Quandaries
and Queries 

My Name is Rob
What bothers me about the calculation is that the standard hours get represented as a percentage of the actual hours and in my opinion changes the focus of the calculation from standard or target, where it should be, to the actual hours. I cannot define why, but this just seems inherently wrong to me.
My calculation is like a %change from standard calculation. However, there is something that also concerns me about my calculation. If you substitute 100 for a and 50 for s, then you come to a quandary, because if you plug those numbers into the second equation the result is of course zero % efficient which doesn't sit right with me either. If you plug them into the first calculation you get 50% efficiency which doesn't really seem to work either, because you require 100% more hours to do the same work in this case. ??? Any help you could provide here would be much appreciated! 

Hi Rob, I agree that the s should be in the denominator. You want the efficiency to be a percentage of the target. (I am going to look at some different expressions for efficiency so I am going to use subscripts to distinguish them.) From the ordinary English usage of the word efficiency I think that an inefficient system should have an efficiency value less than one, and a perfectly working system should have an efficiency of one. You could try
When a = s this gives an efficiency of 1 but for a = 4000 and s = 3000 this gives a value greater than one. You could try
but when a = s this gives an efficiency of zero. You could try
This is equivalent to your expression (1+((sa)/s)).
As you can see I am not going to give you a definitive answer to your question. This is really a question about what you believe efficiency to be and how you model it mathematically. I see s as being fixed, the target, and a measured quantity a that you want to relate to s. Lets try to do something more general. Both expressions E_{1} and E_{2} of the form
where p and q are some numbers. For E_{1}, p = 1 and q = 0. For E_{2}, p = 1 and q = 2. If a system is to have efficiency one when a = s then this says
thus
All that is left is to determine the value of p. For example if a is 2 s then you might want the efficiency is ^{1}/_{2}. In other words, when a = 2 s
In this case
The point here is that if you want efficiency to be expressed as
for some p and q, and if you want the efficiency to be 1 when a = s, then one more stipulation completely determines the expression for E. Harley 

