optimization
 

I will try to explain clearly.

I have a bi-linear curve and I would like to know where the breaking
point is. Looking at a graph of the ordered data (highest to lowest),
it seems to be ~30%. Basically what I need to do is figure out which
ranges of cells would give the greatest difference in average slope if
slope is computed as (Wi-Wi+1)/1. For example, it might be the
average of Wi-Wi+1 for i=1 through i=84 and the average of Wi-Wi+1 for
i=85 through i=244.
Example:
i W m
1 95
2 85 10
3 80 5
4 75 5
5 70 5
6 65 5

Here the solution is easy. The difference in m_avg is greatest
between i=1:2 (10) and i=3:6 (5)

Thanks
ds

optimization
 

On Mar 9, 2:03*am, dschneiderch wrote:
I will try to explain clearly.

I have a bi-linear curve and I would like to know where the breaking
point is. Looking at a graph of the ordered data (highest to lowest),
it seems to be ~30%. *Basically what I need to do is figure out which
ranges of cells would give the greatest difference in average slope if
slope is computed as (Wi-Wi+1)/1. *For example, it might be the
average of Wi-Wi+1 for i=1 through i=84 and the average of Wi-Wi+1 for
i=85 through i=244.
Example:
i *W *m
1 95
2 85 10
3 80 *5
4 75 *5
5 70 *5
6 65 *5

*Here the solution is easy. The difference in m_avg is greatest
between i=1:2 (10) *and i=3:6 *(5)


You have added a column showing the difference to the previous. Your
formula computes the difference to the next.

Having calculated the change, now extend that by calculating the
change in the change. Extending the example (and calculating the
change to the next, and then the change to the change to the next)
would provide :

1 115 10 0
2 105 10 0
3 95 10 5
4 85 5 0
5 80 5 0
6 75 5 0
7 70 5 0
8 65 5 0

3 is obviously the breakpoint.

The breakpoint is the change of the slope. Or if you've done calculus,
the breakpoint is at d2y/dx.

Think Babbages's difference engine. <g

Alan Lloyd