Formulating an LP problem with vectors We have $m$ vectors $v_1,v_2,\dots,v_m\in\mathbb{R}^n$ and $m$ numbers $t_1,t_2,\dots,t_m\in\mathbb{R}$ and we want to find a vector $y\in\mathbb{R}^n$ such that

$$|v_i^Ty-t_i|\leq D$$

for $i=1,\dots,m$ and $D\in\mathbb{R}$ as small as possible.
How do I formulate this as an LP problem of the form
min $c^Tx$
under the condition that $Ax\leq b$,
$x\in\mathbb{R}^n$?
I still find it rather difficult to formulate an LP problem, so maybe you can help me?
Edit: can you tell me explicitly what $c,A,x,b$ are?
 A: $D$ is a variable to be minimized while meeting the constraints which can be translated to
$$-D \le v_i^Ty-t_i\leq D\ \ ,\ D \ge 0\ ,\ y\in\mathbb{R}^n$$
The cost function to minimize is $D$.
Linear programming minimizes/maximizes $F = c^Tx$ where $Ax \le b$ and $x \ge 0$. $x$ is bounded on one side. The $y$ vector in the above problem is not bounded.Each $y_i$ variable needs to be duplicated with $y_{i,1} \ge 0$ and  $y_{i,2} \ge 0$ s.t. $y_i = y_{i,1} - y_{i,2}$.
$x = [y_{i,1}^T,y_{i,2}^T,D]^T $ the unknowns. $x \ge 0$. A column vector with all $y_{i,1}$ first then all $y_{i,2}$, $D$ is the last variable.
The constraints need to be written in the form:
$$v_i^Ty-D\leq t_i$$
$$-v_i^Ty -D \le -t_i$$
Substituting $y_i = y_{i,1} - y_{i,2}$.
$$v_i^Ty_{.,1} - v_i^Ty_{.,2}-D\leq t_i$$
$$-v_i^Ty_{.,1} + v_i^Ty_{.,2} -D \le -t_i$$
Each row of $A$ and $b$ represent one inequality constraint. 
$$A_r = [v_i^T , -v_i^T, -1]\ ,\ b_r = t_i$$
$$A_{r+1} = [-v_i^T, v_i^T, -1]\ ,\ b_{r+1} = -t_i$$
$c$ is the cost. $F_{min}=c^Tx$ is the cost function that will be minimized.
$c^T = [0\dots0,1]$ i.e. $c$ selects $D$ as the cost.
The solution $x$ gives the values of $D$ and all $y_{i,[1,2]}$ then $y_i = y_{i,1} - y_{i,2}$.
LPSolve simplifies matters by allowing unbounded (free) variables and allowing the use of $\le,=,\ge$.
