Hedging Strategy and Electricity Contract Engineering - IFOR

162 Power portfolio optimization
dispatch if the risk constraint (6.54) is inactive. In this case the problem is
equivalent to a pure expected profit maximization and hedging loses its validity,
since risk is not of interest. The only value of the operational flexibility then
comes from the ability to take advantage of the volatile prices and produce at
high prices and to take advantage of the price structure by pumping at low cost
in off-peak periods, store the water and produce electricity in peak periods.
An inflexible plant such as the nuclear plant, producing over long periods,
would produce at the average price of S. We hence believe that S h S is an
appropriate assessment of operational flexibility in a pure profit maximization.
As soon as risk is of concern the actual value of this flexibility is however
larger than S h S. The reason for this is that the flexible hydro storage
plant will allow us to take more risky and hence more profitable positions
in the contract portfolio, since additional risk can be off-set by the hedging
capabilities of the stored water. This hedging value of the stored water can be
motivated by studying the dual problem.
A linear program, such as (6.53)-(6.62) written on its standard primal form
z max c T x
s.t. Ax b,
(6.65)
where x n , A
n m , c n and b m , has an interesting corresponding
dual problem, introduced already in Chapter 6.5, given by
min b T y
s.t. A T y c
y 0.
(6.66)
The dual variables y
m are connected to the primal problem in the sense
that the ith dual variable y i corresponds to the ith constraint Ax b i
.
Given that the primal problem (6.65) has a non-degenerate solution, the ith
dual variable, being the solution of (6.66) yúi
will equal the marginal value of
the ’resource’ b i . For more information on linear programming in general,
see [Chv83]. The interesting resource in this case is the stored water. The
available water can be manipulated by adjusting the right hand side of the
sixth constraint (6.59) stating that V K V end . Given that V max V end one
unit decrease in V end would allow us to utilize one more unit of water during