# Separate impact of change of numerator and denominator on ratio

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?

• b increased, not declined – Yuriy S Feb 14 '18 at 8:41
• true. Sorry my mistake – MRMODELLING Feb 14 '18 at 8:41
• Please correct the question and maybe try making it more clear – Yuriy S Feb 14 '18 at 8:47
• Thank you for advice Yuiry. Is it more clear now? – MRMODELLING Feb 14 '18 at 8:51
• Yes, it is a little more clear. I'll see if I can answer in a way that helps – Yuriy S Feb 14 '18 at 8:55

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.