# Thread: absolute and relative error

1. ## absolute and relative error

Hey just need some help with absolute and relative errors

q) find the absolute and relative error when a computer stores 2/3 (two thirds)

q2)then calculate the absolute and relative error when a computer adds
2/3 + 2/3 + 2/3 (two thirds + two thirds + two thirds)

Some direction to a tutorial or some help would be nice

2. Originally Posted by syner
Hey just need some help with absolute and relative errors

q) find the absolute and relative error when a computer stores 2/3 (two thirds)
In binary:

$\displaystyle \frac{2}{3}=0b10101010101....$

Now how many bits do you have in the mantissa? If you are using normalized floating point you will have an extra bit of precision.

So if we have a 23 bit mantissa we have 24 bits of precision so our floating point representation of $\displaystyle 2/3$ is:

$\displaystyle u=0b101010101010101010101010$

So the error is $\displaystyle \varepsilon=(2/3) 2^{-24}$

(If the representation rounds rather than truncates then:

$\displaystyle u=0b101010101010101010101011$

and you will need to recompute the exact error)

RonL

3. Oh ok, what was the algorithm (method) that you used to get the binary result of $\displaystyle 2/3$ ? (two over three)
and also the algorithm that you used to get the final results etc? sorry just trying to learn this stuff

4. Originally Posted by syner
Oh ok, what was the algorithm (method) that you used to get the binary result of $\displaystyle 2/3$ ? (two over three)
and also the algorithm that you used to get the final results etc? sorry just trying to learn this stuff
No explicit algorithm I just expanded it using the definition of a binary fraction, whuich could be considered a/the greedy algorithm. Then truncating at the effective number of bits.

RonL

5. Lol surley there must be some sort of way that you worked out the binary.

I know how to work from binary to hex, vice versa, floating point representation (now after abit of study)

but this is the one bit that i am stumped at, what working did you use to get the binary value?

Lol still abit lost to how i figure out the relative and absolute error, sorry i think i suffer from maths dislexia

6. Originally Posted by syner
Lol surley there must be some sort of way that you worked out the binary.

I know how to work from binary to hex, vice versa, floating point representation (now after abit of study)

but this is the one bit that i am stumped at, what working did you use to get the binary value?

Lol still abit lost to how i figure out the relative and absolute error, sorry i think i suffer from maths dislexia
Greedy algorithm:

Code:
x: real <1   //input is a real less than 1
max=24     //number of bits required

out=""      //set output to null string

for idx=1 to max
if x>=1/2           //if x>=1/2 tag a 1 onto end of output string
out=out|"1"     //and reduce x by 1/2
x=x-1/2
else
out=out|"0"     //if x<1/2 tag a 0 onto end of output string
endif
endfor

return out
RonL