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Chapter 1 Strategic Problems: Location

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QEM - Chapter 1 Location problems Full truck loadtransportation … CW 1 CW 2 Central warehouse FTL or tourstransportation … DC 1 DC 2 DC 3 DC 4 Distribution centers tourstransportation … C1C1 C2C2 C3C3 C4C4 customers … PS 1 PS 2 PS 3 PS 4 Production site (c) Prof. Richard F. Hartl Kapitel 3 / 2

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QEM - Chapter 1 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 Kapitel 3 / 3

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QEM - Chapter 1 Median Problem Simplest location problem Represent in complete graph. Nodes i 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…): σ(i) = ∑d ij b j → min. undirected graph: σ out (i) = ∑d ij b j → min … out median σ in (i) = ∑d ji b j → min … in median (c) Prof. Richard F. Hartl Kapitel 3 / 4

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QEM - Chapter 1 Example: fromDomschke und Drexl (Logistik: Standorte, 1990, Kapitel 3.3.1) 012210612 203375 10084 425052 869406 11912730 2/0 1/4 3/2 4/3 5/1 6/2 2 2 2 2 3 3 3 4 4 5 D= 4 0 2 3 1 2 b= (c) Prof. Richard F. Hartl Kapitel 3 / 5 Distance between locations weight

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QEM - Chapter 1 Example: Median i\j123456 i/j123456 OUT e.g: emergency delivery of goodsIN e.g.: collection of hazardous waste 4*00*122*23*106*12*1264 4*120*102*03*84*12*1096 4*40*22*53*05*12*2 35 4*80*62*93*40*12*674 4*110*92*123*73*12*092 4*20*02*33*37*12*540 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 0 6 0 24 80 20 6 6 0 0 481 3 4 5 6 2 σ out (i) 1 3 4 5 6 2 σ in (i) City 4 is median since 35+74 = 109 minimal (c) Prof. Richard F. Hartl Kapitel 3 / 6

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QEM - Chapter 1 Related Problem: Center Median Node with min total weighted distance → min. Center Node with min Maximum (weighted) Distance → min. (c) Prof. Richard F. Hartl Kapitel 3 / 7

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QEM - Chapter 1 24 8 15 0 6 0 24 Solution i\j123456 i/j123456 4*00*122*23*106*12*1230 4*120*102*03*84*12*1048 4*40*22*53*05*12*2 16 4*80*62*93*40*12*632 4*110*92*123*73*12*044 4*20*02*33*37*12*510 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 48 20 6 6 0 0 481 3 4 5 6 2 out (i) 1 3 4 5 6 2 in (i) City1 is center since 30+24 = 54 minimal (c) Prof. Richard F. Hartl Kapitel 3 / 8

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QEM - Chapter 1 Uncapacitated (single-stage) Warehouse Location Problem – LP-Formulation single-stage WLP: warehouse customer: W1W1 W2W2 m C1C1 C2C2 C3C3 C4C4 n Deliver goods to n customers each customer has given demand Exist: m potential warehouse locations Warhouse in location i causes fixed costs f i Transportation costs i j are c ij if total demand of j comes from i. (c) Prof. Richard F. Hartl Kapitel 3 / 9

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QEM - Chapter 1 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 Kapitel 3 / 10

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QEM - Chapter 1 Example: from Domschke & Drexl (Logistik: Standorte, 1990, Kapitel 3.3.1) Solution 1: all warehouses i\j1234567fifi 1121096735 2290736 7 376153 55 465 26366 564637265 i\j1234567fifi 1121096735 376153 55 Fixed costs = 5+7+5+6+5 = 28 high Transp. costs = 1+2+0+2+3+2+3 = 13 Total costs = 28 + 13 = 41 Fixed costs = 5+5 = 10 Transp. costs = 1+2+1+5+3+7+3 = 22 Total costs = 10 + 22 = 32 Solution 2: just warehouses 1 and 3 (c) Prof. Richard F. Hartl Kapitel 3 / 11

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QEM - Chapter 1 Formulation as LP (MIP) y i … Binary variable for i = 1, …, m: y i = 1 if location i is chosen for warehouse 0 otherwise when locations are decided: transportation cost easy (closest location) Problem: 2 m -1 possibilities (exp…) 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. (c) Prof. Richard F. Hartl Kapitel 3 / 12

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QEM - Chapter 1 MIP for WLP 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 x ij ≤ y i i = 1, …, m j = 1, …,n i = 1, …, m For all i and j j = 1, …,n (c) Prof. Richard F. Hartl Kapitel 3 / 13

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QEM - Chapter 1 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 Kapitel 3 / 14

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QEM - Chapter 1 ADD for WLP Notation: I:={1,…,m}set of all potential locations I 0 set of (finally) forbidden locations (y i = 0 fixed) I o vl set of preliminary forbidden locations (y i = 0 tentaitively) I 1 set of included (built, realized) locations (y i =1 fixed) reduction in transportation cost, if location i is built in addition to current loc. Ztotal cost (objective) (c) Prof. Richard F. Hartl Kapitel 3 / 15

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QEM - Chapter 1 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 and all customers j as well as row sum ω i … 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 Kapitel 3 / 16

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QEM - Chapter 1 i\j1234567fifi cici f i + c i 11210967353843 22907361073744 37615310553742 46510263663844 5646372653439 i\j1234567 ωiωi fifi 1 2 3 4 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. 11 5213 451 464 26 105 147 115 (c) Prof. Richard F. Hartl Kapitel 3 / 17

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QEM - Chapter 1 Iteration: in each iteration fix as built the location from I o vl, with the largest total saving: I o vl = I o vl – {k} and Z = Z – ω k + f k Fild potential location k from I o vl, where saving in transportation cost minus additional fixed cost ω k – f k is maximum. Also, forbid all locations (finally) where saving in transportation cost are smaller than additional fixed costs Update the savings in transpotrtation cost for all locations I o vl and all customers j : ω ij = max {ω ij - ω kj, 0} For allwith ωi ≤ fi : (c) Prof. Richard F. Hartl Kapitel 3 / 18

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QEM - Chapter 1 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 i\j 1234567 ωiωi fifi 1 2 3 4 Fix k = 2 Forbid i = 4 Beispiel: Iteration 1 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 = {1,3}, I 1 = {2,5}, I o = {4}. Update savings ω ij. 464 11 5213 541 115 147 105 26 (c) Prof. Richard F. Hartl Kapitel 3 / 19

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QEM - Chapter 1 Iteration 2: Location 3 is forbidden, location k = 1 is finally built. i\j1234567ωiωi fifi 1 3 Fix k = 1 Forbid i = 3 123 1 65 15 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. (c) Prof. Richard F. Hartl Kapitel 3 / 20

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QEM - Chapter 1 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 I 1 vl (finally), which reduces total cost most. If deleting would let total cost increase, fix this location as built (c) Prof. Richard F. Hartl Kapitel 3 / 21

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QEM - Chapter 1 Expand matrix C: Row m+1 (row m+2) contains smallest c h1j (second smallest c h2j ) only consider locations not finally deleted → i\j1234567δiδi fifi 1121096735 2290736 7 376153 55 465 26366 564637265 c h1j 6 c h2j 7 h1h1 8 h2h2 9 build delete Example: Initialisation and Iteration 1: I 1 vl ={1,2,3,4,5} Row m+3 (row m+4) contains row number h 1 (and h 2 ) where smallest (second smallest) cost elemet occurs. If location h 1 (from I 1 vl ) is dropped, transportation cost for customer Kunden j increase by c h2j - c h1j 5 1 0 1 1 2 1 4 5 0 2 1 3 2 4 3 5 3 2 3 3 2 5 3 4 3 1 5 32 2 1 1 (c) Prof. Richard F. Hartl Kapitel 3 / 22

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QEM - Chapter 1 For all i from I 1 vl compute increase in transportation cost δ i if I is finally dropped. δ i is sum of differences between smallest and second smallest cost element in rows where i = h 1 contains the smallest element. 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 = {1}, but 1 is no candidate for dropping. Hence do not compute δ i there. 2 examples: If fixed costs savings f i exceed additional transportation cost δ i, finally drop i. In Iteration 1 location 1 is finally built. δ 1 = (c 21 – c 11 ) + (c 52 – c 12 ) + (c 37 – c 17 ) = 5 δ 2 = (c 33 – c 23 ) + (c 35 – c 25 ) = 1 (c) Prof. Richard F. Hartl Kapitel 3 / 23

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QEM - Chapter 1 i\j1234567 112109673 376153 5 465 2636 56463726 δiδi fifi - 5 6 5 build forbid Location 3 is finally built, location 4 finally dropped. c h1j c h2j h1h1 h2h2 2 1 4 5 1 3 6 5 2 4 3 5 3 3 6 1 2 5 3 4 3 1 5 34 6 1 1 - 8 1 1 (c) Prof. Richard F. Hartl Kapitel 3 / 24

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QEM - Chapter 1 Iteration 3: I 1 vl = {5}, I 1 = {1,3}, I 0 = {2,4} Location 5 is finally built i\j1234567 112109673 376153 5 56463726 δiδi fifi - - 5 build c h1j c h2j h1h1 h2h2 2 1 4 5 1 3 6 5 3 5 5 3 3 3 6 1 2 5 7 1 3 1 5 35 6 1 1 - - 7 (c) Prof. Richard F. Hartl Kapitel 3 / 25

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QEM - Chapter 1 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 Kapitel 3 / 26

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QEM - Chapter 1 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 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 Kapitel 3 / 27

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P-Median Number of facilities is fixed … p Typically fixed costs are not needed (but can be considered if not uniform) QEM - Chapter 128 (c) Prof. Richard F. Hartl

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QEM - Chapter 1 MIP for p -Median 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 x ij ≤ y i i = 1, …, m j = 1, …,n i = 1, …, m For all i and j j = 1, …,n (c) Prof. Richard F. Hartl Kapitel 3 / 29 Exactly p facilities

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