Toy Single-Reservoir SDDP Walkthrough

Purpose

This chapter walks through two complete SDDP iterations on a deliberately small system — one reservoir, one thermal unit, one demand block, four stages, and a 0-order (pure seasonal-sampling) inflow model — using numbers chosen so every cut coefficient and every dual variable can be verified by hand. The goal is to make the abstract machinery of forward pass, backward pass, cut construction, and lower-bound update tangible.

Novomodelo ships a reference case, examples/1dtoy/ in the novomodelo repository, that novomodelo init --template 1dtoy <dir> scaffolds and the Quickstart runs end to end. That case has one bus, one hydro plant, two thermal plants, four monthly stages (January to April 2024), ten openings per stage and an annual discount rate of 0.12. The walkthrough keeps the single bus, the single hydro plant and the four stages, and replaces the rest with numbers chosen for hand verification: one thermal unit, three openings per stage, no discounting and idealised units (see the units note in section 1). This walkthrough is a pedagogical caricature, not a reproduction of the shipped case.

The chapter does not explain how the underlying mechanisms work; it shows them working. It does not cover multi-reservoir effects, spatial inflow correlation, the FPHA production model, risk-measure weighting, or autoregressive inflow memory. Those topics belong to the chapters cited in section 10 and to the multi-reservoir companion in Toy Four-Reservoir Walkthrough.

1. The Case in One Picture

The system has one bus, one reservoir, one thermal unit, and one demand block. The reservoir receives a stochastic inflow drawn from a 0-order seasonal-sampling model; the thermal unit and a deficit slack cover any shortfall that the reservoir cannot serve.

Case parameters (chosen for hand-verifiability):

Parameter	Symbol	Value
Stages	$T$	4
Inflow openings per stage	$N_t$	3
Initial storage	$v_0$	30 (storage units)
Reservoir capacity	$\bar{V}$	100
Demand per stage	$D$	40
Inflow seasonal mean	$\mu$	30
Inflow seasonal std deviation	$\sigma$	10
Thermal marginal cost	$c^{th}$	50
Deficit cost	$c^{def}$	1000
Discount factor	$d$	1.0

Demand is set above the mean inflow ( $D = 40 > \mu = 30$ ) so the reservoir depletes over the four stages, eventually forcing thermal dispatch and producing non-trivial dual variables in the backward pass. The thermal unit has no capacity limit, so the deficit $\delta$ stays at zero throughout.

2. Stage LP for This Case

Each stage $t$ solves the following LP (specialised from the general formulation in LP Formulation to one hydro, one thermal, one demand block, and a 0-order inflow):

Objective:

\min \quad 50\, g_{th} + 1000\, \delta + \theta

Load balance:

q + g_{th} + \delta = 40

Water balance:

v = v^{in} + a - q

Incoming-storage pinning (binds incoming storage to the trial value by equal column bounds; the pinned column’s reduced cost becomes the cut coefficient):

\underline{v}^{in} = \bar{v}^{in} = \hat{v}_{t-1}

Inflow (treated as data, not a state variable; see section 3):

a = a_t(\omega) \quad \text{(realised from the 0-order sampler)}

Bounds:

0 \leq v \leq 100, \quad q \geq 0, \quad g_{th} \geq 0, \quad \delta \geq 0, \quad \theta \geq 0

Cost-to-go variable $\theta$ : in the terminal stage 4 there are no cuts, and $\theta$ is bound to zero. As the backward pass runs, cuts of the form $\theta \geq \beta_0 + \beta^v\, v$ are added to earlier stages’ LPs. The cost-to-go $V(v)$ is non-increasing in storage (more water never raises the cost of the remaining stages), so every cut slope $\beta^v$ is zero or negative.

Note the absence of any AR-lag state variable: the 0-order inflow has no memory across stages, so storage is the only state.

3. The 0-Order Inflow Model on This Case

The inflow at every stage is sampled independently from a normal distribution with stage-seasonal mean and standard deviation:

a_t = \mu + \sigma\, \varepsilon_t, \qquad \varepsilon_t \sim \mathcal{N}(0, 1)

with $\mu = 30$ and $\sigma = 10$ for every stage in this walkthrough. The draws across stages are independent. This is the degenerate $p = 0$ case of the PAR(p) Inflow Model: no lag terms, no AR coefficients, no initial-lag state. A novomodelo case expresses it by supplying scenarios/inflow_seasonal_stats.parquet ( $\mu_t$ and $\sigma_t$ per hydro and per stage) with neither inflow_history.parquet nor inflow_ar_coefficients.parquet: novomodelo then applies the seasonal statistics as white noise, $p_m = 0$ for every season, and runs no order selection (the order selection of PAR(p) Inflow Model §3.6 needs an inflow history). The shipped 1dtoy is built this way, and novomodelo run reports its estimation path as user_stats_white_noise.

The three openings used in the backward pass correspond to $\varepsilon \in \{-1, 0, +1\}$ with equal probability $p = 1/3$ , giving inflows $a_t \in \{20, 30, 40\}$ . Because the openings are identical across stages and there is no AR memory, the same three-point distribution applies at every stage.

4. Iteration 1 — Forward Pass

The forward pass samples one trajectory using $\varepsilon_t = 0$ at every stage. Under zero noise the inflow stays at the seasonal mean throughout: $a_t = 30$ for $t = 1, 2, 3, 4$ .

At each stage, the LP minimises $50\, g_{th} + 1000\, \delta + \theta$ with $\theta$ free at zero (no cuts in iteration 1). Pure hydro dispatch dominates whenever water is available.

Stage 1 — incoming storage $\hat{v}_0 = 30$ , inflow $a_1 = 30$ . Available water: $30 + 30 = 60 \geq 40$ . Optimal: $q = 40$ , $g_{th} = 0$ , $\delta = 0$ . End storage $v_1 = 30 + 30 - 40 = 20$ . Stage cost: $0$ .

Stage 2 — $\hat{v}_1 = 20$ , $a_2 = 30$ . Available: $50 \geq 40$ . $q = 40$ , $g_{th} = 0$ , $v_2 = 10$ . Stage cost: $0$ .

Stage 3 — $\hat{v}_2 = 10$ , $a_3 = 30$ . Available: $40 = 40$ . $q = 40$ , $g_{th} = 0$ , $v_3 = 0$ . Stage cost: $0$ .

Stage 4 — $\hat{v}_3 = 0$ , $a_4 = 30$ . Available: $30 < 40$ . The LP sets $q = 30$ (all water turbined), $g_{th} = 10$ (thermal fills the gap), $\delta = 0$ , $v_4 = 0$ . Stage cost: $50 \times 10 = 500$ .

Forward-pass summary:

Stage	$\hat{v}_{t-1}$	$a_t$	$q$	$g_{th}$	$v_t$	Stage cost
1	30	30	40	0	20	0
2	20	30	40	0	10	0
3	10	30	40	0	0	0
4	0	30	30	10	0	500

Upper-bound estimate from this trajectory: $0 + 0 + 0 + 500 = 500$ . This is one realisation of total cost, a statistical estimate. It falls below the iteration-1 lower bound of section 6, $5000/9 \approx 555.6$ , as a single sampled trajectory can; averaging many trajectories reduces the sampling error, and only an exact upper bound certifies the optimality gap (Cut Management — Tier 3).

5. Iteration 1 — Backward Pass

The backward pass walks stages $4 \to 2$ . At each stage $t$ it pins the incoming storage to the forward pass’s trial point $\hat{v}_{t-1}$ , solves all three openings, reads the reduced cost of the pinned incoming-storage column (Reduced-Cost Extraction), computes per-opening intercepts, and aggregates them into one cut on stage $t-1$ (Single-Cut Aggregation). The stage-1 solves are the lower-bound evaluation of section 6.

Kinks. When an opening’s available water exactly meets demand, or its end storage sits where two pieces of the stage’s cost-to-go approximation meet (two cuts, or a cut and the zero floor), the stage cost has a kink at the trial point: one unit less and one unit more of incoming storage change the cost at different rates, and every value between the two rates is a valid subgradient. The reduced cost an LP solver returns there depends on its final basis. This walkthrough takes, at every kink, the rate for one more unit of incoming storage (the right derivative).

Stage 4 (terminal)

Trial point: $\hat{v}_3 = 0$ . Inflows under the three openings: $a_4(\omega) \in \{20, 30, 40\}$ .

Opening	$\varepsilon$	$a_4$	Available $\hat{v}\!+a_4$	$q$	$g_{th}$	$Q_4$
$\omega_1$	$-1$	20	20	20	20	1000
$\omega_2$	$0$	30	30	30	10	500
$\omega_3$	$+1$	40	40	40	0	0

Storage cut coefficient $\beta^v_4(\omega) = \partial Q_4/\partial \hat{v}_3$ , the reduced cost of the pinned incoming-storage column. By the LP envelope theorem applied to the pinned bound $v^{in}_4 = \hat{v}_3$ :

For $\omega_1$ and $\omega_2$ (water-limited, thermal active): one extra unit of $\hat{v}_3$ enables one extra unit of turbining, displacing one unit of thermal worth $50$ . Optimal cost falls by $50$ , so $\beta^v_4(\omega) = -50$ .
For $\omega_3$ (available water exactly meets demand) the stage cost has a kink: one more unit of $\hat{v}_3$ ends in the terminal storage $v_4$ , which has zero value, while one unit less needs one unit of thermal worth $50$ . Every value in $[-50, 0]$ is a valid subgradient, and the walkthrough’s rule gives $\beta^v_4(\omega_3) = 0$ .

Opening	$\beta^v_4$
$\omega_1$	$-50$
$\omega_2$	$-50$
$\omega_3$	$0$

Per-opening intercepts (anchoring each cut at the trial point):

\beta_{0,4}(\omega) = Q_4(\omega) - \beta^v_4(\omega)\,\hat{v}_3

With $\hat{v}_3 = 0$ this simplifies to $\beta_{0,4}(\omega) = Q_4(\omega)$ : $\beta_{0,4} = (1000,\, 500,\, 0)$ .

Single-cut aggregation (uniform probability $p = 1/3$ ; see Cut Management section 3):

\bar{\beta}_0 = \tfrac{1}{3}(1000 + 500 + 0) = 500, \qquad \bar{\beta}^v = \tfrac{1}{3}(-50 - 50 + 0) = -\tfrac{100}{3}

Cut added to stage 3’s LP:

\theta \;\geq\; 500 \;-\; \tfrac{100}{3}\, v

Sanity check. At $v = 0$ the cut gives $\theta \geq 500$ , matching the expected stage-4 cost from the table above. At $v = 15$ the cut gives $\theta \geq 0$ , after which the implicit $\theta \geq 0$ bound takes over.

Stage 3

Trial point: $\hat{v}_2 = 10$ .

The stage-3 LP now carries the cut $\theta \geq 500 - (100/3)\, v_3$ . For each opening $a_3 \in \{20, 30, 40\}$ the optimiser balances spending on thermal now against keeping more water for the future (worth $100/3$ per unit via the cut).

Opening	$a_3$	Available	$q$	$g_{th}$	$v_3$	$\theta^*$	$Q_3$
$\omega_1$	20	30	30	10	0	$500$	$1000$
$\omega_2$	30	40	40	0	0	$500$	$500$
$\omega_3$	40	50	40	0	10	$500/3$	$500/3$

For $\omega_1$ the system is water-limited; the LP turbines all 30 units of available water and dispatches 10 MW of thermal, ending with $v_3 = 0$ and the cut binding at $500$ .

For $\omega_2$ available water exactly meets demand; no thermal, end $v_3 = 0$ , cut value $500$ .

For $\omega_3$ the optimiser must choose how much of the surplus to release. Decreasing $q$ raises $g_{th}$ by one (cost $50$ ) and raises $v_3$ by one (saves $100/3$ on the cut); the marginal cost of holding more is $+50 - 100/3 \approx +16.7$ , so the optimiser pushes $q$ to its load-balance bound at $40$ , leaving $v_3 = 10$ and the cut value at $500/3 \approx 167$ .

Storage cut coefficients $\beta^v_3(\omega) = \partial Q_3/\partial \hat{v}_2$ (reduced costs of the pinned incoming-storage column):

$\omega_1$ (water-limited): one extra unit of $\hat{v}_2$ frees one extra turbine unit, saves $50$ of thermal. $\beta^v_3(\omega_1) = -50$ .
$\omega_2$ (available water exactly meets demand): a kink. One unit less of $\hat{v}_2$ needs one unit of thermal ( $-50$ ); one unit more raises $v_3$ and lowers the cut value by $100/3$ . Every value in $[-50, -100/3]$ is a valid subgradient, and the rule gives $\beta^v_3(\omega_2) = -100/3$ .
$\omega_3$ (water surplus, holding storage): one extra unit of $\hat{v}_2$ raises $v_3$ by one, lowers the cut value by $100/3$ . $\beta^v_3(\omega_3) = -100/3$ .

Per-opening intercepts $\beta_{0,3}(\omega) = Q_3(\omega) - \beta^v_3(\omega)\,\hat{v}_2$ :

Opening	$Q_3$	$\beta^v_3 \cdot \hat{v}_2$	$\beta_{0,3}$
$\omega_1$	$1000$	$-500$	$1500$
$\omega_2$	$500$	$-1000/3$	$2500/3$
$\omega_3$	$500/3$	$-1000/3$	$500$

Aggregation ( $p = 1/3$ ):

\bar{\beta}_0 = \tfrac{1}{3}\Bigl(1500 + \tfrac{2500}{3} + 500\Bigr) = \tfrac{8500}{9}, \qquad \bar{\beta}^v = \tfrac{1}{3}\!\left(-50 - \tfrac{100}{3} - \tfrac{100}{3}\right) = -\tfrac{350}{9}

Cut added to stage 2’s LP:

\theta \;\geq\; \tfrac{8500}{9} \;-\; \tfrac{350}{9}\, v

Sanity check. At $v = \hat{v}_2 = 10$ the cut evaluates to $(8500 - 3500)/9 = 5000/9 \approx 555.6$ , matching the probability-weighted expectation $\bar{Q}_3 = (1/3)(1000 + 500 + 500/3) = 5000/9 \approx 555.6$ .

Stage 2

Trial point: $\hat{v}_1 = 20$ . The stage-2 LP carries the cut $\theta \geq 8500/9 - (350/9)\, v_2$ . A stored unit is worth at most $350/9 \approx 38.9 < 50$ , so every opening turbines up to demand.

Opening	$a_2$	Available	$q$	$v_2$	$\theta^*$	$Q_2$	$\beta^v_2$	$\beta_{0,2}$
$\omega_1$	20	40	40	0	$8500/9$	$8500/9$	$-350/9$	$15500/9$
$\omega_2$	30	50	40	10	$5000/9$	$5000/9$	$-350/9$	$4000/3$
$\omega_3$	40	60	40	20	$500/3$	$500/3$	$-350/9$	$8500/9$

$\omega_1$ is a kink (available water exactly meets demand; every value in $[-50, -350/9]$ is a valid subgradient) and the rule gives $-350/9$ ; in $\omega_2$ and $\omega_3$ one more unit of $\hat{v}_1$ raises $v_2$ along the cut. The intercepts are $\beta_{0,2}(\omega) = Q_2(\omega) - \beta^v_2(\omega)\,\hat{v}_1$ .

Aggregation ( $p = 1/3$ ):

\bar{\beta}_0 = \tfrac{1}{3}\Bigl(\tfrac{15500}{9} + \tfrac{4000}{3} + \tfrac{8500}{9}\Bigr) = \tfrac{4000}{3}, \qquad \bar{\beta}^v = -\tfrac{350}{9}

Cut added to stage 1’s LP:

\theta \;\geq\; \tfrac{4000}{3} \;-\; \tfrac{350}{9}\, v

At $v = \hat{v}_1 = 20$ the cut evaluates to $5000/9 \approx 555.6$ , the probability-weighted expectation $\bar{Q}_2$ .

By the end of the backward pass, stages 1 through 3 each carry one cut. Stage 4 has none (it is the terminal stage and has no future to approximate).

6. Iteration 1 — Lower Bound

After the backward pass installs a cut at stage 1, the iteration-1 lower bound solves the stage-1 LP for every opening with that cut active, from the fixed initial storage $x_0 = v_0 = 30$ , and takes the probability-weighted expectation of the opening objectives (the risk-neutral case of the lower bound):

\underline{z}^1 = \mathbb{E}_{\omega}\bigl[\, Q_1^1(x_0, \omega) \,\bigr]

where $Q_1^1$ is the stage-1 optimal objective under opening $\omega$ with the iteration-1 cut in place.

Opening	$a_1$	Available	$q$	$v_1$	$\theta^*$	$Q_1^1$
$\omega_1$	20	50	40	10	$8500/9$	$8500/9$
$\omega_2$	30	60	40	20	$5000/9$	$5000/9$
$\omega_3$	40	70	40	30	$500/3$	$500/3$

\underline{z}^1 = \tfrac{1}{3}\Bigl(\tfrac{8500}{9} + \tfrac{5000}{9} + \tfrac{500}{3}\Bigr) = \tfrac{5000}{9} \approx 555.6

The lower bound rises from $0$ (before the first iteration, no cuts) to $5000/9$ . It does not decrease across iterations, because cuts are only added; Cut Management — Tier 1 states when it is a valid lower bound. Here it stays below the optimal expected cost of section 9.

7. Iteration 2

The second forward pass draws $\varepsilon = (+1, 0, 0, 0)$ : a wet first stage, then mean inflows. Each stage LP carries its iteration-1 cut.

Stage	$\hat{v}_{t-1}$	$a_t$	$q$	$v_t$	$\theta^*$
1	30	40	40	30	$500/3$
2	30	30	40	20	$500/3$
3	20	30	40	10	$500/3$
4	10	30	40	0	$0$

The trajectory costs $0$ , and every trial point sits 10 units above the iteration-1 one. The backward pass then adds a second cut on each of stages 3, 2 and 1; each non-terminal stage LP it solves carries both of that stage’s cuts, and every kink takes the right derivative:

Stage	Trial point	Opening	$a_t$	$v_t$	$\theta^*$	$Q_t$	$\beta^v_t$	$\beta_{0,t}$
4	$\hat{v}_3 = 10$	$\omega_1$	20	0	$0$	$500$	$-50$	$1000$
4	$\hat{v}_3 = 10$	$\omega_2$	30	0	$0$	$0$	$0$ (kink)	$0$
4	$\hat{v}_3 = 10$	$\omega_3$	40	10	$0$	$0$	$0$	$0$
3	$\hat{v}_2 = 20$	$\omega_1$	20	0	$500$	$500$	$-100/3$ (kink)	$3500/3$
3	$\hat{v}_2 = 20$	$\omega_2$	30	10	$500/3$	$500/3$	$-50/3$ (kink)	$500$
3	$\hat{v}_2 = 20$	$\omega_3$	40	20	$0$	$0$	$0$ (kink)	$0$
2	$\hat{v}_1 = 30$	$\omega_1$	20	10	$5000/9$	$5000/9$	$-350/9$	$15500/9$
2	$\hat{v}_1 = 30$	$\omega_2$	30	20	$2000/9$	$2000/9$	$-50/3$	$6500/9$
2	$\hat{v}_1 = 30$	$\omega_3$	40	30	$500/9$	$500/9$	$-50/3$	$5000/9$

At stage 3 the available water of $\omega_1$ exactly meets demand, the end storage $v_3 = 10$ of $\omega_2$ sits where the two stage-3 cuts meet, and $v_3 = 20$ of $\omega_3$ where the second cut reaches zero, so all three openings are kinks there. Averaging with $p = 1/3$ gives the iteration-2 cuts:

Cut on stage	From trial point	$\bar{\beta}_0$	$\bar{\beta}^v$	At the trial point
3	$\hat{v}_3 = 10$	$1000/3$	$-50/3$	$500/3$
2	$\hat{v}_2 = 20$	$5000/9$	$-50/3$	$2000/9$
1	$\hat{v}_1 = 30$	$1000$	$-650/27$	$2500/9$

With both stage-1 cuts active, the stage-1 openings end at $v_1 = 10$ , $20$ and $30$ with $\theta^* = 8500/9$ , $5000/9$ and $2500/9$ (the second cut is the larger one only at $v_1 = 30$ ), so

\underline{z}^2 = \tfrac{1}{3}\Bigl(\tfrac{8500}{9} + \tfrac{5000}{9} + \tfrac{2500}{9}\Bigr) = \tfrac{16000}{27} \approx 592.6

8. The Cost-to-Go Across Iterations

After iteration 2 each non-terminal stage holds two cuts. In this case the cost-to-go $V_t(v)$ , the expected cost of stages $t$ to $4$ from the storage $v$ left at the end of stage $t-1$ , is a convex, piecewise-linear and non-increasing function of one variable, and each cut on stage $t-1$ is a line in the same variable. The stage- $(t-1)$ LP’s $\theta$ sees the approximation

\underline{V}_t^k(v) \;=\; \max\Bigl\{0,\; \max_{i = 1, \ldots, k} \bigl\{ \bar{\beta}_0^i + \bar{\beta}^{v,i}\, v \bigr\}\Bigr\},

the pointwise maximum (upper envelope) of the cut lines and the floor $\theta \geq 0$ . Every cut lies on or below $V_t$ , so the upper envelope lies on or below $V_t$ too: it is a lower approximation, and each new cut can only raise it.

The table compares $V_3$ , the cost of stages 3 and 4, with the two cuts on stage 2 (trial points $\hat{v}_2 = 10$ in iteration 1 and $\hat{v}_2 = 20$ in iteration 2):

$v$	$V_3(v)$	Iteration-1 cut	Iteration-2 cut	$\underline{V}_3^2(v)$
0	$1000$	$8500/9$	$5000/9$	$8500/9$
10	$5000/9$	$5000/9$	$3500/9$	$5000/9$
15	$3500/9$	$3250/9$	$2750/9$	$3250/9$
20	$2000/9$	$1500/9$	$2000/9$	$2000/9$
30	$500/9$	$-2000/9$	$500/9$	$500/9$
40	$0$	$-5500/9$	$-1000/9$	$0$

The envelope touches $V_3$ at both trial points and on $[20, 30]$ , where the iteration-2 cut coincides with a linear piece of $V_3$ , and equals $V_3$ for $v \geq 40$ , where the floor and $V_3$ are both zero; it stays strictly below $V_3$ elsewhere, as at $v = 0$ and $v = 15$ . Later iterations that visit other trial points, low-storage ones in particular, add cuts that raise the envelope where it is still below $V_3$ .

The figure plots the table: $V_3$ (solid), the iteration-1 and iteration-2 cuts on stage 2 (dashed), and their upper envelope with the floor $\theta \geq 0$ , against the storage $v$ left at the end of stage 2. The envelope meets $V_3$ at the two trial points (marked), on $[20, 30]$ and for $v \geq 40$ , where the floor and $V_3$ are both zero, and lies below it elsewhere. The plotted values come from a tested module that re-runs both iterations of this walkthrough.

9. Convergence on This Case

Two iterations raise the lower bound from $5000/9 \approx 555.6$ to $16000/27 \approx 592.6$ . The optimal expected cost of this case is $17000/27 \approx 629.6$ : no unit of stored water saves more than one unit of thermal output, so turbining up to demand is optimal at every stage, and averaging that policy’s cost over the $3^4 = 81$ equally likely inflow sequences from $v_0 = 30$ gives the value. Later iterations keep or raise the lower bound, and under the hypotheses of Cut Management — Tier 2 it converges to this value.

The two trajectory costs, $500$ and $0$ , are single-sample estimates of the policy’s expected cost; both fall below the lower bound of their iteration and neither certifies anything. A gap certificate needs an exact upper bound from an enumerated forward pass (Tier 3); the percent form of the optimality gap then divides the gap by the magnitude of the lower bound. The shipped 1dtoy runs one sampled forward pass per iteration and stops at its iteration limit of 128; novomodelo rejects a gap stopping rule at setup under a sampled forward pass (see Stopping Rules). The upper-bound mechanisms are compared in Upper Bound Evaluation.

10. What This Example Does Not Show

This case isolates the core SDDP loop in its simplest form. It cannot illustrate:

Multi-reservoir effects: the cut becomes a hyperplane with one storage coefficient per reservoir; the Toy Four-Reservoir Walkthrough shows one on four decoupled reservoirs. Trading water between reservoirs needs a coupling (transmission, a cascade or a shared constraint) that neither walkthrough includes.
Spatial inflow correlation: when multiple plants share a wet/dry signal, their innovations $\varepsilon$ must be drawn from a correlated multivariate normal via the spectral factorisation described in PAR(p) Inflow Model §6.
Autoregressive inflow memory: a PAR(p) model with $p \geq 1$ adds one lag state variable per lag; whether a stage’s cuts carry a coefficient for each lag is that stage’s cut-state projection (State Augmentation §7). See PAR(p) Inflow Model.
FPHA production model: nonlinear head-dependent efficiency approximated by piecewise-linear hyperplanes; one of the planes can bind, contributing to the storage cut coefficient. See Hydro Production Function Models.
Risk-measure effects: CVaR weighting that shifts cut aggregation probabilities away from uniform $p = 1/N_t$ . See Risk Measures.

The Toy Four-Reservoir Walkthrough extends this case to four independent reservoirs at four buses with no transmission, illustrating how the cut hyperplane and the per-bus dispatch mechanics scale up.

Cross-References

SDDP Algorithm — Forward and backward pass structure, lower-bound computation, convergence monitoring
LP Formulation — Complete stage LP: load balance, water balance, objective taxonomy
State Augmentation — Column-bound state pinning
Cut Management — Dual extraction, per-opening intercepts, single-cut aggregation, cut validity, sign convention
PAR(p) Inflow Model — Inflow model definition; the $p = 0$ degenerate case (white noise) used in this walkthrough; stored vs computed quantities
Stopping Rules — Iteration limit, gap threshold, bound-stalling criteria
Upper Bound Evaluation — Simulation-based and inner-approximation upper bounds
Toy Four-Reservoir Walkthrough — Multi-reservoir extension at 4 buses with no transmission and independent 0-order inflows