By a weighted total order on $\mathbb{N}^r$ or $\mathbb{Z}^r$ I mean the relation defined by $$ (m_1,\ldots,m_r) \leq (n_1,\ldots,n_r) \quad\Longleftrightarrow\quad \sum_{i=1}^r c_i m_i \leq \sum_{i=1}^r c_i n_i $$ for some $c_1,\ldots,c_r \in \mathbb{R} _ {\geq 0}$, linearly independent over $\mathbb{Q}$, which are are called the weights defining the order.
By a generalized weighted total order I mean one of the form $$ (m_1,\ldots,m_r) \leq (n_1,\ldots,n_r) \quad\Longleftrightarrow\quad \sum_{i=1}^r c^{(j)}_ i m_i \mathrel{\leq_{\mathrm{lex}(j)}} \sum_{i=1}^r c^{(j)}_ i n_i $$ where $c^{(1)},\ldots,c^{(s)} \in \mathbb{R} _ {\geq 0}^r$ are $s$ weight vectors (I think we can assume w.l.o.g. $s\leq r$) each of length $r$, only the last of which being assumed linearly independent over $\mathbb{Q}$ (that is, $c^{(s)}_ 1,\ldots,c^{(s)}_ r$ are); here the notation $u^{(j)} \mathrel{\leq_{\mathrm{lex}(j)}} v^{(j)}$ means lexicographic comparison, that is, $u^{(j)} < v^{(j)}$ for the smallest $j$ such that $u^{(j)} \neq v^{(j)}$ if it exists.
(For example, the lexicographic order on $m_1,\ldots,m_r$ themselves is a generalized weighted total order but is not a weighted total order. Note that since I require the weights to be nonnegative, any generalized weighted total order extends the product total order on $\mathbb{Z}^r$; also, it is compatible with addition in the sense that $m\leq n$ implies $m+p \leq n+p$.)
Question: can we give a simple necessary and sufficient condition for a total order relation on $\mathbb{N}^r$ or $\mathbb{Z}^r$ for it to be a weighted total order? How about a generalized weighted total order?
(I seem to remember that there are standard results in this direction, so this is more of a reference request than a request to think about the problem. Please don't hesitate to mention any standard result more or less in this spirit, even if it doesn't answer exactly the question as stated.)