ifference between revisions of "EMK:Simplifying River Chains for Dynamic Programming"

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Equations 1 and 4 then give the storages and flows in the original sub-system.
 
Equations 1 and 4 then give the storages and flows in the original sub-system.
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[[File:SRC Fig 3.jpg|140px|thumb|right|Figure 3]]
 +
If M<sub>A</sub> = 0 for the system, then Equations 16 and 17 will not solve.  This will occur only if there is zero storage in all but the last reservoirs.  It is inappropriate to try and simplify this configuration, unless there are no inflows below the first node, in which case the simplified sub-system is shown in figure 3.
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Where
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18. Max = Max<sub>0</sub>
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 +
19. I = I<sub>0</sub>
 +
 +
20. S = S<sub>0</sub>
  
If M<sub>A</sub> = 0 for the system, then Equations 16 and 17 will not solve.  This will occur only if there is zero storage in all but the last reservoirs.  It is inappropriate to try and simplify this configuration, unless there are no inflows below the first node, in which case the simplified sub-system is shown in figure 3.
+
21. M = &sigma;0
[[File:SRC Fig 3.jpg|140px|thumb|right|Figure 3]]
 

Revision as of 09:31, 4 December 2012

Introduction

If too many reservoirs are modelled in the dynamic programming phase of water value calculation, a performance hit may occur. This is the well known ‘problem of dimensionality’ in dynamic programming. It might be necessary to simplify the problem to avoid slow DP (Dynamic Programming) solutions. The following is a method for doing this on a chain of stations.

River Chain Simplification

Figure 1

Figure 1 represents a chain of reservoirs and stations, with the nodes represented by circles.

Si = the storage at node i Ii = the inflows to node i Maxi = the storage capacity at node i fi = the flow from node i to node i + 1 fIN = the flow into node 0 Mi = the generating potential from node i to node i + 1 (MW/cumec)

Instead of modelling each and every reservoir separately on this chain we will apply a constraint which requires all reservoirs to be filled to the same proportion of their capacity. So at any time:

1. Si = α Maxi, for some alpha.

The behaviour of this chain is then determined by the inflows, fIN and the change in α. Change in storage ΔSi is given by change in α.

2. ΔSi = Lamda.jpg Maxi , Lamda.jpg = change in α

Change in storage is also related to flows and inflows:

3. ΔS0 = I0 + fIN – f0

ΔSi = Ii + fi – fi +1 , i > 0

Combining 2. and 3. Gives:

4. SRC E 1.jpg

And Generation is given by

5. SRC E 2.jpg

Defining

6. SRC E 3.jpg

Equation 5 becomes:

7. SRC E 4.jpg

Figure 2

We now have the dynamics of the constrained sub-system of stations. The next step is to represent these dynamics with a simpler sub-system so the DP process can model it. The simpler system is shown in figure 2.

The first node has no storage, so there is only the storage at the second node to model. In order for this sub-system to behave in the same way as the constrained sub-system it is necessary for the generation and outflow to be equal at any time.

In the simplified sub-system flows are given by:

8. fA = fIN + IA

fB = fA + IB - Lamda.jpgMax
= IA + IB + fIN - Lamda.jpgMax

Since fB must equal fn for all inflow, fIN and Lamda.jpg values, comparing Equations 8 and 4 gives:

9. SRC E5.jpg ( from the inflows components )

10. SRC E6.jpg ( from the Lamda.jpg component )

Generation is given by:

11. G* = MA fA + MB fB

= MA IA + MB (IA + IB) + (MA + MB) fIN - MB Lamda.jpgMax

Since G* must equal G for all inflow, fIN and Lamda.jpg values comparing Equations 11 and 7 gives:

12. MA + MB = σ0 ( from the fIN component )

13. SRC E 7.jpg ( from the Lamda.jpg component )

14. SRC E 8.jpg ( from the inflows components )

Combining 12 and 13 gives:

15. SRC E 9.jpg

Combining 14 and 9 gives:

16. SRC E 10.jpg

17. SRC E 11.jpg

Equations 10,13,15 enable a specification of the simplified sub-system to be calculated. Equations 16 and 17 find the inflow figures to use in the system.

Once the results of a simulation with the simplified model are known they can be converted back to results for the original sub-system by calculating Lamda.jpg and α:

Lamda.jpg = ( fA + IA - fB ) / Max

α = S / Max

Equations 1 and 4 then give the storages and flows in the original sub-system.

Figure 3

If MA = 0 for the system, then Equations 16 and 17 will not solve. This will occur only if there is zero storage in all but the last reservoirs. It is inappropriate to try and simplify this configuration, unless there are no inflows below the first node, in which case the simplified sub-system is shown in figure 3.

Where

18. Max = Max0

19. I = I0

20. S = S0

21. M = σ0