I have two sets of large positive numbers $a_1,\ldots,a_n$ and $b_1,\ldots,b_n$. By 'large' I mean of the order of $10^{10}$. I want to calculate the ratio $$R = \frac{a_1 - a_2 + \cdots +(-1)^{n+1}a_n}{b_1 - b_2 + \cdots +(-1)^{n+1}b_n}.$$ It is already known that the exact value of $R$ is between $0$ and $1$. Each of the numerator and denominator is expected to be of the order of $10^{-1}$ to $10^1$.

I would like to know if there is an algorithm to compute ratios like $R$ stably (exactly or approximately), so that the numbers involved in each step are not too large to cause overflow errors using finite-precision arithmetic. I am using the usual 64-bit double data type in Matlab.

I am aware of the Kahan summation algorithm, but using it to compute the numerator and denominator separately may not guarantee that their ratio is also as accurate.

  • 1
    $\begingroup$ There are multiple ways in which division itself is done numerically and some of them employ subtraction as the fundamental operation - restorative division being one of them. You can use Kahan summation on even the division algorithm and maintain the same level of accuracy. More information on web.stanford.edu/class/ee486/doc/chap5.pdf $\endgroup$
    – gpavanb
    Aug 1, 2017 at 21:25
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    $\begingroup$ Why is the division the problem here? Doesn't the alternation in the enumerator and denominator already take care of the problem and lead to moderately sized terms for the division? I would have imagined that any round-off problem appears in the sums themselves. $\endgroup$ Aug 6, 2017 at 22:02

1 Answer 1


Using Kahan summation algorithm should suffice your purpose. Having: $$ R=\frac{\sum\limits_{i=1}^{N}a_i}{\sum\limits_{i=1}^{N}b_i}=\frac{A}{B} $$ if $A$ and $B$ are computed accurate enough, the final ratio $R$ should not be a problem.

If $A$ and $B$ are computed using Kahan summation, you can expect the relative error to be $\mathcal{O}(\epsilon_\text{mach}+N\epsilon_\text{mach}^2)$. For double-precision and $N\approx 10^{10}$ you are safe ($N\ll 1/\epsilon_\text{mach}$), and your error is practically independent of $N$: $\mathcal{O}(\epsilon_\text{mach})$.

It's worth to mention, that the constant in front of the error estimate (condition # of summation $\kappa_{\sum}$) can still be relatively large. If summation condition number:

$$ \kappa_{\sum_A}=\frac{\sum\limits_{i=1}^{N}|a_i|}{\left|\sum\limits_{i=1}^{N}a_i\right|} $$

is large, your best bets would probably be in Shewchuk's algorithm. With that, you will be able to calculate your numerator and denominator to full double precision.

If $\tilde{A}$ and $\tilde{B}$ are calculated in full double precision (say using Shewchuk's algorithm): $$ \tilde{A}=A(1+\delta_A),\quad \tilde{B}=B(1+\delta_B),\quad |\delta_A|,|\delta_B|\le\epsilon_\text{mach} $$ , then

$$ \begin{aligned} \tilde{R}&=\tilde{A}\oslash\tilde{B}&=\frac{A(1+\delta_A)}{B(1+\delta_B)}(1+\delta_{\oslash})=\frac{A}{B}(1+\theta)\\ &&|\theta|\leq\frac{3\epsilon_\text{mach}}{1-3\epsilon_\text{mach}} \end{aligned} $$ Here, $\tilde{A},\tilde{B},\tilde{R}$ are floating point (IEEE-754) representation of $A$, $B$, and $R$, respectively, $\oslash$ - floating point analog of division.

Now, even if $A$ and $B$ are not calculated to the full precision (Kahan), it will just increase $\delta_A$ and $\delta_B$ (not allowing approximation by $\theta$ with such a tight bound). However, I still do not see how the division can be the problem here. Especially, since the true expected values of $A$ and $B$ are in $[10^{-1}, 10^{1}]$.

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    $\begingroup$ I think there is some confusion on the word "large". I think the OP means that his numbers $a_i$ and $b_i$ are large, not that $N$ is $\mathcal{O}(10^{10})$. So I don't know why he just cannot scale his numbers (if they are all the same order of magnitude). $\endgroup$
    – GertVdE
    Apr 20, 2018 at 8:04

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