Evaluating the value of storage facilities for buffering and arbitrage (Joint work with Lisa Flatley, Richard Gibbens, Stan Zachary) James Cruise Maxwell Institute for Mathematical Sciences Edinburgh and Heriot-Watt Universities 26 February 2015
The Problem: Storage facilities are expensive, high capital cost. To facilitate investment we need to understand storage value. Need to also be able to compare to alternatives, for example demand side management. Also need to understand the effect of multiple competing stores (see Lisa’s talk yesterday). Plan: 1 Simple model for a single store and economic environment for studying arbitrage value. 2 Consequences of the model. 3 Extensions to include stochastic prices and buffering.
Model P E E = size of store — capacity constraint P = max input/output rate — rate constraint
Cost function At any ( discrete ) time t , C ( x ) t buy P P x sell C t ( x ) = cost of increasing level of store by x (positive or negative) Assume convex (reasonable). This may model market impact efficiency of store rate constraints
Problem Let S t = level of store at time t , 0 ≤ t ≤ T . Policy S = ( S 0 , . . . , S T ), S 0 = S ∗ 0 (fixed), S T = S ∗ T (fixed). Define also x t ( S ) = S t − S t − 1 ( energy “bought” by store at time t – positive or negative) Problem: minimise cost T � C t ( x t ( S )) t =1 subject to S 0 = S ∗ S T = S ∗ 0 , T and 0 ≤ S t ≤ E , 1 ≤ t ≤ T − 1 .
Small store This is a store whose activities are not so great as to impact upon the market , and which thus has linear buy and sell prices. Thus, for all t , � c ( b ) x if 0 ≤ x ≤ P t C t ( x ) = c ( s ) x if − P ≤ x < 0 t where 0 < c ( s ) ≤ c ( b ) and P is rate constraint. t t Characteristics Optimal control is bang-bang : at each time buy as much as possible, do nothing , or sell as much as possible. If E = ∞ ( no capacity constraint ) then optimal solution is global in time. If P = ∞ ( no rate constraint ) then optimal solution is very local in time.
Example: Periodic Cost functions Consider sinusoidal prices. Interested in what happens as frequency is varied. Assume P = 1. Then for a given value of E , there exists a pair µ b < µ s such that we buy if c t < µ b and sell if c t > µ s . Further µ b is increasing in E and µ s is decreasing in E . Also µ b is increasing and µ s is decreasing in frequency. These are bounded by µ ∗ b and µ ∗ s , the parameters obtained for E = ∞ which does not depend on frequency. For a given E , as frequency increases profit increases upto the unconstrained case, and there is a frequency beyond which you obtain no further benefit.
Example: Real prices with Dinorwig parameters E/P = 5 hrs Efficiency = 0 . 85 (ratio of sell to buy price). Solution is bang-bang : red points buy, blue points sell 10000 8000 Prices (£/MWh) 6000 4000 2000 10.0 7.5 Storage level 5.0 2.5 0.0 Sun Mon Tue Wed Thu Fri Sat Sun 09−Jan 10−Jan 11−Jan 12−Jan 13−Jan 14−Jan 15−Jan 16−Jan
Example: Real prices with Dinorwig parameters E/P = 5 hrs Efficiency = 0 . 65 (ratio of sell to buy price). Solution is bang-bang : red points buy, blue points sell 10000 8000 Prices (£/MWh) 6000 4000 2000 10.0 7.5 Storage level 5.0 2.5 0.0 Sun Mon Tue Wed Thu Fri Sat Sun 09−Jan 10−Jan 11−Jan 12−Jan 13−Jan 14−Jan 15−Jan 16−Jan
General Result: Lagrangian sufficiency Suppose there exists a vector µ ∗ = ( µ ∗ 1 , . . . , µ ∗ T ) and a value S ∗ = ( S ∗ 0 , . . . , S ∗ T ) of S such that S ∗ is feasible , x t ( S ∗ ) minimises C t ( x ) − µ ∗ t x over all x , 1 ≤ t ≤ T , the pair ( S ∗ , µ ∗ ) satisfies the complementary slackness conditions , for 1 ≤ t ≤ T − 1, µ ∗ t +1 = µ ∗ if 0 < S ∗ t < E , t µ ∗ t +1 ≤ µ ∗ if S ∗ t = 0, (1) t µ ∗ t +1 ≥ µ ∗ if S ∗ t = E . t Then S ∗ solves the stated problem.
Comment The above result (essentially an application of the Lagrangian Sufficiency Principle) does not require convexity of the functions C t . However, convexity the functions C t is sufficient to guarantee the existence of a pair ( S ∗ , µ ∗ ) as above. The latter result is a standard application of Strong Lagrangian theory (i.e. the Supporting Hyperplane Theorem ).
Algorithm We need to identify the relevant value of µ ∗ t at each time t . It is important to note that the value of µ ∗ t only changes at times when the store is full or empty. Further µ ∗ t acts as a reference level since the optimal action at time t is given by the x which minimizes: C t ( x ) − µ ∗ t x . We can use this to carry at a search for µ ∗ . Further this method is local in time, as we can ignore times after filling or emptying the store.
Example Again consider real price data but incorporate market impact by the store. Use a quadratic cost function: � p t x (1 + λ x ) if 0 ≤ x ≤ P C t ( x ) = − p t ν x (1 + λν x ) if − P ≤ x < 0 p t is the historic price. λ is a measure of market impact by the store. ν is the round trip efficiency of the store. P is the power constraint.
E = 10 , P = 1 , η = 0.8 , λ = 0.05 E = 10 , P = 1 , η = 0.6 , λ = 0.05 10 10 8 8 Store level Store level 6 6 4 4 2 2 0 0 20 20 Lookahead time (days) Lookahead time (days) 15 15 10 10 5 5 0 0 01/Dec 08/Dec 15/Dec 22/Dec 29/Dec 01/Dec 08/Dec 15/Dec 22/Dec 29/Dec E = 10 , P = 1 , η = 0.8 , λ = 0.5 E = 10 , P = 0.25 , η = 0.8 , λ = 0.05 10 10 8 8 Store level Store level 6 6 4 4 2 2 0 0 20 20 Lookahead time (days) Lookahead time (days) 15 15 10 10 5 5 0 0 01/Dec 08/Dec 15/Dec 22/Dec 29/Dec 01/Dec 08/Dec 15/Dec 22/Dec 29/Dec
Market impact of storage A large store may impact costs (be a price-maker ), and hence the rest of society .
Market impact of storage A large store may impact costs (be a price-maker ), and hence the rest of society . Impact of storage on consumer surplus is in general beneficial , but not necessarily .
Market impact of storage A large store may impact costs (be a price-maker ), and hence the rest of society . Impact of storage on consumer surplus is in general beneficial , but not necessarily . Example ( T = 2 ): Buy x at time 1 (increasing price by p 1 ) and sell at time 2 (decreasing price by p 2 ). Increase in consumer surplus is p 2 d 2 − p 1 d 1 where d 1 and d 2 are the respective demands at times 1 and 2. In general expect d 2 > d 1 and p 2 to be comparable to p 1 , so effect on consumer surplus is beneficial . However, the latter need not be the case.
Optimisation of value of storage Suppose it is desired to maximise store profit plus consumer surplus .
Optimisation of value of storage Suppose it is desired to maximise store profit plus consumer surplus . Then it is straightforward to modify the cost function at each time t so as to reflect total costs: for each time t , the cost c t ( x ) is now the cost to the store of buying x units (positive or negative) plus the cost to society (reduction in consumer surplus) of doing so.
Stochastic Prices Suppose that the cost functions C t evolve randomly in time, and that C t = ξ t ¯ C t , 1 ≤ t ≤ T , where (¯ C 1 , . . . , ¯ C T ) is a sequence of deterministic cost functions and where ( ξ 1 , . . . , ξ T ) is a sequence of strictly positive real-valued random variables such that E ( ξ t | F t − 1 ) = ξ t − 1 , 1 ≤ t ≤ T , and where the deterministic functions ¯ C t are assumed to satisfy the same conditions as the cost functions C t of the deterministic problem. Then the optimal control (cost minimisation) strategy remains exactly as previously , with the optimal sequence of store levels as given in the case where stochastic cost functions C t are replaced by their deterministic counterparts ¯ C t .
Integration of buffering for uncertainty Suppose that the store is also used to provide buffering against relatively rare unexpected events. Then the previous optimization problem may be rewritten as: Problem: minimise T � � � C t ( x t ( S )) + A t ( S t ) t =1 subject to S 0 = S ∗ S T = S ∗ 0 , T and 0 ≤ S t ≤ E , 1 ≤ t ≤ T − 1 . Here the functions A t , such that A t ( S t ) is the cost (negative benefit) of having a level of reserve S t in the store at time t , are independently calculable .
Result - Lagrangian sufficiency Assume, for simplicity, we incorporate the capacity constraints into the functions A t , and the functions C t and A t are differentiable . Suppose there exists a scalar ν ∗ and a value S ∗ = ( S ∗ 0 , . . . , S ∗ T ) (with S ∗ 0 and S ∗ T having their required values) such that T � C ′ A ′ t ( x t ) = ν − u ( S u ) =: µ t u = t Then S ∗ solves the stated problem.
Example E/P = 5 hrs Efficiency = 0 . 85 (ratio of sell to buy price). A t ( S ) = ν/ S (Black: ν = 0 . 02,Red: ν = 0 . 2, Blue: ν = 1) 10 8 Store level 6 4 2 0 Time (March 2011)
Functions A t The function A t is represents the extra cost associated with random deviation. In many cases we can either determine this a priori or use a suitable approximation. Examples: Random deviations are corrected immediately. Failure to be able to provide the service, leads to penalty. Cases where the probability of the buffering service being called upon is small.
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