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find lowest possible x, y and z whole number variables where:

x  =< 2y+2z 
6y =< x+z   
3z =< x+y

I am trying to solve this system of 3 linear equations with 3 unknowns and get the ratio that x, y and z have to be in for the inequalities to work. So far I have been able to get a working ratio

Q1 : but I am trying to get a ratio that satisfies the equations below.

x  = 2y+2z 
6y = x+z   
3z = x+y

Q1, a : And if no such ratio exists how can I mathematically determine that?

The formular I used to get a working ratio for the inequality is by solving the equation. I made the equations all equal to each other as follows:

0.5x + x  =  x + y + z 
  6y + y  =  x + y + z   
  3z + z  =  x + y + z

This way I was able to get the equation below.

1.5x = 7y = 4z  

The equation above satisfies the first inequality but not the corresponding equation.

PS: I am not very conversant with the tags if I have left any out please add. Thank you.

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migrated from mathoverflow.net Apr 11 at 13:32

This question came from our site for professional mathematicians.

  • $\begingroup$ Remark: the Stack Exchange system claims that I migrated this question from Mathoverflow (mathoverflow.net/posts/327772/revisions), but it is false. $\endgroup$ – Federico Poloni Apr 11 at 13:33
  • $\begingroup$ The only solution of the system of equations is $(0,0,0)$. As the coefficients are small, you can try solutions by brute force. [$(5,1,2)$] $\endgroup$ – Yves Daoust Apr 11 at 13:53
  • $\begingroup$ @YvesDaoust Is there no formular to get the suggested solution 5,1,2 I am trying to code an algorithm that solves such equations $\endgroup$ – YulePale Apr 11 at 14:44
  • $\begingroup$ @YulePale: no, use brute force. $\endgroup$ – Yves Daoust Apr 11 at 14:46
  • $\begingroup$ @YvesDaoust: kindly help , how do I start brute forcing? $\endgroup$ – YulePale Apr 11 at 14:54
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This can be formulated as a Mixed Integer Linear Programming problem (MILP). To do so, you need to decide on an objective function to be minimized.

The statement "find lowest possible x, y and z whole number variables" is a little value. So we could separately solve 3 separate problems,, one for each objective function x, y, and z.

For instance, using YALMIP, this can be formulated as

intvar x y z;  % declares that x, y, and z are integer variables
optimize([x <= 2*y+2*z,6*y <= x+z,3*z <= x+y],x) % minimizes x subject to the constraints
value([x y z]) % displays optimal values of x, y, z
optimize([x <= 2*y+2*z,6*y <= x+z,3*z <= x+y],y) % same as above but objective function is y
value([x y z]) % displays optimal values of x, y, z
optimize([x <= 2*y+2*z,6*y <= x+z,3*z <= x+y],z) % same as above but objective function is z
value([x y z]) % displays optimal values of x, y, z

For each of the 3 problems, the optimal values of x, y and z are zero. So we have found the lowest possible x, y and z whole number variables under any reasonable interpretation.

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