Strategic Problems: Location Chapter 1 Strategic Problems: Location

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

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

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

Example: fromDomschke und Drexl (Logistik: Standorte, 1990, Kapitel 3 2 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

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 35+74 = 109 minimal OUT e.g: emergency delivery of goods IN e.g.: collection of hazardous waste (c) Prof. Richard F. Hartl QEM - Chapter 1

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

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 30+24 = 54 minimal (c) Prof. Richard F. Hartl QEM - Chapter 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Iteration 3: I1vl = {5}, I1 = {1,3}, I0 = {2,4} i\j 1 2 3 4 5 6 7 10 9 fi - 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

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

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

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

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