Using the solver

[Siberian]

I've just learned how to use the solver add-in to solve sets of
equations. The problem is that this particular set (let's call the
unknowns x and y) has an infinite amount of solutions as long as x is 5
time bigger than y. This means that 5 and 1 is equal to 5000 and 1000.
My solver for some reason doesn't find the lowest usable values (it
probably doesn't care either), but it would be nice to have it return
more "pleasant" values like 5 and 1 instead of 5000 and 1000. Can it be
"told" to do so, or can I make it happen in some other way?

 

[ScottO]

Can you scale it down with a constraint like X-Y<50 (or 100), or
something like that?
ScottO

"Siberian" <[email protected]>
wrote in message
|
| I've just learned how to use the solver add-in to solve sets of
| equations. The problem is that this particular set (let's call the
| unknowns x and y) has an infinite amount of solutions as long as x
is 5
| time bigger than y. This means that 5 and 1 is equal to 5000 and
1000.
| My solver for some reason doesn't find the lowest usable values (it
| probably doesn't care either), but it would be nice to have it
return
| more "pleasant" values like 5 and 1 instead of 5000 and 1000. Can
it be
| "told" to do so, or can I make it happen in some other way?
|
|
| --
| Siberian
| -------------------------------------------------------------------
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[Bernard Liengme]

What do use for starting values?
Solver generally finds the solution closest to the initial values of the
variables

 

[kcc]

You can usually restructure a problem into one that works.
Depending on exactly what you are trying, this may work.

Lets say z is a cell you want to equal to 0 by changing
x and y. Add a cell that is x+y and solve to minimize
that cell subject to the constraints that:
z is >= -.00000001 and
z is <= .00000001
You could try to constrain z to 0, but because of
binary rounding, you usually get a solution that is
very small rather than 0.

Benard's comment on starting values is always important
regardless of how the problem is structured.

kcc

 

[Dana DeLouis]

...has an infinite amount of solutions as long as x is 5

time bigger than y. This means that 5 and 1 is equal to 5000 and 1000.

I would just add that since you know the relationship between x & y, I would
have Solver only adjust 1 value.
Say only 'x'. The value for 'y' would then by =x/5.
Then add a constraint that x <= 5, or something similar.