Let us show that a required $k$-times bin-packing exists for some $k\le (n+1)^n$.

Let $V=\{(v_1,\dots,v_n)\in\{0,1\}^n:\sum_{i=1}^n d_iv_i\le s\}$ be the set of all admissible "bin packings". Then the convex hull $\operatorname{conv} V$ of $V$ is the set of all possible ``schedules''.

Let $e=(1,1,\dots,1)\in\mathbb R^n$. Then $r_\max=\sup\{r\in [0,1]:re\in\operatorname{conv} V\}$. Since the set $\operatorname{conv} V$ is compact, $r_\max$ is attained, that is $r_\max e\in \operatorname{conv} V$. We assume that the electricity division problem has a solution, so $r_\max>0$.

Here is an illustration, based on the example in the question:

[![enter image description here][1]][1]

The three blue points are the points in $V$, which are $(1,1,0)$, $(1,0,1)$ and $(0,1,1)$. The ray from the origin is $e\cdot \{r:r
\ge 0\}$. The green point is the intersection of this ray with $V$; its coordinates are $(2/3, 2/3, 2/3)$, which correspond to $r_{\max}=2/3$.

By [Carathéodory's theorem](http://en.wikipedia.org/wiki/Carath%C3%A9odory's_theorem_(convex_hull)),
there exists an $n$ dimensional simplex $S$ with vertices in $V$ such that $r_\max e\in S$.

Let $V'$ be the set of the vertices of the face of $S$ of minimal dimension containing $r_\max e$. Clearly, $r_\max e$ is a boundary point of $S$, so $|V'|\le n$. In other words, the optimal electricity division can be constructed using most $n$ different configurations (in the example $V' = V$ and $|V'|=n=3$).

Let $(\lambda_v)_{v\in V'}$ be the barycentric coordinates of $r_\max e$, such that $\lambda_v>0$ for each $v\in V'$ and $\sum_{v\in V'}\lambda_v=1$ and $r_\max e =\sum_{v\in V'}\lambda_v v$. For instance,  in the example $\lambda_v =1/3$ for each $v\in V'$. 

We claim that the set $V'$ is linearly independent. Indeed, suppose for a contradiction that there exist disjoint nonempty subsets $V_1'$ and $V_2'$ of $V$ and positive numbers $(\mu_v)_{v\in V'_1\cup V_2'}$ such that $\sum_{v\in V_1'} \mu_v v=\sum_{v\in V_2'} \mu_v v$ and $\sum_{v\in V_1'} \mu_v \le \sum_{v\in V_2'} \mu_v$. Let $\nu=\min_{v\in V_2'} \lambda_v/\mu_v$. Then
$$r_\max e =\sum_{v\in V'}\lambda_v v=\sum_{v\in V'}\lambda_v v+\nu(\sum_{v\in V_1'} \mu_v v-\sum_{v\in V_2'} \mu_v v)=\sum_{v\in V'}\lambda_v' v,$$
where $$\lambda_v' v=\cases{
\lambda_v+\nu\mu_v, & if $v\in V_1'$,\\
\lambda_v-\nu\mu_v, & if $v\in V_2'$,\\
\lambda_v, & if $v\in V'\setminus (V_1'\cup V_2')$.
}$$
Then $s=\sum_{v\in V'}\lambda_v'=\sum_{v\in V'}\lambda_v+\nu(\sum_{v\in V_1'} \mu_v v-\sum_{v\in V_2'} \mu_v v)\le 1$, $\lambda_v'\ge 0$ for each $v\in V'$, and $\lambda_v'=0$ for some $v\in V_2'$. Then $s^{-1}r_\max e \in\operatorname{conv} V'\setminus\{v\}$, that contradicts to the minimality of $V'$.

Since the set $V''=V'\cup\{e\}$ is linearly dependent, by Lemma from [this](https://math.stackexchange.com/questions/3070925/biggest-coefficients-of-a-linear-combination-between-vectors-of-zeros-and-ones/3085439#3085439)
answer there exist integers $(f_v)_{v\in V''}$ which are not all zeroes such that $|f_v|\le (n+1)^n$ for each $v\in V''$ and $\sum_{v\in V''} f_vv=0$. The linear independence of the set $V'$ implies that $0\ne f_e\le (n+1)^n$ and $\lambda_v/r_\max=f_v/f_e$ for each $v\in V'$.
This provides the required $f_e$-times bin-packing.



  [1]: https://i.sstatic.net/NINp4.png