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Implement multi-period heuristic

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This commit is contained in:
Alinson S. Xavier
2020-10-02 11:25:40 -05:00
parent fe9cacef24
commit a6846be9eb
7 changed files with 194 additions and 30 deletions
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50
test/instance_test.jl
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@@ -11,7 +11,6 @@ using RELOG
centers = instance.collection_centers
plants = instance.plants
products = instance.products
location_name_to_plant = Dict(p.location_name => p for p in plants)
product_name_to_product = Dict(p.name => p for p in products)
@@ -75,5 +74,54 @@ using RELOG
@testset "validate timeseries" begin
@test_throws String RELOG.parsefile("fixtures/s1-wrong-length.json")
end
@testset "compress" begin
basedir = dirname(@__FILE__)
instance = RELOG.parsefile("$basedir/../instances/s1.json")
compressed = RELOG._compress(instance)
product_name_to_product = Dict(p.name => p for p in compressed.products)
location_name_to_facility = Dict()
for p in compressed.plants
location_name_to_facility[p.location_name] = p
end
for c in compressed.collection_centers
location_name_to_facility[c.name] = c
end
p1 = product_name_to_product["P1"]
p2 = product_name_to_product["P2"]
p3 = product_name_to_product["P3"]
c1 = location_name_to_facility["C1"]
l1 = location_name_to_facility["L1"]
@test compressed.time == 1
@test compressed.building_period == [1]
@test p1.name == "P1"
@test p1.transportation_cost ≈ [0.015]
@test p1.transportation_energy ≈ [0.115]
@test p1.transportation_emissions["CO2"] ≈ [0.051]
@test p1.transportation_emissions["CH4"] ≈ [0.0025]
@test c1.name == "C1"
@test c1.amount ≈ [1869.12]
@test l1.plant_name == "F1"
@test l1.location_name == "L1"
@test l1.energy ≈ [0.115]
@test l1.emissions["CO2"] ≈ [0.051]
@test l1.emissions["CH4"] ≈ [0.0025]
@test l1.sizes[1].opening_cost ≈ [500]
@test l1.sizes[2].opening_cost ≈ [1250]
@test l1.sizes[1].fixed_operating_cost ≈ [60]
@test l1.sizes[2].fixed_operating_cost ≈ [60]
@test l1.sizes[1].variable_operating_cost ≈ [30]
@test l1.sizes[2].variable_operating_cost ≈ [30]
@test l1.disposal_limit[p2] ≈ [2.0]
@test l1.disposal_limit[p3] ≈ [2.0]
@test l1.disposal_cost[p2] ≈ [-10.0]
@test l1.disposal_cost[p3] ≈ [-10.0]
end
end