using lagrangian relaxation in optimisation of unit

Fakulteta za Elektrotehniko

Eva Thorin, Heike Brand, Christoph Weber

Using Lagrangian relaxation in optimisation of unit

commitment and planning-Part2

OSCOGENDiscussion Paper No.10

August 2002

Contract No. ENK5-CT-2000-00094 Project co-funded by the European Community under the 5th Framework Programme (1998-2002)

Contract No. BBW 00.0627 Project co-funded by the Swiss Federal Agency for Education and Science

Contract No. CEU-BS-1/2001 Project co-funded by Termoelektrarna toplarna Ljubljana, d.o.o.

IER Stuttgart 09.12.02

1 INTRODUCTION........................................................................................................................................ 1

2 PROCEDURE USED FOR SOLVING THE SIMPLIFIED CHP-PROBLEM ..................................... 1

3 UPDATING OF THE LAGRANGE MULTIPLIERS.............................................................................. 2

4 EXAMPLE WITH FOUR TURBINES...................................................................................................... 4

5 EXTENDED PROBLEM ............................................................................................................................ 7

5.1 Time resolution of 1h ...............................................................................................12 5.2 Introduction of time blocks.......................................................................................13

6 LITERATURE ........................................................................................................................................... 14

IER Stuttgart 1 09.12.02

1 Introduction

In Discussion Paper No.3 , “Using Lagrangian relaxation in optimisation of unit commitment and dispatching” the basic theory behind Lagrangian relaxation was described and a formulation for a simplified CHP-problem was derived. The formulation includes a system of CHP turbines with given heat and electric power demand and the possibility to buy electric power from take-or-pay contracts and the spot market and to sell electric power to the spot market. The problem was relaxed by including the restrictions for the sum of electricity and the sum of heat and was then decomposed into one problem for each turbine and one problem for the terms that are not turbine dependent. In this discussion paper a procedure for solving the relaxed problem is presented and the Lagrange relaxation is then applied for a small system consisting of four turbines. Further the problem is extended to the BEWAG demo-model, including 8 turbines, 2 boilers, 2 district heat systems, restrictions for the heat flow between the district heat system, and restrictions for the sum of electricity produced by two of the turbines. The results for this system is compared to the results previously obtained with the model without Lagrange relaxation. 2 Procedure used for solving the simplified CHP-problem

The iteration procedure for solving a Lagrange problem is shown in Figure 2.1.

Figure 2.1: Procedure for solving the Lagrange problem.

Initialisation of the Lagrange multipliers.

Is the convergence criteria satisfied?

Construct a primal feasible solution from the solution of the relaxed problem.

Solve the relaxed problem (by solving the sub problems)

Update the Lagrange multipliers.

The best found primal solution is the solution of the problem

Yes

No


After the decomposition of the problem into several sub problems the problem can be solved by minimising the dual problem (the original problem with the relaxed restrictions included) which is done by solving the sub problems for different Lagrange multipliers. For the problem presented in Chapter 4 different starting values for the Lagrange multipliers have been tested. Further, different methods to update the multipliers, which are described in Chapter 3, have been tested. The program language used for the problem is GAMS (General Algebraic Modelling System) with the optimisation solver CPLEX 7.5. The primal feasible solution is constructed from the solution of the relaxed problem with the help of CPLEX. The operation variables are first fixed for each turbine for those times when the turbine is committed and then the primal problem is optimised. The solve is stopped after the first integer solution has been found. If the solution of the dual problem is not feasible for the primal problem it means that the relaxed restrictions are not fully fulfilled . To be able to fulfil them more units have to be committed for some time steps. By fixing the operation variable only for the time steps when the units are committed the additional commitment of the units in the time steps when the restrictions are not fulfilled is possible. The dual gap, that is the gap between the best dual and primal solution found so far, is usually used as convergence criteria. In the example described in Chapter 4 the dual gap have been calculated but since the behaviour of the functions have been of interest a predetermined number of iterations have been performed. For the extended problem presented in Chapter 5 the calculations have been stopped when a dual gap lower than 1 % have been reached. 3 Updating of the Lagrange multipliers

One common method to update the Lagrange multipliers is to use the so called sub gradient method.

nnnn sαλλ +=+1 where

λ = the Lagrange multiplier α = the step length s = the sub gradient n = the number of the iteration


The Lagrange relaxation method assumes that the dual function for a maximisation problem (that means for a problem where we would like to know the maximum of the primal problem) is a convex function of the Lagrange multipliers. For a minimisation problem it is assumed to be a concave function. To be able to reach the minimum of the dual function we would like to change the Lagrange multiplier in the direction of the decreasing gradient in the point we are (see Figure 3.1). Since the function may not always be differentiable we instead for the gradient use the sub gradient which is a tangential of the function in the point we are /Huonker 2001/.

00

λλλλ

Φ(λ)Φ(λ)Φ(λ)Φ(λ)

λn λn+1

αnsn

Φ(λn)

Φ(λn+1)

sn

Figure 3.1: An example of a convex dual function and the construction of the next Lagrange multiplier. There are several methods available to calculate the step length and the sub gradient. For the simple problem presented in Chapter 4 the Polyak II rule and the use of a convergent series have been tested for calculating the step length. Polyak rule:

s)()( *n λφλφξα −=

where

)( nλφ = the value of the dual function at the iteration of interest

)( *λφ = the optimal value of the dual function

ξ = a constant which value is halved if the dual function is not improved in a given


number of iterations.

s =Euclidian norm of the sub gradient vector

( eg. 222 21 )timeK(tsubgradien.......)time(tsubgradien)time(tsubgradien ++ )

In most cases the optimal values of the dual function is not known. Often a known value for the primal solution are used instead. /Huonker 2001, Dotzauer,2001/. Convergent series:

( )nψσα += 0

ψσ ,0 = parameters

n = the number of the iteration

4 Example with four turbines

The Lagrangian method was first tried on a small system consisting of four turbines (T2, T3, T4 and T5 in the BEWAG demo-model), all extraction-condensing turbines. The convergent series and Polyak rule methods to update the Lagrangian multipliers were used and different values of the parameters in these methods tested. Also different starting values for the multipliers were tried. In Table 4.1 and 4.2 the different runs made are summarised for the use of the convergent series method and the Polyak rule method, respectively. In Figure 4.1 and 4.2 the iterations for the different runs are presented graphically. For all runs the time period has been 1 week in August (000731-000806). Table 4.1: Runs for the simple example with the convergent series method for updating the Lagrange multipliers.

run nr. start value λλλλ start value µµµµ σσσσ0 ψψψψ

Computational time (s) ( to a dual gap <1 %)

L47 25 3 200 0,89 >1 h 14 min L49 25 3 100 0,89 1 h 10 min L50 25 3 50 0,89 * L62 0 0 100 0,89 1 h 9 min L63 16 0 100 0,89 1 h 13 min L64 0 0 100 0,95 * L65 0 0 100 0,83 > 1h 30 min * a dual gap of 1 % was not reached with 50 iterations


-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50Iterat io n

L47

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L49

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L50

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L62

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L63

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L64

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L65

Dual problemPrimal problem

Figure 4.1: The value of the dual and the primal problem in each iteration for the runs made with the convergent series method for updating the Lagrange multipliers. Table 4.2: Runs for the simple example with the Polyak rule method for updating the Lagrange multipliers.

run nr. start value λλλλ start value µµµµ ξξξξ Limit**

Computational time (s) ( to a dual gap <1 %)

L52 25 3 0,1 5 * L56 25 3 0,3 5 1 h 31 min L57 25 3 0,3 2 * L53 25 3 0,5 5 58 min L55 25 3 0,75 5 1 h 15 min L54 25 3 1 5 * L58 0 0 0,3 5 1 h 8 min L66 0 0 0,5 5 * L59 100 15 0,3 5 * L60 16 0 0,3 5 * L61 0 1,75 0,3 5 1 h 10 min * a dual gap of 1 % was not reached with 50 iterations ** limit = number of iterations without improvement before the ξ ξ ξ ξ parameter is halved.


-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L52

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L56

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L57

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L53

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L55

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L54

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L58

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iteration

L66

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L59

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L60

-580000-560000-540000-520000-500000-480000-460000

0 10 20 30 40 50

Iterat io n

L61

Figure 4.2: The value of the dual and the primal problem in each iteration for the runs made with the Polyak rule method for updating the Lagrange multipliers. (For legend see Figure 4.1) Of the parameters tested for the convergent series method to update the Lagrange multipliers a σ0 of 100 and ψ of 0,89 give the best calculation performance. The sensitivity is low for changes in the σ0 value and also for changes in the starting value for the Lagrange multipliers. The calculation performance is very similar for all runs with the convergent series method for updating the Lagrange multipliers, except the one where the ψ parameter has been set to 0,95 (run L64). With the Polyak rule method to update the Lagrange multipliers the value 0,5 for the ξ parameter gives the best performance when the start value for λ is 25 and the start value for µ


is 3. However, when both start values are set to zero the value 0,3 for the ξ parameter gives the best performance. This method seems to be more sensitive of the initial values of the multipliers and the variation in calculation performance is rather high. Runs was also made with the convergent series method for 1 week in December and 1 week in October. Here the computational times to reach a dual gap lower than 1 % were 3 minutes and 1 hour and 14 minutes, respectively.

5 Extended problem

In the next step the problem was extended to include all units in the BEWAG demo-model, that is 8 turbines (6 extraction–condensing turbines and 2 gas turbines), 2 boilers and 2 district heat systems. The restrictions for the heat flow between the district heat system and the restrictions for the sum of electricity produced by two of the turbines (T7 and T8) was also included. The maximisation problem can then be described as below.


{ } down timesshut andoperation Minimum

01

E E

B 0

0 0

87

8

7

01

0101

2121

1212

122211

211121

1 12112

1 1

1

10

1

110

11

,O

)t(P)t(P

GTuP

PP)t(P

ECTu)t(O)t(P)t(P

ECTu)t(O)t(P)t(P)t(O)t(P

Uu)t(OP)t( P)t( OPUu)t(OP)t(P)t(OP

uP)t(P)t(P

P)t(P)t(P

P)t(PP)t(P

)t(P)t(P)t(P)t(P

)t(P)t(P)t(P)t(P

)t(P)t(P)t(P)t(P)t(Ps. t

tcos)t(Ptcos)t(P

tcos)t(P)t(tcos)t(P

)t(Ptcos))t(P)t(O(tcos

)))t(P)t(P()t(O(tcos)t(price)t(P

max

u

TT,EL,MIN

T

Tuu,EL

u,TH,MAX

MAX,EL,uu,THu,EL

uuuu,THu,EL

uuuu,THu,ELuuuu,TH

uu,TH,MAXu,THuu,TH,MIN

uu,EL,MAXu,ELuu,EL,MIN

u,TH,MAXu,TH

SPOT,SELL,EL

TOP,BUY,EL,MAXTOP,BUY,EL

SPOT,BUY,EL

DHS_TO_DHS,TH,MAXDHS_TO_DHS,TH


DHS_TO_DHS,THDHS_DEMAND,THDHS_TO_DHS,THDHSu

u,TH

DHS_TO_DHS,THDHS_DEMAND,THDHS_TO_DHS,THDHSu

u,TH

SPOT,SELL,ELTu

DEMAND,ELTOP,BUY,ELSPOT,BUY,ELu,EL

T

t

T

tTRANSFERDHS_TO_DHS,THTRANSFERDHS_TO_DHS,TH

T

t

T

tTOPTOP,BUY,ELSPOTSPOT,BUY,EL

u

u,THT

t Buru,ELuuu

T

t GTur

u,THuu,ELuuu

T

t ECTur

T

tSPOTSPOT,SELL,EL

O,P,P THEL

∈

≥

∈≥

∈+≥

∈+≤≤+

∈≤≤∈≤≤

∈≤≥

≤≤≥

≤

≤

+≥+

+≥+

+=++

��

�

��

�

�

��

�

��

�

�

−−

−−

−+−

−+−

+=

∈

∈

∈

= =

= =

= ∈= ∈

= ∈=

�

�

�

�

� �

� �

� ��

� ��

γγ

ββεε

ηαα

βαα

where

)(,, tP SPOTSELLEL = electric power sold at the spot market at time step t

)(tpriceSPOT = the price for the electric power when you sell it at the spot market at time step t ECT = extraction condensing turbines

rtcos = the cost for fuel r )t(P u,EL = the electric power produced by unit u at time step t )t(P u,TH = the heat produced by unit u at time step t


uα = constant for unit u in fuel consumption equation 000uuu ,, εγβ = the sections of the lines in the PQ-chart for unit u 111uuu ,, εγβ = the slopes of the lines in the PQ-chart for unit u

)t(Ou = variable indicating the on-off status for the unit u at time step t GT= gas turbines B = boilers ηu= efficiency for unit u

)t(P SPOT,BUY,EL = electric power bought at the spot market at time step t )t(tcos SPOT = the cost for buying electric power at the spot market at time step t

)t(P TOP,BUY,EL = electric power bought from take-or-pay contract at time step t

TOPtcos = the cost for buying electric power from the take-or-pay contract )t(P DEMAND,EL = the electric power demand at time step t

)t(P DHS_TO_DHS,TH 21 =the heat transferred from district heat system 1 to district heat system 2 at time step t

)t(P DHS_TO_DHS,TH 12 =the heat transferred from district heat system 1 to district heat system 2 at time step t costTRANSFER= the cost for transferring heat from one district heat system to another EU = electric power producing units DHS1= district heat system 1 DHS2= district heat system 2

)t(P DHS_DEMAND,TH 1 = the heat demand for district heat system 1 at time step t )t(P DHS_DEMAND,TH 2 = the heat demand for district heat system 2 at time step t

Due to that the system have two district heat systems two Lagrange multipliers are needed for the heat (µ1 and µ2). The relaxed problem then becomes the following:



01

0 0

0

87

8

7

01

0101

2121

1212

12

21

122

2

11

12

211

1

1

121

112

11

1

10

1

110

1

1

21

,O

)t(P)t(P

GTuP

PP)t(P

ECTu)t(O)t(P)t(P

ECTu)t(O)t(P)t(P)t(O)t(P

EUu)t(OP)t( P)t( OPEUu)t(OP)t(P)t(OP

BuP)t(P

)t(PP)t(PP)t(P

P)t(PP)t(PP)t(Ps. t

)t(P)t(P

)t(P)t(P)t(

)t(P)t(P

)t(P)t(P)t(

)t(P)t(P)t(P

)t(P)t(P)t(

tcos)t(P

tcos)t(P

tcos)t(P)t(tcos)t(P

)t(Ptcos))t(P)t(O(tcos

)))t(P)t(P()t(O(tcos

)t(price)t(P

max),,(

u

TT,EL,MIN

T

Tuu,EL

u,TH,MAX

MAX,EL,uu,THu,EL

uuuu,THu,EL

uuuu,THu,ELuuuu,TH



u,TH,MAXu,TH

TOP,BUY,ELEUu

u,EL,MAXSPOT,SELL,EL


DEMAND,ELSPOT,BUY,EL



T

tDHSu

DHS_TO_DHS,THu,TH

DHS_TO_DHS,THDHS_DEMAND,TH

T

tDHSu

DHS_TO_DHS,THu,TH

DHS_TO_DHS,THDHS_DEMAND,TH

T

t TOP,BUY,ELSPOT,BUY,ELEUu

u,EL

SPOT,SELL,ELDEMAND,EL

T

tTRANSFERDHS_TO_DHS,TH

T

tTRANSFERDHS_TO_DHS,TH

T

tTOPTOP,BUY,EL

T

tSPOTSPOTx,BUY,EL

u

u,THT

t Buru,ELuuu

T

t GTur

u,THuu,ELuuu

T

t ECTur

T

tSPOTSPOT,SELL,EL

O,P,P THEL

∈

≥

∈≥

∈+≥

∈+≤≤+

∈≤≤∈≤≤

∈≤

+≤≤≤≤≤≤

≤

≤

��

�

��

�

�

��

�

��

�

�

��

��

�

−−

+−

��

��

�

−−

+−

��

��

�

−−−

+−

−

−

−−

−+−

−+−

=Φ

+=

∈

=∈

=∈

=∈

=

=

==

= ∈= ∈

= ∈

=

�

�

��

��

��

�

�

��

� ��

� �

�

γγ

ββεε

µ

µ

λ

ηαα

βαα

µµλ


This is then divided into the following sub problems: For boilers (both boilers are connected to District heat system 1) :

111

u,TH,MAXu,TH

T

tu,TH

u

u,THrO,P,P

P)t(Ps. t

)t(P)t()t(P

tcosmaxTHEL

≤

��

��

��

��

��

��

+�

=µ

η

For extraction condensing turbines:


01

87

8

7

01

0101

1 12

11110

,O

)t(P)t(P

)t(O)t(P)t(P

)t(O)t(P)t(P)t(O)t(P

)t(OP)t( P)t( OP)t(OP)t(P)t(OPs. t

)t(P)t()t(P)t()t(P)t()))t(P)t(P()t(O(tcos

max

u

TT,EL,MIN

T

Tuu,EL

uuuu,THu,EL

uuuu,THu,ELuuuu,TH



T

t DHSu,TH

DHSu,THu,ELu,THuu,ELuuur

O,P,P THEL

∈

≥

+≥

+≤≤+

≤≤≤≤

��

��

��

��

��

��

�

+++−+−

+=

= ∈

∈

�

�

γγ

ββεε

µµλβαα

For gas turbines (both gas turbines are connected to District heat system 1):

( )


01

111

10

,O

GTuP

PP)t(P

)t(OP)t( P)t( OP)t(OP)t(P)t(OPs. t

)t(P)t()t(P)t())t(P)t(O(tcosmax

u

u,TH,MAX

MAX,EL,uu,THu,EL



T

tDHSu,THu,ELu,ELuuurO,P,P THEL

∈

∈≥

≤≤≤≤

��

�� +++−�

=∈µλαα

For contract terms:


2121

1212

1

211222

122111

0 0

0



TOP,BUY,ELEUu

u,EL,MAXSPOT,SELL,EL


DEMAND,ELSPOT,BUY,EL

T

t

DHS_TO_DHS,THDHS_TO_DHS,THDHS_DEMAND,TH

DHS_TO_DHS,THDHS_TO_DHS,THDHS_DEMAND,TH

DEMAND,ELTOPTOP,BUY,EL

SPOTSPOT,BUY,ELSPOTSPOT,SELL,EL

P

P)t(PP)t(P

)t(PP)t(PP)t(P

)t(P)t(Pt.s

))t(P)t(P)t(P)(t(

))t(P)t(P)t(P)(t()t(P)t()tcos)t((P

))t(tcos)t()(t(P))t()t(price)(t(P

maxEL

≤

≤

+≤≤≤≤

≤≤

��

�

��

�

�

��

�

��

�

�

��

��

�

−+−

−+−−−+

−+−

�

�

∈

=

µµ

λλλλ

For Turbine 7 and 8 there exist a restriction for the sum of power they produce, therefore these two turbines are put into the same sub problem. Altogether we then have 10 sub problems (one for each boiler, 7 for the turbines and one for the contract terms). The convergent series method, with a σ0 of 100 and a ψ of 0,89, was used to update the Lagrange multipliers. The Lagrange multiplier for the sum of electricity restriction corresponds to the marginal cost for electric power production. In a system where electric power can be both bought and sold on the electric power spot market the spot market prices are the marginal cost for electric power production. Since the LPX prices are given the Lagrange multiplier for the electric power balance is also given and the Lagrange multiplier is fixed to the LPX price. in the program. To be able to do this, the cost and price for selling and buying power at the spot market are set equal. The initial value of Lagrange multipliers for the heat (µ1 and µ2) is set to zero. The values of the variables after each sub problem optimisation is saved and those values are then used as starting values for the optimisation of the sub problems in the next iteration. The three weeks that previously have been calculated for this system with the optimisation model without Lagrangian relaxation were calculated also with the Lagrangian method. The weeks are 1 week in August (000807-000813), 1 week in October (001002-001008), and 1 week in December (001211-001217).

5.1 Time resolution of 1h

In Table 5.1 the calculation performance for the Lagrangian model for the above mentioned three weeks are presented for the case with a time resolution of 1 h. The calculated profits are also compared to the results of the model without Lagrangian relaxation.


Table 5.1: Computational performance and the results of the optimisation model with Lagrange relaxation with the time resolution of 1 h for 1 week in August, October and December, respectively. The results are also compared to the results of the model without Lagrange relaxation.

Week Computational time (s)( to a dual gap <1 %)

Best primal solutionProfit [€ ]

Profit [€] for model without Lagrange

000807-000813 2 h 8 min -1196567 -1230222 001002-001008 48 min -1411528 -1497478 001211-001217 2 min -2764269 -2768228

The results of the model without Lagrange relaxation is 2,8 %, 6,1 % and 0,1 % lower for the August, October and December weeks, respectively compared to the results of the Lagrange model. The calculation times are high for the Lagrange model, at least for the August and October weeks. The results for the model without Lagrange relaxation have been obtained within 5,5 minutes for all three weeks but then the optimisation has been stopped when a solution with a relative gap below 10 % has been found (the actual relative gap is 4,3 %, 7,9 % and 0,2 % for the August, October and December weeks, respectively). Previous experience with the model without Lagrange relaxation shows that only slightly better solutions are found when the program are allowed to run for several hours.

5.2 Introduction of time blocks

In order to shorten the computational time the time resolution was lowered for the operation, start-up and shut-down variables by introducing nine time blocks for each day. The time blocks chosen were the following: 00-05 06-09 10-11 12-13 14-16 17-18 19-20 21 22-23 Table 5.2 shows the results of the calculations with time blocks. It can be seen that the computational time is reduced considerably for the August and October weeks (88% and 81%, respectively). The calculated profits are, however, lower. For the August week the difference is 1,5 %, for the October week 3,5 % and for the December week only 0,1 %. One


reason for the low profit for the October week is that the restriction for the power production of turbines 7 and 8 do not match the time blocks resulting in that Turbine 7 is run more than necessary. Table 5.2 The results of the optimisation program when nine time blocks are used for each day instead of making calculations for each hour.

Week Computational time (s)

Best primal solution Profit [€ ]

000807-000813 16 min ( to a dual gap <1 %)

-1214674

001002-001008 9 min ( to a dual gap of 1,07 %)

-1460226

001211-001217 1,5 min ( to a dual gap <1 %)

-2767644

6 Literature

/Dotzauer, 2001/ Dotzauer, Erik: Energy System Operation by Lagrangian Relaxation, Linköping

Studies in Science and Technology, Dissertations No. 665, Division of Optimization, Departments of Mathematics, Linköpings universitet, Sweden, 2001 /Huonker, 2001/

Huonker, U.: Ein Werkzeug zur Lagrange-Optimierung in der Kraftwerkseinsatzplanung, Dissertationsentwurf, Institut für Energiewirtschaft und Rationelle Energieanwendung, Univeristät Stuttgart,2001

using lagrangian relaxation in optimisation of unit

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