unbounded in linear programming

Hay guys I'm trying to solve this problem using the linprog() function but it keeps telling that the problem is unbounded

%each type number of orders
syms x1 x2 x3 x4 x5 x6 ; 

%profit margins for each type ($/vehicle)
m1=150; %Subcompact 
m2=225; %Compact 
m3=250; %Intermediate
m4=500; %Luxury 
m5=400; %Truck
m6=200; %Van
%=========================================

%fuel constraint simplification
MPG=(40*x1 + 34*x2 + 15*x3 +12*x4 +20*x5 + 25*x6) -28*(x1 + x2 + x3 + x4 + x5 + x6); % >= 0

%Objective function
Rev=(m1*x1+m2*x2+m3*x3+m4*x4+m5*x5+m6*x6);

%========================================

%{The constraints:
%x1 + x2 + x3 + x4 + x5 + x6 <= 1,200,000
%x1 + x2 <= 620,000
%x3 + x4 <= 400,000
%x5 + x6 <= 275,000
% -12*x1 - 6*x2 + 13*x3 + 16*x4 + 8*x5 + 3*x6 <= 0
%}


A=[1 1 1 1 1 1;1 1 0 0 0 0;0 0 1 1 0 0;0 0 0 0 1 1;-12 -6 13 16 8 3];

b= [1200000;620000;400000;275000;0];

f=[-150 -225 -250 -500 -400 -200];

Aeq=[];
beq=[];

LB = [0;0];
%UB = [Inf; Inf]
UB = [];

[X,Z]=linprog(f,A,b,Aeq,beq,LB,UB);

Warning: Length of lower bounds is < length(x); filling in missing lower bounds with -Inf.

Problem is unbounded.

x=X

x =

     []

z=1*Z

z =

     []

[Dyuman Joshi]

If Rev is the objective function then what is f? And what is the use of MPG?

It's not clear from your code what you are solving for. Please attach the mathematical formulation of the problem you are trying to solve.

Answer 1

[John D'Errico]

First, read the warning message!

"Warning: Length of lower bounds is < length(x); filling in missing lower bounds with -Inf."

You gave only TWO lower bounds. But you have 6 unknowns. So MATLAB tells you the lower bounds it will employ are:

LB = [0 0 -inf -inf -inf -inf];

The upper bounds will all be +inf. Warning messages are used to tell you important information. Read them. Think about what they tell you.

Given that, I would not be at all suprised if your problem is indeed unbounded. If instead, I supply a complete set of lower bounds, the problem does indeed have a solution.

A=[1 1 1 1 1 1;1 1 0 0 0 0;0 0 1 1 0 0;0 0 0 0 1 1;-12 -6 13 16 8 3];
b= [1200000;620000;400000;275000;0];
f=[-150 -225 -250 -500 -400 -200];
Aeq=[];
beq=[];
LB = zeros(1,6);
UB = [];
[X,Z]=linprog(f,A,b,Aeq,beq,LB,UB)

Optimal solution found.

X = 6×1

560000 60000 0 305000 275000 0

Z = -360000000