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Chapter 1

Chapter 1. Strategic Problems: Location. PS 1. PS 2. PS 3. PS 4. Production site. …. transportation. Full truck load. CW 1. CW 2. Central warehouse. …. transportation. FTL or tours. DC 1. DC 2. DC 3. DC 4. Distribution centers. …. transportation. tours. C 1. C 2. C 3.

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Chapter 1

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  1. Chapter 1 Strategic Problems: Location

  2. PS1 PS2 PS3 PS4 Production site … transportation Full truck load CW1 CW2 Central warehouse … transportation FTL or tours DC1 DC2 DC3 DC4 Distribution centers … transportation tours C1 C2 C3 C4 customers … Location problems QEM - Chapter 1

  3. 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) QEM - Chapter 1

  4. Median Problem • 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 • 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 QEM - Chapter 1

  5. 3 2/0 4/3 2 2 2 6/2 3 4 1/4 5 3 2 3/2 5/1 4 D= b= Example: fromDomschke und Drexl(Logistik: Standorte, 1990, Kapitel 3.3.1) weight Distancebetween locations QEM - Chapter 1

  6. 4*0 0*12 2*2 3*10 6*1 2*12 64 1 4*0 4*12 4*2 4*10 24 48 4*2 0*0 2*3 3*3 7*1 2*5 40 2 0*2 0*0 0*3 0*3 0 0 4*12 0*10 2*0 3*8 4*1 2*10 96 3 2*12 2*10 2*0 2*8 8 20 4*4 0*2 2*5 3*0 5*1 2*2 4 3*4 3*2 3*5 3*0 15 6 35 4*8 0*6 2*9 3*4 0*1 2*6 74 5 1*8 1*6 1*9 1*4 0 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 σin(i) 66 98 56 74 53 80 Example: Median σout(i) 1 2 3 4 5 6 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

  7. Related Problem: Center • Median • Node with min total weighted distance • → min. • Center • Node with min Maximum (weighted) Distance • → min. QEM - Chapter 1

  8. 4*0 0*12 2*2 3*10 6*1 2*12 30 1 4*0 4*12 4*2 4*10 24 48 4*2 0*0 2*3 3*3 7*1 2*5 10 2 0*2 0*0 0*3 0*3 0 0 4*12 0*10 2*0 3*8 4*1 2*10 48 3 2*12 2*10 2*0 2*8 8 20 4*4 0*2 2*5 3*0 5*1 2*2 4 3*4 3*2 3*5 3*0 15 6 16 4*8 0*6 2*9 3*4 0*1 2*6 32 5 1*8 1*6 1*9 1*4 0 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 Solution out(i) 1 2 3 4 5 6 City1 is center since 30+24 = 54 minimal QEM - Chapter 1

  9. W1 W2 m C1 C2 C3 C4 n Uncapacitated (single-stage) Warehouse Location Problem – LP-Formulation • single-stage WLP: • warehouse • customer: • 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. QEM - Chapter 1

  10. 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 QEM - Chapter 1

  11. Example: from Domschke & Drexl (Logistik: Standorte, 1990, Kapitel 3.3.1) Solution 2: just warehouses 1 and 3 Solution 1: all warehouses 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 QEM - Chapter 1

  12. 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. QEM - Chapter 1

  13. i = 1, …, m j = 1, …,n xij≤ yi j = 1, …,n i = 1, …, m For all i and j MIP for WLP transportation cost+ fixed cost Delivery only from locations i that are built Satisfy total demand of customer j yi is binary xij non negative QEM - Chapter 1

  14. 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) QEM - Chapter 1

  15. 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) QEM - Chapter 1

  16. 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 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} QEM - Chapter 1

  17. 5 2 1 3 11 5 4 6 4 14 7 5 4 1 10 5 1 1 2 6 ωijis 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. QEM - Chapter 1

  18. Iovl = Iovl – {k} and Z = Z – ωk + fk For all with ωi ≤ fi : • 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. • 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 Iovl and all customers j : ωij = max {ωij - ωkj, 0} QEM - Chapter 1

  19. 5 2 1 3 11 5 4 6 4 14 7 5 4 1 10 5 1 1 2 6 • Stiopping criterion: • 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 Fix k = 2 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. QEM - Chapter 1

  20. 1 2 3 6 5 1 5 Fix k = 1 Forbid i = 3 1 • 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. QEM - Chapter 1

  21. DROP for WLP • Die Set Iovl is replaced by I1vl. • I1vlset 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 QEM - Chapter 1

  22. 5 build delete 1 0 1 1 1 2 0 2 3 2 3 2 4 1 3 3 3 5 1 1 2 4 2 5 1 2 5 3 5 3 4 3 • 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 • Expand matrix C: • Row m+1 (row m+2) contains smallest ch1j (second smallest ch2j) only consider locations not finally deleted → • Example: Initialisation and Iteration 1: I1vl ={1,2,3,4,5} QEM - Chapter 1

  23. 2 examples: • For all i from I1vl compute increase in transportation cost δiif 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. • 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. QEM - Chapter 1

  24. - 8 build 1 forbid 1 1 2 1 2 3 2 3 6 4 6 3 6 3 5 3 1 1 4 3 5 1 4 5 5 5 1 4 3 Location 3 is finally built, location 4 finally dropped. QEM - Chapter 1

  25. - - 7 1 2 1 3 3 2 3 6 4 6 5 6 7 5 1 1 3 5 3 5 1 5 5 5 3 1 1 3 • Iteration 3: • I1vl = {5}, I1 = {1,3}, I0 = {2,4} build Location 5 is finally built QEM - Chapter 1

  26. 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) QEM - Chapter 1

  27. 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. QEM - Chapter 1

  28. P-Median Number of facilities is fixed … p Typically fixed costs are not needed (but can be considered if not uniform) QEM - Chapter 1

  29. i = 1, …, m j = 1, …,n xij ≤ yi j = 1, …,n i = 1, …, m For all i and j MIP for p-Median transportation cost+ fixed cost Delivery only from locations i that are built Satisfy total demand of customer j yi is binary xij non negative Exactly p facilities QEM - Chapter 1

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