Numerical Linear Algebra
where the algebra stops being the arithmetic
Theme

The thread: Whose fault is it — page 14

Essays 313 to 321 of the 321 on this theme, in the same order.
110¹110¹the order cut's needfirst-level readingthe kept pairings' ratiodearest kept overcheapest of alldashed: equaldotted: a factor 1.5seven tensors, sixty networksthe offer does not show the need Where the flop count stopped predicting the time

The need is set further on

A beam over contraction orders needs a wide tolerance at its partial-order cut, a median factor of 2.33 at seven tensors, and the beam was predicted to see that need in its own first level — the ratio of the dearest prefix it is offered to the cheapest, within a factor of 1.5 on four networks in five. The best such ratio is within 1.5 on 32 of 60, and none ranks the networks better than chance. A beam that sets its tolerance from its first level lands on the line the fixed tolerances draw. The ratio is exact where the need is set at the first level, and 35 of 60 networks set it later.

 24685055606570758085decay's fall a trace after the turn, hundredthsspread used, % of the truemeasuredthe last ratio, squaredlast ratio over the nextmedians over 400 sequencesthe square holds only when the turn is symmetric Randomised, and the guarantee that changes kind

The fall sets the undershoot

A trend-carried spread meets a turn sharper than the drift before it — the decay rising 0.02 a trace and then falling 0.06 — and the prediction was a spread about 0.6 of the truth on the trace after the turn, coverage under 55 per cent, a fresh pilot there restoring it, and the sequence's average within two points. All four hold: 0.617, 48.5 per cent, 71.0 per cent, and 67.7 against the table's 68.3. What fails is the rule the earlier essay gave for the error, the square of the last ratio. The undershoot is the last ratio over the next one, and the coverage it costs is the normal table's at that fraction.

 051015202530354010⁻³10⁻¹10¹GMRES steprelative residualthe two curves are the same curvea number re-derived, not carriedworst reported/actual factor1.8at step40‖VᵀV − I‖ of the basis1.4reported at the last step0.041actual at the last step0.074the same family of methodsand only one of them lies Iterating, instead of factorising

The number that is re-derived

GMRES prints a residual it never computes from its answer either. On the matrix that sends a conjugate gradient recurrence 7.3·10¹⁰ wrong, and on two others chosen to be worse, its number is never more than a factor of 2.86 out — while the basis it is computed from has lost orthogonality entirely. The disease is not iterative methods, and it is not floating point.

 00.250.50.75100.250.50.751xuexact at ε = 0.005exact at ε(1 + Pe)the upwind answeran exact answer to a different question‖upwind(ε) − central(ε(1+Pe))‖/‖·‖0distance to the problem it solves0.026distance to the problem posed0.36the added diffusion is h/2 = 0.001953, whatever ε isso refining removes it Iterating, instead of factorising

A different equation on every grid

Upwinding is the exact discretisation of a convection–diffusion problem with diffusion ε + h/2, entry for entry, at a relative difference of between 0 and 1.26·10⁻¹⁶ on every mesh from 15 points to 511. The equation it is exact for is chosen by the mesh and not by ε — the added diffusion is 0.01563 on a 31-point grid whether ε is 0.2 or 0.001.

 10²10⁻⁴10⁻³10⁻²10⁻¹grid points nworst nodal errorupwindtunedcentralthe same tuning, another problemtuned ÷ central at n = 3149tuned ÷ central at n = 12796central's error at the finest grid10·10⁻⁵exact on the problem it was derived fromand harmful on the one beside it Iterating, instead of factorising

A parameter that is also a price

ξ = coth(Pe) − 1/Pe is the fraction of h/2 that makes a boundary-layer solution exact at every node. On a problem with no layer in it, the error the same scheme commits is ξ times upwinding's — 0.2511 against a ξ of 0.2504, 0.7461 against 0.7448 — so the number that buys the exactness is also the invoice.

 0183654729010812610⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹products with Asizeresidual boundtrue errorbounded memorybasis vectors kept8products with A140worst error in the k wanted7.1·10⁻¹⁵the bound is free and the error is notand the basis never grows Eigenvalues, singular values, rank

The same budget, spent five ways

A restarted method has one budget — products with A — and two ways to spend it, in many short cycles or a few long ones. At about a hundred and forty products the answer is the same to a factor of seven whichever split is chosen, and the residual bound the method reports spans ten orders of magnitude across the same five runs.

 -213284300.250.50.751rotation of the anisotropy (degrees)factor / couplingusableconvergence factoraxis couplingdiagonal couplingthe standard answer, and the anglefactor at 0°0.19factor at 45°0.8axis ÷ diagonal coupling at 45°2the hierarchy reads the matrixand the matrix lost the direction Iterating, instead of factorising

How much direction there was to lose

At 45° the nine-point stencil hands smoothed aggregation the same wrong hierarchy at every anisotropy — six strong neighbours per interior point, 121 aggregates, the identical partition from ε = 10⁻⁴ to 0.099. The convergence factor that one hierarchy produces runs from 0.802 to 0.581 over the same range.

 10⁻³10⁻²10⁻¹1024681012size of the negative eigenvalue, −λproducts before the test firesharder to find, and milder8 spectra, n = 50products at the largest λ3products at the smallest10smallest share of λ recovered0.14largest0.34the one that hidesis the one that matters least Iterating, instead of factorising

A proof that does not ask how large the matrix is

Proving a Hessian indefinite costs three matrix–vector products when the negative eigenvalue is 3 and nine to eleven when it is a thousandth, and that pair of numbers barely moves across a fourfold range in n. The factorisation that settles the same question costs a third of n³, which grows by a factor of sixty-four over the same range.

 10³10⁶10⁹10¹²10¹⁵10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1largest iterate on the way, ‖x‖difference between the two residualsthe bound, linear in ‖x‖what it does, slope 0.46a bound of one, a walk of a halffitted slope0.46the bound's slope1smallest gap measured3.9·10⁻¹⁵largest gap measured1.5·10⁻¹⁰share of the bound, near end9.1·10⁻⁴share of the bound, far end3.4·10⁻⁹the bound is rightand loose by a square root Iterating, instead of factorising

A walk needs a length

The gap between the two residuals grows as the square root of something, and a square root needs a length. Two quantities are candidates — how far the iterates travelled and how many steps were taken — and only a second sweep separates them. Across a fourfold change in size the iteration count goes from 39 to 96 and the gap goes from 5.04·10⁻¹⁵ to 5.33·10⁻¹⁵.

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