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I really have no idea of how to do these questions - in fact I have no idea of how to do any question in the paper - but I have tried to figure out what's going on in the course called Computational Mathematics but the lecturer's notes are honestly unless to someone who doesn't have a strong background maths.

The course also has a high failure.

I'm trying to find materials online but the course isn't focused on one thing, I even asked the lecturer for a recommended book but he said there isn't one book that covers the whole module, so I'm really stuck. Here a link to the exam paper. Link

Here's the first question from last year's paper.


Please note that I'm not asking for just the solutions but an explanation and probably a link, so that I can have a background knowledge and so that I'll be able to answer similar questions myself. This is not an assignment, I'm just preparing for an exam.

Thank you. :)



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What do you mean by "a floating-point system defined by parameters $\beta,s,m,M$"? Presumably this is defined in your course, but you have to tell us how! – TonyK Apr 17 '14 at 12:14
@TonyK Please, see edit. – Adegoke A Apr 17 '14 at 12:17
@Amzoti Thanks a lot, these links are very useful. – Adegoke A Apr 17 '14 at 13:34

What I can help is to provide an analogue using base 10 floating point numbers.

If it is non-normalized, then it has infinitely many non-unique representations. Examples are:-

6.25 = 0.625 * 10………..…… (1)

6.25 = 0.0625*100…………… (2)

6.25 = 625 * 10^(-2)……..… (3)

This is not a ‘healthy’ environment because a number has so many 'looking different' but in fact equivalent representations. In order to ensure the representation of a number is unique, normalization is necessary.

Normalization requires:-

I. All number should start as $0.d_1d_2d_3…d_s$ where the $d_i’s$ are the extracted digits.

II. The leading digit (i.e. $d_1$) must not be zero and other digits have no such a restriction. This is formally stated as $1 \le d_1 \le 10 – 1$ and $0 \le d_i \le 10 – 1$ for $i = 2, … ,s.$ At this stage, only (1) above can meet the requirement.

III. In order to make the so far representation numerically equivalent to the original, it must be compensated by multiplied a suitable exponent. That is, $*10^e$ for some suitable integer e; and e can be 0, + or –.

IV. If the number is 0, then ……

Thus, the normalized representation of $6.25$ is $0.6250000000...00 * 10^1$; the number of 0s appended depends on the size of the ‘container’ or ‘WORD’.

If the size of the WORD and the m and M (as in $m \le e \le M$) are given, one can find the smallest and largest number that this system can hold.

Example in addition of two floating numbers using a simplified representation

$6.25 + 703.94 = 0.625 * 10^1 + 0.70394 *10^3$

$= 0.00625 *10^3 + 0.70394 *10^3$

$= 0.71019 *10^3$

$= 710.19$

Note-1: Add./sub. must be done when the 2 operands are converted to the ‘same level’ first.

Note-2: It is possible that some data are lost due to conversion.

Note-3: The result might exceed the upper/lower limit (i.e. an overflow or underflow).

Example in multiplication of two floating numbers using a simplified representation

$6.25 * 703.94 = (0.625 * 10^1) * (0.70394 *10^3)$

$= (0.625 * 0.70394) *10^{1 + 3}$

$= 0.4399625 *10^4$

Note-4: Comment in Note-3 applies.

Note-5: Truncation may occur.

Further note:- In evaluating an expression via more than one steps, different orders or operations may yield different results. Example, computing the average of a and b by (1) $(a + b)/2$ and by (2) $a + (b – a)/2$ may yield different results due to errors like truncation.

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Why infinitely many representations if the representation usually must have finite length? – Ruslan Apr 17 '14 at 15:15
@Mick Thank you. – Adegoke A Apr 17 '14 at 16:35
The example (6.25) shown in the introductory part in deed has infinitely many equivalent forms when normalization is not enforced. Varying the exponent can generate that many. Note also that, at that stage, it has not been placed in the WORD of finite length yet. – Mick Apr 17 '14 at 16:37
You are welcome. Hope it helps. – Mick Apr 17 '14 at 16:41

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