EE 553 Security-constrained unit commitment (SCUC) J. McCalley 1

EE 553Security-constrained unit

commitment (SCUC)

J. McCalley

1

Unit Commitment (UC)

2

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.

SCUC Objective Function (no demand bidding)

Costs Shutdown

t iitit

Costs Startup

t iitit

Costs Production

t iitit

Costs load)-(no Fixed

t iitit HxSyCgFz min

• Decision variables are zit, git, yit, xit

• zit, yit, xit are discrete, git is continuous

3

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.

SCUC Problem (no demand bidding)

4

power balance i

itti

it dDg , t (2)

reserve ti

it SDr , t (3)

min generation iitit MINzg ,, ti (4) max generation iititit MAXzrg ,, ti (5) max spinning reserve iitit MAXSPzr ,, ti (6) ramp rate pos limit iitit MxIncgg 1 ,, ti (7) ramp rate neg limit iitit MxDecgg 1 ,, ti (8) start if off-then-on ititit yzz 1 ,, ti (9) shut if on-then-off ititit xzz 1 ,, ti (10) normal line flow limit

ikititki MxFlowdga )( ,, tk (11)

security line flow limits i

jkitit

jki MxFlowdga )()( )( ,,, tjk (12)

Costs Shutdown

t iitit

Costs Startup

t iitit

Costs Production

t iitit



Subject to

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 outputMxDeci is max ramprate (MW/period) for decreasing gen i outputaij is linearized coefficient relating bus i injection to line k flow

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

)( jkia

)( jkMxFlow

Power balance at each period t.

Max gen, min gen, reserves.

Two Categories of Reserves

5

• “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.

Contingency Reserves

6

min generation iitit MINzg ,, ti (4) max generation iititit MAXzrg ,, ti (5) max spinning reserve iitit MAXSPzr ,, ti (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


7

power balance i

itti

it dDg , t (2)

reserve ti

it SDr , t (3)




jkitit


Costs Shutdown

t iitit

Costs Startup

t iitit

Costs Production

t iitit



Subject to





)( jkia

)( jkMxFlow

Max increase and max decrease. This reflects ramp rates.

Max Inc and Max Dec (Ramp Rates)

8

MxInci=RampRateUpi*ΔT

ramp rate pos limit iitit MxIncgg 1 ,, ti (7) ramp rate neg limit iitit MxDecgg 1 ,, ti (8)

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

MxDeci=RampRateDowni*ΔT

MW/Min Min

%/min

Coal 1-5

Nuclear 1-5

NGCC 5-10

CT 20

Diesel 40

Max Inc and Max Dec (Ramp Rates)

9

MxInci=RampRateUpi*ΔT

ramp rate pos limit iitit MxIncgg 1 ,, ti (7) ramp rate neg limit iitit MxDecgg 1 ,, ti (8)

MxDeci=RampRateDowni*ΔT

MW/Min Min

%/min

Coal 1-5

Nuclear 1-5

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 Long-term Infrastructure Planning Software,” to appear in IEEE Transactions on Power Systems.


10

power balance i

itti

it dDg , t (2)

reserve ti

it SDr , t (3)




jkitit


Costs Shutdown

t iitit

Costs Startup

t iitit

Costs Production

t iitit



Subject to





)( jkia

)( jkMxFlow

Start constraint

Shut constraint

Start and Shut Constraintsstart if off-then-on ititit yzz 1 ,, ti (9) shut if on-then-off ititit xzz 1 ,, ti (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


12

power balance i

itti

it dDg , t (2)

reserve ti

it SDr , t (3)




jkitit


Costs Shutdown

t iitit

Costs Startup

t iitit

Costs Production

t iitit



Subject to





)( jkia

)( jkMxFlow

Transmission normal constraint

Transmisison security constraint,

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.

normal line flow limit i

kititki MxFlowdga )( ,, tk (11)


jkitit


aki is linearized coefficient relating bus i injection to line k flow


)( jkia

We have learned how to compute the aij (which are just generation shift factors). The aki

(j) are developed on the next slide.13

Note: Should have absolute value signs around the both summations on the left to account for bi-directional flow.

Modeling transmission security constraints

14

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

iii Paf

1l

(1)l

N

iikikkk Padfdf

1

)1(ll

(2)l

N

iikiki

N

iikik

N

iii PadaPadPa

fff

111

)2()1(

llll

ll(2)l

kikiki adaa lll )(

Effective generation shift factor: linearized coefficient relating bus i injection to line l flow under contingency k.

Key Concept

15

power balance i

itti

it dDg , t (2)

reserve ti

it SDr , t (3)




jkitit


Costs Shutdown

t iitit

Costs Startup

t iitit

Costs Production

t iitit



Subject to

These are inter-temporal constraints!

Problem is a function of t.

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.

Branch & Bound is Best Note the financial implications of changing to MIP.

16Source: M. Rothleder, presentation to the Harvard Energy Policy Group, Dec 7, 2007, available at http://www.hks.harvard.edu/hepg/Rothleder_Mark.pdf.

http://www.hks.harvard.edu/hepg/Rothleder_Mark.pdf

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 Analysis,” Power Engineering Society General Meeting, 2005. IEEE 12-16 June 2005 , pp. 2724 - 2731 Vol. 3.. 19

Illustration – Problem data

y13 =-j10 y14 =-j10

y34 =-j10

y23 =-j10

y12 =-j10

g1

Pd3

Pd2

1 2

3 4

g2

g4

Unit, k

Fixed costs ($/hr)

Startup Costs

($)

Shutdown Costs ($)

Production Costs ($/pu-hr) gk1t gk2t gk4t

1 50 100 20 1246 1307 1358 2 50 100 20 1129 1211 1282 4 50 100 20 1183 1254 1320

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.

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.

Pi,min

Pi3 Pi2 Pi1

Ci $/hr

Pi (MW)

gk1t

gk2t

gk3t

Three offers per gen: (gk1t,Pk1t), (gk2t, Pk2t), (gk3t, Pk3t)

20

Illustration – Problem data

t

g

g

g

g

g

g

g

g

g

t

t

t

t

t

t

t

t

t

,

40.0

50.0

45.0

20.0

60.0

35.0

40.0

60.0

50.0

0

0

0

0

0

0

0

0

0

43

42

41

23

22

21

13

12

11

Constraints on the offers

21

Illustration – Problem data Load curveHour, t Load, Dt (pu)

1 1.50 2 1.40 3 1.30 4 1.40 5 1.70 6 2.00 7 2.40 8 2.80 9 3.20 10 3.30 11 3.30 12 3.20 13 3.20 14 3.30 15 3.35 16 3.40 17 3.30 18 3.30 19 3.20 20 2.80 21 2.30 22 2.00 23 1.70 24 1.60

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!

One Day Load variation

0

50

100

150

200

250

300

350

400

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)

Lo

ad (

MW

)

Series1

22

Illustration 1: CPLEX code for 4 hours

23

Illustration 1: CPLEX code for 4 hours

24

Illustration 1: CPLEX result

t

g

g

g

g

g

g

g

g

g

t

t

t

t

t

t

t

t

t

,

40.0

50.0

45.0

20.0

60.0

35.0

40.0

60.0

50.0

0

0

0

0

0

0

0

0

0

43

42

41

23

22

21

13

12

11

Note that all y- and x-variables are 0, therefore there is no starting up or shutting down.

25

Illustration 1: CPLEX resultWhy did we obtain such a simple solution with unit 1 down, units 2 and 4 up for all four hours?


0

50

100

150

200

250

300

350

400

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)

Lo

ad (

MW

)Series1

This is a result of the fact that the initial solution of

initialu1: z11=0initialu2: z21=1initialu4: 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.

To test this, let’s try a different initial condition: initialu1: z11=1 initialu2: z21=0 initialu4: z41=1

Unit, k

Fixed costs ($/hr)

Startup Costs

($)

Shutdown Costs ($)

Production Costs ($/pu-hr) gk1t gk2t gk4t

1 50 100 20 1246 1307 1358 2 50 100 20 1129 1211 1282 4 50 100 20 1183 1254 1320

Most expensive unit

26

Illustration 2: CPLEX resultPrevious solution was 7020.70. Why was this one more expensive?


0

50

100

150

200

250

300

350

400

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)

Lo

ad (

MW

)

Series1

Because of •U1 shut-down cost •U2 startup cost•Difference in cost between running U1 and U2 during hr 1

27

Illustration 2: CPLEX resultBecause 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). The decision variables are…. 28

Illustration 3: CPLEX resultWe 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.

29

Illustration 4: 24 hoursWe refrain from showing the data in this case because it is extensive, having 426 variables:

72 z-variables69 y-variables69 x-variables216 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 hoursThe 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?

31


0

50

100

150

200

250

300

350

400

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)

Lo

ad (

MW

)

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.

Illustration 4: 24 hours

32


0

50

100

150

200

250

300

350

400

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)

Lo

ad (

MW

)

Unit 1

0

0.2

0.4

0.6

0.8

1

1.2

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

Up

or

do

wn

Unit 1

Unit 2

0

0.2

0.4

0.6

0.8

1

1.2

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

Up

or

do

wn

Unit 2

Unit 3

0

0.2

0.4

0.6

0.8

1

1.2

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

Up

or

do

wn

Unit 3

Unit 2 is always up.


Unit 1 starts at hr 8, shuts at hr 21.


33

One day load variation

0

50

100

150

200

250

300

350

400

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

load


0

50

100

150

200

250

300

350

400

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)

Lo

ad (

MW

)

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?


34


Unit 4 is always up, until hour 24.

Unit 1 starts at hr 8, shuts at hr 19.


0

50

100

150

200

250

300

350

400

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

load

Unit 1

0

0.2

0.4

0.6

0.8

1

1.2

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

0

0.2

0.4

0.6

0.8

1

1.2

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

0

0.2

0.4

0.6

0.8

1

1.2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24


35


0

50

100

150

200

250

300

350

400

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

load

New load curve

What effect do you think this will have on the UC?

Use new load curve but reduce startup costs to $10 and shutdown costs to $2. All other data remains as before.

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.


36


Unit 4 shuts at hr 2, starts at hr 5, and shuts at hr 23.

Unit 1 starts at hr 8, shuts at hr 11, starts at hr 12, shuts at hr 20.


0

50

100

150

200

250

300

350

400

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

load

Unit 1

0

0.2

0.4

0.6

0.8

1

1.2

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

0

0.2

0.4

0.6

0.8

1

1.2

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

0

0.2

0.4

0.6

0.8

1

1.2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

An observation

37

What do low start up and shut down costs do to the UC solution?

They tend to make UC change more.

They tend to make UC change less.

What do high start up and shut down costs do to the UC solution?

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.

Another observation

38

Do we obtain the generation dispatch from the UC solution?

Yes, these are the gkit variables.

The SCED provides the LMPs, the SCUC does not.

Why, then, do we need the SCED?

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.

Co-optimization: what?Electricity markets: simultaneous clearing of two or more commodity markets within one optimization problem.

39

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))

Multiobjective optimization optimizes two (or more) objectives which depend on the same decisions: min f1(x)+f2(x)

Choose copper sheet transmission (y) then optimize generation f1(x) then optimize transmission f2(y).

How to choose operating condition x to achieve good risk-cost f1-f2 trade-off of risk?

Co-optimization (SC-SCUC)

40

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

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.

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.

Operating reserve


41

power balance i

itti

it dDg , t (2)

reserve ti

it SDr , t (3)




jkitit


Costs Reserve

t iitit

Costs Shutdown

t iitit

Costs Startup

t iitit

Costs Production

t i kiktikt


t iitit RrHxSyCgFz min

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.

SCUC Problem (w/demand+reserve bidding)

42

power balance i

itti

it dDg , t (2)

reserve ti

it SDr , t (3)




jkitit


Value Reserve

t iitit

Costs Reserve

t iitit

Costs Shutdown

t iitit

Costs Startup

t iitit

eDemandValu

t i kiktikt

Costs Production

t i kiktikt


t iitit WwRrHxSyUdCgFz min

Subject to

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.

This is interesting….

SCUC Problem (w/demand+reserve bidding)

43

“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.

Documents

EE 553 Security-constrained unit commitment (SCUC) J. McCalley 1