Skip to main content

plotting - Graphically solving a linear programming problem


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



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]

Mathematica graphics


Comments

Popular posts from this blog

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...