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# HELP AQA A2 Computing watch

1. can someone help me with one question... module 4, Floating points numbers

Basically, you've got the number 1011 1010 0000 0011

10 bit mantissa, 6 bit exponent

1011 1010 00 | 000011

exponent +3, to get the number i move the dec point to the right 3 places

101.1101

i get 4+1. 1/2+1/4+1/16= 5 whatever....

But the marking skeems says the answer is 4 3/8 ( or 35/8)

How the hell did they get that!! Plz somebody help me... that was january 2003 paper if someone interested...
2. Lol. I'm glad I dont have to do that sh*t ever again.
3. ... hm, that's very encouraging,
4. does anyone at all doing A2 computing here?
5. this forum sux! Nobody knows the answer to the simple floating point question...
6. I can do IEEE 754 floating point, but you haven't specified in what format the mantissa and exponent are stored - biased representation? unsigned with single sign bit? two's complement?

Never did A level computing so I don't know what craz scheme they made up for that!

Alaric.
7. ops sorry forgot to tell ya... i believe it is in twos complement. any ideas how they got the answer?
8. (Original post by maljosh)
i believe it is in twos complement. any ideas how they got the answer?
In a word, no.

The first number you listed was a 15bit one, with the third block only having three numbers, makesure there isn't a 1 in there some where.
I'm also not entirely sure what their doing about normalising/denormalising and hidden bits. The answer may of course be wrong if you're doing it the way you've been taught and not getting it correct.

sorry i can't be of much more help

Alaric.
9. thanks for helping anyways.. i appreciate it mate...

i've made a correction, i missed a 0 there... it didn't affect my calculations though... it's a very confusing question... the book sux, there is no proper explanation.
10. interestingly I think if you take the first ten bits as 2s complement you get -280.
35/8 is 280/2^6
or 280 * 2^(-6).

so it seems to be fairly close, I can't quite work it out though, it may have something to do with hidden bits, but I don't know how they're dealing with that! (Oh and the negative signs are little messed)

All in all I prefer IEEE754

Alaric.
11. well, that looks like the correct answer... but i don't think they ever mentioned anything about hidded bits anywhere in the book... no way, they can give us something like that...
12. Hey, at last i know the answer...

Firstly, calculate exponent which is +3, then you place decimal point in the right place ignoring the first 1, which is just to tell us that the number is negative.

10 bit mantissa, 6 bit exponent

1 .011 1010 00 | 000011

Secondly, you twos complement mantissa, because it was in twos complement you have to twos complement it back....(don't forget we ignore the very first 1)

011 1010 00= 100011000

Thirdly, you shift decimal place 3 places to the right:

100.011000 and we get 4 + 1/4 + 1/8= -4.375

It wasn't such a difficult question afterall...

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Updated: January 12, 2004
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