Graphically solving a linear programming problem

Question

I try to realize the graphical linear programming method. Here is my code

Clear[GraphicalMethod]
GraphicalMethod[L_?VectorQ, A_?MatrixQ, b_?VectorQ, vars_?VectorQ] := 
 Module[{obj = L.vars, 
   cond = Thread[A.vars <= b]~Join~Thread[vars >= 0], r, sol, x1, x2},
  sol = Maximize[{L.vars, Thread[A.vars <= b]~Join~Thread[0 <= vars]},
     vars];
  x1 = Max[
    First@vars /. Solve[#, First@vars] & /@ 
     Thread[First /@ (A.vars) == b]];
  x2 = Max[
    Last@vars /. Solve[#, Last@vars] & /@ 
     Thread[Last /@ (A.vars) == b]];
  r = Sequence[Evaluate@{First@vars, 0, x1}, 
    Evaluate@{Last@vars, 0, x2}];
  Print[sol];
  Manipulate[
   Show[RegionPlot[And @@ cond, Evaluate@r, 
     BoundaryStyle -> {Blue, Thick}], 
    ContourPlot[Evaluate@Apply[Equal, cond, 1], Evaluate@r, 
     ContourStyle -> {{Blue, Thick}}],
    ContourPlot[obj == k, Evaluate@r, ContourStyle -> {{Red, Thick}}]
    ], {{k, N@(First@sol)/2, "Objective Function"}, 0, First@sol, 
    Appearance -> "Open"}]]

GraphicalMethod[{12, 15}, {{4, 3}, {2, 5}}, {12, 10}, {x[1],x[2]}]

It can work for this example, but if we have conditions for 1 variable

GraphicalMethod[{12, 15}, {{4, 0}, {2, 5}}, {12, 10}, {x[1],x[2]}]

it

x1 = Max[First@vars /. Solve[#, First@vars] & /@ 
    Thread[First /@ (A.vars) == b]];
x2 = Max[Last@vars /. Solve[#, Last@vars] & /@ 
    Thread[Last /@ (A.vars) == b]];

gives False and function will not work. I understand why, but I haven't another ideas how to get the optimal interval r for x[1] and x[2]. Can you help me?

Answer 1

I believe the following does more or less the same and is much easier to read:

Clear[GraphicalMethod];
GraphicalMethod[c_?VectorQ, m_?MatrixQ, b_?VectorQ] :=
 Module[{k, eqs, l2, l1, jeq, x = {x1, x2}},
  k = c.LinearProgramming[-c, m, Thread[{b, -1}]];
  eqs = Reduce /@ Thread[m.x == b];
  jeq = Join[eqs, {k == x.c}];
  {l2, l1} = Max@Flatten[Solve /@ (jeq /. # -> 0)][[All, 2]] & /@ x;
  Manipulate[
   Show[
    RegionPlot [And@@ Thread[m.x <= b], {x1, 0, l1}, {x2, 0, l2}],
    ContourPlot[Evaluate@eqs,           {x1, 0, l1}, {x2, 0, l2}],
    ContourPlot[k1 == x.c,              {x1, 0, l1}, {x2, 0, l2},ContourStyle -> Red]],
   {{k1, k/2, "Objective Function"}, 0, k}]]

c = {12, 15};
m = {{4, 3}, {2, 5}};
b = {12, 8};

GraphicalMethod[c, m, b]

Answer 2

The following is a quick and dirty patch. You can do something more elegant taking it as a template:

Clear[GraphicalMethod]
GraphicalMethod[L_?VectorQ, A_?MatrixQ, b_?VectorQ, vars_?VectorQ] := 
 Module[{obj = L.vars, cond = Thread[A.vars <= b]~Join~Thread[vars >= 0], r, sol, x1, x2},
   sol = Maximize[{L.vars, Thread[A.vars <= b]~Join~Thread[0 <= vars]}, vars];
  x1 = Max[First@vars /. Solve[#, First@vars] & /@ Thread[First /@ (A.vars) == b]];
  x2 = Max[Last@vars /. Solve[#, Last@vars] & /@ Thread[Last /@ (A.vars) == b]];
  x2 = First@ If[NumericQ[x2], {x2}, Evaluate[ Last[vars]] /. (Solve[
        obj == sol[[1]] /. Evaluate[First[vars]] -> 0])];
  x1 = First@ If[NumericQ[x1], {x1}, Evaluate[First[vars]] /. (Solve[
        obj == sol[[1]] /. Evaluate[Last[vars]] -> 0])];
  r = Sequence[Evaluate@{First@vars, 0, x1}, Evaluate@{Last@vars, 0, x2}];

  Print[sol];
  Manipulate[Show[
    RegionPlot[And @@ cond, Evaluate@r, BoundaryStyle -> {Blue, Thick}], 
    ContourPlot[Evaluate@Apply[Equal, cond, 1], Evaluate@r, ContourStyle -> {{Blue, Thick}}],
    ContourPlot[obj == k, Evaluate@r, 
     ContourStyle -> {{Red, Thick}}]], {{k, N@(First@sol)/2, 
     "Objective Function"}, 0, First@sol, Appearance -> "Open"}]]

And then

Row[{
  GraphicalMethod[{12, 15}, {{4, 3}, {2, 5}}, {12, 10}, {x[1], x[2]}],
  GraphicalMethod[{12, 15}, {{4, 0}, {2, 5}}, {12, 10}, {x[1], x[2]}],
  GraphicalMethod[{12, 15}, {{4, 0}, {0, 5}}, {12, 10}, {x[1], x[2]}]}]