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Hedging Strategy and Electricity Contract Engineering - IFOR

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ö<br />

Ü<br />

6.6 Power portfolio optimization with CVaR 159<br />

<strong>and</strong> vice versa, a linear penalty function could be introduced to the loss function<br />

- Û V end V Kã jÝ , where<br />

-<br />

is a positive constant, representing the value of water.<br />

Another possibility to avoid that some scenarios leave a low amount of water<br />

for coming periods is to sharpen the constraint on the end-water level <strong>and</strong> state<br />

that V K j V ã end j 1Ü Ü Ü J where every trajectory of the water level, V Kã j<br />

has to exceed the threshold V<br />

á?á@á<br />

end . The threshold V end can again be determined<br />

by a one-year optimization as described for the static case.<br />

6.6.7. Size of dynamic problem<br />

Compared to the static model, the dynamic model will typically have fewer<br />

variables describing the dispatch. We here only need one weighting factor per<br />

exercise function, which will give us r variables, compared to twice ô the<br />

number of periods in the static case. Normally the number of periods will exceed<br />

the number of exercise functions, why one could expect this optimization<br />

to be faster. However, whereas the constraints on the water level in the static<br />

case were scenario independent, in the dynamic case they are not, why the size<br />

of the dynamic problem is typically substantially larger. To get an idea on the<br />

size of the problem we again study the A matrix. The size of A is given by<br />

A<br />

2K J J 4 r n 2K J ë ë<br />

K J ë<br />

1<br />

. For large problems, i. e. many periods<br />

¢ ë ÿ ë ÿ ë ë ì<br />

<strong>and</strong> many scenarios, the number of variables will essentially be given by K J<br />

<strong>and</strong> the number of constraints by 2K J <strong>and</strong> we can immediately conclude that<br />

the size of the dynamic problem will then be substantial. ¢ The density of the<br />

matrix, defined as the percentage non-zero elements is given by<br />

Û 2Û 3Ý ôkÝ<br />

Û 4Ý?Û J ô 1Ý<br />

K 2 r K 4r n J 4r 1<br />

2K J J r n 2K J K<br />

<strong>and</strong> will for large problems be determined by the ratio<br />

K 2 J<br />

2K 2 J 2 1<br />

2J á<br />

Even though the size of the matrix grows fast for large problems, the density<br />

will obviously be fairly low.

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