EE533 Power Operations Cost Based Unit Commitment

1
EE533 Power OperationsCost Based Unit Commitment

• S. J. Ranade

2
The Unit Commitment Problem

• Given Time Horizon week/month/
• Load Forecast
• Units available for service
• Determine Units that should be placed
• on line each hour ( or day)
• Objective Minimize Cost
• Fuel OM Startup/Shutdown
• Risk ( Probabilistic )
• Constraints Spinning Reserve Emissions
• Network Ramp rates

3
The Unit Commitment Problem

• Tradeoffs
• A unit that is on line may be expensive and
  running at minimum capacity
• Shutting the unit down may save money
• BUT
• Will incur startup cost when needed again
• May not be possible to start by the time it is
  needed

4
The Unit Commitment Problem

• Unit Characteristics Startup/Shutdown
• Nuclear shut down only for refueling
• Hydro zero resource cost
• Large coal ( 250 MW) Very long start time (Days)
• Gas (lt 200 MW or so) 8-24 Hours start time
• Combustion turbines As low as 5 Min

5
The Unit Commitment Problem

• Unit Characteristics Startup/Shutdown Costs
• Fuel to get system to correct temperature/pressure
  
• Crew time
• Electricity for auxiliary system

6
The Unit Commitment ProblemDaily
7
The Unit Commitment ProblemWeekly
Unit 4
Unit 4
Unit 2
Unit 2
Unit 1
Unit 1
8
The Unit Commitment Problem

• Discrete Optimization Problem
• 3 units -- 238 possibilities each hour
• 24 hours -- 8244.722E21 possibility( many
  unrealistic)
• Large Scale Problem
• Discrete Optimization
• -- Convert to continuous
• -- add constraints to force towards discrete
• -- may find approx. answer may miss answer
• -- Develop a search technique

9
The Unit Commitment ProblemPriority Lists (
Wood/Wollenberg)
Common sense solution ( May not give best answer)
Order units by Full load average cost /MWH
Full load average cost Full load Cost (/H) /
Capacity MW This ordering is the Priority
List Commit enough units in priority order to
meet peak loadreserve In each hour see if a
unit can be shut down ( reverse priority)
Total capacity gt Hourly loadreserve
10
The Unit Commitment ProblemPriority Lists (
Continued))
If unit can be shut down in some hour -- When
will it be needed again -- Will this meet
minimum up/down time and startup time
constraints? If time constraints are satisfied
calculate two costs -- C1 Fuel OM cost
change for proposed shut down period -- C2
Startup cost If C1gtC2 shut unit down Repeat for
all hours and units
11
(No Transcript)
12
(No Transcript)
13
(No Transcript)
14
(No Transcript)
15
(No Transcript)
16
(No Transcript)
17
(No Transcript)
18
(No Transcript)
19
(No Transcript)
20
(No Transcript)
21
(No Transcript)
22
(No Transcript)
23
(No Transcript)
24
Unit Commitment Formal Methods

• Dynamic Programming
• Lagrangean Relaxation
• Heuristic/Intelligent Methods

25
Dynamic Programming

• A way to stage-wise decompose problems to reduce
  computational burden
• The problem f(x1,x2,xn), g (x1,x2,xn)0
• Can sometimes be expressed as
• Min f1 (x1) o f2 (x2)ofn(xn)
• y1g1 (x1)
• y2g2(x2,y2)
• .
• .
• yngn(xn, yn-1)

26
Dynamic Programming

• Min f1 (x1) o f2 (x2)ofn(xn)
• -- The objective can be separated into n
  functions fi .
• --Each such function fi . is a function of only
  one of the problem or decision variables
  namely xi
• -- o is the composition operator, simplest
  operator is

27
Dynamic Programming

• y1g1 (x1,y0) y2g2(x2,y2)yngn(xn, yn-1)
• The constraints(and objective) can be separated
  into a system or stage-wise format this is
  called decomposition
• y1g1 (x1,y0) y2g2(x2,y2)yngn(xn, yn-1)
• gt g (x1,x2,xn)0

States
Decisions
Stages
Further,
28
Dynamic Programming
y2
Example Min x12x22 x1x2 5 x1 , x2
integers 3 The objective is separable
with fn(xn) xn2 Let yo 5 y1 yo -
x1 ? g1 y2 y1 - x2 ? g2 y2
0 Add both sides of the above and simplify
5-x1 x2 0 original constraint
29
Dynamic Programming-Visualization
y2
Possible states at the end Of each state (y
values) are shown in the circles (nodes) Initial
state is 5, terminal is 0
0
1
The decisions (x s) at each stage define the
next state Initial state 5 and decision x1 2
yields state 3 after stage 1 There is no
transition from initial state 5 to state 0 at
stage 1 since x1 , 3
x2 2
x1 3
2
5
0
x2 3
x1 2
3
x1 1
x2 4
4
x1 0
x2 5
5
Stage 1 Stage 2
30
Dynamic Programming-Solution
Stage 1 Initial state 5 There is no way to get
to states 0 and 1 since x1 3 Starting at state
0 the cost to states 0 and 1 at stage 1 are 8
8
0
8
1
0
9
x2 2
x1 3
2
5
0
x2 3
x1 2
4
3
x1 1
For state 2 x1 3 cost f1(x1) 9 For state 3
x1 2 cost f1(x1) 4 For state 4 x1 1 cost
f1(x1) 1 For state 5 x1 0 cost f1(x1) 0
x2 4
1
4
x1 0
x2 5
0
5
Stage 1 Stage 2
31
Dynamic Programming-Solution
Stage 2 Final state 0 What is the optimal
path? From 0 total cost 8 0 8 From 1 total
cost 8 1 8 From 2 total cost 9 4 13 From 3
total cost 4 9 13 From 4 total cost 1 16
17 From 5 total cost 0 25 25 So minimum cost
to terminal state is 13. The path can originate
from states 2 or 3 at stage 1
8
0
8
1
9
x2 2
2
0
x2 3
4
3
x2 4
1
4
x2 5
0
5
Stage 1 Stage 2
32
Dynamic Programming-Solution
Stage 2 Final state 0 What is the optimal
path? In turn states 2 and 3 at stage 1
originate from state 0 at stage 0 Thus optimal
policies are X12 x23 or x13 and x22
8
0
8
1
0
9
x2 2
x1 3
2
0
5
x2 3
x1 2
4
3
x1 1
x2 4
1
4
x1 0
x2 5
0
5
Stage 1 Stage 2
33
Dynamic Programming-Solution
Stage 1 Which state is part of the optimal
solution???? AT this point we dont know What
we do know is that If state 5 were part of
the optimal trajectory at stage 1 then it would
originate in state 5 with a total cost of 0 out
to stage 1!!!!
0
1
0
9
x2 2
x1 3
2
5
0
x2 3
x1 2
4
3
x1 1
x2 4
1
4
x1 0
x2 5
0
5
Stage 1 Stage 2
34
Dynamic Programming-Solution
Stage 1 Which state is part of the optimal
solution???? AT this point we dont know What
we do know is that If state 5 were part of
the optimal trajectory at stage 1 then it would
originate in state 5 with a total cost of 0 out
to stage 1!!!!
0
1
0
9
x2 2
x1 3
2
5
0
x2 3
x1 2
4
3
x1 1
x2 4
1
4
x1 0
x2 5
0
5
Stage 1 Stage 2
35
Dynamic Programming-Visualization
A Sequence of decisions e.g., x1 2 , x2 3 is
a policy or solution Policy x1 2 , x2 3
yields a cost of 13 Policy x1 0 , x2 5 yields
a cost of 25 An optimum policy minimizes Total
cost
0
1
0
13
x2 2
x1 3
2
5
0
x2 3
x1 2
4
3
x1 1
x2 4
4
x1 0
x2 5
5
Stage 1 Stage 2