# The factory schedule example

This is a Julia translation of part 5 from "Introduction to to Linear Programming with Python" available at https://github.com/benalexkeen/Introduction-to-linear-programming

For 2 factories (A, B), minimize the cost of production over the course of 12 months while meeting monthly demand. Factory B has a planned outage during month 5.

It was originally contributed by @Crghilardi.

using JuMP
import GLPK
import Test

function example_factory_schedule()
# Sets in the problem:
months, factories = 1:12, [:A, :B]
# This function takes a matrix and converts it to a JuMP container so we can
# refer to elements such as d_max_cap[1, :A].
containerize(A::Matrix) = Containers.DenseAxisArray(A, months, factories)
# Maximum production capacity in (month, factory) [units/month]:
d_max_cap = containerize([
100000	50000;
110000	55000;
120000	60000;
145000	100000;
160000	0;
140000	70000;
155000	60000;
200000	100000;
210000	100000;
197000	100000;
80000	120000;
150000	150000;
])
# Minimum production capacity in (month, factory) [units/month]:
d_min_cap = containerize([
20000	20000;
20000	20000;
20000	20000;
20000	20000;
20000	0;
20000	20000;
20000	20000;
20000	20000;
20000	20000;
20000	20000;
20000	20000;
20000	20000;
])
# Variable cost of production in (month, factory) [$/unit]: d_var_cost = containerize([ 10 5; 11 4; 12 3; 9 5; 8 0; 8 6; 5 4; 7 6; 9 8; 10 11; 8 10; 8 12 ]) # Fixed cost of production in (month, factory) # [$/month]:
d_fixed_cost = containerize([
500	600;
500	600;
500	600;
500	600;
500	0;
500	600;
500	600;
500	600;
500	600;
500	600;
500	600;
500	600
])
# Demand in each month [units/month]:
d_demand = [
120_000,
100_000,
130_000,
130_000,
140_000,
130_000,
150_000,
170_000,
200_000,
190_000,
140_000,
100_000,
]
# The model!
model = Model(GLPK.Optimizer)
# Decision variables
@variables(model, begin
status[m in months, f in factories], Bin
production[m in months, f in factories], Int
end)
# The production cannot be less than minimum capacity.
@constraint(
model,
[m in months, f in factories],
production[m, f] >= d_min_cap[m, f] * status[m, f],
)
# The production cannot be more tha maximum capacity.
@constraint(
model,
[m in months, f in factories],
production[m, f] <= d_max_cap[m, f] * status[m, f],
)
# The production must equal demand in a given month.
@constraint(model, [m in months], sum(production[m, :]) == d_demand[m])
# Factory B is shut down during month 5, so production and status are both
# zero.
fix(status[5, :B], 0.0)
fix(production[5, :B], 0.0)
# The objective is to minimize the cost of production across all time
##periods.
@objective(
model,
Min,
sum(
d_fixed_cost[m, f] * status[m, f] + d_var_cost[m, f] * production[m, f]
for m in months, f in factories
)
)
# Optimize the problem
optimize!(model)
# Check the solution!
Test.@testset "Check the solution against known optimal" begin
Test.@test termination_status(model) == MOI.OPTIMAL
Test.@test objective_value(model) == 12_906_400.0
Test.@test value.(production)[1, :A] == 70_000
Test.@test value.(status)[1, :A] == 1
Test.@test value.(status)[5, :B] == 0
Test.@test value.(production)[5, :B] == 0
end
println("The production schedule is:")
println(value.(production))
return
end

example_factory_schedule()
Test Summary:                            | Pass  Total
Check the solution against known optimal |    6      6
The production schedule is:
2-dimensional DenseAxisArray{Float64,2,...} with index sets:
Dimension 1, 1:12
Dimension 2, Symbol[:A, :B]
And data, a 12×2 Array{Float64,2}:
70000.0   50000.0
45000.0   55000.0
70000.0   60000.0
30000.0  100000.0
140000.0       0.0
60000.0   70000.0
90000.0   60000.0
70000.0  100000.0
100000.0  100000.0
190000.0       0.0
80000.0   60000.0
100000.0       0.0

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