advanced33 min studyLesson 297 in Mathematics

Augmented Lagrangian methods

Combine multiplier correction with a finite penalty and distinguish inner stationarity from outer feasibility.

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What you’ll understand

  • Derive the equality augmented Lagrangian and its multiplier update.
  • Calculate the behavior of a finite-penalty multiplier iteration.
  • Explain why local guarantees require accurate subproblem solutions and regularity.

A quadratic penalty makes a constraint violation expensive, but an accurate constrained answer may then require a very large penalty and a difficult inner problem. An augmented Lagrangian adds a multiplier estimate to carry part of the constraint-balancing work. The outer iteration updates that estimate using the remaining violation, while the inner iteration minimizes a smooth objective for the current estimate. This division can improve conditioning and accuracy, but it creates two distinct questions: was the inner problem solved adequately, and is the original constrained problem becoming feasible?

A penalty plus a linear correction

Consider smooth equalities h(x)=0 and use the sign convention L=f+λ-transpose-h. The augmented Lagrangian adds half rho times the squared Euclidean constraint norm, where rho is positive. For a fixed multiplier and penalty, the inner problem is unconstrained unless the original formulation also has simple bounds. Its gradient contains both the estimated multiplier force and a correction proportional to the current violation. A raw small objective gradient is no longer the appropriate stopping diagnostic.

After an inner approximate minimization, update the multiplier by adding rho times the new residual. With an exact inner stationary point, the original Lagrangian gradient evaluated at this updated multiplier is zero. This identity explains the update sign and the link between inner stationarity and an original KKT residual. It does not make the primal residual zero. If the inner solve stops early, its nonzero gradient transfers directly to the corresponding stationarity residual and must be accounted for in an outer tolerance policy.

Lρ(x,λ)=f(x)+λTh(x)+ρ2∥h(x)∥2,λk+1=λk+ρkh(xk+1)\mathcal L_\rho(x,\lambda)=f(x)+\lambda^{\mathsf T}h(x)+\frac\rho2\|h(x)\|^2,\qquad \lambda_{k+1}=\lambda_k+\rho_k h(x_{k+1})
The plus sign matches L=f+λᵀh. For an exact inner stationary point, ∇ₓL(xₖ₊₁,λₖ₊₁)=0; primal feasibility remains a separate residual.↔ Scroll wide equations horizontally.

Why a finite penalty can suffice locally

At a feasible KKT point, the augmented Hessian equals the Lagrangian Hessian plus rho times the constraint Jacobian Gram matrix. The second term adds curvature in directions that change the constraints, while a second-order sufficient condition supplies positive curvature on tangent directions. With full-row-rank equality Jacobian and suitable smoothness, a sufficiently large finite rho makes the augmented Hessian positive definite near that solution. This is a local curvature mechanism, not a proof that every inner problem has a unique global minimum.

A pure quadratic penalty has no adjustable linear multiplier term, so its residual must generate the missing constraint force by itself. Achieving a finite nonzero force while reducing that residual often drives rho upward. In a multiplier method the stored estimate can approach the correct force while rho stays fixed in favorable local circumstances. Raising rho is still useful when feasibility improvement stalls, but an infinite automatic escalation can destroy the numerical advantage and conceal an inaccurate inner solve.

∇xx2Lρ(x∗,λ∗)=∇xx2L(x∗,λ∗)+ρJh(x∗)TJh(x∗)\nabla_{xx}^2\mathcal L_\rho(x_*,\lambda_*)=\nabla_{xx}^2L(x_*,\lambda_*)+\rho J_h(x_*)^{\mathsf T}J_h(x_*)
The identity is evaluated at h(x*)=0. Away from feasibility the augmented Hessian also contains rho-weighted residual times constraint-Hessian terms.↔ Scroll wide equations horizontally.

Report both loops and their limits

A practical outer stopping rule combines the constraint norm and the Lagrangian gradient norm, with multiplier and variable scales recorded. The inner tolerance is often loose initially and tightened as the outer iteration advances. Solving every early inner problem to extreme accuracy can waste computation, while never tightening can prevent a meaningful final KKT residual. Precise convergence theorems specify the accuracy schedule, penalty updates, regularity, and the local or global subproblem properties; the update formula alone is insufficient.

Inequality constraints require additional structure. One approach introduces nonnegative slacks and treats the resulting equalities together with bounds; another uses a projected multiplier formula derived for inequalities. Simply applying an unrestricted equality multiplier update to an inequality and ignoring sign and complementarity is wrong. For nonconvex objectives, even an excellent feasible stationary result may be only locally optimal. Repeating initializations can supply empirical comparisons but cannot turn those observations into a general global certificate.

Correct the multiplier without sending the penalty to infinity

Minimize f(x)=(x−3)²/2 subject to h(x)=x−1=0. Set ρ=2, λ₀=0, and solve each inner subproblem exactly. Compute two outer iterations.

  1. Inner stationarity gives (x−3)+λ+2(x−1)=0(x-3)+\lambda+2(x-1)=0, hence x=(5−λ)/3x=(5-\lambda)/3. The only feasible optimum is x*=1; its true multiplier satisfies −2+λ*=0, so λ*=2.

  2. With λ₀=0 the inner solution is x₁=5/3 and its residual is 2/3. Update to λ1=0+2(2/3)=4/3\lambda_1=0+2(2/3)=4/3. The updated original stationarity residual is (5/3−3)+4/3=0(5/3-3)+4/3=0, although the point remains infeasible.

  3. The next inner solution is x2=(5−4/3)/3=11/9x_2=(5-4/3)/3=11/9. Its residual is 2/9, and the next multiplier is 4/3+2(2/9)=16/94/3+2(2/9)=16/9. More generally the multiplier error obeys λk+1−2=(λk−2)/3\lambda_{k+1}-2=(\lambda_k-2)/3, so it contracts at fixed penalty.

The result

The multiplier error and feasibility residual contract while rho stays equal to two. Zero stationarity after each exact inner solve does not mean the constraint is already satisfied.

Explore the model

Interactive lab

Interactive lab · function

Inspect the inner curvature cost

Pause, Reset, and set a=1 and x=1. Then repeat from Reset with both a=2 and x=1. Predict the slope doubling from two to four. These centered quadratics represent the inner curvature 1+ρ for ρ=1 and ρ=3 in the worked scalar family; vertical constants and shifted minimizers are omitted. The instrument does not update multipliers or solve constrained problems.

Inspect the inner curvature costThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forIncreasing a penalty can stiffen an inner objective; multiplier correction aims to avoid requiring an arbitrarily large stiffness.

Check your understanding

For the equality augmented Lagrangian with L=f+λᵀh, what does an exact inner stationary point imply after λ⁺=λ+ρh(x⁺)?

Try it yourself

For the same scalar problem, instead take ρ=1 and λ₀=0. Compute x₁, λ₁, x₂, λ₂ and the multiplier-error contraction factor.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Inner stationarity becomes 2x−4+λ=0. Keep the equality residual x−1 separate from the gradient residual.

Compare with the solution

The inner solution is x=(4−λ)/2. Thus x₁=2, residual one, and λ₁=1. Next x₂=3/2, residual one half, and λ₂=3/2. Substituting the inner solution into λ⁺=λ+x−1 gives λ⁺=1+λ/2, so λ⁺−2=(λ−2)/2. The factor is one half, slower than the factor one third at ρ=2, while the inner curvature is smaller.

Take it with you

  • The multiplier update converts inner augmented stationarity into original Lagrangian stationarity.
  • Constraint feasibility remains an independent outer-loop target.
  • Finite-penalty local benefits depend on regularity and controlled inner accuracy.

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