Quandaries and Queries My Name is Rob I am a cost accountant I don't know what the level of the question would be.  I can tell you I didn't go beyond grade 12 algebra in terms of mathematical complexity and most of even that now escapes my flagging memory.   The problem, on the surface, seems very simple and yet has created some controversy among a group of accountants.  The problem itself has to do with labour efficiency rates and only involves two variables; standard working hours, and actual working hours.  The difficulty lies in deriving an efficiency % from these two numbers.   Standard working hours or the targeted number of labour hours required to produce one widget, which I will represent as "s". Actual working hour or the actual number of labour hours require to produce one widget, which I will represent as "a". Labour efficiency I will represent with "E". The prevailing calculation with which I have a problem with is this: s/a=E  or if s=3000, and a=4000 then 3000/4000=75% 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. The calculation that I use: (1+((s-a)/s))=E or if s=3000, and a=4000 then (1+((3000-4000)/3000))=66.67% 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. ???   Is the first calculation correct?  Am I missing something altogether?  Are both calculations off base?  Any help you could provide here would be much appreciated! Thanks for your time. Rob 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 E1 = a/s 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 E2 = 1 -  a/s but when a = s this gives an efficiency of zero. You could try E3 = 2 -  a/s This is equivalent to your expression (1+((s-a)/s)). Aside 1+((s-a)/s = 1 + s/s - a/s = 1 + 1 - a/s = 2 - a/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 E1 and E2 of the form E = p  a/s + q where p and q are some numbers. For E1, p = 1 and q = 0. For E2, p = -1 and q = 2. If a system is to have efficiency one when a = s then this says E = p  s/s + q =1 so p + q = 1 or q = 1 - p thus E = p  a/s + 1 - p 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 E = p  2s/s + 1 - p = 1/2 or 2 p + 1 - p = 1/2 p = -1/2 In this case E = 1/2 (3 - a/s ) The point here is that if you want efficiency to be expressed as E = p  a/s + q 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 Go to Math Central