Worksheet

A7.4 Errors, range and precision

Data representation · A level · OCR H446 1.4.1, AQA 7517 4.5.4.5, Eduqas A500QS 2.3 · about 30 min

What this lesson is about

Rounding errors, absolute and relative error, range against precision, overflow and underflow, with a real sensor reading.

• Rounding errors
• Absolute and relative errors
• Range and precision
• Overflow and underflow
• A real reading

Questions 6 marks in all

1. [1 mark]A true value of 2.7 is stored as 2.625. What is the absolute error?

2. [1 mark]A true value of 2.7 is stored as 2.625. What is the relative error as a percentage, to 2 decimal places?

3. [1 mark]A floating point format keeps 12 bits in total. The mantissa is given more bits and the exponent fewer. What changes?

   1. APrecision increases and range decreases
   2. BRange increases and precision decreases
   3. CBoth increase
   4. DNeither changes

4. [1 mark]What is underflow?

   1. AA result too close to zero for the format to represent
   2. BA result too large for the format
   3. CA negative result from adding two positive numbers
   4. DA loss of the carry bit

5. [1 mark]Why can 0.1 not be stored exactly in binary floating point?

   1. AIts binary fraction repeats for ever, so it must be cut off
   2. BIt is an irrational number
   3. CIt is too small for the exponent
   4. DBinary cannot store fractions

6. [1 mark]What does this program print?

   a = 0.1 + 0.2
   print(a == 0.3)
   print(abs(a - 0.3) < 1e-9)

The task: store a reading

The robot faces a wall. Read distance() once (it returns centimetres) and convert it to metres. Store the metres in a byte of unsigned fixed point with all 8 bits after the point: the stored whole number is int(metres * 256), which truncates. Then print exactly these five lines, calculating every value: - reading: <metres> m, the metres as Python prints the float - stored: <bits>, the stored whole number as 8 binary digits (you may use format(q, "08b")) - stored value: <value> m, the stored whole number divided by 256 - absolute error: <error> m, the absolute error, then round(error, 7) - relative error: <percent>%, the relative error as a percentage, then round(percent, 2), with no space before the %

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

metres = distance() / 100
print("reading:", metres, "m")

Plan your program here, then type it in and press Run.

Do it on the robot
www.bugbotlab.com/learn/a7-4-errors-range-and-precision/
The simulator checks it and tells you when it passes. Nothing to install, no account.

Challenges

1. Round instead of truncating. Does the absolute error go down for this reading? Is that true for every reading?
2. Use 16 bits with 8 after the point. What are the new range and precision?
3. Find the largest reading this byte format can hold, and what happens to a reading of 1.2 m.