Lab 2: Floating-Point Lab · Numerical Methods

Type any number (you can also type fractions such as 1/3). The sign, exponent and fraction bits update instantly, with the exact decimal value that is stored.

Try this: 0.7 (is the stored value above or below?), 1e22 and 1e23 (one is exact), 1/3 in single vs double, 2^53 + 1 = 9007199254740993.

A toy floating-point system with only a few bits. Move x and see which number it is stored as. Change the number of fraction bits.

Decimal to binary

Try this

Run it for 0.1. At which step does the pattern start to repeat?
Change x to 0.625. Why does it stop after 3 steps?
Try 0.3 and 0.2. Which bits repeat?

Machine epsilon

Try this

Run in Python and in C. Same double epsilon?
In C, what is the float epsilon?
Change the test to 1.0 + eps != 1.0 (no /2). What do you get and why is it ε/2?

float vs double

Try this

Run it. How big is the float error?
Change n to 10000000. The true sum is 1000000; why is the float sum about 1087937? (Hint: the gap between floats above 524288 is 0.0625)
Keep fs as float but add 0.1 (a double). Does it help?

Q1 Convert

0.6875_{10}

to binary.

0.6875\times2=1.375\to1

,

0.375\times2=0.75\to0

,

0.75\times2=1.5\to1

,

0.5\times2=1.0\to1

. So

0.1011_2

(exact).

Q2 Write

-0.0003470

in normalised decimal floating-point form. How many significant digits?

-0.3470\times10^{-3}

; 4 significant digits (the trailing zero after 7 is significant, the leading zeros are not).

Q3 Represent

2/3

with 4-digit chopping and 4-digit rounding. Give the absolute and relative error of each.

Chopping:

0.6666

, error

6.67\times10^{-5}

, relative

1.0\times10^{-4}

. Rounding:

0.6667

, error

3.33\times10^{-5}

, relative

5.0\times10^{-5}

.

Q4 Give the IEEE 754 single-precision bit pattern of

5.75

.

5.75 = 101.11_2 = 1.0111_2\times2^{2}

. Sign 0, exponent

2+127 = 129 = 10000001

, fraction

0111000\dots

. Bits: 0 10000001 01110000000000000000000 = 0x40B80000.

Q5 What is machine epsilon, and why can an iteration never meet the tolerance

10^{-20}

in double precision?

\varepsilon = 2^{-52} \approx 2.2\times10^{-16}

is the gap between 1 and the next double. Neighbouring doubles near

x

differ by about

\varepsilon|x|

, so two different iterates near 1 always differ by at least

2.2\times10^{-16} > 10^{-20}

.

Q6 Why is if (x == 0.3) dangerous after computing x = 0.1 + 0.2? What should be written?

Both sides carry different rounding errors (0.30000000000000004 vs 0.29999999999999998890), so the test is false. Write fabs(x - 0.3) < 1e-12, or a relative test fabs(x - y) <= 1e-12 * fabs(y).