time periods

subject to the up-time and down-time constraints and

� this is easily solved as a two-state dynamic programming problem of one variable

( )[ ]∑=

−+T

t

ti

ti

ttiti

tii UPUSPF

1, λ

( ) ( )[ ]{ }∑∑==

−+=T

t

ti

ti

ttiti

tii

N

i

UPUSPFq1

,1

minmin λλ

TtPUPPU iti

tii

ti K1minmin =∀≤≤

Ui = 1

Ui = 0t = 1 t = 2 t = 3 t = 4

Si Si Si



� Minimizing the function with respect to Pit

� at the Uit = 0 state, the minimization is trivial and equals zero

� at the Uit = 1 state, the minimization w.r.t. Pi

t is:

� there are three cases to be considered for Piopt and the limits

� if Piopt ≤ Pi

min then min [Fi(Pi) – λt Pit] = Fi(Pi

min) – λt Pimin

� if Pimin ≤ Pi

opt ≤ Pimax then min [Fi(Pi) – λt Pi

t] = Fi(Piopt) – λt Pi

opt

� if Pimax ≤ Pi

opt then min [Fi(Pi) – λt Pit] = Fi(Pi

max) – λt Pimax

� the two-state DP is solved to minimize the cost of each unit� for Ui

t = 0 the minimum is zero; therefore the only way to have a lower cost is to have [Fi(Pi) – λt Pi

t] < 0

( )[ ]( )[ ] ( ) ( ) tt

iiti

ttiit

i

ti

tiit

i

ti

tii

PFP

PFP

PPFP

PPF

λλλ

λ

=→=+=+

+

d

d0

d

d

d

d

min



� Adjusting λ� λ must be carefully adjusted to maximize q(λ)

� various techniques use a mixture of heuristic strategies and gradient search methods to achieve a rapid solution

� for the unit commitment problem λ is a vector of λt’s to be adjusted each hour

� simple technique� gradient component: where

� heuristic component:� α = 0.01 when is positive

� α = 0.002 when is negative

( )

( ) ∑=

−=

+=

N

i

ti

ti

tload

t

tt

UPPq

q

1d

d

d

d

λλ

αλλ

λλ

( )λλ

qdd

( )λλ

qdd



� The relative duality gap� used as a measure of the closeness to the solution

� for large real-size power-systems unit-commitment calculations, the duality gap becomes quite small as the dual optimization proceeds� the larger the commitment problem, the smaller the gap

� the convergence is unstable at the end� some units are being switched in and out

� the process never comes to a definite end

� there is no guarantee that when the dual solution process stops, it will be at a feasible solution

� the gap equation:( )*

**

q

qJ −



� Lagrange relaxation algorithm using dual optimization

start

pick starting λt

for t = 1 … Tk = 0

J*–q*

q*

loop foreach unit i

two-state dynamic program with Tstages and solvefor Pi

t and Uit

solve for thedual value q*(λt)

solve the economic dispatch for each hourusing committed units

calculate theprimal value (λt)

update λt

for all t

end

make finaladjustmentsto scheduleto achievefeasibility

gap < thresholdgap > threshold



� Example� find the unit commitment for a three-generator system over a

four-hour segment of time� assume no start-up cost and no minimum up-time or down-time

� start with all λt values set to zero

� execute an economic dispatch for each hour when there is sufficient generation committed for that hour� if there is not enough generation, arbitrarily set the cost to 10,000

� the primal value J* is the total generation cost summed over all t

( )( )( ) 20050005.06100

4001000025.08300

600100002.010500

32

3333

22

2222

12

1111

<<++=<<++=<<++=

PPPPF

PPPPF

PPPPF( )MWt

loadPt

1 1702 520 3 11004 330



� Iteration 1:

� q(λ) = 0.0 J* = 40,000 (J* – q*) / q* = undefined� Iteration 2:

� dynamic programming for unit #3

Hour λ u1 u2 u3 P1 P2 P3 dq(λ)/dλ P1edc P2

edc P3edc

1 0.0 0 0 0 0 0 0 170 0 0 02 0.0 0 0 0 0 0 0 520 0 0 03 0.0 0 0 0 0 0 0 1100 0 0 04 0.0 0 0 0 0 0 0 330 0 0 0

U3 = 1

U3 = 0t = 1 t = 2 t = 3 t = 4

3.3247.5

P34 = P3

min

11.0-700.0

P33 = P3

max

5.2152.5

P32 = P3

min

1.7327.5

P31 = P3

min

λ =F(P) – λP =



� Iteration 2:

� q(λ) = 14,982 J* = 40,000 (J* – q*) / q* = 1.67


edc P3edc

1 1.7 0 0 0 0 0 0 170 0 0 02 5.2 0 0 0 0 0 0 520 0 0 03 11.0 0 1 1 0 400 200 500 0 0 04 3.3 0 0 0 0 0 0 330 0 0 0



� Iteration 3:

� q(λ) = 18,344 J* = 36,024 (J* – q*) / q* = 0.965


edc P3edc

1 3.4 0 0 0 0 0 0 170 0 0 02 10.4 0 1 1 0 400 200 -80 0 320 2003 16.0 1 1 1 600 400 200 -100 500 400 2004 6.6 0 0 0 0 0 0 330 0 0 0



� Iteration 4:

� q(λ) = 19,214 J* = 28,906 (J* – q*) / q* = 0.502


edc P3edc

1 5.1 0 0 0 0 0 0 170 0 0 02 10.24 0 1 1 0 400 200 -80 0 320 2003 15.8 1 1 1 600 400 200 -100 500 400 2004 9.9 0 1 1 0 380 200 -250 0 130 200



� Iteration 5:

� q(λ) = 19,532 J* = 36,024 (J* – q*) / q* = 0.844


edc P3edc

1 6.8 0 0 0 0 0 0 170 0 0 02 10.08 0 1 1 0 400 200 -80 0 320 2003 15.6 1 1 1 600 400 200 -100 500 400 2004 9.4 0 0 1 0 0 200 130 0 0 200



� Iteration 6:

� q(λ) = 19,442 J* = 20,170 (J* – q*) / q* = 0.037� Remarks

� the commitment schedule does not change significantly with further iterations� however, the solution is not stable (oscillation of unit 2)

� the duality gap does reduce� after 10 iterations, the gap reduces to 0.027� good stopping criteria would be when the gap reaches 0.05


edc P3edc

1 8.5 0 0 1 0 0 200 -30 0 0 1702 9.92 0 1 1 0 384 200 -64 0 320 2003 15.4 1 1 1 600 400 200 -100 500 400 2004 10.7 0 1 1 0 400 200 -270 0 130 200

EEL 6266 Power System Operation and Control© 2002, 2004 Florida State University EEL 6266 Power...

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Transcript of EEL 6266 Power System Operation and Control© 2002, 2004 Florida State University EEL 6266 Power...