networkx - traverse all paths from node A to node A, perform operation, and find max value

Initial Answer - Based on nx.simple_cycles

There are many ways to achieve what you're looking for, but here's one suggestion:

First, let me replace the original graph with one that has more cycles and another layout.

G = nx.DiGraph()

G.add_edges_from([('A', 'B'),('B', 'A'), ('C','D'),('G','D')], weight=50)
G.add_edges_from([('D','B')], weight=1)
G.add_edges_from([('B','C'),('F','E'),('E','A')], weight=2)
G.add_edges_from([('C','F'),('B','G'),('A','G')], weight=3)

edge_labels=dict([((u,v,),d['weight'])
                 for u,v,d in G.edges(data=True)])

# improve layout and draw node labels

pos=nx.drawing.layout.circular_layout(G,scale=10)
nx.draw_networkx_edge_labels(G,pos,edge_labels=edge_labels)
nx.draw_networkx_labels(G,pos,font_color='white')
nx.draw(G,pos, node_size=1500,edge_cmap=plt.cm.Reds)

Now we find all simple cycles in G that start and end at a given node of interest and calculate n for each cycle.

# node of interest
inode  = 'A'


acycs  = [ cyc for cyc in nx.simple_cycles(G) if inode in cyc ]

# for ease of computation we append the first node to the end of each cycle
acycs = [cyc+[cyc[0]] for cyc in acycs]

mcyc = []
maxn = 0

# for each cycle involving node of interest
for cyc in acycs:
    n = 100
    
    # group the nodes in the cycle by pairs
    for u, v in zip(cyc[:-1], cyc[1:]):
        
        # retrieve the respective edge weigth for the current pair
        print(u,v, end=',\t')
        div = G.get_edge_data(u,v)['weight']
        
        # calculate and update n
        r   = n/div
        print(f"{n}\t/\t{div}\t=\t{r}")
        n=r
        
    print('----------------------------------------------------------------------------')

    if n>maxn:
        maxn=n
        mcyc=cyc

This gives us all loops starting at inode='A':

[out]:

E A,    100 /   2   =   50.0
A G,    50.0    /   3   =   16.666666666666668
G D,    16.666666666666668  /   50  =   0.33333333333333337
D B,    0.33333333333333337 /   1   =   0.33333333333333337
B C,    0.33333333333333337 /   2   =   0.16666666666666669
C F,    0.16666666666666669 /   3   =   0.05555555555555556
F E,    0.05555555555555556 /   2   =   0.02777777777777778
----------------------------------------------------------------------------
E A,    100 /   2   =   50.0
A B,    50.0    /   50  =   1.0
B C,    1.0 /   2   =   0.5
C F,    0.5 /   3   =   0.16666666666666666
F E,    0.16666666666666666 /   2   =   0.08333333333333333
----------------------------------------------------------------------------
B A,    100 /   50  =   2.0
A G,    2.0 /   3   =   0.6666666666666666
G D,    0.6666666666666666  /   50  =   0.013333333333333332
D B,    0.013333333333333332    /   1   =   0.013333333333333332
----------------------------------------------------------------------------
B A,    100 /   50  =   2.0
A B,    2.0 /   50  =   0.04
----------------------------------------------------------------------------

And also the cycle that maximizes n:

print(f"The loop {mcyc} in G maximizes n with {maxn}")

[out]:
The loop ['E', 'A', 'B', 'C', 'F', 'E'] in G maximizes n with 0.08333333333333333

I've uploaded this notebook here in case you prefer to clone it and test this approach locally.

Non-Iterative Alternative - Based on nx.dijkstra_path

As per your comment, it seems like what you're after is not to traverse every path in the network, but if possible traverse only the n-maximizing path and calculate nmax.

This is just a variant of Dijkstra Shortest Path algorithm with an updating weight function.

However, nx.Dijkstra_path has problems finding cycles. nx.Dijkstra_path('A','A') yields A even if w(A,A)=math.inf

But one can use a trick:

1. remove node A and introduce Ai and Ao
2. replace A in all its in-edges with Ai
3. replace A in all its out-edges with Ao
4. link Ai and Ao with an edge weight equal to math.inf.
5. calculate shortest path from Ao to Ai

Here's what the graph I provided in my initial answer would look like:

import math
Gp = nx.DiGraph()

Gp.add_edges_from([('Ao', 'B'),('B', 'Ai'), ('C','D'),('G','D')], weight=50)
Gp.add_edges_from([('D','B')], weight=1)
Gp.add_edges_from([('B','C'),('F','E'),('E','Ai')], weight=2)
Gp.add_edges_from([('C','F'),('B','G'),('Ao','G')], weight=3)

# this edge is strictly speaking not needed.
Gp.add_edges_from([('Ai','Ao'),('Ao','Ai')], weight=math.inf)

edge_labels=dict([((u,v,),d['weight'])
                 for u,v,d in Gp.edges(data=True)])

pos=nx.drawing.layout.circular_layout(Gp,scale=10)
pos['Ai'], pos['B'] = pos['B'], pos['Ai']
nx.draw_networkx_edge_labels(Gp,pos,edge_labels=edge_labels)
nx.draw_networkx_labels(Gp,pos,font_color='white')
nx.draw(Gp,pos, node_size=1500,edge_cmap=plt.cm.Reds)

Using nx.dijkstra_path we have:

nmpath = nx.dijkstra_path(Gp,'Ao','Ai')
nmpath

[out]: ['Ao', 'B', 'C', 'F', 'E', 'Ai']

We can see the weights of that path as follows:

lennmpath=len(nmpath)
datnmpath=[Gp.get_edge_data(u, nmpath[i+1])['weight'] for u,i in zip(nmpath[:-1], range(lnmpath))]

[out]: [50, 2, 3, 2, 2]

And we can calculate nmax trivially:

maxn=100; 
for w in datnmpath: maxn/=w
print(maxn)

[out] 0.08333333333333333

I've updated the notebook so you can clone it and test this approach locally.