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System planning 2013

• Lecture L10: Short term planning of hydro-thermal systems

• Chapter 5.2.5, 5.3.4, 5.4.1, Appendix B• Contents:

– Dual variables– GAMS, example– Home assignment 3

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Dual variables

• LP problem on standard form:

• The solution of the problem will result (except of the optimal solution itself) in one dual variable per constraint

• The dual variables are measurements of how much the objective function will change if the right-hand side of the constraints are changed

min

s.t.

0

Tc x

Ax b

x

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Dual variablesx2

x1

1

2

3

4

1 2 3 4

x1+x2 2

x2

0

4x1+12x2 12

x1 0

z = 20

z = 30

z = 40

12

x1+x2 73

In optimum:1 > 0 (active)2 > 0 (active)3 = 0 (not active)

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Dual variables

• Hydrological constraints:

– Hydrological constraints have units HE– Objective function has unit currency (SEK, $, £ ...)– Increase of V more water available larger

income/more stored water optimal objective value increase

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2

The dual variables corresponds the water value [SEK/HE].

, ,, , 1 , , , , ,j i j i

i i

i t i t i t i t j t j t i tj Κ j Κ

M M Q S Q S V

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Dual variables• Contracted load constraints for hour t:

, ,i t g t ti g

H G D

– Constraints have unit MWh– Objective function has unit SEK– Increase of Dk increased production cost or less

water available at end of planning period optimal objective function value decreases

Dual variables correspond the marginal cost for production, i.e. the power price [SEK/MWh].

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Dual variables - Example• Assume hydropower planning for 6 hours.

– The hydro system consists of 2 power stations. – The problem includes contracted load constraints.

• Solving the problem resulted in the following dual variables:82.4304 hydr. balance for station 1, k = 1,...,642.6585 hydr. balance for station 2, k = 1,...,6 -190.0389 load balance, k = 1,2,3,5,6-200.0410 load balance, k = 4

• Questions:– What can be said about the structure of the hydro

system?– Assume that a customer want to buy another 5 MWh

hour 2. How high must the price be for this to be profitable for the producer?

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2

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Home assignment 2

• Short term hydro-thermal planning• Solved by using the software GAMS

(www.gams.com)

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GAMS

• Software specialized for solving optimization problems.

• Typical structure of GAMS program:1. Define sets2. Define parameters3. Declare optimization variables4. Declare equations5. Define equations6. Define variable limits

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Example

• Two power stations, Degerforsen (1) and Edensforsen (2).

• Have sold power to the power exchange. Contracted load for the next 6 hours: 90, 98, 104, 112, 100,80 MWh.

• Reservoirs filled to 80% at the beginning• Stored water can be used at best efficiency and power

generated in the future can be sold for 185 SEK/MWh.• Neglect the delay time between stations

Known:1

2

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Example

Power plant

Maximal active storage of

reservoir [m3]

Marginal production equivalents [MWh/HE]

Mean flow, wi [m3/s]

Deger-forsen

5*1060.209, 0 Q 225 HE0.199, 225 Q 300

163

Edens-forsen

4*1060.236, 0 Q 202.5 HE0.224, 202.5 Q 270

164

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Example

• Known parameters:

, ,0,, , , , , ,i j i f i t ii jQ M w D M

• Optimization variables:

, ,i j tQ = discharge, station i, segment j, during hour t, i=1,2, j=1,2, t=1,...,6

tiS ,= spillage, station i, during hour t, i=1,2, j=1,2, t=1,...,6

tiM ,= reservoir contents, station i, end of hour t, i=1,2, j=1,2, t=1,...,6

• Calculated parameters:

1 1

2 2 1

V w

V w w

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Example

• Objective:

• Constraints:

6,21,26,11,21,1 ))(max MMf

2

1, 1, 1 1, , 1, 11

2 2

2, 2, 1 2, , 2, 1, , 1, 21 1

2 2

, , ,1 1

, 1, ,6

, 1, ,6

, 1, ,6

t t j t tj

t t j t t j t tj j

i j i j t ti j

M M Q S V t

M M Q S Q S V t

Q D t

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Example

• Variable limits:

, , ,

,

,

0 , 1,2, 1,2, 1, ,6

0 , 1,2, 1, ,6

0 , 1,2, 1, ,6

i j t i j

i t

i t i

Q Q i j t

S i t

M M i t

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GAMS

Setsi power stations

/Degerforsen, Edensforsen/j segments /segment1*segment2/t time /hour1*hour6/

;

ParametersMmax(i) maximum contents

/Degerforsen 5e6 Edensforsen 4e6/

lambdaf future price /185/w(i) mean flow

/Degerforsen 163 Edensforsen 164/

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GAMS

D(t) contracted load

/hour1 90, hour2 98, hour3 104,

hour4 112, hour5 100, hour6 80/

Qmax(i,j) maximum discharge station i segment j

/Degerforsen.segment1 225,

Degerforsen.segment2 75,

Edensforsen.segment1 202.5,

Degerforsen.segment2 67.5/

mu(i,j) prod equivalent station i segmen j

/Degerforsen.segment1 0.209,

Degerforsen.segment2 0.199

Edensforsen.segment1 0.236

Degerforsen.segment2 0.224/

V(i) local inflow

Mstart(i) start contents of reservoir i

;

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GAMS

Mmax(i) = Mmax(i)/3600;

V(i) = w(i)-w(i-1);

Mstart(i) = 0.5*Mmax(i);

Positive variables

Q(i,j,t) discharge station i, segment j, hour t

S(i,t) spillage station i, hour t

M(i,t) reservoir contents station i, hour t

;

Free variable

z objective value of stored water

;

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GAMS equation types

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GAMSEquation

objfnc objective function;

objfnc.. z =e= lambdaf*((mu(”Degerforsen”,”segment1”) +

mu(”Edensforsen”,”segment1”))*M(”Degerforsen”,”hour6”) + mu(”Edensforsen”,”segment1”)*M(”Edensforsen”,”hour6”))

;

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GAMSEquations

hydbal(i,t) Hydrological balance constraints

loadbal(t) Load balance constraints

;

hydbal(i,t).. M(i,t) =e= M(i,t-1) + Mstart(i)$(ord(t)=1) – sum(j,Q(i,j,t))

– S(i,t) + sum(j,Q(i-1,j,t)) – S(i-1,t) + V(i);

loadbal(t).. sum((i,j),mu(i,j)*Q(i,j,t)) =e= D(t);

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GAMSQ.up(i,j,”hour1”) = Qmax(i,j);

M.up(i,”hour1”) = Mmax(i);

etc...

Alternatively:

loop(t,Q.up(i,j,t) = Qmax(i,j);

M.up(i,t) = Mmax(i));

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GAMSmodel hydroplanning /all/;

solve hydroplanning using lp maximizing z;

Structuring output:

Parameters

Qtot(i,t) total discharge per station and hour

H(i,t) power production per station and hour

;

loop((i,t), Qtot(i,t) = sum(j,Q.L(i,j,t));

H(i,t) = sum(j,mu(i,j)*Q.L(i,j,t)));

display M.L, Q.L, Qtot, H, S.L;

display hydbal.M, loadbal.M;

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Discharge plan