Integer aggregation (opt-in)

By default only scalar continuous variables are eliminated, so the default behaviour is MINLP-safe and byte-identical to the continuous-only pipeline. Pass integer_aggregation=True to also eliminate a scalar integer/binary variable — but only when its integrality is implied.

solve(m, method="lm", integer_aggregation=True)   # collapse an integer chain

This is the standard MIP-presolve condition (Achterberg et al. 2020): an integer variable y may be eliminated via coeff*y + rest = 0 only when

• the defining equality is fully affine (no nonlinear term),
• the pivot is ±1,
• every other participating variable is integer/binary, and
• every other coefficient and the constant term are integral.

An all-integer stock balance qualifies:

# s[t] = s[t-1] + q[t] - d[t]  — all integer, unit pivot, integral data
m.subject_to(s[t] - (s[t-1] + q[t] - d[t]) == 0)

What abstains

The flag never eliminates a variable it cannot prove integral:

• a continuous variable in the definition (breaks implied integrality),
• a non-unit pivot — y = 2*i can define y but cannot eliminate i,
• a fractional coefficient or constant (including a variable-free nonlinear term such as sin(2)),
• a nonlinear body.

Recovery and duals

Recovery of an eliminated integer returns an exactly integral value; a non-integral result raises rather than silently rounding a wrong answer. Dual recovery reports available=False over eliminated integers (KKT multipliers are not defined for branching variables).

When it helps

Use it for MINLPs with integer accounting/counting chains, where fewer integers means a smaller branching space. It does nothing for models whose integers appear only in inequalities (e.g. on/off binaries x <= cap*b) — those are not defined by an equality, so they are never candidates.

Default-off, proven no-op

integer_aggregation defaults to False. With the flag off, integer and binary variables are never aggregation candidates, and a test suite pins that the continuous-only results are unchanged.