
MHB Master
#1
February 6th, 2014,
08:12
Hey!!! I have also an other question
Suppose the base $\beta =10$ ,the precision $t=3$, $L=U=10$ and $$A=(317+0.3)(171.499+145.501)$$
I have to find the relative error for $A$.
We don't make rounding.For example,if we have the value $345.924$ it is equal to $0.345924*10^3$ and the corresponding floating number is $0.345*10^3$.
I found that it is equal to
$$\frac{Afl(A)}{A}=\frac{7}{5}=2.33$$
Could you tell me if it is right?

February 6th, 2014 08:12
# ADS
Circuit advertisement

MHB Master
#2
February 6th, 2014,
15:10
Thread Author
Originally Posted by
evinda
Hey!!!
I have also an other question
Suppose the base $\beta =10$ ,the precision $t=3$, $L=U=10$ and $$A=(317+0.3)(171.499+145.501)$$
I have to find the relative error for $A$.
We don't make rounding.For example,if we have the value $345.924$ it is equal to $0.345924*10^3$ and the corresponding floating number is $0.345*10^3$.
I found that it is equal to
$$\frac{Afl(A)}{A}=\frac{7}{5}=2.33$$
Could you tell me if it is right?
I found that A=0.3 and fl(A)=1..

MHB Seeker
#3
February 6th, 2014,
20:09

MHB Master
#4
February 7th, 2014,
07:49
Thread Author
Originally Posted by
I like Serena
Looks good!!
... but doesn't that mean that:
$$\frac{Afl(A)}{A} = \frac{0.3  1}{0.3} = \frac{0.7}{0.3} = 2.33$$
Oh wait! You also got $2.33$... while you shouldn't have.
I accidentally wrote $\frac{7}{5}$ I meant that it is equal to $\frac{7}{3}$..
Thank you very much!!!!!