Internal order of the constraint multipliers

[309815765]

Hi,

I need to get the constraint multipliers for sensitivity analysis. To my knowledge, the constraint with both lower bound and upper bound will be converted to two constraints (correct me if it's wrong). I'd like to know how does Ipopt order multipliers (lambda) for constraints, especially for the constraints with both boundaries.

Thanks

[svigerske]

If lower and upper bounds are equal, then the constraint will be converted from lhs <= g(x) <= rhs to g(x) - lhs = 0.
If lower and upper bounds are different (and possibly infinite), then the constraint will be converted from lhs <= g(x) <= rhs to g(x) - s = 0, lhs <= s <= rhs, where s is a slack variable that gets introduced.
See also https://coin-or.github.io/Ipopt/classIpopt_1_1TNLPAdapter.html#details

The bound multiplier lambda for constraints is taken from the bound multiplier of g(x) - lhs = 0 (if lhs==rhs) or g(x) - s = 0 (if lhs != rhs).
So in the latter case, the sign of lambda tells you whether lhs <= g(x) or g(x) <= rhs is active.

[309815765]

Thanks for your response!

So, lambda[i] is the multiplier for g[i], and lamdda[i] > 0 indicates that g[i] <= rhs is active whereas lambda[i] < 0 means lhs<= g[i] is active.

[svigerske]

Or the other way around. I cannot remember signs.