# With 2 lists of numbers is it possible to get the actual ratio of the sum of both lists with only the ratio of each individual group?

To get the ratio of two lists of numbers, I take the sum of each column and divide the first by the second.

Suppose I only have 'access' to the ratio of each individual number combo. Is it possible to get the 'total' ratio from just the individual ratios?

Taking the average of the individual ratios gets me pretty close but it's no good.

EDIT: Thanks for the re-tagging :)

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Since the ratio of totals can change while the individual ratios stay the same, the answer is no. – Henry Apr 12 '12 at 22:32
oh :( Alright. Thanks for your help. If you submit a answer, I'll accept it for you. – Robo Apr 12 '12 at 22:35

Since the ratio of totals can change while the individual ratios stay the same, the answer is no.

Here is an example where two single ratios (of 3 and 5) can correspond to different ratios of the totals

     num denom ratio
a     3   1      3
b     7   1      7
------------------
sum  10   2      5

num denom ratio
a     9   3      3
b     7   1      7
------------------
sum  16   4      4

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I went the other way around with my answer (one total ratio can be found by two different ratio vectors), but this is more convenient for an answer. +1 – Patrick Da Silva Apr 12 '12 at 22:42

So clearly, you are given $a_1, a_2, a_3, b_1, b_2, b_3$. You're computing the individual ratios $r_i = a_i / b_i$, you sum them to form $R_1 = r_1 + r_2 + r_3$. Then you compute another ratio which is $(a_1 + a_2 + a_3)/(b_1 + b_2 + b_3)$, and you expect that given $r_1, r_2, r_3$, you can compute $(a_1 + a_2 + a_3) / (b_1 + b_2 + b_3)$. If I am wrong on this description correct me.

Let's see what we can do. I'll try to express the total ratio in terms of the $r_i$'s and assume they are real numbers. You can compute $$\frac{a_1 + a_2 + a_3}{b_1 + b_2 + b_3} = \frac{r_1 b_1 + r_2 b_2 + r_3 b_3}{r_1 a_1 + r_2 a_2 + r_3 a_3} = \frac{r \cdot b}{r \cdot a}$$ by letting $r = (r_1, r_2, r_3)$, $a = (a_1, a_2, a_3)$, and $b = (b_1, b_2, b_3)$, with the $\cdot$ meaning scalar product (hence my algebra-precalculus tag). You have access to the vector $r$ and wish to know the ratio $(r \cdot b)/(r \cdot a)$. As you can see from linear algebra, you can find two different ratios that will give you the same total ratio. For instance, if $\lambda \in \mathbb R \backslash \{0 \}$, $$\frac{r \cdot b}{r \cdot a} = \frac{(\lambda r) \cdot ((1/\lambda)b)}{(\lambda r) \cdot ((1/\lambda) a)}$$ Therefore, given the ratios $r_1, r_2, r_3$ or the ratios $\lambda r_1, \lambda r_2, \lambda r_3$, you find the same total ratio. This makes no hope of computing the total ratio with only the "component-wise ratios", since you can take component wise ratios as big as you want and still obtain the same total ratio. But then again it doesn't mean that there exist no such formula ; I just have little faith that such a formula exists. It doesn't seem very useful either, to be honest ; why do you want to compute things this way? With more information perhaps I could help on the problem behind the question.

Hope that helps,

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Unfortunately, the ratio of the components is the only thing that we have access to due to the way the data was processed and stored in a database. I was hoping there was a mathematical solution to this problem so I wouldn't have to re-write the application I'm working on. – Robo Apr 12 '12 at 22:51