Separate impact of change of numerator and denominator on ratio Please help me with this issue:
In period 1, 
   a = 15, b = 10, therefore a/b = 1.5
In period 2, 
   a = 12, b= 12, therefore a/b = 1
As you see, a declined by 20%, and b increased by 20%, as well. However, their ratio declined by 50%. 
What contributed MORE to the decline of 50% in ratio, increase of 20% in a, or decrease of 20% in b? 
How to measure what impacts the ratios most? Change in numerator or change in denominator?
 A: Let's say, we initially had some positive numbers $a$ and $b$ such that:
$$\frac{a}{b}=r$$
Now $a$ and $b$ have changed by a certain percentage, meaning:
$$a_1=a \cdot (1+x)$$
$$b_1=b \cdot (1+y)$$
Where $x$ and $y$ are real numbers (the percentages are given by $100 \cdot x$ and $100 \cdot y$), which could be negative or positive, corresponding to decrease or increase. We can take them to be $ \in(-1,1)$ so the change is less than $100$ %.
Then we have:
$$\frac{a_1}{b_1}=\frac{a \cdot (1+x)}{b \cdot (1+y)}= \frac{1+x}{1+y} \cdot r$$
Speaking about percentage again, write the change in $r$ as:
$$r_1=r  \cdot (1+z)$$
Where:
$$z=\frac{1+x}{1+y}-1=\frac{x-y}{1+y}$$

Your question is "what contributed more to the change in ratio?" It is not really clear to me what you are asking. But there are ways to see how $x$ and $y$ contribute to the change separately.

Let's consider your case, initially:
$$x=0 \\ y=0$$
And:
$$\Delta x=-0.2 \\ \Delta y=0.2$$
Then, applying the changes separately we have:
$$\Delta_x z=\frac{x+\Delta x-y}{1+y}-\frac{x-y}{1+y}=\frac{0-0.2-0}{1+0}-\frac{0-0}{1+0}=-0.2$$
$$\Delta_y z=\frac{x-y-\Delta y}{1+y+\Delta y}-\frac{x-y}{1+y}=\frac{-0.2}{1+0.2}=-0.16666\dots$$

It seems that the change in $x$ contributed more in this case. But you can try some other cases to see what happens.

Applying the changes at the same time we have:
$$\Delta z=\frac{x+\Delta x-y-\Delta y}{1+y+\Delta y}-\frac{x-y}{1+y}=\frac{-0.2-0.2}{1+0.2}=-\frac{0.4}{1.2}=-\frac{1}{3}=-0.33333$$

The change in ratio was actually $ \approx -33.3 $%, not $-50$ % as you seem to believe.

