AMS526: Numerical Analysis I (Numerical Linear Algebra) Lecture 8: Floating Point Arithmetic; Accuracy and Stability

Xiangmin Jiao

Stony Brook University

Xiangmin Jiao Numerical Analysis I 1 / 16 Outline

1 Floating Point Arithmetic

2 Accuracy and Stability

3 Stability of Algorithms

Xiangmin Jiao Numerical Analysis I 2 / 16 Floating Point Representations Computers can only use ﬁnite number of bits to represent a real number

I Numbers cannot be arbitrarily large or small (associated risks of overflow and underflow) I There must be gaps between representable numbers (potential round-off errors) Commonly used computer-representations are floating point representations, which resemble scientific notation −1 −p+1 e ±(d0 + d1β + ··· + dp−1β )β , 0 ≤ di < β where β is base, p is digits of precision, and e is exponent between emin and emax Normalize if d0 6= 0 (except for 0) Gaps between adjacent numbers scale with size of numbers 1−p Relative resolution given by machine epsilon machine = 0.5β For all x, there exists a floating point x0 such that 0 |x − x | ≤ machine|x| Xiangmin Jiao Numerical Analysis I 3 / 16 IEEE Floating Point Representations

Single precision: 32 bits

I 1 sign bit (S), 8 exponent bits (E), 23 signiﬁcant bits (M), (−1)S × 1.M × 2E−127 −24 I machine is 2 ≈ 6e − 8 Double precision: 64 bits

I 1 sign bit (S), 11 exponent bits (E), 52 signiﬁcant bits (M), (−1)S × 1.M × 2E−1023 −53 I machine is 2 ≈ e − 16 Special quantities

I +∞ and −∞ when operation overﬂows; e.g., x/0 for nonzero x I NaN (Not a Number) is returned√ when an operation has no well-deﬁned result; e.g., 0/0, −1, arcsin(2), NaN

Xiangmin Jiao Numerical Analysis I 4 / 16 Machine Epsilon Define fl(x) as closest floating point approximation to x By definition of machine, we have:

For all x ∈ R, there exists with || ≤ machine such that ﬂ(x) = x(1 + )

Given operation +, −, ×, and / (denoted by ∗), floating point numbers x and y, and corresponding floating point arithmetic (denoted by ~), we require that x ~ y = fl(x ∗ y) This is guaranteed by IEEE floating point arithmetic Fundamental axiom of floating point arithmetic:

For all x, y ∈ F, there exists with || ≤ machine such that x ~ y = (x ∗ y)(1 + )

These properties will be the basis of error analysis with rounding errors

Xiangmin Jiao Numerical Analysis I 5 / 16 Potential “Catastrophic” Error 1: Cancellation

Catastrophic cancellation: If x ≈ y and |x − y| |x| + |y|, then the computed result xˆ − yˆ is in general inaccurate.

This is an issue if intermediate result involves cancellation errors This sometimes can be avoided by using alternative equations Example: 2 I Quadratic formula for quadratic roots of ax + bx + c = 0 can be solved as √ −b ± b2 − 4ac x = 2a or 2c x = √ −b ∓ b2 − 4ac depending on the sign of b

Xiangmin Jiao Numerical Analysis I 6 / 16 Potential “Catastrophic” Error 2: Swamping

Swamping: If |x| |y| (more precisely |x| . machine|y|), then x + y ≈ y (or fl(x) ⊕ fl(y) = fl(y)). In other words, the effect of smaller value may be lost

This is an issue if there are intermediate large values in the computation It sometimes can be avoided by reordering computations Examples: PN 1 PN 1 I Approximating e with n=0 n! versus n=0 (N−n)! I in Gaussian elimination, using small values as pivoting can lead to large intermediate values

Xiangmin Jiao Numerical Analysis I 7 / 16 Potential “Catastrophic” Error 3: Overﬂow

Overﬂow: When multiplying two very large numbers x and y, then |xy| can cause unnecessary overﬂow

This is an issue, for example, when computing squares before taking squared root, but it can often be addressed by rescaling Analogously, unnecessary underflow can occur when multiplying two small numbers, but underflow is typically less harmful Examples: 2 I In finding roots of ax + bx + c = 0, it is advisable to rescale the problem by dividing it by max{|a|, |b|, |c|} (this avoids unnecessary overflow and underflow)

x I When computing kxk2, rescale the problem to compute kxk∞ kxk ∞ 2 (i.e., first dividing x by max{|xi |} and multiplying it back. This can avoid unnecessary overflow and underflow)

Xiangmin Jiao Numerical Analysis I 8 / 16 Outline

1 Floating Point Arithmetic

2 Accuracy and Stability

3 Stability of Algorithms

Xiangmin Jiao Numerical Analysis I 9 / 16 Accuracy

Roughly speaking, accuracy means that “error” is small in an asymptotic sense, say O(machine) Notation ϕ(t) = O(ψ(t)) means ∃C s.t. |ϕ(t)| ≤ C|ψ(t)| as t approaches 0 (or ∞) 2 2 I Example: sin t = O(t ) as t → 0 If ϕ depends on s and t, then ϕ(s, t) = O(ψ(t)) means ∃C s.t. |ϕ(s, t)| ≤ C|ψ(t)| for any s as t approaches 0 (or ∞) 2 2 2 I Example: sin t sin s = O(t ) as t → 0

When we say O(machine), we are thinking of a series of idealized machines for which machine → 0

Xiangmin Jiao Numerical Analysis I 10 / 16 More on Accuracy An algorithm f˜ is accurate if relative error is in the order of machine precision, i.e., ˜ kf (x) − f (x)k/kf (x)k = O(machine), i.e., ≤ C1machine as machine → 0, where constant C1 may depend on the condition number and the algorithm itself In most cases, we expect ˜ kf (x) − f (x)k/kf (x)k = O(κmachine), i.e., ≤ Cκmachine as machine → 0, where constant C should be independent of κ and value of x (although it may depends on the dimension of x) How do we determine whether an algorithm is accurate or not?

I It turns out to be an extremely subtle question I A forward error analysis (operation by operation) is often too diﬃcult and impractical, and cannot capture dependence on condition number I An eﬀective solution is backward error analysis

Xiangmin Jiao Numerical Analysis I 11 / 16 Outline

1 Floating Point Arithmetic

2 Accuracy and Stability

3 Stability of Algorithms

Xiangmin Jiao Numerical Analysis I 12 / 16 We say an algorithm is backward stable if it gives “exactly the right answer to nearly the right question” More formally, an algorithm f˜ for problem f is backward stable if (for all x) f˜(x) = f (x˜) for some x˜ with kx˜ − xk/kxk = O(machine) Is stability or backward stability stronger?

I Backward stability is stronger. Does (backward ) stability depend on condition number of f (x)?

I No.

Xiangmin Jiao Numerical Analysis I 13 / 16 Is stability or backward stability stronger?

I Backward stability is stronger. Does (backward ) stability depend on condition number of f (x)?

I No.

Stability We say an algorithm is stable if it gives “nearly the right answer to nearly the right question” More formally, an algorithm f˜ for problem f is stable if (for all x) ˜ kf (x) − f (x˜)k/kf (x˜)k = O(machine) for some x˜ with kx˜ − xk/kxk = O(machine) We say an algorithm is backward stable if it gives “exactly the right answer to nearly the right question” More formally, an algorithm f˜ for problem f is backward stable if (for all x) f˜(x) = f (x˜) for some x˜ with kx˜ − xk/kxk = O(machine)

Xiangmin Jiao Numerical Analysis I 13 / 16 I Backward stability is stronger. Does (backward ) stability depend on condition number of f (x)?

I No.

Xiangmin Jiao Numerical Analysis I 13 / 16 I No.

I Backward stability is stronger. Does (backward ) stability depend on condition number of f (x)?

Xiangmin Jiao Numerical Analysis I 13 / 16 Stability We say an algorithm is stable if it gives “nearly the right answer to nearly the right question” More formally, an algorithm f˜ for problem f is stable if (for all x) ˜ kf (x) − f (x˜)k/kf (x˜)k = O(machine) for some x˜ with kx˜ − xk/kxk = O(machine) We say an algorithm is backward stable if it gives “exactly the right answer to nearly the right question” More formally, an algorithm f˜ for problem f is backward stable if (for all x) f˜(x) = f (x˜) for some x˜ with kx˜ − xk/kxk = O(machine) Is stability or backward stability stronger?

I Backward stability is stronger. Does (backward ) stability depend on condition number of f (x)?

I No. Xiangmin Jiao Numerical Analysis I 13 / 16 Stability of Floating Point Arithmetic Backward stability of floating point operations is implied by these two floating point axioms: 1 ∀x ∈ R, ∃, || ≤ machine s.t. fl(x) = x(1 + ) 2 For floating-point numbers x, y, ∃, || ≤ machine s.t. x ~ y = (x ∗ y)(1 + ) Example: Subtraction f (x1, x2) = x1 − x2 with floating-point operation ˜ f (x1, x2) = fl(x1) fl(x2)

I Axiom 1 implies fl(x1) = x1(1 + 1), fl(x2) = x2(1 + 2), for some |1|, |2| ≤ machine I Axiom 2 implies fl(x1) fl(x2) = (fl(x1) − fl(x2))(1 + 3) for some |3| ≤ machine I Therefore,

ﬂ(x1) ﬂ(x2) = (x1(1 + 1) − x2(1 + 2))(1 + 3)

= x1(1 + 1)(1 + 3) − x2(1 + 2)(1 + 3)

= x1(1 + 4) − x2(1 + 5) where | |, | | ≤ 2 + O(2 ) 4 5 machine machine Xiangmin Jiao Numerical Analysis I 14 / 16 Stability of Floating Point Arithmetic Cont’d

Example: Inner product f (x, y) = xT y using floating-point operations ⊗ and ⊕ is backward stable Example: Outer product f (x, y) = xy T using ⊗ and ⊕ is not backward stable Example: f (x) = x + 1 computed as f˜(x) = fl(x) ⊕ 1 is not backward stable Example: f (x, y) = x + y computed as f˜(x, y) = fl(x) ⊕ fl(y) is backward stable

Xiangmin Jiao Numerical Analysis I 15 / 16 Proof: Backward stability means f˜(x) = f (x˜) for x˜ such that kx˜ − xk/kxk = O(machine) Deﬁnition of condition number gives

kf (x˜) − f (x)k/kf (x)k ≤ (κ(x) + o(1))kx˜ − xk/kxk

where o(1) → 0 as machine → 0. Combining the two gives desired result.

Accuracy of Backward Stable Algorithm

Xiangmin Jiao Numerical Analysis I 16 / 16 Accuracy of Backward Stable Algorithm

Theorem If a backward stable algorithm f˜ is used to solve a problem f with condition number κ using ﬂoating-point numbers satisfying the two axioms, then ˜ kf (x) − f (x)k/kf (x)k = O(κ(x)machine) Proof: Backward stability means f˜(x) = f (x˜) for x˜ such that kx˜ − xk/kxk = O(machine) Deﬁnition of condition number gives

kf (x˜) − f (x)k/kf (x)k ≤ (κ(x) + o(1))kx˜ − xk/kxk

where o(1) → 0 as machine → 0. Combining the two gives desired result.

Xiangmin Jiao Numerical Analysis I 16 / 16