Question : Is there a simple starter example of Periodic VRP with a month of planning period ?

In the documentation, I noticed that just setting the frequency of visits for each node is given as an option. How do I set the duration ( planning period ) and define periodicity in a better way ?

Also - how would I get the routes for each day in the planning period from the solved problem.

I have a central depot where nodes ( customers ) are to be visited a certain number of times every month.

G.nodes[1]["frequency"] = 2
prob.periodic = True

Apologies if the question is very basic. I am new to vehicle routing problems. Any help would be appreciated !

[sumi1234]

Thanks again ! I was able to run it on dummy data using a 10 day planning period and I have started using it for my problem.

Code snippet is provided for future reference below [ Hope I have understood the changes correctly ]

DISTANCES = [
    [0,548,776,696,582,274,502,194,308,194,536,502,388,354,468,776,662,0], # from Source
    [0,0,684,308,194,502,730,354,696,742,1084,594,480,674,1016,868,1210,548],
    [0,684,0,992,878,502,274,810,468,742,400,1278,1164,1130,788,1552,754,776],
    [0,308,992,0,114,650,878,502,844,890,1232,514,628,822,1164,560,1358,696],
    [0,194,878,114,0,536,764,388,730,776,1118,400,514,708,1050,674,1244,582],
    [0,502,502,650,536,0,228,308,194,240,582,776,662,628,514,1050,708,274],
    [0,730,274,878,764,228,0,536,194,468,354,1004,890,856,514,1278,480,502],
    [0,354,810,502,388,308,536,0,342,388,730,468,354,320,662,742,856,194],
    [0,696,468,844,730,194,194,342,0,274,388,810,696,662,320,1084,514,308],
    [0,742,742,890,776,240,468,388,274,0,342,536,422,388,274,810,468,194],
    [0,1084,400,1232,1118,582,354,730,388,342,0,878,764,730,388,1152,354,536],
    [0,594,1278,514,400,776,1004,468,810,536,878,0,114,308,650,274,844,502],
    [0,480,1164,628,514,662,890,354,696,422,764,114,0,194,536,388,730,388],
    [0,674,1130,822,708,628,856,320,662,388,730,308,194,0,342,422,536,354],
    [0,1016,788,1164,1050,514,514,662,320,274,388,650,536,342,0,764,194,468],
    [0,868,1552,560,674,1050,1278,742,1084,810,1152,274,388,422,764,0,798,776],
    [0,1210,754,1358,1244,708,480,856,514,468,354,844,730,536,194,798,0,662],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], # from Sink
    ]
    
# Demands (key: node, value: amount)
DEMAND = {1: 1, 2: 1, 3: 2, 4: 4, 5: 2, 6: 4, 7: 8, 8: 8, 9: 1, 10: 2, 11: 1, 12: 2, 13: 4, 14: 4, 15: 8, 16: 8}

# Frequencies (key: node, value: frequency in planning period)
FREQ = {1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 4, 7: 5, 8: 3, 9: 1, 10: 8, 11: 4, 12: 2, 13: 2, 14: 2, 15: 2, 16: 2}

# The matrix is transformed into a DiGraph
A = matrix(DISTANCES, dtype=[("cost", int)])
G = from_numpy_matrix(A, create_using=nx.DiGraph())

# The demands are stored as node attributes
set_node_attributes(G, values=DEMAND, name="demand")

##Setting the frequency for each node
set_node_attributes(G, values=FREQ, name="frequency") 

# The depot is relabeled as Source and Sink
G = relabel_nodes(G, {0: "Source", 17: "Sink"})

prob = VehicleRoutingProblem(G, load_capacity=15)

# Setting planning horizon to 10 days 
prob.periodic = 10
    
prob.solve()
    
#Note : Some days might have zero routes
for route_num, route_keys in sorted(prob.schedule.items()):
    print("\nRoute for Day " + str(route_num), " : ")
    i = 1
    for k in route_keys:
        print("\tRoute ",i," : ",prob.best_routes[k])
        i=i+1
   
print("Total Cost of set of optimum Route : ", prob.best_value)

[Kuifje02]

Yes exactly, the fixed cost is added on a per route basis, so it is charged for every route, every day of the planning period. It could be the driver’s salary for example.