Path in a graph with durations I have the following problem:

given 


a directed graph $G=(V,E,d)$, where $d:V\to\mathcal{I}(\mathbb{Q}_0^+\cup\{+\infty\})$ (here $\mathbb{Q}_0^+$ denotes the set of non-negative rationals and $\mathcal{I}(\mathbb{Q}_0^+\cup\{+\infty\})$ the set of intervals, bounded or unbounded above, with non-negative rational bounds) is a function associating with each vertex $v\in V$ a "minimum/maximum duration" $d(v)=[a,b]$ for some $a\in \mathbb{Q}_0^+,b\in \mathbb{Q}_0^+\cup\{+\infty\}$ and $a\leq b$,
two vertices $s,t\in V$ and
an integer $h$ encoded in binary,


we have to decide whether or not there exist


a path in $G$, possibly with repeated vertices and edges, $v_0 \cdot v_1 \cdots v_{n-1}\cdot v_n$, with $v_0=s$ and $v_n=t$ and 
a list of values $d_0,\ldots,d_n\in\mathbb{Q}_0^+$,
such that $\sum_{i=0}^n d_i = h$ and for all $i=0,\ldots, n$, $d_i\in d(v_i)$.


Intuitively, we have to find a path in $G$, possibly where we get to the same vertices/edges also more than once, and where we remain in each vertex a non-negative rational amount of time allowed by the minimum/maximum duration function, such that the overall time of the path equals $h$.

This can be solved easily in PSPACE.
We conjecture it to be in NP (we already know it is NP-hard!). This is not trivial to prove, as we may have $h\in\Theta(2^n)$, for instance. Thus the required path may have length exponential in both $|V|$ and in the binary encoding of $h$.

Have you ever seen a similar problem? Can you come up with an NP algorithm? Or do you know some connected literature?
This solution is by Gerhard Woeginger.

In order to prove that this problem belongs to NP,
we provide a polynomial-size certificate,
and then we show how to check it in deterministic polynomial time.

The certificate is just the following:
a set of integers $\{x_{u,v}\mid (u,v)\in E\}$.
Intuitively, $x_{u,v}$ is the number of times the solution path
traverses $(u,v)$.

We now describe the
verification algorithm.


We consider the subset $E'$ of edges of $G$, $E':=\{(u,v)\in E\mid x_{u,v}&gt;0\}$. We check whether $E'$ induces a strongly (undirected)
connected subgraph of $G$.
We check whether 


$\sum_{(u,v)\in E'} x_{u,v}=\sum_{(v,w)\in E'} x_{v,w}$, for all $v \in V\setminus\{s,t\}$;
$\sum_{(u,s)\in E'} x_{u,s}=\sum_{(s,w)\in E'} x_{s,w}-1$;
$\sum_{(u,t)\in E'} x_{u,t}=\sum_{(t,w)\in E'} x_{t,w}+1$.

For all $v \in V\setminus\{s\}$, we define $y_v:=\sum_{(u,v)\in E'} x_{u,v}$, i.e., the number of times the solution path gets into $v$. Moreover, 
$y_s := \sum_{(s,u)\in E'} x_{s,u}$.
We check whether there exist real values $z_v$, for every $v \in V$, such that


$d_{min}(v)\cdot y_v \leq z_v \leq d_{max}(v)\cdot y_v$ (here $d_{min}(v)$ and $d_{max}(v)$ denote resp. the lower and the upper
bound of the rational interval $d(v)$), and
$\sum_{v \in V} z_v = h$.



We now sketch the correctness:

Steps 1. and 2. together check that the values $x_{u,v}$ for the arcs specify
a directed Eulerian path from $s$ to $t$ (we refer to http://www.maths.manchester.ac.uk/~mrm/Teaching/DiscreteMaths/LectureNotes/EulerianMultigraphs.pdf)

Steps 3. and 4. calculate $z_v$, for all $v\in V$, which is the total waiting time of the path on the node $v$. We observe that the (in)equalities of step 4.
form a linear program (LP), which can be solved in deterministic polynomial time (e.g., using the ellipsoid algorithm).