t t t Sample Clauses

t t t. 1 − q¯t )2 S(0, 1) 2q¯t (q¯t − q¯2 − σ2) − (q¯t − q¯2 − σ2 )2 − (q¯2 + σ2 )2
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t t t t n m  p xy q xy t 1 t 2 x1 y1 and rebalancing side constraint formula, for each Reference Tariff: n m t t2  p xy q xy 1  CPI 1  X 1  A 1  PT 1  0.1  x1 y1
t t t. The expectation value of an observable Oˆ is computed as a trace of the state times the observable: Σi
t t t. 0 := X0. Hence, i+1 = 2 i + 1. That is, n > 2n. This shows the exponential dependency of the unification time on the length of the structures. In this example the growth of the argument lengths is caused by duplication of subterms. As a matter of fact, the same check is repeated many times. Something that could be avoided by sharing various instances of the same structure. In the literature one can find linear algorithms but they are sometimes quite elaborate. On the other hand, Prolog systems usually “solve” the problem simply by omitting the occur-check during unification. Roughly speaking such an approach corresponds to a solved form algorithm where case 5a–b is replaced by: 1The size of a term is the total number of constant, variable and functor occurrences in t.
t t t. ¯T ¯ —1 ¯ t t t t t t γ t
t t t. Otherwise, compute Ki,j = eˆ(fi,j H1(IDj), saP2) · eˆ(sH1(IDi), M ) if ci,j = 0, else compute K i,j t i,j = eˆ(f t H1(IDj), sbP2) · eˆ(sH1(IDi), M ), and accept the session. t i,j i,j , Π – Reveal(Πt ): Algorithm A maintains a list L with tuples of the form (IDi, IDj, Xi, Yj, Kt i,j ). The algorithm A proceeds in the following way to respond: i,j • Get the tuple of oracle Πt from Ω. i,j • If Πt has not accepted, return ⊥. x,x x,x a,b • If the Test(Πw ) query has been issued and if Πt = Πw , or XXx = IDj and IDb = IDj and two oracles have the same session ID, then disallow the query (this should not happen if the adversary obey the rules of the game).
t t t the ceiling is binding, part of the emission flow is abated and pt < cr . The derivative of (t with respect to time cannot be signed for ID, is zero for SD and negative for DD. t r r e a 0 Defineθ( as the time at which the above price p = c so thatθ = ρ −1[log(c − c − ζc ) − logλ ] . Let τ( be the time at which pt = pet , if such a time exists and is unique over [0,∞). If two such instants of time exist, we denote them byτ1 and τ 2 , with τ1 < τ 2 . When the ceiling is not binding but will in the future, Zt < Z and by (5), νt = 0 so that from (10) the absolute value of μt must grow at the rate ρ + α. From (6), γ et = 0 . Then for λ0 ∈[0, cr − ce ] and t μ0 ∈(− [cr − (ce + λ0 )] / ζ , 0), define pˆt as the corresponding marginal cost of coal where from (2), we have pˆt = ce + λ0eρt − ζμe(ρ +α )t . Let xˆ be the extraction rate for coal at which the marginal utility equals this marginal cost, i.e., that is, xˆt is the solution to u1 (x,t ) = pˆt . Thus xˆt is the optimal extraction rate of coal if at time t the ceiling is not binding, has never been earlier, but will be in the future. The time derivative of DD. xˆt cannot be signed for ID but is negative for SD and Denote byθˆ the time at which pˆ = c . Then θˆ = ρ −1[log(c − c ) − log(λ − ζμ )] . Define Zˆ as t r r e 0 0 t the pollution stock generated by xˆt , starting from Z0, so that ∂Zˆt / ∂t = ζ xˆt − α Zˆt and Zˆ0 = Z0 . For any λ0 ∈[0, pe0 − ce ] and μ0 ∈ (− [pe0 − (ce + λ0 )] / ζ , 0), letτˆ be the time at which pˆt = pet . Lastly, for any λ0 ∈[0, pe0 − ce ] and μ0 ∈(− [cr − (ce + λ0 )] / ζ , 0), denote by δ the time at which pˆt = pt . Under the regularity assumptions, δ is well-defined and unique.
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t t t wS ≥ wT for i = t + 1, t + 2, ..., maxT .
t t t. To obtain the optimiser, we notice that the drift rate is a quadratic function in risky asset πtSt. This yields the maximised risky position πˆtSt µJx = − σ2J
t t t. ≈ X0,x X0,x X0,x + U1,k X0,x x=0 Xx ⊗ Xx X0,x X0,x = X0,x .X0,x + . R Σ R Σ T U 1,k U (Ai ⊗ Xx)X0,xX X,x T V 1,k Σ
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