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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
– Exam 2 (take home)
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
Winston-Salem GreensboroDurham
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
Winston-Salem GreensboroDurham
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
2. Facility location
3. Freight transport
– Exam 1 (take home)
4. Network models
5. Routing
– Exam 2 (take home)
6. Warehousing
– Final exam (in class)
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
Logistics Engineering Design Constants
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
"
Logistics Engineering Design Constants
7. TL Revenue per Loaded Truck-Mile: $2/mi in 2004 ( r )– TL revenue for the carrier is your TL cost as a shipper
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
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
Truck Shipment Example: One-Time
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
Truck Shipment Example: One-Time
• 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
Truck Shipment Example: One-Time
• (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
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 $2
8. Each truck’s cubic and weight capacity is 2,750 ft3 and 25 tons, respectively
163
-83 -82 -81 -80 -79 -78
34
34.5
35
35.5
36
36.5
Asheville
Statesville
Winston-Salem GreensboroDurham
Raleigh
Wilm
ingto
n
50
150
190 220270
295
420
40
Ex 12: FTL Location
($/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
*
Asheville DurhamStatesville Winston-Salem Wilmington
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
*
Asheville DurhamStatesville Winston-Salem Wilmington
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
2. Facility location
3. Freight transport
– Exam 1 (take home)
4. Network models
5. Routing
– Exam 2 (take home)
6. Warehousing
– Final exam (in class)
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
Graph Representations
• 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
Graph Representations
• 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
Graph Representations
• 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)
A* Path Planning Example 2
• 3-D (x,y,t) A* used for planning path of each container in a DC
• Each container assigned unique priority that determines planning sequence
– Paths of higher-priority containers become obstacles for subsequent containers
188
A* Path Planning Example 2
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
Road Network Modifications
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
Road Network Modifications
3. Add connector
– Given new nodes, add arcs that connect the new nodes to the existing nodes in a graph and to each other
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
Ex 15: Coupled Networks via Truck Capacity
• 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
Ex 15: Coupled Networks via Truck Capacity
• 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
Production and Inventory: Multiple Products
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
Production and Inventory: Multiple Products
203
Flowbalance
Capacity
Setup
Linking
MILP
dummy
Production and Inventory: Multiple Products
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
2. Facility location
3. Freight transport
– Exam 1 (take home)
4. Network models
5. Routing
– Exam 2 (take home)
6. Warehousing
– Final exam (in class)
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)
Shipment Size (tons)
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)
Shipment Size (tons)
Tran
spo
rt C
har
ge (
c, $
)
MC
TL
Full-Truckload
max min , ,LTL TLc c c MC
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
max min , ,LTL TLc c c MC
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
2. Facility location
3. Freight transport
– Exam 1 (take home)
4. Network models
5. Routing
– Exam 2 (take home)
6. Warehousing
– Final exam (in class)
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
2-D Multiple Region Expected Distance
294
2-D Multiple Region Expected Distance
295
2-D Multiple Region Expected Distance
• If more than two regions:– For regions below diagonal (D),
start with region closest to I/O
– For regions above diagonal (A+C), start with regions closest to I/O’ (C)
– For region in the middle (B), solve using whole area less other regions
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
StorageReceive Putaway Replenish
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
StorageReceive Putaway Replenish
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
StorageReceive Putaway Replenish
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
StorageReceive Putaway Replenish
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
StorageReceive Putaway Replenish
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
StorageReceive Putaway Replenish
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
StorageReceive Putaway Replenish
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