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From: rabbahs on 21 Mar 2010 00:56
Thanks Mr. Walter for your time.

I m looking fwd to see the solution.

If you want then i can send you my script file ? (it might help you to figure out the problem easily).

have a nice day

bye
From: Walter Roberson on 21 Mar 2010 03:47
rabbahs wrote:
> Thanks Mr. Walter for your time.
> I m looking fwd to see the solution.
>
> If you want then i can send you my script file ? (it might help you to
> figure out the problem easily).

I have determined that it is not _necessarily_ the case that some
variables must come out negative. It depends on the initial constants.

For the constants
c1(1) = 5/2
c1(2) = 9/5
c1(3) = 33/10
c1(4) = 10
c1(5) = 7/5
c1(6) = 17/10
c1(7) = 17/5
c1(8) = 23/10

I reach the solution

a = 1.103817340
b = 8.258157698
d = .4327670834
c = 1.800258210
e = 12.784
f = .33420762
g = .205257879
p = 24.71320795

The solution process involved reducing it down to a problem in 3
expressions, with the other 5 reducing to 0 by linear combinations of
the 3 remaining variables. Those expressions (which have to be solved to
equal 0) are:

c11*a*(a + b + 1/2*c15 - d + 94/25*c17 + c18) - b^2,

c12*(c16 + 2*c17 + c18 - 2*a - b)*(a + b + 1/2*c15 - d + 94/25*c17 +
c18) - b*(1/2*c15 - 2*d - c16 - 2*c17 + 2*a+b),

c13*(1/2*c15 - 2*d - c16 - 2*c17 + 2*a + b)^2 - d*(a + b + 1/2*c15 - d +
94/25*c17 + c18)

As before, c18 is what you wrote as the constant c1(8) and so on.

The solution to this trio of equations involves finding the roots of a
quartic. The generalized solution is pretty long and messy, so it took
me numerous false starts to find a tractable approach. What I did was
generate random values for all of the constants *except* c1(4), and
substituted those values in; c1(4) is not present in the trio of
equations, so it is possible to find values for the four tuples of
solutions for a, b, and d. Four tuples because a quartic has four
solutions. With the particular random values I had, two of the tuples
had negative values for b, so those two tuples could be eliminated
immediately. The other two tuples were all positive for a, b, and d. I
then back-substituted a, b, and d in to find the values for c, e, f, and
p, and to find the expression for g, which is c14 - a - b - d . One of
the two tuples gave negative values for c and f; the other gave all
positive values for the variables, with g = c14 minus a constant. From
there it was just a matter of selecting any c14 larger than the constant
to arrive at a set of all-positive constants that will produce a set of
all-positive variables.

This is proof by construction: I constructed a set of constants that was
solvable for solutions in the required range. As such, it is only a
proof that desirable solutions exist for _some_ combinations of
constants: other combinations of constants will have no valid solutions.


The values that can be deduced linearly from a, b, and d are:

c = 1/2*c15 - 2*d - c16 - 2*c17 + 2*a + b
e = 94/25*c17
f = c16 + 2*c17 + c18 - 2*a - b
g = c14 - a - b - d
p = a + b + c + d + e + f

p can also be expressed as p = a + b - d + c15/2 + 94/25*c17 + c18
by combining the other equations.

By examination, we can see that e is the only variable which is
guaranteed to come out positive based upon the constants; everything
else depends upon a combination of the values of the constants and the
solutions found to the 3 non-linear equations. And make sure to check
all four roots of the quartic, as they might not meet your constraints.

When you are calculating the roots of the quartic, it is common that
most or all of the roots you get out will be imaginary numbers. If the
imaginary components are on the order of 10^(-6) or less, then those are
being introduced by round-off error and you should just use real() to
extract the real portions. But you do have to be careful, because with
some values of the constants, you can get roots whose imaginary portions
are on the order of .05 to -21 (values observed in practice), and those
values are much too large for round off: those are true imaginary
numbers and taking the real portion of them will seriously mislead you.
From: rabbahs on 21 Mar 2010 19:40
Walter Roberson <roberson(a)hushmail.com> wrote in message <ho4itl$msf$1(a)canopus.cc.umanitoba.ca>...
> rabbahs wrote:
> > Thanks Mr. Walter for your time.
> > I m looking fwd to see the solution.
> >
> > If you want then i can send you my script file ? (it might help you to
> > figure out the problem easily).
>
> I have determined that it is not _necessarily_ the case that some
> variables must come out negative. It depends on the initial constants.
>
> For the constants
> c1(1) = 5/2
> c1(2) = 9/5
> c1(3) = 33/10
> c1(4) = 10
> c1(5) = 7/5
> c1(6) = 17/10
> c1(7) = 17/5
> c1(8) = 23/10
>
> I reach the solution
>
> a = 1.103817340
> b = 8.258157698
> d = .4327670834
> c = 1.800258210
> e = 12.784
> f = .33420762
> g = .205257879
> p = 24.71320795
>
> The solution process involved reducing it down to a problem in 3
> expressions, with the other 5 reducing to 0 by linear combinations of
> the 3 remaining variables. Those expressions (which have to be solved to
> equal 0) are:
>
> c11*a*(a + b + 1/2*c15 - d + 94/25*c17 + c18) - b^2,
>
> c12*(c16 + 2*c17 + c18 - 2*a - b)*(a + b + 1/2*c15 - d + 94/25*c17 +
> c18) - b*(1/2*c15 - 2*d - c16 - 2*c17 + 2*a+b),
>
> c13*(1/2*c15 - 2*d - c16 - 2*c17 + 2*a + b)^2 - d*(a + b + 1/2*c15 - d +
> 94/25*c17 + c18)
>
> As before, c18 is what you wrote as the constant c1(8) and so on.
>
> The solution to this trio of equations involves finding the roots of a
> quartic. The generalized solution is pretty long and messy, so it took
> me numerous false starts to find a tractable approach. What I did was
> generate random values for all of the constants *except* c1(4), and
> substituted those values in; c1(4) is not present in the trio of
> equations, so it is possible to find values for the four tuples of
> solutions for a, b, and d. Four tuples because a quartic has four
> solutions. With the particular random values I had, two of the tuples
> had negative values for b, so those two tuples could be eliminated
> immediately. The other two tuples were all positive for a, b, and d. I
> then back-substituted a, b, and d in to find the values for c, e, f, and
> p, and to find the expression for g, which is c14 - a - b - d . One of
> the two tuples gave negative values for c and f; the other gave all
> positive values for the variables, with g = c14 minus a constant. From
> there it was just a matter of selecting any c14 larger than the constant
> to arrive at a set of all-positive constants that will produce a set of
> all-positive variables.
>
> This is proof by construction: I constructed a set of constants that was
> solvable for solutions in the required range. As such, it is only a
> proof that desirable solutions exist for _some_ combinations of
> constants: other combinations of constants will have no valid solutions.
>
>

Thanks Mr. Walter,

I m working on the problem, will let you know the results as soon as i done with it.

thanks for all the help !!!

If i have any query, i will let you know.
> The values that can be deduced linearly from a, b, and d are:
>
> c = 1/2*c15 - 2*d - c16 - 2*c17 + 2*a + b
> e = 94/25*c17
> f = c16 + 2*c17 + c18 - 2*a - b
> g = c14 - a - b - d
> p = a + b + c + d + e + f
>
> p can also be expressed as p = a + b - d + c15/2 + 94/25*c17 + c18
> by combining the other equations.
>
> By examination, we can see that e is the only variable which is
> guaranteed to come out positive based upon the constants; everything
> else depends upon a combination of the values of the constants and the
> solutions found to the 3 non-linear equations. And make sure to check
> all four roots of the quartic, as they might not meet your constraints.
>
> When you are calculating the roots of the quartic, it is common that
> most or all of the roots you get out will be imaginary numbers. If the
> imaginary components are on the order of 10^(-6) or less, then those are
> being introduced by round-off error and you should just use real() to
> extract the real portions. But you do have to be careful, because with
> some values of the constants, you can get roots whose imaginary portions
> are on the order of .05 to -21 (values observed in practice), and those
> values are much too large for round off: those are true imaginary
> numbers and taking the real portion of them will seriously mislead you.
From: Walter Roberson on 21 Mar 2010 23:20
rabbahs wrote:
> Walter Roberson <roberson(a)hushmail.com> wrote in message
> <ho4itl$msf$1(a)canopus.cc.umanitoba.ca>...

>> The values that can be deduced linearly from a, b, and d are:
>>
>> c = 1/2*c15 - 2*d - c16 - 2*c17 + 2*a + b
>> e = 94/25*c17
>> f = c16 + 2*c17 + c18 - 2*a - b
>> g = c14 - a - b - d
>> p = a + b + c + d + e + f
>>
>> p can also be expressed as p = a + b - d + c15/2 + 94/25*c17 + c18
>> by combining the other equations.

If you examine the expressions for c, g, and p, you can see that d
always appears in the negative for them. Thus those expressions have the
greatest chance of being positive if d is as small as possible. To
explore that possibility, evaluate the system of three non-linear
equations at d = 0, leaving a system of three equations in two
variables. The second of the equations is the most complex of them, so
solve the first against the third. The result is a relatively simple
quadratic which lead to "reasonable" expressions for a and b -- indeed,
one of the two solutions to the quadratic has b = 0. Choosing the other
root and substituting back in to the system of three equations, sure
enough the first and third go to 0, and the second goes to something
that is not all that long -- and which factors moderately well.

If you then check which constants that expression involve, and solve the
expression for each of them in turn (one at a time), then you find that
each in turn is the sum of negatives of some of the constants. Under the
assumption that the constants are positive, the implication is that the
constant being solved for would end up negative; and since this happens
for all the constants involved, there is no possibility to solve the
system to end up with all positive constants. From this, we deduce that
in fact d must be non-zero and large enough to interact with the other
constants to produce magnitudes comparable to the other constants.

I proceeded to experiment with the same solving method for d=1, but the
equations it spits out are horribly long, and have my symbolic
computation engine pretty much locked up trying to simplify them... the
fourth common subexpression is somewhere over 6000 lines long, and the
main expression and earlier common sub-expressions have scrolled off my
10000 line scrollback...
From: rabbahs on 22 Mar 2010 03:54
Walter Roberson <roberson(a)hushmail.com> wrote in message <ho6nln$bnj$1(a)canopus.cc.umanitoba.ca>...
> rabbahs wrote:
> > Walter Roberson <roberson(a)hushmail.com> wrote in message
> > <ho4itl$msf$1(a)canopus.cc.umanitoba.ca>...
>
> >> The values that can be deduced linearly from a, b, and d are:
> >>
> >> c = 1/2*c15 - 2*d - c16 - 2*c17 + 2*a + b
> >> e = 94/25*c17
> >> f = c16 + 2*c17 + c18 - 2*a - b
> >> g = c14 - a - b - d
> >> p = a + b + c + d + e + f
> >>
> >> p can also be expressed as p = a + b - d + c15/2 + 94/25*c17 + c18
> >> by combining the other equations.
>
> If you examine the expressions for c, g, and p, you can see that d
> always appears in the negative for them. Thus those expressions have the
> greatest chance of being positive if d is as small as possible. To
> explore that possibility, evaluate the system of three non-linear
> equations at d = 0, leaving a system of three equations in two
> variables. The second of the equations is the most complex of them, so
> solve the first against the third. The result is a relatively simple
> quadratic which lead to "reasonable" expressions for a and b -- indeed,
> one of the two solutions to the quadratic has b = 0. Choosing the other
> root and substituting back in to the system of three equations, sure
> enough the first and third go to 0, and the second goes to something
> that is not all that long -- and which factors moderately well.
>
> If you then check which constants that expression involve, and solve the
> expression for each of them in turn (one at a time), then you find that
> each in turn is the sum of negatives of some of the constants. Under the
> assumption that the constants are positive, the implication is that the
> constant being solved for would end up negative; and since this happens
> for all the constants involved, there is no possibility to solve the
> system to end up with all positive constants. From this, we deduce that
> in fact d must be non-zero and large enough to interact with the other
> constants to produce magnitudes comparable to the other constants.
>
> I proceeded to experiment with the same solving method for d=1, but the
> equations it spits out are horribly long, and have my symbolic
> computation engine pretty much locked up trying to simplify them... the
> fourth common subexpression is somewhere over 6000 lines long, and the
> main expression and earlier common sub-expressions have scrolled off my
> 10000 line scrollback...



Hi Mr. Walter,

can u give me the expression to calculate the positive value of a,b and d ?
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