ISE 754: Logistics Engineering - Nc State University · Management Logistics Engineering Supply Chain Analysis (military logistics , humanitarian logistics , school bus routing )

ISE 754: Logistics Engineering

Michael G. Kay

Fall 2019

Topics

1. Introduction

2. Facility location

3. Freight transport

– Exam 1 (take home)

4. Network models

5. Routing


6. Warehousing

– Final exam (in class)

Inside the box

Ou

tsid

e th

e b

ox

2

SCM vs Logistics Engineering

3

Supply Chain

Management

Logistics

Engineering

Supply Chain

Analysis

(military logistics,

humanitarian

logistics, school

bus routing)

(contracting,

procurement,

freight terms)

Analysis Triangle

4

Data

Model

Computer

Symbolic

Analysis

Machine

Learning

Simulation

Analysis

Logistics

Engineering

Scope

• Strategic (years)

– Network design

• Tactical (weeks-year)

– Multi-echelon, multi-period, multi-product production and inventory models

• Operational (minutes-week)

– Vehicle routing

5

Strategic: Network Design

-120 -110 -100 -90 -80 -7015

20

25

30

35

40

45

50

1

2

3

4

5

Optimal locations for five DCs serving 877 customers throughout the U.S.

6

Tactical: Production-Inventory Model

7

pc

Flow balance

x y

ic

Capacity

x

Setup

z

–K

k 0

1

k 0

sc 0

Flow balance

x y

Capacity

x

Setup

z

–K

k 0

1

k 0

pc ic sc 0

Linking 1k 2k 1

0

0

Pro

du

ct

1

Pro

du

ct

2

Constraint matrix for a 2-product, multi-period model with setups

Vehicle RoutingEight routes served from DC in Harrisburg, PA

8

-80 -79 -78 -77 -76 -75 -74 -73 -72 -71 -70

36

37

38

39

40

41

42

43

44

1

2

3 4

5

6

7 8

9

10

11

12

13

14

15

16

17

18

19

20

21

22 23

24

25

26

27

28

29

30

31

32

33

34

Route

Load

Weight

Route

Time

Customers

in Route

Layover

Required

1 12,122 18.36 4 1

2 4,833 16.05 2 1

3 9,642 17.26 3 1

4 25,957 13.77 6 0

5 12,512 9.90 2 0

6 15,156 13.70 5 0

7 29,565 11.30 6 0

8 32,496 8.84 5 0

109.18 3

Route Summary Information

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

• 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

9

Geometric Mean: 1 100 10X LB UB

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 = 350 18.71 (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)

10

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)

– Bad control can result in higher values than no control

11

8Estimated wait time 8 5.67 min

2LB UB

http://www.nextbuzz.gatech.edu/

12

Levels of Modeling

0. Guesstimation (order of magnitude)

1. Mean value analysis (linear, ±20%)

2. Nonlinear models (incl. variance, ±5%)

3. Simulation models (complex interactions)

4. Prototypes/pilot studies

5. Build/do and then tweak it

13

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)

14

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 ‘10-15 times less than the cost of current last-mile delivery alternatives.’”

15

Ex 1: Geometric Mean• If, during the morning rush, there are three buses operating

on Wolfline Route 13 and it takes them 45 minutes, on average, to complete one circuit of the route, what is the estimated waiting time for a student who does not use TransLoc for real-time bus tracking?

16

3 bus/circuit 1 1Frequency (TH) = bus/min, Headway = 15 min/bus

45 min/circuit 15 Freq.

15Estimated wait time 15 10.61 min

2

WIP

CT

LB UB

Answer :

Ex 2: Fermi Problem• Estimate the average amount spent per trip to a grocery store.

Total U.S. supermarket sales were recently determined to be $649,087,000,000, but it is not clear whether this number refers to annual sales, or monthly, or weekly sales.

17

$6.5 11$2,000 / person-yr, 1 trips/wk, 7 trips/wk

3 8

$2,0001(7) 52 2 52 100 trips/yr $20 / person-trip Annual

100

eLB UB

e

Answer :

Computational Tools

18

Calculator Spreadsheet

Scripting

Language

Hybrid(toolbox,

addon)

Data

Pro

ce

ss

ing

Structured Unstructured

Co

mp

lex

Sim

ple

Matrix Multiplication

19

m n n p m p

m

n

n

p

× A B m

p

C=

Arrays must have sameinner dimensions

Element-by-Element Multiplication

20

m

nn

1.× Ab

b expanded to have compatible size with A

m

n

m

n

A b.× m

n

C=

1m n n m n

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

21

.m n m pA B

1.m n nA b

1 1.m na b

1 . p nma B

2-D Euclidean Distance

2 21 1,1 2 1,2

12 2

2 1 2,1 2 2,2

3 2 21 3,1 2 3,2

2

1 1

2 3 , 7 1

4 5

2 3 1 1

2 3 7 1

2 3 4 5

x p x pd

d x p x p

dx p x p

x P

d

d

22

Logistics Software Stack

23

• New Julia (1.0) scripting language– (almost?) as fast as C and Java (but not FORTRAN)

– does not require compiled standard library for speed

– uses multiple dispatch to make type-specific versions of functions

MIP Solver

(Gurobi,Cplex,etc.)

Standard Library

(in compiled C,Java)

User Library

(in script language)

MIP Solver

(Gurobi, etc.)

Standard Library

(C,Java)Data

(csv,Excel,etc.)

Report

(GUI,web,etc.)

Commercial

Software

(Lamasoft,etc.)

Scripting

(Python,Matlab,etc.)

Basic Matlab Workflow• Given problem to solve:

1. Test critical steps at Command Window2. 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)

24

Why Are Cities Located Where They Are?

25

Taxonomy of Location Problems

Location Decision

Cooperative

Location

Competitive

Location

Minisum Location“Nonlinear”

Location

Resource Oriented

Location

Market Oriented

Location

Transport Oriented

Location

Local-Input Oriented

Location

Minimax Cost

Maximin Cost

Center of Gravity

Minimize Sum of Costs

Sum of Costs = SC = TC +LC

LC > TC

Local Input Costs = LC = labor

costs, ubiquitous input costs, etc.

Minimize Individual Costs

PC > DC

Procurement Costs = PC

“Weight-losing” activities

DC > PC

Distribution Costs = DC

“Weight-gaining” activities

Minimize System Costs

TC > LC

Transport Costs = TC = PC + DC

26

Hotelling's Law

0 134

0 134

0 112

0 134

14

27

1-D Cooperative Location

0 30

1 2

w1 = 1 w2 = 2

Durham RaleighUS-70 (Glenwood Ave.)

Min ki iTC w d

Min i iTC w d

2Min i iTC w d

28

1 2

22

*

0, 30

2 0

1(0) 2(30)20

1 2

i i i i

i i

i i i

i i

i

a a

TC w d w x a

dTCw x a

dx

x w w a

w ax

w

Squared−Euclidean Distance Center of Gravity:

“Nonlinear” Location

ki iTC w d

29

Minimax and Maximin Location

• Minimax

– Min max distance

– Set covering problem

• Maximin

– Max min distance

– AKA obnoxious facility location

30

2-EF Minisum Location

3010

-8

+8

+5

-3

+2

-5+3

1 2

25 x

TC

90

+w1

+w2

+w1+w2

-w2

-w1

-w1-w2

+w1-w2

1 1 2 2

1 2

, if ( ) ( ) ( ), where

, if

(25) (25 10) ( )(25 30)

5(15) ( 3)( 5) 90

i ii i i

i i

w x xTC x w d x x x x

w x x

TC w w

31

Median Location: 1-D 4 EFs

wi

-5-3-2-4 = -14 +5-3-2-4 = -4 +5+3+2-4 = +6

Minimum at point where

TC curve slope switches

from (-) to (+)

5

TC

3 2 4

1 2 3 4

-14

-4 +2

+6

+14

+5+3-2-4 = +2 +5+3+2+4 = +14

5 < W/2 5+3=8 > W/2

4 < W/24+2=6 < W/24+2+3=9 > W/2

32

Median Location: 1-D 7 EFs

33

-83 -82 -81 -80 -79 -78

34

34.5

35

35.5

36

36.5

Asheville

Statesville

Winston-Salem GreensboroDurham

Raleigh

Wilm

ington

50

150

190 220270

295

420

40

3 24 3 56 1

13>1210<126<12

1

:2

j

i

i

Ww

:iw

14>12 11<12 5<129<12 8<12

*

Median Location: 2-D Rectilinear Distance 8 EFs

5 15 60 70 90

15

25

60

70

95

1

2

3

4

5

6

7

8

X

Y

19

53

82

42

9

8

39

6

101

101 < 129

50

151 > 129

157 > 129

*

48

107 < 129

53

59 < 129

6

6 < 129

62

62

< 1

29

19

81

< 1

29

48

12

9 =

12

9

*

39

39

< 1

29

90

12

9 =

12

9*

Optimal location

anywhere along line

: wi

: x

wi :

y :

1 1 2 1 2 1 2

2 22 1 2 1 2 1 2

( , )

( , )

d P P x x y y

d P P x x y y

34

Ex 3: 2D Loc with Rect Approx to GC Dist

• It is expected that 25, 42, 24, 10, 24, and 11 truckloads will be shipped each year from your DC to six customers located in Raleigh, NC (36N,79W), Atlanta, GA (34N,84W), Louisville, KY (38N,86W), Greenville, SC (35N, 82W), Richmond, VA (38N,77W), and Savannah, GA (32N,81W). Assuming that all distances are rectilinear, where should the DC be located in order to minimize outbound transportation costs?

35

136, 682

i

WW w

Optimal location (36N,82W)

(65 mi from opt great-circle location)

Answer :

24

10

42

11

25

2448

25

10

42

11

24 42 10 11 25 24

24<68 66<68 76>68

*

11<68

53<68

63<68

* 88<68

Logistics Network for a Plant

DDDD

EEEE

FFFF

GGGG

CCCC

BBBB

AA

AA

A

A

A

A

Customers

DCsPlantTier One

Suppliers

Tier Two Suppliers

Re

so

urc

e

Ma

rke

t

vs.

vs.

vs.

vs.

Distribution Network

Distribution

Outbound Logistics

Finished Goods

Assembly Network

Procurement

Inbound Logistics

Raw Materials

downstream

upstream

A = B + C

B = D + E

C = F + G

36

Basic Production System

Supplier Customer

raw

material

finished

goods

ubiquitous

inputs

scrap

4 ton 3 ton

1 ton

2 ton

Production

System

Inbound

FOB Origin

title transfer

Se

ller you pay

Bu

ye

r

Se

ller

FOB Destination

supplier pays

title transfer

FOB Destination

title transfer

you pay

Bu

ye

r

FOB Origin

customer pays

title transfer

Outbound

FOB (free on board)

37

FOB and Location• Choice of FOB terms (who directly pays for transport) usually

does not impact location decisions:

– 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

38

Procurement Landed costcost at supplier

Production Procurement Local resource cost cost cost (labor, etc.)

Total delivered Production

Inbound transport cost

Outbound transport cocost cost

Transport

s

cos

t

t (T

Inbound transport Outbound transport C) cost cost

Monetary vs. Physical Weight

39

in out

in out

(Montetary) Weight Gaining:

Physically Weight Losing:

w w

f f

1 1

min ( ) ( , ) ( , )

where total transport cost ($/yr)

monetary weight ($/mi-yr)

physical weight rate (ton/yr)

transport rate ($/ton-mi)

( , ) distance between NF at an

m m

i i i i i

i i

i

i

i

i

iw

TC X w d X P f r d X P

TC

w

f

r

d X P X

d EF at (mi)

NF = new facility to be located

EF = existing facility

number of EFs

i iP

m

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:

40

1 1

min ( ) ( , ) ( , )

where total transport distance (mi/yr)

monetary weight (trip/yr)

trips per year (trip/

transport rate =

yr)

( , ) per-trip distance between NF an E

1

d

m m

i i i i i

i

i

i

i

i

i

iw

r

TD X w d X P f r d X P

TD

w

f

d X P

F (mi/trip)i

Ex 4: Single Supplier and Customer Location

• The cost per ton-mile (i.e., the cost to ship one ton, one mile) for both raw materials and finished goods is the same (e.g., $0.10).

1. 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 (fi)?

41

Asheville Durham

raw

material

finished

goods

scrap

2 ton 1 ton

1 ton

Product A

-83 -82 -81 -80 -79 -78

34

34.5

35

35.5

36

36.5

Asheville

Statesville


Raleigh

Wilm

ingto

n

50

150

190 220270

295

420

40

WilmingtonWinston-

Salem

raw

material

finished

goods

ubiquitous

inputs

1 ton 3 ton

2 ton

Product B

1

( ) ( , )m

i i i

iiw

TC X f r d X P

Ex 5: 1-D Location with Procurement and Distribution Costs

Assume: all scrap is disposed of locally

42

Asheville unit of

finished

good

1 ton

Production

System

Durham

A product is to be produced in a plant that will be located along I-40. Two tons of raw materials from a supplier in Ashville and a half ton of a raw material from a supplier in Durham are used to produce each ton of finished product that is shipped to customers in Statesville, Winston-Salem, and Wilmington. The demand of these customers is 10, 20, and 30 tons, respectively, and it costs $0.33 per ton-mile to ship raw materials to the plant and $1.00 per ton-mile to ship finished goods from the plant. Determine where the plant should be located so that procurement and distribution costs (i.e., transportation costs to and from the plant) are minimized, and whether the plant is weight gaining or weight losing.

Ex 5: 1-D Location with Procurement and Distribution Costs

($/yr) ($/mi-yr) (mi)

monetary physicalweight weight

($/mi-yr) ($/ton-mi)(ton/yr)

i i

i i i

TC w d

w f r

in out

in out

(Montetary) Weight Gaining: 50 60

Physically Weight Losing: 150 60

w w

f f

2010 3040 10

70>5550<5540<55

1

:2

j

i

i

Ww

:iw

60>55 30<5540<55

*

Asheville DurhamStatesville Winston-Salem Wilmington

Assume: all scrap is disposed of locally

43

Asheville unit of

finished

good

1 ton

Production

System

Durham

NF 4

3

5

1

2

out $1.00/ton-mir

3 3 3 out10, 10f w f r

4 4 4 out20, 20f w f r

5 5 5 out30, 30f w f r

in $0.33/ton-mir

1 1 out 1 1 in2 60 120, 40f BOM f w f r

2 2 out 2 2 in0.5 60 30, 10f BOM f w f r

2-D Euclidean Distance

2 21 1,1 2 1,2

12 2

2 1 2,1 2 2,2

3 2 21 3,1 2 3,2

1 1

2 3 , 7 1

4 5

x p x pd

d x p x p

dx p x p

x P

d

44

Minisum Distance Location

2 2

1 ,1 2 ,2

3

1

*

* *

1 1

7 1

4 5

( )

( ) ( )

arg min ( )

( )

i i i

i

i

d x p x p

TD d

TD

TD TD

x

P

x

x x

x x

x

45

120°

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°)

Minisum Weighted-Distance Location

• Solution for 2-D+ andnon-rectangular distances:

– Majority Theorem: Locate NF at

EFj if

– Mechanical (Varigon frame)

– 2-D rectangular approximation

– Numerical: nonlinear unconstrained optimization

• Analytical/estimated gradient (quasi-Newton, fminunc)

• Direct, gradient-free (Nelder-Mead, fminsearch)

1

*

* *

number of EFs

( ) ( )

arg min ( )

( )

m

i i

i

m

TC w d

TC

TC TC

x

x x

x x

x

Varignon Frame

1

, where2

m

j i

i

Ww W w

46

Convex vs Nonconvex Optimization

47

Gradient vs Direct Methods

• Numerical nonlinear unconstrained optimization:

– Analytical/estimatedgradient

• quasi-Newton

• fminunc

– Direct, gradient-free• Nelder-Mead

• fminsearch

48

Nelder-Mead Simplex Method

• AKA amoeba method

• Simplex is triangle in 2-D (dashed line in figures)

reflection expansion

outsidecontraction

insidecontraction

a shrink

49

Feasible Region

50

( ), if is in , if is true( , , ), otherwise , otherwise

TCb aiff a b cc

x x R

if a is true

return b

else

return c

end

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

51

Convex Hull

• Find the points that enclose all points

– Most important data structure

– Calculated, via Graham’s scan in

52

( log ), pointsO n n n

Delaunay Triangulation

• Find the triangula-tion of points that maximizes the minimum angle of any triangle

– Captures proximity relationships

– Used in 3-D animation

– Calculated, via divide and conquer, in

53

( log ), pointsO n n n

Voronoi Diagram

• Each region defines area closest to a point

– Open face regions indicate points in convex hull

54

Voronoi Diagram

• Voronoi diagram from smooshing paint between glass

– https://youtu.be/yDMtGT0b_kg

55

Delaunay-Voronoi

• Delaunay triangula-tion is straight-line dual of Voronoidiagram

– Can easily convert from one to another

56

Minimum Spanning Tree

• Find the minimum weight set of arcs that connect all nodes in a graph

– Undirected arcs:calculated, via Kruskal’s algorithm,

– Directed arcs:calculated, via Edmond’s branching algorithm, in

57

( log ), arcs, nodesO m n m n

( ), arcs, nodesO mn m n

Kruskal’s Algorithm for MST• Algorithm:

1. Create set F of single node trees

2. Create set S of all arcs3. While S nonempty and F is

not yet spanning4. Remove min arc from S5. If removed arc connects two

different trees, then add to F, combining two trees into single tree

6. If graph connected, F forms single MST; otherwise, forms multi-tree min spanning forest

• Optimal “greedy” algorithm, runs in O(m logn)

• If directed arcs, O(mn)– useful in VRP to min vehicles– harder to code

58

m = 15 arcs, n = 8 nodes

Min Spanning vs Steiner Trees

• Steiner point added to reduce distance connecting three existing points compared to min spanning tree

59

ba

b

b

a a

Steiner

Point

13 3 , 30 60 90 triangle

2 2

Min spanning tree distance > Steiner tree distance

2 3

2 3 3

2 3

4 3

ba b a

b a

a a

Steiner Network

60

Metric Distances

61

x

y

Chebychev Distances

62

Proof

Great Circle Distances

13.35

miR

63

Metric Distances using dists

3 2 2 2

3 2

4 5

1 'mi' 'km'dists( , , ),

2 1 2 Inf

3 n d m d

n m

p pD X1 X2

, 2X1 X

, 2X1 X

, 2X1 X

d

d

D

, 2X1 X

D

d = 2 d = 1

, 2X1 X Error

2

1

3

4

5

64

Mercator Projection

proj

1proj

1proj

sinh tan

tan sinh

x x

y y

y y

deg radrad deg

180and

180

x xx x

65

Circuity Factor

66

road

road 1 2 1 2

: , where usually 1.15 1.5, weight of sample

( , ), estimated road distance from to

i

i

i i

GC

GC

dCircuity Factor g v g v i

d

d g d P P P P

Estimating Circuity Factors

• Circuity factor depends on both the trip density and directness of travel network

– Circuity of high trip density areas should be given more weight when estimating overall factor for a region

– Obstacles (water, mountains) limit direct road travel

67

60k pop. 30k pop.

10k pop.

1,3

1,3

1.2

0.22

g

v

1,2

1,2

1.1

0.67

g

v

2,32

,3

1.40.11g

v

3

1 2

Allocation• Example: given n DCs and m customers, with customer j

receiving wj TLs per week, determine the total distance per week assuming each customer is served by its closest DC

68

2 4 6 8

10 20 30 40

45 35 25 15

2(10) 4(20) 6(25) 8(15)

370

w

D

TD

2

1

3

4

Customers

DCs

ijd

1

2

2

1

3

4

Customers

DCs

j ijw d

allocate

1

2

Pseudocode

• Different ways of representing how allocation and TD can be calculated– High-level pseudocode most concise, but leaves out many

implementation details (sets don’t specify order, initial starting points)

– Low-level pseudocode includes more implementation details, which can hide/obscure the core idea, and are usually not essential

69

,

1, , ,

1, , ,

arg min

j

j iji N

j j

j M

N n n N

M m m M

d

TD w d

Low-level Pseudocode High-level Pseudocode Matlab/Matlog

Minisum Multifacility Location

1 1 1

no. of NFs, no. of EFs

NF locations, EF locations

NF-NF 1 2 3 4 5

1 0 0

2 0 0

3 0 0 0 0 0

4 0 0 0 0 0

5 0 0 0 0 0

NF-EF 1 2 3 4 5 6 7

1 0 0 0 0 0

2 0 0 0 0 0

3 0 0 0 0 0

4 0 0 0 0 0

5 0 0 0 0 0 0

( ) ( , )

n d m d

n n

n m

n n m

jk j k ji

j k j i

n m

TC v d w

X P

V

W

X X X1

*

* *

( , )

arg min ( )

( )

n

j id

TC

TC TC

X

X P

X X

X

70

Su

pp

liers

Manufacturing

Cu

sto

me

rs

5

4

6

7

1

2

3

Distribution

NFs

EFs

EFs

4

5

3

1

2

Majority Theorem for Minisum Location

• Single-facility:

• Multifacility: can be used to reduce and sometimes solve

71

1

Locate NF at EF if , where 2

m

j i

i

Wj w W w

1 1

Given EF and NF, let

11. While any , co-locate NF and NF and

2

( ) add row to row of , remove row from

( ) add row to row and column to column of

(

m n

ik ij ij

j j

m n

v w v i k

a k i k

b k i k i

c

V V V

W W

V

1 1

) remove row and column from , and set 0

12. Locate all NF at EF if ,

2

where any NF co-located with NF are also located at EF .

ii

m n

ik ij ij

j j

k k v

i k w w v

j i k

V

Ex 6: Multifacility Majority Theorem

72

1,1

2,3

0 2 0 23 EF, 2 NF, = = :

0 0 2 0

1. No solutions or reductions possible

2 1 0 0 2 5[ ]

4 0 5 2 0 11

2. Modified solution

3.5 2 1.75 NF1 at EF12 1 0 0 0.5[ ]

9.5 5 4.75 NF4 0 5 0.5 0

w

w

V V

W V

V

W V

1,2

1

2 at EF3

3. Modified reduction solution

7 4 3.5 NF1 and NF2 co-located2 1 0 0 4[ ]

13 all , 6.54 0 5 4 0

Reduced , no : 6 1 5 12 6 6 NF1 (and NF2) at EF1

ij ij

v

w v

w

V

W V

W V W

Ex 7: Location of Production Processes

73

2,3

1,1

2,2

4 0 0 3 0 0 3 0

2 EF, 3 NF, = 0 0 , = 0 0 5 = 3 0 5

0 4 0 0 0 0 5 0

4 0 0 3 0

1. [ ] 0 0 3 0 5 8 5 4 NF2 and NF3 co-located

0 4 0 5 0

7 4 3.5 NF1 at EF14 0 0 32.

7 4 3.5 NF2 (and0 4 3 0

v

w

w

W V V

W V

W V NF3) at EF2

1 2$4/mi $3/mi $5/mi $4/mi

1 2 3 4 5

DetroitNagoya

Drop Forge Heat Treat Pressing Finishing Painting

1 2 3

Multiple Single-Facility Location

1 1 1 1

1

( ) ( , ) ( , )

( )

n n n m

jk j k ji j i

j k j i

n

j

j

TC v d w d

TC

X X X X P

X

74

Su

pp

liers

Manufacturing

Cu

sto

me

rs

7

6

8

9

4

5

1

2

3

Distribution

EFs

EFs NFs

2

3

1

Facility Location–Allocation Problem

• Determine both the location of n NFs and the allocation of flow requirements of m EFs that minimize TC

1 1

* *

,1

* * *

(1) flow between NF and EF

total flow requirememts of EF

( , ) ( , )

, arg min ( , ) : , 0

( , )

ji ji ji ji

i

n m

ji j i

j i

n

ji i ji

j

w r f f j i

w i

TC w d

TC w w w

TC TC

X W

X W X P

X W X W

X W

75

2

1

3

4

EFs

NFs

2

3

1

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

( )

1

*

* *

( ) arg min ( , )

( ) ( , )

arg min ( )

( )

i

i j ij

m

i i

i

d

TC w d

TC

TC TC

X

X

X X P

X X P

X X

X

76

2

1

3

4

EFs

NFs

2

3

1

Alternating Formulation

• Alternate between finding locations and finding allocations until no further TC improvement

• Requires n d-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

1 1

, if arg min ( , )( )

0, otherwise

( , ) ( , )

( , ) arg min ( , )

i k ik

ji

n m

ji j i

j i

w d jallocate w

TC w d

locate TC

X

X PX

X W X P

W X X W

77

2

1

3

4

EFs

NFs

2

3

1

ALA: Alternate Location–Allocation

78

2

1

3

4

EFs

NFs

2

3

1

Best Retail Warehouse Locations

79

Optimal Number of NFs

1 52 3 4 6

Facility Fixed + Transport Cost

Facility Fixed Cost

Transport Cost

TC

Number of NFs

80

Uncapacitated Facility Location (UFL)• NFs can only be located at discrete set of sites

– Allows inclusion of fixed cost of locating NF at site opt number NFs

– Variable costs are usually transport cost from NF to/from EF

– Total of 2n – 1 potential solutions (all nonempty subsets of sites)

81

1,..., , existing facilites (EFs)

1,..., , sites available to locate NFs

, set of EFs served by NF at site

variable cost to serve EF from NF at site

fixed cost of locating NF at site

i

ij

i

M m

N n

M M i

c j i

k i

Y

*

*

, sites at which NFs are located

arg min :

min cost set of sites where NFs located

number of NFs located

i

i ij iY

i Y i Y j M i Y

N

Y k c M M

Y

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 construct-then-improve multiple times

• Hard construction:– almost no chance of generating a random feasible solution due

to constraints on what is a feasible solution– need to include randomness at decision points as solution is

generated in order to construct multiple different solutions (which “might” then be able to be improved)

82

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 in a single

step that is feasible, need to include randomness at decision points as solution is constructed

83

1 2 3 4 5 6

1 2 4 5 6

1 2 4 6

2 4 6

2 4

2

4

2

3

5

1

6

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

84

Ex 8: UFL ADD

85

150 200 150 150 200

(1)(1)

1 2 3 4 5

Asheville: 1 0 100 170 245 370

Statesville: 2 100 0 70 145 270

Greensboro: 3 170 70 0 75 200

Raleigh: 4 245 145 75 0 125

Wilmington: 5 370 270 200 125 0

ij j ij j ij ij ij

ij

k

c w d f rd d d

c

-83 -82 -81 -80 -79 -78

34

34.5

35

35.5

36

36.5

Asheville

Statesville

Greensboro

Raleigh

50

150

220

295

420

40

1 2 3 4 5

1 0 100 170 245 370 885 150 1,035

2 100 0 70 145 270 585 200 785

3 170 70 0 75 200 515 150 665

4 245 145 75 0 125 590 150 740

5 370 270 200 125 0 965 200 1,165

Yj Y Yj YY c k c k

1 2 3 4 5

3,1 0 70 0 75 200 345 300 645

3,2 100 0 0 75 200 375 350 725

3,4 170 70 0 0 125 365 300 665

3,5 170 70 0 75 0 315 350 665

Yj Y Yj YY c k c k

1 2 3 4 5

3,1,2 0 0 0 75 200 275 500 775

3,1,4 0 70 0 0 125 195 450 645

3,1,5 0 70 0 75 0 145 500 645

Yj Y Yj YY c k c k

Y

3Y

3,1Y

* 3,1Y

UFLADD: Add Construction Procedure

86

UFLXCHG: Exchange Improvement Procedure

87

Modified UFLADD• Y input can be used to

start UFLADD with Y NFs

– Used in hybrid heuristic

• p input can be used to keep adding until number of NFs = p

– Used in p-median heuristic

88

UFL: Hybrid Algorithm

89

P-Median Location Problem• Similar to UFL, except

– Number of NF has to equal p (discrete version of ALA)

– No fixed costs

90

*

number of NFs

arg min : ,

i

ij iY

i Y j M i Y

p

Y c M M Y p

Bottom-Up vs Top-Down Analysis

• Bottom-Up: HW3 Q3

1

2

NEW

NEW

3 2

1 2 2 3 3 1

1 2 2 3 3 1

3

1

*

*

lon-lat of EFs

48,24,35 (TL/yr)

2 ($/TL-mi)

1 ( , ) ( , ) ( , )

3 ( , ) ( , ) ( , )

( ) ( , ) (outbound trans. costs)

arg min ( )

RD RD RD

GC GC GC

i GC i

i

TC

r

TC

r

d d dg

d d d

TC f r gd

TC

TC

x

P

f

P P P P P P

P P P P P P

x x P

x x

*

cary

cary cary

cary *

( )

lon-lat of Cary

( )

TC

TC TC

TC TC TC

x

x

x

• Top-Down: estimate r(circuity factor cancels, so not needed, HW4 Q6)

OLD nom NEW

cary

cary

nom 3cary

1

3

nom

1

*

* *

cary *

current known

10 ton /TL= known tons per truckload

480,240,350 (ton /yr)

($/ton-mi)

( , )

( ) ( , )

arg min ( )

( )

i GC i

i

i GC i

i

TC r TC

TC TC

TCr

f d

TC f r d

TC

TC TC

TC TC TC

x

f

x P

x x P

x x

x

91

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 statisticalareas (CSAs)

3. Core-based statisticalareas (CBSAs)

4. Metropolitian/micropolitan statisticalareas (MSAs)

5. County (rural)

92

Aggregate Demand Point Data Sources• Aggregate demand point: centroid of population + area + population• Good rule of thumb: use at least 10x number of NFs ( 100 pts provides

minimum coverage for locating 10 NFs)

1. City data: ONLY USE FOR LABELING!, not as aggregate demand points2. 3-digit ZIP codes: 1000 pts covering U.S., = 20 pts NC3. 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= 37183 = 37 NC FIPS + 183 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 NC5. Census Block Group: > 220K pts U.S., 1000 pts Raleigh-Durham-Chapel

Hill, NC CSA– Grouped by state, county, or CSA– Finest resolution aggregate demand data source

93

City vs CSA Population Data

94

Demand Point Aggregation• Existing facility (EF): actual physical location of demand source

• Aggregate demand point: single location representing multiple demand sources

95

-83 -82 -81 -80 -79 -78

34

34.5

35

35.5

36

36.5

Asheville

Statesville


Raleigh

Wilm

ingto

n

50

150

190 220270

295

420

40

Demand Point Aggregation• Calculation of aggregate point depends on objective

• For minisum location, would like for any location x:

• For squared distance:

96

10 40

1 2

w1 = 1 w2 = 2

Durham Raleigh

0

x1 x2x

1 2 agg 1 1 2 2 1 2

1 2 agg 1 1 2 2

1 1 2 2agg

1 2

1 2 agg

( , ) ( , ) ( , ), let 0, , 0

centroid

Note: if , then x not centroid

w w d x x w d x x w d x x x x x

w w x w x w x

w x w xx

w w

x x x

30

xagg

wagg = 3

1 2

1 2

2 2 21 2 agg 1 2

2 21 2

agg

1 2

not centroid

w w x w x w x

w x w xx

w w

Transport Cost if NF at every EF

1 52 3 4 6

Facility Fixed + Transport Cost

Facility Fixed Cost

Transport Cost

TC

Number of NFs

NF = EF

0?

97

transport costfixed cost

i

i ij

i Y i Y j M

TC k c

Area Adjustment for Aggregate Data Distances

• 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

2

0

0

0

0 0 0

2

0.402

320.51

5152

0.45

LB

LB

UB

LB UB

ad

ad a

ad a

d d d a

Mathai, A.M., An Intro to Geo Prob, p. 207 (2.3.68)

98

2

a

a0

LB

d

0UB

d

a

1 2 1 2 local 1 2

1 2 1 2

( , ) max ( , ), 0.45max ,

max 1.2 ( , ), 0.675max ,

a GC

GC

d gd g a a

d a a

X X X X

X X

Fixed Cost and Economies of Scale• Cost data from existing facilities

can be used to fit linear estimate– Economies of scale in production

k > 0 and < 1

99

max

act min 0

0

est

act 0 1act

0

est

max ,

0.62, Hand tool mfg.

0.48, Construction

0.41, Chemical processing

0.23, Medical centers

fixed cost

f f

p

p

p

fTPC TPC TPC

f

TPC c f

TPC TPCAPC f

f f

kAPC c

f

k

k

c

min max

MES

0 0

constant unit production cost

/ min/max feasible scale

/ base cost/rate

f f

f Minimum Efficient Scale

TPC f

Ex 9: Popco Bottling Company• 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’sproduction, procurement, and distribution costs.

100

Ex 9: Popco Bottling Company• Following representative information is available for each of N

current plants (DC) i:

• Assuming plants are (monetarily) weight gaining since they are bottling plants, so UFL can ignore inbound procurement costs related to location

101

location

aggregate production (tons)

total production and procurement cost

total distribution cost

i

DCi

i

i

xy

f

TPC

TDC

Supplier Customer

raw

material

finished

goodsProduction

System

Procurement DistributionProduction

Ex 9: Popco Bottling Company

1. Use plant (DC) production costs to find UFL fixed costs via linear regression

– variable production costs cp do not change and can be cut

only keep for UFL

i

DCi p

i N i N

TPC TPC c fk

k

102

• 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.

Ex 9: Popco Bottling Company2. Allocate all 3-digit ZIP

codes to closest plant (up to 200 mi max) to serve as aggregate customer demand points.

max

max

: arg min and

200 mi

a ai hj ij

h

i

i N

M j d i d d

d

M M

103

Ex 9: Popco Bottling Company3. Allocate each plant’s demand (tons of product) to each of its

customers based on its population.

55 2

5 6

population of EF

i

i

jDCj M i

h

h M

j

DC

qf f

q

q j

qf f

q q

5

4

6

7

1

2

3

104

Ex 9: Popco Bottling Company4. 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).

nom

j

i

i N

aj ij

i N j M

TDC

rf d

105

Ex 9: Popco Bottling Company5. Calculate UFL variable transportation cost cij ($) for each

possible NF site i (all customer and plant locations) and EF site j (all customer locations) as the product of customer jdemand (ton), distance from site i to j (mi), and the nominal transport rate ($/ton-mi).

6. Solve as UFL, where TC returned includes all new distribution costs and the fixed portion of production costs.

noma

i M Nij j ij i M Nj M j M

c r f d

C

106

transport costfixed cost

, number of potential NF sites

, number of EF sites

i

i ij

i Y i Y j M

TC k c

n M N

m M

MILP

LP: max '

s.t.

0

MILP: some integer

ILP: integer

BLP: 0,1

ix

c x

Ax b

x

x

x

107

1 42 63 5

1

2

3

0

4

1x

2x

1 2

1 2

1

1 2

max 6 8

s.t. 2 3 11

2 7

, 0

x x

x x

x

x x

6 8

2 3 11,

2 0 7

c

A b

* *

13

22 , 311 3

13

x c x

Branch and Bound

1 42 63 5

1

2

3

0

4

231

3

1 2

1 2

1

1 2

1 2

max 6 8

s.t. 2 3 11

2 7

, 0

, integer

x x

x x

x

x x

x x

1x

6 8

2 3 11,

2 0 7

c

A b

2x

0

1

2

131

3

26

31

230

3

28

30

3

4

5

6

0

1 8

32

74

65

231 , 0

3UB LB

1 3x 1 4x

2 1x 2 2x

1 2x 1 3x

2 2x 2 3x

LP

131 , 0

3UB LB

131 , 26

3UB LB

Incumbent 31, 26UB LB

230 , 26

3UB LB

Incumbent

230 , 30

32

30 30 13

UB LB

gap

230 , 28

3UB LB

Incumbent

Fathomed,infeasible

Fathomed,infeasible

STOP

108

MILP Solvers

109

MILP Solvers

1 42 63 5

1

2

3

0

4

231

3

1x

2x

1 2 1 2

1 2

2 2 1, , 0 and integer

0

x x x x

x x

• Presolve: eliminate variables

• Cutting planes: keeps all integer solutions and cuts off LP solutions (Gomory cut)

• Heuristics: find good initial incumbent solution (Hybrid UFL)

• Parallel: use separate cores to solve nodes in B&B tree

• Speedup from 1990-2014:– 320,000 computer speed

– 580,000 algorithm improvements

10 days of 24/7 processing 1 sec110

Gomory cut

• Cplex (IBM, comm first solver)

• Gurobi (dev Robert Bixby)

• Xpress (used by LLamasoft)

• SAS/OR (part of SAS system)

• Symphony (open source)

• Matlab’s intlinprog

MILP Formulation of UFL

min

s.t. 1,

,

0 1, ,

0,1 ,

i i ij ij

i N i N j M

ij

i N

i ij

j M

ij

i

k y c x

x j M

my x i N

x i N j M

y i N

, ,i ijy x i N j M

where

fixed cost of NF at site 1,...,

variable cost from to serve EF 1,...,

1, if NF established at site

0, otherwise

fraction of EF demand served from NF at site .

i

ij

i

ij

k i N n

c i j M m

iy

x j i

111

Capacitated Facility Location (CFL)

min

s.t. 1,

,

0 1, ,

0,1 ,

i i ij ij

i N i N j M

ij

i N

i i j ij

j M

ij

i

k y c x

x j M

K y f x i N

x i N j M

y i N

where

fixed cost of NF at site 1,...,

variable cost from to serve EF 1,...,

capacity of NF at site 1,...,

demand EF 1,...,

1, if NF established at site

0, otherwise

i

ij

i

j

i

k i N n

c i j M m

K i N n

f j M m

iy

x

fraction of EF demand served from NF at site .ij j i112

• CFL does not have simple and effective heuristics, unlike UFL

• Other types of constraints:• Fix NF i at site j: set LB and UB of

xij to 1• Convert UFL to p-Median: set all

k to 0 and add constraint sum{yi} = p

Matlog’s Milp• Executing mp = Milp creates a Milp object that can be used

to define a MILP model that is then passed to a Solver

– Similar syntax to math notation for MILP

– AMPL and OPL algebraic modeling languages provide similar capabilities, but Milp integrated into MATLAB

113

Illustrating Milp syntax

114

3 2,4

2,4

addobj 'min', ,

addobj 'min', ,

addcstr , , ' ',7

addcstr { ,{3}},{ ,{2,4}}, ' ',7

addcstr 0 ,1 , ' ',7

addcstr 0,{2,4}, ' ',7

my nx

m n

x

k C

y X

y

Ex 10: UFL MILP

115

min

s.t. 1,

,

0 1, ,

0,1 ,

i i ij ij

i N i N j M

ij

i N

i ij

j M

ij

i

k y c x

x j M

my x i N

x i N j M

y i N

(Weighted) Set Covering

*

1,..., , objects to be covered

, 1,..., , subsets of

cost of using in cover

arg min : , min cost covering of

i

i i

i iI

i I i I

M m

M M i N n M

c M

I c M M M

116

*

1 2 3

4 5

*

1,...,6

1,...,5

1,2 , 1,4,5 , 3,5

2,3,6 , 6

1, for all

arg min :

2,4

2

i

i iI

i I i I

i

i I

M

i N

M M M

M M

c i N

I c M M

c

1

4

2

5

3

6

M2

M1

M3

M4

M5

(Weighted) Set Covering

min

s.t. 1,

0,1 ,

i i

i N

ji i

i N

i

c x

a x j M

x i N

*

1,..., , objects to be covered

, 1,..., , subsets of

cost of using in cover

arg min : , min cost covering of

i

i i

i iI

i I i I

M m

M M i N n M

c M

I c M M M

where

1, if is in cover

0, otherwise

1, if

0, otherwise.

ii

iji

Mx

j Ma

117

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)

118

max

s.t. 1,

0,1 ,

i

i N

ji i

i N

i

x

a x j M

x i N

1

4

2

5

3

6

M1

M3

M5

Bin Packing

min

s.t. ,

1,

0,1 ,

0,1 , ,

i

i M

i j ij

j M

ij

i M

i

ij

y

Vy v x i M

x j M

y i M

x i M j M

*

1,..., , objects to be packed

volume of object

volume of each bin max

arg min : , , min packing of

i i

j

i j

j iB

j B B B

M m

v j

V B v V

B B v V B M M

where

1, if bin is used in packing

0, otherwise

1, if object packed into bin

0, otherwise.

ii

iij

By

j Bx

119

Topics

1. Introduction




4. Network models

5. Routing


6. Warehousing


120

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• 50-250 mi return possible (11 HOS)

• > 250 mi always one-way transport

• > 500-750 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)

121


4.

5. TL Weight Capacity: 25 tons ( Kwt )– (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 ft3 ( Kcu )– Trailer physical capacity = 3,332 ft3

– Effective capacity = 3,332 × 0.80 2,750 ft3

– Cube capacity = 80% of physical capacity

3

3 $2,620 Shanghai-LA/LB shipping cost

2,400

Value1:

Transport Cost ft 40’ ISO container capa$1 ft

city

122

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

In

terio

r H

eig

ht:

(8'6

" -

9'2

" =

10

2"

- 1

10

")

Width:

8'6" = 102"

(8'2" = 98")

Ma

x H

eig

ht:

13

'6"

= 1

62

"


7. TL Revenue per Loaded Truck-Mile: $2/mi in 2004 ( r )– TL revenue for the carrier is your TL cost as a shipper

532 mi

Raleigh Gainesville

LL

U

L

U

Greensboro Jacksonville

15%, average deadhead travel

$1.60, cost per mile in 2004

$1.60$1.88, cost per loaded-mile

1 0.15

6.35%, average operating margin for trucking

$1.88$2.00, revenue per loaded-mile

1 0.0635

123

One-Time vs Periodic Shipments

• One-Time Shipments (operational decision): know shipment size q

– Know when and how much to ship, need to determine if TL and/or LTL to be used

– Must contact carrier or have agreement to know charge• Can/should estimate charge before contacting carrier

• Periodic Shipments (tactical decision): know demand rate f, must determine size q

– Need to determine how often and how much to ship

– Analytical transport charge formula allow “optimal” size (and shipment frequency) to be estimated

• U.S. Bureau of Labor Statistic's Producer Price Index (PPI) for TL and LTL used to estimate transport charges

124

Truck Shipment Example

• Product shipped in cartons from Raleigh, NC (27606) to Gainesville, FL (32606)

• Each identical unit weighs 40 lband occupies 9 ft3 (its cube)

– Don’t know linear dimensions of each unit for TL and LTL

• Units can be stacked on top of each other in a trailer

• Additional info/data is presented only when it is needed to determine answer

125

Truck Shipment Example: One-Time1. Assuming that the product is to be shipped P2P TL, what is

the maximum payload for each trailer used for the shipment?

max

3

3

3

maxmax

max max max

25 ton

2750 ft

40 lb/unit4.4444 lb/ft

9 ft /unit

2000

2000

min , min ,2000

4.4444(2750)min 25, 6.1111 ton

2000

wtwt

cu

cucucu

cu

cuwt cuwt

q K

K

s

q sKK q

s

sKq q q K

126

Truck Shipment Example: One-Time2. On Jan 10, 2018, 320 units of the product were shipped.

How many truckloads were required for this shipment?

3. Before contacting the carrier (and using Jan 2018 PPI ), what is the estimated TL transport charge for this shipment?

max

40 6.4320 6.4 ton, 2 truckloads

2000 6.1111

qq

q

Jan 2018

20042004

max

532 mi

$2.00 / mi102.7

131.0$2.00 / mi $2.5511/ mi

102.7

6.4(2.5511)(532) $2,714.39

6.1111

TL TLTL

TL

TL TL

d

PPI PPIr r

PPI

qc r d

q

127

Truck Shipment Example: One-Time

128

Truck Shipment Example: One-Time4. Using the Jan 2018 PPI LTL rate estimate, what was the

transport charge to ship the fractional portion of the shipment LTL (i.e., the last partially full truckload portion)?

frac max

2

1 1527 29

frac

2

1 1527 29

frac

6.4 6.1111 0.2889 ton

148

72 14

2

4.4414

8177.4 $3.8014 / ton-mi

74.44 2(4.44) 140.2889 532

2

3.8014(0.28

LTL LTL

LTL LTL

q q q

s

r PPI

s sq d

c r q d

89)(532) $584.23

129

Truck Shipment Example: One-Time5. What is the change in total charge associated with the

combining TL and LTL as compared to just using TL?

1

frac

max max

$772.96

TL TL LTL

TL TL LTL

c c c c

q qr d r d r q d

q q

130

Truck Shipment Example: One-Time6. What would the fractional portion have to be so that the TL

and LTL charges are equal?

max

2

1 1527 29

( )

148

( )7

2 142

( ) ( )

arg min ( ) ( )

0.7960 ton

TL TL

LTL LTL

LTL LTL

I TL LTLq

qc q r d

q

s

r q PPI

s sq d

c q r q qd

q c q c q

131

Truck Shipment Example: One-Time7. What are the TL and LTL minimum charges?

• 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?

28

19

28

19

45 $57.402

45104.2 1625

177.4 53245 $87.51

104.2 1625

TLTL

LTLLTL

rMC

PPI dMC

132

Truck Shipment Example: One-Time• Independent Transport Charge ($):

0 ( ) min max ( ), ,max ( ),TL TL LTL LTLc q c q MC c q MC

133

Truck Shipment Example: One-Time8. Using the same LTL shipment, find online one-time (spot) LTL

rate quotes using the FedEx LTL website

3

3

40 lb/unit4.

0

4444 lb/ft9 ft

2

/unit

Class 0

s

Class-Density Relationship

frac 0.2889 ton

0.2889(2000) 578 lb

0.2889(2000)no. = 15 cartons

units 40

q

• Most likely freight class:

• What is the rate quote for the reverse trip from Gainesville (32606) to Raleigh (27606)?

134

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

135


Tariff (in $/cwt) from Raleigh, NC (27606) to Gainesville, FL (32606)

(532 mi, CzarLite DEMOCZ02 04-01-2000, minimum charge = $95.23)

• CzarLite tariff table for O-D pair 27606-32606

100 1hundredweight 100 lb ton

2000 20cwt

136

Truck Shipment Example: One-Time9. Using the same LTL shipment, what is the transport cost

found using the undiscounted CzarLite tariff?

0.2889, 200

0, 95.23

q class

disc MC

1

2 21

2

arg

arg

arg 0.2889 0.5 20.25

B BBi ii

B BB

B

i q q qq

q q qq

q

tariff 1 max ,min ( , ) 20 , ( , 1) 20

1 0 max 95.23,min (200,2) 20(0.2889), (200,3) 20(0.5)

max 95.23,min (127.69) 20(0.2889), (99.92)20(0.5)

max 95.23,min 737.76, 999.20 $737.76

Bic disc MC OD class i q OD class i q

OD OD

137

Truck Shipment Example: One-Time10. What is the implied discount of the estimated charge from

the CzarLite tariff cost?

tariff

tariff

737.76 584.23

737,76

20.81%

LTLc cdisc

c

( , 1)

( , )

99.92(1) 0.3913 ton

127.69

W Bi i

OD class iq q

OD class i

• What is the weightbreak betweenthe rate breaks?

138


• PX: Package Express– (Undiscounted) charge cPX based

rate tables, R, for each service (2-day ground, overnight, etc.)

– Rate determined by on chargeable weight, wtchrg, and zone

– All PX carriers (FedEX, UPS, USPS, DHL) use dimensional weight, wtdim

– wtdim > 150 lb is prorated per-lb rate– Actual weight 1–70 lb (UPS, FedEx

home), 1–150 lb (FedEx commercial) – Carrier sets a shipping factor, which

is min cubic volume per pound– Zone usually determined by O-D

distance of shipment– Supplemental charges for home

delivery, excess declared value, etc.

139

chrg

chrg act dim

act

3

dim 3

3

3

,

max , (lb)

actual weight (1 to 150 lb)

(in )(lb)

(in / lb)

, , length, width, depth (in)

, actual cube

shipping factor (in / lb)

12 , invers

PXc R wt zone

wt wt wt

wt

l w dwt

sf

l w d

l w l w d

sf

s

3

3

e of density

139 FedEx (2019)

12.43 lb/ft (Class 85)

194 USPS 8.9 lb/ft

s

s


• (Undisc.) charge to ship a single carton via FedEx?

140

3act

3 3

dim

chrg act dim

chrg

40 lb, 9 ft

532 mi 4

carton actual cube

9 12 15,552 in 32 27 18

15,552111.9 lb

139

max ,

max 40,111.9 112 lb

,

112,4 $64.2

PX

wt cu

d zone

l w d

l w d

l w dwt

sf

wt wt wt

c R wt zone

R

7

FedEx Standard List Rates (eff. Jan. 7, 2019)

Note: No Zone 1(usually < 50 mi local)

Truck Shipment Example: Periodic11. Continuing with the example: assuming a constant annual

demand for the product of 20 tons, what is the number of full truckloads per year?

max max

20 ton/yr

6.1111 ton/ TL (full truckload )

203.2727 TL/yr, average shipment frequency

6.1111

f

q q q q

fn

q

• Why should this number not be rounded to an integer value?

141

Truck Shipment Example: Periodic12. What is the shipment interval?

1 6.11110.3056 yr/TL, average shipment interval

20

qt

n f

• How many days are there between shipments?

365.25 day/yr

365.25365.25 111.6042 day/TLt

n

142

Truck Shipment Example: Periodic13. What is the annual full-truckload transport cost?

max

532 mi, $2.5511/ mi

2.5511$0.4175 / ton-mi

6.1111

, monetary weight in $/mi

3.2727(2.5511)532 $4,441.73/yr

TL

TLFTL

FTL FTL TL

d r

rr

q

TC f r d n r d wd w

• What would be the cost if the shipments were to be made at least every three months?

max min

max min

min

3 1yr/TL 4 TL/yr

12 max ,

max ,

max 3.2727, 4 2.5511(532) $5,428.78/yr

FTL TL

ft n q

t n n

TC n n r d

143

Truck Shipment Example: Periodic• Independent and allocated full-truckload charges:

Transport Charge for a Shipment

max 0, c ( ), FTLq q UB LB q qr d

144150/2000

87.51

4072

2714

1357

0.7960 6.11 12.22

Shipment Size (tons)

Tra

nsp

ort C

harg

e (

$)

MC

1 TL

2 TL

3 TL

Truck Shipment Example: Periodic• Total Logistics Cost (TLC) includes all costs that could change

as a result of a logistics-related decision

cycle pipeline safety

transport cost

inventory cost

purchase cost

TLC TC IC PC

TC

IC

IC IC IC

PC

• 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

145

Truck Shipment Example: Periodic• Same units of inventory can serve multiple roles at each

position in a production process

• Working stock: held as part of production process• (in-process, pipeline, in-transit, presentation)

• Economic stock: held to allow cheaper production• (cycle, anticipation)

• Safety stock: held to buffer effects of uncertainty• (decoupling, MRO (maintenance, repair, and operations))

146

Truck Shipment Example: Periodic14. 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, what is a “reasonable estimate” for the total annual cost for this cycle inventory?

cycle (annualcost of holding one ton)(average annual inventory level)

( )( )

unit value of shipment ($/ton)

inventory carrying rate, the cost per dollar of inventory per year (1/yr)

average int

IC

vh q

v

h

er-shipment inventory fraction at Origin and Destination

shipment size (ton)q

147

Truck Shipment Example: Periodic• Inv. Carrying Rate (h) = interest + warehousing + obsolescence

• Interest: 5% per Total U.S. Logistics Costs

• Warehousing: 6% per Total U.S. Logistics Costs

• Obsolescence: default rate (yr) h = 0.3 hobs 0.2 (mfg product)

– Low FGI cost (yr): h = hint + hwh + hobs

– High FGI cost (hr): h hobs, can ignore interest & warehousing• (hint+hwh)/H = (0.05+0.06)/2000 = 0.000055 (H = oper. hr/yr)

– Estimate hobs using “percent-reduction interval” method: given time th

when product loses xh-percent of its original value v, find hobs

– Example: If a product loses 80% of its value after 2 hours 40 minutes:

– Important: th should be in same time units as tCT

148

obs obs obs

obs

, andh h

h h h h h

h

x xh t v x v h t x h t

t h

40 0.82 2.67 hr 0.3

60 2.67

hh

h

xt h

t

Truck Shipment Example: Periodic• Average annual inventory level

Origin

In-

Transit

Destination

2

q

0

q q

2

q

0

1 1(1) 1

2 2 2 2

q qq q

149

Truck Shipment Example: Periodic• Inter-shipment inventory fraction alternatives:

Constant

Production

Constant

Consumption

2

q

2

q

Batch

Production

Constant

Consumption

0

2

q

Constant

Production

Immediate

Consumption

2

q

0

Batch

Production

Immediate

Consumption

0 0

1 11

2 2

1 10

2 2

1 10

2 2

0 0 0

O D

150

Truck Shipment Example: Periodic• “Reasonable estimate” for the total annual cost for the

cycle inventory:

cycle

max

(1)(25,000)(0.3)6.1111

$45,833.33 / yr

where

1 1at Origin + at Destination 1

2 2

$25,000 unit value of shipment ($/ton)

0.3 estimated carrying rate for manufactured products (1/yr)

= 6.

IC vhq

v

h

q q

111 FTL shipment size (ton)

151

Truck Shipment Example: Periodic15. What is the annual total logistics cost (TLC) for these full-

truckload TL shipments?

cycle

3.2727(2.5511)532 (1)(25,000)(0.3)6.1111

4,441.73 45,833.33

$50,275.06 /

FTL FTL

TL

TLC TC IC

n r d vhq

yr

152

Truck Shipment Example: Periodic16. What is minimum possible annual total logistics cost for TL

shipments, where the shipment size can now be less than a full truckload?

( ) ( ) ( ) ( )TL TL TL

f fTLC q TC q IC q c q vhq rd vhq

q q

*( ) 20(2.5511)5320 1.9024 ton

(1)25000(0.3)

TL TLTL

dTLC q f r dq

dq vh

* *

*( )

20(2.5511)532 (1)25000(0.3)1.8553

1.8553

14,268.12 14,268.12

$28,536.25 / yr

TL TL TL TL

TL

fTLC q r d vhq

q

153

Truck Shipment Example: Periodic• Including the minimum charge and maximum payload

restrictions:

• What is the TLC if this size shipment could be made as an allocated full-truckload?

*max

max ,min ,

TL TL TLTL

f r d MC f r dq q

vh vh

154

* * * *

*max

( )

2.551120 532 (1)25000(0.3)1.9024

6.1111

4,441.73 14,268.12

$18,709.85 / yr vs. $28,536.25 as independent P2P TL

TLAllocFTL TL TL FTL TL TL

TL

f rTLC q q r d vhq f d vhq

qq

Truck Shipment Example: Periodic17. What is the optimal LTL shipment size?

( ) ( ) ( ) ( ) LTL LTL LTL

fTLC q TC q IC q c q vhq

q

• Must be careful in picking starting point for optimization since LTL formula only valid for limited range of values:

2

1 15

27 29 3

37 3354 (dist)

14 150 10,000 (wt)8

, 2,000 2,0007

2 14 2000 650 ft (cube)2

LTL LTL

ds

qr PPI

qs sq d

s

155

* arg min ( ) 0.7622 ton LTL LTLq

q TLC q

150 10,000 650min , 0.075 1.44

2000 2,000 2000

sq q

Truck Shipment Example: Periodic18. Should the product be shipped TL or LTL?

* * *( ) ( ) ( ) 34,349.19 5,716.40 $40,065.59 / yr LTL LTL LTL LTL LTLTLC q TC q IC q

156

Truck Shipment Example: Periodic19. If the value of the product increased to $85,000 per ton,

should the product be shipped TL or LTL?

157

Truck Shipment Example: Periodic• Better to pick from separate optimal TL and LTL because

independent charge has two local minima:

*0 arg min ( ), ( ) TL LTL

qq TLC q TLC q

*0 0arg min ( )

q

fq c q vhq

q

!

158

Truck Shipment Example: Periodic20. 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?

3

*0

* *0 0

32.16 lb/ft

arg min ( ), ( ) 8.5079 ton

( ) $25,523.60 / yr ( )

TL LTLq

TL LTL

s

q TLC q TLC q

TLC q TLC q

159

Truck Shipment Example: Periodic21. What is the optimal shipment size if both shipments will

always be shipped together on the same truck (with same shipment interval)?

1 2 1 2 1 2

agg 1 2

agg 3agg

31 2

1 2

1 2agg 1 2

agg agg

, ,

20 80 100 ton

aggregate weight, in lb 10014.31 lb/ft

20 80aggregate cube, in ft4.44 32.16

20 8085,000 5000 $21,000 / ton

100 100

d d h h

f f f

fs

f f

s s

f fv v v

f f

agg*

agg

100(2.5511)5324.6414 ton

(1)21000(0.3) TL

f rdq

v h

160

Truck Shipment Example: Periodic• Summary of results:

161

Ex 11: FTL vs Interval Constraint• On average, 200 tons of components are shipped 750 miles from your fabrication

plant to your assembly plant each year. The components are produced and consumed at a constant rate throughout the year. Currently, full truckloads of the material are shipped. What would be the impact on total annual logistics costs if TL shipments were made every two weeks? The revenue per loaded truck-mile is $2.00; a truck’s cubic and weight capacities are 3,000 ft3 and 24 tons, respectively; each ton of the material is valued at $5,000 and has a density of 10 lb per ft3; the material loses 30% of its value after 18 months; and in-transit inventory costs can be ignored.

162

obs max

1 1200, 750, 1, 2, 3000, 24, 5000, 10

2 2

0.30.2 0.05 0.06 0.2 0.31, min , 15

1.5 2000

TL cu wt

h cuFTL wt

h

f d r K K v s

x sKh h q q K

t

max min 2wk 2wk min 2wk

min

2 726.09, 7.67, 51,016

365.25TL

ft n q TLC n r d vhq

n

2wk $7,766 per year increase with two-week interval constraintFTLTLC TLC TLC

2-wk TL LTL not considered13.33, 43,250,FTL FTL FTL TL FTL

FTL

fn TLC n r d vhq

q

Ex 12: FTL Location• Where should a DC be located in order to minimize

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

3. 100 tons/yr of A shipped FTL P2P to DC from supplier in Asheville

4. 380 tons/yr of B shipped FTL P2P to DC from Statesville

5. Each carton of A weighs 30 lb, and occupies 10 ft3

6. Each carton of B weighs 120 lb, and occupies 4 ft3

7. Revenue per loaded truck-mile is $2

8. Each truck’s cubic and weight capacity is 2,750 ft3 and 25 tons, respectively

163

-83 -82 -81 -80 -79 -78

34

34.5

35

35.5

36

36.5

Asheville

Statesville


Raleigh

Wilm

ingto

n

50

150

190 220270

295

420

40

Ex 12: FTL Location

($/yr) ($/mi-yr) (mi)

,($/mi-yr) (TL/yr) ($/TL-mi)(ton/yr) ($/ton-mi)

,

($/ton-mi) max max

,

i i

i i FTL i i i

FTL i

TC w d

w f r n r

r fr n

q q

in out in out

in out

(Montetary) Weight Losing: 79 67 39 33

Physically Weight Unchanging (DC): 480 480

w w n n

f f

164

Winston-

Salem

Statesville

Wilmington

DC35%

Asheville

Durham

1330 3348 20

78 > 7348 < 73

:iw

*


146, 732

WW

DC 4

3

5

1

232 max

2 2 2

120 30(2750)30 lb/ft , min 25, 25 ton

4 2000

380380, 15.2, 15.2(2) 30.4

25

s q

f n w

3 agg 3 3

4 agg 4 4

5 agg 5 5

960.20 96, 6.69, 6.69(2) 13.38

14.3478

1440.30 144, 10.04, 10.04(2) 20.07

14.3478

2400.50 240, 16.73, 16.73(2) 33.45

14.3478

f f n w

f f n w

f f n w

31 max

1 1 1

30 3(2750)3 lb/ft , min 25, 4.125 ton

10 2000

100100, 24.24, 24.24(2) 48.48

4.125

s q

f n w

agg 3agg agg max

480 10.4348(2750)$2 / TL-mi, 100 380 480 ton/yr, 10.4348 lb/ft , 25, 14.3478

100 380 2000

3 30

A BA B

A B

fr f f f s q

f f

s s

Ex 12: FTL Location• Include monthly outbound frequency constraint:

– Outbound shipments must occur at least once each month

– Implicit means of including inventory costs in location decision

165

max min

max

min

3 3

4 4

5 5

1 1yr/TL 12 TL/yr

12

max ,

max 6.69,12 12, 12(2) 24

max 10.04,12 12, 12(2) 24

max 16.73,12 16.73, 16.73(2) 33.45

FTL

t nt

TC n n rd

n w

n w

n w

2430 3348 24

78 < 8048 < 80

:iw

*


102 > 80

160, 802

WW

in out in out

in out

(Montetary) Weight : 79 81 39 41

Physically Weight Unchanging (DC)

Ga

: 480 48

ining

0

w w n n

f f

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

166

1 1

1 1

max max

max1 1

( ) ( ) ( ) ( ) ( ) ( )( )

( ) 1( ) ( )

( )

( ) ( ) ( ) ( ) con

m mi

TL i i i i i

ii i

m mi i i

i i i

i ii i

m mi

FTL i i i FTL

i i

fTLC w d vhq rd vhq

q

f f rdrd vh f rd vh

vhf rd vh

vh

fTLC rd vhq w d vhq TC

q

x x x x x xx

xx x

x

x x x x stant

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

– (Perfect) Cross-dock: use single shipment interval for all inbound and outbound shipments no inventory at DC(usually only cross-dock a selected subset of shipments)

Su

pp

liers

3

4

1

2

Cu

sto

me

rs

A

A

B

B

3

4

DC

1

2

Cu

sto

me

rs

Su

pp

liers

AA

BB AB

AB

167

Uncoordinated Inventory

• Average pipeline inventory level at DC:

1

2

3

0

4

1

2

0

1.552

q

1.112

q

Su

pp

lier 2

Customer 4

Su

pp

lier 1

Customer 3

1, inbound

20 , outbound

O D

O

D

168

3

4

DC

1

2

Cu

sto

me

rs

Su

pp

liers

AA

BB AB

AB

TLC with Transshipment

• Uncoordinated:

• Cross-docking:

*

* *

of supplier/customer

arg min ( )

i

i iq

i i

TLC TLC i

q TLC q

TLC TLC q

00

0

*

* *

, shipment interval

( )( ) cf. ( ) ( )

( ) independent transport charge as function of

0, inbound

0 , outbound

arg min ( )

i i

O

D

it

i

qt

f

c t fTLC t vhf t TLC q c q vhq

t q

c t t

t TLC t

TLC TLC t

169

Ex 13: Direct vs Transhipment• 3 different products supplied to 4 customers, compare:

1. Direct shipments2. Uncoordinated at

existing DC in Memphis3. Cross-docking at

Memphis4. Uncoordinated at

optimal DC location5. Cross-docking at

optimal location

170

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

171

Ex 14: 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 = αvhqmax (number of suppliers 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)

172

Topics

1. Introduction




4. Network models

5. Routing


6. Warehousing


173

Graph Representations

• Complete bipartite directed (or digraph):

– Suppliers to multiple DCs, single mode of transport

10

6

0

13

9

16

1

2

4

5

3

C: 1 2

-:-------

1: 6 10

2: 0 13

3: 9 16

W = 0 0 0 6 10

0 0 0 NaN 13

0 0 0 9 16

0 0 0 0 0

0 0 0 0 0

Interlevel matrix

Weighted adjacency matrix

174


• Bipartite:

– One- or two-way connections between nodes in two groups

Arc list matrix

10

6

0

13

16

1

2

4

3

5

3

W = 0 0 0 6 10

0 0 0 NaN 13

0 0 0 0 16

6 0 0 0 0

0 0 3 0 0

IJC =

4 1 6

5 3 3

1 4 6

2 4 0

1 5 10

2 5 13

3 5 16

i jc

175


• Multigraph:

– Multiple connections, multiple modes of transport

10

6

0

13

16

2

4

5

33

1

18

IJC = 1 -4 6

1 -4 18

1 5 10

2 4 0

2 5 13

3 5 16

3 5 3

no_W =

0 0 0 24 10

0 0 0 NaN 13

0 0 0 0 19

24 0 0 0 0

0 0 0 0 0

Can’t represent using adjacency matrix176


• Complete multipartite directed:

– Typical supply chain (no drop shipments)

Level 1(suppliers)

1

2

4

5

3

12

4

10

15

11

14

6

7

8

10

6

0

13

9

16

Level 2(DCs)

Level 3(customers)

Drop Shipment

177

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

Su

pp

ly

De

ma

nd

55

50

45

20

30

10

6

8

9

9

12

137

149

16

5

40

30

1

2

5

4

6

3

7

Trans 4 5 6 7 Supply

1 8 6 10 9 55

2 9 12 13 7 50

3 14 9 16 5 40

Demand 45 20 30 30

178

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 4 5 6 7 Supply

1 8 6 10 9 55

2 9 12 13 7 50

3 14 9 16 5 40

Demand 45 20 30 30

179

0

-30 = 10

-20 = 35

010

-35 = 0

0

-10 = 40

0

-30 = 10

5(30) 6(20) 8(35) 9(10) 13(30) 1, vs 970 optima0 l30TC

Min Cost Network Flow (MCNF) Problem

• Most general network problem, can solve using any type of graph representation

MCNF: lhs C C C C C C rhs

----:--------------------------------

Min: 2 3 4 5 1 3

1: 6 1 1 0 0 0 0 6

2: 2 -1 0 1 1 0 0 2

3: 0 0 -1 0 0 1 0 0

4: 0 0 0 -1 0 0 1 0

lb: 0 0 0 0 0 0

ub: Inf Inf Inf Inf Inf Inf

Row for node 5 is redundant

Arc cost: 2 3 4 5 1 3

Net node supply : 6 2 0 0 8

1 1 0 0 0 0

1 0 1 1 0 0

Incidence Matrix : 0 1 0 0 1 0

0 0 1 0 0 1

0 0 0 1 1 1

c

s

A

4

5

1

3

8

2

6

2

3

0

0

2

3

1

4

5

TransshipmentSupply

De

ma

nd

Supply

Transshipment

Arc Cost

MCNF: max '

s.t.

0

c x

Ax s

x

net supply of node

0, supply node

0, demand node

0, transshipment node

is i

180

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

net supply of node

0, supply node

0, demand node

0, transshipment node

is i

4,3

5,9

1,3

3

8

2

6,3

6,1

3,∞

,1

0,2,4

0,0,∞,3

2

3

1

4

5

i jc,u,l

c,∞,0

i

s,nc,nu,nl

s,0,∞,0

181

Expanded-Node Formulation of MCNF

• Node cost/constraints converted to arc cost/constraints– Dummy node (8) added so that supply = demand

4,3

5,9

1,3

3

8

2

6,3

6,1

3,∞

,1

0,2,4

0,0,∞,3

2

3

1

4

5

5,36,9

3+3=6

8

0

6

60

0

2

3

1

4

5

6

7

08

4

2+3=5 0,4

5,∞

,1

0

0

0,∞,3 1,

3

i

s,nc,nu,nl

c,u,l

i

0

0,nu,nliʹ

s

c+nc,u,l

182

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 usedfor shortest pathproblem (Dijkstra)

Su

pp

ly

65

80

Transshipment

De

ma

nd

45

20

30

10

6

8

9

9

12

137

149

16

5

30

7

6

8

9

4

6

5

3

8

2

3

4

5

1

2

183

Dijkstra Shortest Path Procedure

2

4 6

38

2

3

1

4

5

5

10

2 61s t

∞

∞ ∞

∞

∞ ∞

0,1

4,1

2,1 12,3

10,3

3,3 8,2

14,4

10,4

13,5

Path: 1 3 2 4 5 6 : 13 184

Dijkstra Shortest Path Procedure

4

3

2

2 Simplex (LP)

Ellipsoid (LP)

Hungarian (transportation)

Dijkstra (linear min)

log Dijkstra (Fibonocci heap)

no. arcs

nO

O n

O n

O n

O m n

m

Orderimportant

Index to indexvector nS

185

Other Shortest Path Procedures• Dijkstra requires that all arcs have nonnegative lengths

– It is a “label setting” algorithm since step to final solution made as each node labeled

– Can find longest path (used, e.g., in CPM) by negating all arc lengths

• Networks with only some negative arcs require slower “label correcting” procedures that repeatedly check for optimality at all nodes or detect a negative cycle– Requires O(n3) via Floyd-Warshall algorithm (cf., O(n2) Dijkstra)

– Negative arcs used in project scheduling to represent maximum lags between activities

• A* algorithm adds to Dijkstra an heuristic LB 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 LB estimate of remaining distance

186

A* Path Planning Example 1

187

* (Raleigh, Dallas) (Raleigh, ) ( , Dallas), for each node dijk GCAd d i d i i

A* looks at a fraction of the nodes (in ellipse) seen by Dijkstra (in green)


• 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

188


189

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)

190

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

191

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= 37183 = 37 NC FIPS + 183 Wake FIPS

192

20 15

10

15

15

1010

15

Road Network Modifications

1. Thin

– Remove all degree-2 nodes from network

– Add cost of both arcs incident to each degree-2 node

– If results in multiple arcsbetween pair of nodes, keepminimum cost

193

Thinned I-40 Around Raleigh

70


2. Subgraph

– Extract portion of graph with only those nodes and/or arcs that satisfy some condition

194

Subgraph of Arcs < 35

Subgraph of Nodes in Rectangle


3. Add connector

– Given new nodes, add arcs that connect the new nodes to the existing nodes in a graph and to each other

195

– 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 2nd

and 3rd closest < threshold (0.1)

– Distance shorter using other connector and graph

1

2

3

4

5

6

7

1

2

3

4

5

6

7

1

2

3

4

5

6

7

1

2

3

4

5

6

7New to existing node

New to new node

New node

Production and Inventory: One Product

196

1

1

2

$0.3 0.3

$-yr

0.3 $0.025

12 $-month

0.3200 5

12

0.3200 800

12

25

mi pm j

j

i

i

h

h

T

hc c

T

c

c

Period (month)

1 2 3

28

00

pc

01,1 1

0

y y

1200

pc

1

1,2

5ic

y

1 20D

02,1 2

0

y y

2200

pc

2

2,2

25ic

y

2 10D

1

1,3

5ic

y

3200

pc

2

2,3

25ic

y

3 15D

41,4 1

0

y y

42,4 2

0

y y

1,1

1,1

50

xK

1,2

1,2

0

xK

1,3

1,3

50

xK

2,1

2,1

60

xK

28

00

pc

2,2

2,2

0

xK

28

00

pc

2,3

2,3

0

xK

Demand

Sta

ge

1

2

Init

ial In

ve

nto

ry

Supply

Fin

al In

ve

nto

ry

Production and Inventory: One Product

197

Flow balance

Initial/Final inventory

Capacity

Use var. LB & UB instead of constraints

Ex 15: Coupled Networks via Truck Capacity

• Facility that extracts two different raw materials for pharmaceuticals1. 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 0.15

4. 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

198


• Separate networks for each raw material are coupled via sharing the same trucks (added as constraint to model)

199

max2,2,1 2,2,2 2 2,2

2,210

x x Q z

z

Couples each network

Numberof trucks

2,1

0p

c

1,1

120

pc

1,2,1 20y

1,1D 2,1D

1,1,

126

x

3,1

0p

c

0

0

0

1,2,2 20y

1,2D 2,2D

0

0

3,1,2 5y


• Math programming model:

200

Production and Inventory: Multiple Products

201

pc

Flow balance

x y

ic

Capacity

x

Setup

z

–K

k 0

1

k 0

sc 0

Flow balance

x y

Capacity

x

Setup

z

–K

k 0

1

k 0

pc ic sc 0

Linking 1k 2k 1

0

0

Pro

du

ct

1

Pro

du

ct

2


202

0,1 , production indicator

1 2 3 4 5 6 7

1 0 1 1 0 1 1 1

2 0 0 0 1 0 0 0

0,1 , setup indicator

1 2 3 4 5 6 7

1 0 0 0 0 0

2 0 0 0 0

1

1 0

1

0

mtg

mtg

mtg

mtg

k

k

z

z

Don’t want(not feasible)

Want(feasible)

Feas

ible

, bu

t n

ot

min

co

st

1 0

0 0 0 0

0 0 1 1

0 1 0 1

0 1 1 0

1 0 0 1

1 0 1 2

1 1 0 0

1 1 1 1

t t tz k k


203

Flowbalance

Capacity

Setup

Linking

MILP

dummy


204

Example of Logistics Software Stack

205

• 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

MIP Solver

(Gurobi,Cplex,etc.)

Standard Library

(in compiled C,Java)

User Library

(in script language)

MIP Solver

(Gurobi, etc.)

Standard Library

(C,Java)Data

(csv,Excel,etc.)

Report

(GUI,web,etc.)

Commercial

Software

(Lamasoft,etc.)

Scripting

(Python,Matlab,etc.)

Topics

1. Introduction




4. Network models

5. Routing


6. Warehousing


206

Routing Alternatives

207

Local P/D

Local P/D

Linehaul

Terminal

P DPickup Delivery

P D D D

P P P D

P P P DD D

(a) Point-to-point (P2P)

(b) Peddling (one-to-many)

(c) Collecting (many-to-one)

(d) Many-to-many

P D P DP D

(e) Interleaved

P D P D P D

empty

(f) Multiple routes

TSP an

d V

RP

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)

208

1

2

3

4

5

6

1 2 3 4 5 6 1

Node sequence = permutation + start node

Depot 6 1 ! 120 possible solutionsn n

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

• Symmetric: solved to optimal using BIP

• Euclidean: arcs costs = distance between nodes

209

1

1 ! 13 billion solutions2

n n

1 !solutions

2ij ji

nc c

TSP Construction

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

210

1 2 3 4 5 6

1 2 4 5 6

1 2 4 6

2 4 6

2 4

2

4

2

3

5

1

6

Spacefilling Curve

211

1.0 0.250 0.254 0.265 0.298 0.309 0.438 0.441 0.452 0.485 0.496 0.500

0.9 0.246 0.257 0.271 0.292 0.305 0.434 0.445 0.458 0.479 0.493 0.504

0.8 0.235 0.229 0.279 0.283 0.333 0.423 0.417 0.467 0.471 0.521 0.515

0.7 0.202 0.208 0.158 0.154 0.354 0.390 0.396 0.596 0.592 0.542 0.548

0.6 0.191 0.180 0.167 0.146 0.132 0.379 0.618 0.604 0.583 0.570 0.559

0.5 0.188 0.184 0.173 0.140 0.129 0.375 0.621 0.610 0.577 0.566 0.563

0.4 0.059 0.070 0.083 0.104 0.118 0.871 0.632 0.646 0.667 0.680 0.691

0.3 0.048 0.042 0.092 0.096 0.896 0.860 0.854 0.654 0.658 0.708 0.702

0.2 0.015 0.021 0.971 0.967 0.917 0.827 0.833 0.783 0.779 0.729 0.735

0.1 0.004 0.993 0.979 0.958 0.945 0.816 0.805 0.792 0.771 0.757 0.746

0.0 0.000 0.996 0.985 0.952 0.941 0.813 0.809 0.798 0.765 0.754 0.750

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

2: 0.0213: 0.1541: 0.4714: 0.783

Sequence determined by sorting position along 1-D line covering 2-D space

1

2

3

4

Two-Opt Improvement

1 2 3 4 5 6 1

212

1

2

3

4

5

6

a

b c

d

ef

a b c d e f

a-c 1 3 2 4 5 6 1

a-d 1 4 3 2 5 6 1

a-e 1 5 4 3 2 6 1

b-d

b-e

b-f

c-e

c-f

d-f

arcs to 2first arc

1 2

3 2 2

3Sequences considered at end to verify local optimum: nodes (1) (1) 9 for 6

2

b na

n n n

j i j i

n nn n

Ex 16: Two-Opt Improvement

• Order in which twoopt considers each sequence:

213

1 2

3 2 2

Sequences considered at end to verify

local : nodes

3(1) (1) 9 for

o ti u

6

p m

2

m

n n n

j i j i

n

n nn

Local optimal sequence

D: 1 2 3 4 5 6

-:---------------------

1: 0 8 6 9 1 5

2: 3 0 1 5 4 2

3: 9 2 0 3 1 1

4: 8 2 1 0 10 6

5: 6 7 10 1 0 10

6: 6 2 5 2 1 0

Note: Not symmetric

TSP Comparison

TSP Procedure Total Cost

1 Spacefilling curve 482.7110

2 1 + 2-opt 456

3 Convex hull insert + 2-opt 452

4 Nearest neighbor + 2-opt 439.6

5 Random construction + 2-opt 450, 456

6 Eil51 in TSPLIB 426* optimal

214

Multi-Stop Routing

• Each shipment might have a different origin and/or destination node/location sequence not adequate

215

300

275

60

4

2

65

250

30

30

60

Shipment 2

Shipment 3

3

1

Shipment 1

1

1 2

1 2

, , = 1,2,3 -element shipment sequence

, , = 3,1,2,2,1,3 2 -element route sequence

, , = 5,1,3,4,2,6 2 -element location (node) sequence

n

n

n

L y y n

R z z n

X x x n

1

2 1

,

1

cost of route

cost between locations and

( ) 60 30 250 30 60 430, total i i

ij

n

x x

i

c i j

c R Rc

5-Shipment Example

216

-81 -80.5 -80 -79.5 -79 -78.5 -78 -77.535

35.5

36

36.5

L3

L2

L5

L4

U1,2,4

Shmt 3

Shmt 5

Shmt 1

Shmt 2

Shmt 4

25

6

4

3

7

1

U3

U5

L1

Route sequence: = 3,2,5,5,1,4,3,1,2,4

Location sequence: = 4,3,7,1,1,6,5,2,2,2

R

X

Route Sequencing Procedures

• Online procedure: add a shipment to an existing route as it becomes available

– Insert and Improve: for each shipment, insert where it has the least increase in cost for route and then improve (mincostinsert twoopt)

• Offline procedure: consider all shipments to decide order in which each added to route

– Savings and Improve: using all shipments,determine insert ordering based on “savings,” then improve final route (savings twoopt)

217

Min Cost Insert

218

*3

2

4

5

1 1

1 2 2

2 2 2

3 2 2

4 2 2

5 2 2

c

c

c

c

×

1 2 2 1

1 3 3

2 3 3

3 3 3

4 3 3

5 3 3

6 3 3

7 3 3

Insert and Improve Online Procedure

219

• To route each shipment added to load:– Minimum Cost

Insertion

– Two-opt improvement

• Different shipment sequences L can result in different routes– Order shipment

joins load important

219

First improvement(cf. steepest descent)

Pairwise Savings

220

1,2

pairwise savings between shipments and

0

300 250 310

240

ij

i j ij

s i j

c c c

s

300

4

2

25030

30

2

3

1

1

1+2

Clark-Wright (Offline) Savings Procedure

221

• First (1964), and still best, offline routing procedure if only have vehicle capacity constraints (vrpsavings)

• Pairs of shipments ordered in terms of their decreasing (positive) pairwise savings

• Given savings pair i-j, without exceeding capacity constraint, either:

1. Create new route if i and j not in any existing route

2. Add i to route only if j at beginning or end of route

3. Combine routes only if i and j are endpoints of each route

Ex 17: Clark-Wright Savings Procedure

222

2 3 40 48 87 1

2 4 40 38 46 32

2 5 8

2 6 13

3 4 19

3 5 40

3 6 49

4 5 1

4 6 52

5 6 12

iji j s

• Node 1 is depot, nodes 2-6 customers

• Customer demands 8, 3, 4, 7, 6, resp.

• Vehicle capacity is 15

• Symmetric costs

1

2

3

4

5

6

8

5

4

7

6

Multi-Stop (Offline) Savings Procedure

223

• Pairs of shipments ordered by their decreasing pairwise savings to create i and j (pairwisesavings)

• Creates set of multi-shipment routes (savings)

– Shipments with no pairwise savings are not included (use sh2rte to add)

• Clark-Wright only adds to beginning or end of a route– Multi-stop savings considers adding

anywhere in route via min cost insert

– More computation required, but can include sequence-dependent constraints like time windows(capacity not sequence dependent)

1,..., mR R R

1. Form new route

2. Add shipment to route

3. Combine two routes

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 (time or distance)

• VRP solution can be one time or periodic– One time (operational) VRP minimizes TC

– Periodic (tactical) VRP minimizes TLC (sometimes called a “milk run”)

224

Ex 18: VRP with Time Windows[0,24] hr; Loading/unloading time = 0; Capacity = ∞; LB = 5 hr

225

Depot

1 hr

1 hr2 hr

1 hr

2

4

1 3

[18,24]

[8,11]

[12,14]

[15,18]

[6,18]

(return window)

(depart window)

(earliest start) a = 6

a = 7 – 8 (arrive at 7 wait to 8)

a = 10 – 12

a = 13 – 15

a = 16 – 18

Earliest Finish – Latest Start = 18 – 10 = 8 hr = 5 travel + 3 delay

Earliest Finish: b = 18

b = 17

b = 16 – 14

b = 12 – 11

Latest Start: b = 10

(move delays to end) c = 10

c = 11

c = 13

c = 14 – 15

c = 16 – 18

Periodic Multi-Stop Routing

226

1 2 3 3 2 1

1 1 1 1 1

2 2 2

3

1 2 1 3 2 3

1 1 2 2 3

2 3

Single load mix

First load mix Second load mix

• Periodic consolidated shipments that have the same frequency/interval

• 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 (to keep common frequency)

Load Mix Example

227

Single Load-Mix Instance Two Load-Mix Instance

TLC Calculation for Multi-Stop Route

• How minTLC determines TLC for a route:

228

max ,(no truck capacity constraints, only min charge)

(allocate based on demand)

(aggregate density of shipments in load-mix )

min ,

min 1,

j

j j

j

agg TL agg

agg

agg

ii agg

agg

iL i j

ii L i L

Lwt

L

f r d MCq

v h

fq q

f

fs f L

s

sK

k

*

* *

*

2000(min ratio of max payload to size of shipments in load-mix)

(apply truck capacity deduction factor)

( = distance of entire rout

j

j

cu

i

i L

i i

aggTL agg agg i agg

i

K

q

q kq

fTLC r d v h q d

q

e)

Ex 19: Periodic Two Load-Mix Instance

229

1

1

1

1

1

max , 350max 2(848),13572.9875, 20.85, 41.71,10.43

1 (371.4286) 0.3

1,2 : 12.8571, 1

min ,min 252000min min,

1,

j

agg TL agg iagg i agg

agg agg

iL i

ii L i L

L cuwt

i

i L

f r d MC fq q q

v h f

fL s f k

s

s KK

k kq

22

* *

12.8571(2750),

min 1, 0.2826 0.28262000

62.5607

12.2093(2750)min 25,

2,3 : 12.2093, min min 0.2826, 0.3220 0.282620000.2826,

52.1339

5.8929,11.7857, 2.9464,

L

aggi i

L s k

fq kq TLC

*

*$31,079TL agg agg i

i

r d v h qq

Ex 20: 30 Periodic NC Shipments

230

Independent Shipments Consolidated Shipments

Ex 21: Minimize Number of Trucks• Given begin-end times for 10 routes, determine minimum

number of trucks needed– Trucks begin and end at the depot

– Optimal solution via directed-arc minimum spanning tree

– Greedy procedure usually works fine for small instances

231

Cost Allocation for Routing

• Allocation Problem: If shipments from different firms are sharing the same vehicle, how much should each shipment contribute to the total cost paid to carrier?

– What is a “fair” allocation?

– Allocated cost should not exceed cost as an independent shipment (its reservation price)

– Examples:

232

600 ft 800 ft 1,200 ft

300

4

2

250

Shmt 2 (TL)

3

1

Shmt 1 (TL)

Shmt 1 & 2 (TL)

30021

Shmt 1 (TL)

Shmt 2 (TL)

Ex 22: TL + TL Same O/D• Shipment 1

– sets r = 1, d = 300, TL, max c = 300

• Shipment 2– same O/D, TL, max c = 300

233

30021

Shmt 1 (TL)


Tran

spo

rt C

har

ge (

c, $

)

MC

TL

Full-Truckload

max min , ,LTL TLc c c MC

21300

Shmt 2 (TL)

1 300c c

1 2 1 2300 , 1502

cc c c c c

Ex 23: TL + LTL Same O/D• Shipment 1

– sets r = 1, d = 300, TL, max c = 300

• Shipment 2– same O/D, LTL, max c = 100

234

30021

Shmt 1 (TL)


Tran

spo

rt C

har

ge (

c, $

)

MC

TL

Full-Truckload


2

1300

1 2300c c c

Shmt 2 (LTL)

100

1 2

1 2 300 0

2 1 200 100

250 50

1 2250, 50c c

(Shapley value allocation)

Ex 24: TL + TL Different O/D• Shipment 1

– sets r = 1, d = 300, TL, max c = 300

• Shipment 2– different O/D, TL, max c = 250

235Shipment Size (tons)

Tran

spo

rt C

har

ge (

c, $

)

MC

TL

Full-Truckload


21310

1 2310c c c

1 2

1 2 300 10

2 1 60 250

180 130

1 2180, 130c c

300

4

2

250

Shmt 2 (TL)

3

1

Shmt 1 (TL)

Shmt 1 & 2 (TL)

Shapley Value Approximation• Shapley value

– Average additional cost each shipment imposes by joining route

– Exact value requires n!

– Use n2 pairwise savings approximation:

236

300

4

2

250

2

3

1

1

6275

3

5

{ }

0 1 \| |

! 1 !

!i M i M

m n M N iM m

m n m

n

sav savsavsav sav

1 1 1

sav 0 0 sav 0,

1

1 1

1 2 1

,

n n nij jiL

i jk

j j k

n

ij i j L i Li j

i

c ccc c

n n n n

c c c c c c c

sav 0

1

300 250 275 430

825 430 $395 savings for load

n

L i L

i

c c c

Ex 25: Intercity Trucking• 4 out 30 available shipments form consolidated load

– Savings of 824.81 – 452.47 = 372.34 from consolidation

– Pairwise approximation differs from exact Shapley value

237

-83 -82 -81 -80 -79 -78 -77

34

34.5

35

35.5

36

36.5

37

North Carolina

Raleigh

Charlotte

-81 -80.5 -80 -79.5 -79 -78.5

34.8

35

35.2

35.4

35.6

35.8

36

Shmt : c0 c_equal (%) c_eq_sav (%) c_Shap_exact (%) c_Shap_approx

-----:--------------------------------------------------------------------

1: 130 113 13 37 72 62 52 52

2: 119 113 5 25 79 53 55 50

3: 254 113 56 161 37 117 54 123

4: 322 113 65 229 29 220 32 227

Total: 825 452 452 452 452

Topics

1. Introduction




4. Network models

5. Routing


6. Warehousing


238

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.

239

Types of Warehouses

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.

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.

Design Trade-Off

• Warehouse design involves the trade-off between building and handling costs:

243

min Building Costs vs. min Handling Costs

max Cube Utilization vs. max Material Accessibility

Shape Trade-Off

244

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

W = D

I/O

W

DW = 2 D

I/O

W

D

Storage Trade-Off

245

vs.

Maximizes cube utilization, but minimizes material accessibility

Making at least one unit of each item accessible decreases cube utilization

A

A

B

B

B

C

C

D

E

A

A

B

B

B C

C D E

Honeycomb

loss

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

246

Dedicated Storage• 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.

247

I/O

A

BC C

Randomized Storage• Each SKU can be stored in

any available slot.

• Total capacity of all the slots must equal the storage space corresponding to the maximum aggregateinventory level of all of the SKUs.

• Maximizes handling cost.

• Minimizes building cost.

248

I/O

ABC

Class-based Storage

A

BC

I/O

249

• 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.

Individual vs Aggregate SKUs

250

Dedicated Random Class-Based

Time A B C ABC AB AC BC

1 4 1 0 5 5 4 1

2 1 2 3 6 3 4 5

3 4 3 1 8 7 5 4

4 2 4 0 6 6 2 4

5 0 5 3 8 5 3 8

6 2 5 0 7 7 2 5

7 0 5 3 8 5 3 8

8 3 4 1 8 7 4 5

9 0 3 0 3 3 0 3

10 4 2 3 9 6 7 5

Mi 4 5 3 9 7 7 8

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.

251

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

252

Heig

ht of 5 L

evels

(Z)

Wall

Depth of 4 Rows (Y)Cross Aisle

Vertical Honeycomb Loss

of 3 Loads

Width of 5 Lanes (X)

Down Aisle

Horizontal Honeycomb Loss

of 2 Stacks of 5 Loads Each

Estimating Cube Utilization

• The (3-D) cube utilization for dedicated and randomized storage can estimated as follows:

253

1

1

item space item spaceCube utilization

honeycomb down aisletotal space item spaceloss space

, dedicated( )(3-D)

, randomized( )

, dedicated( )

(2-D)

N

ii

N i

i

x y z M

TS DCUx y z M

TS D

Mx y

H

TA DCU

x

, randomized( )

My

H

TA D

Unit Load

• Unit load: single unit of an item, or multiple units restricted to maintain their integrity

• Linear dimensions of a unit load:

• Pallet height (5 in.) + load height gives z:

254

Depth (stringer length) Width (deckboard length)

(Stringer length) Depth Width (Deckboard length)

x

DeckboardsStringer

Notch

y x

y x z

Cube Utilization for Dedicated Storage

Storage Area at Different Lane Depths

Item

Space

Lanes

Total

Space

Cube

Util.

A A A A C C CB B B B BD = 1

A/2 = 1

12 12 24 50%

A A C CB B B

A/2 = 1

A A CB B

D = 2

12 7 21 57%

A A CB B

A/2 = 1

A CB BD = 3

A CB

12 5 20 60%

255

Total Space/Area

• The total space required, as a function of lane depth D:

256

Eff. lane depth

Total space (3-D): ( ) ( )2 2

A ATS D X Y Z xL D yD zH

eff( )Total area (2-D): ( ) ( )

2

TS D ATA D X Y xL D yD

Z

y

A

A

x

A A B

B

B

B

B

C

C

C

X = xL

Yef

f = Y

+A

/2

AY

= y

D

Down Aisle Space

Storage Area on Opposite

Side of the Aisle

Ho

ne

yc

om

b

Lo

ss H

CL

Number of Lanes

• Given D, estimated total number of lanes in region:

• Estimated HCL:

257

1

, dedicated

Number of lanes: ( ) 1 1

, randomized ( 1)2 2

Ni

i

M

DH

L D D HM NH N

N

DH

1

1

11 1 1 1 12 1 1 2 1

2 2

D

i

D D DD D i

D D D D

Unit Honeycomb Loss:

1

0

D

A

A

A

A

A

A

Probability:

1

2

D

D

1

1

D

D

Expected Loss:

3D

doesn’t occur

because slots are

used by another

SKU

Optimal Lane Depth

• Solving for D in results in:

258

* 2 1Optimal lane depth for randomized storage (in rows):

2 2

A M ND

NyH

( ) 0dTS D dD

Max Aggregate Inventory Level

• Usually can determine max inventory level for each SKU:

– Mi = 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

• Can add include safety stock for each item, SSi

– For example, if the order size of three SKUs is 50 units and 5 units of each item are held as safety stock

259

1

1

2 2

Ni

i

MM

1

1 50 13 5 90

2 2 2 2

Ni i

i

i

M SSM SS

Steps to Determine Area Requirements

1. For randomized storage, assumed to knowN, H, x, y, z, A, and all Mi

– 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*)

260

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)

261

9 - 11 ft 7 - 8 ft 8 - 10 ft

Stand-Up CB NA Straddle NA Reach

Ex 26: Area Requirements

Units of items A, B, and C are all received and stored as 42 36 36 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.

262

363' 31 10 '

12

3.5 ' 62 3

3' 42 4

3

A

B

C

x M A

y M D

z M H

N

Ex 26: Area Requirements1. If a dedicated policy is used to store the items, what is the 2-

D cube utilization of this storage region?

263

1

2

1

31 62 42( ) (3) 3 6 4 13 lanes

3(4) 3(4) 3(4)

10(3) ( ) 3(13) 3.5(3) 605 ft

2 2

31 623 3.5

4 4item space(3)

(3) (3)

Ni

i

N i

i

ML D L

DH

ATA xL D yD

Mx y

HCU

TA TA

42

461%

605

Ex 26: Area Requirements2. 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?

264

1

1 31 62 42 168

2 2 2 2

Ni

i

MM

Ex 26: Area Requirements3. If a randomized policy is used to store the items, what is

total 2-D area needed for the storage region?

265

2

3

1 1

(3) 2 2

3 1 4 168 3(4)

2 2 8 lanes

3(4)

10(3) ( ) 3(8) 3.5(3) 372 ft

2 2

D

D HM NH N

L

DH

N

ATA xL D yD

Ex 26: Area Requirements4. What is the optimal lane depth for randomized storage?

5. What is the change in total area associated with using the optimal lane depth as opposed to storing the items three deep?

266

* 2 10 2(68) 31 14

2 2 2(3)3.5(4) 2

A M ND

NyH

2

2

4 1 4 168 3(4)

2 24 (4) 6 lanes

3(4)

10(4) 3(6) 3.5(4) 342 ft

2

3 (3) 372 ft

ND L

TA

D TA

Ex 27: Trailer Loading

How many identical 48 42 36 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.

267

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

In

terio

r H

eig

ht:

(8'6

" -

9'2

" =

10

2"

- 1

10

")

Width:

8'6" = 102"

(8'2" = 98")

Ma

x H

eig

ht:

13

'6"

= 1

62

"

Max of 83 units per TL

X 98/12 = 8.166667 8.166667 ft

Y 53 53 ft

Z 110/12 = 9.166667 9.166667 ft

x [48,42]/12 = 4 3.5 ft

y [42,48]/12 = 3.5 4 ft

z 30/12 = 2.5 2.5 ft

L floor(X/x) = 2 2

D floor(Y/y) = 15 13

H min(6,floor(Z/z)) = 3 3

LDH L*D*H = 90 78 units

wt 600 600 lb

unit/TL min(LDH, floor(50000/wt)) = 83 78

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

268

Single-Command S/R Cycle

store

empty

empty

retrieve

I/Oslot

269

• 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

/2SC SC

SC L U L U

d dt t t t

v v

Expected time for each SC S/R cycle:

Industrial Trucks: Walk vs. RideWalk (2 mph = 176 fpm) Ride (7 mph = 616 fpm)

Pallet Jack Pallet Truck

Walkie Stacker Sit-down Counterbalanced Lift Truck

270

Dual-Command S/R Cycle

store

empty

retrieve

I/Oslot1 slot2

271

• 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”

/2 2 4DC DC

DC L U L U

d dt t t t

v v

Expected time for each SC S/R cycle:

Multi-Command S/R Cycle

empty

retrieve

I/O

272

• 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

1-D Expected Distance

1

1 1

11

12 2

( 1)

2 2

2 2

2

L L

way

i i

wayway

X X X XTD i i

L L L L

X L L XL

L L

XL X X XL

TD XED

L

273

• 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 = 12

I/O 3 6 9 X = 12

X X

L

0

2Lx =

1-D Storage Region

12( )SC wayd ED X

Off-set I/O Port

I/O 3 6 9 X = 120

offset

274

• 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

offset2( )SCd d X

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:

275

rectSCd X Y

1

1-way

1 1

2

1-way1-way

2 2

2 3 16

2 2, as

( 1) 3 3 3

2

2 2 41 12 2 , if

3 3 33 3

L L i

i j

triSC

X X X XTD i j

L L L L

XL L

TD XED X X L

L L L

d X X X YX Y X Y

I/O XX

xL

Y

Yy

D

I/O-to-Side Configurations

Rectangular Triangular

276

21

2

2 2

42 1.886

3SC

TA X

X TA TA

d TA TA

2

2SC

TA X

X TA

d TA

TA

I/O

0 X

X

TA

I/O

0 X

X

I/O-at-Middle Configurations

Rectangular Triangular

277

21

2 2

41.333

3SC

TAX

X TA

d TA TA

2

2

2 2

2 1.414SC

TAX

TA TAX

d TA TA

TA/2

I/O

0 X

X

TA/2TA/2TA/2

I/O

0 X

X

Ex 28: 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

278

transport load

empty

Receiving

Dock

AS/RS

500 ft

Ex 28: 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 single-command S/R cycle?

279

/

2(500) 1000 ft/mov

1000 ft/mov 302 2 min/mov

616 ft/min 60

2.622.62 min/mov hr/mov

60

SC

SCSC L U

d

dt t

v

(616 fpm because operator rides on a pallet truck)

Ex 28: Handling Requirements2. 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?

280

80,000 mov/yr40 mov/hr

50(40) hr/yr

1

2.6240 1 1.75 1

60

2 trucks

avg

avg SC

r

m r t

Ex 28: Handling Requirements3. How many trucks are needed to handle a peak expected

demand of 80 moves per hour?

281

80 mov/hr

1

2.6280 1 3.50 1

60

4 trucks

peak

peak SC

r

m r t

Ex 28: Handling Requirements4. 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?

282

Receiving

Dock 3 6 X = 900

offset = 30 ft

8784

. . .

offset

/

2( ) 2(30) 90 150 ft

150 ft/mov 302 2 min/mov

616 ft/min 60

1.241.24 min/mov hr/mov

60

SC

SCSC L U

d d X

dt t

v

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.

283

Ex 29: Estimating Handing Cost

284

I/O

TA = 20,000 /

peak

year

Expected Distance: 2 2 20,000 200 ft

Expected Time: 2

200 ft2(0.5 min) 2 min per move

200 fpm

Peak Demand: 75 moves per hour

Annual Demand: 100,000 moves per year

Number of T

SC

SCSC L U

d TA

dt t

v

r

r

peak

hand truck year labor

rucks: 1 3.5 3 trucks60

Handling Cost:60

23($2,500 / tr-yr) 100,000 ($10 / hr)

60

$7,500 $33,333 $40,833 per year

SC

SC

tm r

tTC mK r C

2

* *

Add 20% Cross aisle:

1.2

20,000 ft

Total Storage Area:

( )

TA TA

D L D TA

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

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]

285

[1] [2] [ ]

[1] [1] [2] [2] [ ] [ ]

N

N N

f f f

M s M s M s

Ex 30: 1-D Slotting

286

Flow

Density

1-D Slot Assignments

Expected

Distance Flow

Total

Distance

217.00

3

C C CI/O

30

2(0) + 3 = 3 21 = 63

246.00

4

A A A AI/O

-3 0 4

2(3) + 4 = 10 24 = 240

71.40

5

B BI/O

B B B

-7 50

2(7) + 5 = 19 7 = 133

C C C A A A A B B

I/OB B B

0 7 123

436

A B C

Max units M 4 5 3 Space/unit s 1 1 1 Flow f 24 7 21 Flow Density f/(M x s) 6.00 1.40 7.00

Ex 30: 1-D Slotting

Dedicated Random Class-Based

A B C ABC AB AC BC

Max units M 4 5 3 9 7 7 8 Space/unit s 1 1 1 1 1 1 1 Flow f 24 7 21 52 31 45 28 Flow Density f/(M x s) 6.00 1.40 7.00 5.78 4.43 6.43 3.50

287

1-D Slot Assignments

Total

Distance

Total

Space

Dedicated

(flow density) C C C A A A A B B

I/OB B B

436 12

Dedicated

(flow only) A A A A C C C B B

I/OB B B

460 12

Class-based C C C AB AB AB AB AB ABI/O

AB

466 10

Randomized ABC ABC ABC ABC ABC ABC ABC ABC ABCI/O

468 9

Ex 31: 2-D Slotting

A B C

Max units M 4 5 3 Space/unit s 1 1 1 Flow f 24 7 21 Flow Density f/(M x s) 6.00 1.40 7.00

288

8 7 6 5 4 5 6 7 8

7 6 5 4 3 4 5 6 7

6 5 4 3 2 3 4 5 6

5 4 3 2 1 2 3 4 5

4 3 2 1 0 1 2 3 4

Original Assignment (TD = 215) Optimal Assignment (TD = 177)

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

Distance from I/O to Slot

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

289

i

j ij ijkl ij kl

iij ijc x

fd x DSAP LAP LP QAP c x x

MTSP

Ex 32: 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. 40 36 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

290

Ex 32: 1-D DSAP

291

• Randomized:

14

1 94 64 50 1104

2 2 2 2

1 1

2 2

3 1 4 1104 3(4)

2 2 11 lanes

3(4)

3(11) 33 ft

33 ft

160 140 130 33

A B C

rand

rand

SC

rand A B C

M M MM

D HM NH N

L

DH

N

X xL

d X

TD f f f X

,190 ft

ABCI/O

0 33

Ex 32: 1-D DSAP

292

• Dedicated:

160 140 1301.7, 2.19, 2.6

94 64 50

94 64 508, 6, 5

3(4) 3(4) 3(4)

3(5) 15, 3(6) 18, 3(8) 24

3(5) 15 ft

A B C

A B C

A B CA B C

C C B B A A

CSC C

S

f f fC B A

M M M

M M ML L L

DH DH DH

X xL X xL X xL

d X

d

2( ) 2(15) 18 48 ft

2( ) 2(15 18) 24 90 ft

160(90) 140(48) 130 23,01 0( 75) ft

BC C B

ASC C B A

A B Cded A SC B SC C SC

X X

d X X X

TD f d f d f d

I/O

0

C B A

15 33 57

1-D Multiple Region Expected Distance

• In 1-D, easy to determine the offset

• In 2-D, no single offset value for each region

293

/O to /

22 2

3

2 2 10

2 2 7 4 10

7

3, 4, 7

,

2

ASC A A

B offset B A A B A A B

I I O B

AB

A A B B A AB A B

A A A B B B

AB AB AB B A B A A B A A B A

A A B B A B

B

d d X

d d X X X X X X X

d X

d

TA X TA X X TA TA TA

TM TA d TM TA d

TM TA d X X X X X X X X X X

TA d TA d TM TM

Td

7(7) 3(3)

104

B AB A AB AB A A

B B B

M TM TM TA d TA d

TA TA TA

A A A B B B B

I/O0 3 7

I/Oʹ

BX,A BX X


294


295


• 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

296

I/O X

I/O' AX CX

DX

A

B

D

C

Given , - - -

24

23

42 4

34

4 ,3

42 4

3

ii i

i

ii i

D D D D

C C C C

C C C

A AC AC A

A AC CA

A A

AC AC C C

A

B AC DB

B B

fTA f D B A C

TA

TATA TA TA X TA

X TA d X

X TA d X X

X X X X X X

X TA d X X

TM TM TMd

TA TA

TA d TA d

TA

TM TM TM TMd

TA TA

Warehouse Operations

Order Picking

Replenish

Pu

taw

ay

Order Picking

Puta

way

Forward PickingReserve

Storage

Packing, Sorting

& Unitizing

Receiving ShippingCross-docking

297

Carton Flow Rack

Receiving Staging Area

(5 lanes)Secure

Storage

Area

Bin Shelving and

Storage Drawers

Horizontal

Carousel

(2 pods)

Takeaway Conveyor (top level return)

Double-Deep Pallet Racks

Block Stacking (20 lanes)

Unitizing

Area

Receiving Dock Doors (5)

Shipping Staging Area

(5 lanes)

Pallet Rack

Sortation

Conveyor

Single-Deep

Selective

Pallet Racks

Shipping Dock Doors (5)

Packing

Area

A-Frame

Dispenser

Warehouse Management System

• WMS interfaces with a corporation’s enterprise resource planning (ERP) and the control software of each MHS

298

ASN

PurchaseOrder

CustomerOrder

ASN

Material Handling Systems

WMS

ERP

Item

Master

File

Carrier

Master

File

Customer

Master

File

Cu

sto

me

r

Su

pp

lie

r

Location

Master

File

Inventory Master

File

• Advance shipping notice (ASN) is a standard format used for communications

Item

On-Hand

Balance

In-Transit

Qty. Locations

A 2 1 11,21

B 4 0 12,22

Inventory Master File

Location Item

On-Hand

Balance

In-Transit

Qty.

11 A 1 0

12 B 3 0

21 A 1 1

22 B 1 0

Location Master File

A

B B B

A

B

11

12

21

22

A

Logistics-related Codes Commodity Code Item Code Unit Code

Level Category Class Instance

Description Grouping of

similar objects

Grouping of identical

objects

Unique

physical object

Function Product

classification

Inventory control Object

tracking

Names — Item number, Part number,

SKU, SKU + Lot number

Serial number,

License plate

Codes UNSPSC,

GPC

GTIN, UPC,

ISBN, NDC

EPC,

SSCC

299

UNSPSC: United Nations Standard Products and Services CodeGPC: Global Product Catalogue

GTIN: Global Trade Item Number (includes UPC, ISBN, and NDC)UPC: Universal Product CodeISBN: International Standard Book NumberingNDC: National Drug CodeEPC: 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.))

Identifying Storage Locations

300

0103

0911

0507

A

B

C

D

E

Bay (X)

Tie

r (Z

)

Aisle (Y)

AAB

AAC

AAA

Cross AisleDown Aisle

Wall

Compartment

12

AB

Position

Location: 1 -AAC - 09 - D - 1 - B

Building

Aisle

Bay

Tier

Pos

ition

Com

partm

ent

Receiving

301

• 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

Reserve

StorageReceive Putaway Replenish

Forward

Pick

Order

Pick

Sort &

PackShip

Putaway

302

• A putaway algorithm is used in WMS to search for and validate locations where each movable unit in the putawayqueue 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)

ReserveStorage

Receive Putaway ReplenishForward

PickOrderPick

Sort &Pack

Ship

Replenishment

303

• 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


Forward

Pick

Order

Pick

Sort &

PackShip

• 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

Reserve Storage Area

Order Picking

304

• Order picking is at the intersection of warehousing and order processing

WH Operating Costs

Reserve


Forward

Pick

Order

Pick

Sort &

PackShip

Receiving10%

Storage15%

Order Picking55%

Shipping20%

Info

rma

tion

Pro

ce

ssin

g

Material Handling

Putaway Storage Order Picking Shipping

Order Entry

Order

Transmittal

Order Status

Reporting

Order Processing

Wa

reh

ou

sin

g

Receiving

Order

Preparation

Order Picking

305

Levels of Order Picking

Reserve


Forward

Pick

Order

Pick

Sort &

PackShip

Case Picking

Pallet Picking

Piece Picking

Pallet and Case Picking Area

Forward Piece Picking Area

Order Picking

306

Reserve


Forward

Pick

Order

Pick

Sort &

PackShip

Voice-Directed Piece and Case Picking

Pallet Flow Rack

for Case Picking

Carton Flow

Racks for Piece

Picking

Static Pallet Rack

for Reserve

Storage and

Pallet Picking

Pick

Conveyor

Tote

Voice Directed

Order Selection

Pick-to-BeltTakeaway

Conveyor

Pick 24 Pack 14

Pack 1

0

Carton Flow Rack Picking Cart

Confirm

Button

Increment/

Decrement

Buttons

Count Display

Pick-to-Light Piece Picking

Order Picking

307

Reserve


Forward

Pick

Order

Pick

Sort &

PackShip

Methods of Order Picking

D

1

2

3

4

5

6

7 A3

A B C E F G H

Picker 1

C4

G1 E5

Zone 1 Zone 2

DA B C E F G H

1

2

3

4

5

6

7 A3

Picker 2Picker 1

C4

E5G1

DA B C E F G H

1

2

3

4

5

6

7 A3

G1

C7

E8

D5

B4

F2

Picker 1

DA B C E F G H

1

2

3

4

5

6

7 A3

Picker 1

Zone 1

Picker 2

Zone 2

C7

D5

G1

B4

F2

E8

Method

Pickers

per

Order

Orders

per

Picker

Discrete Single Single

Zone Multiple Single

Batch Single Multiple

Zone-Batch Multiple Multiple

Discrete

Batch

Zone

Zone-Batch

Sortation and Packing

308

Wave zone-batch piece picking, including

downstream tilt-tray-based sortation

Reserve


Forward

Pick

Order

Pick

Sort &

PackShip

Takeaway Conveyor

G

Downstream

Sortation

Zone 1 Zone 2

Bin

Sh

elv

ing

Induction

Station

Did Not

Read

Packing

Station

Tilt

Tray

Reader

Packing

Station

Order

Consolidation

Chutes

Case Sortation System

Shipping

309

• 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

Reserve


Forward

Pick

Order

Pick

Sort &

PackShip

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 orders in which both items appear 310

SKU B D E

A 0.2 0.2 0.0

C 0.4 0.2

Demand Correlation

Distibution

D 0.2

E

A

B 0.2 0.0

C

0.4

0.2

SKU

Cube

Movement

A 330

C 720

D 576

E 720

Lines per

Item

3

3

2

1

B 2 120

Flow per

Item

11

5

4

18

6

Total Lines = 11

Lines per Order = 11/5 = 2.2

Cube per Order = 493.2

SKU Width Cube Weight

A 3 30 1.25

C 6 180 9.65

D 4 32 6.35

E 4 120 8.20

Length

5

8

4

6

B 3 2 24 4.75

Depth

2

4

5

3

5

Item Master

SKU

A

B

Order: 1

C

D

Qty

5

3

2

6

SKU

C

D

Order: 5

E

Qty

1

12

6

SKU

A

Order: 3

Qty

2

SKU

A

Order: 2

C

Qty

4

1

SKU

B

Order: 4

Qty

2

Customer

Orders

Pallet Picking Equipment

311

Single-Deep Selective Rack

Double-Deep

Rack

Push-Back

Rack

Sliding

Rack

Block

Stacking /

Drive-In

Rack

Pallet Flow Rack

Flow per Item

Cu

be M

ov

em

en

t

Drive-In Rack

Sliding Rack Single-Deep

Selective Rack

Double-Deep Rack

Push-BackRack

Case Picking

312

Flow

Delivery

Lanes

Single-Deep

Selective Rack

Pallet Flow

Rack

Lines per Item

Cu

be M

ovem

ent

Case

Dispensers

Push Back

Rack

Manual Automated Case PickingEquipment

Unitizing

and

Shipping

So

rta

tio

n C

on

ve

yo

r

InductInduct

Single-Deep

Selective

Racks

Zone-Batch Pick to Pallet

Floor- vs. Multi-levelPick to Pallet

Case Picking

Re

ple

nis

h

Re

se

rve

Sto

rag

e

Fo

rwa

rd P

ick

Sto

rag

e

Case

Picking

Fo

rwa

rd P

ick

Sto

rag

e

Case

Picking

Case

Picking

Case

Picking

Floor-level Pick Multi-level Pick

Piece Picking Equipment

313

Bin Shelving Horizontal

Carousel

Storage

Drawers

Carton Flow Rack

Lines per Item

Cu

be M

ovem

ent

A-Frame

Vertical Lift

Module

A-FrameDispenser

Carousel

Carton Flow Rack

Drawers/BinsPick Cart

Vertical Lift Module

Methods of Piece Picking

314

Batch

(Ex: Pick Cart)

Zone-Batch

(Ex: Wave Picking)

Discrete

(Infrequently Used)

Zone

(Ex: Pick-and-Pass)

Lin

es

per

Ord

er,

Cu

be

per

Ord

er

Total Lines

Packing

and

Shipping

Bin

Sh

elv

ing

Pick Cart

Takeaway Conveyor

G

Do

wn

str

ea

m

So

rta

tio

n

Zone 1 Zone 2

Bin

Sh

elv

ing

Packing and

Shipping

Ta

ke

aw

ay

Co

nv

ey

or

Zone 1 Zone 2 Zone 3

Ca

rton

Flo

w R

ac

k

Pick Pick

PassPass Pick

ConveyorNo Pick

Scan

Pick-cart Batch Piece PickingWave Zone-Batch Piece Picking

Pick-and-Pass Zone Piece Picking

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: median moving machine operator annual wage + benefits

– 1.7% average real interest rate 2005-2009 (real = nominal – inflation)

– 5-year service life with no salvage (service life for Custom Software)

5

1

1 1.017$45,432 $45,432 4.83 $219,692

1 1.017

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

ISE 754: Logistics Engineering - Nc State University · Management Logistics Engineering Supply Chain Analysis (military logistics , humanitarian logistics , school bus routing )

Documents