"""
The American Steel Problem for the PuLP Modeller

Authors: Antony Phillips, Dr Stuart Mitchell  2007
"""

# Import PuLP modeller functions
from pulp import *

# List of all the nodes
Nodes = [
    "Youngstown",
    "Pittsburgh",
    "Cincinatti",
    "Kansas City",
    "Chicago",
    "Albany",
    "Houston",
    "Tempe",
    "Gary",
]

nodeData = {  # NODE        Supply Demand
    "Youngstown": [10000, 0],
    "Pittsburgh": [15000, 0],
    "Cincinatti": [0, 0],
    "Kansas City": [0, 0],
    "Chicago": [0, 0],
    "Albany": [0, 3000],
    "Houston": [0, 7000],
    "Tempe": [0, 4000],
    "Gary": [0, 6000],
}

# List of all the arcs
Arcs = [
    ("Youngstown", "Albany"),
    ("Youngstown", "Cincinatti"),
    ("Youngstown", "Kansas City"),
    ("Youngstown", "Chicago"),
    ("Pittsburgh", "Cincinatti"),
    ("Pittsburgh", "Kansas City"),
    ("Pittsburgh", "Chicago"),
    ("Pittsburgh", "Gary"),
    ("Cincinatti", "Albany"),
    ("Cincinatti", "Houston"),
    ("Kansas City", "Houston"),
    ("Kansas City", "Tempe"),
    ("Chicago", "Tempe"),
    ("Chicago", "Gary"),
]

arcData = {  #      ARC                Cost Min Max
    ("Youngstown", "Albany"): [0.5, 0, 1000],
    ("Youngstown", "Cincinatti"): [0.35, 0, 3000],
    ("Youngstown", "Kansas City"): [0.45, 1000, 5000],
    ("Youngstown", "Chicago"): [0.375, 0, 5000],
    ("Pittsburgh", "Cincinatti"): [0.35, 0, 2000],
    ("Pittsburgh", "Kansas City"): [0.45, 2000, 3000],
    ("Pittsburgh", "Chicago"): [0.4, 0, 4000],
    ("Pittsburgh", "Gary"): [0.45, 0, 2000],
    ("Cincinatti", "Albany"): [0.35, 1000, 5000],
    ("Cincinatti", "Houston"): [0.55, 0, 6000],
    ("Kansas City", "Houston"): [0.375, 0, 4000],
    ("Kansas City", "Tempe"): [0.65, 0, 4000],
    ("Chicago", "Tempe"): [0.6, 0, 2000],
    ("Chicago", "Gary"): [0.12, 0, 4000],
}

# Splits the dictionaries to be more understandable
(supply, demand) = splitDict(nodeData)
(costs, mins, maxs) = splitDict(arcData)

# Creates the boundless Variables as Integers
vars = LpVariable.dicts("Route", Arcs, None, None, LpInteger)

# Creates the upper and lower bounds on the variables
for a in Arcs:
    vars[a].bounds(mins[a], maxs[a])

# Creates the 'prob' variable to contain the problem data
prob = LpProblem("American Steel Problem", LpMinimize)

# Creates the objective function
prob += lpSum([vars[a] * costs[a] for a in Arcs]), "Total Cost of Transport"

# Creates all problem constraints - this ensures the amount going into each node is at least equal to the amount leaving
for n in Nodes:
    prob += (
        supply[n] + lpSum([vars[(i, j)] for (i, j) in Arcs if j == n])
        >= demand[n] + lpSum([vars[(i, j)] for (i, j) in Arcs if i == n])
    ), f"Steel Flow Conservation in Node {n}"

# The problem data is written to an .lp file
prob.writeLP("AmericanSteelProblem.lp")

# The problem is solved using PuLP's choice of Solver
prob.solve()

# The status of the solution is printed to the screen
print("Status:", LpStatus[prob.status])

# Each of the variables is printed with it's resolved optimum value
for v in prob.variables():
    print(v.name, "=", v.varValue)

# The optimised objective function value is printed to the screen
print("Total Cost of Transportation = ", value(prob.objective))