Hi. If I understand correctly...
You have data at discreet time intervals.

(11.4L^0.5)(K^0.52)(S^-0.318) is a constant ( k )

T = k * Table(T) ^ -.38

Where T is a discreet integer time, and Table is your lookup Table

It would appear to me that you take each table value 'n, and make a
helper column = k*n ^ -.38

The table value that is closest to its index value would be the closest
solution.

For example, suppose your equation from the 3 constants a

T = 18.66 * Table(T) ^ - .38

and your table had 10 values for time 1,2...10

{2, 4, 8, 16, 32, 64, 128, 256, 512, 1024}

Make a helper column 18.66 * n ^ -.38

{14.339, 11.0187, 8.46715, 6.50648, 4.99982, 3.84205, 2.95238,
2.26872, 1.74337, 1.33967}

Time 1 isn't too close to 1 at 14.33
Time 2 isn't too close to 2 at 11.01
But Time 5 is very close to 5 at 4.999

I would say for this example T = 5 would be the solution.

At time 5, the table value is 32.

18.66*32 ^ -.38

returns 4.99982

Which I think you want.

= = = = =
HTH
Dana DeLouis



On 3/8/10 11:56 AM, Chris wrote:
I am trying to solve the equation T=(11.4L^0.5)(K^0.52)(S^-0.318)(i^-0.38),
where "L, K& S" are constants and "i" is determined from a table based on
time. "T" is time in minutes. To solve the equation fully, once "T" is
solved for it is compared with the time value used in the table that "i" was
determined from. If necessary, a new value of "i" is determined and the
equation is solved again. This is repeated until an acceptable tolerance
between "T" and the value used to determine "i" is reached. Is there an easy
way to set this up in excel to run an iterative process with a table? Any
help would be greatly appreciated.