Kapitel 3

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Transcript Kapitel 3

Chapter 1

Strategic Problems: Location

Location problems

Production site

transportation

PS 1 Central warehouse

transportation

CW 1 Distribution centers DC 1

transportation

customers C 1 PS DC C 2 2 CW 2 2 PS C 3 3 DC 3 PS C 4 4 DC 4 …

Full truck load

…

FTL or tours

…

tours

… QEM - Chapter 1 Kapitel 3 / 2 (c) Prof. Richard F. Hartl



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 Kapitel 3 / 3

Median Problem



Simplest location problem

   Represent in complete graph. Nodes

are customers with weights

b i

Choose one node as location of warehouse Minimize total weighted distance from warehouse 

Definition: Median

 directed graph (one way streets…): 

(

) = ∑

d ij b j

→ min.

 undirected graph: 

out (

) = ∑

d ij b j



in (

) = ∑

d ji b j

→ min … out median → min … in median (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 4

Example: fromDomschke und Drexl (Logistik: Standorte, 1990, Kapitel 3.3.1) 1/4 2 2/0 3 2 3/2 3 2 5 4/3 4 5/1 4 2 3 6/2 Distance between locations (c) Prof. Richard F. Hartl

0 12 2 12 0 10 4 8 11 2 6 9 5 9 12 2 3 0 10 3 8 0 4 7 6 7 4 5 0 3 12 5 10 2 6 0 QEM - Chapter 1 weight

4 0 2 3 1 2 Kapitel 3 / 5

Example: Median

i\j 1 2 1 4*0 0*12 3 2*2 4 5 3*10 6*1 6 2*12 σ out (i) 64 2 4*2 0*0 3 4*12 0*10 2*3 2*0 3*3 7*1 2*5 3*8 4*1 2*10 40 96 i/j 1 2 3 1 2 3 4 5 4*0 4*12 4*2 4*10 24 0*2 0*0 0*3 2*12 2*10 2*0 0*3 2*8 0 8 4 4*4 5 4*8 0*2 0*6 2*5 2*9 3*0 5*1 2*2 3*4 0*1 2*6 35 74 4 5 3*4 1*8 3*2 1*6 3*5 1*9 3*0 1*4 15 0 6 6 6 4*11 0*9 2*12 3*7 3*1 2*0 92 6 σ in (i) 2*11 2*9 2*12 2*7 66 98 56 74 6 53 0 80

City 4 is median since 35+74 = 109 minimal OUT e.g: emergency delivery of goods

IN e.g.: collection of hazardous waste

QEM - Chapter 1 Kapitel 3 / 6 6 48 0 20

Uncapacitated (single-stage) Warehouse Location Problem – LP-Formulation



single-stage WLP:

 warehouse W 1 W 2 m  customer: C 1 C 2 C 3 C 4 n      Deliver goods to

n customers

each customer has given demand Exist:

m potential warehouse

locations Warhouse in location

causes fixed costs

f i

Transportation costs



are

c i

j if total demand of j comes from i.

 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 Kapitel 3 / 10

Example: from Domschke & Drexl (Logistik: Standorte, 1990, Kapitel 3.3.1) Solution 1: all warehouses i\j 1 2 3 4 5 1 1 2 7 6 6 2 2 9 6 5 4 3 10 0 1 10 6 4 9 7 5 2 3 5 6 3 3 6 7 6 7 6 10 3 2 5 6 6 7 3 10 f i 5 7 5 6 5 Fixed costs = 5+7+5+6+5 = 28 high Transp. costs = 1+2+0+2+3+2+3 = 13 Total costs = 28 + 13 = 41 Solution 2: just warehouses 1 and 3 i\j 1 3 1 1 7 2 2 6 3 10 1 4 9 5 5 6 3 6 7 10 7 3 5 f i 5 5 Fixed costs = 5+5 = 10 Transp. costs = 1+2+1+5+3+7+3 = 22 Total costs = 10 + 22 =



when locations are decided:

  transportation cost easy (closest location) Problem: 2 m 1 possibilities (exp…) 

Formulation as LP (MIP)



y i

… Binary variable for

i = 1, …, m

y i

= 1 if location i is chosen for warehouse 0 otherwise 

x ij

… real „assignment“ oder transportation variable für

i = 1, …,m

and

j = 1, …, n

x ij

= fraction of demand of customer j devivered from location i.

transportation cost + fixed cost Delivery only from locations i that are built Satisfy total demand of customer j y i is binary x ij non negative (c) Prof. Richard F. Hartl

MIP for WLP

(

) 

 1

j n m

  1

c ij x ij



i m

  1

f i y i

 min

x ij

≤

y i

i = 1, …, m j = 1, …,n

i m

  1

x ij

 1

y i

 { 0 , 1 }

x ij

 0 j = 1, …,n i = 1, …, m For all i and j QEM - Chapter 1 Kapitel 3 / 13



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 Kapitel 3 / 14

ADD for WLP

 Notation: I 0 I o vl I 1 

Z I:={1,…,m} set of all potential locations set of (finally) forbidden locations (y i = 0 fixed) set of preliminary forbidden locations (y i = 0 tentaitively) set of included (built, realized) locations (y i =1 fixed) reduction in transportation cost, if location i is built in addition to current loc.

total cost (objective) (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 15



Initialzation

:  Determine, which location to build if just one location is built:  row sum of cost matrix c i := ∑c ij … transportation cost  choose location k with minimal cost c k + f k  set I 1 = {k}, I o vl = I – {k} und Z = c k + f k … incumbent solution  compute savings of transportation cost ω ij = max {c kj – c ij , 0} for all locations i from I o vl row sum ω i and all customers j as well as … choose maximum ω i  Example: first location k=5 with Z:= c 5 + f 5 = 39, I 1 = {5}, I o vl = {1,2,3,4} (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 16

i\j 1 2 3 4 5 1 1 2 7 6 6 2 2 9 6 5 4 3 10 0 1 10 6 4 9 7 5 2 3 5 6 3 3 6 7 6 7 6 10 3 2 7 3 10 5 6 6 f i 5 7 5 6 5 c i 38 37 37 38 34 f i + c i 43 44 42 44 39 i\j 1 2 3 4 1 5 4 2 2 3 6 5 1 4 5 1 4 4 1 6 7 3 1 ω i 11 14 10 2 5 f i 7 5 6 (c) Prof. Richard F. Hartl QEM - Chapter 1 ω ij is saving in transportation cost when delivering to custonmer j, by opening additional location i. → row sum ω i is total saving in transportation cost when opening additional location i. Kapitel 3 / 17



Iteration:

 in each iteration fix as built the location from I o vl , with the largest total saving:  Fild potential location k from I o vl , where saving in transportation cost minus additional fixed cost ω k – f k is maximum.

1 

1  {

}, I o vl = I o vl – {k} and Z = Z – ω k + f k  Also, forbid all locations (finally) where saving in transportation cost are smaller than additional fixed costs For all



with ωi ≤ fi :

0 

0  {

}

and I



 {

}  : Update the savings in transpotrtation cost for all locations I o vl and all customers j ω ij = max { ω ij ω kj , 0} (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 18



Stiopping criterion:

   Stiop if no more cost saving are possible by additional locations from I o vl Build locationd from set I 1 .

Total cost Z  assignment: x ij = 1 iff

c ij

 min {

c hj h



1 } 

Beispiel: Iteration 1

i\j 1 2 3 1 2 3 4 5 4 2 6 5 4 4 4 1 5 1 6 7 3 1 ω i 11 14 10 2 f i 5 7 5 6 Fix k = 2 1 Forbid i = 4   Because of ω 4 < f 4 location 4 is forbidden finally. Location k=2 is built. Now Z = 39 – 7 = 32 and I o vl Update savings ω ij.

= {1,3}, I 1 = {2,5}, I o = {4}. (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 19

i\j 1 3 1 1 2 2 3 4 5 6 7 3 1 ω i 6 1 f i 5 5 

Iteration 2:

 Location 3 is forbidden, location k = 1 is finally built. Fix k = 1 Forbid i = 3 

Ergebnis:

   Final solution I 1 = {1,2,5}, I o = {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

DROP for WLP

 Die Set I o vl is replaced by I 1 vl .

 I 1 vl set of preliminarily built locations (y i =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: I 1 vl = I, I 0 = I 1 = { }  Iteration   In each Iteration delete that location from most. I 1 vl (finally), which reduces total cost If deleting would let total cost increase, fix this location as built QEM - Chapter 1 Kapitel 3 / 21 (c) Prof. Richard F. Hartl

 Expand matrix C:  Row m+1 (row m+2) contains smallest c h1j (second smallest c h2j ) only consider locations not finally deleted →



0   Row m+3 (row m+4) contains row number h 1 smallest) cost elemet occurs. If location h 1 (and h 2 ) where smallest (second (from I 1 vl ) is dropped, transportation cost for customer Kunden j increase by c h2j - c h1j Example: Initialisation and Iteration 1: I 1 vl ={1,2,3,4,5} i\j 1 2 3 4 5 6 7 δ i f i 1 1 2 10 9 6 7 3 5 5 build 2 2 9 0 7 3 6 10 1 7 delete 3 4 7 6 6 5 1 10 5 2 3 6 10 3 5 6 0 1 5 6 5 6 4 6 3 7 2 6 1 5 c h1j c h2j h 1 h 2 6 7 8 9 1 2 1 2 2 4 1 5 0 1 2 3 2 3 4 5 3 3 2 3 2 3 5 4 3 5 1 3 (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 22

 For all i from I 1 vl δ i compute increase in transportation cost δ i if I is finally dropped. is sum of differences between smallest and second smallest cost element in rows where i = h 1 contains the smallest element.

  2 examples:   δ 1 = (c 21 δ 2 = (c 33 – c 11 ) + (c 52 – c 23 ) + (c 35 – c 12 ) + (c 37 – c 25 ) = 1 – c 17 ) = 5 If fixed costs savings f i exceed additional transportation cost δ i , finally drop i. In Iteration 1 location 1 is finally built.

 Iteration 2:  I 1 vl = {3,4,5}, I 1 = {1}, I 0 = {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 I 1 compute δ i there.

= {1}, but 1 is no candidate for dropping. Hence do not (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 23

i\j 1 3 4 5 c h1j c h2j h 1 h 2 1 6 1 4 1 1 7 6 6 2 4 1 5 2 2 6 5 4 1 6 3 5 3 10 1 10 6 2 3 4 5 4 9 5 2 3 3 6 3 1 5 6 3 6 7 2 3 5 4 6 7 10 3 2 3 5 1 3 7 3 5 6 6 8 1 δ i 1 f i 5 6 5 build forbid Location 3 is finally built, location 4 finally dropped.

 Iteration 3:  I 1 vl = {5}, I 1 = {1,3}, I 0 = {2,4} i\j 1 3 5 c h1j c h2j h 1 h 2 1 6 1 5 1 1 7 6 2 4 1 5 2 2 6 4 3 10 1 6 1 6 3 5 3 5 5 3 4 9 5 3 3 6 3 1 5 6 3 7 2 7 5 1 6 7 10 2 3 5 1 3 7 3 5 6 7 δ i f i 5 build Location 5 is finally built (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 25



Result

 Build locations I 1 = {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 Kapitel 3 / 26

Improvement for WLP

 In each iteration you can do:  Replace a built location (from I 1 ) by a forbidden location (from I 0 ). Choose

first improvement

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. QEM - Chapter 1 Kapitel 3 / 27 (c) Prof. Richard F. Hartl

P-Median

 

Number of facilities is fixed … p Typically fixed costs are not needed (but can be considered if not uniform)

transportation cost + fixed cost Delivery only from locations i that are built

MIP for

-Median

(

) 

i m n

   1 1

j c ij x ij

 min

x ij ≤ y i

i = 1, …, m j = 1, …,n Satisfy total demand of customer j y i is binary x ij non negative Exactly p facilities (c) Prof. Richard F. Hartl

i m

  1

x ij

 1

y i

 { 0 , 1 }

x ij

 0

i m

  1

y i



QEM - Chapter 1 j = 1, …,n i = 1, …, m For all i and j Kapitel 3 / 29

Kapitel 3

Transcript Kapitel 3

Chapter 1

Strategic Problems: Location

Location problems

More levels possible (regional warehouses)

Can be delegated to logistics service providers

Decision problems

Median Problem

Simplest location problem

Definition: Median

Example: Median

Related Problem: Center

Median

Center

Solution

Uncapacitated (single-stage) Warehouse Location Problem – LP-Formulation

single-stage WLP:

Goal:

when locations are decided:

Formulation as LP (MIP)

MIP for WLP

Problem:

Heuristics:

ADD for WLP

Initialzation

DROP for WLP

Result

Improvement for WLP

P-Median

Number of facilities is fixed … p Typically fixed costs are not needed (but can be considered if not uniform)

MIP for

-Median