Chapter 1 Strategic Problems Location Location problems Production
Chapter 1 Strategic Problems: Location
Location problems Production site PS 1 PS 2 PS 3 PS 4 transportation Central warehouse Full truck load CW 1 CW 2 … transportation Distribution centers FTL or tours DC 1 DC 2 DC 3 DC 4 transportation customers (c) Prof. Richard F. Hartl … … tours C 1 C 2 QEM - Chapter 1 C 3 C 4 … Kapitel 3 / 2
n More levels possible (regional warehouses) n Can be delegated to logistics service providers n Decision problems n n n 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 n Simplest location problem Represent in complete graph. Nodes i are customers with weights bi ¨ Choose one node as location of warehouse ¨ Minimize total weighted distance from warehouse ¨ n Definition: Median ¨ directed graph (one way streets…): ¨σ(i) = ∑dijbj → min. ¨ undirected graph: ¨σout(i) = ∑dijbj → min ¨σin(i) = ∑djibj → min (c) Prof. Richard F. Hartl … out median … in median QEM - Chapter 1 Kapitel 3 / 4
Example: from. Domschke und Drexl (Logistik: Standorte, 1990, Kapitel 3. 3. 1) 2 2/0 3 2 3 1/4 2 3/2 5 4 Distance between locations (c) Prof. Richard F. Hartl 4/3 2 6/2 4 5/1 3 D= weight 0 12 2 10 6 12 4 2 0 3 3 7 5 0 12 10 0 8 4 10 4 2 5 0 5 2 8 6 9 4 0 6 1 11 9 12 7 3 0 2 QEM - Chapter 1 b= 2 3 Kapitel 3 / 5
Example: Median ij 1 2 3 4 5 6 σout(i) i/j 1 2 3 4 5 6 1 4*0 0*12 2*2 3*10 6*1 2*12 64 1 4*0 4*12 4*10 24 48 2 4*2 0*0 2*3 3*3 7*1 2*5 40 2 0*0 0*3 0 0 3 4*12 0*10 2*0 3*8 4*1 2*10 96 3 2*12 2*10 2*8 8 20 4 4*4 0*2 2*5 3*0 5*1 2*2 35 4 3*2 3*5 3*0 15 6 5 4*8 0*6 2*9 3*4 0*1 2*6 74 5 1*8 1*6 1*9 1*4 0 6 6 4*11 0*9 2*12 3*7 3*1 2*0 92 6 2*11 2*9 2*12 2*7 6 0 66 98 56 74 53 80 City 4 is median since 35+74 = 109 minimal OUT e. g: emergency delivery of goods (c) Prof. Richard F. Hartl σin(i) IN e. g. : collection of hazardous waste QEM - Chapter 1 Kapitel 3 / 6
Related Problem: Center n Median ¨ Node with min total weighted distance ¨ n → min. Center ¨ Node with min Maximum (weighted) Distance ¨ (c) Prof. Richard F. Hartl → min. QEM - Chapter 1 Kapitel 3 / 7
Solution ij 1 2 3 4 5 6 out(i) i/j 1 2 3 4 5 6 1 4*0 0*12 2*2 3*10 6*1 2*12 30 1 4*0 4*12 4*10 24 48 2 4*2 0*0 2*3 3*3 7*1 2*5 10 2 0*0 0*3 0 0 3 4*12 0*10 2*0 3*8 4*1 2*10 48 3 2*12 2*10 2*8 8 20 4 4*4 0*2 2*5 3*0 5*1 2*2 16 4 3*2 3*5 3*0 15 6 5 4*8 0*6 2*9 3*4 0*1 2*6 32 5 1*8 1*6 1*9 1*4 0 6 6 4*11 0*9 2*12 3*7 3*1 2*0 44 6 2*11 2*9 2*12 2*7 6 0 in(i) 24 48 24 40 24 48 City 1 is center since 30+24 = 54 minimal (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 8
Uncapacitated (single-stage) Warehouse Location Problem – LP-Formulation n single-stage WLP: ¨ warehouse ¨ customer: W 1 C 1 ¨ ¨ ¨ m W 2 C 3 C 4 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 Kapitel 3 / 9
n n Problem: n How many warehouses? (many/few high/low fixed costs, low/high transportation costs n Where? Goal: n Satisfy all demand n minimize total cost (fixed + transportation) n 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 2: just warehouses 1 and 3 Solution 1: all warehouses ij 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 Fixed costs = 5+7+5+6+5 = 28 high Fixed costs = 5+5 = 10 Transp. costs = 1+2+0+2+3 = 13 Transp. costs = 1+2+1+5+3+7+3 = 22 Total costs = 28 + 13 = 41 Total costs = 10 + 22 = 32 (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 11
n when locations are decided: transportation cost easy (closest location) ¨ Problem: 2 m-1 possibilities (exp…) ¨ n 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 Kapitel 3 / 12
MIP for WLP transportation cost + fixed cost Delivery only from locations i that are built xij ≤ yi Satisfy total demand of customer j j = 1, …, n yi is binary xij non negative (c) Prof. Richard F. Hartl i = 1, …, m For all i and j QEM - Chapter 1 Kapitel 3 / 13
n Problem: ¨ n m*n real Variablen und m binary → for a few 100 potential locations exact solution difficult → 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 n Notation: I: ={1, …, m} set of all potential locations I 0 Iovl I 1 Z (c) Prof. Richard F. Hartl set of (finally) forbidden locations (yi = 0 fixed) set of preliminary forbidden locations (yi = 0 tentaitively) set of included (built, realized) locations (yi =1 fixed) reduction in transportation cost, if location i is built in addition to current loc. total cost (objective) QEM - Chapter 1 Kapitel 3 / 15
n Initialzation: ¨ 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 I 1 = {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: = c 5 + f 5 = 39, I 1 = {5}, Iovl = {1, 2, 3, 4} (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 16
ij 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 ωij is saving in transportation cost ij 1 2 1 5 2 2 4 3 3 4 5 1 6 4 5 4 4 1 (c) Prof. Richard F. Hartl 1 6 7 ωi fi 3 11 5 14 7 10 5 2 6 1 QEM - Chapter 1 when delivering to custonmer j, by opening additional location i. → row sum ωi is total saving in transportation cost when opening additional location i. Kapitel 3 / 17
n Iteration: ¨ 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 Kapitel 3 / 18
n Stiopping criterion: Stiop if no more cost saving are possible by additional locations from Iovl ¨ Build locationd from set I 1. ¨ Total cost Z ¨ assignment: xij = 1 iff ¨ n Beispiel: Iteration 1 ij 1 2 1 5 2 2 4 3 4 5 6 1 6 4 5 4 1 1 7 ωi fi 3 11 5 14 7 10 5 2 6 1 Fix k = 2 Forbid i = 4 Because of ω4 < f 4 location 4 is forbidden finally. Location k=2 is built. ¨ Now Z = 39 – 7 = 32 and Iovl = {1, 3}, I 1 = {2, 5}, Io = {4}. Update savings ωij. ¨ (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 19
ij 1 2 1 1 2 3 4 5 6 3 n ωi fi 3 6 5 Fix k = 1 1 1 5 Forbid i = 3 Iteration 2: ¨ n 7 Location 3 is forbidden, location k = 1 is finally built. Ergebnis: Final solution I 1 = {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 Kapitel 3 / 20
DROP for WLP n Die Set Iovl is replaced by I 1 vl. ¨ I 1 vl set of preliminarily built locations (yi =1 tentatively) n DROP works the other way round compared to ADD, i. e. start with all locations temporarily built; in each iteration remove one location… n Initialisation: I 1 vl = I, I 0 = I 1 = { } n 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 QEM - Chapter 1 Kapitel 3 / 21
n Expand matrix C: ¨ Row m+1 (row m+2) contains smallest ch 1 j (second smallest ch 2 j) only consider locations not finally deleted → 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 ch 2 j - ch 1 j Example: Initialisation and Iteration 1: I 1 vl ={1, 2, 3, 4, 5} ¨ n ij 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 delete 3 7 6 1 5 3 10 5 4 6 5 10 2 6 3 6 1 6 5 6 4 6 3 7 2 6 1 5 ch 1 j 6 1 2 0 2 3 ch 2 j 7 2 4 1 3 3 3 5 h 1 8 1 1 2 4 2 5 1 h 2 9 2 5 3 4 3 (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 22
¨ For all i from I 1 vl 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 = h 1 contains the smallest element. ¨ 2 examples: n n ¨ n 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 If fixed costs savings fi exceed additional transportation cost δi, 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 = {1}, but 1 is no candidate for dropping. Hence do not compute δi there. (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 23
ij 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 forbid 5 6 4 6 3 7 2 6 1 5 ch 1 j 1 2 3 2 3 ch 2 j 6 4 6 3 5 h 1 1 1 3 4 3 5 1 h 2 4 5 5 5 1 4 3 Location 3 is finally built, location 4 finally dropped. (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 24
n Iteration 3: I 1 vl = {5}, I 1 = {1, 3}, I 0 = {2, 4} ¨ ij 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 ch 1 j 1 2 1 3 3 2 3 ch 2 j 6 4 6 5 6 7 5 h 1 1 1 3 5 1 h 2 5 5 5 3 1 1 3 build Location 5 is finally built (c) Prof. Richard F. Hartl QEM - Chapter 1 Kapitel 3 / 25
n 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 n 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 QEM - Chapter 1 Kapitel 3 / 27
P-Median n n 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 28
MIP for p-Median 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 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 Kapitel 3 / 29
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