Mixed Integer linear programming - absolute value of a variable not involved n the objective function I'm looking to find the absolute value of the expression s-t. I have begun by introducing the following constraints:

Where A is the absolute value. Unfortunately, A is not involved in the objective function, so I need to constrain it from above as well. What other constraints do I need to introduce to achieve this.
 A: You are indeed missing some equations. Here is a way.
Define $x$ the Boolean variable equal to 1 iff $s-t \geq 0$. If you know an upper bound $B$ on $|s-t|$, $x$ is defined by
$s-t \leq Bx$
$t-s \geq  B(x-1)$
Then you can define $A$ as
$s-t \leq A$
$t-s \leq A $
$A \leq s-t + B(1-x) $
$A \leq t-s + Bx $
Note that the bigger $B$ is, the less efficient the linear relaxation will be.
There are some other ways, but all the other ones I know are way more expensive in terms of number of variables. 
A: This is based on Vincent's answer, but I lack the ability to comment...
Remember that you must find a set of constraints that are all true if $A$ has the proper value of $|s-t|$.  Further, we want to "hot-box" $A$ so that it can only take on either $s-t$ or $t-s$, depending on which is the correct absolute value.  This is what $x$ tells us.
As Vincent laid out, define $x$ as a Boolean variable equal to 1 iff $s-t \geq 0$, i.e. $s \geq t$. Determine an upper bound $B$ on $|s-t|$, the tighter the better. 
The constraints:
$s-t \leq Bx $
$t-s \leq  B(1-x) $
force $x$ to 1 when $s-t$ is positive.  Note this is slightly different than Vincent's above, but tested with truth tables.  Now define a variable $A$ with range (0, $B$).  The constraints
$s-t \leq A$
$t-s \leq A $
will make the linear relaxation choose a value for $A$ that just satisfies them both.  In once case, $A$ is easily within the half-plane because the subtraction gives a negative value.  In the other, $A$ is at the very edge of the half-plane, because it is defined by (the real) $A$.  The constraints:
$A \leq s-t + 2B(1-x) $
$A \leq t-s + 2Bx $
attempt to move $A$ as far away from the wrong subtraction order as possible (though the previous constraints must still hold.)  The previous constraints make $A$ at least the value of the right subtraction order; these make $A$ at most the value of the right subtraction order if $B$ is zero'ed out, and an allowable value if $B$ is doubled but we subtract off the correct value.  All of this means that two of these constraints will have the value of $A$, one from above and one from below.  The other two will have enough slop so they are not violated.
Vincent's answer missed the doubling of bounds.  To see this, evaluate $(s,t,B,x) = (39,80,42,0)$ and notice that $A \leq s-t + B(1-x) $ fails--you need $2B$.
