![Notations
© University of Washington 19
u(i,t): Status of unit i at period t
p(i,t): Power produced by unit i during period t
Unit i is ON during period t
u(i,t) = 1:
Unit i is OFF during period t
u(i,t) = 0 :
Ci[p(i,t)]: Running cost of unit i during period t
SCi[u(i,t)]: Startup cost of unit i during period t
N : Number of available generating units
T : Number of periods in the optimization horizon](https://image.slidesharecdn.com/unitcommitmentupdated2-250510132408-a8a3dbc5/85/Unit-Commitment-updated-lecture-slidesides-20-320.jpg)
Unit Commitment updated lecture slidesides
- 3. Economic Dispatch: Problem Definition 2 • Given load • Given set of units on-line • How much should each unit generate to meet this load at minimum cost? A B C L
- 4. Unit Commitment • Given load profile (e.g. values of the load for each hour of a day) • Given set of units available • When should each unit be started, stopped and how much should it generate to meet the load at minimum cost? © University of Washington 3
- 5. Typical summer and winter loads © University of Washington 4
- 6. Load variations • Significant difference between peak load and minimum load • Need different number of generating units at the peak and the minimum • Some rapid changes in the load © University of Washington 5
- 7. A Simple Example • Unit 1: • PMin = 250 MW, PMax = 600 MW • C1 = 510.0 + 7.9 P1 + 0.00172 P1 2 $/h • Unit 2: • PMin = 200 MW, PMax = 400 MW • C2 = 310.0 + 7.85 P2 + 0.00194 P2 2 $/h • Unit 3: • PMin = 150 MW, PMax = 500 MW • C3 = 78.0 + 9.56 P3 + 0.00694 P3 2 $/h • What combination of units 1, 2 and 3 will produce 550 MW at minimum cost? • How much should each unit in that combination generate? © University of Washington 6
- 8. Constant Terms in the Cost
- 9. Cost of the various combinations © University of Washington 8
- 10. Cost of the various combinations © University of Washington 9 1 2 3 Pmin Pmax P1 P2 P3 Ctotal Off Off Off 0 0 Infeasible Off Off On 150 500 Infeasible Off On Off 200 400 Infeasible Off On On 350 900 0 400 150 5418 On Off Off 250 600 550 0 0 5389 On Off On 400 1100 400 0 150 5613 On On Off 450 1000 295 255 0 5471 On On On 600 1500 Infeasible -
- 11. Observations on the example: © University of Washington 10
- 12. Effect of the no-load cost © University of Washington 11
- 13. Another Example • Optimal generation schedule for a load profile • Decompose the profile into a set of periods • Assume load is constant over each period • For each period, which units should be committed to generate at minimum cost during that period? © University of Washington 12 Load Time 12 6 0 18 24 500 1000
- 14. Optimal combination for each hour Load Unit 1 Unit 2 Unit 3 1100 On On On 1000 On On Off 900 On On Off 800 On On Off 700 On On Off 600 On Off Off 500 On Off Off © University of Washington 13 Repeat calculation from previous example for each period
- 15. Matching the combinations to the load © University of Washington 14
- 16. Operating costs of generating units • Running cost • Start-up cost © University of Washington 15
- 17. Effect of the start-up cost • Need to “balance” start-up and running costs © University of Washington 16
- 18. Unit commitment as an optimization problem • Minimize total cost over time horizon • Total cost = running cost + startup cost © University of Washington 17
- 19. Notations © University of Washington 18
- 20. Notations © University of Washington 19 u(i,t): Status of unit i at period t p(i,t): Power produced by unit i during period t Unit i is ON during period t u(i,t) = 1: Unit i is OFF during period t u(i,t) = 0 : Ci[p(i,t)]: Running cost of unit i during period t SCi[u(i,t)]: Startup cost of unit i during period t N : Number of available generating units T : Number of periods in the optimization horizon
- 21. Objective function © University of Washington 20
- 22. Unit Constraints © University of Washington 21
- 23. System Constraints © University of Washington 22
- 24. Load/Generation Balance Constraint © University of Washington 23 u(i,t)p(i,t) i=1 N ∑ = L(t) At all times, the power produced by the generating units must be equal to the load
- 25. Reserve Constraint • Unanticipated loss of a generating unit or an interconnection causes unacceptable frequency drop if not corrected • Need to increase production from other units to keep frequency drop within acceptable limits • Rapid increase in production only possible if committed units are not all operating at their maximum capacity • Some of the capacity of the generating units must be kept “in reserve” © University of Washington 24
- 26. Reserve Constraint © University of Washington 25
- 27. How much reserve? • Protect the system against “credible outages” • Reserve requirement: • Capacity of largest unit or interconnection • Percentage of peak load © University of Washington 26
- 28. Why can’t we treat each period separately? © University of Washington 27
- 29. Typical summer and winter loads © University of Washington 28
- 30. The “California Duck Curve” © 2011 D. Kirschen and University of Washington 29 Typical March day
- 31. Economic Dispatch vs. Unit Commitment • Generation scheduling or unit commitment is a more general problem than economic dispatch • Economic dispatch is a sub-problem of generation scheduling • Unit commitment must strike a balance between cheaper inflexible units and more expensive flexible units © University of Washington 30
- 32. Solving the Unit Commitment Problem • Decision variables: © University of Washington 31
- 33. Optimization with integer variables • Continuous variables • Discrete variables © University of Washington 32
- 34. How many combinations are there? © University of Washington 33 • Examples • 3 units: 8 possible states • N units: 2N possible states 111 110 101 100 011 010 001 000
- 35. How many solutions are there anyway? © University of Washington 34 1 2 3 4 5 6 T=
- 36. How many solutions are there anyway? © University of Washington 35 1 2 3 4 5 6 T= Optimization over a time horizon divided into intervals A solution is a path linking one combination at each interval How many such path are there? Answer: 2N ( ) 2N ( )… 2N ( ) = 2N ( )T
- 37. The Curse of Dimensionality • Example: 5 units, 24 hours • Processing 109 combinations/second, this would take 1.9 1019 years to solve • There are 100’s of units in large power systems... • Many of these combinations do not satisfy the constraints © University of Washington 36 2N ( ) T = 25 ( ) 24 = 6.21035 combinations
- 38. How do you Beat the Curse? Brute force approach wonʼt work! • Need to be smart • Try only a small subset of all combinations • Canʼt guarantee optimality of the solution • Try to get as close as possible within a reasonable amount of time © University of Washington 37
- 40. Solving the Unit Commitment Problem • State of the art: • Mixed Integer Linear Programming (MILP) • Efficient MILP solvers © University of Washington 38
- 42. A Simple Unit Commitment Example
- 43. Unit Data © University of Washington 40 Unit Pmin (MW) Pmax (MW) Min up (h) Min down (h) No-load cost ($) Marginal cost ($/MWh) Start-up cost ($) Initial status A 150 250 3 3 0 10 1,000 ON B 50 100 2 1 0 12 600 OFF C 10 50 1 1 0 20 100 OFF
- 44. Cost curves © University of Washington 41 p C(p) A B C
- 45. Demand Data © University of Washington 42 Hourly Demand 0 50 100 150 200 250 300 350 1 2 3 Hours Load Reserve requirements are not considered
- 46. Feasible Unit Combinations (states) © University of Washington 43 Combinations Pmin Pmax A B C 1 1 1 210 400 1 1 0 200 350 1 0 1 160 300 1 0 0 150 250 0 1 1 60 150 0 1 0 50 100 0 0 1 10 50 0 0 0 0 0 1 2 3 150 300 200
- 47. Transitions between feasible combinations © University of Washington 44 A B C 1 1 1 1 1 0 1 0 1 1 0 0 0 1 1 1 2 3 Initial State
- 48. Infeasible transitions: Minimum down time of unit A © University of Washington 45 A B C 1 1 1 1 1 0 1 0 1 1 0 0 0 1 1 1 2 3 Initial State TD TU A 3 3 B 1 2 C 1 1
- 49. Infeasible transitions: Minimum down time of unit A © University of Washington 46 A B C 1 1 1 1 1 0 1 0 1 1 0 0 0 1 1 1 2 3 Initial State TD TU A 3 3 B 1 2 C 1 1
- 50. Infeasible transitions: Minimum up time of unit B © University of Washington 47 A B C 1 1 1 1 1 0 1 0 1 1 0 0 0 1 1 1 2 3 Initial State TD TU A 3 3 B 1 2 C 1 1
- 51. Feasible transitions © University of Washington 48 A B C 1 1 1 1 1 0 1 0 1 1 0 0 0 1 1 1 2 3 Initial State
- 52. Operating costs © University of Washington 49 1 1 1 1 1 0 1 0 1 1 0 0 1 4 3 2 5 6 7
- 53. Economic dispatch © University of Washington 50 State Load PA PB PC Cost 1 150 150 0 0 1500 2 300 250 0 50 3500 3 300 250 50 0 3100 4 300 240 50 10 3200 5 200 200 0 0 2000 6 200 190 0 10 2100 7 200 150 50 0 2100 Unit Pmin Pmax No-load cost Marginal cost A 150 250 0 10 B 50 100 0 12 C 10 50 0 20
- 54. Operating costs © University of Washington 51 1 1 1 1 1 0 1 0 1 1 0 0 1 4 3 2 5 6 7 $1500 $3500 $3100 $3200 $2000 $2100 $2100
- 55. Start-up costs © University of Washington 52 1 1 1 1 1 0 1 0 1 1 0 0 1 4 3 2 5 6 7 $1500 $3500 $3100 $3200 $2000 $2100 $2100 Unit Start-up cost A 1000 B 600 C 100
- 56. Accumulated costs © University of Washington 53 1 1 1 1 1 0 1 0 1 1 0 0 1 4 3 2 5 6 7 $1500 $3500 $3100 $3200 $2000 $2100 $2100 $1500 $5100 $5200 $5400 $7300 $7200 $7100 $0 $0 $0 $0 $0 $600 $100 $600 $700
- 57. Total costs © University of Washington 54 1 1 1 1 1 0 1 0 1 1 0 0 1 4 3 2 5 6 7 $7300 $7200 $7100 Lowest total cost
- 58. Optimal solution © University of Washington 55 1 1 1 1 1 0 1 0 1 1 0 0 1 2 5 $7100
- 59. Notes • This example is intended to illustrate the principles of unit commitment • Some constraints have been ignored and others artificially tightened to simplify the problem and make it solvable by hand • Therefore it does not illustrate the true complexity of the problem • The solution method used in this example is based on dynamic programming. This technique is no longer used in industry because it only works for small systems (< 20 units) © University of Washington 56