Sep 13, 2012 ... and again we could think that there is no solution since 2/6 is not an integer, but if we allow x[2] to be non-zero, we get the solution x[1] = ...
May 20, 2014 ... 3} x[j] = 2; ampl: fix {j in 1..3} x[j] := 1; ampl: solve; presolve ... Thus when x[1], x[2], and x[3] are all fixed to 1, the constraint becomes ...
What does this mean and what I do to solve it? And the original question is: minimize: z=4x1+4x2+x3 constrains: x1+x2+x3<=2 2x1+x2<=3 2x1+x2+3x3>=3 bound ...
Jul 31, 2011 ... 4 - (Abs[x] - 1)^2] - y)==0. where g[a_]:=Sqrt[Abs[a]/a]. Note that g ... 46rom the definition of g we find that this has real solutions for x
Jun 10, 2021 ... This is y′sinx+ycosx=2cosxsinx or (ysinx)′=sin(2x) so ysin(x)=−12cos(2x)+c so y=−cos(2x)−c2sin(x)=−1−2sin2(x)−c2sin(x)=sin(x)+c1sin(x).
The messages after "solve" are telling you that AMPL's presolve phase determined your problem to be infeasible -- that is, it found that there is no way to ...
Jul 12, 2005 ... ... solution": f(2 cos(t/2)) = 2 cos(SQRT(2)*t/2) A bit of algebra shows that the two fixed points of g(x): g(x) = x are x = 2,-1. Furthermore ...
May 20, 2013 ... both {x:1, y:2} and (1, 2), so that sol[0] and sol[x] both give 1. For issue 3667, let's just do it. Make dict=True the default. Let's not ...
I was able to have 1, 2, or 3 inputs, converging the output to the solution, but seams failing on having multiple outputs with inputs. Trying simple, I have ...
x2 >= 1. x3 + x4 >= 0. x1 + x2 >= 1. which can be expressed as G*x + s = h with. G = [ -1 0 0 0 ] h = [ -1 ] ... solution for a much simpler network with 4 nodes ...