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OCR H446 1.4.1 Floating-point arithmetic
Part 7 of 11 · H446 1.4.1 · Data types
Once the fields can be read and normalised, H446 1.4.1 expects students to calculate with them and to show every stage. This is the arithmetic worksheet in the floating-point sequence, set in a digital mixer combining signed adjustments: exponents are aligned, signed mantissas are added or subtracted, results are normalised, and precision loss is diagnosed rather than ignored.
Students will:
- align two values to a common exponent before any mantissa arithmetic
- add and subtract signed mantissas at a fixed width
- normalise a result and adjust its exponent to match
- explain why bits pushed out of the field during alignment cost precision
- recognise when a correctly normalised result needs an exponent the field cannot hold
Inside: 7 explanation cells, 2 multiple-choice questions, 1 fill-in-the-blanks cell, 1 written answer and 2 number answers. 26 marks, about 50 to 60 minutes.
Series: H446 1.4.1 · Data types, part 7 of 11.
Shared by Coding PathwayVerified teacher
- 13 cells
- About 60 minutes
- CC BY-SA 4.0
- Shared 31 Aug 2026
- Updated 3 Sept 2026
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The whole resource, exactly as a class sees it. Answers and marking are held back.
Floating-point arithmetic
A digital mixer combines signed signal adjustments. OCR rewards visible stages: exponent interpretation, alignment, mantissa arithmetic and the final normalised fields.
By the end, you will be able to
- align two floating-point values to a common exponent
- add or subtract signed mantissas
- normalise the result
- explain precision loss caused by discarded bits
Reactivate: reading and normalising OCR-format values.
Four-stage method
- Interpret and compare the signed exponents.
- Align the values to a common exponent by shifting the mantissa of one value.
- Add or subtract the aligned two's-complement mantissas.
- Normalise the result and adjust its exponent.
Worked addition: 010000 0010 represents 2 and 010000 0001 represents 1. Align to exponent 2 by changing the second value to 001000 0010. Add the aligned mantissas: 010000 + 001000 = 011000. The result 011000 0010 is normalised and represents 3. The exponent was matched before the mantissas were combined.
Bits discarded during alignment cannot be recovered; this can reduce precision.
Worked model: complete an aligned addition
Suppose A = 0.10100 × 2³ and B = 0.11000 × 2¹. To use exponent 3 for both, shift B’s mantissa right by two places:
A = 0.10100 × 2³
B = 0.00110 × 2³
--------
0.11010 × 2³
The mantissa sum 0.11010 is already normalised because it begins 01, so the final fields are mantissa 011010 and exponent 0011. A denary sense check gives 5 + 1.5 = 6.5, while 0.11010₂ × 2³ = 6.5.
Record any non-zero bits that leave the fixed-width mantissa during alignment: they are lost precision, not bits that normalisation can recover.
Before adding two floating-point mantissas, what must be true?
- ABoth mantissas begin with 01.
- BBoth exponents represent the same value.
- CBoth bit patterns are positive.
- DBoth exponents are zero.
Add each pair of OCR-format floating-point values. Record exponent alignment and the resulting fields.
a)Add the floating point numbers 011110 (exponent 0001) and 010000 (exponent 0010). Give a normalised answer.
b)Add the floating point numbers 011000 (exponent 0001) and 010111 (exponent 0011). Give a normalised answer.
Subtraction
Treat subtraction as signed mantissa arithmetic after the exponents are aligned. Do not convert the whole question to denary and expect to earn every binary-method mark: keep exponent conversion, alignment and mantissa working visible.
Guided four-line layout
- exponent values: e₁ = ___, e₂ = ___;
- aligned operands at exponent ___;
- signed mantissa calculation;
- normalised mantissa and adjusted exponent.
After calculating, check sign and approximate magnitude against the original operands. A positive plus a small negative should not unexpectedly become a very large positive value. If the required exponent is above the maximum, exponent overflow occurs; if a non-zero magnitude is too small for the minimum exponent and mantissa precision, it may underflow towards zero.
Subtract each second floating-point value from the first. Show the staged method.
a)Work out 011111 (exponent 0010) − 010100 (exponent 0001) in floating point. Give a normalised answer.
b)Work out 011100 (exponent 1111) − 011110 (exponent 1110) in floating point. Give a normalised answer.
During alignment, a mantissa is shifted right by three places and non-zero bits leave the fixed-width field. Explain why the final result may be imprecise.
Trace what happens to the discarded low-value bits and whether normalisation can restore them.
Students type their answer here.
A correctly normalised result needs an exponent larger than the field can represent. What has occurred?
- AMantissa underflow
- BA character-set error
- CExponent overflow
- DSign-magnitude cancellation
Closed-book checkpoint
Complete each sentence from memory. There is no answer bank and correctness is held for teacher review.
Retrieval check
Why must exponents match? Which operand shifts? Where can precision be lost? Why can normalisation not recreate discarded bits?