Decomposing change in a ratio I am looking for some help on the decomposition of effects. I am forming a ratio (R) of a to b at t=1 and t=0. I want to express the change in this ratio (Delta R) as the sum of a "contribution of a" and a "contribution of b" component. These two components should be additive in nature.
Here is a numeric example:  
    t=1   t=0   
a   200   150   
b   175   100   

R   1.14  1.5   

Delta R -0.36 = "contribution of a" + "contribution of b"

Is it possible to solve this without an approximation? If not, which would be the easiest approximation to solve these calculations irrespective of magnitude? 
Many thanks! 
 A: For the exact change in the ratio we have, $$\Delta\left(\frac ab\right) = {a+\Delta a \over b+\Delta b} - \frac ab = {b \Delta a - a \Delta b \over b(b-\Delta b)}.$$ Although you can get a term that only contains $\Delta b$ from this, there’s no way that I can see to get one that involves only $\Delta a$.  
For small values of $\Delta a$ and $\Delta b$, the change in the ratio can be approximated by the differential: $$D\left(\frac ab\right)[\Delta a,\Delta b] = \left(\frac1b,-\frac a{b^2}\right)\cdot\left(\Delta a,\Delta b\right) = {b\Delta a-a\Delta b\over b^2}.$$ For your example, this approximation predicts a change in the ratio of $0.625$, almost double the actual change, so is unfortunately not likely to be of use to you.  
A possible way to separate the effects of $\Delta a$ and $\Delta b$ is to look at the change in the logarithm of the ratio instead: $$\ln\left({a+\Delta a \over b+\Delta b}\right)-\ln\left(\frac ab\right) = \ln\left(1+\frac{\Delta a}a\right)-\ln\left(1+\frac{\Delta b}b\right).$$
