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Paul Smith...
Posted: Mon May 05, 2008 7:36 am
Guest
Dear All,

Define

f1(x) = x * (1 - x - y)

and

f2(y) = y * (1 - x - y).

Consider now the following coupled optimization problem:

max f1(x) and max f2(y) (simultaneously).

I know that the analytical solution is

x = 1/3 and y = 1/3,

but how can one solve this numerically? My question is motivated by
the fact that I have to solve similar problems, but large-scale ones.

Thanks in advance,

Paul
Hans Mittelmann...
Posted: Mon May 05, 2008 9:08 am
Guest
On May 5, 10:36 am, Paul Smith <phh... at (no spam) gmail.com> wrote:
Quote:
Dear All,

Define

f1(x) = x * (1 - x - y)

and

f2(y) = y * (1 - x - y).

Consider now the following coupled optimization problem:

max f1(x) and max f2(y) (simultaneously).

I know that the analytical solution is

x = 1/3 and y = 1/3,

but how can one solve this numerically? My question is motivated by
the fact that I have to solve similar problems, but large-scale ones.

Thanks in advance,

Paul

max t
s.t. t <= f1(x)
t <= f2(x)
Paul Smith...
Posted: Mon May 05, 2008 10:37 am
Guest
On May 5, 8:08 pm, Hans Mittelmann <mittelm... at (no spam) asu.edu> wrote:
Quote:
Define

f1(x) = x * (1 - x - y)

and

f2(y) = y * (1 - x - y).

Consider now the following coupled optimization problem:

max f1(x) and max f2(y) (simultaneously).

I know that the analytical solution is

x = 1/3 and y = 1/3,

but how can one solve this numerically? My question is motivated by
the fact that I have to solve similar problems, but large-scale ones.

max t
s.t. t <= f1(x)
t <= f2(x)

Thanks, Hans. Do you mean to run your optimization program on the 3
variables, x, y and t? If so, it does not lead to the correct
solution; by optimizing numerically, I get the following solution:

x = 0.25
y = 0.25

Paul
...
Posted: Mon May 05, 2008 10:38 am
Guest
On May 5, 10:36 am, Paul Smith <phh... at (no spam) gmail.com> wrote:
Quote:
Dear All,

Define

f1(x) = x * (1 - x - y)

and

f2(y) = y * (1 - x - y).

Consider now the following coupled optimization problem:

max f1(x) and max f2(y) (simultaneously).

Both f1 and f2 depend on both x and y. Do you mean that you only want
to maximize f1 wrt x and f2 wrt y, i.e. find (x,y) such that
x (1 - x - y) >= x' (1 - x' - y) for all x' and
y (1 - x - y) >= y' (1 - x - y') for all y'?

Or do you want to somehow maximize both objectives with respect to
both variables?

Quote:
I know that the analytical solution is

x = 1/3 and y = 1/3,

I guess this supports the first interpretation...

Quote:
but how can one solve this numerically? My question is motivated by
the fact that I have to solve similar problems, but large-scale ones.

You could use numerical methods to look for solutions to the equations
df1/dx = 0, df2/dy = 0.

Robert Israel israel at (no spam) math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Paul Smith...
Posted: Mon May 05, 2008 10:49 am
Guest
On May 5, 9:38 pm, isr... at (no spam) math.ubc.ca wrote:
Quote:
f1(x) = x * (1 - x - y)

and

f2(y) = y * (1 - x - y).

Consider now the following coupled optimization problem:

max f1(x) and max f2(y) (simultaneously).

Both f1 and f2 depend on both x and y. Do you mean that you only want
to maximize f1 wrt x and f2 wrt y, i.e. find (x,y) such that
x (1 - x - y) >= x' (1 - x' - y) for all x' and
y (1 - x - y) >= y' (1 - x - y') for all y'?

Or do you want to somehow maximize both objectives with respect to
both variables?

I know that the analytical solution is

x = 1/3 and y = 1/3,

I guess this supports the first interpretation...

but how can one solve this numerically? My question is motivated by
the fact that I have to solve similar problems, but large-scale ones.

You could use numerical methods to look for solutions to the equations
df1/dx = 0, df2/dy = 0.

Yes, Robert, your first interpretation corresponds to what I wanted to
mean.

I had already had the idea of solving numerically

df1/dx = 0, df2/dy = 0,

but, in my large-scale problem, it can be messy. Therefore it would be
of great help if one could convert the problem into another one
solvable by the available large-scale optimization numeric algorithms.

Paul
 
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