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CIVIL-557
Decision Aid Methodologies
In Transportation
Transport and Mobility Laboratory (TRANSP-OR)
École Polytechnique Fédérale de Lausanne EPFL
Virginie Lurkin
Lab VI:
Case study – Freight Trains Management
Solution of the previous lab
F1
Customer (service costs 𝒄𝒊𝒋) Fixed cost (𝒇𝒊)
Facility 1 2 3
1 1000 20 30 200
2 1000 30 40 200
3 1000 160 150 200
4 50 140 120 200
Which formulation is better?
F2
UFLP problem - solution
max 30𝑥1 + 17𝑥2 + 14𝑥3 + 11𝑥4 + 9𝑥5
𝑠. 𝑡. 29𝑥1 + 20𝑥2 + 16𝑥3 + 12𝑥4 + 10𝑥5 ≤ 37
𝑥𝑖 ∈ 0,1 , ∀𝑖 = 1,…5.
IP(1)
Knapsack problem – example of solution
𝒙𝟏∗ 𝒙𝟐
∗ 𝒙𝟑∗ 𝒙𝟒
∗ 𝒙𝟓∗ 𝒛∗ CT
0 1 1 0 0 31 12s
max 30𝑥1 + 17𝑥2 + 14𝑥3 + 11𝑥4 + 9𝑥5
𝑠. 𝑡. 29𝑥1 + 20𝑥2 + 16𝑥3 + 12𝑥4 + 10𝑥5 ≤ 37
𝑥𝑖 ≥ 0, ∀𝑖 = 1,…5.
LP(1)
Knapsack problem - example of solution
𝒙𝟏∗ 𝒙𝟐
∗ 𝒙𝟑∗ 𝒙𝟒
∗ 𝒙𝟓∗ 𝒛∗ CT
1.2759 0 0 0 0 38.28 8s
max 30𝑥1 + 17𝑥2 + 14𝑥3 + 11𝑥4 + 9𝑥5
𝑠. 𝑡. 29𝑥1 + 20𝑥2 + 16𝑥3 + 12𝑥4 + 10𝑥5 ≤ 37
𝑥𝑖 ≥ 0, ∀𝑖 = 1,…5.
LP(1)
Knapsack problem - example of solution
𝒙𝟏∗ 𝒙𝟐
∗ 𝒙𝟑∗ 𝒙𝟒
∗ 𝒙𝟓∗ 𝒛∗ CT
1.2759 0 0 0 0 38.28 8s
𝒙𝟏 ≤ 𝟏 is the Gomory cut you obtained
max 30𝑥1 + 17𝑥2 + 14𝑥3 + 11𝑥4 + 9𝑥5
𝑠. 𝑡. 29𝑥1 + 20𝑥2 + 16𝑥3 + 12𝑥4 + 10𝑥5 ≤ 37
𝑥𝑖 ≥ 0, ∀𝑖 = 1,…5.
LP(2)
Knapsack problem - example of solution
𝒙𝟏∗ 𝒙𝟐
∗ 𝒙𝟑∗ 𝒙𝟒
∗ 𝒙𝟓∗ 𝒛∗ CT
1 0 0 0.667 0 37.33 7s
𝒙𝟏 ≤ 𝟏
max 30𝑥1 + 17𝑥2 + 14𝑥3 + 11𝑥4 + 9𝑥5
𝑠. 𝑡. 29𝑥1 + 20𝑥2 + 16𝑥3 + 12𝑥4 + 10𝑥5 ≤ 37
𝑥𝑖 ≥ 0, ∀𝑖 = 1,…5.
LP(2)
Knapsack problem - example of solution
𝒙𝟏∗ 𝒙𝟐
∗ 𝒙𝟑∗ 𝒙𝟒
∗ 𝒙𝟓∗ 𝒛∗ CT
1 0 0 0.667 0 37.33 7s
𝒙𝟏 + 𝒙𝟒 ≤ 𝟏 is a cover cut
𝒙𝟏 ≤ 𝟏
max 30𝑥1 + 17𝑥2 + 14𝑥3 + 11𝑥4 + 9𝑥5
𝑠. 𝑡. 29𝑥1 + 20𝑥2 + 16𝑥3 + 12𝑥4 + 10𝑥5 ≤ 37
𝑥𝑖 ≥ 0, ∀𝑖 = 1,…5.
LP(3)
Knapsack problem - solution
𝒙𝟏∗ 𝒙𝟐
∗ 𝒙𝟑∗ 𝒙𝟒
∗ 𝒙𝟓∗ 𝒛∗ CT
1 0 0 0 0.8 37.2 7s
𝒙𝟏 + 𝒙𝟒 ≤ 𝟏
𝒙𝟏 ≤ 𝟏
max 30𝑥1 + 17𝑥2 + 14𝑥3 + 11𝑥4 + 9𝑥5
𝑠. 𝑡. 29𝑥1 + 20𝑥2 + 16𝑥3 + 12𝑥4 + 10𝑥5 ≤ 37
𝑥𝑖 ≥ 0, ∀𝑖 = 1,…5.
LP(4)
Knapsack problem - example of solution
𝒙𝟏∗ 𝒙𝟐
∗ 𝒙𝟑∗ 𝒙𝟒
∗ 𝒙𝟓∗ 𝒛∗ CT
0.724 0.276 0.276 0.276 0.276 35.79 7s
𝒙𝟏 + 𝒙𝟓 ≤ 𝟏
𝒙𝟏 + 𝒙𝟒 ≤ 𝟏
𝒙𝟏 ≤ 𝟏
𝒙𝟏 + 𝒙𝟑 ≤ 𝟏
𝒙𝟏 + 𝒙𝟐 ≤ 𝟏
max 30𝑥1 + 17𝑥2 + 14𝑥3 + 11𝑥4 + 9𝑥5
𝑠. 𝑡. 29𝑥1 + 20𝑥2 + 16𝑥3 + 12𝑥4 + 10𝑥5 ≤ 37
𝑥𝑖 ≥ 0, ∀𝑖 = 1,…5.
LP(4)
Knapsack problem - example of solution
𝒙𝟏∗ 𝒙𝟐
∗ 𝒙𝟑∗ 𝒙𝟒
∗ 𝒙𝟓∗ 𝒛∗ CT
0.724 0.276 0.276 0.276 0.276 35.79 7s
𝒙𝟏 + 𝒙𝟓 ≤ 𝟏
𝒙𝟏 + 𝒙𝟒 ≤ 𝟏
𝒙𝟏 ≤ 𝟏
𝒙𝟏 + 𝒙𝟑 ≤ 𝟏
𝒙𝟏 + 𝒙𝟐 ≤ 𝟏
Gomory cut on the optimal tableau of LP(4) gives me:1.24137931*x1 + 1.068965517*x2 + 0.413793103*x3 + 0.413793103*x4 <=1;
max 30𝑥1 + 17𝑥2 + 14𝑥3 + 11𝑥4 + 9𝑥5
𝑠. 𝑡. 29𝑥1 + 20𝑥2 + 16𝑥3 + 12𝑥4 + 10𝑥5 ≤ 37
𝑥𝑖 ≥ 0, ∀𝑖 = 1,…5.
LP(5)
Knapsack problem - example of solution
𝒙𝟏∗ 𝒙𝟐
∗ 𝒙𝟑∗ 𝒙𝟒
∗ 𝒙𝟓∗ 𝒛∗ CT
0.252 0.063 0.747 0.747 0.747 34.07 7s
𝒙𝟏 + 𝒙𝟓 ≤ 𝟏
𝒙𝟏 + 𝒙𝟒 ≤ 𝟏
𝒙𝟏 ≤ 𝟏
𝒙𝟏 + 𝒙𝟑 ≤ 𝟏
𝒙𝟏 + 𝒙𝟐 ≤ 𝟏
1.24137931*x1 + 1.068965517*x2 + 0.413793103*x3 + 0.413793103*x4 <=1;
Lab 6
The Locomotive Fleet Fueling Problem
Overview
The locomotive fleet fueling problem
Problem description
MILP formulation
Simple example (2010 RAS Competition)
Assignment #6
Larger instance
Valid inequalities
Evalution of three models
2010 RAS Competition
Problem description
Problem description
Yards
o Tanker truck contracting cost
o Amount of dispensable fuel per tanker truck
o Fuel cost per gallon
Locomotives
o Schedule – sequence of stops
o Delay cost per stop
Problem description
Sets
o 𝐼: set of locomotives (1 locomotive = 1 train)
o 𝐽: set of stops
o 𝐾: set of yards
o 𝑇: set of days
Parameters
o 𝐷𝐶𝑖𝑗: delay cost ($) for locomotive 𝑖 and stop 𝑗
o 𝐶𝐶: contracting cost ($/truck) for a truck for the entire planning horizon,
same for all yards
o 𝑃𝑘: fuel cost ($/gallon) at yard 𝑘
Problem description
Parameters
o 𝑇𝑇: Tanker truck capacity (in gallons) per day, same for all trucks
o 𝐿𝑇: Capacity (in gallons) of fuel tank of a locomotive, same for all
locomotives
o 𝐹𝐶𝑖𝑗: fuel consumption (in gallons) for a locomotive 𝑖 to go for stop 𝑗 to
stop 𝑗 + 1
o 𝑁𝑖: set of stops for locomotive 𝑖 in the planning horizon
o 𝑌𝑎𝑟𝑑𝑖𝑗: Yard used if locomotive 𝑖 refuels at stop 𝑗
o 𝐷𝑎𝑦𝑖𝑗: Day in which locomotive 𝑖 is at stop 𝑗
Problem description
Decision variables
o 𝑥𝑖𝑗 = ቊ1 𝑖𝑓 𝑙𝑜𝑐𝑜𝑚𝑜𝑡𝑖𝑣𝑒 𝑖 𝑟𝑒𝑓𝑢𝑒𝑙𝑠 𝑎𝑡 𝑖𝑡𝑠 𝑠𝑡𝑜𝑝 𝑗0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
o 𝑦𝑘 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑢𝑒𝑙 𝑡𝑎𝑛𝑘𝑠 𝑡𝑜 𝑏𝑒 𝑐𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑒𝑑 𝑎𝑡 𝑦𝑎𝑟𝑑 𝑘
o 𝑓𝑖𝑗 = 𝑔𝑎𝑙𝑙𝑜𝑛𝑠 𝑜𝑓 𝑓𝑢𝑒𝑙 𝑎𝑞𝑢𝑖𝑟𝑒𝑑 𝑏𝑦 𝑙𝑜𝑐𝑜𝑚𝑜𝑡𝑖𝑣𝑒 𝑖 𝑎𝑡 𝑖𝑡𝑠 𝑠𝑡𝑜𝑝 𝑗
o 𝑣𝑖𝑗 = 𝑔𝑎𝑙𝑙𝑜𝑛𝑠 𝑜𝑓 𝑓𝑢𝑒𝑙 𝑖𝑛 𝑡ℎ𝑒 𝑡𝑎𝑛𝑘 𝑜𝑓 𝑙𝑜𝑐𝑜𝑚𝑜𝑡𝑖𝑣𝑒 𝑖 𝑢𝑝𝑜𝑛 𝑎𝑟𝑟𝑖𝑣𝑎𝑙 𝑎𝑡 𝑖𝑡𝑠 𝑠𝑡𝑜𝑝 𝑗
MILP formulation
𝑚𝑖𝑛 delay costs + contracting costs + fueling costs
𝑠. 𝑡.
𝑔𝑎𝑙𝑙𝑜𝑛𝑠 𝑜𝑓 𝑓𝑢𝑒𝑙 𝑖𝑛 𝑡ℎ𝑒 𝑡𝑎𝑛𝑘 𝑜𝑓 𝑙𝑜𝑐𝑜𝑚𝑜𝑡𝑖𝑣𝑒 𝑖 𝑢𝑝𝑜𝑛 𝑎𝑟𝑟𝑖𝑣𝑎𝑙 𝑎𝑡 𝑖𝑡𝑠 𝑠𝑡𝑜𝑝 𝑗
𝑔𝑎𝑙𝑙𝑜𝑛𝑠 𝑜𝑓 𝑓𝑢𝑒𝑙 𝑖𝑛 𝑡ℎ𝑒 𝑡𝑎𝑛𝑘 𝑜𝑓 𝑙𝑜𝑐𝑜𝑚𝑜𝑡𝑖𝑣𝑒 𝑖 𝑢𝑝𝑜𝑛 𝑎𝑟𝑟𝑖𝑣𝑎𝑙 𝑎𝑡 𝑖𝑡𝑠 𝑠𝑡𝑜𝑝 𝑗 − 1
𝑔𝑎𝑙𝑙𝑜𝑛𝑠 𝑜𝑓 𝑓𝑢𝑒𝑙 𝑎𝑞𝑢𝑖𝑟𝑒𝑑 𝑏𝑦 𝑙𝑜𝑐𝑜𝑚𝑜𝑡𝑖𝑣𝑒 𝑖 𝑎𝑡 𝑖𝑡𝑠 𝑠𝑡𝑜𝑝 𝑗 − 1
𝑓𝑢𝑒𝑙 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑖𝑛 𝑔𝑎𝑙𝑙𝑜𝑛𝑠 𝑓𝑜𝑟 𝑎 𝑙𝑜𝑐𝑜𝑚𝑜𝑡𝑖𝑣𝑒 𝑖 𝑡𝑜 𝑔𝑜 𝑓𝑜𝑟 𝑠𝑡𝑜𝑝 𝑗 − 1 𝑡𝑜 𝑠𝑡𝑜𝑝 𝑗
=
+
-
1 𝑓𝑢𝑒𝑙 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛
𝑠. 𝑡.
𝑔𝑎𝑙𝑙𝑜𝑛𝑠 𝑜𝑓 𝑓𝑢𝑒𝑙 𝑖𝑛 𝑡ℎ𝑒 𝑡𝑎𝑛𝑘 𝑜𝑓 𝑙𝑜𝑐𝑜𝑚𝑜𝑡𝑖𝑣𝑒 𝑖 𝑎𝑡 𝑎𝑛𝑦 𝑡𝑖𝑚𝑒
≤ Capacity (in gallons) of fuel tank of a locomotive
2 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑓𝑢𝑒𝑙 𝑡𝑎𝑛𝑘
MILP formulation
𝑚𝑖𝑛 delay costs + contracting costs + fueling costs
𝑠. 𝑡.
𝑇ℎ𝑒 𝑔𝑎𝑙𝑙𝑜𝑛𝑠 𝑜𝑓 𝑓𝑢𝑒𝑙 𝑎𝑞𝑢𝑖𝑟𝑒𝑑 𝑏𝑦 𝑙𝑜𝑐𝑜𝑚𝑜𝑡𝑖𝑣𝑒 𝑖 𝑎𝑡 𝑖𝑡𝑠 𝑠𝑡𝑜𝑝 𝑗 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛
𝑤ℎ𝑒𝑡ℎ𝑒𝑟 𝑡ℎ𝑒 𝑙𝑜𝑐𝑜𝑚𝑜𝑡𝑖𝑣𝑒 𝑖 𝑠𝑡𝑜𝑝𝑠 𝑎𝑡 𝑗
3 𝑟𝑒𝑓𝑢𝑒𝑙𝑖𝑛𝑔
MILP formulation
𝑚𝑖𝑛 delay costs + contracting costs + fueling costs
Simple example (2010 RAS Competition)
4 yards:Y1, Y2, Y3, Y4
2 trains:T1,T2
T1: [Y1,Y2,Y3,Y4]
T2: [Y4,Y2,Y1]
Distances between yards (miles):
d(Y1,Y2) = 106
d(Y2,Y3) = 146
d(Y2,Y4) = 162
d(Y3,Y4) = 16
Y1 Y2 Y3 Y4
Simple example (2010 RAS Competition)
Simple example (2010 RAS Competition)
2 locomotives: L1, L2
Each train needs to be powered by exactly one locomotive
L1 and L2 take staggered turns everyday at pulling the trains
Simple example (2010 RAS Competition)
Locomotive
Consumption: 3.5 gallons of fuel per mile
Tank capacity: 4500 gallons of fuel
Distance that can run on full tank: 1285.71 miles
Fueling truck
Capacity: 25000 gallons
Weekly operating cost: $4000
Fixed refueling cost: $250
Fuel price ($/gallon):
Y1: 3.25
Y2: 3.05
Y3: 3.15
Y4: 3.15
MILP formulation: data
Sets
𝐼 = 2 (number of locomotives)
𝑁 = 35 (number of potential stops, same for both
locomotives 𝑁𝑖 = 𝑁)
7 times T1 x 3 potential stops = 21 potential stops
7 times T2 x 2 potential stops = 14 potential stops
𝐾 = 4 (number of yards)
𝑇 = 14 (number of days)
MILP formulation: data
Parameters
𝐷𝑖𝑗 = $250 (delay cost of locomotive 𝑖 at stop 𝑗)
𝐿𝑇 = 4500 𝑔𝑎𝑙𝑜𝑛𝑠 (capacity of fuel tank of a locomotive)
𝑇𝑇 = 25000 𝑔𝑎𝑙𝑜𝑛𝑠 (tanker truck capacity of yard)
𝐶𝐶 = $8000 (contracting cost of a truck for 2 weeks)
𝐹𝐶𝑖𝑗 : fuel consumption of locomotive 𝑖 on its journey from
stop 𝑗 to 𝑗 + 1 L1:T2 T1 T2 T1 …
L1:Y1 Y2 Y3 Y4 Y2 Y1 …
CO
NST
AN
T
T1 T2
Y1 Y2 Y3 Y4
371 gal 511 gal 56 gal
567 gal371gal
MILP formulation: data
Parameters
Y𝑎𝑟𝑑 𝑖, 𝑗 : yard of stop 𝑗 of
locomotive 𝑖 D𝑎𝑦 𝑖, 𝑗 : day in which stop
𝑗 of locomotive 𝑖 occurs
𝑃𝑘 =
3.25, 3.05, 3.15, 3.15$/gal (fuel price at yard 𝑘)
Provided data: fuel_prices_small.dat: fuel prices at yards (size
𝐾) number_of_stops_small.dat: number of stops for
each locomotive (size 𝐼) list_of_events_small.csv: table with all the
possible events
Read from CSV files
1 1 1 1 371 250
1 2 2 1 511 250
1 3 3 1 56 250
1 4 4 2 567 250
1 5 2 2 371 250
Locomotive
Stop
Yard
DayConsumption to next stop Delay costs
Read from CSV files
SheetConnectionsheet1("fuel_prices_small.csv");SheetConnectionsheet2("number_of_stops_small.csv");SheetConnectionsheet3("list_of_events_small.csv");
P fromSheetRead(sheet1,"'fuel_prices_small'!A1:A4"); // fuel price at yard k (in dolars/gallon)N fromSheetRead(sheet2,"'number_of_stops_small'!A1:A2"); // number of stops for each locomotive iData fromSheetRead(sheet3,"'list_of_events_small'!A1:F70"); // LocoNumber,LocoStopNumber,YardNumber,Day,ConsumptionStops,StopDelayCost
Model formulation and small instance
Write a mathematical model (MILP) designed to find the most
cost effective plan to fuel the locomotives.
Implement your model in OPL and solve it for the small
instance.
Provide an interpretation of the solution.
MILP formulation
minimize
𝑘
𝐶𝐶𝑘 ⋅ 𝑦𝑘 +
𝑖,𝑗
𝐷𝑖,𝑗 ⋅ 𝑥𝑖,𝑗 + 𝑃𝑦𝑎𝑟𝑑 𝑖,𝑗 ⋅ 𝑓𝑖,𝑗
𝑣𝑖,𝑗 = 𝑣𝑖,𝑗−1 + 𝑓𝑖,𝑗−1 − 𝐹𝐶𝑖,𝑗−1 ∀𝑖, 𝑗 = 2,… , 𝑁𝑖
𝑣𝑖,𝑗 + 𝑓𝑖,𝑗 ≤ 𝐿𝑇𝑖 ∀𝑖, 𝑗
𝑓𝑖,𝑗 ≤ 𝐿𝑇𝑖𝑥𝑖,𝑗 ∀𝑖, 𝑗
(𝑖,𝑗)∶ 𝑦𝑎𝑟𝑑(𝑖,𝑗) = 𝑘,𝑑𝑎𝑦(𝑖,𝑗)=𝑑
𝑓𝑖,𝑗 ≤ 𝑇𝑇𝑘 ⋅ 𝑦𝑘 ∀𝑘, 𝑑
𝑦𝑘 integer, 𝑥𝑖,𝑗 binary, 𝑓𝑖,𝑗 , 𝑣𝑖,𝑗 ≥ 0
MILP formulation
minimize
𝑘
𝐶𝐶𝑘 ⋅ 𝑦𝑘 +
𝑖,𝑗
𝐷𝑖,𝑗 ⋅ 𝑥𝑖,𝑗 + 𝑃𝑦𝑎𝑟𝑑 𝑖,𝑗 ⋅ 𝑓𝑖,𝑗
𝑣𝑖,𝑗 = 𝑣𝑖,𝑗−1 + 𝑓𝑖,𝑗−1 − 𝐹𝐶𝑖,𝑗−1 ∀𝑖, 𝑗 = 2,… , 𝑁𝑖
𝑣𝑖,𝑗 + 𝑓𝑖,𝑗 ≤ 𝐿𝑇𝑖 ∀𝑖, 𝑗
𝑓𝑖,𝑗 ≤ 𝐿𝑇𝑖𝑥𝑖,𝑗 ∀𝑖, 𝑗
(𝑖,𝑗)∶ 𝑦𝑎𝑟𝑑(𝑖,𝑗) = 𝑘,𝑑𝑎𝑦(𝑖,𝑗)=𝑑
𝑓𝑖,𝑗 ≤ 𝑇𝑇𝑘 ⋅ 𝑦𝑘 ∀𝑘, 𝑑
𝑦𝑘 integer, 𝑥𝑖,𝑗 binary, 𝑓𝑖,𝑗 , 𝑣𝑖,𝑗 ≥ 0
Results of the basic instance
Loc Yard StopNumber Day Gallons
2 4 1 1 0
2 2 2 1 0
2 1 3 2 0
2 2 4 2 0
2 3 5 2 0
2 4 6 3 0
2 2 7 3 0
2 1 8 4 0
2 2 9 4 0
2 3 10 4 0
2 4 11 5 0
2 2 12 5 4313
2 1 13 6 0
2 2 14 6 0
2 3 15 6 0
2 4 16 7 0
2 2 17 7 0
2 1 18 8 0
2 2 19 8 0
2 3 20 8 0
2 4 21 9 0
2 2 22 9 0
2 1 23 10 0
2 2 24 10 4263
2 3 25 10 0
2 4 26 11 0
2 2 27 11 0
2 1 28 12 0
2 2 29 12 0
2 3 30 12 0
2 4 31 13 0
2 2 32 13 0
2 1 33 14 0
2 2 34 14 0
2 3 35 14 0
Loc Yard StopNumber Day Gallons
1 1 1 1 0
1 2 2 1 0
1 3 3 1 0
1 4 4 2 0
1 2 5 2 0
1 1 6 3 0
1 2 7 3 0
1 3 8 3 0
1 4 9 4 0
1 2 10 4 0
1 1 11 5 0
1 2 12 5 2633
1 3 13 5 0
1 4 14 6 0
1 2 15 6 0
1 1 16 7 0
1 2 17 7 0
1 3 18 7 0
1 4 19 8 0
1 2 20 8 4500
1 1 21 9 0
1 2 22 9 0
1 3 23 9 0
1 4 24 10 0
1 2 25 10 0
1 1 26 11 0
1 2 27 11 0
1 3 28 11 0
1 4 29 12 0
1 2 30 12 0
1 1 31 13 0
1 2 32 13 1128
1 3 33 13 0
1 4 34 14 0
1 2 35 14 0
Results of the basic instance
Interpretation of the results:
Fueling trucks are contracted only inYard 2
Locomotive 1 stops three times:
Day 5, stop 12: 2633 gallons
Day 8, stop 20: 4500 gallons
Day 13, stop 32: 1128 gallons
Locomotive 2 stops two times:
Day 5, stop 12: 4313 gallons
Day 10, stop 24: 4263 gallons
Objective value: 60602.85
Assignment #6
The following inequalities are valid for our problem:
V1: 𝑥𝑖,𝑗 ≤ 𝑦𝑦𝑎𝑟𝑑 𝑖,𝑗 ∀𝑖, 𝑗
V2:
Using the large instance, run three versions of the model
separately:
M1: Standard model
M2: Standard model + V1
M3: Standard model + V2
𝑗=1
𝑁𝑖
𝑥𝑖,𝑗 ≥ 𝑀𝑖𝑛𝐹𝑢𝑒𝑙𝑠 𝑖 ∀𝑖
𝑀𝑖𝑛𝐹𝑢𝑒𝑙𝑠 𝑖 =σ𝑗=1𝑁𝑖 𝐹𝐶𝑖,𝑗
𝐿𝑇𝑖
Assignment #6
Stop the running after 15 minutes (900 seconds): cplex.tilim = 900;
Report the following information in a table: getobjectivevalue(m)
getobjbound(m)
getobjgap(m)
getsolvetime(m)
M1 M2 M3
Best known solution
Lower bound
Optimality gap
Solution time (s) Analyze the solutions and provide conclusions
Assignment #6
Send a single file containing all the answers including your
code (format is up to you) AND mathematical formulations of
the different models.
Group/Individual work
Send your reports to virginie.lurkin(at)epfl.ch
Send your reports by 8:00 P.M. next Monday