Reduce Math in Formula Algebraically: BestCost

[1plus1is10]

Is there a way to reduce the math in my formula algebraically:
BestCost = lowest+((((highest+lowest)/2)-lowest)*0.6667);

I want to use BestCost to filter out any Cost that is higher than 2/3 of the low range of the average. I hope that makes sense.

Thanks

 

[Mark44]

Is there a way to reduce the math in my formula algebraically:
BestCost = lowest+((((highest+lowest)/2)-lowest)*0.6667);

I want to use BestCost to filter out any Cost that is higher than 2/3 of the low range of the average. I hope that makes sense.

"Filter out" implies to me that you want to discard or ignore some values. Based on your use of a semicolon at the end of your formula, I suspect that this is part of some code, possibly C or a language derived from C. If that's the case, to filter out values, you need an if statement.

What do you mean by "low range of the average?" The average (or mean) is a single number. What you're calculating in your formula with (highest + lowest)/2 is called the midrange.

It would help if you provided a reasonably small list of numbers, and showed what your calculation is supposed to do.

 

[1plus1is10]

You are very perceptive and bright - yes, it is code and after messing with it for quite some time now - I stopped to think about it. And "midrange" is exactly what I want. Unfortunately, I was getting confused in my head with median, mean, and price range. In the end, I will have to ditch my formula, and instead, simply sort my Costs to determine the correct 1/3 lowest.

Random costs after sorting: 1,2,3,4,5,6,7,8,9,10,11,12
BestCost = anything below the 4th one in the list;

Thanks for your help.

 

[mfb]

Your formula uses only linear operations. You can simplify it step by step until you get something like BestCost = a*lowest+b*highest with some coefficients a and b.
If a+b=1 (which is the case here) then you calculate a weighted average of the two numbers. BestCost is then always at the same relative place in the interval from lowest to highest.