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Two-agent consensus least squares with exact ADMM iterates

Statement

Take f(x)=12(x−1)2, g(z)=12(z+1)2, constraint x−z=0, ρ=1, and scaled initial values z0=u0=0. The primal optimizer is x∗=z∗=0 with value p∗=1; y∗=1 is a saddle and dual optimizer. The first two exact scaled ADMM sweeps are (x1,z1,u1)=(12,−14,34),(x2,z2,u2)=(0,−18,78).

Facts & Assumptions

Given: The scalar functions, equality constraint, penalty and initial state in the Statement.

[A1]

The scalar functions, equality constraint, penalty, and initial state are exactly those in the Statement; the equality row (1,−1) is nonzero and hence has full row rank.

[F1]

Scaled two-block ADMM with explicit capped-solve statuses: xk+1 minimizes f(x)+12(x−zk+uk)2, zk+1 minimizes g(z)+12(xk+1−z+uk)2, and uk+1=uk+xk+1−zk+1.

[F2]

Saddle points of extended-real Lagrangians: saddle means L((x∗,z∗),y)≤L((x∗,z∗),y∗)≤L((x,z),y∗) for all arguments.

[F3]

Finite-dimensional convex program, Lagrangian dual, and Slater condition: the dual function is the infimum of the Lagrangian and the dual value is its supremum over multipliers.

[F4]

Every Lagrangian dual feasible pair is a lower-bound certificate: every dual value is at most every feasible primal objective value.

Proof

1.1A1givenF2algebra

On the feasible set z=x, the objective is f(x)+g(x)=12(x−1)2+12(x+1)2=x2+1, which is uniquely minimized at x=z=0 with value 1. For y∗=1, L((x,z),1)=f(x)+g(z)+x−z=1+12x2+12z2≥1, with equality at (0,0); feasibility makes L((0,0),y)=1 for every y. These are exactly the saddle inequalities [F2].

1.2A1F1givenalgebra

Differentiating the first strictly convex quadratic in [F1] gives 2xk+1=1+zk−uk. At k=0 this yields x1=1/2. Differentiating the second gives 2zk+1=xk+1+uk−1, hence z1=−1/4 and then u1=u0+x1−z1=3/4.

2.1A1F3F4step 1.1algebra

The equality matrix is A=(1,−1), whose single row is linearly independent, so the dual definition [F3] applies. Its dual function is d(y)=inf⁡x{12(x−1)2+yx}+inf⁡z{12(z+1)2−yz}=(y−12y2)+(y−12y2)=1−(y−1)2. Thus its maximum is 1 at y=1, consistent with weak duality [F4] and the saddle value.

3.1A1F1step 1.2algebra∎

The same two equations give x2=(1−1/4−3/4)/2=0, z2=(0+3/4−1)/2=−1/8, and u2=3/4+0−(−1/8)=7/8. Each subproblem is a strictly convex quadratic, so both minimizers in each of these sweeps are attained and unique. The recurrence values are exact rationals.

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