[ReSolved]Limits, both Lower and Upper...

[Something Else]

Here's the problem. Lets say I have a series of variables, F1 thru Fn.
Each one of these, individually, are >= B1. They also are individually <= T1. { For any Fn, B1 <= Fa <= T1 }
Also, If you sum two of them together, any combination of two are >= B2, and <= T2.

{ For any 2 Fn, B2 <= Fa + Fb <= T2 }
So, more explixitely, if you had F1 + F2, B2<= (F1+F2) <= T2.
And More Generically, if you sum together N elements of {F1, F2, F3, ...}, then
These N elements are <= Tn, and >= Bn.

{ For any K Fn, Bk <= Fa + Fb + ... <= Tk }

Now, Some of these F's combine together in such a way to produce other F's.
Given these equations, I would like to produce as many minima and maxima as possible.

For example, lets say I had the equations as seen in the attached picture.




I can build the limits as seen by hand thru visual inspection.

Question #1) Can anyone add or improve upon the illustrated limits.
Please feel free to build some upper limits.

Question #2) Can anyone help me with how I should approach creating a progie that,
given equations, variables, and elemental limits {All The Bn's and Tn's}
produces the optimal Lower and Upper limits of the control variables?

-Thanks,
-Lou

[Something Else]

I forgot to Mention. I'm not sure how to say it, but:

Lets say I had a sum of 3 F's, where they did NOT each have a multiplier of 1.
For example, Lets say I had Fa+Fb+2Fc. The following would be true:

B3 + B1 <= (Fa + Fb + Fc) + Fc <= T3 + T1 or:
B3 + B1 <= Fa+Fb+2Fc <= T3 + T1

Also, if we had 2Fa + 3Fb + 5Fc + 6Fd,the following would be true:
Since 2Fa + 3Fb + 5Fc + 6Fd ==>
2(Fa + Fb + Fc + Fd) + (Fb + 3Fc + 4Fd) ==>
2(Fa + Fb + Fc + Fd) + (Fb + Fc + Fd) + (2Fc + 3Fd) ==>
2(Fa + Fb + Fc + Fd) + (Fb + Fc + Fd) +2 (Fc + Fd) +Fd ==>

Then:
2B4 + B3 + 2B2 + B1 <= 2Fa + 3Fb + 5Fc + 6Fd <= 2T4 + T3 + 2T2 + T1

Of course, it could also be said that:
2Fa + 3Fb + 5Fc + 6Fd = 2Fa + 3(Fb + Fc + Fd) + 2Fc + 3Fd==>
2(Fa+Fc + Fd) + 3(Fb + Fc + Fd) + Fd
So:

5B3 + B1 <= 2Fa + 3Fb + 5Fc + 6Fd <= 5T3 + T1

Which complicates matters furthur.

-Lou

[NotLKH]

Also from Eq1 & Eq3:

2F1 + F3 = F9 – 12
B2 + B1 <= F9 – 12
B2 + B1 + 12 <= F9
219 <= F9

-Lou

[NotLKH]

OOPS!
I Believe B3 = 279.

[sql_lall]

have you heard of the Simplex method of solving linear equations? It deals with all these sorts of things nicely and easily. Google it for more info, possibly ith "tableau" or "linear equations" included

[NotLKH]

Thanks SQL. I didn't google, but that was the direction I was going in anyways.
I'll add Resolved to the thread as soon as I get home.

-Lou