Need construction for coequalizer in $\mathbf{Poset}$ My question can be stated quickly:

I would like to see a construction of the coequalizer of two arbitrary Poset morphisms (along with a proof of its correctness, of course).

Thanks!

(The stuff beyond this point is of dubious value; I provide it mostly to show why I've come to regard the question above as a sufficiently non-trivial one, certainly not a trivial extension of the analogous construction for Set.  If you don't need convincing of any of this, you won't miss anything if you skip it.)
In general, the coequalizer object for the corresponding set functions cannot necessarily be ordered so as to ensure that the coequalizing map is order-preserving.  (In case my reasoning is wrong, I give an example of what I mean at the end of this post).  I've come up with "improved" constructions that attempt to remedy the shortcomings of the Set construction when applied to Poset, but proving their correctness involves exposure to lethal doses of tedium, which I'd prefer to avoid.
Googling for the problem of constructing coequalizers in Poset has turned up surprisingly little (which, of course, I attribute to the lethality of tedium).  If anyone can point me to one such construction, and more importantly, to a proof of that the resulting coequalizing map is indeed order-preserving, I'd appreciate it.
BTW, in ch. 6 of Arbib & Manes, the authors give a theorem that ensures the admissibility of a map if there's an "optimal lift" for (in this case) its domain.  (I post this in case their terminology is sufficiently standard to be informative to some of the readers of this post, since A&M's definition of this concept depends on a fair amount of preliminary groundwork.)  If I understand their argument at all, the problem of finding an "optimal lift", in this case at least, doesn't look any easier than the problem of constructing the coequalizer object in the first place.  (Perhaps what they have in mind is that one may be able to prove the existence of such an optimal lift without actually having to construct it, but this I would find very unsatisfying.)
For a simple example of a Set coequalizer that cannot be made order-preserving, consider the automorphisms $i \mapsto i$ and $i \mapsto i + 2$ on $\mathbb{Z}$, equipped with its standard order (hence, these are Poset morphisms).  Their standard Set coequalizer is the quotient of $\mathbb{Z}$ by the equivalence closure of $\{\;(i, i + 2) \;|\; i \in \mathbb{Z}\;\}$.  If $q$ is the canonical projection of $\mathbb{Z}$ onto this quotient, then we have $q(i+1) \neq q(i) = q(i+2), \forall i \in \mathbb{Z}$, which rules out the existence of any order for the quotient that would render $q$ order-preserving.
 A: Brian M. Scott's comment is more or less a complete answer to the question (any cocone admits a map from the set-theoretic coequalizer with the obvious preorder and also necessarily collapses equivalence classes). 
I'd just like to point out that this is an argument for working in the category of preorders instead of the category of posets, since in the former category you no longer need to perform the quotient. Note that the forgetful functor from posets to sets has a left adjoint (take the antichain on a set) but does not have a right adjoint; consequently it preserves limits but can't be expected to preserve colimits. But the forgetful functor from preorders to sets has the same left adjoint as above but also has a right adjoint (make every element of a set less than or equal to every other element), hence it preserves both limits and colimits. 
A: There is already a more or less complete answer in the comments, but I just thought I'd add my attempt to flesh out the details, since the same problem has come up in my studies.
Suppose we have two parallel arrows $f,g:P\rightrightarrows Q$ in $\mathsf{Poset}$. If $\rho\subset P\times P$ the (nonstrict) ordering relation on $P$, then it follows that any coequalizer $h:Q\to E$ with $\tau\subset E\times E$ the ordering on $E$, $\tau$ must contain the following as a subset: $$\tau^\ast = \{(f(x),f(y)) ~ : ~ x,y\in P, ~ x\leq y\}\cup\{(g(x),g(y)) ~ : ~ x,y\in P, ~ x\leq y\}$$ We can now form a kind of "transitive closure" to deduce the ordering on $E$. If $\alpha\subset A\times A$ is a relation, let us denote by $T\alpha$ the set $$T\alpha = \{(x,z) ~ : ~ (x,y),(y,z)\in \alpha\}$$ Then let us denote by $\overline\alpha$ the following relation: $$\overline\alpha = \alpha\cup T\alpha\cup T^2\alpha\cup\cdots$$ so that $\overline\alpha$ is the least transitive relation containing $\alpha$ as a subrelation.
We shall use this to define the coequalizer $h:Q\to E$. Let $\tau^\ast$ be as defined before, so that any coequalizer of $f,g$ must have an ordering that contains $\tau^\ast$, and therefore $\overline{\tau^\ast}$. Now define an equivalence relation on $Q$ as follows: let $x\sim y$ if and only if both $(x,y)$ and $(y,x)$ are contained in $\overline{\tau^\ast}$.  (These are the elements whose images under any coequalizer are "forced to be equal".) Let $E=Q/\sim$, with a partial ordering $\tau$ on $E$ defined by $\tau=\{([x],[y]) ~ : ~ (x,y)\in\overline{\tau^\ast}\}$ where $[x]$ denotes the equivalence class of $x\in Q$ under $\sim$. We should then have that the projection map $h:Q\to E=Q/\sim$ sending $x\mapsto [x]$ for each $x\in Q$ is a coequalizer of $f,g$.
