# Standard Form

\left\{ \begin{aligned} max(z)&=CX\\ AX&=b\\ x_i>0\\ \end{aligned} \right.

$$b\geq0$$, $$m\leq n$$, $$rank(A)=m$$, but Why $$rank(A)=m$$?

If not, the constraint $$AX=b$$ would have only one solution.

# Transform a general form into standard form

• Maxmium the objective function. $$min(z)=CX\rightarrow max(w)=-CX(w=-z)$$
• Transform inequality into equality(Same thing for greater inequality)
$a_{11}x_1+a_{12}x_2+...a_{1n}x_n\leq b_1$ $\downarrow$ $a_{11}x_1+a_{12}x_2+...a_{1n}x_n+y_k= b_1(y_k>0)$
• Change unconstrait variable into two constraint variable.
$x_i=y_k'-y_k''(y_k'>0,y_k''>0)$
• When $$b_j<0$$ multiply (-1) in both side.

# Now we have the standard form. How to solve for the optimal solution?

Remember the $$y$$ variables we added for constriant?

These variables all have the same coefficient $$1$$, so these coefficient form a identity matrix $$I$$.

So we can write something like this $$[A\space I]\begin{bmatrix}X\\Y\end{bmatrix}=b$$ We use what is called Simplex Tableau

# Example

$max (x_1+x_2)$ s.t. \left\{ \begin{align} 2x_1+x_2+x_3=12 \\ x_1+2x_2+x_4=9 \\ x_i>0 \end{align} \right| \quad i=1,2,3,4

$$x_3,x_4$$是基变量（basic variable也就是上面的y，辅助变量），$$x_1,x_2$$是非基变量，也就是要求的值。

## Simple Tableau

$$x_1$$ $$x_2$$ $$x_3$$ $$x_4$$ $$b$$
$$x_3$$ 2 1 1 0 12
$$x_4$$ 1 2 0 1 9
$$c$$ 1 1 0 0 0

We introduce a operation called povit(d,e).

We choose a non-basic variable $$x_e$$ $$s.t.\space c_e>0(Max\space c_e)$$

Then choose a basic variable $$x_d$$ $$s.t.\space A_{(d,e)}>0\space and \space min(\frac{b_d}{A_{(d,e)}})$$

Here we have the same $$c_e$$for both $$x_1$$ and $$x_2$$, we just go with $$x_1$$

So here we see that for $$x_3$$, $$b_3=12$$, $$A_{(3,1)}=2$$, for $$x_4$$, $$b_4=9$$, $$A_{(4,1)}=1$$

Because, $$\frac{12}{2}<\frac{9}{1}$$ we choose $$x_3$$

## After Row Exchange

$$x_1$$ $$x_2$$ $$x_3$$ $$x_4$$ $$b$$
$$x_3$$ 1 1/2 1/2 0 6
$$x_4$$ 0 3/2 -1/2 1 3
$$c$$ 0 1/2 -1/2 0 -6

## Repeat until we have all $$c$$ variable into non-positive ones

$$x_1$$ $$x_2$$ $$x_3$$ $$x_4$$ $$b$$
$$x_3$$ 1 0 2/3 -1/3 5
$$x_4$$ 0 1 -1/3 2/3 2
$$c$$ 0 0 -1/3 -1/3 -7

Now $$x_3=5,x_4=2$$ is the optimal solution.