storage and warehousing in cleanroom slides

Storage and Warehousing
Chapter 12

Phoenix Pharmaceuticals
• German company founded in 1994
• Receives supplies from 19 plants across Germany
and distributes to drugstores
• $400 million annual turnover
• 30% market share
• Fill orders in < 30 minutes
• 87,000 items
• 61% pharmaceutical, 39% cosmetic

Phoenix Pharmaceuticals (cont.)
• 150-10,000 picks per month
• Three levels of automation
– Manual picking via flow-racks
– Semi-automated using dispensers
– Full automation via robotic AS/RS

Warehouse Functions
• Provide temporary storage of goods
• Put together customer orders
• Serve as a customer service facility
• Protect goods
• Segregate hazardous or contaminated materials
• Perform value-added services
• Inventory

Elements of a Warehouse
• Storage Media
• Material Handling System
• Building

Storage Media
• Block Stacking
• Stacking frames
– Stool like frames
– Portable (collapsible) frames
• Cantilever Racks

Storage Media (Continued)
• Selective Racks
– Single-deep
– Double-deep
– Multiple-depth
– Combination (Fig 12.5)
• Mobile Racks
• Flow Racks
• Push-Back Rack

Storage Media (Continued)
• Racks for AS/RS
• Combination Racks
• Modular drawers (high density storage)
• Racks for storage and building support

Storage and Retrieval Systems
• Person-to-item
• Item-to-person
• Manual S/RS
• Semi-automated S/RS
• Automated S/RS
• Aisle-captive AS/RS
• Aisle-to-aisle AS/RS

Storage and Retrieval Systems
(cont)
• Storage Carousels
– Vertical
– Horizontal
• Miniload AS/RS
• Robotic AS/RS
• High-rise AS/RS (two motors)

Warehouse Problems
• Design
• Operational or Planning

Warehouse Design
• Location
– How many?
– Where?
– Capacity
• Overall Layout (Figures 12.23 and 12.24)
• Layout and Location of Docks
– Pickup by retail customers?
– Combine or separate shipping and receiving?

Warehouse Design (cont)
– Layout of road/rail network
– Room available for maneuvering trucks?
– Similar trucks or a variety of them?
• Number of Docks
– Shipping and receiving combined or separated?
– Average and peak number of trucks or rail
cars?
– Average and peak number of items per order?

Warehouse Design (cont)
– Seasonal highs and lows
– Types of load handled? Sizes? Shapes?
Cartons? Cases? Pallets?
– Protection from weather elements
• Rack Design
– LP Model

Model for Rack Design
Minimize
x(a 1)  y(b 1)
2
Subject to xyz  n
x, y, integer
• x, y are # of columns, rows of rack spaces
• a, b are aisle space multipliers in x, y directions

Model for Rack Design (Cont)
xyz  n  x 
n
yz
• In the relaxed problem,
• The unconstrained objective is

• Taking derivative wrt y, setting equation to
zero and solving, we get
n(a 1)
yz
 y(b 1)
2
.

y  n(a 1)
z(b 1)
and x  n(b 1)
z(a 1)
.

Rack Design Example
• Consider warehouse shown in figure 12.26
• Assume travel originates at lower left
corner
• Assume reasonable values for the aisle
space multipliers a, b

Rack Design Example (Cont)
• Determine length and width of the
warehouse so as to accommodate 2000
square storage spaces of equal area in:
– 3 levels
– 4 levels
– 5 levels

Rack Design Example Solution
• Reasonable values for a, b are 0.5, 0.2
• For the 3-level case,
y 
2000(0.5 1)
3(0.2 1)
 29 and x 
2000(0.2 1)
3(0.5 1)
 24.

(Cont)
• Previous solution gives a total storage of
24x29x3=2088
• Due to rounding, we get 88 more spaces
• If inadequate to cover the area required for
lounge, customer entrance/exit and other
areas, the aisle space multipliers a, b must be
increased appropriately and the x, y values
recalculated

(Cont)
• For the 4 level and 5 level case, the building
dimensions are 25x20 units and 18x23 units,
respectively
• Easy to calculate the average distance
traveled - simply substitute a, b, x and y
values in the objective function
• For 3-level case, average one-way distance =
35.4 units

Block Stacking
• Simple formula to determine a near-optimal
lane depth assuming
– goods are allocated to storage spaces using the
random storage operating policy
– instantaneous replenishment in pre-determined lot
sizes
– replenishment done only when inventory excluding
safety stock has been fully depleted
– lots are rotated on a FIFO basis

Block Stacking (Cont)
– withdrawal of lots takes place at a constant rate
– empty lot is available for use immediately
• Let Q, w and z denote lot size in pallet
loads, width of aisle (in pallet stacks) and
stack height in pallet loads, respectively

• Kind’s (1975) formula for near-optimal lane
depth, d
d 
Q w
z

w
2

• E.g., if lot size is 60 pallets, pallets are stacked 3
pallets high and aisle width is 1.7 pallet stacks, then
• Optimality of this answer can be verified by
checking the utilization for all possible lane depths
(a finite number)
d  (60)(1.7)
3
 1.7
2
 5 pallets

• Several issues omitted in Kind’s formula.
Some examples
– What if pallets withdrawn not at a constant rate
but in batches of varying sizes?
– What if lots are relocated to consolidate pallets
containing similar items?

Storage Policies
• Random
– In practice, not purely random
• Dedicated
– Requires more storage space than random, but
throughput rate is higher because no time is lost
in searching for items
• Cube-per-order index (COI) policy
• Class-based storage policy

Storage Policies (Cont)
• Shared storage policy
• Class based and shared storage policies are
between the two “extreme” policies -
random and dedicated
• Class based policy variations
– if each item is a class, we have dedicated policy
– if all items in one class, we have random policy

Design Model for Dedicated
Policy
• Warehouse has p I/O points
• m tems are stored in one of n storage spaces
or locations
• Each location requires the same storage
space
• Item i requires Si storage spaces

Policy (Cont)
• Ideally, we would like
• However, if LHS < RHS, add a dummy
product (m+1) to take up remaining spaces

m
i 1
Si  n.
n  
m
i 1
Si

Policy (Cont)
• So, assume that the above equality holds
• But, if RHS < LHS, no feasible solution
• Model Parameters
– fik trips of item i through I/O point k
– cost of moving a unit load of item i to/from I/O
point k is cik
– distance of storage space j from I/O point k is dkj

Policy (Cont)
• Model Variable
– binary decision variable xij specifying whether
or not item i is assigned to storage space j

Policy (Cont)
• Model
• Minimize
• S.T.

m
i 1

n
j 1

p
k 1
cikfik dkj
Si

n
j 1
xij  Si, i 1,2,...,m

m
i 1
xij  1 j 1,2,...,n

Policy (Cont)
• Substituting
• Objective function is
• Minimize
xij  0 or 1 i 1,2,...,m, j 1,2,...,n
wij 

p
k 1
cikfikdkj
Si
,

m
i 1

n
j 1
wijxij

Policy (Cont)
• Model is generalized QAP
• Can be solved via transportation algorithm
• No need for binary restrictions in the model

Policy - Example
• Warehouse layout
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16

Policy - Example (Cont)
• 3 I/O points located in middle of south,
west and north walls
• 4 items

Policy - Example (Cont)
1 2 3 Si
1 150(5) 25(5) 88(5) 3
[fik(cik)] = 2 60(7) 200(3) 150(6) 5
3 96(4) 15(7) 85(9) 2
4 175(15) 135(8) 90(12) 6

Policy - Example Solution (Cont)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 5 4 4 5 4 3 3 4 3 2 2 3 2 1 1 2
[dkj] = 2 2 3 4 5 1 2 3 4 1 2 3 4 2 3 4 5
3 2 1 1 2 3 2 2 3 4 3 3 4 5 4 4 5

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 1626.7 1271.7 1313.3 1751.7 1481.7 1126.7 1168.3 1606.7 1378.3 1023.3 1065.0 1503.3 1316.7 961.7 1003.3 1441.7
2 1020.0 876.0 996.0 1380.0 996.0 852.0 972.0 1356.0 1092.0 948.0 1068.0 1452.0 1308.0 1164.0 1284.0 1668.0
[wij] 3 1830.0 1308.0 1360.5 1987.5 1968.0 1446.0 1498.5 2125.5 2158.5 1636.5 1689.0 2316.0 2401.5 1879.5 1932.0 2559.0
4 2907.5 2470.0 2650.0 3447.5 2470.0 2032.5 2212.5 3010.0 2212.5 1775.0 1955.0 2752.5 2135.0 1697.5 1877.5 2675.0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 1626.7 1271.7 1313.3 1751.7 1481.7 1126.7 1168.3 1606.7 1378.3 1023.3 1065.0 1503.3 1316.7 961.7 1003.3 1441.7
2 1020.0 876.0 996.0 1380.0 996.0 852.0 972.0 1356.0 1092.0 948.0 1068.0 1452.0 1308.0 1164.0 1284.0 1668.0
[wij] 3 1830.0 1308.0 1360.5 1987.5 1968.0 1446.0 1498.5 2125.5 2158.5 1636.5 1689.0 2316.0 2401.5 1879.5 1932.0 2559.0
4 2907.5 2470.0 2650.0 3447.5 2470.0 2032.5 2212.5 3010.0 2212.5 1775.0 1955.0 2752.5 2135.0 1697.5 1877.5 2675.0

2 3 3 2
2 2 1 2
4 4 4 1
4 4 4 1

Design Model for COI Policy
• Consider special case of dedicated storage
policy model
– All items use I/O points in same proportion
– Cost of moving a unit load of item i is independent
of I/O point
• Define Pk as % trips through I/O point k
• No need for the first subscript in fik as well as cik

(Cont)
Minimize 
m
i 1

n
j 1

p
k 1
cifiPkdkj
Si

(Cont)
• S.T.

n
j 1
xij  Si, i 1,2,...,m

m
i 1
xij  1 j 1,2,...,n
xij  0 or 1 i 1,2,...,m, j 1,2,...,n

(Cont)
• Substitute
• Rewrite objective function
wj  
p
k 1
Pkdkj

m
i 1

n
j 1
ci fi
Si
wj

Design Model for COI Policy -
Solution
• COI model easier than Dedicated Model
• Rearrange “cost”, “distance” terms (cifi/Si), wj
in non-increasing and non-decreasing order
• Match
– Item corresponding to 1st
element in ordered
“cost” list with storage spaces corresponding to
1st
Si elements in ordered “distance” list

Solution
– Second item with storage spaces corresponding
to next Sl elements, and so on …
• COI policy calculates inverse of the “cost”
term and orders elements in non-decreasing
order, of their COI values, thereby
producing the same result as above

Solution
• Arranging cost and distance vectors in non-
increasing and non-decreasing order and
taking their product provides a lower bound
on cost function
• Above algorithm is optimal

Example
• Consider dedicated policy example
• Ignore cik and fik data
• Assume
• all 4 items use 3 I/O points in same proportion
• pallets moved/time period are 100, 80, 120 and 90
• cost to move unit load through unit distance is $1.00
• Determine optimal assignment of items to
storage spaces

Example Solution
j 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
wj 3.0 2.6 3.0 4.0 2.6 2.3 2.6 3.6 2.6 2.3 2.6 3.6 3.0 2.6 3.0 4.0
Sorting the above values in non-decreasing order, we get
j 6 10 2 5 7 9 11 14 1 3 13 15 8 12 4 16
wj 2.3 2.3 2.6 2.6 2.6 2.6 2.6 2.6 3.0 3.0 3.0 3.0 3.6 3.6 4.0 4.0

Example Solution
• Sort [cifi/Si] values in non-increasing order
– [60, 33.33, 16, 15], corresponding to items 3, 1,
2 and 4
• Optimal storage space assignment
– Item 1 to Storage Spaces 2, 5, 7
– Item 2 to Storage Spaces 1, 3, 9, 11, 14
– Item 3 to Storage Spaces 6, 10
– Item 4 to Storage Spaces 4, 8, 12, 13, 15, 16

Example Solution
2 1 2 4
1 3 1 4
2 3 2 4
4 2 4 4

Design Model for Random Policy
• Items stored randomly in empty and
available storage spaces
• Each empty space has an equal probability
of being selected
• Storage or retrieval may not be purely
random, but we assume so for model

(Cont)
• Problem Definition
– Determine storage space layout so total
expected travel distance between each of n
storage spaces and p I/O points is minimized
• Sum of distances of each storage space
from each I/O point

p
k 1
dkj ,

Design Model for Random
Policy- Solution
• Arrange spaces in non-decreasing order of
the sum of above distances
• Pick the n closest storage spaces
• n depends upon inventory levels of all items
• n is less than that required under dedicated
policy

- Example
• Determine storage space layout for 56
storage spaces in a 40x70 feet warehouse
• Random storage policy
• Minimize total distance traveled
• Each storage space is a 10x10 feet square
• I/O point located in middle of south wall

- Example (Cont)

- Example Solution
• Calculate distance of all potential storage
spaces to the I/O point
• Arrange them in non-decreasing order

- Example Solution (Cont)
• Largest distance traveled is 70 feet
• Sum total distance traveled (2800) by
number of storage spaces (56) to get
average distance traveled = 50 feet

- Example Solution (Cont)
70 70
70 60 60 70
70 60 50 50 60 70
70 60 50 40 40 50 60 70
70 60 50 40 30 30 40 50 60 70
70 60 50 40 30 20 20 30 40 50 60 70
70 60 50 40 30 20 10 10 20 30 40 50 60 70

Travel Time Models
• For random policy, average distance
traveled
• When number of storage spaces are large,
calculating average distance can be tedious

p
k 1

n
j 1
dkj
n
.

Travel Time Models (Cont)
• If storage spaces are small relative to total
area, approximate average distance traveled
– assume spaces are continuous points on a plane
– use the integral

X
0

Y
0
1
A
(x y)dxdy

• We assume in previous integral that
– warehouse is in 1st quadrant
– only one I/O point (at origin and SW corner)
– distance metric of interest is rectilinear
• Previous integral can be easily modified if
– two or more I/O points
– distance metric is not rectilinear
– no restrictions on location of warehouse

• Suppose designer interested in shape that
minimizes travel time
• Then, depending upon number and
location of I/O points, distance metric,
warehouse shape can range from diamond
to circle to trapezium !!!

• Models minimizing construction costs and
travel distance
• Consider following assumptions
– Warehouse shape is fixed - rectangle
– Warehouse area = A
– Construction cost is function of warehouse
perimeter - r[2(a+b)]
• r is unit (perimeter) distance construction cost
• a and b are warehouse dimensions

– One I/O point at origin and SW corner
• coordinates are (p, q)
– cost for each unit distance traveled = c
• Model
2r(a b)  c

p a
p

q b
q
1
A
(|x| |y|)dxdy

• Optimal value of a and b, given that
– I/O point must be on or outside exterior
walls, i.e., p  0
– warehouse area must be A square units
a  A
c  8r
2c  8r
and b  A
2c  8r
c  8r

• Single command cycle
• Dual or multiple command cycles

Warehouse Operations
• Warehouse operational problems
– Sequence in which orders to be picked
– How frequently orders picked from high-rise
storage area?
– Batch picking or pick when order comes in?
– Limit on number of items picked?
• If so, what is the limit?
• Operator assignment to stacker cranes

Warehouse Operations (Cont)
• How to balance picking operator’s
workload?
• Release items from stacker crane into
sorting stations in batches or as soon as
items are picked?
• Order picking consumes over 50% of the
activities in warehouse

Warehouse Operations (Cont)
• Not surprising that order picking is the
single largest expense in warehouse
operations
• Since construction and operation of AS/RS
are very high,managers interested in
maximizing throughput capacity

Order Picking Sequence
• Two basic picking methods
– Order picking
– Zone picking
• Consider this:
– An AS/R machine has two independent
motors
– Movement in horizontal and vertical
directions simultaneously

Order Picking Sequence (Cont)
– Time to travel from (xi, yi) to (xj, yj)
max
|xi  xj|
h
,
|yi  yj|
v

Order Picking Sequence Model
Minimize 
n
i 1

n
j 1,j i
dij wij
subject to 
n
i 1,i j
wij 1, for each j

n
j 1,j i
wij 1, for each i
ui  uj  nwij  n 1, for 2 i j n
dij  max
|xi xj|
h
,
|yi yj|
v
, for 1 i j n
wij  0 or 1, for each i,j i j
ui are arbitrary real numbers, for each i

Algorithms
• Construction
• Improvement
• Hybrid

Algorithms (Cont)
• 2-opt
• 3-opt
• Branch-and-bound
• Simulated Annealing
• Convex Hull

Convex Hull Algorithm - Phase 1
• Find xmax and ymax
• Delete points inside polygon formed by xmax,
ymax and origin
• For each region, construct convex path
between extreme points
– Sort points in regions 1 and 2 in ascending
order of x-coordinate

(Cont)
– Sort region 3 points in descending order
– Starting with 1st
extreme point, compute V
for three consecutive points i, i+1, i+2
– V= (yi+1-yi)(xi+2-xi+1)+(xi-xi+1)(yi+2-yi+1).
– Repeat until other extreme point is reached
– If V  0, no convex hull with i, i+1, i+2
– Otherwise, convex hull possible

(Cont)

(Cont)
– Using some or all of the sorted points in
regions 1, 2, and 3, three at a time, generate
convex hull (sub-tour)
– Points not in sub-tour are considered in
phases 2 and 3.
– If xmax = ymax following explanation still good

• Insert points that maybe included in sub-
tour without increasing cost
• Such free insertion points lie on a
parallelogram with two adjacent points in
the sub-tour as its corner

• Insert points not included in the sub-tour in
phases 1 and 2 using minimal insertion
cost criteria
– greedy hull
– steepest descent hull
• If no points left for insertion in phase 2 or
3, phase 1 sub-tour is optimal

Simulated Annealing Algorithm
• Set S, z, r, Tin, T= Tin; Tfin= 0.1Tin
• Randomly select points i and j in S and
exchange their positions
• If new solution S' has z' z, set S = S', and z
= z’
• Otherwise, set S= S' with probability e-/T

Simulated Annealing Algorithm
(Cont)
• Repeat Step 1 until number of new
solutions = 16 times the number of
neighbors
• Set T= rT. If T > Tfin, go to Step 1
• Otherwise return S, and STOP

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