VEHICLE ROUTING PROBLEM
Readings: E&M 350-353
Topics:
• VRP versus TSP
• Solution methods
• Decision support systems for VRP
Relationship between TSP and VRP
Vehicle routing problem (VRP) is similar to the Traveling salesman problem (TSP) but with more
constraints such as vehicle capacity. Consider a TSP problem:
Depot
Suppose that the vehicle has a capacity of 10 and each customer has an order size as shown
below, then the TSP solution is no longer feasible.
5
1
3
2
2
2
Depot
1
4
3
4
3
1
VRP
5
1
A simple method :
ƒ
Use the TSP as usual
ƒ
When exceeding the vehicle capacity,
go back to the depot
ƒ
Start another tour with the nearest
customer as the first customer
ƒ
Repeat b. and c. until all customers are
served.
2
3
2
2
1
Depot
4
3
3
4
Can we use only three tours to serve all
customers?
5
1
3
2
2
2
1
Depot
4
3
4
3
As a Decision Problem
ƒ
Decisions
–
–
ƒ
Assignment of orders to vehicles
The sequence of orders to be served
Constraints
–
Vehicle capacity (in terms of volume or weight normally)
ƒ
Objectives
ƒ
Minimizing mileages or time
Considerations
–
–
–
Known demand
Delivery or pick-up
2
VRP
Solution Strategy
Tour construction heuristics
• Insertion methods (e.g. nearest insertion, min cost insertion)
• Savings methods (e.g. Clarke-Wright savings)
• Generalized assignment methods (e.g Fisher-Jaikumar method)
Tour improvement
• Intra-route improvement methods (e.g. 2-Opt, 3-Opt)
• Inter-route improvement methods (e.g. Relocate)
Tour construction method 1: Clarke-Wright savings method
Given
1 depot
4 customers with known demand
Vehicle capacity (100)
Costs for traveling between customers
35
3
4
40
2
45
52
43
57
60
5
29
25
3
72
Demand for
customer 5
43
45
43
1
Depot
Basic Steps
ƒ
ƒ
ƒ
Identify the distance matrix
Identify the savings matrix
Combine routes
3
VRP
4
VRP
5
VRP
s(2, 3) = 25 + 43 – 29 = 39
s(2, 4) = c(1, 2) + c(4, 1) – c(2, 4)
= 25 + 57 – 34 = 48
1
2
3
4
5
1
-
-
-
-
-
2
25
-
39
48
25
3
43
29
-
48
-
4
57
34
52
-
-
5
43
43
72
45
-
s(2, 5) = c(1, 2) + c(5, 1) – c(2, 5)
= 25 + 43 – 43 = 25
s(3, 4) = c(1, 3) + c(4, 1) – c(3, 4)
= 43 + 57 – 52 = 48
Begin combining tours: starting with the highest savings and (a) the
combined tour cannot exceed vehicle capacity, (b) insertion cannot be
in the interior of a tour.
1
2
3
4
5
1
-
-
-
-
-
2
25 -
3
43 29 -
4
57 34 52 -
5
43 43 72 45 -
4
2
4
39 48 25
5
48 14
55
2
3
5
4
Demand (5) = 60
Demand (4) = 35
Vehicle capacity = 100
Decision: combine the tour
4
1
Depo
6
VRP
Next highest savings:
Demand (2) = 40
Demand (4+5) = 95
Vehicle capacity = 100
Decision: cannot combine the
1
2
3
4
5
1
-
-
-
-
-
2
25 -
3
43 29 -
4
57 34 52 -
5
43 43 72 45 -
4
2
45
39 48 25
48 14
55
34
5
25
3
5
4
43
1
Depot
7
VRP
All nodes are visited once and the total cost is 242.
How to speed up the Clarke-Wright Savings Method?
We can use Heap to store the savings (from the largest to the smallest). The use of Heap can speed
up the operations.
Tour construction method 2: Generalized assignment method
(also called Fisher – Jaikumar Method)
ƒ
ƒ
ƒ
Assign seed points for each route
Evaluate insertion cost for each customer
Assign customers to routes
Parameters
di
uk
: demand of customer i
: capacity of vehicle k
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VRP
Decision variables
1 if customer i is assigned to route k
yik = 
0 otherwise
1 if i is visited followed by visiting j on route k
xijk = 
0 otherwise
Define:
N(yk) = all nodes on route k
f(yk) = the cost of route k
Our objective is to minimize the route cost while satisfying the demand. Therefore, we can write
the problem as follows:
Example
y41 = 1
Vehicle 1
y51 = 0
3
N ( y1 ) = {1, 3, 4}
1
2
4
0
5
6
min
Vehicle 2
y42 = 0
y52 = 1
N ( y2 ) = {2, 5, 6}
∑ f ( yk )
k
subject to
∑ di yik
i
∑ yik
k
yik
≤
uk
k i = 0
= 
1 i = 1,..., n
∈ {0,1}
The cost of route f (yk) is defined as follows:
9
VRP
f ( yk ) = min
subject to
∑ cij xijk
ij
∑ xijk
=
yik
∑ xijk
=
yik
∑ ∑ xijk
≥
i
j
i∈Q j∉Q
j = 0,..., n
1, Q ⊂ N ( yk )
A big problem: f (yk) is a complex, nonlinear function.
Solution idea: Approximate f (yk) by some linear function:
f (yk) ~
∑ cik yik
i
The coefficient, cik , represents the approximated cost of putting customer i on route k.
How do we get cik ?
Idea 1: cheapest insertion cost
Idea 2: use the concept of opportunity cost
Steps of the generalized assignment
Step 1: Assign a seed to each vehicle (or route) where the seeds are farthest away from each other
(Suppose that we have 3 vehicles)
D
10
VRP
Step 2: Evaluate the cost of inserting the customers into the existing routes.
Option 1: Approximate by the cheapest insertion cost
a
Customer 1
9
Best choice for customer 1: c
Best choice for customer 2: b
Best choice for customer 3: b
Second best for customer 3: a
Total cost: 38
cost = 7
cost = 12
not available
cost = 19
7
c
D
19
22
13
14
13
Customer 2
12
Customer 3
b
Option 2: Approximate by Opportunity cost or Regret
Regret = Cost for second best choice – cost for the best choice
Customer with the highest opportunity cost should be inserted first
a
Opportunity cost for customer 1: 2
Opportunity cost for customer 2: 1
Opportunity cost for customer 3: 6
Customer 1
9
7
c
D
19
22
14
13
13
Customer 3
12
b
Customer 2
Opportunity cost for customer 1: 2
Opportunity cost for customer 2: 1
customer 3 to b: 13
customer 1 to c: 7
customer 2 to a: 14
Total: 34
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VRP
Step 3: Rank customers from highest to lowest regret.
Example: Regret (3) > Regret (2)
Customer 2
a
From 3 to c route: 11
From 3 to b route: 17
From 3 to a route: 18
Regret(3) = 6
From 2 to c route: 10
From 2 to a route: 11
From 2 to b route: 12
c
D
Customer 3
Regret(2) = 1
b
Step 4: Insert the customers according to the rank of regrets one at a time. Notice that the regrets
may need to be updated.
Tour improvement Methods
There are general types of methods:
1. Intra-route improvement (changing the sequence of customers within a route)
Examples:
ƒ
2-Opt
ƒ
3-Opt
2. Inter-route improvement
ƒ
ƒ
Exchanging customers between routes
Exchanging segments of routes between routes
Example of Intra-route improvement: 2-opt algorithm
Do a swap of arc pairs in a route if it can reduce the cost (or time). Below we consider a case
where customers may impose delivery time-windows.
12
VRP
2-opt
Note: the indices i and j are in the order of the nodes being visited in a given tour. For simplicity,
we represent the route as a chain and ignore the physical shape of the route
2-opt swapping illustration
2-opt
i
i+1
j
i
i+1
j
j+1
j+1
13
VRP
o Flow chart:
For the node pairs
of i, j, consider swapping?
Y
d(i,j)+d(i+1, j+1)<
d(i, i+1) +d(j, j+1) ?
N
Y
After the swap, is the
resulting tour feasible?
N
Y
Swap the arcs (Note that the
arcs from i+1 to j are reversed
Stop?
N
Y
Return new solution
We can continue the swapping until the two cost savings of two consecutive swaps is smaller than
some small values or the number of swapping has reached a pre-defined maximum number of
swaps. In implementing 2-Opt algorithm, there are several possible ways to select which pairs of
arcs to be considered for swapping.
•
Select the pair of arcs with the best cost savings and then make swap;
•
Select several pair of arcs with the best cost savings and swap the best pair. From the
second swap on, this method will check the feasibility before making it since the route has
been changed after the first swap.
14
VRP
•
Randomly consider a pair of arcs and makes swap if it is feasible and can produce cost
savings;
•
Traverse the route from the starting point and make the swap once an eligible pair of arcs
(feasible and cost savings) is found.
Example of Inter-route improvement: Relocate
Consider the swapping between two existing routes. One way is to “relocate” a node in a route
(say route B) to another route (say route A)
i-1
Route A
i
j
i+1
j+1
Route B
Depot
We need to choose the route from which one node will be removed (route B) and the route to
which the node will be inserted (route A). Then, we can follow the procedure below:
15
VRP
Expanding route A,
Shrinking route B
For each node i in B
Y
Find a new feasible position for
i in Route A, say between j and j+1
Y
d(j,i)+d(i, j+1) +d(i-1, i+1)<
d(i, i+1) +d(j, j+1) +d(i-1, i)?
N
Y
Insert i between (j, j+1)
& Connect i-1, i+1
Y
N
Finish expanding/shrinking
Return
In implementing the relocate algorithm, we have several options:
ƒ
For each node in the shrinking route, find the first position in the expanding route after
which the node can be inserted and the relocation of the node results in cost saving.
ƒ
Similar to the above, except that for each node in the shrinking route, it will find the
position in the expanding route that the relocation will result in the largest saving.
ƒ
For each node in the shrinking route, find the best position as well as the best expanding
route
Other Implementation issues
– Time window constraints
Each customer has a specified time period during which the vehicle can visit
16
VRP
– Backhaul (pickup and delivery)
Not just delivery goods from the depot, but also pick up goods from customers
– Multiple depots
Vehicles can be coming from different depots
– Split orders
May have more than one delivery to a customer
The case of a chemical product manufacturer:
orders
Customer
service
Production
unit
Warehouse
Information
system
Fleet unit
delivery
Sites
– Multiple factories in Pearl River Delta
– Supply petrol chemical products to customers in PRD
17
VRP
Production
– Just-in-time production
Distribution
– 20 - 50 trucks per factory
– Order cut-off at 4 pm
– More than 1 trip per vehicle
– Divide region in zones and one route per zone
Process
– Receive an order
– Estimate the order finishing time
– Sort the orders to be finished on tomorrow
– Estimate the delivery schedules
– According to the delivery schedules, re-estimate the production schedules
– When the orders are ready, ask warehouse people prepare for the loading
– Ask fleet unit to send trucks to warehouse for loading
Operation Issues
– Information is not synchronized
– Paper work between units
– Spend 3 hours in determining initial delivery schedules
– Low truck utilization
– Difficult to entertain customer’s request on exact/approximate delivery time
– Urgent orders (some trucks are reserved to handle urgent order)
DECISION SUPPORT SYSTEMS FOR VRP
What is a decision support system?
What are the differences between Databases, Enterprise systems, Decision support systems?
ƒ
ƒ
ƒ
Database
– Organized storage of data
– Minimize data redundancy and inconsistency
– Data retrieval, search, and sort
Enterprise system
– Minimize inconsistency across databases
– Real-time data
– Integrating databases
Decision support system
– A system to support decision making
– Extract data and process data intelligently
18
VRP
The Components of DSS:
1. Database
2. Model-base
3. User interface
Decision support system development
•
Analyzing the needs
o Focus on causes, not symptoms
o What will be the reasonable expectation?
•
Data Development
o Data administration is a critical requirement for long-term success
o DSS requires their own database to produce effective results
o Is historical data available? Is the data telling the truth?
•
Model-base
o Model contents should be consistent with its goals: strategic, tactical, or
operational?
o What class of models do you want? Descriptive? Prescriptive?
•
Performance – how to measure?
•
What assumptions do you make? Are we assuming:
o Perfect data
o Dispatchers follow instruction exactly
o Drivers follow instruction exactly
o Shippers with perfectly predictable demands
o Personnel from different department have the same goal
19
VRP
Source: P.72, Mazza, Faircloug, Melton, de Pablo, Scheffer, Stevens, Software Engineering
Standards, Prentice Hall
Major phases in DSS development (and examples of activities)
User Requirement Definition
Capture, determine, specify and verify user requirements
Software Requirement Definition
Construct logical model
Specify functional, performance, documentation, acceptance testing requirements, etc.
Architectural Design
Decompose software into components
Define data input and output requirement, data structure, etc.
Select programming languages
Detailed Design and Production
Detailed design of lower-level components
Coding
Integrating components
Testing
Transfer
Installation
Acceptance tests
Operations and Maintenance
Final acceptance
20
VRP
Data Preparation for DSS:
Customer
independent
Customer
dependent
Static
Dynamic
Raw location data:
Road, speed limit, tolls, major
intersections, landmarks, etc.
Processed data:
Distances, costs, times, … etc. between
points of interests (via shortest path
methods on raw location data)
Locations
Trucks’ Details (e.g. capacity, loading
and unloading time, running cost)
Drivers’ Details (e.g. salary, overtime)
Products’ Details (e.g. size, weight,
price)
Travel flow pattern
Road restriction
Demand data:
Orders’ Details (e.g. daily
demand, amount)
Items’ Details (e.g. product to be
ordered, quantity)
Processing on preliminary static data:
Use shortest path engine to compute the distance, cost, or time, etc. between customers or between
depot and customers.
21
VRP
VRP Engine for the DSS
A combination of Saving Heuristics, Generalized assignment (Fisher-Jaikumar method), tour
improvement (2-Opt, Or-Opt), and different insertion methods.
Resulting handling and tools for analysis
•
•
•
•
•
Visualizing results
o Gantt charts
o Static route display
o Animation
Allow users to make changes
Measure performance
Compare scenarios
What-if analysis
Typical decision support
•
•
•
•
•
What are some good suggested solutions?
What is the impact of changing the sequence of orders being served?
What orders should be outsourced?
Can an urgent pickup order be accepted? By what vehicle? At what time?
What are the costs if different numbers of vehicles are used?
22
VRP