Routing

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Routing

• John H. Vande Vate
• Spring, 2001

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Shared Transportation Capacity

• Large shipments reduce transportation costs but
  increase inventory costs
• EOQ trades off these two costs
• Reduce shipment size without increasing
  transportation costs?
• Combine shipments on one vehicle

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TL vs LTL
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Shared Loads

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Inventory
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Issues

• Design Routes that
• Minimize the transportation cost
• Respect the capacity of the vehicle
• This may require several routes
• Consider inventory holding costs
• This may require more frequent visits

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Our Approach

• Minimize Transportation Cost (Distance)
• Traveling Salesman Problem
• Respect the capacity of the Vehicle
• Multiple Traveling Salesmen
• Consider Inventory Costs
• Estimate the Transportation Cost
• Estimate the Inventory Cost
• Trade off these two costs.

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The Traveling Salesman Problem

• Minimize Distance
• s.t.
• start at the depot,
• visit each customer exactly once,
• return to the depot
• A single vehicle no capacity
• Only issue is distance.

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IP Formulation

• Set Cities
• param dCities, Cities
• var xCities, Cities binary
• minimize Distance
• sumf in Cities, t in Citiesdf,txf,t
• s.t. DepartEachCityf in Cities
• sumt in Citiesxf,t 1
• s.t. ArriveEachCityt in Cities
• sumf in Citiesxf,t 1

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SubTours
3 cities 3 edges
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Eliminating Subtours

• S.t. SubTourElimination
• S subset Cities
• card(S) gt 0
  and
• card(S) lt
  card(Cities)
• sumf in S, t in S xf,t lt card(S) -
  1

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An Equivalent Statement
3 cities No edges out
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An Equivalent Formulation

• S.t. SubTourElimination
• S subset Cities
• card(S) gt 0
  and
• card(S) lt
  card(Cities)
• sumf in S, t not in S xf,t gt 1

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How Many Constraints?

• How many subsets of N items?
• 2N
• Omit 2 subsets
• All N items
• The empty set
• 2N - 2

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OK, Half of that...

• If we have an edge out of 1, 2, 3, we must have
  an edge out of 4, 5, 6
• Why? The edge out of 1, 2, 3 is an edge into
  4, 5, 6. But there are exactly 3 edges into
  cities in 4, 5, 6. Since one of them comes from
  1, 2, 3, not all the edges from cities in 4,
  5, 6 can lead to cities in 4, 5, 6. At least
  one must go to 1, 2, 3

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Still Lots!

• How many subsets of N items?
• 2N
• Omit 2 subsets
• All N items
• The empty set
• 2N - 2
• Half of that 2N-1 - 1

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How Many?
N 2N-1 - 1
Too Many!
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Optimization is Possible But...
It is difficult Few 100 cities is the limit Is it
appropriate? Other approaches.
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Heuristics

• The Strip Heuristic
• Partition the region into narrow strips
• Routing in each strip is easy 1-Dimensional
• Paste the routes together

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The Strip Heuristic
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Nearest Neighbor
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Clark-Wright

• Shortcut a tour by finding the greatest
  savings

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Clark-Wright
Shortcut a tour by finding the greatest
savings
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Nearest Insertion
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Nearest Insertion
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Nearest Insertion
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Nearest Insertion
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Nearest Insertion
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Nearest Insertion
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Nearest Insertion
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Nearest Insertion
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Nearest Insertion
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Nearest Insertion
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Nearest Insertion
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Nearest Insertion
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Nearest Insertion
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Nearest Insertion
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Nearest Insertion
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Improvement Heuristics

• 2-Opt

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Local Minima

• No improvement found, but
• Tour still isnt good

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Probabilistic Methods

• Simulated Annealing
• With probability that reduces over time, accept
  an exchange that makes things worse (gets you out
  of local minima).

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Optimization-Base Heuristics

• Minimum Spanning Tree Heuristic
• Build a minimum spanning tree on the edges
  between customers
• Double the tree to get a Eulerian Tour (visits
  everyone perhaps several times and returns to the
  start)
• Short cut the Eulerian Tour to get a Hamilton
  Tour (Traveling Salesman Tour)

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The Spanning Tree

• Is Easy to construct
• Use the Greedy Algorithm
• Add edges in increasing order of length
• Discard any that create a cycle
• Is a Lower bound on the TSP
• Drop one edge from the TSP and you have a
  spanning tree
• It must be at least as long as the minimum
  spanning tree

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The Spanning Tree
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Double the Spanning Tree

• Duplicate each edge in the Spanning Tree
• The resulting graph is connected
• The degree at every node must be even
• Thats an Eulerian Graph (you can start at a
  city, walk on each edge exactly once and return
  to where you started)
• Its no more than twice the length of the
  shortest TSP

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The Spanning Tree
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Short Cut the Eulerian Tour
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Short Cut the Eulerian Tour
Short Cut the Eulerian Tour
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Spacefilling Curves

• There are no more points in the unit square than
  in the interval from 0 to 1!?

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Proof

• Each point (X,Y) on the map
• Express X string of 0s and 1s
• X 16.5 10000.10
• 12402302202102012-1
  02-2
• Express Y string of 0s and 1s
• Y 9.75 01001.11
• 02412302202112012-1
  12-2
• Space Filling Number - interleave bits
• ?(X,Y) 1001000001.1101

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So,...

• Each pair of points
• X 16.5 10000.10
• Y 9.75 01001.11
• maps to a unique point
• ?(X,Y) 1001000001.1101

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How to Use this?

• A mapping of ?(X,Y) into the unit interval
• Think of this as the inverse mapping of the unit
  interval onto the square (our super tour)
• For a given customer ?(X,Y) is the fraction of
  the way along the super tour where it lies
• Visit the customers in the order of ?(X,Y) (short
  cut the super tour to visit our customers)

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The TSP

• For More on SpaceFilling Curves visit
• http//www.isye.gatech.edu/faculty/John_Bartholdi/
  mow/mow.html
• There are several books on the TSP
• .

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Our Approach

• Minimize Transportation Cost (Distance)
• Traveling Salesman Problem
• Respect the capacity of the Vehicle
• Multiple Traveling Salesmen
• Consider Inventory Costs
• Estimate the Transportation Cost
• Estimate the Inventory Cost
• Trade off these two costs.

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Different Approaches

• Route First - Cluster Second
• Build a TSP tour
• Partition it to meet capacity
• Cluster First - Route Second
• Decide who gets served by each route
• Then build the routes

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Route First
Vehicle Cap 15
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Cluster First

• Sweep Heuristic

Vehicle Cap 15
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Our Approach

• Minimize Transportation Cost (Distance)
• Traveling Salesman Problem
• Respect the capacity of the Vehicle
• Multiple Traveling Salesmen
• Consider Inventory Costs
• Estimate the Transportation Cost
• Estimate the Inventory Cost
• Trade off these two costs.