Get the analytical solution of inequalities in Matlab.

Hello,

x + 2 y + 3 c + 4 d == 3/2 && x + y + c + d == 2 && x >= 2 && 
  y < 2 && c < 3 && d > 5

How to solve this inequations to get the solution of x, y, c, and d in Matlab?

Best regards.

Answer 1

[Walter Roberson]

Most of the time, MATLAB is not able to solve systems of inequalities.

See the recent discussion at https://www.mathworks.com/matlabcentral/answers/2007972-why-do-solutions-get-lost-when-i-add-a-assumption-where-i-know-that-the-correct-solution-would-sati#answer_1287092

[Walter Roberson]

syms x y c d 
eqns = [x + 2 * y + 3 * c + 4 * d == 3/2
        x + y + c + d == 2
        x >= 2 
        y < 2
        c < 3
        d > 5]

eqns = 

sol = solve(eqns, 'returnconditions', true)

sol = struct with fields:
             c: 13/2 - 2*z1 - 3*z
             d: 2*z + z1 - 9/2
             x: z
             y: z1
    parameters: [z    z1]
    conditions: 7 < 6*z + 4*z1 & 19 < 4*z + 2*z1 & z1 < 2 & 2 <= z

%now try to find the boundaries
parts = children(sol.conditions)

parts = 1×4 cell array
    {[7 < 6*z + 4*z1]}    {[19 < 4*z + 2*z1]}    {[z1 < 2]}    {[2 <= z]}

sol2 = solve([lhs(parts{1})==rhs(parts{1}), lhs(parts{2})==rhs(parts{2})])

sol2 = struct with fields:
     z: 31/2
    z1: -43/2

subs(sol, sol2)

ans = struct with fields:
             c: 3
             d: 5
             x: 31/2
             y: -43/2
    parameters: [31/2    -43/2]
    conditions: 7 < 7 & 19 < 19 & -43/2 < 2 & 2 <= 31/2

After that you have to explore which side of the bounds the values need to be on

Note that z effectively stands in for x and z1 effectively stands in for y -- that once you have particular x and y values then the c and d are pinned down.

[zhenyu zeng]

The output of this way is still too complicate. Is there another way just like Maple of Julia do?

[Sam Chak]

Can you describe your expectation of the results? There are 4 variables but only 2 equations.

[zhenyu zeng]

This one:

((15/4 < x <= 31/2 && 1/2 (19 - 4 x) < y < 2 && 
     c == 1/2 (13 - 6 x - 4 y)) || (x > 31/2 && 
     1/4 (7 - 6 x) < y < 2 && c == 1/2 (13 - 6 x - 4 y))) && 
 d == 1/8 (3 - 6 c - 2 x - 4 y)

[zhenyu zeng]

Or this one

{[{31/2 < x}, {y < 2, 7/4 - (3*x)/2 < y}, {c = 13/2 - 3*x - 2*y}, {d = -9/2 + 2*x + y}], [{x <= 31/2, 15/4 < x}, {y < 2, 19/2 - 2*x < y}, {c = 13/2 - 3*x - 2*y}, {d = -9/2 + 2*x + y}]}

[Walter Roberson]

The Maple output looks like this

Answer 2

[Bruno Luong]

The real question is how would you get as formal decription of the so called "solution" of the equality linear system?

There is no mathematical semantic AFAIK to describe the solution beside what is similar to you question, a set S that satisfies

A*x <= b
Aeq*x = beq
lo <= x <= up

One can show to linear transform the variables x to y so as the above can be reduced to the so called canonical form of

C*y = d
y <= 0

but there is no way to express it shorter.

Another way is if the region S is bounded, it is a polytope and one can express as a "sub-convex" combination of a finite set of vertexes.

S = sum wi * Vi
sum wi <= 1 

In other word a convex hull where the vetices are {Vi}. There is a numerical method in FEX made by Matt J. It works OK for toy dimension of 2-3, but I wouldn't trust it for dimension >= 10.

In short the most convenient to "solve" linear inequalitues is to let it as it is. You want to know a point is a solution? Just replace and check it.