Ensure a vertex has the highest flow in max-flow algorithm
Add auxiliary nodes 8a, 9a, 10a between nodes 8, 9, 10 and node 11, so that for example we have 8 -&gt; 8a -&gt; 11. Set all the capacities on the (8 -&gt; 8a, 9 -&gt; 9a, and 10 -&gt; 10a) edges to $c = 0$, and run max flow. If there is a solution (if the maximum flow through the graph is equal to the number $n$ of packages to be shipped) then congratulations - we are done. Otherwise, set the capacities to $c = 1$, and go again. And so on until you find the smallest capacity $c^*$ such that the maximum flow is $n$. If the flow through vertex 7 is larger than this capacity $c^*$ then we have a solution, but if not there is no solution. If you continue to increase up to $c = n$ and there is no solution, then the original problem has no solution.

As requested by D.W. I will elaborate on the correctness and efficiency of this solution. The simple method of increasing $c$ by $1$ each time to find $c^*$ can be improved by binary searching instead: as $c$ increases the maximum flow will also increase up to $n$, and therefore we can binary search on the possible values of $c$ in order to find the smallest $c^*$ such that the maximum flow is $n$. This improves the number of calls to a max-flow subroutine from $n$ to $\log n$.
In order to show correctness I will set up some more notation. In any allocation of flow to the graph, let $f_0$ be the flow through the special warehouse, and $f_1, \ldots, f_k$ the flow through the other warehouses. We will call an allocation of flow &quot;feasible&quot; if $f_0 + f_1 + \cdots + f_k = n$ and $f_0 &gt; f_i$ for all $i = 1, \ldots, k$ and &quot;infeasible&quot; otherwise.
Suppose there is no feasible flow. Then no matter how we set $c$ in our algorithm, it will never find a feasible flow either (introducing capacities to vertices only limits the set of solutions). Therefore the algorithm gets the correct answer in this case.
Suppose there is a feasible flow. Out of all possible feasible flows, choose one which minimises $c' = \max_{i = 1, \ldots, k} f_i$. Then there is no feasible flow where $f_i &lt; c'$ for all $i$, and there is a feasible flow where $f_i \leq c'$ for all $i$, and hence our algorithm will arrive at $c^* = c'$. However, the algorithm may not arrive at the same flows: in particular if $g_0, g_1, \ldots, g_k$ are the flows found by the algorithm, we are guaranteed that $g_i \leq c'$ for all $i = 1, \ldots, k$, but we may not have $g_0 &gt; g_i$ for all $i$.
I think this is a problem with the algorithm actually: it finds $c^* = c'$ correctly, but needs some help in order to figure out whether flow through the special node can be made strictly larger. There are some things you could do with the residual graph here, or you could use a reduction to a circulation problem with lower bounds on capacities: use the algorithm above, but as a final step introduce the constraint that the flow through the special node must be at least $c^* + 1$, and check for a solution.