An obvious lower bound for the number of bowls that is needed is the maximum number of operations one operation directly depends on, plus one. This is directly implied by the problem statement.

However, this figure is not always attainable. By way of contradiction, let operation 1 be dependent on operations 2 and 3. If we know a priori that both operations need 10 bowls to complete, then we actually need 11 bowls to complete operation 1. We arbitrarily pick one of the operations to complete, and then we need to allocate one bowl to store that intermediate result, and then we need 10 more bowls to complete the other operation.

This leads us to our first critical realization - given operations $x_1$ through operations $x_k$ that operation 1 directly depends on, we should not interleave intermediate operations from two distinct ones. Therefore, we should select some operation and commit to finishing it before switching to another one that operation 1 depends on.

What order should we commit to finishing them in? Intuitively, we should commit to the operations which use the most bowls first. Formally, let $f(n)$ be the minimum number of bowls needed to complete operation $n$. If $n$ does not depend on any operations, $f(n)$ is equal to $1$. Otherwise, consider the operations $x_1$ through $x_k$ that operation $n$ directly depends on, and compute $f(x_1)$ through $f(x_k)$. Assume without loss of generality that $f(x_i)\ge f(x_{i+1})$ for $1\le i\le k-1$. $f(n)$ is therefore equal to $max(k+1,f(x_1),f(x_2)+1,\dots ,f(x_k)+(k-1))$.

This runs in $\mathcal{O}(N\log N)$ if done naively, and can be optimized to $\mathcal{O}(N)$ though that improvement is not necessary.