# Calculating realization ratio of a value related to a target where lower values are more interesting

I want to calculate realization ratio of a value related to a target when lower values are considered better, in one special case the goal is zero. To make my problem more clear, there is a complete use case here: http://jhealthscope.com/en/articles/80342.html Where author wants to calculate the ratio in table 2: http://jhealthscope.com/en/articles/80342.html#A80342TBL2 Author have solved the same question by dividing the target by the performance where lower is better, e.g. "The job complications of personnel" index. But I have doubt about this approach and want to find the best approach and a complete answer.

If the actual value ($$a$$) is never less than the goal ($$g\ge 0$$), you could look at $$100 \frac{a-g}{a}$$. For a goal of 10 and an actual value of 22 (difference of 12), this would evaluate to about 54.55%. For a goal of 0 and an actual value of 12 (the same difference), it would evaluate to 100%. The figure represents how much you would need to improve, as a fraction of the current value, to reach your goal.

UPDATE: To avoid having the ratio default to 100% when $$g=0$$, regardless of the value of $$a$$, you can adjust the denominator as follows. Make the deviation ratio $$\frac{a - g}{\max\{a,\epsilon\}}$$ for some $$\epsilon >0$$. If the $$g=0$$, the deviation is 1 (and realization is 0) until $$a$$ gets small enough that $$a<\epsilon$$, at which point deviation is $$\frac{a}{\epsilon}$$ and realization is $$\frac{\epsilon - a}{\epsilon}$$.

• I think when kpi meets the target, we have to get 100% for the realization ratio and 0 deviation, and your answer is closer to the deviation than the realization ratio. Oct 2 '19 at 22:25
• And with this approach, I cant calculate the ratio when actual value is zero. Oct 3 '19 at 7:34
• Your question said "lower is better" and mentioned goal zero. If lower is better and outcomes cannot be negative, actual = 0 would imply goal = 0. As I said in the answer, we would just define this as 100% (which is the limit of the formula I gave as $a\rightarrow g$ when $g=0$. There is one catch: using my formula, any $a$ results in 100% when $g=0$. So the problem is not $a=0$, but you may not be happy with the formula when $g=0$. Oct 3 '19 at 18:11
• OK, if we use $(a-g) /a$ to calculate deviation then with $1-deviation=realization$ will get $realization=g/a$ that is the same formula we are using already. Oct 3 '19 at 18:45
• You could use $\frac{a-g}{\max\{a,\epsilon\}}$ for some arbitrary $\epsilon >0$ (maybe $\epsilon = 1$). If $g=0$ and $a$ gets close to 0, it will depart from the previous formula. I've seen variations of this in other contexts. Oct 3 '19 at 20:40