OST: Chapter 2
Strategic Planning Problems Location
Content
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2.1 Location problems as strategic planning problems
2.2 Continuous single location planning in the plane
2.3 Median problem (single location)
2.4 Center problem (single location)
2.5 Warehouse location problem (WLP)
2.6 ADD and DROP for WLP
2.7 Branch and Bound (B&B) for WLB
2.8 p-median problem
2.9 Transportation problem
2.10 Capacitated WLP
2 2
2.1 Location Problems (Multi Stage)
Production site
PS1
PS2
PS3
PS4
transport
Central warehouses
Full truck load (FTL)
CW1
CW2
CW3
CW4
transport
Regional warehouses
…
FTL or tours
RW1
RW2
RW3
RW4
transport
Customers
…
…
LTL tours
C1
C2
C3
C4
…
3
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
More or less stages possible
Some stages (inventory holding and/or transportation) can
be delegated to logistics service providers

Transportation „less than truck load“ (LTL) will be covered
in course → Transportation Logistics

Location decisions:



Number of warehouses
Which warehouse locations?
Assignment of customers to warehouses
4
2.2 Continuous location in the plane
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
A single location in the
plane (with beeline
distance) can be
determined by an analog
model „Varignon frame“
(Varignonscher Apparat)
Not very relevant in
reality
5
2.3 Median Problem (Single Location)

Given:




Define



bj … demand of customer j
dij … distance from location i to customer j
Total delivery cost if location i is built is


Set of customers which must be serviced
Set of potential locations (often same as customer locations)
One of these locations must be selected
σ(i) = ∑dijbj → min
Can be solved by complete enumeration
6
Example: Median
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Example from Domschke und Drexl
(Logistik: Standorte, 1990, Kapitel 3.3.1)
Numbers in table are already the cij = dijbj
Which location should be built (just one)
i\j
1
2
3
4
5
6
7
1
1
2
10
9
6
7
3
2
2
9
0
7
3
6
10
3
7
6
1
5
3
10
5
4
6
5
10
2
6
3
6
5
6
4
6
3
7
2
6
7
Example: Median (continued)


Total delivery cost if location i is built is ci
(sum in row)
i\j
1
2
3
4
5
6
7
ci
1
1
2
10
9
6
7
3
38
2
2
9
0
7
3
6
10
37
3
7
6
1
5
3
10
5
37
4
6
5
10
2
6
3
6
38
5
6
4
6
3
7
2
6
34
Minimum is c5=34 if location i=5 is built (median)
8
2.4 Center Problem (Single Location)


A related problem is the center problem – “Fairness”
cij is now interpreted as disutility of customer j visiting
facility i



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
Set of customers which must be serviced
Set of potential locations
One of these locations must be selected
Minimize the maximum disutility! (not total cost!)
Can be solved by complete enumeration


Similar to with median
Replace „sum“ by „max“ operator
9
Example: Center
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
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Example adapted from Domschke und Drexl
(Logistik: Standorte, 1990, Kapitel 3.3.1)
Numbers in table are the disutilities cij
Which location should be built (just one)
i\j
1
2
3
4
5
6
7
1
1
2
10
9
6
7
3
2
2
9
0
7
3
6
10
3
7
6
1
5
3
10
5
4
6
5
10
2
6
3
6
5
6
4
6
3
7
2
6
10
Example: Center (continued)
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
Maximum disutility if i is built is i
(sum in row)
i\j
1
2
3
4
5
6
7
i
1
1
2
10
9
6
7
3
10
2
2
9
0
7
3
6
10
10
3
7
6
1
5
3
10
5
10
4
6
5
10
2
6
3
6
10
5
6
4
6
3
7
2
6
7
Minimum is 5 =7 if location i=5 is built (center)
11
2.5 Multiple Locations:
Warehouse Location Problem (WLP)

Here only single layer WLP (facility location problem FLP)
L1
K1
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
m
L2
K2
K3
K4
n
Assume at the moment that the capacities are unlimited
Each location has a certain fixed cost
How many and which locations should be built?
Minimize Total cost = transportation cost + fixed cost
12
Example: from Domschke und Drexl
(Logistik: Standorte, 1990, Kapitel 3.3.1)

Same example as before



Just fixed costs fi are introduced for each location i
As before, numbers cij in table are delivery cost, if customer j
gets all his demand from location i
i\j
1
2
3
4
5
6
7
fi
1
1
2
10
9
6
7
3
5
2
2
9
0
7
3
6
10
7
3
7
6
1
5
3
10
5
5
4
6
5
10
2
6
3
6
6
5
6
4
6
3
7
2
6
5
Which locations should be built (and how many)?
13
Example WLP (continued)
two typical solutions
Solution 2: just locations 1 and 3 are built
Solution 1: all locations are built
i\j
1
2
3
4
5
6
7
fi
i\j
1
2
3
4
5
6
7
fi
1
1
2
10
9
6
7
3
5
1
1
2
10
9
6
7
3
5
2
2
9
0
7
3
6
10
7
3
7
6
1
5
3
10
5
5
3
7
6
1
5
3
10
5
5
4
6
5
10
2
6
3
6
6
5
6
4
6
3
7
2
6
5
Fixed costs = 5+7+5+6+5 = 28
Fixed costs = 5+5 = 10
Transp. Cost = 1+2+0+2+3+2+3 = 13
Transp. Cost = 1+2+1+5+3+7+3 = 22
Total Cost = 28 + 13 = 41
Total Cost = 10 + 22 = 32

Here, solution 2 is better. But what is the optimal solution?
14
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Complete enumeration of all possible configurations<



2 possibilities for each location  exponential:
2m-1 possble solutions (if m=10 → 1023 solutions)
Formulation as mixed integer LP-Model (MIP)

yi … binary variable für i = 1, …, m
yi = 1 if a warehouse is built at potential location i
yi = 0 otherwise

xij … real „assignment“ oder transportation variable
for i = 1, …,m and j = 1, …, n:
xij = fraction of demand of customer j coming from
location i.
15
MIP Formulation for the Uncapacitated WLP
Transportation cost +
fixed cost
Only locations i which are
built (yi=1) can be used for
delivery (xij > 0)
Total demand of customer j
must be satisfied
m
n
m
Z ( x, y)   cijxij   fiyi  min
i 1 j 1
xij ≤ yi
m
x
i 1
for i = 1, …, m
and j = 1, …,n
1
for j = 1, …,n
yi {0,1}
for i = 1, …, m
for all i and j
ij
i 1
yi is binary
and xij non-negative
xij  0
16
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Complexity problem:
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m*n real variables und m binary variables
→ for about 100 and more potential locations exact solution
extremely time consuming / impossible → heuristics
Different types of heuristics:

Construction heuristics (determine a feasible starting solution)
e.g. ADD and DROP in what follows

Improvement heuristics (improve a starting solution using
exchenges, etc.)
17
2.6 Heuristics for the WLP
ADD Heuristic

Notation used
I:={1,…,m} set of all possible locations
I0
Iovl
I1
i
Z
set of finally closed locations (yi fixed to 0)
set of preliminarily closed locations, „don‘t know“
set of finally open locations (yi fixed to 1)
reduction in transportation cost, if i is opened in addition to current
ones
total cost (objective)
18

Initialization:

Determine which location should be built first (best if just one location)
 row sum ci := ∑cij of cost matrix

determine location k with minimal total cost ck + fk
 Set I1 = {k}, Iovl = I – {k} and Z = ck + fk

For all i in Iovl determine reduction in transp cost ωij = max {ckj – cij, 0}
for all customers j and row sum ωi.

Example: first location k=5 with Z:= c5 + f5 = 39, I1 = {5}, Iovl = {1,2,3,4}
similar to mewdian problem (just add fixed costs)
19
i\j
1
2
3
4
5
6
7
fi
ci
fi + c i
1
1
2
10
9
6
7
3
5
38
43
2
2
9
0
7
3
6
10
7
37
44
3
7
6
1
5
3
10
5
5
37
42
4
6
5
10
2
6
3
6
6
38
44
5
6
4
6
3
7
2
6
5
34
39
i\j
1
2
1
5
2
2
4
3
4
3
4
5
1
6
4
5
4
1
1
6
7
ωi
fi
3
11
5
14
7
10
5
2
6
1
 matrix: ωij is the reduction in
transportation / delivery cost for
ciustomer j if location i is built in
addition to the already open
locations.
 Row sum ωi is the total reduction
in transportation / delivery cost , by
additional location i .
20

Iteration:

in each iteration exactly one location out of the „don‘t know“ set Iovl is finally built.
It is the one with the maximam saving in total cost

Find potential location in Iovl, for which (reduction in transport cost minus
additional fixed cost) ωk – fk is maximal
I 1  I 1  {k},

Iovl = Iovl – {k} and Z = Z – ωk + fk
Furthermore, all those locations can be finally closed which where the reduction
in transport cost is smaller than the additional fixed cost →
Close all i  I 0vl with ωi ≤ fi finally:

I 0  I 0  {i} and I 0vl  I 0vl  {i}
For all i in Iovl and all customers j compute the updated reduction in transport
cost :
ωij = max {ωij - ωkj, 0}
21

Termination condition:
The procedure stops, as soon as all additional locations from Iovl would lead to a
cost increase if built. i.e. Iovl = { }.
 Bulid all locations from I1.
 Total cost Z
cij  min { chj / h  I 1}
 Optimal assignment: xij = 1 if


Example: Iteration 1
i\j
1
2
1
5
2
2
4
3
4
3
4
5
1
6
4
5
4
1
1
6
7
ωi
fi
3
11
5
14
7
10
5
2
6
1
Bild k = 2
Close i = 4
Because ω4 < f4 location 4 can be finally closed. Location k=2 is finally open.
 Now total cost Z=39 – 7 = 32 and Iovl = {1,3}, I1 = {2,5}, Io = {4}.
 Determine new cost reductions ωij.

22
i\j
1
2
1
1
2
3

3
4
5
6
7
ωi
fi
3
6
5
Build k = 1
1
1
5
Close i = 3
Iteration 2:

the above table shows the max possible reduction in transportation cost for the
don‘t know locations 1 und 3.
 Location 1 is finally built. Location 3 is finally closed.

Final result:
The ADD algorithm stops with solution I1 = {1,2,5}, Io = {3,4}, Iovl = { },
and Z = 32 – 1 = 31
 Locations 1, 2 and 5 are built
 Customers {1,2,7} are assigned to location 1, {3,5} to location 2, and {4,6} to
location 5. Total cost is Z = 31.

23
DROP Heuristic

The don‘t know set is now I1vl:
 I1vl … set of preliminarily open locations (yi = 1 currently, but can still change)

The Drop Algorithm works the other way round compared to ADD, i.e. it starts with all
alle potential locations to be open (preliminarily).

Initialisation: I1vl = I, I0 = I1 = { }

Iteration
 In each iteration one location out of I1vl is finally closed. It is the one where the
closing causes the highest cost reduction..
 If closing a location from I1vl would increase total cost, it can be finally built, i.e.
move to set I1
24

For efficient implementation 4 additional lines are added to transp cost matrix C:
 For each column j = 1, …, n, rows m+1 and m+2 of C hold the smallest (c h1j) and
second smallest (ch2j) element of this column. Onlky rows which are not finally
deleted are considered → i I 0
 In rows m+3 and m+4 we put the row numbers, where the smallest (h1) and second
smallest (h2) cost elements occur. If location h1 (from I1vl) is closed, the
transportation cost increase for customer j by ch2j - ch1j

Example: Initialisation and Iteration 1: I1vl ={1,2,3,4,5}
i\j
1
2
3
4
5
6
7
δi
fi
1
1
2
10
9
6
7
3
5
5
build
2
2
9
0
7
3
6
10
1
7
close
3
7
6
1
5
3
10
5
0
5
4
6
5
10
2
6
3
6
1
6
5
6
4
6
3
7
2
6
1
5
ch1j
6
1
2
0
2
3
2
3
ch2j
7
2
4
1
3
3
3
5
h1
8
1
1
2
4
2
5
1
h2
9
2
5
3
5
3
4
3
25

For all i from I1vl compute increase in transportation cost δi if I is finally dropped.
δi is sum of differences between smallest and second smallest cost element in
rows where i = h1 contains the smallest element.

2 examples:




δ1 = (c21 – c11) + (c52 – c12) + (c37 – c17) = 5
δ2 = (c33 – c23) + (c35 – c25) = 1
If fixed costs savings fi exceed additional transportation cost δi, finally drop i. In
Iteration 1 location 1 is finally built.
Iteration 2:

I1vl = {3,4,5}, I1 = {1}, I0 = {2}

Omit row 2 because finally dropped. Update remaining 4 rows, where changes
are only possible where smallest or second smallest element occurred

Keep row 1 since I1 = {1}, but 1 is no candidate for dropping. Hence do not
compute δi there.
26
i\j
1
2
3
4
5
6
7
δi
fi
1
1
2
10
9
6
7
3
-
-
3
7
6
1
5
3
10
5
8
5
build
4
6
5
10
2
6
3
6
1
6
close
5
6
4
6
3
7
2
6
1
5
ch1j
1
2
1
2
3
2
3
ch2j
6
4
6
3
6
3
5
h1
1
1
3
4
3
5
1
h2
4
5
5
5
1
4
3
Location 3 is finally built, location 4 finally dropped.
27

Iteration 3:
I1vl = {5}, I1 = {1,3}, I0 = {2,4}

i\j
1
2
3
4
5
6
7
δi
fi
1
1
2
10
9
6
7
3
-
-
3
7
6
1
5
3
10
5
-
-
5
6
4
6
3
7
2
6
7
5
ch1j
1
2
1
3
3
2
3
ch2j
6
4
6
5
6
7
5
h1
1
1
3
5
3
5
1
h2
5
5
5
3
1
1
3
build
Location 5 is finally built, since closing would only save fixed cost of f5 = 5 while transp cost
would increase by δ5 = 7.
28

Result:

Build locations I1 = {1,3,5}

Deliver customers {1,2,7} from 1, customers {3,5} from 3, and customers {4,6}
from 5.

Total cost Z = 30 (slightly better than ADD – can be the other way round)
29
Improvement for WLP

In each iteration you can do:

Replace a built location (from I1) by a forbidden location (from I0).
Choose first improvement of best improvement

Using rules of DROP-Algorithm delete 1 or more locations, so that cost decrease
most (or increase least) and then apply ADD as long as cost savings are
possible.

Using rules of ADD-Algorithm add 1 or more locations, so that cost decrease
most (or increase least) and then apply DROP as long as cost savings are
possible.
30
2.7 Branch and Bound for WLB




Branch and Bound is an exact solution procedure for
(mixed) integer programming problems (MIPs)
It is e.g. used by general MIP solvers like XPRESS,
CLPEX, GUROBI, GLPK, etc. to solve MIPs
For various combinatorial optimization problems (e.g. in
logistics) tailored B&B algorithms are available
(independent of MIP formulation)
Basic idea:



Starting from a root node construct a tree by branching
Constructing the whole tree would mean complete enumeration
Cutting away whole branches (bounding) is essential
31
Example Thonemann (2015)


A firm has to deliver to 4 customers and has 3 potential
locations for warehouses
Given dij … unit cost i→j, bj … demand, fi … fixed cost
Unit cost of
transport dij

Customer j
Warehouse i
1
2
3
4
Fixed
cost fi
1
0,83
0,69
0,86
0,23
85
2
1,23
1,05
0,39
0,43
60
3
0,52
1,13
0,68
2,03
65
Demand bj
42
39
51
45
Multiply distance dij and demand bj demand  cij = dij * bj
32
Example continued 1

Table shows now the cij i.e. the distribution cost if
customer j gets all his demand from location i
total cost of
transport cij
Warehouse i

Customer j
Fixed
cost fi
1
1
39
2
31
3
49
4
15
2
56
45
25
24
60
3
26
48
40
96
65
85
Determine which locations to choose
33
Example: Notation

xxx … state of the warehouses: possible values x=0/1/?





111 … all are built
01? … 1 is closed, 2 is built, 3 don’t know
01? … would correspond to I1 = {2}, I0 = {1}, I1/0vl = {3},
Root node ??? represents the whole set of possible
solutions (no decisions yet)
In each branching step,


one chooses one node still having some ?
this node is split in two sub-nodes (children, subsets):
 choose a ?
 put 1 in one child and 0 in the other
34
Example: B&B Tree

Complete enumeration of the 2n = 8 solutions gives
complete tree
???
0??
1??
11?
111



110
01?
10?
101
100
011
010
00?
001
000
In each level one more decision is fixed (000 infeasible)
At the “leaves”, all decisions made  complete solutions
B&B avoids building the whole tree  bounding
35
Example: B&B Bounding


In each node a lower bound for the optimal total cost is
computed: LB = LBfix = LBtr
LBfix = treat all ? locations as closed = 0, e.g.





LBfix(???) = LBfix(000) = 0,
LBfix(0??) = LBfix(000) = 0,
LBfix(1??) = LBfix(100) = 85,
LBfix(101) = 85 + 65 = 150
cij
Customer j
Fixed
Warehouse
i
1
1
2
3
4
cost fi
39
31
49
15
85
2
56
45
25
24
60
3
26
48
40
96
65
LBtr = treat all ? locations as open = 1

LBtr(???) = LBtr(111) = 26 + 31 + 25 + 15 = 97
LBtr(0??) = LBtr(011) = 26 + 45 + 25 + 24 = 120
LBtr(1??) = LBtr(111) = 26 + 31 + 25 + 15 = 97
LBtr(101) = 26 + 31 + 40 + 15 = 112
 LB = 97
 LB = 120
 LB = 182
 LB = 262
36
Example: B&B Bounding





If many ?  LB looks promising (low value) but LB is poor
approximation of real obj value (total cost)
The leaves contain just 1 solution and LB = real obj value
LB is computed for each node
A single upper bound UB for the optimal total cost of the
complete problem is kept → best known solution e.g. from
heuristic
E.g. choose solution 100
total cost of
Customer j



Fixed cost = 85
Transp cost = 134
 UB = 219
transport cij
Warehouse i
Fixed
cost fi
1
1
39
2
31
3
49
4
15
2
56
45
25
24
60
3
26
48
40
96
65
85
37
Example: B&B




???
97
UB = 219
0??
1??
Root ???, LB(???) = 97
120
182
1st branching:
01?
00?
11?
10?
180
210
242
197
??? → 1??, 0??
LB(1??) = 182  node alive (node can contain opt sol)
LB(0??) = 120  node alive (node can contain opt sol)
Both LB < UB  nodes alive
total cost of
2nd branching on 1??
Customer j
transport c
LB(11?) = 242>UB  delete Warehouse i 1
2
3
4
39
31
49
15
1
LB(10?) = 197  node alive
56
45
25
24
2
rd
3 branching on 0??
26
48
40
96
3
ij

38
Fixed
cost fi
85
60
65
Example: B&B continued






After 3 branchings: UB = 219
4th branching on 10?
LB*(101) = 262  delete
LB*(100) = 219  delete
At leaves LB* is not only LB
but solution with this value
5th branching on 01?
LB*(011) = 245  delete
LB*(010) = 210 < UB
New UB = 210
00? only contains 001
LB*(001) = 275  delete
???
97
0??
120
1??
182
11?
242
01?
180
10?
197
101
262
100
219
total cost of
transport cij
Warehouse i
011
245
00?
210
010
210
001
275
000

Customer j
Fixed
cost fi
1
1
39
2
31
3
49
4
15
2
56
45
25
24
60
3
26
48
40
96
65
85
39
UB by ADD

Heuristic ADD for upper bound: 010 with cost 210
total cost of
transport cij
Customer j
Transp
Fixed
total
Warehouse i
1
2
3
4
cost
cost fi
cost
1
39
31
49
15
85
219
2
56
45
25
24
60
210
3
26
48
40
96
134
150
210
Tran c
reduct
40
30
65
275
1
17
14
-
9
3
30
-
-
-
Fixed c
85
delete
65
delete
4040
B&B with UB from ADD


In case the better UB = 210 from ADD is used, the tree
gets smaller: 00? can be fathomed
For B&B is important




good starting heuristic /UB
good LB for the nodes
rule where to branch
Branching rule



Here binary sorting
Heuristics useful
E.g. 2→3→1 from ADD
???
97
0??
120
1??
182
11?
242
01?
180
10?
197
101
262
100
219
011
245
00?
210
010
210
4141
2.8 P-Median

Number of facilities is fixed … p
Typically fixed costs are not needed (but can be
considered if not uniform)

Exact solution:



MIP with MIP solver or B&B
Heuristics


ADD, DROP
Lagrangean relaxation, …
42
MIP for p-Median
transportation cost
+ fixed cost
m
Z ( x, y)   cijxij  min
i 1 j 1
Delivery only from
locations i that are built
xij ≤ yi
m
Satisfy total demand of
customer j
yi is binary
xij non negative
x
ij
1
i = 1, …, m
j = 1, …,n
j = 1, …,n
i 1
yi {0,1}
xij  0
m
Exactly p facilities
n
y
i 1
i
i = 1, …, m
For all i and j
p
43
2.9 Transportation Problem

In order to evaluate the transportation cost for a capacitated version of the WLP, the
solution of the so called „transportation problem“ is needed.
A set of customers has to be serviced, i.e. their demand must be fulfilled
A set of locations (suppliers) is available, and all their supply must be used
Total demand = total supply
Transportation cost per unit from each location to each customer is given
Find the optimal delivery plan where the total transportation cost is minimized

Example with 3 locations and 4 customers:





Verkaufsstellen
Fabrik
V1
V2
V3
V4
Produktion
F1
10
5
6
11
25
F2
2
2
7
4
25
F3
9
1
4
8
50
Nachfrage
15
20
30
35
100
44


General formulation:

m suppliers with capacity/supply
si, i = 1, …, m

n customers with demand
dj, j = 1, …, n

Unit transportation cost cij from i to j,
i = 1, …, m; j = 1, …, n
LP-Formulation

Define variables: xij transported quantity from i to j
Transport cost
m n
K
  cij xij  min
i 1 j 1
n
Supply
si 
Demand
dj 
i = 1, …, m
 xij
j = 1, …, n
j 1
m
i 1
xij  0
Non-negativity
Demand = supply
 xij
m
n
s   d
i
i =1
j 1
j
i = 1, …, m; j = 1, …, n
M
Otherwise contradiction
45

Transportation problem in the above example::

K = (10x11+5x12+6x13+11x14) + (x21+2x22+7x23+4x24) + (9x31+x32+4x33+8x34)  min

Supply constraints:
 x11 + x12 + x13 + x14




x21 + x22 + x23 + x24
x31 + x3 2+ x33 + x34
Demand constraints :
 x11
+ x21
+ x31

x12
+ x22
+ x32

x13
+ x23
+ x33

x14
+ x24
+ x34
= 25
= 25
= 50
(i=1)
(i=2)
(i=3) :
= 15
= 20
= 30
= 35
(j=1)
(j=2)
(j=3)
(j=4)
Non-negativity:
 xij  0 für i = 1, … , 3; j = 1, … , 4
46

Solution:

Simplex-Method → not very efficient if the whole matrix is used

General structure of coefficient matrix:
1
...
1
1
...
1
...
1
1
1
.
...
.
.
1
.
.
1
1
.
.
.
...
.
1
1

In each column exactly 2 of the m + n elemants are ≠ 0
 2-phase (or big M) simplex method applicable but inefficient

Efficient solution (course Management 2):


Use starting heuristic such as Vogel approximation Use modi (stepping
stone) to find optimal solution
Exactly same computations as simplex method but more efficient notation
47
2.10 Capacitated WLP

The capacitated (single stage) WLP is different from the
uncapacitated WLP only in the following respect:





The potential locations i = 1, ..., m have a maximum capacity of
s1, ..., sm quantity units (per unit time)
The transportation cost cij from i to j are now per unit cost
The demand quantities of the customers are explicitly given: d1, ..., dn
xij is now the quantity that is sent from location i to customer j
(per unit time)
Because of capacity constraints, 2 things can/will happen, that were
not possible for uncapacitated WLPs:

Some customers receive quantities from more than one supplier
 Some customers are not (only) serviced from the closest supplier

48
LP – Formulation
Transportation cost +
Fixed cost
Only if location i is built, it can be used
for delivery; Only supply quantity can be
used for delivery
The quantity that goes from i to j cannot
exceed demand
m
Z ( x, y ) 
n
m
 c x   f y  min .
ij
i 1 j 1
ij
i
i
i 1
n
x
ij
 si y i
for i = 1,…,m
j 1
xij  d j yi
for i = 1,…,m
for j = 1,…,n
n
Total demand must be satisfied
x
ij
 dj
for j = 1,…,n
i 1
yi is binary and xij is non-negative
yi {0;1}
xij  0
for i = 1,…,m
for all i and j
49
ADD and DROP for the capacitated WLP

Basic idea same as for uncapacitated problems:



ADD → start with 1 location and iteratively add locations
DROP → start with all locations open and iteratively delete locations
Different


Each time a configuration (which locations are open/closed) is
evaluated, a transportation problem has to be solved
In order to overcome imbalances between supply and demand a dummy
node must be introduced:


If total demand < total supply, a dummy customer is introduced. His demand
is set to (total supply) – (total demand). Quantities „transported to“ this
dummy customer mean unused capacity
If total demand > total supply, a dummy supplyer is introduced. His supply is
set to (total demand) – (total supply). Quantities „transported from“ this
dummy supplyer mean unsatisfied demand  penalize with M
50

Example: (DROP für capacitated WLP):

Given are 4 potential locations with capacities 20, 20, 10, and 10
and 5 customers with demand values 8, 9, 10, and 11.
Total capacity = 60 > total demand = 38
We introduce a dummy customer with demand 60 – 38 = 22
Determine which locations should be opened and which quantities should
go from which location to which customer, so that
total transportation cost + fixed cost is minimized
all demand is satisfied, and no capacity is exceeded



i\j
1
2
3
4
5
si
fi
1
8
3
5
4
0
20
10
2
1
2
3
4
0
20
10
3
6
5
7
3
0
10
7
4
8
4
7
5
0
10
7
dj
8
9
10
11
22
60
51

Example continued

Before we solve this example by DROP, we observe that cost matrices of
many logistical problems (e.g. TP) can be reduced without changing the
optimal solution
If an arbitrary number is added to each element of some row (or column)
then the optimal solution does not change. Just the objective value changes
by this constant times si (or dj) – „reduction constant“



i\j
1
2
3
4
5
si
fi
i\j
1
2
3
4
5
si
fi
1
8
3
5
4
0
20
10
1
7
1
2
1
0
20
10
2
1
2
3
4
0
20
10
2
0
0
0
1
0
20
10
3
6
5
7
3
0
10
7
3
5
3
4
0
0
10
7
4
8
4
7
5
0
10
7
4
7
2
4
2
0
10
7
dj
8
9
10
11
22
60
dj
8
9
10
11
22
60
Subtract the smallest cij from each column
 reduction constant = 8*1 + 9*2 + 10*3 + 11*3 + 22*0 = 89
52

Starting solution with DROP:
all locations are open
i\j
1



Transportation cost =
89 + 7*1 + 1*1 = 97
Fixed cost = 10+10+7+7 = 34
Total cost = 131
2
3
4





Iteration: check whether some
location i should be closed
 save fixed cost fi
 Increase in transportation cost
evaluated by solving a TP
Transportation cost =
89 + 7*2 + 1*2 = 105
Fixed cost = 10+7+7 = 24
Total cost = 129
Improvement by 2
 candidate for dropping
1
7
0
2
1
8
0
7
2
3
2
0
4
1
10
1
5
3
4
0
7
2
4
2
dj
8
9
1
5
0
12
0
10
0
0
10
si
fi
20
10
20
10
10
7
10
7
10
11
22
60
3
4
5
si
fi
20
10
10
7
10
7
Drop location 1?
i\j
2
3
4
dj
1
0
8
2
0
5
3
7
2
8
2
7
9
0
10
1
4
0
4
2
10
0
10
1
11
0
0
2
2
40
53
Drop location 2?
i\j
1
3
4
1
7
5
1
8
7
dj
2
9
3
2
10
4
1
3
4
0
2
4
2
8
9
1
2
8
5
0
0
si
fi

20
10

10
7


0
2
10
7
10
11
2
40
3
4
5
si
fi
20
10

20
10

Transportation cost = 89 +
9*1+10*2+1*1+8*5 +8*2 = 175
Fixed cost = 10+7+7 = 24
Total cost = 199
Cost increase!
 fix as open
Drop location 3?
i\j
1
2
4
dj
1
7
0
2
1
8
7
0
7
2
2
8
2
0
1
10
4
9
10
11
0
1
0
2
0
11
2


10
12
10
7
Transportation cost =
89 + 7*1+11*1 = 107
Fixed cost = 10+10+7 = 27
Total cost = 134
Cost increase!
 fix as open
50
54
Drop location 4?
i\j
1
2
4
dj

1
2
7
1
0
0
8
5
7
2
3
8
3
2
1
0
1
10
4
9
4
0
10
1
5
0
12
0
10
11
0
12
si
fi
20
10
20
10
10
7




Transportation cost =
89 + 7*1 + 7 = 97
Fixed cost = 10+10+7 = 27
Total cost = 124
Improvement by 7
 candidate for dropping
50
Result of iteration 2:

Fix locations 2 and 3 as open

Fix location 4 as closed

I0 = {4}, I1 = {2,3} und I1vl = {1}.

Location 1 cannot be deleted, since otherwise there is not enough
capacity to serve all customers
  DROP finished with I1 = {1,2,3} → Total cost = 124
55