Location and Allocation problems
Location problems primarily address location of facilities that have capacity to meet the
demand placed on them from customers or other locations. These optimize cost of setting
up the facilities or distance associated with the transportation of items from the chosen
locations. Location decisions are strategic in nature and involve high costs. Location of
public facilities is very important considering that the society at large is benefitted. The
objective shifts to maximizing or enabling reach as against minimizing cost.
In many instances the demand from the customer is to be met and this involves transporting
items from the located facilities to the customers. The transportation is sometimes direct to
the customer while in many instances it is through intermediate facilities such as
warehouses which also have to be located. These result in what are called “Location and
Allocation problems” where the decisions are to identify locations as well as the
transportation quantities.
Single facility location in a plane.
Consider n customer locations in a plane. Customer j is located at (xj, yj). It is desired to
locate a facility at (X, Y) such that the sum of squares of distance is minimized.
2
2
Sum of squares of distance = ∑𝑛𝑗=1(𝑥𝑗 − 𝑋) + (𝑦𝑗 − 𝑌) . This is minimized by partially
differentiating the distance squared with respect to X and Y and setting to zero. This gives
∑𝑛 𝑥
∑𝑛 𝑦
us the solution 𝑋 = 𝑗=1 𝑗⁄𝑛 and 𝑌 = 𝑗=1 𝑗⁄𝑛.
∑𝑛𝑗=1 𝑤𝑗 𝑥𝑗
⁄𝑤
If there is a weight wj associated with customer j, the solution becomes 𝑋 =
𝑗
∑𝑛𝑗=1 𝑤𝑗 𝑦𝑗
⁄𝑤 .
and 𝑌 =
𝑗
Illustration 4.6
Consider 5 customers located at (3,4), (5,2), (7,8), (1,5) and (6,6). Locate a facility at (X,Y)
such that the sum of squares of distance is minimized. What happens when the weights for
five customers are 1, 3, 2, 4, and 2.5?
When the weights are equal, the location (X,Y) is given by ((
3+5+7+1+6
4+2+8+5+6
5
5
), (
(4.4, 5),
3𝑥1+5𝑥3+7𝑥2+1𝑥4+6𝑥2.5
With weights the location for (X, Y) becomes ((
(
4𝑥1+2𝑥3+8𝑥2+5𝑥4+6𝑥2.5
1+3+2+4+2.5
)) = (4.08, 4.88).
1+3+2+4+2,5
),
)) =
Minimizing absolute distance
Instead of minimizing the sum of squares of distances we can locate points that minimize
the sum of absolute distance from (X, Y). Here we minimize ∑𝑛𝑗=1(|𝑥𝑗 − 𝑋| + |𝑦𝑗 − 𝑌|). Let
uj be the absolute distance between xj and X and vj be the absolute distance between yj and
Y.
The problem reduces to
𝑛
𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 ∑(𝑢𝑗 + 𝑣𝑗 )
𝑗=1
Subject to
𝑢𝑗 ≥ 𝑥𝑗 − 𝑋
𝑢𝑗 ≥ 𝑋 − 𝑥𝑗
𝑣𝑗 ≥ 𝑦𝑗 − 𝑣𝑗 ≥ 𝑌 − 𝑦𝑗
uj, vj ≥ 0.
If there is a weight wj associated with customer j, the objective function becomes
𝑛
𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 ∑ 𝑤𝑗 (𝑢𝑗 + 𝑣𝑗 )
𝑗=1
When emergency facilities such as fire station, hospital, petrol station etc are located, the
objective is to minimize maximum distance instead of minimizing total distance. The
objective is to
Minimize P + Q; where P ≥ uj and Q ≥ vj. The constraints become P ≥ wj uj and Q ≥ wj vj when
emergency facility is considered.
Illustration 4.7
Consider 5 customers located at (3,4), (5,2), (7,8), (1,5) and (6,6). Locate a facility at (X,Y)
such that the sum of absolute distance is minimized. What happens when the weights for
five customers are 1, 3, 2, 4, and 2.5? Consider also the case of emergency facility?
The problem is to minimize u1 + u2 + u3 + u4 + u5 + v1 + v2 + v3 +v4 + v5
Subject to
u1 ≥ X-3; u1 ≥ 3 – X; u2 ≥ X-5; u2 ≥ 5 – X; u3 ≥ X-7; u3 ≥ 7 – X; u4 ≥ X-1; u4 ≥ 1 – X; u5 ≥ X-6; u5 ≥
6-X
v1 ≥ Y-4; v1 ≥ 4 – Y; v2 ≥ Y-2; v2 ≥ 2 – Y; v3 ≥ Y-8; v3 ≥ 8 – Y; v4 ≥ Y-5; v4 ≥ 5 –Y; v5 ≥ Y-6; v5 ≥ 6 –
Y
X, Y, uj, vj ≥ 0
The optimum solution is given by X =5, Y = 5, u1 = 2, u3 = 2, u4 = 4, u5 = 1, v1 = 1, v2 = 3, v3 = 3
and v5 = 1. The location is (5, 5) with distance 17.
We fix each of the xj values as X and compute the sum of weighted absolute distance. The
five values are 29.5, 24.5, 36.5, 38.5 and 28 which gives X = 5. We fix each of the y j values as
Y and compute the sum of weighted absolute distance. The five values are 23, 36, 39, 18.5
and 22 which gives Y = 5. The facility is located at (5, 5) with sum of absolute distances = 43.
We fix each of the xj values as X and compute the sum of weighted absolute distance. The
five values are 12, 9, 13, 17 and 10 which gives X = 5. We fix each of the yj values as Y and
compute the sum of absolute distance. The five values are 9, 15, 15, 8 and 9 which gives Y =
5. The facility is located at (5, 5) with sum of absolute distances = 17.
If we consider emergency facility the optimum solution is given by (4, 5)
It is also observed that (5, 5) and (5, 6) give maximum distance of 5. If we consider the given
weights, (5, 5) is optimal.
1 – median problem (Hakimi, 1964)
Given a graph where every node is connected to every other node, the 1- median problem is
to find a node (or vertex) such that the distance (on the network) from all other nodes to
the median node is minimized.
The mathematical programming formulation for the 1-median problem is as follows:
Let Yj = 1 if node j is the median. The objective function is to minimize ∑𝑛𝑖=1 ∑𝑛𝑗=1 𝑑𝑖𝑗 𝑌𝑗
subject to ∑𝑛𝑗=1 𝑌𝑗 = 1; Yj = 0,1. If there is a weight wj associated with node j, the objective
function minimizes ∑𝑛𝑖=1 ∑𝑛𝑗=1 𝑤𝑗 𝑑𝑖𝑗 𝑌𝑗
Illustration 4.8
Consider 5 customers located at (3,4), (5,2), (7,8), (1,5) and (6,6). Solve a 1-median
problem? What happens when the weights for five customers are 1, 3, 2, 4, and 2.5?
Compute dij based on rectangular distance between nodes I and j
The distances are as follows: d12 = 4; d13 = 8; d14 = 3; d15 = 5; d23 = 8; d24 = 7; d25 = 5; d34 = 9;
d35 = 3 and d45 = 6. The distances are symmetric and djj = 0. The problem is to
Minimize 20Y1 + 24Y2 + 28Y3 + 25Y4 + 19Y5. Subject to Y1 + Y2 + Y3 + Y4 + Y5 = 1; Yj = 0,1.
The optimum solution is Y5 = 1 with total distance 19.
If we include weights, the objective function is to minimize 52.5Y1 + 60.5Y2 + 75.5Y3 + 57Y4 +
50Y5. The optimum solution is Y5 = 1.
Illustration 4.9
Consider 5 customers located at (3,4), (5,2), (7,8), (1,5) and (6,6). The connections between
the nodes are 1-3, 2-3, 2-4 and 2-5. Solve a 1-median problem? What happens when the
weights for five customers are 1, 3, 2, 4, and 2.5? Compute dij based on rectangular distance
between the given coordinates for the given arcs.
The arc distances for the given arcs are d13 = 8, d23 = 8, d24 = 7 and d25 = 5. The given
network is a tree. We find the distances when each node becomes the median. The sum of
distances if nodes 1 to 5 are treated as median are 68, 36, 44, 57, 51. The optimum solution
is to choose node 2 as the median.
It is also observed that node 2 which is not a leaf node which is connected to more than one
node is a candidate. Nodes that are connected to only node do not become the median.
When weights are considered, the weighted distances are 208.5, 72.5, 124.5, 104 and 110
respectively. Node 2 is the median.
If we consider emergency facility, we would locate it in a node such that maximum distance
is minimized. Here node 3 with minimum value of 15 is chosen. If we consider the give n
weights, node 2 becomes the median.
Multiple facilities
Here we create locations for multiple facilities to meet the requirements of the customers.
Let there be n customers where customer j is located at (xj, yj). We wish to create k (k < n)
facilities. We wish to find the locations of these such that each customer is attached to one
facility and the sum of distance between the customers and the assigned facility is
minimized.
Let the coordinates of the new facilities be (Xk, Yk). Distance d(j,k) between customer j and
facility k is given by 𝑑𝑗𝑘 = ∑𝑛𝑗=1(|𝑥𝑗 − 𝑋𝑘 | + |𝑦𝑗 − 𝑌𝑘 |) . Let Zjk = 1 if customer j is allotted to
facility k.
We minimize ∑𝑛𝑗=1 ∑𝑝𝑘=1 𝑑𝑗𝑘 𝑍𝑗𝑘
Subject to ∑𝑛𝑗=1 ∑𝑝𝑘=1 𝑍𝑗𝑘 = 1; Zjk = 0,1.
The formulation is not linear since djk depends of (Xk, Yk).
Illustration 4.10
Consider 5 customers located at (3,4), (5,2), (7,8), (1,5) and (6,6). Create two locations such
that each customer is attached to one location and the total distance between the
customers and location is minimized?
In the earlier example we have located the facility at (5, 5) such that the minimum distance
of 17. Let (5, 5) be one location and (3,4) be another. Facility 1 would attract customers 1
and 4 with distance 0 and 3. Facility 2 will attract customers 2, 3 and 5 with distance 3, 5 and
2. The total distance is 13. The solution need not be optimum.
A heuristic solution would be to group the points into two and solve a single facility location
problem for each. Points 1, 2 and 4 can form a group and points 3 and 5 form another
group. We can consider the centroids or midpoints of the groups. The two locations can be
(3,4) and (6.5, 7). The distances are 0, 4 and 3 for points 1, 2 and 4 and 1.5 and 1.5 for points
1 and 3. The sum of distances is 10. We will revisit this example after we solve a 2 median
problem.
Covering problem
Given a connected network with nodes and arcs, the problem is to find minimum nodes
such that each of the other nodes has an arc connecting one of the chosen nodes. The
formulation is as follows:
Let there be m nodes and n arcs. aij =1 if there is an arc connecting nodes i and j. Let Yj = 1 if
node j is chosen. We minimize ∑𝑛𝑗=1 𝑌𝑗 subject to 𝑌𝑖 + ∑𝑗 𝑎𝑖𝑗 𝑌𝑗 ≥ 1; for each arc i-j. Yj = 0,1.
We can also have a restriction on the number of nodes attached to a selected node.
Illustration 4.11
Consider a network with 6 nodes and 8 arcs. The arcs are 1-2, 2-3, 3-4, 4-5, 5-6, 1-6, 1-3 and
1-4. Find out minimum number of vertices such that each edge (arc) is incident to at least
one vertex?
The formulation is to Minimize Y1 + Y2 + Y3 + Y4 + Y5 + Y6; subject to Y1 + Y2 + Y3 + Y6 ≥ 1,Y1 +
Y2 + Y3 ≥ 1,Y1 + Y2 + Y3 + Y4 ≥ 1,Y3 + Y4 + Y5 + Y6 ≥ 1, Y4 + Y5 + Y6 ≥ 1, Y1 + Y4 + Y5 + Y6 ≥ 1, Yj =
0
The optimum solution is Y1 = Y5 = 1.
A heuristic method would be as follows:
1. Choose the node with maximum number of arcs and all the incident nodes to the
set.
2. If the set does not contain all nodes, add a node such that the additional new nodes
incident to the chosen node is maximized.
3. Repeat step 2 till all nodes are in the set.
The number of arcs incident to the six nodes is 3, 2, 3, 3, 2 and 3. We choose node 1 and add
nodes 1, 2, 3 and 6. We choose node 4 and add 4 and 5 to the set. We have two nodes.
If we place a restriction that each chosen node should not cater to more than 3 nodes, the
solution would still be 2 nodes. Y1 = Y4 = 1 with node 1 catering for nodes 2 and 3 and node 4
for nodes 5 and 6.
k-median problem
The k-median problem is an extension of the 1-median problem where we fix k nodes in the
given network as median points and minimize the distance between every non median and
the median to which it is assigned. Each non median point is assigned to one median point
only.
The formulation of the k-median problem is as follows:
Let Xjj = 1 if point j is a median and Xij = 1 if point i is attached to median j. The objective is to
n
Minimize
n
 d
i 1 j 1
ij
X ij
Subject to ∑𝑛𝑗=1 𝑋𝑗𝑗 = 𝑘
n
X
i 1
ij
1
X ij  X jj
Xij = 0,1
The objective function minimizes the sum of distance between the median points and the
points allotted to that median. There are as many median points as the required number of
facilities. Each point is either a median or is allotted to a median. A non median point can be
allotted only to a median point.
The k-median problem is a zero one problem that can be solved using binary Integer
Programming. The problem is “hard” and the computational time increases exponentially as
the problem size increases. Given n points, the formulation has n2 decision variables (binary)
and n2+1 constraints.
The medians chosen by the k-median problem are the places where the factory or
warehouse is located. The medians are identified based on minimum distance among the
points. In the above formulation there are no restrictions on number of non median points
assigned to a median. Capacity and demand are not explicitly modeled here.
In the above formulation k out of the n given points act as medians. The k-median problem
can also be used to determine p groups out of n given points, an idea used in grouping
machines in Cellular Manufacturing Systems.
Illustration 4.12
________________________________________________
Consider six points whose distances are given in Table 4.9. Solve a k median problem to
locate two facilities among these six points. Solve a k median problem to locate three
facilities among these six points.
Table 4.9 – Data for Illustration 4.12
1
2
3
4
5
6
1
-20
18
14
16
12
2
20
-22
18
30
26
3
18
22
-32
20
22
4
14
18
32
-20
22
5
16
30
20
20
-30
6
12
26
22
22
30
--
The k-median formulation for 6 points and 2 medians (facilities) has 36 binary decision
variables and 37 constraints. The formulation is to minimize 20X12 + 18X13 + 14X14 + 16X15 +
12X16 + 22X23 + 18X24 + 30X25 + 26X26 + 32X34 + 20X35 + 22X36 + 20X45 + 22X46 + 30X56
The first constraint is X11 + X22 + X33 + X44 + X55 + X66 = 2. The second set has 6 constraints and
the third has 30 constraints. The optimal solution is given by X11 = X44 = 1 and X24 = X31 = X54
= X61 = 1 with Z = 68. Here points 1 and 4 act as medians where the plants have to be
located. Points 2 and 5 are attached to median 4 and points 3 and 6 are attached to median
1.
If we wish to use the k-median model to get two groups out of six points, the corresponding
solution would be {2, 4, 5} and {1, 3, 6}.
A quick heuristic would be to consider two farthest points as medians and allot the rest of
the points to the medians. This would give us the solution X33 = X44 = 1 and X13 = X24 = X54 =
X63 = 1 with Z = 78. Here points 3 and 4 act as medians where the plants have to be located.
Points 1 and 6 are attached to median 3 and points 2 and 5 are attached to median 4. Points
5 and 6 are equidistant with respect to both the medians and can be allotted to any of the
medians resulting in several alternate solutions with the same distance.
The k-median formulation for 6 points and 3 medians (facilities) also has 36 binary decision
variables and 37 constraints. The optimal solution is given by X11 = X22 = X55 = 1 and X31 = X42
= X61 = 1 with Z = 48. Here points 1, 2 and 5 act as medians where the three plants have to
be located. Points 3 and 6 are attached to median 1 and point 4 is attached to median 2.
Median 5 does not attract any point. Since median 5 does not attract any point, we may not
eventually create a facility there. A p-median problem with two medians is more
appropriate. However, if we use the model to get three groups, the solution is {1, 3, 6}, {2,
4} and {5}.
A quick heuristic would be to consider two farthest points as medians. Points 3 and 4 qualify
as the first two medians. The third median is chosen such that its distance with the two
chosen medians is the largest. Point 6 qualifies as the third median. The rest of the points
are allotted to the median with minimum distance. This would give us the solution X33 = X44
= X66 = 1 and X16 = X24 = X53 =1 with Z = 50. Here points 3, 4 and 6 act as medians where the
plants have to be located. Points 1 is attached to median 6, point 2 is attached to median 3
and point 5 is attached to median 4. Point 5 is equidistant with respect to both 3 and 4 and
can be allotted to either of the medians resulting in an alternate solution with the same
distance. We would prefer the given solution because all the medians attract a point. The
solution to the corresponding grouping problem is {1, 6}, {2, 3} and {4, 5}.
Illustration 4.13
Consider 5 customers located at (3,4), (5,2), (7,8), (1,5) and (6,6). Solve a 2 median problem.
Create two locations such that each customer is attached to one location and the total
distance between the customers and location is minimized?
The distances are as follows: d12 = 4; d13 = 8; d14 = 3; d15 = 5; d23 = 8; d24 = 7; d25 = 5; d34 = 9;
d35 = 3 and d45 = 6. The distances are symmetric and djj = 0.
The optimum solution to the 2 median problem is to have X11 = X33 = 1 and X21 = X41 = X53 = 1
with total distance = 10. Points 1 and 3 act as median points. Points 2 and 4 are attached to
median 1 and point 5 is attached to point 3 (median). If we make two groups of points from
the above solution, the groups are {1, 2, 4} and {3, 5}. The locations are (3,4) and (7,8). The
distances are 4, 3 and 3 and the minimum distance is 10.
The formulation for 3 medians has the optimum solution X22 = X44 = X55 = 1 and X14 = X35 = 1
with total distance = 6. Points 2, 4 and 5 act as median points. Points 1 and 3 are attached
to medians 4 and 5 respectively.
Alternate formulations of the k-median problem
Over a period of time, researchers have worked on developing formulations for the kmedian model involving fewer variables and/or constraints. One such formulation is
described below:
Minimize: ∑𝑛𝑖=1 ∑𝑛𝑗=𝑖+1 𝑑𝑖𝑗 𝑋𝑖𝑗
Subject to ∑𝑛𝑗=1 𝑋𝑗𝑗 = 𝑘
n
i
 X X
j i 1
ij
k 1
ki
 U B X ii  1
i = 1, 2, 3… n and k  i .
X ij  X ii  X jj  0

i = 1, 2, 3… n
j i 1 …n
(3.19)
X ij  X ii  X jj  2

i = 1, 2, 3… n
j i 1 …n
(3.20)
X ij  1 if point i is assigned to median j otherwise 0 (i = 1,…,n; j=i+1,…n)
U B  Maximum number of points attached to a median.
The above formulation for the k-median formulation we have 𝑛𝐶2 + n variables (binary) and
2 x 𝑛𝐶2 + n + 1 constraints. This formulation has fewer variables and constraints than the
original formulation.
The basic idea behind this formulation is that median can be either i or j in the variable X ij .
In the k-median formulation only j can become median machine. The objective function
looks similar to p-median formulation, but only half the variables of k-median are added in
this expression. The first constraint is similar to k-median which controls the number of
groups or median to be formed. The second constraint serves as upper bound to the
number of points attached to a median when i becomes a median and it also it forces to
associate the non-median points to a median point when i is not a median. The third
constraint makes sure that non-median points are not associated with each other. The
fourth constraint makes sure that median points are not associated with each other.
Illustration 4.14
Consider 5 customers located at (3,4), (5,2), (7,8), (1,5) and (6,6). Solve a 2 median problem.
Create two locations such that each customer is attached to one location and the total
distance between the customers and location is minimized? Use the alternate formulation.
The distances are as follows: d12 = 4; d13 = 8; d14 = 3; d15 = 5; d23 = 8; d24 = 7; d25 = 5; d34 = 9;
d35 = 3 and d45 = 6. The distances are symmetric and djj = 0. In this formulation we have
variables X11, X12, X13, X14, X15, X22, X23, X24, X25, X34, X35 and X45. There are 26 constraints.
The first constraint in set 2 is X12+X13+X14+ X15 – 4X11≤1. There are 10 such constraints. The
first constraint in set 3 is X12 - X11 - X22 ≤ 0. There are 10 such constraints. The first constraint
in set 4 is X12 + X11 + X22 ≤ 2. There are 10 such constraints.
The optimum solution to the 2 median problem is to have X11 = X33 = 1 and X12 = X14 = X53 = 1
with total distance = 10. Points 1 and 3 act as median points. Points 2 and 4 are attached to
median 1 and point 5 is attached to point 3 (median). If we make two groups of points from
the above solution, the groups are {1, 2, 4} and {3, 5}. The locations are (3,4) and (7,8). The
distances are 4, 3 and 3 and the minimum distance is 10. Since we do not have variables X41
and X21 the formulation has variables X12 = X14 = 1 conveying that points 2 and 4 are
attached to median point 1.
Reducing the number of constraints and solving for optimum k.
In the original k-median formulation we have n2 –n constraints of the type 𝑋𝑖𝑗 ≤ 𝑋𝑗𝑗 . These
can be reduced to the set ∑𝑛𝑖=1 𝑋𝑖𝑗 ≤ 𝑈𝑋𝑗𝑗 for j = 1,…,n and i ≠ j (Wang). This reduced the
number of constraints from n2 –n to n constraints.
In the k-median it is often assumed that the value of k is known. If we keep k as a variable,
we will have k = n where each point becomes a median and the distance becomes zero. We
can treat k as a variable and prevent k from becoming n by adding a penalty to the objective
function. Ben suggested that a value
𝑛
𝛼𝑘 ∑𝑛
𝑖=1 ∑𝑗=1 𝑑𝑖𝑗
𝑛2
be added. Here α is known and fixed by
the user.
Illustration 4.15
Consider 5 customers located at (3,4), (5,2), (7,8), (1,5) and (6,6). Solve a 2 median problem.
Create two locations such that each customer is attached to one location and the total
distance between the customers and location is minimized? Use the alternate formulation.
The distances are as follows: d12 = 4; d13 = 8; d14 = 3; d15 = 5; d23 = 8; d24 = 7; d25 = 5; d34 = 9;
d35 = 3 and d45 = 6. We use α = 1. The objective function becomes Minimize 4X12 + 8X13 +
3X14 + 5X15 + 8X23 + 7X24 + 5X25 +9X34 +3X35+ 6X45 + 4X21 + 8X31 + 3X41 + 5X51 + 8X32 + 7X42 +
5X52 +9X43 +3X53+ 6X54 + 4.64k. The 20 constraints Xij ≤ Xjj are replaced by 5 constraints of the
type X21+X31+X41+ X51-2X11 ≤ 0. Here was force a maximum of two non median points to be
attached to a median point.
The optimum solution to the MILP problem (p is defined as an integer while the Xij are
binary variables) is X11 = X33 = 1 and X21 = X41 = X53 = 1 and k = 2 with an objective function
value of 19.28. When we have α = 0.8, we add 3.712k to the objective function. The
optimum solution is X11 = X22 = X33 = 1 and X41 = X53 = 1 and k = 3 with an objective function
value of 17.136.
k-centre problem
The k-centre problem is similar to the k-median problem where instead of minimizing the
total distance, we minimize the maximum distance. This problem is important when
emergency facilities are located. This problem is useful when we locate public systems. The
formulation is as follows
Minimize 𝑢
Subject to ∑𝑛𝑗=1 𝑋𝑗𝑗 = 𝑘
n
X
i 1
ij
1
X ij  X jj
𝑑𝑖𝑗 𝑋𝑖𝑗 ≤ 𝑢
Xij = 0,1
Illustration 4.16
Consider 5 customers located at (3,4), (5,2), (7,8), (1,5) and (6,6). Solve a 2 centre problem.
Create two locations such that each customer is attached to one location and the maximum
distance between the customers and location is minimized?
The distances are as follows: d12 = 4; d13 = 8; d14 = 3; d15 = 5; d23 = 8; d24 = 7; d25 = 5; d34 = 9;
d35 = 3 and d45 = 6. We use α = 1. The objective function becomes Minimize u such that u ≥
dijXij. We also have the constraint X11 + X22 + X33 + X44 + X55 + X66 = 2 and the constraints
n
X
i 1
ij
 1 and X ij  X jj
The optimum solution to the 2 centre problem is X11 = X33 = 1 and X21 = X41 = X53 = 1 with an
objective function value of 4. The three distances are 3, 4, and 4 and the maximum is 4.
Fixed charge problem – Location Allocation Problem
The fixed charge problem (Hirsch and Dantzig, 1954) locates facilities to potential sites to
meet customer demand. There are n demand points where dj is the demand from customer
j. There are m possible sites to locate facilities (factories). There is a fixed cost f i of locating
a facility in site i. There is a capacity Ki if a facility is located in site i. There is a demand dj in
point j and there is a transportation cost of Cij between i and j. The formulation is as follows:
Let Yi = 1 if a facility is located in site i.
Let Xij be the quantity transported from site ito customer (demand point) j.
The objective is to Minimize
Subject to
m
Y
i 1
i
p
n
X
j 1
ij
 K i Yi
ij
 dj
m
X
i 1
m
m
n
i 1
i 1 j 1
 f iYi   Cij X ij
Yi = 0,1 and Xij ≥ 0.
The objective function minimizes the sum of the fixed cost (of setting up the facilities) and
the transportation costs. The first constraint ensures that exactly p facilities are created. The
second constraint ensures that items can be transported only from facilities that are created
and that the total quantity leaving a facility is less than or equal to its capacity. The third
constraint ensures that the demand of all the customers is met.
It is not absolutely necessary to fix the number of facilities created. The formulation
otherwise will decide the correct number of facilities that minimizes total cost.
The above formulation is called the fixed charge problem or location-allocation problem and
is used widely in practice. It is also a well researched problem in the literature. This problem
is also a “hard” problem involving an exponential number of computations as the size
increases. Here it is assumed that fiYi and CijXij have the same units. The fixed cost is now
given as cost/period and the CijXij represent the transportation cost for meeting the demand
of that period.
4
1
2
5
1
2
7
3
3
6
8
Figure 4.2 – Network for Illustration 4.17
Illustration 4.17
_________________
Consider the network shown in Figure 4.2. The fixed costs of locating facilities in the three
potential locations are as follows:
1. Location 1
Rs 5000 000
2. Location 2
Rs 4000 000
3. Location 3
Rs 4500 000
The capacities of the three locations are 1000 000, 800 000 and 1250000. The demand at
the eight demand points are 200 000 for the first four points and 250000 for the remaining
points. The unit transportation costs given in Table 4.10
Table 4.10 – Unit transportation costs
D1
D2
D3
Location 1
4
5
5
Location 2
2.5
3.5
4.5
Location 3
2
4
5
D4
4
3
2.5
D5
4
2.2
2.6
D6
4.2
4
3.8
D7
3.3
2.6
2.9
D8
5
5
5.5
Solve the fixed charge problem assuming that all demand points are connected to all the
facilities. Also solve considering only those connections shown in Figure 2.2
Solution
The integer programming formulation would involve three variables Y1 to Y3 indicating
whether facility at i is selected or not. It would involve another 24 Xij variables indicating the
quantities transported from facility i to demand j. There are 3 supply constraints and eight
demand constraints. We assume that we do not fix the number of facilities created and let
the solution tell us the optimal number of facilities located.
The formulation is to Minimize 5000000Y1 + 4000000Y2 + 4500000Y3 + 4X11 + 5X12 + 5X13 +
4X14 + 4X15 + 4.2X16 + 3.3X17 + 5X18 + 2.5X21 + 3.5X22 + 4.5X23 + 3X24 + 2.2X25 + 4X26 + 2.6X27 +
5X28 +2X31 + 4X32 + 5X33 + 2.5X34 + 2.6X35 + 3.8X26 + 2.9X37 + 5.5X38
Subject to Y1 + Y2 + Y3 = 2
X11 + X12 + X13 + X14 + X15 + X16 + X17 + X18 ≤ 1000000Y1
X21 + X22 + X23 + X24 + X25 + X26 + X27 + X28 ≤ 800000Y2
X31 + X32 + X33 + X34 + X35 + X26 + X37 + X28 ≤1250000Y3
X11 + X21 + X31 ≥ 200000
X12 + X22 + X32 ≥ 200000
X13 + X23 + X33 ≥ 200000
X14 + X24 + X34 ≥ 200000
X15 + X25 + X35 ≥ 250000
X16 + X26 + X36 ≥ 250000
X17 + X27 + X37 ≥ 250000
X18 + X28 + X38 ≥ 250000
Yj = 0,1; Xij ≥ 0.
The optimal solution to the Integer programming formulation is Y2 = Y3 = 1 and X22 = X23 = X31
= X34 = 200, X28 = X36 = X37 = 250, X25 = 150 and X15 = 100. (The quantities are in thousands).
The total fixed cost is 9500000. The transportation cost is 6015000 and the total cost is Rs
15515000. Two facilities are to be located in sites 2 and 3.
An easy heuristic would be to choose the minimum number of facilities that can meet the
demands at minimum cost. This gives us Y1 = Y2 = 1 (supply of 1800000 to meet the total
demand of 1800000). The fixed cost is 9000000. Now it is enough to solve a transportation
problem to determine the quantities to be supplied to the eight demand points. The
optimum solution to the transportation problem is given by X13 = X21 = X22 = 200, X16 = X17 =
X18 = X25 = 250, X14 = 50, X24 = 150 with a transportation cost of 6525000. The total cost is
15525000.
So far we assumed that all the customers (demand points) can be served by all the plants.
From figure 2 we observe that certain customers cannot be served by certain plants. Plant 1
can handle the demands of customers 1, 2, 4, 5; Plant 2 can meet the demand of customers
4, 5, 7, 8 and Plant 3 can meet the demands of customers 2, 3, 5, 6, 7, 8. We therefore have
C13 = C16 = C17 = C18 = C21 = C22 = C23 = C26 = C31 = C34 = M where M is large and positive and
tends to infinity. Wherever the plant cannot meet the customer requirement we replace the
cost by M. The other alternative is to leave out those variables and solve a smaller sized
problem with fewer variables.
The optimal solution is given by Y1 = Y3 = 1 and X11 = X14 = X33 = 200; X12 = 150, X32 = 50, X35 =
X36 = X37 = X38 = 250. (The quantities are in thousands). The fixed cost is 10500000. The
transportation cost is 7250000 and the total cost is Rs 17750000. Two facilities at 1 and 3
are chosen.
A heuristic solution could work as follows:
From the customer plant incidence data, it becomes necessary to choose plants 1 and 3 to
meet the demand of customers 1 and 3. The available capacities are 800 and 1250
respectively (in thousands). Based on quantities, minimum transportation costs and
marginal increase, we can allot the quantities for customers 5, 6, 7, and 8 to plant 3. We
have to allot customer 3 to plant 3. Total demand of 1200000 is consumed. Customer 1 goes
to plant 1. The available 50000 in plant 3 goes to customer 4 because of the marginal
difference in costs. The rest of the demand is met from plant 1. This is the same solution
given by the integer programming formulation.
Model 3 – Fixed Charge with dedicated facilities
In the formulation in Model 2, we assumed that the demand for a customer can be met
from more than one created facility. In Illustration 2 we observe that the demand of
customer 5 was met by transporting 150 units from facility 2 and 100 from facility 1.
We may consider situations where all the demand of a customer may be met by a single
facility. The formulation becomes slightly different. The formulation where all the demand is
met from a single facility is as follows:
We now define
Xij = 1 if the entire demand of customer j is met from facility i. The objective function is to
Minimize
m
fY
i 1
i i
n
d
j 1
j
i 1
m
n
i 1
i 1 j 1
 fiYi   d jCij X ij
p
X ij  K i Yi
m
X
m
ij
1
Yi , X ij  0,1
Illustration 4.18
________________________________________________
Consider the data given in Figure 2.2 for Illustration 4.17. There is an additional condition
that all the demand from a customer has to be met from only one facility. Solve the problem
to identify sites to facilities to minimize total cost?
The formulation is to Minimize 5000000Y1 + 4000000Y2 + 4500000Y3 + 8000X11 + 10000X12 +
10000X13 + 8000X14 + 10000X15 + 10500X16 + 8250X1 7 + 12500X18 + 10000X21 + 7000X22 +
9000X23 + 6000X24 + 5500X25 + 10000X26 + 6400X27 +12500X28 +4000X31 + 8000X32 +
10000X33 + 5000X34 + 6400X35 + 9500X26 + 14500X37 + 27500X38
Subject to Y1 + Y2 + Y3 = 2
2000X11 + 2000X12 + 2000X13 +2000X14 + 2500X15 + 2500X16 + 2500X17 + 2500X18 ≤ 1000000Y1
2000X21 + 2000X22 + 2000X23 +2000X24 + 2500X25 + 2500X26 + 2500X27 + 2500X28 ≤ 800000Y2
2000X31 + 2000X32 + 2000X33 + 2000X34 + 2500X35 + 2500X26 + 2500X37 + 2500X28 ≤1250000Y3
X11 + X21 + X31 =1
X12 + X22 + X32 =1
X13 + X23 + X33 =1
X14 + X24 + X34 =1
X15 + X25 + X35 =1
X16 + X26 + X36 =1
X17 + X27 + X37 = 1
X18 + X28 + X38 =1
Yj = 0,1; Xij = 0,1.
The optimal solution to the integer programming problem is given by Y2 = Y3 = 1 and X22 =
X23 = X28 =1, X31 = X34 = X35 = X36 = X37 = 1. The fixed cost is 9500000. The transportation cost
is 6075000 and the total cost is Rs 15575000. Two facilities at 2 and 3 are chosen.
Fixed charge, budget constraints and unmet demand
In this problem we have a fixed charge associated with locating facilities in each node. We
also have a limit on the number of facilities. We are given the distances among the points
connected on the network. A located facility can serve customers who are within a certain
distance from the facility. There is a weight associated with each node which gives the
importance given to the customer. Due to all these constraints it may happen that some
customer demand may not be met. We minimize the weighted demand. This type of
problems happen when we locate public systems such as hospitals etc. The unmet demand
has to be met from other available facilities or from facilities to be created soon.
The formulation of this problem is as follows:
Yj = 1 if the facility is located; Xj = 1 if the demand from customer j is not met. There is a
fixed cost fj to locate a facility at j and there is a budget restriction of B. The weight
associated with customer j is wj. The objective is to Minimize ∑𝑛𝑗=1 𝑤𝑗 𝑋𝑗 subject to
𝑛
∑ 𝑓𝑗 𝑌𝑗 ≤ 𝐵
𝑗=1
𝑛
∑ 𝑌𝑗 ≤ 𝑘
𝑗=1
𝑛
∑ 𝑎𝑖𝑗 𝑌𝑗 + 𝑋𝑗 ≥ 1
𝑗=1
Xj, Yj = 0,1
Illustration 4.19
________________________________________________
Consider the data given below for a network with 6 nodes and 10 arcs. The arcs and the
distances (shown in brackets) are 1-2 (5), 1-3 (7), 1-4 (6), 2-3 (8), 2-6 (6), 3-4 (4), 3-5 (6), 3-6
(7), 4-5 (10), 5-6 (9). The fixed costs are 100, 80, 175, 60, 90 and 100 and the budget
restriction is 160. The weights for the nodes are 1, 3, 3, 2, 4, 2 respectively. There is a
distance restriction of 8 units. Solve the problem to minimize the weighted unmet demand?
We remove arcs 4-5 and 5-6 from the network because the distances are more than 8. We
now have only 8 arcs. The formulation is to
Minimize X1 + 3X2 + 3X3 +2X4 + 4X5+ 2X6
subject to
100Y1 + 80Y2 + 175Y3 + 60Y4 + 90Y5 + 100Y6 ≤ 160
Y1 + Y2 + Y 3 + Y4 + Y 5 + Y6 ≤ 2
X1 + Y1 + Y2 + Y3 + Y4 ≥ 1
X2 + Y2 + Y1 + Y3 + Y6 ≥ 1
X3 + Y3 + Y1 + Y2 + Y4 + Y5 + Y6 ≥ 1
X4 + Y4 + Y1 + Y3 + Y5 ≥ 1
X5 + Y 5 + Y 3 ≥ 1
X6 + Y6 + Y2 + Y3 ≥ 1
Xj, Yj = 0,1
The optimum solution to the problem is given by Y2 = Y4 = X5 = 1 with objective function
value = 3. Facilities are created at locations 2 and 4. The demand of location 5 is not met.
From the data we observe that location 3 is connected to all the other locations but its fixed
cost exceeds 160. Therefore we could not locate at 3. Points 1 and 3 are attaced to either 2
or 4 and point 6 is attached to point 4.
Supply Chain – Location and allocation in multiple stages
Let us consider a two stage location allocation problem. Here we assume that the product is
made in factories (or plants), transported to warehouses from which they are transported to
the customers. A typical 3 plant, 3 warehouse 4 customer supply chain is shown in Figure x.x
Each plant has a capacity of Pi. Each warehouse has a capacity of Aj. There is a fixed cost fi of
creating and operating plant i and a fixed cost gj of creating and operating warehouse j.
There is a unit transportation cost Cij between plant i and warehouse j and a unit
transportation cost of Cjk between warehouse j and customer k. the problem is to locate
plants and warehouses and transport the product such that the total cost of location and
allocation is minimized.
The mathematical programming formulation of the problem is as follows.
Let Yi = 1 if plant i is opened
Let Wj = 1 if warehouse j is opened.
Let Xij be the quantity of the product transported from plant i to warehouse j.
Let Tjk be the quantity of the product transported from warehouse j to customer k.
The objective function is to minimize the total cost of location and allocation. This is to
p
m
Minimize
p
p
n
 f Y   g W   C X   C
i 1
Subject to
m
i i
j 1
j
j
i 1 j 1
ij
ij
j 1 k 1
jk
Tjk
p
X
j 1
ij
 PY
i i i
ij
  T jk j
m
X
i 1
n
p
T
j 1
jk
k 1
 d k k
Yi , W j  0,1
X ij , T jk  0
Illustration 4.20 _______________________________________________________
Consider the two stage network given in Figure 4.3. The unit cost of transportation from the
plants to the warehouses is given in Table 4.11. The unit cost of transportation from the
warehouses to the customers is given in Table 4.12. The capacities of the three potential
plants are 3000, 2000 and 2600 units. The fixed costs for the plants are Rs 8000, Rs 7000
and Rs 9000 respectively. The capacities of the three potential warehouses are 2500, 2400
and 2000 units. The fixed costs for the warehouses are Rs 5000, Rs 6000 and Rs 4000
respectively. The demands at the customers are 1000, 800, 1200 and 900 units.
Solve a two stage location allocation model to minimize the sum of fixed costs and the
transportation costs?
1
1
1
2
2
2
3
3
3
Warehouses
Plants
4
Customers
Figure 4.3 – Network for Illustration 4.20
Table 4.11 – Transportation cost between plant and warehouse
W1
W2
W3
Plant1
Plant 2
Plant 3
4
3
4.2
5
3.6
5
4.5
4
4.5
Table 4.12 – Transportation cost between warehouse and customer
C1
Warehouse 1
Warehouse 2
Warehouse 3
2
4
2.4
C2
1.8
3.8
2
C3
2.2
3.2
2.3
C4
3
3.6
2
The mathematical programming formulation has 27 variables out of which six location
variables are binary and 21 allocation variables are continuous. The problem has 9
transportation variables from plants to warehouses and 12 transportation variables from
warehouses to customers. There are 13 constraints out of which three are plant capacity
constraints, three are warehouse capacity constraints, four are customer demand
constraints and three are flow constraints at the warehouses.
The optimal solution to the integer programming problem is given by Y1 = Y2 = 1; Z1 = Z3 = 1;
X11 = 500, X13 = 1400, X21 = 2000, Y11 = 1000, Y12 = 800, Y13 = 700, Y33 = 500, Y34 = 900. Plants
1 and 2 are opened and warehouses 1 and 3 are opened. The fixed costs are 15000 and
9000 respectively. The total transportation cost from plants to warehouses is 14300. The
total transportation cost from warehouses to customers is 7930. The total cost is 46230.
A heuristic solution would be to open plants 1 and 2 based on minimum fixed costs and to
open warehouses 1 and 3 based on minimum fixed costs. Solving the relevant
transportation problem would give us the same solution given by integer programming in
this case.
Supply Chain – Location and allocation in multiple stages and dedicated supply
In the earlier formulation, we assumed that the demand for a customer can be met from
more than one warehouse. In Illustration X we observe that the demand of customer 3 was
met by transporting 700 units from warehouse 1 and 500 units from warehouse 3.
We may consider situations where all the demand of a customer may be met from a single
warehouse. The formulation becomes slightly different. The formulation where all the
demand of a customer is met from a single warehouse is as follows:
Let Yi = 1 if plant i is opened
Let Wj = 1 if warehouse j is opened.
Let Tjk = 1 if the demand for customer k is met entirely from warehouse j.
Let Xij be the quantity of the product transported from plant i to warehouse j.
The objective function is to minimize the total cost of location and allocation. This is to
p
m
m
p
p
n
 f Y   g W   C X   d C
Minimize
i 1
i i
j 1
j
j
i 1 j 1
ij
ij
j 1 k 1
k
jk
T jk
Subject to
p
X
j 1
ij
 PY
i i i
ij
  d k T jk j
m
X
i 1
n
p
T
j 1
jk
k 1
 1 k
Yi , W j , T jk  0,1
X ij  0
Illustration 4.21________________________________________________________
Apply the model with dedicated supply to the network shown in Figure 4.3 and the data in
Tables 2.11 and 2.12?
The mathematical programming formulation has 27 variables out of which six location
variables and 12 allocation variables between warehouses and customers are binary and 9
allocation variables between plants and warehouses are continuous. The problem has 9
transportation variables from plants to warehouses and 12 transportation variables from
warehouses to customers. There are 13 constraints out of which three are plant capacity
constraints, three are warehouse capacity constraints, four are customer demand
constraints and three are flow constraints at the warehouses.
The optimal solution to the integer programming problem is given by
Y1 = Y2 = 1; Z1 = Z3 = 1; X13 = 1900, X21 = 2000, Y12 = Y13 = Y31 = Y34 = 1.
Plants 1 and 2 are opened and warehouses 1 and 3 are opened. The fixed costs are 15000
and 9000 respectively. The total transportation cost from plants to warehouses is 14550.
Customers 1 and 4 get all their demand from warehouse 3 and customers 2 and 3 get all
their demand from warehouse 1. The total transportation cost from warehouses to
customers is 8280. The total cost is 46830.
A heuristic solution would be to open plants 1 and 2 based on minimum fixed costs and to
open warehouses 1 and 3 based on minimum fixed costs. Solving the relevant
transportation problem would give us the same solution given by integer programming in
this case also.