# Solutions of the given linear programming problem

#### mathmari

##### Well-known member
MHB Site Helper
Hello!!
Given the following linear programming problem
$$\min(2x + 3y + 6z + 4w)$$
$$x+2y+3z+w \geq 5$$
$$x+y+2z+3w \geq 3$$
$$x,y,z,w \geq 0$$
I am asked to find all the solutions using the simplex method.

To solve this problem we use the Two-Phase method, don't we?
Then I found that there are infinite many solutions, $(x,y,z,w)=(1-m, 2+0.4m, 0, 0.2m), 0≤m≤1$ with $min=8$.
Could you tell me if this is right?

#### Opalg

##### MHB Oldtimer
Staff member
Hello!!
Given the following linear programming problem
$$\min(2x + 3y + 6z + 4w)$$
$$x+2y+3z+w \geq 5$$
$$x+y+2z+3w \geq 3$$
$$x,y,z,w \geq 0$$
I am asked to find all the solutions using the simplex method.

To solve this problem we use the Two-Phase method, don't we?
Then I found that there are infinite many solutions, $(x,y,z,w)=(1-m, 2+0.4m, 0, 0.2m), 0≤m≤1$ with $min=8$.
Could you tell me if this is right?
Your answer is certainly correct, because if you add the two inequalities you get $2x + 3y + 5z + 4w \geqslant 5+3=8$. Therefore $2x + 3y + 6z + 4w \geqslant 8+z$, which is minimised by taking $z=0$. Your solutions, with $z=0$, clearly satisfy all the given conditions, so they must be right.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hello!!
Given the following linear programming problem
$$\min(2x + 3y + 6z + 4w)$$
$$x+2y+3z+w \geq 5$$
$$x+y+2z+3w \geq 3$$
$$x,y,z,w \geq 0$$
I am asked to find all the solutions using the simplex method.

To solve this problem we use the Two-Phase method, don't we?
Then I found that there are infinite many solutions, $(x,y,z,w)=(1-m, 2+0.4m, 0, 0.2m), 0≤m≤1$ with $min=8$.
Could you tell me if this is right?
Looks good!

EDIT: Aargh, overtaken by Opalg.

#### mathmari

##### Well-known member
MHB Site Helper
Your answer is certainly correct, because if you add the two inequalities you get $2x + 3y + 5z + 4w \geqslant 5+3=8$. Therefore $2x + 3y + 6z + 4w \geqslant 8+z$, which is minimised by taking $z=0$. Your solutions, with $z=0$, clearly satisfy all the given conditions, so they must be right.