1. 1 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

2. 2 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

3. 3 Dual variables x 2 x 1 1 2 3 4 1234 x 1 +x 2  2 x 2  0 4x 1 +12x 2  12 x 1  0 z = 20 z = 30 z = 40 1 2 x 1 +x 2  7 3 In optimum: 1 > 0 (active) 2 > 0 (active) 3 = 0 (not active)

4. 4 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 1 2 The dual variables corresponds the water value [SEK/HE].

5. 5 Dual variables Contracted load constraints for hour t: –Constraints have unit MWh –Objective function has unit SEK –Increase of D k  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].

6. 6 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,...,6 42.6585 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? 1 2

7. 7 Home assignment 2 Short term hydro-thermal planning Solved by using the software GAMS (www.gams.com)

8. 8 GAMS Software specialized for solving optimization problems. Typical structure of GAMS program: 1.Define sets 2.Define parameters 3.Declare optimization variables 4.Declare equations 5.Define equations 6.Define variable limits

9. 9 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

10. 10 Example Power plant Maximal active storage of reservoir [m 3 ] Marginal production equivalents [MWh/HE] Mean flow, w i [m 3 /s] Deger- forsen 5*10 6 0.209, 0  Q  225 HE 0.199, 225  Q  300 163 Edens- forsen 4*10 6 0.236, 0  Q  202.5 HE 0.224, 202.5  Q  270 164

11. 11 Example Known parameters: Optimization variables: = discharge, station i, segment j, during hour t, i=1,2, j=1,2, t=1,...,6 = spillage, station i, during hour t, i=1,2, j=1,2, t=1,...,6 = reservoir contents, station i, end of hour t, i=1,2, j=1,2, t=1,...,6 Calculated parameters:

12. 12 Example Objective: Constraints:

13. 13 Example Variable limits:

14. 14 GAMS Sets i power stations /Degerforsen, Edensforsen/ j segments /segment1*segment2/ t time /hour1*hour6/ ; Parameters Mmax(i)maximum contents /Degerforsen5e6 Edensforsen4e6/ lambdaffuture price /185/ w(i)mean flow /Degerforsen163 Edensforsen164/

15. 15 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 ;

16. 16 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 zobjective value of stored water ;

17. 17 GAMS equation types

18. 18 GAMS Equation objfncobjective function; objfnc..z =e= lambdaf*((mu(”Degerforsen”,”segment1”) + mu(”Edensforsen”,”segment1”))*M(”Degerforsen”, ”hour6”) + mu(”Edensforsen”,”segment1”)* M(”Edensforsen”,”hour6”)) ;

19. 19 GAMS Equations 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);

20. 20

21. 21 GAMS Q.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));

22. 22 GAMS model 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;

23. 23 Discharge plan