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From: Joe Riel <joer@san.rr.com>
Newsgroups: comp.soft-sys.math.maple
Subject: Re: Syrup Problem
Date: Wed, 04 Jan 2012 18:08:36 -0800
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Thomas Dean <tomdean@speakeasy.org> writes:

> I am getting up-to-speed, again.
>
> I have a problem from a text book schematic that syrup cannot solve.
>
> restart;with(Syrup);
> ckt:="
> e1 3 1 4 0 4
> h1 1 0 r3 10
> i1 2 5 2
> r1 2 1 4
> r2 3 2 5
> r3 3 4 8
> r4 4 0 2
> r5 5 4 2
> v1 3 0 15
> .end";
> volts:=syrup(ckt,dc,'current');
> evalf(volts); evalf(current);
>
> I am not certain of the value for r5, but, as long as it is reasonable, it 
> should not prevent solution.
>
> I get maple cannot solve the equations.
>
> I1 is     node2 --> node5 so, I choose node 5 as +
> e1 is     node3 +-  node1 4*V4
> h1 is     node1 +-  node0 10*Ir3

The problem is that the circuit has no solution.
Note that there is a new version of syrup that has a slightly
different calling convention.  I released it on MaplePrimes;
it is also in the Maple Application Center.  However, the
versions there are not the very latest and have a few bugs.

To analyze this circuit with the new version I changed
the v1 line to

v1 3 0 V1
.VARS V1

and made V1 a variable (this is a new feature):

Syrup could still not solve the circuit---actually it did, however, it
recognized that the result was not unique and so rejected it.  

Using the modified deck, I did

sol := Solve(ckt,'dc'):
Solve:   Could not solve equations; circuit may not be
well-defined. Returning equations and variables as [eqs,vars].

solve(op(sol));
{V1 = -11, i[e1] = 1/6 - i[v1], i[h1] = - 1/2 - i[v1], i[i1] = 2, i[r1] = -2/3,
    i[r2] = 4/3, i[r3] = -3/2, i[r4] = 1/2, i[r5] = 2, i[v1] = i[v1],
    v[1] = -15, v[2] = -53/3, v[3] = -11, v[4] = 1, v[5] = 5, v[e1] = 4,
    v[h1] = -15, v[i1] = -68/3, v[r1] = -8/3, v[r2] = 20/3, v[r3] = -12,
    v[r4] = 1, v[r5] = 4, v[v1] = -11
}

Note that i[i1] is arbitrary and that V1 must equal -11 for there to be
a solution.

I'll email you the latest version of Syrup.


-- 
Joe Riel