Gravity location problem locates a warehouse at $(x,y)$ to serve $N$ retailers. Retailer $n$ is already

located at $(a_n,b_n).$ The objective in the gravity location problem is

$mi{n}_{x,y}{\displaystyle \underset{n=1}{\overset{N}{}}}\sqrt[\phantom{2}]{(a_n-x{)}^2+(b_n-y{)}^2}.$

When we measure the distances in terms of rectilinear distances as opposed to Euclidean distances,

we arrive at the new rectilinear objective

$mi{n}_{x,y}{\displaystyle \underset{n=1}{\overset{N}{}}}|a_n-x|+|b_n-y|,$

where $|*|$ denotes the absolute value. In comparison to the gravity location objective, the rectilinear

objective seems more "linear". The rectilinear objective is not totally linear, because absolute value

function is not; draw $f(x)=|x|$ to convince yourself. Despite this, it is possible to provide an LP

formulation.

Towards an LP formulation, first let $d_n^x$ and $d_n^y$ be new decision variables denoting the rectilinear

distance from the warehouse to retailer $n$ respectively in $x-$ and $y-$axis directions. Then we can write

the following formulation

$mi{n}_{x,y}{\displaystyle \underset{n=1}{\overset{N}{}}}{d}_n^x+d_n^y$

s$.$t$.$

$d_n^x|a_n-x|$

$d_n^y|b_n-y|$

First convince yourself that in the optimal solution, the inequalities in the constraints will be satisfied

as equalities. This formulation, due to absolute values in the constraints, is not an LP$.$ However,

there is a slick way to write absolute values with linear constraints. First observe that $|a_n-x|=$

max$\{a_n-x,x-a_n\}.$ Second, note that if $z$max$\{u,v\},$ then $zu$ and $zv.$ Using these ideas

provide an LP formulation to solve the rectilinear distance problem.