Facility Location

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Facility Location

• Chapter 2 Applied Mathematics of Logistics

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An instance of facility location

• 5 markets
• 3 candidate locations of the warehouses
• The annual fixed costs of warehouses are known.
• The transportation costs (per unit volume) from
  warehouses to markets are known.
• The annual market demands of markets are known.
• The objective is to minimize the total cost that
  is the sum of the annual fixed cost and the
  transportation cost.
• The constraints are
• all the demands must be satisfied,
• the transport is allowed from the opened
  warehouse(s) only.

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The date of the instance
Demand (unittons), transportation cost
(unit10000 yen/ton), and fixed cost to open
warehouse (unit10000 yen)
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Mosel model (declaration)
declare as an array of real numbers
Warehouse 1..3 Market 1..5
Demand array (Market) of real TransCost
array (Warehouse, Market) of real FixedCost
array (Warehouse) of real x array(Warehouse,
Market) of mpvar y array(Warehouse) of mpvar
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Mosel model (data input)

• Demand 180,80,200,160,220Remark for new
  version of model, we have to write Demand
  180,80,200,160,220
• TransCost
• 1000 ,800 ,600 ,500 ,400,
• 600 ,500 ,400 ,300 ,600,
• 300 ,400 ,500 ,500 , 900
• FixedCost 100000 ,100000 ,100000

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Objective function
name of the constraint
total_cost sum(i in Warehouse) FixedCost(i)
y(i) sum (i in Warehouse, j in Market)
TransCost(i,j) x(i,j)
sum represents the summation (S) that compute the
sum
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Demand constraints
Transportation volume that flows into market j
Total demand of market j
forall(j in Market) sum(i in Warehouse) x(i,j)
Demand(j)
We have a constraint for each market.
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Forcing constraints (capacity of warehouse)
Transportation volume from warehouse i ?
Capacity of warehouse i (if warehouse i is
open) 0
(otherwise)
forall(i in Warehouse) sum (j in Market)
x(i,j) lt 999999y(i)
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Output
obj. function value 508000 y(1) 0 y(2) 1
y(3) 0 x(2,1) 180 x(2,2) 80 x(2,3) 200
x(2,4) 160 x(2,5) 220
writeln(obj. function value ", getobjval)
forall(i in Warehouse) writeln(" y(", i, ") ",
getsol(y(i))) forall(i in Warehouse, j in
Market) do if (getsol(x(i,j))gt0.01) then
writeln(" x(", i, ",",j,") ", getsol(x(i,j)))
end-if end-do
output the solution when the transportation
volume is positive
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Forcing constraints (week and strong
formulations)

• The markets are supplied only from open
  warehouses.

Week (but compact) formulation (number of
constraints is few)
Strong formulation (number of constraints is
large)
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Cutting plane

• Cutting planes add the strong constraints if
  necessarily(when the constraints are violated by
  an LP solution)

Declare the constraints to be add as an array of
inequalities (linctr)
declarations cut array(Warehouse,Market) of
linctr end-declarations
Add to the cut pool using setmodcut(
constraints) command
forall(i in Warehouse,j in Market) do
cut(i,j) x(i,j) lt Demand(j)y(i)
setmodcut(cut(i,j)) end-do
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Weber problem

• Customers are on the plane. The X,Y coordinates
  of customer i is (Xi,Yi).
• Customer i has a given demand wi .
• The distance between the points is calculated by
  an lp norm (1 ?p). Distance between (Xi,Yi) and
  (Xj,Yj) is
• Objective is to find a location (X,Y) of the
  facility that minimizes the total weighted
  distance from the customers to the facility,
  i.e.,

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Proximal point algorithm(1)

• f(X,Y) is a convex functionthe necessarily and
  sufficient condition that (X,Y) is an optimal
  location is
• and

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Proximal point algorithm (2)

• The derivative with respect to X is set to 0
• This equation cannot be solved directly.
• The iterative method below find a proximal point.
• Similar formula can be obtained with respect to
  variable Y.
• Converges to a globally optimal solution when 1 ?
  p ? 2.

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Euclid k-median problem

• For a given set of points (customers) Pn xi,
  i1,?,n , find the location of k facilities
  that minimizes the total travel distances of
  customers.
• Points are on two-dimensional plane, i.e.,
  xi(Xi,Yi)
• xi-xj Euclidean distance between two
  points xi, xj
• L (k Pn) the optimal value of the Euclidean
  k-median problem for a given customer set Pn

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Aggregation heuristics

• 1.Divide the region 0,12 into t2 square
  sub-regions Ri, i1,?,t2 .
• Let the points in the region Ri be
  . Define the distance d(Ri,Rj) between
  regions Ri,Rj as the minimum distance between two
  points in each regions.
• Let N be the set of regions that have a
  non-empty .
• 2.Solve the following weighted k-median problem
• 3. Select an arbitrary point from the region
  selected in Step 2 and output the points as a
  heuristic solution.

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Worst case analysis of aggregation heuristics

• L(k Pn)the optimal value of the Euclid
  k-median problem for Pn
• Lah(k Pn) the approximate solution value of the
  aggregation heuristics with parameter t

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Honeycomb heuristics

• 1.Cover the region by hexagons with area 1/k .
• 2.Select the nearest points to the centers of
  hexagons and output the points as a heuristic
  solution.
• Assume that all the points are randomly
  distributed on the plane.
• L(k Pn) the optimal value of the Euclid
  k-median problem for Euclid k-median
• Lhc(kPn) the approximate solution value of the
  honeycomb heuristics
• If k ? (log n), ko (n /log n), we have
• where ? is a constant (approximately 0.3771967)

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k-facility maximization problem

• cij (?0) the profit when customer i 2 I is
  served by facility j 2 J
• Select k facilities from the set J of candidate
  points
• Maximize the total profit
• The total profit v(S) when we open a subset S ?
  J of facilities

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Greedy algorithm for k-facility maximization
problem

• S0
• for t0 to k do

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Worst case analysis of greedy algorithm

• Agreedy(I)the approximate solution value
  obtained by the greedy algorithm
• OPT(I)the optimal value