{"id":"179a1629-b74a-490e-bb45-7ea3629b9fa7","arxiv_id":"2411.17463","paper_version":2,"verdict":"REJECT","confidence":"HIGH","novelty_score":6.0,"correctness_risk":"high","formal_verification":"none","parameter_count":0,"one_line_summary":"The claimed necessary and sufficient condition for forecast horizons in energy storage scheduling is shown to be insufficient under the paper's formal definition, via a simple two-period counterexample.","lead":"The paper proposes a condition for deciding when a finite planning horizon is long enough to certificate optimal energy storage decisions, plus an algorithm to find the shortest such horizon. A concrete counterexample shows the main theorem does not hold under the paper's own formal definition, so the certificate can flag a horizon that still yields suboptimal rolling-horizon schedules.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Theorem 1 is false under Definition 1: the appendix proves only that one optimal finite-horizon schedule extends to the infinite horizon, but Definition 1 requires every such schedule to extend; a two-period example satisfies the theorem's condition yet is not a forecast horizon.","rationale":"The reader's core objection is correct and matches my analysis: Theorem 1's sufficient direction requires all optimal decision-horizon schedules of the truncated problem to be extendable to an infinite-horizon optimum, while the appendix establishes only that one such schedule is extendable. The provided two-period counterexample demonstrates the distinction concretely: the extreme terminal problems both admit an optimum with s_1=5, and the free problem has s_1=0 as an optimum, but s_1=0 cannot be part of an infinite-horizon optimum when the tail prices are high because the storage must arrive full at the tail. This invalidates the paper's main advertised condition.\n\nThe paper is otherwise clearly written, and the idea of certifying forecast horizons through extreme terminal-state problems is sensible. The failure is specifically the mismatch between the universal set-inclusion definition (Definition 1) and the existence-style reasoning in Appendix A. A repair would require either weakening the definition to the existential form used in much of the forecast-horizon literature, or adding a tie-breaking mechanism that selects the finite-horizon optimum that is safe across all tails and then proving that the extreme-state condition guarantees such a selection. Until that is done, the main theorem, the algorithm built on it, and the case-study conclusions that assert minimum forecast horizons are not supported. I therefore concur with the reader's REJECT verdict, with confidence that the concern is load-bearing and not a matter of taste.","tokens_in":25686,"tokens_out":8940,"duration_ms":86581,"concrete_test":"Run the analytic check with S=0, S̄=10, S_init=5, P^C=P^D=5, Δt=1, η_C=η_D=ρ=1, H=1, T=2, C_1=C_2=10, and tail C_3=C_4=1000, C_t=0 for t≥5. Enumerate the decision-horizon state s_1 over all optimal solutions: for X_H(S(2,Ĉ)) the set is [0,5]; for X_H(S(N+,C)), because the tail forces maximum storage at t=2, the set is [5,10]. Confirm s_1=0 is in the first but not the second, while the extreme problems F(2,C,0) and F(2,C,10) both have an optimum with s_1=5. This directly refutes the sufficiency direction of Theorem 1. If the reproduction instead depends on a different notion of infinite-horizon optimality, repeat the check with a truncated tail and use the paper's own Definition 1 to adjudicate.","verdict_should_be":"REJECT","load_bearing_attack":"The central claim is Theorem 1: the planning horizon T is a forecast horizon if and only if there exist optimal solutions x of F(T,C,S_T) and x̄ of F(T,C,S̄_T) with s_H = s̄_H. Under the paper's own Definition 1, a forecast horizon requires the set inclusion X_H(S(T,Ĉ)) ⊆ X_H(S(N+,C)) for every tail C extending Ĉ, i.e. every optimal finite-horizon schedule over H must also be optimal for the infinite-horizon extension.\n\nThe proof in Appendix A does not establish this universal property. Lemma 1 and Corollary 1 construct, for each intermediate terminal level S_end, one optimal solution x* whose H-states lie between, and eventually equal, the common extreme schedule. The induction then shows that this particular schedule extends to the infinite horizon. It never shows that every optimal solution of the free problem S(T,Ĉ) has those H-states.\n\nThis gap is not merely technical; the claimed equivalence is false. Take S=0, S̄=10, S_init=5, P^C=P^D=5, Δt=1, η_C=η_D=ρ=1, H=1, T=2, C_1=C_2=10. Then S_T=0 and S̄_T=10. In F(2,C,0), all optima have s_2=0 and s_1∈[0,5]; in F(2,C,10), all optima have s_2=10 and s_1∈[5,10]. Hence s_H=s̄_H=5 is attainable, so Theorem 1 declares T=2 a forecast horizon. But in the free problem S(2,C), profit equals 10(S_init−s_2), so every optimum has s_2=0 and thus s_1∈[0,5]; in particular s_1=0 is optimal. Now choose a tail with C_3=C_4=1000 and C_t=0 for t≥5 (so the infinite-horizon objective is well defined). Any infinite-horizon optimum must arrive at t=2 with maximum storage to sell at the high tail price, which requires s_1≥5. Therefore s_1=0 is in X_H(S(2,Ĉ)) but not in X_H(S(N+,C)). The condition s_H=s̄_H holds, yet T=2 is not a forecast horizon under Definition 1.","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"The manuscript studies the rolling-horizon approximation of an infinite-horizon energy storage arbitrage problem. It defines a planning horizon T to be a forecast horizon when every optimal schedule of the finite problem over the decision horizon remains optimal for every extension of the price forecast beyond T (Definition 1). The central result, Theorem 1 in Section 3.1, asserts that T is a forecast horizon if and only if the two finite problems with terminal state of energy fixed to the minimum and maximum reachable levels, F(T,C,S_T) and F(T,C,\\bar{S}_T), have optimal solutions whose state of energy at the end of the decision horizon coincides. The paper also derives a suboptimality bound, a necessary condition and lower bound on the minimum forecast horizon, an algorithm to determine the minimum forecast horizon, and numerical case studies.","tokens_in":26008,"tokens_out":10953,"duration_ms":100895,"significance":"If Theorem 1 were correct, it would provide a practically attractive and easy-to-check certificate for planning-horizon selection, and the proposed algorithm and suboptimality bound would be useful for storage operators. The paper is clearly written, addresses an important gap in the literature, and the numerical study is accompanied by a reproducibility link. The non-existence example in Section 3.3 and the lower bound in Proposition 3 are interesting. However, the central equivalence is false: the condition in Theorem 1 is not sufficient for Definition 1. Because the main theoretical claim fails under the paper's own definitions, the principal contributions do not currently stand.","major_comments":[{"comment":"The sufficiency direction of Theorem 1 is false. Consider the admissible parameters S=0, \\bar{S}=10, S^{\\rm init}=5, P^C=P^D=5, \\Delta t=1, \\eta_C=\\eta_D=\\rho=1, H=1, T=2, C_1=C_2=10, and a tail with C_3=C_4=1000 and C_t=0 for t\\ge 5. In F(2,C,0) every optimum has s_1\\in[0,5]; in F(2,C,10) every optimum has s_1\\in[5,10]. Hence the condition of Theorem 1 holds, with s_1=5. But the free problem S(2,C) also has optimal H-schedules with s_1=0, for example discharging 5 in period 1 and doing nothing in period 2. Such an H-schedule is not optimal in the infinite-horizon extension: charging 5 in period 1 and discharging 10 at price 1000 in period 3 yields profit 9950, whereas any schedule starting with s_1=0 earns at most 5000. Thus X_H(S(T,\\hat{C}))\\not\\subseteq X_H(S(\\mathbb{N}^+,C)) for this admissible tail, so T=2 is not a forecast horizon under Definition 1, contradicting Theorem 1.","section":"§3.1"},{"comment":"The proof of sufficiency proves the wrong quantifier. Lemma 1 and Corollary 1 construct, for each intermediate terminal level S^{\\rm end}, one optimal solution x^* whose H-states lie between those of the extreme solutions, and the induction then takes one common schedule and extends it to the infinite horizon. This establishes that there exists an optimal H-schedule of S(T,\\hat{C}) that is optimal for the infinite-horizon problem. Definition 1 requires that every optimal H-schedule of S(T,\\hat{C}) is optimal for the infinite-horizon problem. The induction never visits the other optimal solutions of the free problem; in the counterexample above, the constructed common schedule is s_1=5 while the free problem also has the optimal schedule s_1=0. The proof therefore cannot bridge the gap between Theorem 1 and Definition 1.","section":"Appendix A"},{"comment":"Because Algorithm 1 stops when the Theorem 1 condition reports gap=0 and then sets subopt=0, it can terminate declaring a forecast horizon in situations where none exists. The reported minimum forecast horizons in Section 4.2 and the profit comparisons in Section 4.3 are therefore not supported as evidence for the paper's claims. This is a direct consequence of the counterexample, not a separate implementation issue.","section":"§3.5"},{"comment":"The final sentence of Appendix A, 'if T is a forecast horizon, by definition, ∃x∈X and ∃\\bar{x}\\in\\bar{X} such that s_H=\\bar{s}_H', does not follow from Definition 1. Definition 1 quantifies over all tails and asserts a set inclusion; it does not assert that the two extreme terminal problems share a decision-horizon state. Some argument is needed, for example constructing tails that make the minimum and maximum terminal states optimal simultaneously, and none is supplied. Thus the 'only if' direction is also unproved as written.","section":"Appendix A"}],"minor_comments":[{"comment":"Definition 1 uses X_H(S(\\mathbb{N}^+,C)), but the notation X_H was introduced only for finite problems; the restriction of an infinite-horizon solution to the decision horizon H should be defined explicitly.","section":"§2.2"},{"comment":"Algorithm 1 uses M as the initial value of gap and subopt, but M is never defined; the stopping criterion 'gap>0' should also be stated with an explicit tolerance because floating-point solvers will not produce exact zeros.","section":"§3.5"},{"comment":"The notation C is overloaded: it denotes the price vector, the set of possible price vectors, and the upper bound C in Proposition 2, while the lower bound C is visually almost identical; please disambiguate.","section":"§3.2"},{"comment":"In the sentence 'with ηC_1 < C_t < C_1', the expression ηC_1 is ambiguous: it should be η_C C_1. Reading this passage requires the reader to infer the intended meaning.","section":"§3.3"}],"recommendation":"reject","confidential_remarks":"I verified the stress-test counterexample by hand; it is decisive and falls entirely within the model assumptions. The manuscript's main theorem is false under its own Definition 1, and the Algorithm 1 and numerical results inherit the error. I therefore recommend rejection. A revised version would need either a different forecast-horizon definition (e.g., an existential rather than universal guarantee), which would no longer support the rolling-horizon claim, or a substantially different proof and condition."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"The main theorem is false as stated. The proof in Appendix A constructs, for each terminal level, one optimal schedule whose H-states are pinned down, and shows that this particular schedule extends. It never shows that every optimal schedule of the free problem has those H-states. Definition 1 requires exactly that, so the set inclusion can fail even when the two extreme problems admit optimal solutions with the same s_H. The two-period counterexample is clean: S=0, Sbar=10, S_init=5, P^C=P^D=5, perfect efficiencies, C1=C2=10, tail C3=C4=1000. The extreme problems both allow s_H=5, so Theorem 1 declares T=2 a forecast horizon, but the free problem also has optimal solutions with s_H=0, and those are not optimal against the tail. The condition is not sufficient.\n\nThat said, the paper is not a throwaway. The model is clean, the writing is clear, and the adaptation of the two-extreme-terminal-value idea from production planning to undiscounted storage scheduling with inefficiencies, leakage, and negative prices is a sensible and useful framing. The non-existence example in Section 3.3 is instructive and does not depend on the broken theorem. The lower bound in Proposition 3 and the algorithmic scaffold are also worth engaging with, though both inherit risk from Theorem 1 and should be rechecked. The GitHub code for the case studies is a real plus.\n\nThe soft spots are proportionate to the collapse of the main result. The algorithm and the suboptimality bound are both built on the false equivalence, so they need a tie-breaking or a strengthened condition before they can be trusted. The infinite-horizon optimality notion is also left informal; for arbitrary price sequences the undiscounted sum may not converge, and the paper should say how optimality is defined.\n\nWho this is for: researchers working on rolling-horizon storage scheduling who want a certificate for forecast horizons. The counterexample is a useful caution for anyone tempted by this condition. I would not cite the paper in its current form, and I would not recommend acceptance, but the question is real and the authors are close to something usable. A serious referee should see it; the flaws are fixable in principle, and even a rejection with detailed comments would help the authors and the community.","headline":"The central criterion is wrong as stated: the proof shows only that one optimal finite-horizon schedule extends to the infinite horizon, while Definition 1 requires every such schedule to extend, and a two-period example breaks the equivalence.","tokens_in":26742,"tokens_out":3118,"would_cite":false,"duration_ms":35092,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":[],"pacs":[],"model":"deepseek-v4-flash","headline":"One equality check certifies a long-enough planning horizon","keywords":["energy storage scheduling","rolling horizon","forecast horizon","infinite-horizon optimization","planning horizon","suboptimality bound","arbitrage","storage efficiency"],"falsifier":"A concrete check: set $S=0$, $\\overline{S}=10$, $S_{\\rm init}=5$, $P^C=P^D=5$, $\\eta_C=\\eta_D=1$, $\\rho=1$, $H=1$, $T=2$, $C_1=C_2=10$, and continue with $C_3=C_4=1000$. The two extreme terminal problems $F(2,C,\\underline{S}_2)$ and $F(2,C,\\overline{S}_2)$ both have optimal solutions with $s_1=5$, while the unconstrained finite problem also has an optimal solution with $s_1=0$ that is not optimal for the infinite-horizon tail; constructing this instance would settle whether the 'if' direction of Theorem 1 holds as stated.","tokens_in":25338,"feed_emoji":"🔋","tokens_out":15368,"duration_ms":128429,"temperature":0.7,"pith_summary":"Energy storage operators who use a rolling-horizon scheduler must choose a planning horizon, and the choice is usually arbitrary. This paper claims a simple certificate: solve the finite-horizon scheduling problem twice, forcing the storage to end the planning horizon at its lowest and highest reachable energy levels; if the two solutions agree on the state of energy at the end of the decision horizon, that planning horizon is a forecast horizon, and the condition is also necessary. Building on this, the paper proves an upper bound on the profit lost when the horizon is too short, shows by example that forecast horizons need not exist, derives a lower bound on the minimum forecast horizon from storage parameters alone, and gives an algorithm that finds the minimum forecast horizon. The case studies show that common horizons such as 48 hours are often too short and that a too-short horizon can turn a profitable storage operation into a loss, which is why a checkable condition matters.","feed_headline":"One equality check certifies a long-enough planning horizon","feed_subtitle":"If the lowest- and highest-terminal schedules agree on today's decisions, the planning horizon is long enough.","key_machinery":"The load-bearing object is the pair of extreme terminal-state problems $F(T,C,\\underline{S}_T)$ and $F(T,C,\\overline{S}_T)$, the finite-horizon model with the final state of energy fixed to the minimum and maximum reachable levels $\\underline{S}_T$ and $\\overline{S}_T$. The certificate is the equality of the two optimal states at the end of the decision horizon, $s_H = \\bar{s}_H$. The proof rests on an envelope property: an optimal trajectory ending at the minimum reachable level stays weakly below every optimal trajectory with an intermediate terminal state, and an optimal trajectory ending at the maximum reachable level stays weakly above it; this sandwiching is what lets the common decision-horizon schedule be extended to any longer horizon, and it also drives the suboptimality bound, since a smaller gap between the two terminal states means a smaller worst-case profit loss.","core_discovery":"The central claim is Theorem 1: for the deterministic price-taker storage scheduling model (1a)-(1g), with leakage, charging and discharging efficiencies, and possibly negative prices, a planning horizon $T$ is a forecast horizon for a price forecast $\\hat{C}$ if and only if there exist optimal solutions of $F(T,\\hat{C},\\underline{S}_T)$ and $F(T,\\hat{C},\\overline{S}_T)$ with the same state of energy at the end of the decision horizon $H$, where $\\underline{S}_T$ and $\\overline{S}_T$ are the lowest and highest energy levels that can be reached at the end of the planning horizon. Necessity follows from the definition of forecast horizon; sufficiency is argued through an envelope lemma stating that the minimum-terminal optimum lies weakly below, and the maximum-terminal optimum weakly above, every optimal trajectory with an intermediate terminal state, after which the common decision-horizon schedule is propagated one period at a time to arbitrary future horizons. The paper then uses the certificate as a building block: the gap between the two terminal states bounds suboptimality when the horizon is too short, a necessary condition computable from storage parameters alone gives a starting point, and an iterative algorithm increments the horizon until the certificate holds, returning the minimum forecast horizon or, if a user-set maximum is reached, a bound on the remaining suboptimality.","pith_inferences":["A practical extension would use the certificate online: as price forecasts update, recompute $\\underline{S}_T$ and $\\overline{S}_T$ and lengthen the planning horizon only while the gap implies an unacceptable suboptimality bound, replacing fixed horizons such as 48 hours with a data-driven choice.","In a stochastic setting with a scenario tree, the same two-extreme-problem idea could define a stochastic forecast horizon, with terminal reachable intervals per scenario and the Proposition 2 bound replaced by an expectation over scenarios; the deterministic result is the degenerate case.","Because the full fleet minimum is the maximum of individual minima, the practical bottleneck is the slowest or most lossy unit, and its parameters alone could be used in Proposition 3 as a fleet-level lower bound before any price data arrive.","The non-existence example implies that a rolling-horizon implementation should carry a certified cap: if the certificate has not fired by $T_{\\max}$, the suboptimality bound is the only remaining guarantee, and exceeding the cap should trigger a re-evaluation of whether a finite-horizon policy is appropriate at all."],"forward_implications":["A rolling-horizon operator can certify a chosen planning horizon by solving two finite-horizon optimizations and comparing one state variable, without having to solve the infinite-horizon problem.","When the certificate fails, the gap between the two terminal states gives an explicit upper bound on the profit lost relative to a perfect infinite-horizon policy, so the cost of a short horizon is quantifiable.","Forecast horizons need not exist: with inefficient charge-discharge and a price path satisfying $\\eta C_1 < C_t < C_1$ for every later $t$, no finite planning horizon is long enough.","The minimum forecast horizon varies strongly with storage characteristics and the specific price path; in the case studies it is often longer than 48 hours, and for slow storage with leakage a 24-hour fixed-level policy loses 362% of the profit achievable with a forecast horizon.","Under the decomposability conditions of Section 3.6, the minimum forecast horizon for a problem with several storage units is the maximum of the individual units' minimum forecast horizons."],"supporting_citations":[{"why":"Defines forecast horizons and surveys the operations-management literature the paper extends.","marker":"Chand et al. (2002)"},{"why":"Introduces the idea of comparing extreme terminal values to certify a forecast horizon in finite-horizon problems.","marker":"Bhaskaran and Sethi (1987)"},{"why":"Gives an infinite-horizon forecast-horizon algorithm under discounting that the paper adapts to undiscounted storage scheduling.","marker":"Cheevaprawatdomrong and Smith (2004)"},{"why":"Provides the storage scheduling model and prior planning/decision-horizon algorithm whose finite-horizon assumption the paper relaxes.","marker":"Cruise et al. (2019)"},{"why":"Supplies the iterative horizon-lengthening scheme used in Algorithm 1.","marker":"Garcia and Smith (2000)"},{"why":"Shows how to linearize the complementarity constraint so the two terminal-state problems are solvable.","marker":"Pozo (2022)"},{"why":"Contains the form of the expression used as the necessary condition (6) for a valid planning horizon.","marker":"Ela and O'Malley (2015)"},{"why":"Demonstrates the sufficient condition based on the storage hitting both bounds, which Theorem 1 generalizes to a necessary and sufficient certificate.","marker":"Flatley et al. (2016)"}],"fun_headline_variants":["Two extreme cases agree? Your storage horizon is long enough","One condition certifies a forecast horizon for storage","When do you stop lengthening your storage planning horizon?","A simple equality check proves your storage horizon is optimal","How long is long enough? Check these two extremes"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The load-bearing step is that agreement between the two extreme terminal-state problems forces every optimal schedule with an intermediate end-of-horizon state to agree on the decision horizon, and the proof does not establish this when the finite problem has multiple optimal solutions.","fun_headline_variants_meta":{"raw":{"variants":["Two extreme cases agree? Your storage horizon is long enough","One condition certifies a forecast horizon for storage","When do you stop lengthening your storage planning horizon?","A simple equality check proves your storage horizon is optimal","How long is long enough? Check these two extremes"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000911,"raw_usage":{"total_tokens":3954,"prompt_tokens":1028,"completion_tokens":2926,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":644,"completion_tokens_details":{"reasoning_tokens":2849}},"tokens_in":644,"tokens_out":2926,"duration_ms":19230,"temperature":1.0,"reasoning_tokens":2849,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-12T12:11:04.119389+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"A concrete check: set $S=0$, $\\overline{S}=10$, $S_{\\rm init}=5$, $P^C=P^D=5$, $\\eta_C=\\eta_D=1$, $\\rho=1$, $H=1$, $T=2$, $C_1=C_2=10$, and continue with $C_3=C_4=1000$. The two extreme terminal problems $F(2,C,\\underline{S}_2)$ and $F(2,C,\\overline{S}_2)$ both have optimal solutions with $s_1=5$, while the unconstrained finite problem also has an optimal solution with $s_1=0$ that is not optimal for the infinite-horizon tail; constructing this instance would settle whether the 'if' direction of Theorem 1 holds as stated.","supporting_citations":[{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"Introduces the idea of comparing extreme terminal values to certify a forecast horizon in finite-horizon problems."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"Gives an infinite-horizon forecast-horizon algorithm under discounting that the paper adapts to undiscounted storage scheduling."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"Provides the storage scheduling model and prior planning/decision-horizon algorithm whose finite-horizon assumption the paper relaxes."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"Supplies the iterative horizon-lengthening scheme used in Algorithm 1."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"Demonstrates the sufficient condition based on the storage hitting both bounds, which Theorem 1 generalizes to a necessary and sufficient certificate."}],"review_version":1}