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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

(Optimization Group) TUC Problem 07 January 2012 1 / 35

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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


3



6 Results

7 Questions


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Contents

1 Introduction


3



6 Results

7 Questions


O li

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Contents

1 Introduction


3



6 Results

7 Questions


O tli

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Outline

Contents

1 Introduction


3



6 Results

7 Questions


Outline

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Outline

Contents

1 Introduction


3



6 Results

7 Questions


Outline

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Outline

Contents

1 Introduction


3



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


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.


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.


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


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.


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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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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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.


Heuristic Solution

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Heuristic Approach


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.


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


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


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


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 .


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


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

.(Optimization Group) TUC Problem 07 January 2012 23 / 35

Deterministic Solution

Deterministic Method


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Deterministic Method

Branch and Bound Method



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



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



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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Figure: BB Tree



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



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



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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Figure: Time against Units



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Figure: Power against Period



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


Q and A

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Thank You!!!

Any Questions?

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