Numerical Linear Algebra
where the algebra stops being the arithmetic
Ladder

Reuse — the ladder

3 distinct arguments against one idea, from the one that introduces it to the one that assumes the rest.
  1. 024681012141618202210⁶10⁷10⁸members served by one factorisationmultiplications for the whole sequencedoes not convergethe contraction ruledrift 0.01 a memberevery member1.6·10⁷every 5 members9.3·10⁶contraction rule9.2·10⁶its factorisations4cliff at a period of20a factorisation has a shelf lifeand the cliff is past the optimum

    A factorisation kept past its date

    One Cholesky factor can serve five members of a drifting sequence and save 44 per cent of the work. Kept for twenty it does not lose accuracy — it stops converging altogether. The optimum and the cliff are four members apart, both move with the drift, and a rule written in a ratio the iteration has already computed finds them without being told what the drift is.

    rung 1 · sequence
  2. 10⁻⁶10⁻⁵10⁻⁴10⁻³10⁻²10⁻¹10112233445566relative drift ‖E‖ / ‖A₀‖preconditioned conjugate gradient iterationson the small eigenvalueson the large onessame drift, two placessmall end at 10⁻²38large end at 10⁻²9κ(M⁻¹A), small end36κ(M⁻¹A), large end1.3‖E‖/‖A₀‖, both0.01how much the matrix changedis not what the preconditioner cares about

    Where the drift lands

    The standing rule for when a preconditioner has gone stale is to rebuild it once the matrix has changed by more than some fraction of itself. Two drifts of exactly the same relative size cost 19 iterations and 5 on the same matrix, and the quantity that separates them is not in the rule at all — the perturbation is divided by the eigenvalue it lands on.

    rung 2 · sequence
  3. 10¹10²10³03691215what one rebuild is worth, in preconditioned iterationscheapest number of members between rebuildsa dense Cholesky here is worth 6.7 iterationsringed: where the growth rule beats every fixed periodcheapest period, by setup costsetup worth 51setup worth 102setup worth 203setup worth 506setup worth 1008setup worth 20012how long to keep itis a question about what it cost to build

    What a rebuild is worth

    One sequence, one drift, one preconditioner — and six different right answers, because the cheapest rebuild period depends on what a rebuild cost to build and on nothing else. The optimum walks from every member to every twelfth as the setup gets dearer, and the free rule that reads the iteration count beats it in the middle of that range and loses at both ends.

    rung 3 · sequence

All ladders