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EE/Econ 458 SCUC

EE/Econ 458 SCUC. J. McCalley. 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. Unit Commitment (UC). In the day-ahead market, it is always 24 hours.

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EE/Econ 458 SCUC

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  1. EE/Econ 458SCUC J. McCalley

  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. Unit Commitment (UC) 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.

  3. Decision variables are zit, git, yit, xit • zit, yit, xit are discrete, git is continuous SCUC Objective Function (no demand bidding) 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.

  4. Subject to Power balance at each period t. SCUC Problem (no demand bidding) 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

  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 Two Categories of Reserves 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.

  6. 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 MAXSPi=MAXi-MINi

  7. Subject to SCUC Problem (no demand bidding) 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

  8. MxInci=RampRateUpi*ΔT Max Inc and Max Dec (Ramp Rates) MW/Min Min MxDeci=RampRateDowni*ΔT 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

  9. Subject to SCUC Problem (no demand bidding) 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

  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 Start and Shut Constraints 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. Subject to SCUC Problem (no demand bidding) 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

  12. 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.

  13. Problem is a function of t. Subject to Key Concept 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.

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

  15. “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.” Some good industry quotes in notes 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.

  16. “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….” Some good industry quotes in notes 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..

  17. “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]…” Some good industry quotes in notes 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. 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. Illustration – Problem data Three offers per gen: (gk1t,Pk1t), (gk2t, Pk2t), (gk3t, Pk3t) 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.

  19. Constraints on the offers Illustration – Problem data

  20. Load curve Illustration – Problem data

  21. Illustration 1: CPLEX code for 4 hours

  22. Illustration 1: CPLEX code for 4 hours

  23. Note that all y- and x-variables are 0, therefore there is no starting up or shutting down. Illustration 1: CPLEX result

  24. 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 • that the initial solution of • initialu1: z11=0 • initialu2: z21=1 • 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. Illustration 1: CPLEX result To test this, let’s try a different initial condition: initialu1: z11=1 initialu2: z21=0 initialu4: z41=1

  25. Previous solution was 7020.70. Why was this one more expensive? Illustration 2: CPLEX result • Because of • U1 shut-down cost • U2 startup cost • Difference in cost between running U1 and U2 during hr 1

  26. 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. Illustration 2: CPLEX result 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….

  27. 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. Illustration 3: CPLEX result

  28. 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 Illustration 4: 24 hours • 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.

  29. 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? Illustration 4: 24 hours 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.

  30. Illustration 4: 24 hours Unit 1 starts at hr 8, shuts at hr 21. Unit 2 is always up. Unit 4 is always up.

  31. 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?

  32. 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.

  33. 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.

  34. 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.

  35. What do low start up and shut down costs do to the UC solution? They tend to make UC change more. An observation 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.

  36. Do we obtain the generation dispatch from the UC solution? Yes, these are the gkit variables. Another observation 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.

  37. 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 • MxInci=MxDeci=0.3 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. Exam Problem 1 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.

  38. Co-optimization, in the context of electricity markets, refers to the simultaneous clearing of two or more commodity markets within the same optimization problem. Co-optimization (SC-SCUC) • 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.

  39. Subject to SCUC Problem (no demand bidding) 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.

  40. Subject to SCUC Problem (w/demand+reserve bidding) 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.

  41. “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.” SCUC Problem (w/demand+reserve bidding) 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.

  42. Optimization • Linear programming • How SCUC/SCED/RAC are used in the market • LPOPF (SCED) • Branch & bound • SCUC Exam 2 Coverage 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.

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