decision aid methodologies in transportation · lab vi: case study –freight trains management ....

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)


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)


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)


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)


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)


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)


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


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



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


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


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)


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


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

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



Locomotive 2 stops two times:



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

decision aid methodologies in transportation · lab vi: case study –freight trains management ....

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