Question on Pyomo Transport/Shipping issue - pyomo

Question

I am learning pyomo and practicing an exercise about shipping problem with conditions below:
The shipment can either directly deliver from source cities to the destination or through a depot (source city -> depot -> destination)
The way of shipping will be selected based on least cost consumed
Every Depot has its own capacity limitation
I tried to use pyomo to solve this question, however, no solution can be figured out.
Please give me some advice and direction why the codes didnt work. Thank you!
import pyomo.environ as pyomo
import numpy as np
supply = dict({'Liverpool': 150000,
'Brighton': 200000})
through = dict({'Newcastle': 70000,
'Birmingham': 50000,
'London': 100000,
'Exeter': 40000,
'Liverpool': 150000,
'Brighton': 200000
})
demand = dict({'C1': 50000,
'C2': 10000,
'C3': 40000,
'C4': 35000,
'C5': 60000,
'C6': 20000})
cost = dict({
('Liverpool', 'Newcastle'): 0.5,
('Liverpool', 'Birmingham'): 0.5,
('Liverpool', 'London'): 1.0,
('Liverpool', 'Exeter'): 0.2,
('Liverpool', 'C1'): 1.0,
('Liverpool', 'C3'): 1.5,
('Liverpool', 'C4'): 2.0,
('Liverpool', 'C6'): 1.0,
('Brighton', 'Birmingham'): 0.3,
('Brighton', 'London'): 0.5,
('Brighton', 'Exeter'): 0.2,
('Brighton', 'C1'): 2.0,
('Newcastle', 'C2'): 1.5,
('Newcastle', 'C3'): 0.5,
('Newcastle', 'C5'): 1.5,
('Newcastle', 'C6'): 1.0,
('Birmingham', 'C1'): 1.0,
('Birmingham', 'C2'): 0.5,
('Birmingham', 'C3'): 0.5,
('Birmingham', 'C4'): 1.0,
('Birmingham', 'C5'): 0.5,
('London', 'C2'): 1.5,
('London', 'C3'): 2.0,
('London', 'C5'): 0.5,
('London', 'C6'): 1.5,
('Exeter', 'C3'): 0.2,
('Exeter', 'C4'): 1.5,
('Exeter', 'C5'): 0.5,
('Exeter', 'C6'): 1.5
})
cost_total = {}
for city_from in supply.keys():
for city_through in through.keys():
for city_to in demand.keys():
if city_from == city_through:
cost_total[(city_from , city_through , city_to)] = 0 + cost.get((city_through , city_to) , 9999)
else:
cost_total[(city_from , city_through , city_to)] = cost.get((city_from , city_through) , 9999) + cost.get((city_through , city_to) , 9999)
supplier = supply.keys()
througher = through.keys()
demander = demand.keys()
model = pyomo.ConcreteModel()
model.i = pyomo.Set(initialize = supplier , doc = 'City From')
model.j = pyomo.Set(initialize = througher , doc = 'City Through')
model.k = pyomo.Set(initialize = demander , doc = 'City To')
model.s = pyomo.Param(model.i , initialize=supply, doc='Supply by City')
model.t = pyomo.Param(model.j , initialize = through , doc = 'through / warehouse')
model.d = pyomo.Param(model.k , initialize=demand , doc='Demand by City')
model.cost = pyomo.Param(model.i , model.j , model.k , initialize=cost_total , doc = 'total cost')
model.x = pyomo.Var(model.i , model.j , model.k , bounds = (0 , None))
def supply_rule(model, i):
return sum(model.x[i,j,k] for j in model.j for k in model.k) <= model.s[i]
model.supply = pyomo.Constraint(model.i, rule=supply_rule, doc='Observe supply limit at plant i')
def demand_rule(model, k):
return sum(model.x[i,j,k] for i in model.i for j in model.j) == model.d[k]
model.demand = pyomo.Constraint(model.k, rule=demand_rule, doc='Satisfy demand at market j')
def depot_cont(model , j):
return sum(model.x[i , j , k] for i in model.i for k in model.k) <= model.t[j]
model.through_cont2 = pyomo.Constraint(model.j , rule = depot_cont)
def objective_rule(model):
return sum(model.cost[i,j,k]*model.x[i,j,k] for i in model.i for j in model.j for k in model.k)
model.objective = pyomo.Objective(rule=objective_rule, sense=pyomo.minimize, doc='Define objective function')

Answer 1

Your code (unmodified) appears to be working for me.
Realize there are a couple ways to set this up, which is a larger discussion... In your case you enumerated all possible routes, and snuck in a zero-cost option for the direct routes and added high cost for infeasible routes, which is fine. On a larger problem, or more complicated network, the standard way to approach this is as a network flow problem with flow-balance constraints at all the nodes (cities).
Anyhow, your question isn't too clear on what "isn't working". I added this to the end of your code:
solver = pyomo.SolverFactory('glpk')
result = solver.solve(model)
print(result)
for ijk in model.x.index_set():
if model.x[ijk]:
print(f'ship : {ijk} qty: {model.x[ijk].value}')
and it produced this result, which passes 1st level sanity check, but I didn't look at it too closely...
Problem:
- Name: unknown
Lower bound: 198500.0
Upper bound: 198500.0
Number of objectives: 1
Number of constraints: 15
Number of variables: 73
Number of nonzeros: 217
Sense: minimize
Solver:
- Status: ok
Termination condition: optimal
Statistics:
Branch and bound:
Number of bounded subproblems: 0
Number of created subproblems: 0
Error rc: 0
Time: 0.007380008697509766
Solution:
- number of solutions: 0
number of solutions displayed: 0
ship : ('Liverpool', 'Liverpool', 'C1') qty: 50000.0
ship : ('Liverpool', 'Liverpool', 'C6') qty: 20000.0
ship : ('Brighton', 'Birmingham', 'C2') qty: 10000.0
ship : ('Brighton', 'Birmingham', 'C4') qty: 35000.0
ship : ('Brighton', 'Birmingham', 'C5') qty: 5000.0
ship : ('Brighton', 'London', 'C5') qty: 55000.0
ship : ('Brighton', 'Exeter', 'C3') qty: 40000.0
[Finished in 579ms]