Misg12 Tuc
Transcript of Misg12 Tuc
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Thermal Unit Commitment Problem
Moshe Potsane, Luyanda Ndlovu, Simphiwe SimelaneChristiana Obagbuwa, Jesal Kika, Nadine Padayachi, Luke O. Joel
Lady Kokela, Michael Olusanya, Martins Arasomwa, Sunday Ajibola
07 January 2012
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Outline
Contents
1 Introduction
2 Problem description
3
Constraints4 Heuristic Solution
5 Deterministic Solution
6 Results
7 Questions
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Outline
Contents
1 Introduction
2 Problem description
3
Constraints4 Heuristic Solution
5 Deterministic Solution
6 Results
7 Questions
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Outline
Contents
1 Introduction
2 Problem description
3
Constraints4 Heuristic Solution
5 Deterministic Solution
6 Results
7 Questions
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O li
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Outline
Contents
1 Introduction
2 Problem description
3
Constraints4 Heuristic Solution
5 Deterministic Solution
6 Results
7 Questions
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O tli
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Outline
Contents
1 Introduction
2 Problem description
3
Constraints4 Heuristic Solution
5 Deterministic Solution
6 Results
7 Questions
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Outline
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Outline
Contents
1 Introduction
2 Problem description
3
Constraints4 Heuristic Solution
5 Deterministic Solution
6 Results
7 Questions
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Outline
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Outline
Contents
1 Introduction
2 Problem description
3
Constraints4 Heuristic Solution
5 Deterministic Solution
6 Results
7
Questions
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Problem Description
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Problem Description
Problem Description
minx
F (P i ,t ,U i ,t ) =i ∈I
i ∈T
C i ,t (P i ,t ,U i ,t )
s .t . i ∈I
P i ,t = Lt , ∀t ∈ T ,
i ∈I
U i ,t P̄ i ≥ Lt + R t ∀t ∈ T ,
P i ,t ∈ Π(i , t ),∀t ∈ T , t ∈ T ,
Ui , t ∈ (0, 1),
P i ,t ∈ R
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Constraints
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Constraints
1 Maximum and Minimum Power generation.
2 Minimum Up Time.
3 Minimum Down Time.
4 Shut Down Cost.
Maximum and Minimum power generation
P mini ≤ P i (t ) ≤ P max (t ), U i (t ) > 0,
P i (t ) = 0, U i (t )
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The minimum and maximum power generation reduces to:
U i ,t P min
i (t ) ≤ P i (t ) ≤ U i ,t P max
i (t ), U i (t ) > 0,
where U i ,r
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Minimum Up time
This constraint signifies the minimum time for which a committed unitshould be turned on.Note : once the unit is running, it cannot be turned off immediately.
U i ,t ≤ U i ,r − U i ,r −1,
where r = t − τ + + 1, ..., t − 1, ∀t ∈ T ,∀i ∈ I .
t = 1, ..., |T |, T is the time horizon committed.
i = 1, ..., |I |, I is the number of units.
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Constraints
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Minimum Up time continued
Example:Assuming that we have t = 5, minimum up time(MU)=3,then, r = 3.4.First units binary variable:
U 1,t =
U 1,5 ≥ U 1,3 − U 1,2U 1,5 ≥ U 1,4 − U 1,3.
Second unit variable:
U 2,t =
U 2,5 ≥ U 2,3 − U 2,2U 2,5 ≥ U 2,4 − U 2,3.
Thus such observation shows the general behaviour to be:
U i ,t =
U i ,5 ≥ U i ,3 − U i ,2U i ,5 ≥ U i ,4 − U i ,3.
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Constraints
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Minimum up time continued
Figure below goes further to show the period that the possiblecombinations of minimum up time are
Figure: Period of possible combinations
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Constraints
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Minimum Down Time
This constraint signifies the minimum time for which a de-committed unitshould be turned off.Note : Once the unit is de-committed, there is a minimum time before itcan be recommitted.
1 − U i ,t ≥ (1 − U 1,r ) − (1 − U 1,r −1)U i ,t ≤ U i ,r −1 + U i ,r
U i ,r ≤ 0; U i ,t ∈ [0, 1]
where r = t − τ + + 1, ..., t − 1, ∀t ∈ T ,∀i ∈ I .
t = 1, ..., |T |, T is the time horizon committed.i = 1, ..., |I |, I is the number of units.
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Minimum Down Time Continued
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Assuming that we have, t = 4, minimum down time(MD ) = 2.From the above we get that, r = 3.Observing the first unit binary variable:
U 1,t = {U 1,4 ≤ 1 − U 1,2 − U 1,3
Second variable:
U 2,t = {U 2,4 ≤ 1 − U 2,2 − U 2,3
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Minimum Down Time Continued
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Minimum Down Time Continued
Thus such observation shows the general behaviour to be:
U 2,t = {U i ,4 ≤ 1 − U i ,2 − U i ,3Figure below goes further to show the period that the possiblecombinations of minimum down time are
Figure: Period of possible combinations
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Minimum Down Time Continued
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Shut Down Cost
Objective function was,
F (P i ,t , U i ,t ) =i ∈I
i ∈T
C i ,t (P i ,t , U i ,t )
F (P i ,t ,U i ,t ) =i ∈I
i ∈T
C i ,t (P i ,t ,U i ,t ) + SD i ,t
SD i ,t =
0, if not shut downSD cost , if shut down.
t = 1, ..., |T |, T is the time horizon committed.i = 1, ..., |I |, I is the number of units.
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Heuristic Solution
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Heuristic Approach
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Heuristic Solution
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Available Methods
Dynamic programming
Benders decomposition
mixed integer programming
Lagrangian relaxation
Simulated annealing
Tabu search
The high dimensionality and combinatorial nature of the unit commitment
problem curtail attempts to develop any rigorous mathematicaloptimization method capable of solving the whole problem for any real-sizesystem.
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Heuristic Solution
L R l Al h
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Lagrangian Relaxation Algorithm
Why choose the LR algorithm
1 Specific for the UCP.
2 Flexible in dealingg with different types of constraints.
3 Flexible to incorporating additional coupling constraints that have notbeen considered so far.
4 Flexible because no priority ordering is imposed
5 Computationally much more attractive for large system since the
amount of computation varies with the number of units
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Heuristic Solution
H h l i h k
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How the algorithm works:
The problem has three components;
1 Cost function
2 Set of constraints involving a single unit3 Set of coupling constraints, one for each hour in the study period
involving all unit.
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Heuristic Solution
N
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Now
1 loading contraints
P t load −N
i =1
P t i U t i = 0
2 unit limits
U t i P min
i ≤ P t
i ≤ U t
i P max
i
3 unit minimum up- and down-time constraints
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Heuristic Solution
Obj ti f ti
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Objective function
minx
F (P i ,t ,U i
,t ) =
i ∈I
i ∈T
C i ,t (P i
,t ,U i
,t )
s .t .i ∈I
P i ,t = Lt , ∀t ∈ T ,
i ∈I
U i ,t P̄ i ≥ Lt + R t ∀t ∈ T ,
P i ,t ∈ Π(i , t ),∀t ∈ T , t ∈ T ,
Ui , t ∈ (0, 1),
P i ,t ∈ R
The procedure attempts to reach the constrained optimum by maximizingthe lagrange multipliers, while minimizing with respect to the othervariables in the problem. That is
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Heuristic Solution
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q ∗(λ) = maxλt
q (λ)]
whereq (λ) = minP t i ,U t
i
L(P ,U , λ). (1)
This is achieved in two basic steps
1 Find a value for each λt which moves q (λ) towards a larger value
2 Assuming that the λt found are now fixed, find the minimum of L byadjusting the values of P t and U t .
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Heuristic Solution
W it th bj ti f ti b t ki th li t i t d
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We rewrite the objective function by taking the coupling constraints andadding them into the objective function to come up with the lagrangianfunction
L =i ,t
C i ,t (P i ,t ,U i ,t ) +t
λt
(Lt −i
P i ,t ) +t
u t
(Lt +R t −i
U i ,t P̄ i )
Now drop the constant terms, thus the equation above simplifies to
L =i ,t
C i ,t (P i ,t ,U i ,t ) −t
λt i
P i ,t −t
u t i
U i ,t P̄ i ).
=i
(t
C i ,t (P i ,t , U i ,t ) −t
λt i
P i ,t −t
u t i
U i ,t P̄ i )
λ(t ) − demand lagrange multiplier
U(t) - spinning reserve langrange multiplier
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Heuristic Solution
Inner System
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Inner System
Low-level: i = 1, 2, ..., I
minP i ,t ,U i ,t
L
(2)
with,L =i
(t
C i ,t (P i ,t , U i ,t ) −t
λt i
P i ,t −t
u t i
U i ,t P̄ i )
(3)
subject to,
P i ,t ∈ (i , t )
U u ,t ∈ [0, 1].
thus Li
(λ, u
) is the optimal lagrangian for low level with given and u
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Deterministic Solution
Deterministic Method
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Deterministic Method
Branch and Bound Method
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Deterministic Solution
What is a deterministic solution?
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What is a deterministic solution?
One which guarantees the optimal solution
The current state of the solution determines the next stateIt is a more reliable method
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Deterministic Solution
Some general solution methods considered for solving
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Some general solution methods considered for solving
MIQPs
Benders Decomposition
Outer Approximation
Lagrangian Decomposition
Branch and Bound Method
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Deterministic Solution
Why Branch and bound?
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Why Branch and bound?
BB algorithm searches the complete space for optimal solution
For a convex problem, the convergence to a global optimum can be
provedCan be used for general discrete and continuous problems
Most popular in Optimization Literature
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Deterministic Solution
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Figure: BB Tree
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Deterministic Solution
General Procedure
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General Procedure
Choosing the branching variable:
Randomly
Choosing a value U from the continuous relaxation closest to an
integerBounding:
Lower Bound- Continuous relaxation of the objective function
Upper Bound-Using a heuristic
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Deterministic Solution
General Procedure
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General Procedure
3 rules for fathoming the nodes:
If the problem is infeasible
If the lower bound of node A is greater than or equal to the upper
bound of node BThe solution is an integer
Stopping Condition:
|ub − lb | <
When all the nodes have been fathomed
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Deterministic Solution
Constraints
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Maximum and minimum power generated
The power generated while the machine is switched on must satisfy
the loadThe maximum power while the machine is switched on must exceedthe sum of the load and the reserve at each time period
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Deterministic Solution
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Figure: Time against Units
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Deterministic Solution
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Figure: Power against Period
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Deterministic Solution
Remarks
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In this instance, the bigger generators were utilized first
In reality, a combination of both big and small generators will ensureefficiency
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Q and A
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Thank You!!!
Any Questions?
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