1. ## Linear programming

Maximize $\displaystyle f(x,y)= 2x+y$

subject to
$\displaystyle 2x-y-8 \leq 0$
$\displaystyle x+2y-14 \leq 0$
$\displaystyle -x+y-4 \leq 0$
$\displaystyle -x \leq 0$
$\displaystyle -y \leq 0$

a)with Graphical method of Linear programming
b)with Simplex method of Linear programming

2. Originally Posted by abstar
Maximize $\displaystyle f(x,y)= 2x+y$

subject to
$\displaystyle 2x-y-8 \leq 0$
$\displaystyle x+2y-14 \leq 0$
$\displaystyle -x+y-4 \leq 0$
$\displaystyle -x \leq 0$
$\displaystyle -y \leq 0$

a)with Graphical method of Linear programming
b)with Simplex method of Linear programming

Please show us your work and let us know where you get stuck; we are not here to do out the entire problem for you.

To get you started, denote $\displaystyle 2x+y$ as $\displaystyle 2x_1 + x_2$. Declare your objective function $\displaystyle z=2x_1 + x_2$.

$\displaystyle 2x_1-x_2-8 \leq 0$
$\displaystyle x_1+2x_2-14 \leq 0$
$\displaystyle -x_1+x_2-4 \leq 0$

The last two constraints can be ignored, as the Simplex Method itself requires all variables to be nonnegative. In other words, as you carry out your pivots correctly, the minimal ratio test will ensure that your variables say nonnegative.

For these three constraints, introduce slack variables $\displaystyle x_3,x_4,x_5$ and use that as your initial basis for a feasible tableau:

maximize $\displaystyle z - 2x_1 - x_2 = 0$
$\displaystyle 2x_1-x_2 + x_3 = 8$
$\displaystyle x_1+2x_2 + x_4 = 14$
$\displaystyle -x_1+x_2 + x_5 = 4$

Your basis is currently [3,4,5] with an objective value of 0. Since both of them have positive reduced costs, choose one of $\displaystyle x_1,x_2$ as the entering variable k.

Can you take it from there? Check your answers by comparing the graphical and the Simplex method - they should agree with each other.

3. i've started to do this and got as far as the formulas part, i'm trying to work outthe new values in the new table

where R3 is the pivot row

row1 = (R1-R3)
row2 = (r2-2(r3)
row4 = (R4 + 2(r3)

also is it always Rx - Pivot row, i get confused on the formula section, the rest i think im ok with. Could someone please explain what to do here?

my pivot table is attached to this thread, could someone have a look and check for me please?

4. Originally Posted by abstar
Maximize $\displaystyle f(x,y)= 2x+y$

subject to
$\displaystyle 2x-y-8 \leq 0$
$\displaystyle x+2y-14 \leq 0$
$\displaystyle -x+y-4 \leq 0$
$\displaystyle -x \leq 0$
$\displaystyle -y \leq 0$

a)with Graphical method of Linear programming
b)with Simplex method of Linear programming

a) with Graphical method of Linear programming

$\displaystyle \overrightarrow n$ is a normal-vector