Optimization Problem on a Directed Graph
lemma 1. The posted problem (as I understand it) is NP-hard, even on DAGs, 

Proof. The proof is by reduction from Subset Sum.  Given a Subset Sum instance with $n$ positive integers $x_1, x_2, \ldots, x_n$ and target $T$, the reduction produces the following instance of your problem.  The instance will have a minimal solution (one with objective equal to the sum of the activation values -- the minimum possible) if and only if some subset of the $x_i$'s sums to $T$.

Let the $x_i$'s be $b$-bit integers for some $b$, so the input instance has size polynomial in $n$ and $b$ and each $x_i$ is an integer in $\{1,2,\ldots, 2^b\}$.

Construct a "tower" gadget with two leaves $u_0, w_0$ with activation 1.  For each $i\in\{1,2,\ldots, b\}$, add two vertices $u_i, w_i$ each with activation threshold $2^i$ and two incoming edges: $(u_{i-1}, u_i)$, $(u_{i-1}, w_{i})$, $(w_{i-1}, u_i)$, $(w_{i-1}, w_i)$.  In any feasible solution to the instance (so far), each such vertex $u_i$ and $w_i$ will have to use both its incoming edges, and will have incoming activation value exactly equal to its activation threshold $2^i$.

Next, for each integer $x_i$ in the given Subset Sum instance, construct a vertex $X_i$ with activation threshold $x_i$, with edges from every $u_j$.  In any minimal solution, each $X_i$ will use just those incoming edges that represent the binary representation of $x_i$, thereby achieving incoming activation value equal to $x_i$ (and this is always possible).

Finally, construct a "root" vertex $R$ with activation value $T$ (the Subset Sum target) and edges from every vertex $X_i$.  There will be a subset of edges into $R$ to use to achieve activation value $T$ if and only if there is a subset of the $x_i$'s that sums to $T$.  $~~~\Box$

What am I missing?  As I understand the problem, showing NP-hardness seems like a nice homework exercise...  

Given that Graphman was last active in 2016, perhaps clarification is not to be had.