Using the solver

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Siberian - 12 Jan 2006 04:13 GMT

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?

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Siberian

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ScottO - 12 Jan 2006 04:50 GMT

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

| 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
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Bernard Liengme - 12 Jan 2006 13:31 GMT

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

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Bernard V Liengme
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> 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
[quoted text clipped - 4 lines]
> 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?

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kcc - 13 Jan 2006 03:15 GMT

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

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Dana DeLouis - 13 Jan 2006 03:33 GMT

> ...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.

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Dana DeLouis
Win XP & Office 2003

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