The Quadratic Assignment Problem (QAP)
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Layout and Design Kapitel 4 / 1(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
common mathematical formulation for intra-company location problems
cost of an assignment is determined by the distances and the material flows between all given entities
each assignment decision has direct impact on the decision referring to all other objects
Layout and Design Kapitel 4 / 2(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
Activity relationship charts: graphical means of representing the desirability of locating pairs
of machines/operations near to each other common letter codes for classification of “closeness” ratings:
A Absolutely necessary. Because two machines/operations use the same equipment or facilities, they must be located near each other.
E Especially important. The facilities may for example require the same personnel or records.
I Important. The activities may be located in sequence in the normal work flow.
Nahmias, S.: Production and Operations Analysis, 4th ed., McGraw-Hill, 2000, Chapter 10
Layout and Design Kapitel 4 / 3(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
common letter codes for classification of “closeness” ratings:
O Ordinary importance. It would be convenient to have the facilities near each other, but it is not essential.
U Unimportant. It does not matter whether the facilities are located near each other or not.
X Undesirable. Locating a wedding department near one that uses flammable liquids would be an example of this category.
In the original conception of the QAP a number giving the reason for each closeness rating is needed as well.
In case of closeness rating “X” a negative value would be used to indicate the undesirability of closeness for the according machines/operations.
Nahmias, S.: Production and Operations Analysis, 4th ed., McGraw-Hill, 2000, Chapter 10
Layout and Design Kapitel 4 / 4(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
Example: Met Me, Inc., is a franchised chain of fast-food hamburger restaurants. A new restaurant is being located in a growing suburban community near Reston, Virginia. Each restaurant has the following departments:
1. Cooking burgers2. Cooking fries3. Packing and storing burgers4. Drink dispensers5. Counter servers6. Drive-up server
Nahmias, S.: Production and Operations Analysis, 4th ed., McGraw-Hill, 2000, Chapter 10
Layout and Design Kapitel 4 / 5(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
Nahmias, S.: Production and Operations Analysis, 4th ed., McGraw-Hill, 2000, Chapter 10
Activity relationship diagram for the example problem:
Layout and Design Kapitel 4 / 6(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
Mathematical formulation: we need both distances between the locations and material flow
between organizational entities (OE) n organizational entities (OE), all of them are of same size and
can therefore be interchanged with each other n locations, each of which can be provided withe each of the OE
(exactly 1) thi ... transp. intensity, i.e. material flow between OE h and OE i djk ... distance between j and location k (not implicitly symmetric) Transportation costs are proportional to transported amount and
distance.
Layout and Design Kapitel 4 / 7(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
If OE h is assigned to location j and OE i to location k the transportation cost per unit transported from OE h to OE i is
determined by djk we determine the total transportation cost by multiplying djk with
the material flow between OE h zu OE i which is thi
j k... ... ... locations
h i... ... ... OE
djk.
thi.
Cost = thi djk
Layout and Design Kapitel 4 / 8(c) Prof. Richard F. Hartl
Similar to the LAP:binary decision variables
If OE h location j (xhj = 1)and OE i location k (xik = 1)
Transportation cost per unit transported from OE h to OE i :
Total transportation cost:
The Quadratic Assignment Problem (QAP)
otherwise 0
location toassigned is OE if 1 jhxhj
n
j
n
kikhjjk xxd
1 1
j k... ... ...locations
h i... ... ...OE
djk.
thi.
Cost = thi djk xhj = 1 xik = 1
n
h
n
i
n
j
n
kikhjjkhi xxdt
1 1 1 1
Layout and Design Kapitel 4 / 9(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
Objective: Minimize the total transportation costs between all OE
min1 1 1 1
n
h
n
i
n
j
n
kikhjjkhi xxdt
Constraints
11
n
jhjx
11
n
hhjx
hjx
für h = 1, ... , n ... each OE h assigned to exactly 1 location j
für j = 1, ... , n ... each location j is provided with exactly 1 OE h
= 0 or 1 ... binary decision variable
Similar to LAP!!!
Quadratic function QAP
Layout and Design Kapitel 4 / 10(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
Example: Calculate cost for 3 OE (1 ,2 ,3) and 3 locations (A, B, C)
A
B C
013101210
CBA
D
CBA
Distances between locations djk
013202110
321
321
T
Material flow thi
1 possible solution: 1 A, 2 B, 3 C, i.e. x1A = 1, x2B = 1, x3C = 1, all other xij = 0
All constraints are fulfilled.Total transportation cost: 0*0 + 1*1 + 2*1 + 1*2 + 0*0 + 1*2 + 3*3 + 1*1 + 0*0 = 17
Layout and Design Kapitel 4 / 11(c) Prof. Richard F. Hartl
013101210
CBA
D
CBA
011102130
BAC
D
BAC
The Quadratic Assignment Problem (QAP)
This solution is not optimal since OE 1 and 3 (which have a high degree of material flow) are assigned to locations A and C (which have the highest distance between them).
A better solution would be.: 1 C, 2 A and 3 B, i.e. x1C = 1, x2A = 1, x3B = 1.
with total transportation cost: 0*0 + 3*1 + 1*1 + 2*2 + 0*0 + 2*1 + 1*3 + 1*1 + 0*0 = 14
2 A
3 B 1 C
Distances Material flow
013202110
321
321
T
Layout and Design Kapitel 4 / 12(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
We resorted the matrix
such that row and columns appear the following sequence 1 C, 2 A and 3 B, i.e C, A, B(it is advisable to perform the resorting in 2 steps: first rows than columns or the other way round)
011102130
BAC
D
BAC
013101210
CBA
D
CBA
013101210
CBA
D
CBA
130011102
CBA
D
BAC
Layout and Design Kapitel 4 / 13(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
Starting heuristics: refer to the combination of one of the following
possibilities to select an OE and a location. the core is defined by the already chosen OE After each iteration another OE is added to the core due
to one of the following priorities
Layout and Design Kapitel 4 / 14(c) Prof. Richard F. Hartl
1. Selection of (non-assigned) OE
A1 those having the maximum sum of material flow to all (other) OE
A2 a) those having the maximum material flow to the last-assigned OEb) those having the maximum material flow to an assigned OE
A3 those having the maximum material flow to all assigned OE (core)
A4 random choice
Layout and Design Kapitel 4 / 15(c) Prof. Richard F. Hartl
1. Selection of (non-assigned) locations
B1 those having the minimum total distance to all other locations
B2 those being neighbouring to the last-chosen locationB3 a) those leading to the minimum sum of transportation
cost to the core b) like a) but furthermore we try to exchange the location with neigboured OEc) a location (empty or allocated) such that the sum of transportation costs within the new core is minimized (in case an allocated location is selected, the displaced OE is assigned to an empty location)
B4 random choice
Layout and Design Kapitel 4 / 16(c) Prof. Richard F. Hartl
ExampleCombination of A1 and B1: Arrange all OE according to decreasing sum of material
flow Arrange all locations according to increasing distance to
all other locations
Manhatten-distance between locations. (matrix is symmetric -> consideration of the matrix triangle is sufficient)
A B C D E F G H I
Layout and Design Kapitel 4 / 17(c) Prof. Richard F. Hartl
Example – Material flow
OE 1 2 3 4 5 6 7 8 9
1 - - - - 3 - - - -
2 - 3 1 2 - 4 - -
- 3 5 2 - 3 4
4 - - - 1 - -
5 - 2 2 1 -
6 - - - -
7 - - -
8 - -
9 -
310205
154744
3
Sequence of OE (according to decreasing material flow): 3, 5, 2, 7, 4, 6, 8, 9, 1
Sum of material flow between 15 and 51
Layout and Design Kapitel 4 / 18(c) Prof. Richard F. Hartl
Example - Distances
L A B C D E F G H I
A - 1 2 1 2 3 2 3 4
B - 1 2 1 2 3 2 3
C - 3 2 1 4 3 2
D - 1 2 1 2 3
- 1 2 1 2
F - 3 2 1
G - 1 2
H - 1
I -
181518151215181518
E
Sequence of locations (according to increasing distances): E, B, D, F, H, A, C, G, I
Layout and Design Kapitel 4 / 19(c) Prof. Richard F. Hartl
Sequence of OE: 3, 5, 2, 7, 4, 6, 8, 9, 1
Sequence of locations: E, B, D, F, H, A, C, G, I
Assignment:
Example - Assignment
OE 1 2 3 4 5 6 7 8 9Loc. I D E H B A F C G
Layout and Design Kapitel 4 / 20(c) Prof. Richard F. Hartl
Example – Total costs
OE 1 2 3 4 5 6 7 8 91 - - - - 3*3 - - - -
2 - 3*1 1*2 2*2 - 4*2 - -
3 - 3*1 5*1 2*2 - 3*2 4*2
4 - - - 1*2 - -
5 - 2*1 2*2 1*1 -
6 - - - -
7 - - -
8 - -
9 -
OE 1 and 5 are assigned to locations I and B
3 (Distance 1-5) * 3 (Flow I-B)
Total cost = 61
Layout and Design Kapitel 4 / 21(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
Improvement heuristics: Try to improve solutions by exchanging OE-pairs (see the
introducing example) Check if the exchange of locations of 2 OE reduces costs. Exchange of OE-triples only if computational time is acceptable. There are a number of possibilities to determine OE-pairs
(which should be checked for an exchange of locations):
Layout and Design Kapitel 4 / 22(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
Selection of pairs for potential exchanges:C1 all n(n - 1)/2 pairsC2 a subset of pairsC3 random choice Selection of pairs which finally are exchanged:D1 that pair whose exchange of locations leads to
the highest cost reduction. (best pair)D2 the first pair whose exchange of locations leads to
a cost reduction (first pair)
Layout and Design Kapitel 4 / 23(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
Solution quality Combination of C1 and D1:
Quite high degree of computational effort. Relatively good solution quality A common method is to start with C1 and skip to D1 as soon as the
solution is reasonably good. Combination of C1 and D1 is the equivalent to 2-opt method for the
TSP CRAFT :
Well-known (heuristic) solution method For problems where OE are of similar size CRAFT equals a
combination of C1 and D1
Layout and Design Kapitel 4 / 24(c) Prof. Richard F. Hartl
The Quadratic Assignment Problem (QAP)
Random Choice (C3 and D2): Quite good results the fact that sometimes the best exchange of all exchanges
which have been checked leads to an increase of costs is no disadvantage, because it reduces the risk to be trapped in local optima
The basic idea and several adaptions/combinations of A, B, C, and D are discussed in literature
Layout and Design Kapitel 4 / 25(c) Prof. Richard F. Hartl
„Umlaufmethode“
Heurisitic method
Combination of starting and improvement heuristics
Components: Initialization (i = 1):
Those OE having the maximum sum of material flow [A1] is assigned to the centre of locations (i.e. the location having the minimum sum of distances to all other locations [B1]).
Iteration i (i = 2, ... , n): assign OE i
Layout and Design Kapitel 4 / 26(c) Prof. Richard F. Hartl
„Umlaufmethode“
Part 1: Selection of OE and of free location:
select those OE with the maximum sum of material flow to all OE assigned to the core [A3]
assign the selected OE to a free location so that the sum of transportation costs to the core (within the core) is minimized [B3a]
Layout and Design Kapitel 4 / 27(c) Prof. Richard F. Hartl
„Umlaufmethode“
Part 2: Improvement step in iteration i = 4:
check pair wise exchanges of the last-assigned OE with all other OE in the core [C2]wenn eine Verbesserung gefunden ist, führe diese Änderung durch und beginne wieder mit Teil 2 [D2]
if an improvement is found, the exchange is conducted and we start again with Part 2 [D2]
Layout and Design Kapitel 4 / 28(c) Prof. Richard F. Hartl
Example – Part 1
Initialization (i = 1):E = centreAssign OE 3 to centre.
A B C
D E 3 F
G H I
Layout and Design Kapitel 4 / 29(c) Prof. Richard F. Hartl
Sequence of assignment
i = 1 2 3 4 5 6 7 8 9OE 3
1 0
2 3
3
4 3
5 5
6 2
7 0
8 3
9 4
53
2
0
22
10
0 0 00 0 0
0
0
0
0 0 0
0 00
0
0 0
1 1
4
2 7 4 6 8 9 1
i = 1: assign 3 firsti = 5
i = 2: 5 highest mat.flow to 3
i = 9i = 3: 2 highest mat.flow to core (3,5)
i = 6i = 4i = 7i = 8
Layout and Design Kapitel 4 / 30(c) Prof. Richard F. Hartl
Example – PArt 1Iteration i = 2
The maximum material flow to OE 3 is from OE 5
Distances dBE = dDE = dFE = dHE = 1 equally minimal select D,
In iteration i = 2 OE 5 is assigned to D-5.
A B C
D 5 E 3 F
G H I
Layout and Design Kapitel 4 / 31(c) Prof. Richard F. Hartl
Example – Part 1Iteration i = 3
The maximum material flow to the core (3,5) is from OE 2 Select location X, such that
dXEt23 + dXDt25 = dXE3 + dXD2 is minimal: (A, B, G od. H)
X = A dAE3 + dAD2 = 23 + 12 = 8X = B dBE3 + dBD2 = 13 + 22 = 7X = F dFE3 + dFD2 = 13 + 22 = 7X = G dGE3 + dGD2 = 23 + 12 = 8X = H dHE3 + dHD2 = 13 + 22 = 7
B, F or H B is selected In iteration i = 3 we assign OE to B
A B 2 C
D 5 E 3 F G H I
Layout and Design Kapitel 4 / 32(c) Prof. Richard F. Hartl
Example – Part 1Iteration i = 4
The maximum material flow to the core (2,3,5) is from OE 7 (2, 3, 5)
Select location X, such thath dXEt73 + dXDt75 + dXBt72 = dXE0 + dXD2 + dXB4 is minimal
according to the given map -> A is the best choice In iteration i = 4 we tentatively assign OE 7 to location A
Layout and Design Kapitel 4 / 33(c) Prof. Richard F. Hartl
Example – Part 2
Try to exchange A with E, B or D and calculate the according costs:
Original assignement (Part 1): E-3, D-5, B-2, A-7 Cost =15+13+20+22+12+14 = 18try E-3, D-5, A-2, B-7 Cost = 15+23+10+12+22+14 = 21try E-3, A-5, B-2, D-7 Cost = 25+13+10+12+12+24 = 25try A-3, D-5, B-2, E-7 Cost = 15+13+20+22+12+14 = 18
Exchanging A with E would be possible but does not lead to a reduction of costs. Thus, we do not perform any exchange but go on with the solution determined in part 1…and so on…
Layout and Design Kapitel 4 / 34(c) Prof. Richard F. Hartl
Example – Part 2
After 8 iterations without part 2: Cost = 54
With part 2 (last-assigned OE (9) is to be exchanged with OE 4): Cost = 51
While a manual calculation of larger problems is obviously quite time consuming an implementation and therefore computerized calculation is relatively simple