Lumber Transportation Problem

10. Reeb and S. Leavengood

Millco has three wood mills and is planning three new logging sites. Each mill has a maximum capacity and each logging site can harvest a certain number of truckloads of lumber per day. The cost of a haul is $2/mile of distance. If distances from logging sites to mills are given below, how should the hauls be routed to minimize hauling costs while meeting all demands?

Logging Site	Mill A	Mill B	Mill C	Max loads per day
1	8	15	50	20
2	10	17	20	30
3	30	26	15	45
Mill demand	30	35	30

from gamspy import Container, Set, Alias, Parameter, Variable, Equation, Model, Sum, Sense
import numpy as np

m = Container()

# Model sets and parameters
sites = Set(container=m, name="sites", records=["1", "2", "3"])
mills = Set(container=m, name="mills", records=["Mill A", "Mill B", "Mill C"])
dist = Parameter(
    container=m,
    name="dist",
    domain=[sites, mills],
    records=np.array([[8, 15, 50], [10, 17, 20], [30, 26, 15]]),
)
supply = Parameter(
    container=m, name="supply", domain=sites, records=np.array([20, 30, 45])
)
demand = Parameter(
    container=m, name="demand", domain=mills, records=np.array([30, 35, 30])
)
cost_per_haul = 4

# Model variables
ship = Variable(container=m, name="ship", type="positive", domain=[sites, mills])

defcost = cost_per_haul * Sum([sites, mills], ship[sites, mills] * dist[sites, mills])

defsupply = Equation(container=m, name="defsupply", domain=sites)
defsupply[sites] = Sum(mills, ship[sites, mills]) == supply[sites]

defdemand = Equation(container=m, name="defdemand", domain=mills)
defdemand[mills] = Sum(sites, ship[sites, mills]) == demand[mills]

millco = Model(
    container=m,
    name="millco",
    equations=m.getEquations(),
    problem="LP",
    sense=Sense.MIN,
    objective=defcost,
)

millco.solve()

	Solver Status	Model Status	Objective	Num of Equations	Num of Variables	Model Type	Solver	Solver Time
0	Normal	OptimalGlobal	5760	7	10	LP	CPLEX	0

ship.records.pivot(index='sites', columns='mills', values='level')

mills	Mill A	Mill B	Mill C
sites
1	0.0	20.0	0.0
2	30.0	0.0	0.0
3	0.0	15.0	30.0

print(f'Total cost will be {millco.objective_value}')

Total cost will be 5760.0

m = Container()

# Generic Model sets and parameters
i = Set(
    container=m,
    name="i",
    records=["i" + str(i) for i in range(6)],
    description="equation index",
)
j = Set(
    container=m,
    name="j",
    records=["j" + str(j) for j in range(9)],
    description="variable index",
)
A = Parameter(
    container=m,
    name="A",
    domain=[i, j],
    records=np.array(
        [
            [1, 1, 1, 0, 0, 0, 0, 0, 0],
            [0, 0, 0, 1, 1, 1, 0, 0, 0],
            [0, 0, 0, 0, 0, 0, 1, 1, 1],
            [1, 0, 0, 1, 0, 0, 1, 0, 0],
            [0, 1, 0, 0, 1, 0, 0, 1, 0],
            [0, 0, 1, 0, 0, 1, 0, 0, 1],
        ]
    ),
)
b = Parameter(
    container=m, name="b", domain=i, records=np.array([20, 30, 45, 30, 35, 30])
)
c = Parameter(
    container=m,
    name="c",
    domain=j,
    records=np.array([8, 15, 50, 10, 17, 20, 30, 26, 15]),
)
x = Variable(container=m, name="x", type="positive", domain=j)

obj = 4 * Sum(j, c[j] * x[j])

e = Equation(container=m, name="e", domain=i)
e[i] = Sum(j, A[i, j] * x[j]) == b[i]

generic_model = Model(
    container=m,
    name="generic_model",
    equations=m.getEquations(),
    problem="LP",
    sense=Sense.MIN,
    objective=obj,
)

generic_model.solve()

	Solver Status	Model Status	Objective	Num of Equations	Num of Variables	Model Type	Solver	Solver Time
0	Normal	OptimalGlobal	5760	7	10	LP	CPLEX	0

x.records

	j	level	marginal	lower	upper	scale
0	j0	0.0	-0.0	0.0	inf	1.0
1	j1	20.0	0.0	0.0	inf	1.0
2	j2	0.0	184.0	0.0	inf	1.0
3	j3	30.0	0.0	0.0	inf	1.0
4	j4	0.0	0.0	0.0	inf	1.0
5	j5	0.0	56.0	0.0	inf	1.0
6	j6	0.0	44.0	0.0	inf	1.0
7	j7	15.0	0.0	0.0	inf	1.0
8	j8	30.0	0.0	0.0	inf	1.0

print(f'Total cost will be {generic_model.objective_value}')

Total cost will be 5760.0