JMEE7311
Power Systems Operation and
Management
lecture 5
Unit Commitment
Dr. Nassim Iqteit
16-3-2019
Daniel Kirschen
1
Economic Dispatch: Problem Definition
• Given load
L
A B C
• Given set of units on-line
• How much should each unit generate to meet this load at minimum cost?
Typical summer and winter loads
2
Unit Commitment
• Given load profile
? ? ?
(e.g. values of the load for each hour of a day)
Load Profile
• Given set of units available G G G
• When should each unit be started, stopped and how much should it generate to
meet the load at minimum cost?
Example Cost of the various combinations
• Unit 1:
• PMin = 250 MW, PMax = 600 MW
• C1 = 510.0 + 7.9 P1 + 0.00172 P12 $/h
• Unit 2:
• PMin = 200 MW, PMax = 400 MW
• C2 = 310.0 + 7.85 P2 + 0.00194 P22 $/h
• Unit 3:
• PMin = 150 MW, PMax = 500 MW
• C3 = 78.0 + 9.56 P3 + 0.00694 P32 $/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?3
Observations on the example:
• Far too few units committed:
Can’t meet the demand
• Not enough units committed:
Some units operate above optimum
• Too many units committed:
Some units below optimum
• Far too many units committed:
Minimum generation exceeds demand
• No-load cost affects choice of optimal combination
4
A more ambitious example
• Optimal generation schedule for a load profile
• Decompose the profile into a set of period
• Assume load is constant over each period
• For each time period, which units should be
committed to generate at minimum cost during
that period?
Optimal combination for each hour Matching the combinations to the load
Load Unit 3
Unit 2
Unit 1
0 12 24
Time
6 18
5
Issues Unit Constraints
Must consider constraints Constraints that affect each unit individually:
Unit constraints • Maximum generating capacity
System constraints • Minimum stable generation
Some constraints create a link between periods
• Minimum “up time”
Start-up costs
• Minimum “down time”
Cost incurred when we start a generating unit
• Ramp rate
Different units have different start-up costs
Curse of dimensionality
6
Notations
u(i,t) : Status of unit i at period t
u(i,t) = 1: Unit i is on during period t
u(i,t) = 0 : Unit i is off during period t
x(i,t) : Power produced by unit i during period t
Minimum up- and down-time
• Minimum up time
Once a unit is running it may not be shut down
immediately: If u(i,t) = 1 and t up < tiup,min then u(i,t + 1) = 1
i
• Minimum down time
Once a unit is shut down, it may not be started
immediately: If u(i,t) = 0 and tidown < tidown,min then u(i,t + 1) = 0 7
Ramp rates
Maximum ramp rates
To avoid damaging the turbine, the electrical output of a unit cannot change by
more than a certain amount over a period of time:
Maximum ramp up rate constraint: x ( i,t + 1) − x ( i,t ) ≤ ∆Pi up,max
Maximum ramp down rate constraint: x(i,t) − x(i,t + 1) ≤ ∆Pi down,max
System Constraints
Constraints that affect more than one unit
• Load/generation balance
• Reserve generation capacity
• Emission constraints
N
• Network constraints
Load/Generation Balance Constraint
∑ u(i,t)x(i,t) = L(t)
i=1
N : Set of available units
Reserve Capacity Constraint
• Unanticipated loss of a generating unit or an interconnection causes unacceptable
frequency drop if not corrected rapidly
• 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
N
∑ u(i,t)Pi
max
≥ L(t) + R(t)
i=1
R(t): Reserve requirement at time t
9
How much reserve?
• Protect the system against “credible outages”
• Deterministic criteria:
Capacity of largest unit or interconnection
Percentage of peak load
• Probabilistic criteria:
Takes into account the number and size of the committed units as well as their outage rate
Types of Reserve
Spinning reserve
Primary
Quick response for a short time
Secondary
Slower response for a longer time
Tertiary reserve
Replace primary and secondary reserve to protect against another outage
Provided by units that can start quickly (e.g. open cycle gas turbines)
Also called scheduled or off-line reserve
• Positive reserve
Increase output when generation < load
• Negative reserve
Decrease output when generation > load
• Other sources of reserve:
Pumped hydro plants
Demand reduction (e.g. voluntary load shedding)
• Reserve must be spread around the network
Must be able to deploy reserve even if the network is congested
Cost of Reserve
• Reserve has a cost even when it is not called
• More units scheduled than required
Units not operated at their maximum efficiency
Extra start up costs
• Must build units capable of rapid response
• Cost of reserve proportionally larger in small systems
Important driver for the creation of interconnections between systems
Environmental constraints
• Scheduling of generating units may be affected by environmental constraints
• Constraints on pollutants such SO2, NOx
– Various forms:
• Limit on each plant at each hour
• Limit on plant over a year
• Limit on a group of plants over a year
• Constraints on hydro generation
– Protection of wildlife
12
– Navigation, recreation
Network Constraints
• Transmission network may have an effect on the commitment of units
Some units must run to provide voltage support
The output of some units may be limited because their output would exceed the transmission
capacity of the network
A B
Cheap generators More expensive generator
May be “constrained off” May be “constrained on”
13
Start-up Costs
• Thermal units must be “warmed up” before they can be brought on-line
• Warming up a unit costs money
• Start-up cost depends on time unit has been off
t iOFF
τi: thermal time constant for the unit
−
SCi (tiOFF ) = αi + βi (1− e τ i
) α i :fixed cost includes crew expense,
maintenance expenses (in $)
αi + βi
tiOFF :time h the unit was cooled
βi = Cc F
αi
tiOFF
14
Start-up Costs….contine.…
Need to “balance” start-up costs and running costs
Example:
• Diesel generator: low start-up cost, high running cost
• Coal plant: high start-up cost, low running cost
Issues:
• How long should a unit run to “recover” its start-up cost?
• Start-up one more large unit or a diesel generator to cover the peak?
• Shutdown one more unit at night or run several units part-loaded?
Summary
• Some constraints link periods together
• Minimizing the total cost (start-up + running) must be done over the whole period of study
• Generation scheduling or unit commitment is a more general problem than economic dispatch
• Economic dispatch is a sub-problem of generation scheduling
Flexible Plants Inflexible Plants
• Power output can be adjusted (within limits) • Power output cannot be adjusted for
technical or commercial reasons
• Examples:
• Coal-fired • Examples:
• Oil-fired • Nuclear
• Open cycle gas turbines Thermal units • Run-of-the-river hydro
• Combined cycle gas turbines • Renewables (wind, solar,…)
• Hydro plants with storage • Combined heat and power (CHP, cogeneration)
• Status and power output can be optimized • Output treated as given when optimizing
16
Solving the Unit Commitment Problem
• Decision variables:
• Status of each unit at each period:
u(i,t) ∈{0,1} ∀ i,t
• Output of each unit at each period:
{
x(i,t) ∈ 0, Pi min ; Pi max } ∀ i,t
• Combination of integer and continuous variables
Optimization with integer variables
Continuous variables
Can follow the gradients or use LP
Any value within the feasible set is OK
Discrete variables
There is no gradient
Can only take a finite number of values
Problem is not convex
Must try combinations of discrete values
How many combinations are there? How many solutions are there anyway?
111
110
• Examples
101
• 3 units: 8 possible states
100
• N units: 2N possible states
011
010
001
000
T= 1 2 3 4 5 6
• Optimization over a time horizon divided into intervals
• A solution is a path linking one combination at each interval
• How many such paths are there?
How many solutions are there anyway?...continue
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?
T
( )( ) ( ) ( )
Answer: 2 N 2 N … 2 N = 2 N
Example: 5 units, 24 hours
N T 5 24
T= 1 2 3 4 5 6
(2 ) = (2 ) = 6.2 10 35 combinations
• Processing ૢ combinations/second, this would
take . ૢ × ૢ years to solve
• There are 100’s of units in large power systems...
The Curse of Dimensionality
• Many of these combinations do not satisfy the
constraints
How do you Beat the Curse?
Brute force approach won’
’t work!
• Need to be smart
• Try only a small subset of all combinations
• Can’t guarantee optimality of the solution
• Try to get as close as possible within a reasonable amount of time
Main Solution Techniques
Characteristics of a good technique
• Solution close to the optimum
• Reasonable computing time
• Ability to model constraints
Priority list / heuristic approach
Dynamic programming
Lagrangian relaxation
Mixed Integer Programming State of the art
Unit Data
Min Min No-load Marginal Start-up
Pmin Pmax Initial
Unit up down cost cost cost
(MW) (MW) status
(h) (h) ($) ($/MWh) ($)
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
Hourly Demand
Demand Data
350
300
Reserve requirements are not considered 250
200
Load
150
100
50
0
1 2 3
Hours
Feasible Unit Combinations (states)
Combinations 1 2 3
Pmin Pmax
A B C 150 300 200
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
24
Transitions between feasible combinations
1 2 3
A B C
1 1 1
1 1 0
1 0 1
1 0 0 Initial State
0 1 1
25
Infeasible transitions: Minimum down time of unit A
1 2 3
A B C
1 1 1
1 1 0
1 0 1
1 0 0 Initial State
0 1 1
TD TU
A 3 3
B 1 2
C 1 1 26
Infeasible transitions: Minimum up time of unit B
1 2 3
A B C
1 1 1
1 1 0
1 0 1
1 0 0 Initial State
0 1 1
TD TU
A 3 3
B 1 2
C 1 1 27
Feasible transitions
1 2 3
A B C
1 1 1
1 1 0
1 0 1
1 0 0 Initial State
0 1 1
28
Operating costs
1 1 1 4
1 1 0 3 7
1 0 1
2 6
1 0 0 1
5
29
Economic dispatch
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
30
Operating costs
1 1 1 4
$3200
1 1 0 3 7
$3100 $2100
1 0 1 2 6
$3500 $2100
1 0 0 1 5
$1500 $2000
31
Start-up costs
1 1 1 4
$3200
$0
1 1 0 $700 3 $0 7
$3100 $600 $2100
$600
1 0 1 2 $0 6
$3500 $2100
$100
$0
1 0 0 $0 1 5
$1500 $2000
Unit Start-up cost
A 1000
B 600
C 100 32
Accumulated costs
$5400
1 1 1 4
$3200
$0
$5200 $7300
1 1 0 $700 3 $0 7
$3100 $600 $2100
$600 $5100 $7200
1 0 1 2 $0 6
$3500 $2100
$100
$0
$1500 $7100
1 0 0 $0 1 5
$1500 $2000
33
Total costs
1 1 1 4
$7300
1 1 0 3 7
$7200
1 0 1 2 6
$7100
1 0 0 1 5
Lowest total cost
34
Optimal solution
1 1 1
1 1 0
1 0 1 2
$7100
1 0 0 1 5
35
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)
HW#1
Let us consider a plant having three units. The cost characteristics and minimum
and maximum limits of power generation (MW) of each unit are as follows:
To supply a total load of 600 MW most economically,
Find the combinations of units and their generation status.
Determine the optimal status.
HW#2
A three-generator, four-hour unit commitment problem will be solved. The data for
this problem are as follows. Given the three generating units in the following,
No start-up costs, no minimum up- or downtime constraints.
Find the optimal solution.
Thank you
