1. Strategic Problems: Location
Chapter 1
Strategic Problems: Location

2. Location problems … PS1 PS2 PS3 PS4 Production site Full truck load
transportation …
CW1
CW2
Central warehouse
FTL or tours
transportation …
DC1
DC2
DC3
DC4
Distribution centers
tours
transportation … C1 C2 C3 C4
customers
(c) Prof. Richard F. Hartl
QEM - Chapter 1

3. More levels possible (regional warehouses)
Can be delegated to logistics service providers
Decision problems
Number and types of warehouses
Location of warehouses
Transportation problem (assignment of customers)
(c) Prof. Richard F. Hartl
QEM - Chapter 1

4. Median Problem Simplest location problem Definition: Median
Represent in complete graph. Nodes i are customers with weights bi
Choose one node as location of warehouse
Minimize total weighted distance from warehouse
Definition: Median
directed graph (one way streets…):
σ(i) = ∑dijbj → min.
undirected graph:
σout(i) = ∑dijbj → min … out median
σin(i) = ∑djibj → min … in median
(c) Prof. Richard F. Hartl
QEM - Chapter 1

5. Example: fromDomschke und Drexl (Logistik: Standorte, 1990, Kapitel 32 3 4 5
2/0
1/4
3/2
4/3
5/1
6/2
weight D= 12 2 10 6 3 7 5 8 4 9 11 b= 4 2 3 1
Distance between
locations
(c) Prof. Richard F. Hartl
QEM - Chapter 1

6. Example: Median City 4 is median since 35+74 = 109 minimal
i\j 1 2 3 4 5 6
i/j 1 2 3 4 5 6
σout(i) 1
4*0
0*12
2*2
3*10
6*1
2*12 64 1 3 4 5 6 2
σin(i) 66
2*12
3*4
1*8
2*11
0*2
4*0 98
2*10
3*2
1*6
2*9
0*0
4*12 56
2*0
3*5
1*9
2*12
0*3
4*2 74
2*8
3*0
1*4
2*7
0*3
4*10 53 8 15 6 24 80 20 6 48 2
4*2
0*0
2*3
3*3
7*1
2*5 40 3
4*12
0*10
2*0
3*8
4*1
2*10 96 4
4*4
0*2
2*5
3*0
5*1
2*2 35 5
4*8
0*6
2*9
3*4
0*1
2*6 74 6
4*11
0*9
2*12
3*7
3*1
2*0 92
City 4 is median since = 109 minimal
OUT e.g: emergency delivery of goods
IN e.g.: collection of hazardous waste
(c) Prof. Richard F. Hartl
QEM - Chapter 1

7. Related Problem: Center
Median
Node with min total weighted distance
→ min.
Center
Node with min Maximum (weighted) Distance
(c) Prof. Richard F. Hartl
QEM - Chapter 1

8. Solution City1 is center since 30+24 = 54 minimal i\j 1 2 3 4 5 6 i/j
out(i) 1
4*0
0*12
2*2
3*10
6*1
2*12 30 1 3 4 5 6 2
 in(i) 24
2*12
3*4
1*8
2*11
0*2
4*0 48
2*10
3*2
1*6
2*9
0*0
4*12 24
2*0
3*5
1*9
2*12
0*3
4*2 40
2*8
3*0
1*4
2*7
0*3
4*10 24 8 15 6 48 20 6 2
4*2
0*0
2*3
3*3
7*1
2*5 10 3
4*12
0*10
2*0
3*8
4*1
2*10 48 4
4*4
0*2
2*5
3*0
5*1
2*2 16 5
4*8
0*6
2*9
3*4
0*1
2*6 32 6
4*11
0*9
2*12
3*7
3*1
2*0 44
City1 is center since = 54 minimal
(c) Prof. Richard F. Hartl
QEM - Chapter 1

9. Uncapacitated (single-stage) Warehouse Location Problem – LP-Formulation
single-stage WLP:
warehouse
customer: W1 W2 m C1 C2 C3 C4 n
Deliver goods to n customers
each customer has given demand
Exist: m potential warehouse locations
Warhouse in location i causes fixed costs fi
Transportation costs i  j are cij if total demand of j comes from i.
(c) Prof. Richard F. Hartl
QEM - Chapter 1

10. Problem:
How many warehouses? (many/few  high/low fixed costs, low/high transportation costs
Where?
Goal:
Satisfy all demand
minimize total cost (fixed + transportation)
transportation to warehouses is ignored
(c) Prof. Richard F. Hartl
QEM - Chapter 1

11. Example: from Domschke & Drexl (Logistik: Standorte, 1990, Kapitel 3.3.1)
Solution 1: all warehouses
Solution 2: just warehouses 1 and 3
i\j 1 2 3 4 5 6 7 fi 10 9
i\j 1 2 3 4 5 6 7 fi 10 9
Fixed costs = = 28 high
Transp. costs = = 13
Total costs = = 41
Fixed costs = 5+5 = 10
Transp. costs = = 22
Total costs = = 32
(c) Prof. Richard F. Hartl
QEM - Chapter 1

12. when locations are decided:
transportation cost easy (closest location)
Problem: 2m-1 possibilities (exp…)
Formulation as LP (MIP)
yi … Binary variable for i = 1, …, m:
yi = 1 if location i is chosen for warehouse
0 otherwise
xij … real „assignment“ oder transportation variable für i = 1, …,m and j = 1, …, n:
xij = fraction of demand of customer j devivered from location i.
(c) Prof. Richard F. Hartl
QEM - Chapter 1

13. MIP for WLP xij ≤ yi j = 1, …,n transportation cost + fixed cost
Delivery only from locations i that are built
xij ≤ yi
i = 1, …, m
j = 1, …,n
Satisfy total demand of customer j
j = 1, …,n
i = 1, …, m
For all i and j
yi is binary
xij non negative
(c) Prof. Richard F. Hartl
QEM - Chapter 1

14. Problem:
m*n real Variablen und m binary → for a few 100 potential locations exact solution difficult → Heuristics
Heuristics:
Construction or Start heuristics (find initial feasible solution)
Add
Drop
Improvement heuristics (improve starting or incumbent solution)
(c) Prof. Richard F. Hartl
QEM - Chapter 1

15. ADD for WLP Notation: I:={1,…,m} set of all potential locations
I0 set of (finally) forbidden locations (yi = 0 fixed)
Iovl set of preliminary forbidden locations (yi = 0 tentaitively)
I1 set of included (built, realized) locations (yi =1 fixed)
reduction in transportation cost, if location i is built in addition to current loc.
Z total cost (objective)
(c) Prof. Richard F. Hartl
QEM - Chapter 1

16. Initialzation: row sum of cost matrix ci := ∑cij … transportation cost
Determine, which location to build if just one location is built:
row sum of cost matrix ci := ∑cij … transportation cost
choose location k with minimal cost ck + fk
set I1 = {k}, Iovl = I – {k} und Z = ck + fk … incumbent solution
compute savings of transportation cost ωij = max {ckj – cij, 0} for all locations i from Iovl and all customers j as well as row sum ωi … choose maximum ωi
Example: first location k=5 with Z:= c5 + f5 = 39, I1 = {5}, Iovl = {1,2,3,4}
(c) Prof. Richard F. Hartl
QEM - Chapter 1

17. i\j 1 2 3 4 5 6 7 fi ci
fi + ci 10 9 38 43 37 44 42 34 39
ωij is saving in transportation cost when delivering to custonmer j, by opening additional location i.
→ row sum ωi is total saving in transportation cost when opening additional location i.
i\j 1 2 3 4 5 6 7 ωi fi 5 2 1 3 11 5 4 6 14 7 4 5 1 10 5 1 2 6
(c) Prof. Richard F. Hartl
QEM - Chapter 1

18. Iteration: For all with ωi ≤ fi :
in each iteration fix as built the location from Iovl, with the largest total saving:
Fild potential location k from Iovl, where saving in transportation cost minus additional fixed cost ωk – fk is maximum.
Iovl = Iovl – {k} and Z = Z – ωk + fk
Also, forbid all locations (finally) where saving in transportation cost are smaller than additional fixed costs
For all
with ωi ≤ fi :
Update the savings in transpotrtation cost for all locations Iovl and all customers j :
ωij = max {ωij - ωkj, 0}
(c) Prof. Richard F. Hartl
QEM - Chapter 1

19. Stiopping criterion: Beispiel: Iteration 1
Stiop if no more cost saving are possible by additional locations from Iovl
Build locationd from set I1.
Total cost Z
assignment: xij = 1 iff
Beispiel: Iteration 1
i\j 1 2 3 4 5 6 7 ωi fi 5 2 1 3 11 5 4 6 14 7
Fix k = 2 5 4 1 10 5 1 2 6
Forbid i = 4
Because of ω4 < f4 location 4 is forbidden finally. Location k=2 is built.
Now Z = 39 – 7 = 32 and Iovl = {1,3}, I1 = {2,5}, Io = {4}. Update savings ωij.
(c) Prof. Richard F. Hartl
QEM - Chapter 1

20. Iteration 2: Ergebnis: i\j 1 2 3 4 5 6 7 ωi fi Fix k = 1 1 2 3 6 5
Forbid i = 3 1 1 5
Iteration 2:
Location 3 is forbidden, location k = 1 is finally built.
Ergebnis:
Final solution I1 = {1,2,5}, Io = {3,4} and Z = 32 – 1 = 31.
Build locations 1, 2 and 5
Customers {1,2,7} are delivered from location 1, {3,5} from location 2, and {4,6} from location 5. Total cost Z = 31.
(c) Prof. Richard F. Hartl
QEM - Chapter 1

21. DROP for WLP Die Set Iovl is replaced by I1vl.
I1vl set of preliminarily built locations (yi =1 tentatively)
DROP works the other way round compared to ADD, i.e. start with all locations temporarily built; in each iteration remove one location…
Initialisation: I1vl = I, I0 = I1 = { }
Iteration
In each Iteration delete that location from I1vl (finally), which reduces total cost most.
If deleting would let total cost increase, fix this location as built
(c) Prof. Richard F. Hartl
QEM - Chapter 1

22. Example: Initialisation and Iteration 1: I1vl ={1,2,3,4,5} i\j 1 2 3 4
Expand matrix C:
Row m+1 (row m+2) contains smallest ch1j (second smallest ch2j) only consider locations not finally deleted →
Row m+3 (row m+4) contains row number h1 (and h2) where smallest (second smallest) cost elemet occurs. If location h1 (from I1vl) is dropped, transportation cost for customer Kunden j increase 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 10 9
build
delete 5 1
ch1j 6
ch2j 7 h1 8 h2 9 1 2 1 2 2 4 3 2 2 5 3 1 2 4 5 1 3 3 5 3 3 4 5 3
(c) Prof. Richard F. Hartl
QEM - Chapter 1

23. 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.
(c) Prof. Richard F. Hartl
QEM - Chapter 1

24. Location 3 is finally built, location 4 finally dropped.
i\j 1 2 3 4 5 6 7 10 9 δi fi - 5 6 - 8 1
build
forbid
ch1j
ch2j h1 h2 1 2 1 1 3 2 4 3 2 5 3 1 4 6 4 5 6 5 3 5 6 1 3 4 5 3
Location 3 is finally built, location 4 finally dropped.
(c) Prof. Richard F. Hartl
QEM - Chapter 1

25. Iteration 3: I1vl = {5}, I1 = {1,3}, I0 = {2,4} i\j 1 2 3 4 5 6 7 10 9fi - 5 - 7
build
ch1j
ch2j h1 h2 1 2 1 1 3 3 5 3 2 5 3 1 5 6 4 5 6 5 5 3 6 1 7 1 5 3
Location 5 is finally built
(c) Prof. Richard F. Hartl
QEM - Chapter 1

26. 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)
(c) Prof. Richard F. Hartl
QEM - Chapter 1

27. 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.
(c) Prof. Richard F. Hartl
QEM - Chapter 1

28. P-Median Number of facilities is fixed … p
Typically fixed costs are not needed (but can be considered if not uniform)
(c) Prof. Richard F. Hartl
QEM - Chapter 1

29. MIP for p-Median xij ≤ yi j = 1, …,n transportation cost + fixed cost
Delivery only from locations i that are built
xij ≤ yi
i = 1, …, m
j = 1, …,n
j = 1, …,n
Satisfy total demand of customer j
yi is binary
xij non negative
i = 1, …, m
For all i and j
Exactly p facilities
(c) Prof. Richard F. Hartl
QEM - Chapter 1