The plan is a mixed-integer linear programme: a linear cost to minimise, linear rules, and some decisions that must be whole numbers. Below is the core, simplified for reading (buying, for instance, is mostly of components, not finished goods); the code adds the details named at the end.

Indices. Materials $m$ , sites $p$ , time buckets $t$ , lanes $ℓ$ (from one site to another), customers $c$ .

Decisions.

Symbol	Meaning	Kind
$x_{m p t}$	units made	continuous, $\geq 0$
$y_{m p t}$	a changeover (setup) happens	binary
$b_{m p t}$	units bought	continuous, $\geq 0$
$s_{ℓ m t}$	units shipped on a lane	continuous, $\geq 0$
$q_{ℓ m t}$	pallets on a lane	integer
$k_{ℓ t}$	trucks on a lane	integer
$I_{m p t}$	stock at the end of a bucket	continuous, $\geq 0$
$u_{m p c t}$	a customer's demand not served	continuous, $\geq 0$

Objective. Minimise total cost plus the price of shortfall:

min \sum_{m, p, t} (v_{m p} x_{m p t} + f_{m p} y_{m p t} + π_{m p t} b_{m p t} + h_{m p t} I_{m p t})

+ \sum_{ℓ, t} F_{ℓ t} k_{ℓ t} + \sum_{m, p, c, t} (g_{m} + ρ_{m}) w_{c} u_{m p c t}

where $v$ is the variable cost of making a unit (with its carbon), $f$ the cost of a changeover, $π$ the purchase price, $F$ the cost of a truck (freight, toll and carbon), $h$ the cost of holding a unit for the bucket (warehousing and capital), $g$ the gross margin, $ρ$ the shortage penalty and $w_{c}$ the customer's weight.

Stock balance. What a site holds at the end of a bucket is what it held before, plus what it made and received, minus what it sent and served:

I_{m p t} = I_{m p, t - 1} + x_{m p t} + b_{m p t} + \sum_{ℓ \in in (p)} s_{ℓ m, t - L_{ℓ}} - \sum_{ℓ \in out (p)} s_{ℓ m t} - d_{m p t}^{served}

with $L_{ℓ}$ the lane's lead time in buckets, and demand either served or short:

d_{m p t}^{served} + \sum_{c} u_{m p c t} = D_{m p t}

Line hours. Running and changeovers must fit in the hours the line really gives, after its efficiency (OEE):

\sum_{m} (r_{m p} x_{m p t} + σ_{m p} y_{m p t}) \leq H_{p t} \cdot {OEE}_{p}

Lot sizes. Nothing is made without a changeover, and a run is at least its minimum lot:

Q_{m p}^{min} y_{m p t} \leq x_{m p t} \leq Q_{m p}^{max} y_{m p t}

Pallets and trucks. A part pallet takes a whole slot; pallets ride on whole trucks; no truck leaves empty, and dense freight runs out of payload before slots:

s_{ℓ m t} \leq n_{m} q_{ℓ m t} \sum_{m} q_{ℓ m t} \leq P_{ℓ} k_{ℓ t} \sum_{m} κ_{m} s_{ℓ m t} \leq W_{ℓ} k_{ℓ t}

with $n_{m}$ units per pallet, $P_{ℓ}$ pallets per truck, $κ_{m}$ kilograms per unit and $W_{ℓ}$ the maximum payload.

How sure: the gap. With $z$ the cost of the plan found and $\underset{―}{z}$ the bound the solver proved,

gap = \frac{z - \underset{―}{z}}{z}

and the plan is proven when the gap is at or below the target: 5% for the tactical profile, 0.5% for the published plans.

Layers. The year layer solves the same model with each family $f$ as one product. Its stock at the hand-over, ${\tilde{I}}_{f p}$ , is split between the family's products by their share of its demand, $α_{m p}$ , and becomes a target for the near weeks, priced like a shortage if missed:

I_{m p, T} + e_{m p} \geq α_{m p} {\tilde{I}}_{f p}, cost (g_{m} + ρ_{m}) e_{m p}

A target the near weeks cannot reach is a finding, not an infeasible model.

Also in the code: components exploded from recipes (bill of materials), shelf life (stock older than its customers accept is written off at standard cost), DDMRP buffer floors whose breach is priced, co-packer rules (they may only send back to their parent plant), dangerous goods only on licensed carriers, the empty run home where charged, a committed first bucket that is not reopened, and optionally a time fence beyond which changeovers, pallets and trucks are planned as averages.