ISE 754: Logistics Engineering. Michael G. Kay Spring 2018

Transcription

1 ISE 754: Logistics Engineering Michael G. Kay Spring 2018

2 Topics 1. Introduction 2. Facility location 3. Freight transport Exam 1 (take home) 4. Network models 5. Routing Exam 2 (take home) 6. Warehousing Final project Final exam (in class) Outside the box Inside the box 2

3 Strategic (years) Network design Tactical (weeks year) Scope Multi echelon, multi period, multi product production and inventory models Operational (minutes week) Vehicle routing 3

4 Strategic: Network Design Optimal locations for five DCs serving 877 customers throughout the U.S. 4

5 Tactical: Production Inventory Model p c i c s c 0 p c i c s c 0 Product Product 2 k1 k Constraint matrix for a 2 product, multi period model with setups 5

6 Vehicle Routing Eight routes served from DC in Harrisburg, PA Route Summary Information Route Load Weight Route Time Customers in Route Layover Required 1 12, , , , , , , ,

7 Geometric Mean How many people can be crammed into a car? Certainly more than one and less than 100: the average (50) seems to be too high, but the geometric mean (10) is reasonable Geometric Mean: X LBUB Often it is difficult to directly estimate input parameter X, but is easy to estimate reasonable lower and upper bounds (LB and UB) for the parameter Since the guessed LB and UB are usually orders of magnitude apart, use of the arithmetic mean would give too much weight to UB Geometric mean gives a more reasonable estimate because it is a logarithmic average of LB and UB 7

8 Fermi Problems Involves reasonable (i.e., +/ 10%) guesstimation of input parameters needed and back of the envelope type approximations Goal is to have an answer that is within an order of magnitude of the correct answer (or what is termed a zeroth order approximation) Works because over and under estimations of each parameter tend to cancel each other out as long as there is no consistent bias How many McDonald s restaurants in U.S.? (actual 2013: 14,267) Parameter LB UB Estimate Annual per capita demand 1 1 order/person day x 350 day/yr = (order/person yr) U.S. population 300,000,000 (person) Operating hours per day 16 (hr/day) Orders per store per minute (in store + drive thru) 1 (order/store min) Analysis Annual U.S. demand (person) x (order/person yr) = 5,612,486,080 (order/yr) Daily U.S. demand (order/yr)/365 day/yr = 15,376,674 (order/day) Daily demand per store (hrs/day) x 60 min/hr x (order/store min) = 960 (order/store day) Est. number of U.S. stores (order/day) / (order/store day) = 16,017 (store) 8

9 System Performance Estimation Often easy to estimate performance of a new system if can assume either perfect or no control Example: estimate waiting time for a bus 8 min. avg. time (aka headway ) between buses Customers arrive at random assuming no web based bus tracking Perfect control (LB): wait time = half of headway No control (practical UB): wait time = headway assuming buses arrive at random (Poisson process) 8 Estimated wait time LBUB min 2 Bad control can result in higher values than no control 9

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11 Crowdsourcing Obtain otherwise hard to get information from a large group of online workers Amazon s Mechanical Turk is best known Jobs posted as HITs (Human Information Tasks) that typically pay $1 2 per hour Main use has been in machine learning to create tagged data sets for training purposes Has been used in logistics engineering to estimate the percentage homes in U.S. that have sidewalks (sidewalk deliveries by Starship robots) 11

12 Starship Technologies Started by Skype co founders 99% autonomous Goal: deliver two grocery bags worth of goods (weighing up to 20lbs) in 5 30 minutes for times less than the cost of current last mile delivery alternatives. 12

13 Matrix Multiplication mnn p m p Arrays must have same inner dimensions 13

14 Element by Element Multiplication b expanded to have compatible size with A mn1n mn 14

15 Compatible Sizes Two arrays have compatible sizes if, for each respective dimension, either has the same size, or size of one of arrays is one, in which case it is automatically duplicated so that it matches the size of the other array A A a a mn.b mp. b 1 mn n 1. m Bpn. m1 bn1 15

16 d 1 2 D Euclidean Distance x , P y d 3 d d 1 d x p x p d x1 p1,1 x2 p1, ,1 2 2,2 x1 p3,1 x2 p3,

17 Logistics Software Stack MIP Data Solver (Gurobi, etc.) (csv,excel,etc.) Standard Report Library (C,Java) (GUI,web,etc.) Scripting (Python,Matlab,etc.) User Library (in script language) MIP Solver (Gurobi,Cplex,etc.) Standard Library (in compiled C,Java) Commercial Software (Lamasoft,etc.) New Julia (0.6.2) scripting language almost as fast as C and Java (but not FORTRAN) does not require compiled standard library for speed 17

18 Basic Matlab Workflow Given problem to solve: 1. Test critical steps at Command Window 2. Copy working critical steps to a cell (&&) in script file (myscript.m) along with supporting code (can copy selected lines from Command History) Repeat using new cells for additional problems Once all problems solved, report using: >> diary hw1soln.txt Evaluate each cell in script: To see code + results: select text then Evaluate Selection on mouse menu (or F9) To see results: position cursor in cell then Evaluate Current Section (Cntl+Enter) >> diary off Can also report using Publish (see Matlab menu) as html or Word Submit all files created, which may include additional Data files (myscript.mat) or spreadsheet files (myexcel.xlsx) Function files (myfun.m) that can allow use to re use same code used in multiple problems All code inside function isolated from other code except for inputs/outputs: [out1,out2] = myfun(input1,input2) 18

19 Taxonomy of Location Problems 19

20 Hotelling's Law

21 1 D Cooperative Location Min TC wi di Min TC wi di Min TC wi d 2 k i Squared Euclidean Distance Center of Gravity: a 1 2 * 0, a 30 2 i i i i TC w d w x a dtc dx 2 w xa 0 x w w a x i i i i i wa i i 1(0) 2(30) 20 w 1 2 i 2 21

22 Nonlinear Location TC wi d k i k = 1 k = 1.4 k = 2 k = 4 55 Normalized TC mile 22

23 Minimax and Maximin Location Minimax Min max distance Set covering problem Maximin Max min distance AKA obnoxious facility location 23

24 2 EF Minisum Location +w 1 +w 2 TC w 2 -w 1 -w 2 +w 1 -w w 1 -w w x i i i i i, if TC( x) w d ( x x ) ( x x ), where TC(25) w (25 10) ( w )(25 30) 1 2 5(15) ( 3)( 5) 90 w, if x x w x x i i 24

25 Median Location: 1 D 4 EFs TC Minimum at point where TC curve slope switches from (-) to (+) w i = = = = = < W/2 5+3=8 > W/ =9 > W/2 4+2=6 < W/2 4 < W/2 25

26 Median Location: 1 D 7 EFs j i1 w i w i : W : 2 26

27 Median Location: 2 D Rectilinear Distance 8 EFs wi : y : 62 < < d ( P, P ) x x y y ( 1, 2) d P P x x y y 129 = 129 * 48 Y * = < X 27

28 Logistics Network for a Plant Tier Two Suppliers Tier One Suppliers Plant DCs Customers DDDD BBBB B = D + E A = B + C C = F + G downstream upstream Assembly Network Procurement Inbound Logistics Raw Materials vs. vs. vs. vs. Distribution Network Distribution Outbound Logistics Finished Goods 28

29 Basic Production System FOB (free on board) 29

30 FOB and Location Choice of FOB terms (who directly pays for transport) usually does not impact location decisions: Procurement cost Landed cost at supplier Production Procurement Local resource cost cost cost (labor, etc.) Total delivered cost Production cost Inbound transport cost Outbound transport cost Transport cost Inbound transport Outbound transport (TC) cost cost Purchase price from supplier and sale price to customer adjusted to reflect who is paying transport cost Usually determined by who can provide the transport at the lowest cost Savings in lower transport cost allocated (bargained) between parties 30

31 Monetary vs. Physical Weight m min TC( X ) wi d( X, Pi) fi ri d( X, Pi) i1 i1 where TC total transport cost ($/yr) w f i i r i m d( X, P) distance between NF at X and EF at P (mi) i w i monetary weight ($/mi-yr) physical weight rate (ton/yr) transport rate ($/ton-mi) NF = new facility to be located EF = existing facility m number of EFs i i (Montetary) Weight Gaining: Physically Weight Losing: w f in in w f out out 31

32 Minisum Location: TC vs. TD Assuming local input costs are same at every location or insignificant as compared to transport costs, the minisum transport oriented single facility location problem is to locate NF to minimize TC Can minimize total distance (TD) if transport rate is same: min TD( X ) wi d( X, Pi) fi ri d( X, Pi) where TD w f i m i1 i1 total transport distance (mi/yr) monetary weight (trip/yr) transport rate = 1 m d( X, P) per-trip distance between NF an d EF (mi/trip) i i r i w i trips per year (trip/ yr) i 32

33 Example: Single Supplier/Customer Asheville 50 Statesville Winston-Salem Greensboro Durham RaleighWilmington Wilmington Winston- Salem The cost per ton mile (i.e., the cost to ship one ton, one mile) for both raw materials and finished goods is $ Where should the plant for each product be located? 2. How would location decision change if customers paid for distribution costs (FOB Origin) instead of the producer (FOB Destination)? In particular, what would be the impact if there were competitors located along I 40 producing the same product? 3. Which product is weight gaining and which is weight losing? 4. If both products were produced in a single shared plant, why is it now necessary to know each product s annual demand (f i )? raw material ubiquitous inputs 2 ton finished goods Product B 1 ton 3 ton 33

34 1 D Location with Procurement and Distribution Costs Asheville Durham Production System Assume: all scrap is disposed of locally unit of finished good 1 ton 5 TC w d i i ($/yr) ($/mi-yr) (mi) monetary weight physical weight w f r i i i ($/mi-yr) (ton/yr) ($/ton-mi) rin $0.33/ton-mi rout $1.00/ton-mi 3 3 f3 10, w3 f3rout 10 f 1 1 BOM1 fi , w1 f1rin 40 i1 NF 4 f , w4 f4rout 20 f2 BOM2 fi , w2 f2rin 40 2 f 30, w f r 30 i out j i1 w i w i W 2 : : (Montetary) Weight Gaining: w 50 w 60 Physically Weight Losing: f 150 f 60 in in out out 34

35 d 1 2 D Euclidean Distance x , P y d 3 d d 1 d x p x p d x1 p1,1 x2 p1, ,1 2 2,2 x1 p3,1 x2 p3,

36 Minisum Distance Location * d P ,1 2,2 2 2 i( x) i i d x p x p TD( x) d ( x) TD * 3 i1 x argmin TD( x) * * TD( x ) i x x Fermat s Problem (1629): Given three points, find fourth (Steiner point) such that sum to others is minimized (Solution: Optimal location corresponds to all angles = 120 ) 36

37 Minisum Weighted Distance Location Solution for 2 D+ and non rectangular distances: Majority Theorem: Locate NF at m W EFj if w, wherew w j 2 i1 Mechanical (Varigon frame) 2 D rectangular approximation Numerical: nonlinear unconstrained optimization Analytical/estimated gradient (quasi Newton, fminunc) Direct, gradient free (Nelder Mead, fminsearch) i m number of EFs TC( x) w d ( x) x TC * m i1 arg min TC( x) * * TC( x ) i x Varignon Frame i 37

38 Convex vs Nonconvex Optimization 38

39 Gradient vs Direct Methods Numerical nonlinear unconstrained optimization: Analytical/estimated gradient quasi Newton fminunc Direct, gradient free Nelder Mead fminsearch 39

40 Nelder Mead Simplex Method reflection expansion AKA amoeba method Simplex is triangle in 2 D (dashed line in figures) outside contraction inside contraction a shrink 40

41 Feasible Region if a is true return b else return c end b, if a is true TC( ), if is in iff ( a, b, c) c, otherwise x x R, otherwise 41

42 Computational Geometry Design and analysis of algorithms for solving geometric problems Modern study started with Michael Shamos in 1975 Facility location: geometric data structures used to simplify solution procedures

43 Convex Hull Find the points that enclose all points Most important data structure Calculated, via Graham s scan in On ( log n), npoints

44 Delaunay Triangulation Find the triangulation of points that maximizes the minimum angle of any triangle Captures proximity relationships Used in 3 D animation Calculated, via divide and conquer, in On ( log n), npoints

45 Voronoi Diagram Each region defines area closest to a point Open face regions indicate points in convex hull

46 Delaunay Voronoi Delaunay triangulation is straight line dual of Voronoi diagram Can easily convert from one to another

47 Minimum Spanning Tree Find the minimum weight set of arcs that connect all nodes in a graph Undirected arcs: calculated, via Kruskal s algorithm, Om ( log n), marcs, nnodes Directed arcs: -2 calculated, via Edmond s branching -3 algorithm, in Omn ( ), marcs, nnodes 47

48 Steiner Network 48

49 Metric Distances 49

50 Chebychev Distances Proof 50

51 Metric Distances using dists D 'mi' 'km' dists( X1, X2, p), p Inf nd md 3 2 n m d = 2 d = 1 X1, X2 X1, X2 X1, X2 d d D X1, X2, 2 D X1 X Error 51

52 Heuristic Solutions Most problems in logistics engineering don t admit optimal solutions, only Within some bound of optimal (provable bound, opt. gap) Best known solution (stop when need to have solution) Heuristics computational effort split between Construction: construct a feasible solution Improvement: find a better feasible solution Easy construction: any random point or permutation is feasible can then be improved Hard construction: almost no chance of generating a random feasible solution is possible in a single step need to include randomness at decision points as solution is constructed 52

53 Heuristic Construction Procedures Easy construction: any random permutation is feasible and can then be improved Hard construction: almost no chance of generating a random solution is a single step that is feasible, need to include randomness at decision points as solution is constructed

54 Great Circle Distances mi R 54

55 Circuity Factor droadi Circuity Factor: g, where usually 1.15 g 1.5 d GC i d gd ( P, P ), estimated road distance from P to P road GC From High Point to Goldsboro: Road = 143 mi, Great Circle = 121 mi, Circuity = High Point Goldsboro

56 Mercator Projection x proj x x rad xdeg 180 and x deg x rad 180 y proj sinh y tan 1 1 tan y sinh y proj 56

57 Allocation Example: given n DCs and m customers, with customer j receiving w j TLs per week, determine the total distance per week assuming each customer is served by its closest DC d ij 1 w d j ij 1 w allocate 1 2 D TD 2(10) 4(20) 6(25) 8(15) DCs DCs Customers Customers 57

58 Pseudocode Different ways of representing how allocation and TD can be calculated High level pseudocode most concise, but leaves out many implementation details (e.g., sets don t specify order) Low level pseudocode includes more implementation details, which can hide the core idea, and are usually not essential Low level Pseudocode High level Pseudocode Matlab/Matlog N 1,, n, n N M 1,, m, m M j TD wjd j jm arg min d, j in ij 58

59 Minisum Multifacility Location nno. of NFs, mno. of EFs Sup pliers EFs Manufactu ring 1 2 NFs Distribu tion EFs Customers X V W nd nn nm NF locations, P md EF locations n n n m TC( X) v d( X, X ) w d( X, P) jk j k ji j i j1 k1 j1 i1 X * arg min TC( X) X TC * * TC( X ) 59

60 Multiple Single Facility Location Sup pliers 1 2 Manufactu ring 4 5 Distribu tion Customers n n n m TC( X) v d( X, X ) w d( X, P) jk j k ji j i j1 k1 j1 i1 n j1 TC( X ) j 3 3 EFs NFs 9 EFs 60

61 Facility Location Allocation Problem NFs EFs Determine both the location of n NFs and the allocation of flow requirements of m EFs that minimize TC w r f (1) f flow between NF j and EFi w ji ji ji ji i total flow requirememts of EFi n TC( XW, ) w d( X, P) m j1 i1 n * * X, W arg min TC( X, W) : wji wi, wji 0 XW, j1 TC * * * TC( X, W ) ji j i 61

62 Integrated Formulation If there are no capacity constraints on NFs, it is optimal to always satisfy all the flow requirements of an EF from its closest NF Requires search of (n x d) dimensional TC that combines location with allocation ( X) arg min d( X, P) i j i j NFs 4 EFs TC( X) w d( X, P) X TC * m i1 * * ( X) arg min TC( X) TC( X ) i X i i 62

63 Alternating Formulation NFs EFs Alternate between finding locations and finding allocations until no further TC improvement Requires nd dimensional location searches together with separate allocation procedure Separating location from allocation allows other types of location and/or allocation procedures to be used: Allocation with NF with capacity constraints (solved as minimum cost network flow problem) Location with some NFs at fixed locations wi, if argmin d( Xk, Pi) j allocate( X) wji k 0, otherwise n TC( XW, ) w d( X, P) j1 i1 locate( WX, ) argmin TC( XW, ) m X ji j i 63

64 ALA: Alternate Location Allocation NFs 4 EFs 64

65 Best Retail Warehouse Locations 65

66 Aggregate Demand Point Data Sources Aggregate demand point: centroid of population Good rule of thumb: use 100x number of NFs ( 1000 pts provides good coverage for locating 10 NFs) 1. City data: ONLY USE FOR LABELING!, not as demand points 2. 3 digit ZIP codes: 1000 pts covering U.S., = 20 pts NC 3. County data: 3000 pts covering U.S., = 100 pts NC Grouped by state or CSA (Combined Statistical Area) CSA = defined by set of counties (174 CSAs in U.S.) FIPS code = 5 digit state county FIPS code = 2 digit state code + 3 digit county code = = 37 NC FIPS Wake FIPS CSA List: www2.census.gov/programs surveys/metro micro/geographies/ reference files/2017/delineation files/list1.xls 4. 5 digit ZIP codes: > 35K pts U.S., 1000 pts NC 5. Census Block Group: > 220K pts U.S., 1000 pts Raleigh Durham Chapel Hill, NC CSA Grouped by state, county, or CSA 66

67 City vs CSA Population Data 67

68 Demand Point Aggregation Existing facility (EF): actual physical location of demand source Aggregate demand point: single location representing multiple demand sources Asheville 50 Statesville Winston-Salem Greensboro Durham RaleighWilmington

69 Demand Point Aggregation Calculation of aggregate point depends on objective For minisum location, would like for any location x: w w d( x, x ) wd( x, x ) w d( x, x ), w.l.o.g., let x0, x, x agg w w x w x w x 1 2 agg x agg wx w wx w centroid For squared distance: w1w2xagg w1x w 1 2x2 x agg wx 2 wx w 1 2 w 1 2 not centroid 69

70 Optimal Number of NFs 70

71 Uncapacitated Facility Location (UFL) NFs can only be located at discrete set of sites Allows inclusion of fixed cost of locating NF at site Variable costs are usually transport cost from NF to EF Total of 2 n 1 potential solutions (all nonempty subsets of sites) M N 1,..., m, existing facilites (EFs) 1,..., n, sites available to locate NFs M M, set of EFs served by NF at site i i c variable cost to serve EF j from NF at site i ij k i Y * * fixed cost locating NF at site i N, sites at which NFs are located Y arg min k c : M M Y Y i ij i iy iy jm iy min cost set of sites where NFs located number of NFs located i 71

72 UFL Solution Techniques Being uncapacitated allows simple heuristics to be used to solve ADD construction: add one NF at a time DROP construction: drop one NF at a time XCHG improvement: move one NF at a time to unoccupied sites HYBRID algorithm combination of ADD and DROP construction with XCHG improvement, repeating until no change in Y Use as default heuristic for UFL See Daskin [2013] for more details UFL can be solved as a MILP Easy MILP, LP relaxation usually optimal (for strong formulation) MILP formulation allows constraints to easily be added e.g., capacitated facility location, fixed number of NFs, some NF at fixed location Will model UFL as MILP mainly to introduce MILP, will use UFL HYBRID algorithm to solve most problems 72

73 UFL ADD Example Asheville 50 Statesville 150 Greensboro k cij wj dij f j rdij (1)(1) dij dij cij Asheville: Statesville: Greensboro: Raleigh: Wilmington: Raleigh 420 Y Y cyj ky cyj ky , ,165 Y 3 Y cyj ky cyj ky 3, , , , Y 3,1 Y cyj ky cyj ky 3,1, ,1, ,1, * Y 3,1 73

74 UFLADD: Add Construction Procedure 74

75 UFLXCHG: Exchange Improvement Procedure 75

76 Modified UFLADD 76

77 UFL: Hybrid Algorithm 77

78 P Median Location Problem Similar to UFL, except Number of NF has to equal p (discrete version of ALA) No fixed costs p number of NFs * Y arg min c : M M, Y p Y ij iy jm iy i i 78

79 Bottom Up vs Top Down Analysis Bottom Up: HW 3 Q 3 P 32 * cary cary lon-lat of EFs 3 i1 * * 48, 24,35 (TL/yr) r 2 ($/TL-mi) 1 drd ( P1, P2) drd( P2, P3) drd( P3, P1) g 3 dgc ( 1, 2) dgc ( 2, 3) dgc ( 3, 1) P P P P P P TC() x f rgd (, x P) TC x TC x f arg min TC( x) TC( x ) lon-lat of Cary TC( x i GC i x cary ) cary * TC TC TC Top Down: estimate r (circuity factor cancels, so not needed, HW 4 Q 4) TC r cary nom 3 * current known TC, 10 ton /TL 480,240,350 (ton /yr) i1 3 i1 * * cary cary i GC x Pi fd i nom GC i cary * (, ) TC() x f r d (, x P) x TC f arg min TC( x) TC( x ) TC TC TC TC x ($/ton-mi) 79

80 U.S. Geographic Statistical Areas Defined by Office of Management and Budget (OMB) Each consists of one or more counties Top to bottom: 1. Metropolitan divisions 2. Combined statistical areas (CSAs) 3. Core based statistical areas (CBSAs) 4. Metropolitian/ micropolitan statistical areas (MSAs) 5. County (rural) 80

81 Transport Cost if NF at every EF Facility Fixed + Transport Cost Facility Fixed Cost Transport Cost TC 0? Number of NFs NF = EF transport cost fixed cost TC k c i ij iy iy jm i 81

82 Area Adjustment for Aggregate Data Distances a a a 2 d 0 LB d 0 UB LB: avg. dist. from center to all points in area UB: avg. dist. between all random pairs of points Local circuity factor = 1.5, regular non local = 1.2 a 2 LB d0 2 d d LB 0 UB 0 a a 5 2 LB UB a 0.51 d d d 0.45 a a Mathai, A.M., An Intro to Geo Prob, p. 207 (2.3.68) d ( X, X ) max gd ( X, X ), g 0.45max a, a a 1 2 GC 1 2 local 1 2 max 1.2 d ( X, X ), 0.675max a, a GC

83 Popco Bottling Company Example Problem: Popco currently has 42 bottling plants across the western U.S. and wants to know if they should consider reducing or adding plants to improve their profitability. Solution: Formulate as an UFL to determine the number of plants that minimize Popco s production, procurement, and distribution costs. 83

84 Popco Bottling Company Example Following representative information is available for each of N current plants (DC) i: xy i location f DC i TPC TDC i i aggregate production (tons) total production and procurement cost total distribution cost Assuming plants are (monetarily) weight gaining since they are bottling plants, so UFL can ignore inbound procurement costs related to location 84

85 Popco Bottling Company Example Difficult to estimate fixed cost of each new facility because this cost must not include any cost related to quantity of product produced at facility. 1. Use plant (DC) production costs to find UFL fixed costs via linear regression variable production costs c p do not change and can be cut in i TPC TPC k c f in only keep k for UFL p i DC 85

86 Popco Bottling Company Example Allocate all 3 digit ZIP codes to closest plant (up to 200 mi max) to serve as aggregate customer demand points max a a i hj ij h M j:argmind iandd d d M 200 mi M in i max

87 Popco Bottling Company Example 3. Allocate each plant s demand (tons of product) to each of its customers based on its population f q f DC jm fi j i population of EFj DC 5 f2 q q hm q5 q 5 6 i j q h 7 87

88 Popco Bottling Company Example 4. Estimate a nominal transport rate ($/ton mi) using the ratio of total distribution cost ($) to the sum of the product of the demand (ton) at each customer and its distance to its plant (mi). r nom in in jm TDC j j i f d a ij 88

89 Popco Bottling Company Example 5. Calculate UFL variable transportation cost c ij ($) for each possible NF site i (all customer and plant locations) and EF site j (all customer locations) as the product of customer j demand (ton), distance from site i to j (mi), and the nominal transport rate ($/ton mi). a ij imn nom j ij imn jm jm C c r f d 6. Solve as UFL, where TC returned includes all new distribution costs and the fixed portion of production costs. transport cost fixed cost TC k c i iy iy jm n M N i ij, number of potential NF sites m M, number of EF sites 89

90 MILP LP: max cx ' s.t. Ax b x 0 MILP: some integer x i max 6x 8x 1 2 s.t. 2x 3x x 7 1 x, x c A , 2 0 b 7 ILP: x integer BLP: x 0,1 x x , 31 1 cx * * x 1 90

91 Branch and Bound x max 6x1 8x2 s.t. 2x13x2 11 2x1 7 x1, x2 0 x, x integer x 1 c A, 2 0 b 7 1 UB 31, LB LP 6 2 UB 31, LB 0 3 x1 3 x1 4 x2 1 x2 2 Incumbent 2 3 UB 31, LB 26 1 UB 31, LB 26 x1 2 3 x1 3 2 UB 30, LB 26 3 Incumbent x2 2 x2 3 2 UB 30, LB Incumbent 2 UB 30, LB gap STOP 3 Fathomed, infeasible Fathomed, infeasible 91

92 MILP Solvers 92

93 MILP Solvers Presolve: eliminate variables 2x 2x 1, x, x 0 and integer x 2 x Gomory cut x x Cutting planes: keeps all integer solutions and cuts off LP solutions (Gomory cut) Heuristics: find good initial incumbent solution Parallel: use separate cores to solve nodes in B&B tree Speedup from : 320,000 computer speed 580,000 algorithm improvements 93

94 MILP Formulation of UFL min i i ij ij in in jm s.t. x 1, jm where in my x, i N i ky 0 x 1, in, j M ij jm ij ij y 0,1, in i cx k fixed cost of NF at site in 1,..., n i c variable cost from i to serve EF j M 1,..., m ij 1, if NF established at site i yi 0, otherwise x fraction of EF j demand served from NF at site i. ij y x, in, j M i ij 94

95 1 4 M 2 M * (Weighted) Set Covering 1,..., m, objects to be covered M M, in 1,..., n, subsets of M i c i cost of using M in cover i I arg min c : M M, min cost covering of M I ii i ii M M 1 M 3 M i ii M i * 1,...,6 N 1,...,5 1, 2, 1, 4, 5, 3, 5 2,3,6, M 6 M M M M c 1, for all in i * I arg min c : M M c i 2 2, 4 I ii i ii i 95

96 min in i (Weighted) Set Covering i s.t. a x 1, j M in M ji i x 0,1, in i * cx 1,..., m, objects to be covered M M, in 1,..., n, subsets of M i c i cost of using M in cover i I arg min c : M M, min cost covering of M I ii i ii i where a x i ji 1, if M i is in cover 0, otherwise 1, if j Mi 0, otherwise. 96

97 Set Packing Maximize the number of mutually disjoint sets Dual of Set Covering problem Not all objects required in a packing Limited logistics engineering application (c.f. bin packing) 1 M M 3 M 5 max in i s.t. a x 1, j M in ji i x 0,1, in i x

98 min im y i s.t. Vy v x, i M where y x i ij i j ij jm im x 1, jm ij y 0,1, im i x 0,1, im, jm ij M v j * Bin Packing 1,..., m, objects to be packed volume of object j V volume of each bin B max v V B arg min B : v V, B M, min packing of M B j jb BB i i 1, if bin Bi is used in packing 0, otherwise 1, if object j packed into bin Bi 0, otherwise. 98 i j i

99 Topics 1. Introduction 2. Facility location 3. Freight transport Exam 1 (take home) 4. Network models 5. Routing Exam 2 (take home) 6. Warehousing Final project Final exam (in class) 99

100 Logistics Engineering Design Constants 1. Circuity Factor: 1.2 ( g ) 1.2 GC distance actual road distance 2. Local vs. Intercity Transport: Local: < 50 mi use actual road distances Intercity: > 50 mi can estimate road distances mi return possible (11 HOS) > 250 mi always one way transport > mi intermodal rail possible 3. Inventory Carrying Cost ( h ) = funds + storage + obsolescence 16% average (no product information, per U.S. Total Logistics Costs) (16% 5% funds + 6% storage + 5% obsolescence) 5 10% low value product (construction) 25 30% general durable manufactured goods 50+% computer/electronic equipment >> 100% perishable goods (produce) 100

101 Logistics Engineering Design Constants 4. Value Transport Cost 1: $1 ft 3 $2,620 Shanghai LA/LB shipping cost 3 2,400 ft 40 ISO container capacity 5. TL Weight Capacity: 25 tons ( K wt ) (40 ton max per regulation) (15 ton tare for tractor trailer) = 25 ton max payload Weight capacity = 100% of physical capacity 6. TL Cube Capacity: 2,750 ft 3 ( K cu ) Trailer physical capacity = 3,332 ft 3 Effective capacity = 3, ,750 ft 3 Cube capacity = 80% of physical capacity Truck Trailer Cube = 3,332-3,968 CFT Max Gross Vehicle Wt = 80,000 lbs = 40 tons Max Payload Wt = 50,000 lbs = 25 tons Length: 48' - 53' single trailer, 28' double trailer Max Height: 13'6" = 162" Interior Height: (8'6" - 9'2" = 102" - 110") Width: 8'6" = 102" (8'2" = 98") 101

102 Logistics Engineering Design Constants 7. TL Revenue per Loaded Truck Mile: $2/mi in 2004 ( r ) TL revenue for the carrier is your TL cost as a shipper 15%, average deadhead travel $1.60, cost per mile in 2004 $ $ $1.88, cost per loaded mile 6.35%, average operating margin for trucking $2.00, revenue per loaded mile 102

103 Truck Shipment Example Product is to be shipped in cartons from Raleigh, NC (27606) to Gainesville, FL (32606). Each unit weighs 40 lb and occupies 9 ft 3, and units can be stacked on top of each other in a trailer. One Time Shipments (operational decision): know q Know when and how much to ship, need to determine if TL and/or LTL to be used Periodic Shipments (tactical decision): know f, determine q Need to determine how often and how much to ship 103

104 Truck Shipment Example: One Time 1. Assuming that the product is to be shipped P2P TL, what is the maximum payload for each trailer used for the shipment? q wt max K wt 25 ton K cu 2750 ft 3 s 40 lb/unit 3 9ft /unit lb/ft 3 K cu cu max q cu qmax s 2000 skcu 2000 wt cu skcu qmax min qmax, qmax min Kwt, (2750) min 25, ton

105 Truck Shipment Example: One Time 2. Next Monday, 350 units of the product are to be shipped. How many truckloads are required for this shipment? 40 q 7 q ton, 2 truckloads 2000 q max Using the most recent rate estimate available, what is the TL transport charge for this shipment? d 532 mi c r TL July 2017 TL 2004 TL PPI PPITL r2004 $2.00 / mi PPI $2.00 / mi $ / mi q 7 rd ( )(532) $2, q max

106 Truck Shipment Example: One Time 4. Using the most recent LTL rate estimate, what is transport charge to ship the fractional portion of the shipment LTL (i.e., the last partially full truckload portion)? q qq ton r frac LTL max LTL frac frac d 2 s s 2s $ / ton-mi (4.49) 14 2 c r q d ( )(532) $1, LTL PPI LTL q 106

107 Truck Shipment Example: One Time 5. What is the change in total charge associated with the combining TL and LTL as compared to just using TL? cc c c TL TL1 LTL q q rd rd rltl qfrac d q max q max $

108 Truck Shipment Example: One Time 6. What would the fractional portion have to be so that the TL and LTL charges are equal? q ctl( q) rd q max rltl( q) PPILTL q d c ( q) r ( q) qd LTL LTL q arg min c ( q) c ( q) I TL LTL q ton 2 s s 2s

109 Truck Shipment Example: One Time 7. What are the TL and LTL minimum charges? MC MC TL LTL r 45 $ PPILTL d $ Why do these charges not depend on the size of the shipment? Why does only the LTL minimum charge depend of the distance of the shipment? 109

110 Truck Shipment Example: One Time Independent Transport Charge ($): TL TL LTL LTL c0( q) min max c ( q), MC,max c ( q), MC Independent shipment charge: Class 200 from to Transport Charge ($) Shipment Size (ton) 110

111 Truck Shipment Example: One Time 8. Using the same LTL shipment, find online one time (spot) LTL rate quotes using the FedEx LTL website q frac ton (2000) 1778 lb Most likely freight class: 40 lb/unit s 3 9 ft /unit Class Density Relationship lb/ft Class 200 What is the rate quote for the reverse trip from Gainesville (32606) to Raleigh (27606)? 3 111

112 Truck Shipment Example: One Time The National Motor Freight Classification (NMFC) can be used to determine the product class Based on: 1. Load density 2. Special handling 3. Stowability 4. Liability 112

113 Truck Shipment Example: One Time CzarLite tariff table for O D pair cwt hundredweight 100 lb ton Tariff (in $/cwt) from Raleigh, NC (27606) to Gainesville, FL (32606) (532 mi, CzarLite DEMOCZ , minimum charge = $95.23) $ 113

114 Truck Shipment Example: One Time 9. Using the same LTL shipment, what is the transport cost found using the undiscounted CzarLite tariff? q , class 200 disc 0, MC B i arg qi q qq B i1 B B B 3 q2 3 arg q qq B 3 B i arg q B ctariff 1disc max MC,min OD( class, i)20 q, OD( class, i 1)20q i 10max 95.23, min OD(200,3) 20(0.8889), OD(200,4) 20(1) max95.23, min (99.92) 20(0.8889), (81.89)20(1) max 95.23,min 1,776.23,1, $1,

115 Truck Shipment Example: One Time 10. What is the implied discount of the estimated charge from the CzarLite tariff cost? disc ctariff c LTL c tariff 1, , , % What is the weight break between the rate breaks? q W i OD( class, i 1) OD( class, i) q B i $ (1) ton TC tariff w/o Break TC tariff ton 115

116 Truck Shipment Example: Periodic 11. Continuing with the example: assuming a constant annual demand for the product of 20 tons, what is the number of full truckloads per year? f 20 ton/yr q q ton/ TL (full truckload q q ) max f 20 n TL/yr, average shipment frequency q max Why should this number not be rounded to an integer value? 116

117 Truck Shipment Example: Periodic 12. What is the shipment interval? 1 q t yr/tl, average shipment interval n f 20 How many days are there between shipments? day/yr t day/tl n 117

118 Truck Shipment Example: Periodic 13. What is the annual full truckload transport cost? r d 532 mi, r $ / mi FTL r $0.3961/ ton-mi q max TC f r d n rd wd, w monetary weight in $/mi FTL FTL ( )532 $4, /yr What would be the cost if the shipments were to be made at least every three months? t max 3 1 yr/tl nmin 4 TL/yr 12 t TC max n, n rd FTL min max max , (532) $5,151.13/yr 118

119 Truck Shipment Example: Periodic Independent and allocated full truckload charges:, c ( ), FTL qq UB LB q qr d max 0 Transport Charge for a Shipment TL LTL 2 TL 3 TL Transport Charge ($) MC /2000 Independent Allocated Truckload Allocated Full Truckload Shipment Size (tons) 119

120 Truck Shipment Example: Periodic Total Logistics Cost (TLC) includes all costs that could change as a result of a logistics related decision TLC TC IC PC TC transport cost IC PC inventory cost IC IC IC cycle pipeline safety purchase cost Cycle inventory: held to allow cheaper large shipments Pipeline inventory: goods in transit or awaiting transshipment Safety stock: held due to transport uncertainty Purchase cost: can be different for different suppliers 120

121 Truck Shipment Example: Periodic 14. Since demand is constant throughout the year, one half of a shipment is stored at the destination, on average. Assuming that the production rate is also constant, one half of a shipment will also be stored at the origin, on average. Assuming each ton of the product is valued at $25,000 and loses 5% of its value after 3 months, what is a reasonable estimate for the total annual cost for this cycle inventory? IC cycle v (annualcost of holding one ton)(average annual inventory level) ( vh)( q) unit value of shipment ($/ton) h inventory carrying rate, the cost per dollar of inventory per year (1/yr) average inter-shipment inventory fraction at Origin and Destination q shipment size (ton) 121

122 Truck Shipment Example: Periodic Rate (h) = sum of interest + warehousing + obsolescence rate Interest: 5% per Total U.S. Logistics Costs Warehousing: 6% per Total U.S. Logistics Costs Obsolescence: default rate h annual = 0.3 h obs 0.2 Hourly vs Annual: h hour = h annual /H = 0.3/2000 = (H =operhr/yr) High FGI cost: h h obs, can ignore interest & warehousing Estimate h obs using percent reduction interval method: given time t h when product loses x h percent of its original value v, find h obs x h t v x vh t x h t h obs h h obs h h obs, and h th Example: If a product loses 5% of its value after 3 months: 3 xh 0.05 th 0.25 yr hobs 0.2 h t 0.25 h x h h obs 122

123 Truck Shipment Example: Periodic Average annual inventory level q q 1 1 q(1) q q q q 2 0 q

124 Truck Shipment Example: Periodic Inter shipment inventory fraction alternatives: O D q 2 q 2 q q Batch Production Immediate Consumption

125 Truck Shipment Example: Periodic Reasonable estimate for the total annual cost for the cycle inventory: IC cycle where vhq (1)(25,000)(0.3) $45, / yr 1 1 at Origin + at Destination v $25,000 unit value of shipment ($/ton) h 0.3 estimated carrying rate (1/yr) q q = FTL shipment size (ton) max 125

126 Truck Shipment Example: Periodic 15. What is the annual total logistics cost (TLC) for these fulltruckload TL shipments? TLC TC IC FTL nrd FTL vhq cycle ( )532 (1)(25,000)(0.3) , , $50, / yr 126

127 Truck Shipment Example: Periodic 16. What is the minimum possible annual total logistics cost for independent TL shipments, where the shipment size can now be less than a full truckload? f f TLCTL( q) TCTL( q) IC( q) ctl( q) vhq rd vhq q q dtlctl( q) * frd 20( )532 0 qtl ton dq vh (1)25000(0.3) f TLC q rd vhq * * TL( TL) * TL qtl 20 ( )532 (1)25000(0.3) , , $27, / yr 127

128 Truck Shipment Example: Periodic Including the minimum charge and maximum payload restrictions: q * TL f max rd, MC TL frd min, qmax vh vh What is the TLC if this size shipment could be made as an allocated full truckload? f r TLC q q r d vhq f d vhq * * * * AllocFTL( TL) * TL FTL TL TL qtl qmax (1)25000(0.3) , , $18, / yr vs. $27, as independent P2P TL 128

129 Truck Shipment Example: Periodic 17. What is the optimal LTL shipment size? f TLCLTL( q) TCLTL( q) IC( q) cltl( q) vhq q * LTL q arg min TLC ( q) ton q LTL Must be careful in picking starting point for optimization since r LTL PPI LTL q d s s 2s ,000 q, , ft 3 q d 2,000 2,000 s 129

130 Truck Shipment Example: Periodic 18. Should the product be shipped TL or LTL? * * * LTL LTL LTL LTL LTL TLC ( q ) TC ( q ) IC( q ) 32, , $38, / yr TLC LTL $ per year TC LTL TLC TL TC TL IC Shipment weight (tons) 130

131 Truck Shipment Example: Periodic 19. If the value of the product increased to $85,000 per ton, should the product be shipped TL or LTL? TLC LTL TLC LTL $ per year TLC TL $ per year IC TC LTL TC TL IC TLC TL TC TL TC LTL Shipment weight (tons) Shipment weight (tons) (a) $25000 value per ton (b) $85000 value per ton 131

132 Truck Shipment Example: Periodic Better to pick from separate optimal TL and LTL because independent charge has two local minima: * q0 arg min TLC ( q), TLC ( q) TL LTL q! * f q0 arg min c0( q) vhq q q 132

133 Truck Shipment Example: Periodic 20. What is optimal independent shipment size to ship 80 tons per year of a Class 60 product valued at $5000 per ton between Raleigh and Gainesville (assuming the same carrying rate)? * 0 3 q arg min TLC ( q), TLC ( q) ton q * * 0 LTL 0 TLC ( q ) $24, / yr TLC ( q ) TL s lb/ft TL LTL 133

134 Truck Shipment Example: Periodic 21. What is the optimal shipment size if both shipments will always be shipped together on the same truck (with same shipment interval)? d d, h h, f f f ton s agg 1 2 agg 1 2 agg 1 2 fagg fagg aggregate weight, in lb f lb/ft 3 aggregate cube, in ft f1 f s s agg f f v v v 85, $21, 000 / ton q * TL faggrd 100(2.3953)532 v h (1)21000(0.3) agg ton 134

135 Truck Shipment Example: Periodic Summary of results: 135

136 Location and Transport Costs Monetary weights w used for location are, in general, a function of the location of a NF Distance d appears in optimal TL size formula TC & IC functions of location Need to minimize TLC instead of TC FTL (since size is fixed at max payload) results in only constant weights for location Need to only minimize TC since IC is constant in TLC m m fi TL i i i i i i1 i1 qi( x) TLC ( x) w ( x) d ( x) vhq ( x) rd ( x) vhq ( x) m m fi firdi( x) 1 rdi( x) vh firdi( x) vh i1 frd i i( x) vh i1 vh vh m m fi FTL i max i i max FTL i1 qmax i1 TLC ( x) rd ( x) vhq w d ( x) vhq TC ( x) constant 136

137 FTL Location Example Where should a DC be located in order to minimize 36.5 transportation costs, given: 1. FTLs containing mix of products A and B shipped P2P from DC to customers in Winston Salem, Durham, and Wilmington 2. Each customer receives 20, 30, and 50% of total demand tons/yr of A shipped FTL P2P to DC from supplier in Asheville tons/yr of B shipped FTL P2P to DC from Statesville 5. Each carton of A weighs 30 lb, and occupies 10 ft 3 6. Each carton of B weighs 120 lb, and occupies 4 ft 3 7. Revenue per loaded truck mile is $2 8. Each truck s cubic and weight capacity is 2,750 ft 3 and 25 tons, respectively Asheville 50 Statesville Winston-Salem Greensboro Durham RaleighWilmington

138 FTL Location Example Asheville DC 35% Statesville Winston- Salem TC w d i i ($/yr) ($/mi-yr) (mi) w f r n r i i FTL, i i i ($/mi-yr) (ton/yr) ($/ton-mi) (TL/yr) ($/TL-mi) Durham Wilmington r r FTL, i ($/ton-mi) qmax qmax, n f fagg (2750 r $2 / TL-mi, fagg fa fb ton/yr, sagg lb/ft, qmax 25, fa f B (2750) sa sb 3 30 s1 3 lb/ft, qmax min 25, ton f fagg 96, n3 6.69, w3 6.69(2) f1 100, n , w (2) DC 4 f fagg 144, n , w (2) (2750) 2 s2 30 lb/ft, qmax min 25, 25 ton f fagg 240, n , w (2) f2 380, n2 15.2, w2 15.2(2) W W 146, 73 2 w i : Asheville Statesville Winston-Salem Durham Wilmington < 73 * (Montetary) Weight Losing: w 79 w 67 n 39 n 33 in 78 > 73 in out in out Physically Weight Unchanging (DC): f 480 f 480 out 138

139 FTL Location Example Include monthly outbound frequency constraint: Outbound shipments must occur at least once each month Implicit means of including inventory costs in location decision t max 1 1 yr/tl nmin 12 TL/yr 12 t TC max n, n rd FTL n n n min max max 6.69, 12 12, w 12(2) 24 max 10.04, 12 12, w 12(2) 24 max 16.73, , w 16.73(2) W W 160, 80 2 w i : Asheville Statesville Winston-Salem Durham Wilmington < < > 80 * (Montetary) Weight Gaining: w 79 w 81 n 39 n 41 in in out in out Physically Weight Unchanging (DC): f 480 f 480 out 139

140 Transshipment Direct: P2P shipments from Suppliers to Customers Transshipment: use DC to consolidate outbound shipments Uncoordinated: determine separately each optimal inbound and outbound shipment hold inventory at DC Cross dock: use single shipment interval for all inbound and outbound shipments no inventory at DC 140

141 Uncoordinated Inventory Average pipeline inventory level at DC: 4 O D 1 O, inbound 2 0 D, outbound Supplier q 2 Supplier q 2 Customer 3 Customer 4 141

142 TLC with Transshipment Uncoordinated: Cross docking: TLC TLC of supplier/customer i q i * i arg min TLC ( q) * * i i TLC TLC q 0 * q q t, shipment interval f c0() t TLCi () t vhf t t * * i i c ( t) independent transport charge as function of t O 0, inbound 0 D, outbound t arg min TLC ( t) TLC TLC t t i 142

143 TLC and Location TLC should include all logistics related costs TLC can be used as sole objective for network design (incl. location) Facility fixed costs, two options: 1. Use non transport related facility costs (mix of top down and bottom up) to estimate fixed costs via linear regression 2. For DCs, might assume public warehouses to be used for all DCs Pay only for time each unit spends in WH No fixed cost at DC Transport fixed costs: Costs that are independent of shipment size (e.g., $/mi vs. $/ton mi) Costs that make it worthwhile to incur the inventory cost associated with larger shipment sizes in order to spread out the fixed cost Main transport fixed cost is the indivisible labor cost for a human driver Why many logistics networks (e.g., Walmart, Lowes) designed for all FTL transport 143

144 Example: Optimal Number DCs for Lowe's Example of logistics network design using TLC Lowe s logistics network (2016): Regional DCs (15) Costal holding facilities Appliance DCs and Flatbed DCs Transloading facilities Modeling approach: Focus only on Regional DCs Mix of top down (COGS) and bottom up (typical load/tl parameters) FTL for all inbound and outbound shipments ALA used to determine TC for given number of DCs IC = αvhq max x (number of suppliers x number of DCs + number stores) Assume uncoordinated DC inventory, no cross docking Ignoring max DC to store distance constraints, consolidation, etc. Determined 9 DCs min TLC (15 DCs 0.87% increase in TLC) 144

145 Topics 1. Introduction 2. Facility location 3. Freight transport Exam 1 (take home) 4. Network models 5. Routing Exam 2 (take home) 6. Warehousing Final project Final exam (in class) 145

146 Graph Representations Complete bipartite directed (or digraph): Suppliers to multiple DCs, single mode of transport C: 1 2 -: : : : 9 16 Interlevel matrix W = NaN Weighted adjacency matrix 146

147 Bipartite: Graph Representations One or two way connections between nodes in two groups W = NaN IJC = Arc list matrix

148 Multigraph: Graph Representations Multiple connections, multiple modes of transport IJC = no_w = NaN Can t represent using adjacency matrix 148

149 Graph Representations Complete multipartite directed: Typical supply chain (no drop shipments) Drop Shipment 14 8 Level 1 (suppliers) Level 2 (DCs) Level 3 (customers) 149

150 Transportation Problem Satisfy node demand from supply nodes Can be used for allocation in ALA when NFs have capacity constraints Min cost/distance allocation = infinite supply at each node Trans Supply Demand

151 Greedy Solution Procedure Procedure for transportation problem: Continue to select lowest cost supply until all demand is satisfied Fast, but not always optimal for transportation problem Dijkstra s shortest path and simplex method for LP are optimal greedy procedures Trans Supply Demand = = = 0 10 = = 10 TC 5(30) 6(20) 8(35) 9(10) 13(30) 1, 030 vs 970 optimal 151

152 Min Cost Network Flow (MCNF) Problem Most general network problem, can solve using any type of graph representation Row for node 5 is redundant MCNF: lhs C C C C C C rhs ----: Min: : : : : lb: ub: Inf Inf Inf Inf Inf Inf si net supply of node i Arc cost: c Net node supply : s Incidence Matrix : A MCNF: max s.t. 0, supply node 0, demand node 0, transshipment node cx ' Ax x s 0 152

153 MCNF with Arc/Node Bounds and Node Costs Bounds on arcs/nodes can represent capacity constraints in a logistic network Node cost can represent production cost or intersection delay 6,1 2 4,3 4 0,2,4 i c,,0 j 6,3 1 5,9 3 0,0,,3 3 1,3 5 8 si net supply of node i 0, supply node 0, demand node 0, transshipment node 153

154 Expanded Node Formulation of MCNF Node cost/constraints converted to arc cost/constraints Dummy node (8) added so that supply = demand 6,1 2 4,3 4 0,2,4 6,3 1 5, , ,0,,3 s,nc,nu,nl i c,u,l 0 s i 0,nu,nl iʹ c+nc,u,l 154

155 Solving an MCNF as an LP Special procedures more efficient than LP were developed to solve MCNF and Transportation problems e.g., Network simplex algorithm (MCNF) e.g., Hungarian method (Transportation and Transshipment) Now usually easier to transform into LP since solvers are so good, with MCNF just aiding in formulation of problem: Trans MCNF LP Special, very efficient procedures only used for shortest path problem (Dijkstra) Transshipment

156 2 Dijkstra Shortest Path Procedure 3,3 4,1 10,3 4 6 s 1 8 t 0,1 2 14,4 3 13,5 2,1 12,3 10,4 Path: :

157 Dijkstra Shortest Path Procedure Index to index vector ns n Order important O 2 Simplex (LP) 4 O n 3 O n 2 O n Ellipsoid (LP) Hungarian (transportation) Dijkstra (linear min) O mlog n m Dijkstra (Fibonocci heap) no. arcs 157

158 Other Shortest Path Procedures Dijkstra requires that all arcs have nonnegative lengths Is a label setting algorithm since step to final solution made as each node labeled Can find longest path (used in CPM) by making by negating arc lengths Networks with some negative arcs require slower label correcting procedures that repeatedly check for optimality at all nodes or detect a negative cycle Negative arcs used in project scheduling to represent maximum lags between activities A* algorithm adds to Dijkstra an heuristic estimate of each node s remaining distance to destination Used in AI search for all types of applications (tic tac toe, chess) In path planning applications, great circle distance from each node to destination could be used as estimate of remaining distance 158

159 A* Path Planning Example 1 d * (Raleigh, Dallas) d (Raleigh, i) d ( i, Dallas), for each node i A dijk GC A* looks at a fraction of the nodes (in ellipse) seen by Dijkstra (in green) 159

160 A* Path Planning Example 2 3 D (x,y,t) A* used for planning path of each container in a DC Each container assigned unique priority that determines planning sequence Paths of higher priority containers become obstacles for subsequent containers 160

161 A* Path Planning Example 2 161

162 Minimum Spanning Tree Find the minimum cost set of arcs that connect all nodes Undirected arcs: Kruskal s algorithm (easy to code) Directed arcs: Edmond s branching algorithm (hard to code)

163 U.S. Highway Network Oak Ridge National Highway Network Approximately 500,000 miles of roadway in US, Canada, and Mexico Created for truck routing, does not include residential Nodes attributes: XY, FIPS code Arc attributes: IJD, Type (Interstate, US route), Urban U.S. Interstate Road Network From High Point to Goldsboro: Distance mi, Time = 2 hr 28 min Rural Roads Urban Roads Interstate Roads Connector Roads Cities Shortest Path High Point Goldsboro

164 FIPS Codes Federal Information Processing Standard (FIPS) codes used to uniquely identify states (2 digit) and counties (3 digit) 5 digit Wake county code = 2 digit state + 3 digit county = = 37 NC FIPS Wake FIPS 164

165 1. Thin Road Network Modifications Remove all degree 2 nodes from network Add cost of both arcs incident to each degree 2 node If results in multiple arcs between pair of nodes, keep minimum cost Thinned I 40 Around Raleigh 165

166 2. Subgraph Road Network Modifications Extract portion of graph with only those nodes and/or arcs that satisfy some condition Subgraph of Arcs < Subgraph of Nodes in Rectangle 3 166

167 Road Network Modifications 3. Add connector Given new nodes, add arcs that connect the new nodes to the existing nodes in a graph and to each other Distance of connector arcs = GC distance x circuity factor (1.5) New node connected to 3 closest existing nodes, except if Ratio of closest to 2 nd and 3 rd closest < threshold (0.1) Distance shorter using other connector and graph 167

168 Production and Inventory: One Product y y 0 0 1,1 1 h h T m i p m j T j 1 i 1 i 2 $ $-yr 0.3 $ 12 $-month h c c c c p c2 800 x K p c1 200 x K 1,1 1,1 50 p c2 200 x K 1,2 1,2 0 p c3 200 x K 1,3 1, ,1 2 y y D c i y 2 1 c i y 1,2 2, D c i y 2 1 c i y 1,3 2, D y y 0 4 1,4 1 2,1 2,1 60 p c2 800 x K 2,2 2,2 0 p c2 800 x K 2,3 2, ,4 2 y y 0 168

169 Production and Inventory: One Product Flow balance Capacity Initial/Final inventory Use var. LB & UB instead of constraints 169

170 Example: Coupled Networks via Truck Capacity Facility that extracts two different raw materials for pharmaceuticals 1. Extracted material to be sent over rough terrain in a truck to a staging station where it is then loaded onto a tractor trailer for transport to its final destination 2. Facility can extract up to 26 and 15 tons per week of each material, respectively, at a cost of $120 and $200 per ton 3. Annual inventory carrying rate is Facility can store up to 20 tons of each material on site, and unlimited amounts of material can be stored at the staging station and the final destination 5. Currently, five tons of the second material is in inventory at the final destination and this same amount should be in inventory at the end of the planning period 6. Costs $200 for a truck to make the roundtrip from the facility to the staging station, and it costs $800 for each truckload transported from the station to the final destination 7. Each truck and tractor trailer can carry up to 10 and 25 tons of material, respectively, and each load can contain both types of material Determine the amount of each material that should be extracted and when it should be transported in order to minimize total costs over the planning horizon Separate networks for two products are coupled via sharing truck capacity 170

171 Example: Coupled Networks via Truck Capacity Separate networks for each raw material are coupled via sharing the same trucks (added as constraint to model) 0 0 c c p 1,1 120 p 2,1 0 1,1,1 26 x y1,2, y1,2,2 20 Couples each network max 2,2,1 2,2,2 2 2,2 x x Q z 10z 2,2 c p 3,1 0 Number of trucks 0 y3,1,2 5 D1,1 D2,1 D1,2 D2,2 171

172 Example: Coupled Networks via Truck Capacity Math programming model: 172

173 Production and Inventory: Multiple Products p c i c s c 0 p c i c s c 0 Product Product 2 k1 k

174 Production and Inventory: Multiple Products k k z z mtg 0,1, production indicator mtg mtg mtg 0,1, setup indicator Don t want (not feasible) Want (feasible) zt kt kt Feasible, but not min cost 174

175 Production and Inventory: Multiple Products dummy Flow balance Capacity Setup Linking MILP 175

176 Production and Inventory: Multiple Products 176

177 Example of Logistics Software Stack MIP Data Solver (Gurobi, etc.) (csv,excel,etc.) Standard Report Library (C,Java) (GUI,web,etc.) Scripting (Python,Matlab,etc.) User Library (in script language) MIP Solver (Gurobi,Cplex,etc.) Standard Library (in compiled C,Java) Commercial Software (Lamasoft,etc.) Flow: Data Model Solver Output Report reports are run on a regular period to period, rolling horizon basis as part of normal operations management model only changed when logistics network changes 177

178 Topics 1. Introduction 2. Facility location 3. Freight transport Midterm exam 4. Network models 5. Routing 6. Warehousing Final project Final exam 178

179 Routing Alternatives Pickup P D Delivery (a) Point-to-point (P2P) TSP and VRP P D D D (b) Peddling (one-to-many) P P P D (c) Collecting (many-to-one) P P P D D D (d) Many-to-many P D P P D D (e) Interleaved empty P D P D P D (f) Multiple routes 179

180 TSP Problem: find connected sequence through all nodes of a graph that minimizes total arc cost Subroutine in most vehicle routing problems Node sequence can represent a route only if all pickups and/or deliveries occur at a single node (depot) 3 Node sequence = permutation + start node Depot 1 5 n n 6 1! 120 possible solutions 6 180

181 TSP TSP can be solved by a mix of construction and improvement procedures BIP formulation has an exponential number of constraints to eliminate subtours ( column generation techniques) Asymmetric: only best known solutions for large n 1 n1! n13 billion solutions 2 Symmetric: solved to optimal using BIP c ij c ji n 1! 2 solutions Euclidean: arcs costs = distance between nodes 181

182 TSP Construction Construction easy since any permutation is feasible and can then be improved

183 Spacefilling Curve Sequence determined by sorting position along 1 D line covering 2 D space 2: : : :

184 Two Opt Improvement a b c d e f a c a d a e b d b e b f c e c f d f first arcs to 2 arc a b n n1 n2 n n n 3 Sequences considered at end to verify local optimum: n nodes (1) (1) 9 for n6 2 j3 i2 ji2 184

185 TSP Comparison TSP Procedure Total Cost 1 Spacefilling curve opt Convex hull insert + 2 opt Nearest neighbor + 2 opt Random construction + 2 opt 450, Eil51 in TSPLIB 426* optimal 185

186 Multi Stop Routing Each shipment might have a different origin and/or destination node/location sequence not adequate 1 n L y,, y = 1, 2,3 n shipments 1 2n R z,, z = 3,1, 2, 2,1,3 2n-element route sequence 1 2n X x,, x = 5,1,3, 4, 2,6 2n-element location (node) sequence c cost between locations i and j ij 2n1 cr ( ) c , total i1 x, x i i1 cost of route R 186

187 Min Cost Insert c c c c 2 *

188 Route Sequencing Procedures Online procedure: add a shipment to an existing route as it becomes available Insert and Improve: for each shipment, mincostinsert + twoopt Offline procedure: consider all shipments to decide order in which each added to route Savings and Improve: determine insert ordering based on savings, then improve final route 188

189 Insert and Improve Online Procedure 189

190 5 Shipment Example R X = 3,2,5,5,1,4,3,1,2,4 = 4,3,7,1,1,6,5,2,2,2 190

191 Pairwise Savings s s pairwise savings between shipments i and j ij 1,2 c c c i j ij

192 Savings Route Construction Procedure Pairs of shipments are ordered in terms of their decreasing pairwise savings to create i and j Creates set of multi shipment routes R R R 1,..., m Shipments with no pairwise savings are not included (use sh2rte to add) Given savings pair i,j, the original Clark Wright savings procedure (vrpsavings): adds i to route only if j at beginning or end of route routes combined only if i and j are endpoints of each route 1. Form new route 2. Add shipment to route 3. Combine two routes 192

193 Vehicle Routing Problem VRP = TSP + vehicle constraints Constraints: Capacity (weight, cube, etc.) Maximum TC (HOS: 11 hr max) Time windows (with/without delay at customer) VRP uses absolute windows that can be checked by simple scanning Project scheduling uses relative windows solved by shortest path with negative arcs Maximum number of routes/vehicles (hard) Criteria: 1. Number of routes/vehicles 2. TC 193

194 Time Window Example [0,24] hr; Loading/unloading me = 0; Capacity = ; LB = 5 hr (earliest start) a = 6 (depart window) a = 7 8 (arrive at 7 wait to 8) (return window) a = a = a = Earliest Finish Latest Start = = 8 hr = 5 travel + 3 delay 194

195 Min TLC Multi Stop Routing Periodic consolidated shipments that have the same frequency/interval Similar to a milk run Min TLC of aggregate shipment may not be feasible Different combinations of shipments (load mix) may be on board during each segment of route Minimum TLC of unconstrained aggregate of all shipments first determined If needed, all shipment sizes reduced in proportion to load mix with the minimum max payload Single load mix First load mix Second load mix 195

196 Load Mix Example 196

197 TLC Calculation for Multi Stop Route How mintlc determines TLC for a route: q agg f max rd, MC agg v agg agg h (no truck capacity constraints, only min charge) q i q agg (allocate based on demand) fi sl (aggregate density of shipments in load-mix ) j fi Lj s q il k min 1, L j * i j kq i f f i agg il j i sl K j min Kwt, 2000 q il f TLC rd v h q ( d = distance of entire route) * agg * * agg agg i agg qi j i cu (min ratio of max payload to size of shipment in load-mix) (apply truck capacity deduction factor) 197

198 Cost Allocation for Routing Shmt Shmt ft 800 ft 1,200 ft 198

199 Cost Allocation for Routing

200 Shapley Value n sav 0 L i L i1 c c c sav 0 0 ij i j i, j c c c c m! nm1! {} n! i M i M 0mn1 MN\ i M m c sav i n sav sav n n sav cl 1 cij cji 1 n n1 2 nn1 j1 j1 k1 c sav jk 200

201 Topics 1. Introduction 2. Facility location 3. Freight transport Midterm exam 4. Network models 5. Routing 6. Warehousing Final project Final exam 201

202 Warehousing Warehousing are the activities involved in the design and operation of warehouses A warehouse is the point in the supply chain where raw materials, work in process (WIP), or finished goods are stored for varying lengths of time. Warehouses can be used to add value to a supply chain in two basic ways: 1. Storage. Allows product to be available where and when its needed. 2. Transport Economies. Allows product to be collected, sorted, and distributed efficiently. A public warehouse is a business that rents storage space to other firms on a month to month basis. They are often used by firms to supplement their own private warehouses. 202

203 Types of Warehouses

204 Warehouse Design Process The objectives for warehouse design can include: maximizing cube utilization minimizing total storage costs (including building, equipment, and labor costs) achieving the required storage throughput enabling efficient order picking In planning a storage layout: either a storage layout is required to fit into an existing facility, or the facility will be designed to accommodate the storage layout.

205 Warehouse Design Elements The design of a new warehouse includes the following elements: 1. Determining the layout of the storage locations (i.e., the warehouse layout). 2. Determining the number and location of the input/output (I/O) ports (e.g., the shipping/receiving docks). 3. Assigning items (stock keeping units or SKUs) to storage locations (slots). A typical objective in warehouse design is to minimize the overall storage cost while providing the required levels of service.

206 Design Trade Off Warehouse design involves the trade off between building and handling costs: min Building Costs vs. min Handling Costs max Cube Utilization vs. max Material Accessibility 206

207 Shape Trade Off vs. Square shape minimizes perimeter length for a given area, thus minimizing building costs Aspect ratio of 2 (W = 2D) min. expected distance from I/O port to slots, thus minimizing handling costs 207

208 Storage Trade Off B A A C B B E D C vs. A A B B B Honeycomb C C D loss E Maximizes cube utilization, but minimizes material accessibility Making at least one unit of each item accessible decreases cube utilization 208

209 Storage Policies A storage policy determines how the slots in a storage region are assigned to the different SKUs to the stored in the region. The differences between storage polices illustrate the trade off between minimizing building cost and minimizing handling cost. Type of policies: Dedicated Randomized Class based 209

210 Dedicated Storage C B A I/O C Each SKU has a predetermined number of slots assigned to it. Total capacity of the slots assigned to each SKU must equal the storage space corresponding to the maximum inventory level of each individual SKU. Minimizes handling cost. Maximizes building cost. 210

211 Randomized Storage ABC I/O Each SKU can be stored in any available slot. Total capacity of all the slots must equal the storage space corresponding to the maximum aggregate inventory level of all of the SKUs. Maximizes handling cost. Minimizes building cost. 211

212 Class based Storage BC A I/O Combination of dedicated and randomized storage, where each SKU is assigned to one of several different storage classes. Randomized storage is used for each SKU within a class, and dedicated storage is used between classes. Building and handling costs between dedicated and randomized. 212

213 Individual vs Aggregate SKUs A B C ABC Dedicated Random Class-Based Time A B C ABC AB AC BC M i Time 213

214 Cube Utilization Cube utilization is percentage of the total space (or cube ) required for storage actually occupied by items being stored. There is usually a trade off between cube utilization and material accessibility. Bulk storage using block stacking can result in the minimum cost of storage, but material accessibility is low since only the top of the front stack is accessible. Storage racks are used when support and/or material accessibility is required. 214

215 Honeycomb Loss Honeycomb loss, the price paid for accessibility, is the unusable empty storage space in a lane or stack due to the storage of only a single SKU in each lane or stack Vertical Honeycomb Loss of 3 Loads Wall Height of 5 Levels (Z) Down Aisle Width of 5 Lanes (X) Cross Aisle Depth of 4 Rows (Y) Horizontal Honeycomb Loss of 2 Stacks of 5 Loads Each 215

216 Estimating Cube Utilization The (3 D) cube utilization for dedicated and randomized storage can estimated as follows: Cube utilization item space total space item space item space honeycomb down aisle loss space N x y z M 1 i i, dedicated CU (3-D) TS( D) x y z M, randomized TS( D) N M i xyi 1 H, dedicated TA( D) CU (2-D) M x y H, randomized TA( D) 216

217 Unit Load Unit load: single unit of an item, or multiple units restricted to maintain their integrity Linear dimensions of a unit load: (Stringer length) Depth x Width (Deckboard length) Deckboards Stringer Notch Depth (stringer length) Width (deckboard length) y x Pallet height (5 in.) + load height gives z: y x z 217

218 Cube Utilization for Dedicated Storage Storage Area at Different Lane Depths Item Space Lanes Total Space Cube Util. D = 1 A/2 = 1 A A A A B B B B B C C C % D = 2 A/2 = 1 A A B B C A A B B B C C % A B C D = 3 A B B C A A B B C % A/2 = 1 218

219 Total Space/Area The total space required, as a function of lane depth D: A A Total space (3-D): TS( D) X Y Z xl( D) yd zh 2 2 Eff. lane depth TS( D) eff Total area (2-D): TA( D) X Y xl( D) yd Z X = xl A 2 A B C A B B C A A B B C x Down Aisle Space Storage Area on Opposite Side of the Aisle 219

220 Number of Lanes Given D, estimated total number of lanes in region: N M i, dedicated i1 DH Number of lanes: LD ( ) D1 H 1 M NH N 2 2, randomized ( N 1) DH Estimated HCL: D 3 1 D 0 D 1 1 D D 2 D 1 D D1 D D 1 D 2 D i 1 D D D D 2 2 i1 220

221 Optimal Lane Depth Solving for D in dts( D) dd 0 results in: Optimal lane depth for randomized storage (in rows): 70,000 * A 2M N 1 D 2NyH 2 60,000 50,000 40,000 30,000 20,000 10, Item Space 24,000 24,000 24,000 24,000 24,000 24,000 24,000 24,000 24,000 24,000 Honeycomb Loss 1,536 3,648 5,376 7,488 9,600 11,712 13,632 15,936 17,472 20,160 Aisle Space 38,304 20,736 14,688 11,808 10,080 8,928 8,064 7,488 6,912 6,624 Total Space 63,840 48,384 44,064 43,296 43,680 44,640 45,696 47,424 48,384 50,784 Lane D ep t h ( in R o ws) 221

222 Max Aggregate Inventory Level Usually can determine max inventory level for each SKU: M i = maximum number of units of SKU i Since usually don t know M directly, but can estimate it if SKUs inventory levels are uncorrelated Units of each item are either stored or retrieved at a constant rate M N i1 Mi Can add include safety stock for each item, SS i For example, if the order size of three SKUs is 50 units and 5 units of each item are held as safety stock M N Mi SSi SSi i

223 Steps to Determine Area Requirements 1. For randomized storage, assumed to know N, H, x, y, z, A, and all M i Number of levels, H, depends on building clear height (for block stacking) or shelf spacing Aisle width, A, depends on type of lift trucks used 2. Estimate maximum aggregate inventory level, M 3. If D not fixed, estimate optimal land depth, D * 4. Estimate number of lanes required, L(D * ) 5. Determine total 2 D area, TA(D * ) 223

224 Aisle Width Design Parameter Typically, A (and sometimes H) is a parameter used to evaluate different overall design alternatives Width depends on type of lift trucks used, a narrower aisle truck reduces area requirements (building costs) costs more and slows travel and loading time (handling costs) 9-11 ft 7-8 ft 8-10 ft Stand-Up CB NA Straddle NA Reach 224

225 Example 1: Area Requirements Units of items A, B, and C are all received and stored as in. (y x z) pallet loads in a storage region that is along one side of a 10 foot wide down aisle in the warehouse of a factory. The shipment size received for each item is 31, 62, and 42 pallets, respectively. Pallets can be stored up to three deep and four high in the region. 36 x 3' MA 31 A 10' 12 y 3.5' M 62 D3 B z 3' M 42 H 4 C N 3 225

226 Example 1: Area Requirements 1. If a dedicated policy is used to store the items, what is the 2 D cube utilization of this storage region? N Mi LD ( ) L(3) lanes i1 DH 3(4) 3(4) 3(4) A 10 TA(3) xl( D) yd 3(13) 3.5(3) 605 ft 2 2 N Mi xy 33.5 item space i1 H 4 4 CU (3) 4 61% TA(3) TA(3)

227 Example 1: Area Requirements 2. If the shipments of each item are uncorrelated with each other, no safety stock is carried for each item, and retrievals to the factory floor will occur at a constant rate, what is an estimate the maximum number of units of all items that would ever occur? M N Mi i

228 Example 1: Area Requirements 3. If a randomized policy is used to store the items, what is total 2 D area needed for the storage region? D 3 D1 H 1 M NH N L(3) 2 2 DH (4) N lanes 3(4) A 10 TA(3) xl( D) yd 3(8) 3.5(3) 372 ft

229 Example 1: Area Requirements 4. What is the optimal lane depth for randomized storage? * A 2M N (68) 3 1 D 2NyH 2 2(3)3.5(4) What is the change in total area associated with using the optimal lane depth as opposed to storing the items three deep? (4) N D4 L(4) lanes 3(4) 10 TA(4) 3(6) 3.5(4) 342 ft 2 D3 TA(3) 372 ft

230 Example 2: Trailer Loading How many identical in. four way containers can be shipped in a full truckload? Each container load: 1. Weighs 600 lb 2. Can be stacked up to six high without causing damage from crushing 3. Can be rotated on the trucks with respect to their width and depth. X 98/12 = ft Y ft Z 110/12 = ft x [48,42]/12 = ft y [42,48]/12 = ft z 30/12 = ft L floor(x/x) = 2 2 D floor(y/y) = H min(6,floor(z/z)) = 3 3 LDH L*D*H = units wt lb unit/tl min(ldh, floor(50000/wt)) = Truck Trailer Cube = 3,332-3,968 CFT Max Gross Vehicle Wt = 80,000 lbs = 40 tons Max Payload Wt = 50,000 lbs = 25 tons Length: 48' - 53' single trailer, 28' double trailer Max of 83 units per TL Max Height: 13'6" = 162" Width: 8'6" = 102" (8'2" = 98") Interior Height: (8'6" - 9'2" = 102" - 110") 230

231 Storage and Retrieval Cycle A storage and retrieval (S/R) cycle is one complete roundtrip from an I/O port to slot(s) and back to the I/O Type of cycle depends on load carrying ability: Carrying one load at a time (load carried on a pallet): Single command Dual command Carrying multiple loads (order picking of small items): Multiple command 231

232 Single Command S/R Cycle empty store I/O slot empty retrieve Expected time for each SC S/R cycle: d d t t t t v v SC SC SC L U 2 L/ U Single command (SC) cycles: Storage: carry one load to slot for storage and return empty back to I/O port, or Retrieval: travel empty to slot to retrieve load and return with it back to I/O port 232

233 Industrial Trucks: Walk vs. Ride Walk (2 mph = 176 fpm) Ride (7 mph = 616 fpm) Pallet Jack Pallet Truck Walkie Stacker Sit down Counterbalanced Lift Truck 233

234 Dual Command S/R Cycle Expected time for each SC S/R cycle: d d t t t t v v DC DC DC 2 L 2 U 4 L/ U Dual command (DC): Combine storage with a retrieval: store load in slot 1, travel empty to slot 2 to retrieve load Can reduce travel distance by a third, on average Also termed task interleaving 234

235 Multi Command S/R Cycle I/O empty retrieve Multi command: multiple loads can be carried at the same time Used in case and piece order picking Picker routed to slots Simple VRP procedures can be used 235

236 1way L 1 D Expected Distance X X X X TD1 way i i L 2L L 2L ED I/O X = 12 X 2L x = X L i1 i1 X L( L1) X L L 2 2L XL X X XL 2 2 TD1 way X L 2 d 2( ED ) X SC 1 D Storage Region 1way L 1 Assumptions: All single command cycles Rectilinear distances Each slot is region used with equal frequency (i.e., randomized storage) Expected distance is the average distance from I/O port to midpoint of each slot e.g., [2(1.5) + 2(4.5) + 2(6.5) + 2(10.5)]/4 =

237 Off set I/O Port I/O X = 12 offset d 2( d ) X SC offset If the I/O port is off set from the storage region, then 2 times the distance of the offset is added the expected distance within the slots 237

238 2 D Expected Distances Since dimensions X and Y are independent of each other for rectilinear distances, the expected distance for a 2 D rectangular region with the I/O port in a corner is just the sum of the distance in X and in Y: For a triangular region with the I/O port in the corner: rect SC L L i 1 X X X X TD1-way i j i1 j1 L 2L L 2L tri dsc 2 X 2 X Y X Y d X Y X 2 2L 3L1 6 ED1-way TD1-way 2 X 2 X X, LL ( 1) 3 3L 3 as L 2 Y y D Y I/O x X L 238 X

239 I/O to Side Configurations Rectangular Triangular X TA 0 X TA X 2 X d SC TA 2 TA TA 1 2 X I/O 2 X 2TA 2 TA 4 dsc 2 TA TA 3 239

240 I/O at Middle Configurations Rectangular Triangular TA X 2 2 TA TA X 2 2 d 2 TA TA SC TA 1 X X TA 4 dsc TA TA 3 240

241 Example 3: Handling Requirements Pallet loads will be unloaded at the receiving dock of a warehouse and placed on the floor. From there, they will be transported 500 feet using a dedicated pallet truck to the in floor induction conveyor of an AS/RS. Given a. It takes 30 sec to load each pallet at the dock b. 30 sec to unload it at the induction conveyor c. There will be 80,000 loads per year on average d. Operator rides on the truck (because a pallet truck) e. Facility will operate 50 weeks per year, 40 hours per week 241

242 Example 3: Handling Requirements 1. Assuming that it will take 30 seconds to load each pallet at the dock and 30 seconds to unload it at the induction conveyor, what is the expected time required for each singlecommand S/R cycle? d t SC SC 2(500) 1000 ft/mov d v 1000 ft/mov min/mov 616 ft/min 60 SC tl/ U min/mov hr/mov 60 (616 fpm because operator rides on a pallet truck) 242

243 Example 3: Handling Requirements 2. Assuming that there will be 80,000 loads per year on average and that the facility will operate for 50 weeks per year, 40 hours per week, what is the minimum number of trucks needed? r avg 80,000 mov/yr 50(40) hr/yr 40 mov/hr mr t avg SC trucks 243

244 Example 3: Handling Requirements 3. How many trucks are needed to handle a peak expected demand of 80 moves per hour? r peak 80 mov/hr mr t peak SC trucks 244

245 Example 3: Handling Requirements 4. If, instead of unloading at the conveyor, the 3 foot wide loads are placed side by side in a staging area along one side of 90 foot aisle that begins 30 feet from the dock, what is the expected time required for each single command S/R cycle? d 2( d ) X 2(30) ft t SC SC d v offset 150 ft/mov min/mov 616 ft/min 60 SC tl/ U min/mov hr/mov

246 Estimating Handling Costs Warehouse design involves the trade off between building and handling cost. Maximizing the cube utilization of a storage region will help minimize building costs. Handling costs can be estimated by determining: 1. Expected time required for each move based on an average of the time required to reach each slot in the region. 2. Number of vehicles needed to handle a target peak demand for moves, e.g., moves per hour. 3. Operating costs per hour of vehicle operation, e.g., labor, fuel (assuming the operators can perform other productive tasks when not operating a truck) 4. Annual operating costs based on annual demand for moves. 5. Total handling costs as the sum of the annual capital recovery costs for the vehicles and the annual operating costs. 246

247 Example 4: Estimating Handing Cost TA = 20,000 I/O Expected Distance: d 2 TA 2 20, ft dsc Expected Time: tsc 2t v Peak Demand: Annual Demand: r peak r year SC L/ U 200 ft 2(0.5 min) 2 min per move 200 fpm 75 moves per hour 100,000 moves per year tsc Number of Trucks: mrpeak trucks 60 Handling Cost: t TC mk r C 60 SC hand truck year labor 2 3($2,500 / tr-yr) 100,000 ($10 / hr) 60 $7,500 $33,333 $40,833 per year 247

248 Dedicated Storage Assignment (DSAP) The assignment of items to slots is termed slotting With randomized storage, all items are assigned to all slots DSAP (dedicated storage assignment problem): Assign N items to slots to minimize total cost of material flow DSAP solution procedure: 1. Order Slots: Compute the expected cost for each slot and then put into nondecreasing order 2. Order Items: Put the flow density (flow per unit of volume) for each item i into nonincreasing order f f f M s M s M s [1] [2] [ N ] [1] [1] [2] [2] [ N] [ N] 3. Assign Items to Slots: For i = 1,, N, assign item [i] to the first slots with a total volume of at least M [i] s [i] 248

249 1 D Slotting Example A B C Max units M Space/unit s Flow f Flow Density f/(m x s) Flow Density 1-D Slot Assignments Expected Distance Flow Total Distance C C C I/O 0 3 2(0) + 3 = 3 21 = A A A A I/O (3) + 4 = = B B B B B I/O (7) + 5 = 19 7 = 133 I/O C C C A A A A B B B B B

250 1 D Slotting Example (cont) Dedicated Random Class-Based A B C ABC AB AC BC Max units M Space/unit s Flow f Flow Density f/(m x s) D Slot Assignments Total Distance Total Space Dedicated (flow density) I/O C C C A A A A B B B B B Dedicated (flow only) I/O A A A A C C C B B B B B Class-based C C C AB AB AB AB AB AB I/O AB Randomized ABC ABC ABC ABC ABC ABC ABC ABC ABC I/O

251 2 D Slotting Example A B C Max units M Space/unit s Flow f Flow Density f/(m x s) Distance from I/O to Slot C C B C A A B B A A I/O B B B B B B A C A B A C I/O C A Original Assignment (TD = 215) Optimal Assignment (TD = 177) 251

252 DSAP Assumptions 1. All SC S/R moves 2. For item i, probability of move to/from each slot assigned to item is the same 3. The factoring assumption: a. Handling cost and distances (or times) for each slot are identical for all items b. Percent of S/R moves of item stored at slot j to/from I/O port k is identical for all items Depending of which assumptions not valid, can determine assignment using other procedures c x DSAP LAP LP QAP c x x cx ij ij TSP i ij ijkl ij kl 252

253 Example 5: 1 D DSAP What is the change in the minimum expected total distance traveled if dedicated, as compared to randomized, block stacking is used, where a. Slots located on one side of 10 foot wide down aisle b. All single command S/R operations c. Each lane is three deep, four high d in. two way pallet used for all loads e. Max inventory levels of SKUs A, B, C are 94, 64, and 50 f. Inventory levels are uncorrelated and retrievals occur at a constant rate g. Throughput requirements of A, B, C are 160, 140, 130 h. Single I/O port is located at the end of the aisle 253

254 Randomized: Example 5: 1 D DSAP L MA MB MC M rand d X SC D1 H 1 M NH N 2 2 DH (4) N lanes 3(4) xl 3(11) 33 ft rand X 33 ft TD f f f X,190 ft rand A B C 254

255 Example 5: 1 D DSAP Dedicated: fa 160 fb 140 fc , 2.19, 2.6 C B A M 94 M 64 M 50 A B C MA 94 MB 64 MC 50 LA 8, LB 6, LC 5 DH 3(4) DH 3(4) DH 3(4) X xl 3(5) 15, X xl 3(6) 18, X xl 3(8) 24 C C B B A A d C SC X 3(5) 15 ft C d B S C X C X B 2( ) 2(15) ft A SC C B A d 2( X X ) X 2(15 18) ft A B C ded A SC B SC C SC TD f d f d f d 160(90) 140(48) 130( 15) 23, 070 ft 255

256 1 D Multiple Region Expected Distance A A A B B B B I/O I/Oʹ X A, X B X B A SC A A d d X 3 d di/o to I/ O XB In 1 D, easy to determine the offset In 2 D, no single offset value for each region d 2d X X 2X X X X X 10 B offset B A A B A A B AB TA X 3, TA X X 4, TA TA TA 7 A A B B A AB A B TM TA d, TM TA d A A A B B B AB AB AB B A B A A B A A B A TM TA d X X X X X X X X X X d B TA d TA d TM TM A A B B A B TMB TMAB TMA TAAB dab TAA da 7(7) 3(3) 10 TA TA TA 4 B B B 256

257 2 D Multiple Region Expected Distance 257

258 2 D Multiple Region Expected Distance 258

259 2 D Multiple Region Expected Distance If more than two regions For regions below diagonal (D), start with region closest to I/O For regions above diagonal (A+C), start with regions closest to I/O (C) For region in the middle (B), solve using whole area less other regions X A X C X D 259

260 Warehouse Operations 260 Order Picking Putaway Putaway

261 Warehouse Management System WMS interfaces with a corporation s enterprise resource planning (ERP) and the control software of each MHS Inventory Master File Item On-Hand Balance In-Transit Qty. Locations A ,21 B ,22 Location Master File Location Item On-Hand Balance In-Transit Qty. 11 A B A B 1 0 A 11 A 21 A B B B B Advance shipping notice (ASN) is a standard format used for communications 261

262 Logistics related Codes Commodity Code Item Code Unit Code Level Category Class Instance Description Function Grouping of similar objects Product classification Grouping of identical objects Inventory control Names Item number, Part number, SKU, SKU + Lot number Codes UNSPSC, GPC GTIN, UPC, ISBN, NDC Unique physical object Object tracking Serial number, License plate EPC, SSCC UNSPSC: United Nations Standard Products and Services Code GPC: Global Product Catalogue GTIN: Global Trade Item Number (includes UPC, ISBN, and NDC) UPC: Universal Product Code ISBN: International Standard Book Numbering NDC: National Drug Code EPC: Electronic Product Code (globally unique serial number for physical objects identified using RFID tags) SSCC: Serial Shipping Container Code (globally unique serial number for identifying movable units (carton, pallet, trailer, etc.)) 262

263 Identifying Storage Locations 263 Building Aisle Bay Tier Position Compartment Location: 1 -AAC D B Compartment A B 2 1 Wall Position E D C B A AAC AAB AAA Aisle (Y) Cross Aisle Bay (X) Down Aisle Tier (Z)

264 Receiving Reserve Receive Putaway Replenish Storage Forward Pick Order Pick Sort & Pack Ship Basic steps: 1. Unload material from trailer. 2. Identify supplier with ASN, and associate material with each moveable unit listed in ASN. 3. Assign inventory attributes to movable unit from item master file, possibly including repackaging and assigning new serial number. 4. Inspect material, possibly including holding some or all of the material for testing, and report any variances. 5. Stage units in preparation for putaway. 6. Update item balance in inventory master and assign units to a receiving area in location master. 7. Create receipt confirmation record. 8. Add units to putaway queue 264

265 Putaway A putaway algorithm is used in WMS to search for and validate locations where each movable unit in the putaway queue can be stored Inventory and location attributes used in the algorithm: Environment (refrigerated, caged area, etc.) Container type (pallet, case, or piece) Product processing type (e.g., floor, conveyable, nonconveyable) Velocity (assign to A, B, C based on throughput of item) Preferred putaway zone (item should be stored in same zone as related items in order to improve picking efficiency) 265

266 Replenishment Reserve Receive Putaway Replenish Storage Forward Pick Order Pick Sort & Pack Ship Replenishment is the process of moving material from reserve storage to a forward picking area so that it is available to fill customer orders efficiently Other types of in plant moves include: Consolidation: combining several partially filled storage locations of an item into a single location Rewarehousing: moving items to different storage locations to improve handling efficiency Reserve Storage Area 266

267 Order Picking Reserve Receive Putaway Replenish Storage Forward Pick Order Pick Sort & Pack Ship Order picking is at the intersection of warehousing and order processing Information Processing Order Processing Order Preparation Order Transmittal Storage 15% WH Operating Costs Receiving 10% Shipping 20% Material Handling Order Entry Warehousing Receiving Putaway Storage Order Picking Shipping Order Status Reporting Order Picking 55% 267

268 Order Picking Reserve Receive Putaway Replenish Storage Forward Pick Order Pick Sort & Pack Ship Pallet and Case Picking Area Levels of Order Picking Pallet Picking Case Picking Piece Picking Forward Piece Picking Area 268

269 Order Picking Reserve Receive Putaway Replenish Storage Forward Pick Order Pick Sort & Pack Ship Voice Directed Piece and Case Picking Pick to Light Piece Picking Increment/ Decrement Buttons Count Display Confirm Button Pick 24 Pack 14 Pack 10 Carton Flow Rack Picking Cart 269

270 Order Picking Reserve Receive Putaway Replenish Storage Forward Pick Order Pick Sort & Pack Ship Discrete Batch Methods of Order Picking Method Pickers per Order Orders per Picker Discrete Single Single Zone Multiple Single Batch Single Multiple Zone-Batch Multiple Multiple A B C D E F G H 7 A C7 G1 D5 Picker 1 Zone 1 Zone Zone Batch B4 F2 Zone E8 Picker 2

271 Sortation and Packing Reserve Receive Putaway Replenish Storage Forward Pick Order Pick Sort & Pack Ship Case Sortation System Wave zone batch piece picking, including downstream tilt tray based sortation 271

272 Shipping Reserve Receive Putaway Replenish Storage Forward Pick Order Pick Sort & Pack Ship Staging, verifying, and loading orders to be transported ASN for each order sent to the customer Customer specific shipping instructions retrieved from customer master file Carrier selection is made using the rate schedules contained in the carrier master file Shipping Area 272

273 Activity Profiling Total Lines: total number of lines for all items in all orders Lines per Order: average number of different items (lines/skus) in order Cube per Order: average total cubic volume of all units (pieces) in order Flow per Item: total number of S/R operations performed for item Lines per Item (popularity): total number of lines for item in all orders Cube Movement: total unit demand of item time x cubic volume Demand Correlation: percent of Customer Orders Order: 1 SKU A B C D Qty Order: 2 SKU Qty A 4 C 1 Order: 3 SKU Qty A 2 Order: 4 SKU Qty B 2 Order: 5 SKU Qty C 1 D 12 E 6 Item Master SKU Length Width Depth Cube Weight A B C D E Total Lines = 11 Lines per Order = 11/5 = 2.2 Cube per Order = SKU Flow per Item Lines per Item Cube Movement A B C D E Demand Correlation Distibution SKU A B C D E A B C D 0.2 E orders in which both items appear 273

274 Pallet Picking Equipment Cube Movement Block Stacking / Drive-In Rack Pallet Flow Rack Double-Deep Rack Push-Back Rack Push Back Rack Sliding Rack Single-Deep Selective Rack Double Deep Rack Flow per Item Sliding Rack Drive In Rack Single Deep Selective Rack 274

275 Case Picking Manual Automated Case Picking Cube Movement Push Back Rack Pallet Flow Rack Flow Delivery Lanes Equipment Induct Induct Single-Deep Selective Rack Lines per Item Case Dispensers Sortation Conveyor Floor-level Pick Multi-level Pick Case Picking Unitizing and Shipping Reserve Storage Replenish Forward Pick Storage Case Picking Case Picking Single-Deep Selective Racks Zone Batch Pick to Pallet Forward Pick Storage Case Picking Case Picking Floor vs. Multi level Pick to Pallet 275

276 Piece Picking Equipment Carton Flow Rack Cube Movement Storage Drawers Bin Shelving Vertical Lift Module Horizontal Carousel A-Frame Carton Flow Rack Lines per Item Vertical Lift Module Drawers/Bins A Frame Dispenser Carousel 276

277 Methods of Piece Picking Lines per Order, Cube per Order Discrete Zone (Infrequently Used) (Ex: Pick-and-Pass) Batch Zone-Batch (Ex: Pick Cart) (Ex: Wave Picking) Total Lines Scan Zone 1 Zone 2 Zone 3 Pick Pass No Pick Pass Pick Pick and Pass Zone Piece Picking Pick Conveyor Packing and Shipping Pick Cart Pick cart Batch Piece Picking Packing and Shipping Wave Zone Batch Piece Picking 277

278 Warehouse Automation Historically, warehouse automation has been a craft industry, resulting highly customized, one off, high cost solutions To survive, need to adapt mass market, consumer oriented technologies in order to realize to economies of scale replace mechanical complexity with software complexity How much can be spent for automated equipment to replace one material handler: $45, 432 $45, $219, $45,432: median moving machine operator annual wage + benefits 1.7% average real interest rate (real = nominal inflation) 5 year service life with no salvage (service life for Custom Software)

279 KIVA Mobile Robotic Fulfillment System Goods to man order picking and fulfillment system Multi agent based control Developed by Peter Wurman, former NCSU CSC professor Kiva now called Amazon Robotics purchased by Amazon in 2012 for $775 million