I would like to convert this problem into a Linear Programming Problem : 


$\min |x|+|y|+|z|$


subject to $x+y \leq 1$ 


$2x+z=3$.


The solution to this problem is given [chapter](https://www.calvin.edu/~pribeiro/courses/Power%20Systems%20Interim/Sakis-Book/PSA_Appendix_B.pdf) and [here](http://orinanobworld.blogspot.in/2012/07/modeling-absolute-values.html). But I still do not get it. Could anyone please explain it explicitly the problem solution. Especially the part how the objective value is formed.


Thanks in advance.



convert absolute form into linear programming problem
Let's call the following model model $A$:


$\min |x|+|y|+|z|$


subject to $x+y \leq 1$ 


$2x+z=3$.


and the following model model $B$:


$\min x\_1+x\_2+|y|+|z|$


subject to $x\_1-x\_2+y \leq 1$ 


$2x\_1-2x\_2+z=3$.


$x\_1,x\_2 \geq 0$


I claim that the two models are equivalent. 


To see this, suppose the optimal solution from model $A$ that is $(x\_A, y\_A, z\_A)$. We can let 


$$(x\_{1B},x\_{2B},y\_B,z\_B)=\begin{cases} (x\_A,0,y\_A,z\_A)& x\_A\geq0 \\ (0,-x\_A,y\_A,z\_A)& x\_A <0\end{cases}$$


and we can see that it is feasible for model $B$ and obtain the same value. Hence the optimal value for model $B$ cannot be worse than model $A$. the question is can it do better?


To verify that this is not the case, suppose we have an optimal solution for model $B$. $(x\_{1B},x\_{2B},y\_B,z\_B)$, we want to construct a solution to model $A$ that shares the same value.


I claim that $x\_{1B}x\_{2B}=0$, i.e., one of them is zero. Otherwise, we can construct another feasible solution $(x\_{1B}-\min(x\_{1B},x\_{2B}),x\_{2B}-\min(x\_{1B},x\_{2B}),y\_B,z\_B)$ that is feasible and has a smaller objective value, contradicting optimality of $(x\_{1B},x\_{2B},y\_B,z\_B)$.


Now, we can verify that 


$$(x\_A,y\_A,z\_A)=(x\_{1B}-x\_{2B},y\_B,z\_B)$$
is indeed feasible for model $A$.


If $x\_{1B}=0$, then $x\_A=x\_{1B}-x\_{2B}=-x\_{2B}$, and $$|x\_A|=-(-x\_{2B})=x\_{2B}=0+x\_{2B}=x\_{1B}+x\_{2B}$$


If $x\_{2B}=0$, then $x\_A=x\_{1B}-x\_{2B}=x\_{1B}$, and $$|x\_A|=x\_{1B}=x\_{1B}+0=x\_{1B}+x\_{2B}$$


and hence the two optimization problem is equivalent.


To answer your original problem, repeat the same procedure for $y$ and $z$.


