A MINIMUM COST DESIGN FOR AN AUTOMATED WAREHOUSE
A THESIS Presented to Faculty of the Division of Graduate Studies by Yavuz Ahmet Bozer
In Partial Fulfillment of the Requirements for the Degree Master of Science in Industrial Engineering
Georgia Institute of Technology December, 1978
A MINIMUM COST DESIGN FOR AN AUTOMATED WAREHOUSE
Approved:
John A. Whi" , v,udirman
KCDer
J.^five's/
'
Date approved by Chairman
ii
ACKNOWLEDGMENTS
I would like to express my deep appreciation to my thesis advisor, Dr. John A. White, for his invaluable guidance and support in carrying out this study.
His knowledge and creative thinking have
contributed towards every single conclusion reached throughout the thesis. I would also like to thank Dr. Robert J. Graves and Dr. Gunter P. Sharp for serving on my reading committee and for their helpful comments in the early stages of the study. My special thanks go to Dr. Donovan Young for his significant contributions in developing many of the expressions presented in Chapter IV.
I am also very thankful to Mr. Howard Zollinger of Rapistan,
Inc. for sharing his priceless experience with me in constructing the cost model presented in Chapter II. In addition, I would like to thank Ms. Pam Walker for typing the thesis.
She worked unselfishly under tight deadlines and provided
useful suggestions regarding format and grammatical matters. This acknowledgment would have never been complete without expressing my deep admiration and respect to my sister and my parents who have supported me in every stage of my life and career with their love and encouragement.
iii
TABLE OF CONTENTS Page ACKNOWLEDGMENTS
ii
LIST OF TABLES
v i
LIST OF ILLUSTRATIONS NOMENCLATURE SUMMARY
vii ix x
Chapter I.
INTRODUCTION
i
Obj ective Importance of Study Foundation Material Description of the Problem Literature Search Summary II.
DEFINITION OF THE TOTAL PRESENT WORTH COST Introduction Cost of Land Cost of Building Cost of Racks Cost of S/R Machines The Annual Labor Cost The Annual Maintenance Cost Present Worth Cost of the System Summary
III.
DETERMINING SPACE REQUIREMENTS . Introduction Definition of the System The Simulation Model Regression Analysis Conclusions Summary
2 9
iv
Page
Chapter IV.
ANALYSIS OF THE STORAGE/RETRIEVAL MACHINE TRAVEL TIME
62
Introduction Randomized Storage Dedicated Storage Statistical Distribution of Travel Times Summary V.
THE OPTIMIZATION MODEL
1 2 1
Introduction The Mathematical Model Analysis of Cost Elements Analysis of Throughput Description of the Solution Procedure Summary VI.
DISCUSSION OF SAMPLE RUNS AND SENSITIVITY ANALYSIS . . .
1 4 8
Introduction Sample Runs Sensitivity Analysis Summary VII.
CONCLUSIONS AND RECOMMENDATIONS
164
Introduction Conclusions Recommendat ions APPENDICES 1.
2.
3.
4.
5.
Program listing for simulating the storage space requirement under dedicated and randomized storage . . .
169
Output of the simulation runs obtained from the program listing in Appendix 1
172
Output of all possible regressions based on simulation runs
185
Program listing for computing the expected S/R travel time by complete enumeration and the conventional method
202
Program listing for simulating dual and single command trips over a continuous normalized rack
205
V
Page 6.
7.
8.
9.
10.
11.
Program listing for finding the expected mean and variance of travel time by enumeration under dedicated storage
207
Program listing for simulating dual command trips over a discrete rack under dedicated storage
210
Outputs obtained from those programs listed in Appendices 6 and 7
213
Sample runs to demonstrate that the expected mean and variance of travel time under dedicated storage are minimized when b is maximized
220
Program listing from which sample runs presented in Appendix 9 were obtained
230
Program Listing of the algorithm developed to find an optimum design
234
12.
Outputs related to the sensitivity analysis on "DUAL" .
244
13.
Outputs related to the sensitivity analysis on A (TPUT)
257
14.
Detailed Flow-chart of the program listing presented in Appendix 1. BIBLIOGRAPHY
266 271
vi
LIST OF TABLES
Table
Page
1-1
Storage Assignment/Interleaving Policies
24
1-2
Expected Round Trip Times
25
2-1
Minimum Building Length Addition
32
2-2
Conversion Factors for Unit Building Cost
32
2-3
Factor Numbers to Determine Rack Cost
32
2-4
The Annual Depreciation on the Building, Rack and S/R's
3-1
Results of the Simulation Runs
55
3-2
All Possible Regressions
57
4-1
Travel Times Obtained by Complete Enumeration and the Conventional Method
...
.
41
71
4-2
Simulation Results for Different b Values
92
4-3
Sample Runs Data
101
4-4
Enumeration vs. Simulation Results
102
4-5
Percent Differences Between Enumeration and Simulation Results 103
5-1
Sample Run Showing That Travel Time and Its Variance are Minimized at b - 1
142
6-1
Sensitivity Analysis Results for "DUAL"
161
6-2
Sensitivity Analysis Results for the Required Throughput Level
161
vii
LIST OF ILLUSTRATIONS
Figure
Page
1-1
One Aisle of Storage
1-2
Two Class Storage Assignment . . . . .
16
1-3
Representation of Continuous Rack Locations
16
3-1
Flow-Chart of the Simulation Program
51
3-2
The Effect of the Difference Between Total Number of Parts Received (in two consecutive weeks) on the Relative Space Requirement
52
The Conventional Method of Computing S/R Travel Time for Single and Dual Command Trips
67
Partitioning of a Rack Where Horizontal Travel Time is Dominating
74
4-1
4-2
4-3
4
Partitioning of a Rack Where Vertical Travel Time is Dominating
74
4-4
Location of the Square ABCD
87
4-5
Location of the Rectangle XYWZ
8 7
4-6
Scatter Diagram of Simulation Results
4-7
Alternative Configurations of a Rack with Three Product Classes The Statistical Distribution of Randomized Storage Single Command Travel Time
4-8 4-9
......
92
96 1 Q
6
The Statistical Distribution of Randomized Storage Dual Command Travel Time
H3
5-1
The Total Cost Function for Fixed Number of Aisles
131
5-2
Allocation of Three Product Classes Over a Rack that is Square-in-time Flow-chart of the Optimization Algorithm
1^0 145
5-3
viii
Figure
Page
6-1
Data of Sample Run Number 1
149
6-2
Solution to Sample Run Number 1
15°
6-3
Data of Sample Run Number 2
152
6-4
Solution to Sample Run Number 2
153
6-5
Data of Sample Run Number 3
156
6-6
Solution to Sample Run Number 3
157
6-7
Data of Sample Run Number 4
158
6-8
Solution to Sample Run Number 4
159
ix
NOMENCLATURE
AS/RS - Automated Storage Retrieval System BHEHT - Building height BLTH - Building length BWTH - Building width Class Based TOR Storage - Dedicated Storage DC - dual command (an S/R trip performed on a dual command basis) DEPTH - unit load depth Full TOR Storage - An extreme case of dedicated storage where every pallet is assigned to a particular opening HEHT - unit load height hv - horizontal travel velocity of the S/R ICOL - denotes the rack length in number of columns NOAI - number of storage aisles NOL - denotes the rack height in number of levels. SC - single command (an S/R trip performed on a single command basis) S/R - Storage/Retrieval Machine TNOA - Total number of openings in the warehouse w
- vertical travel velocity of the S/R
WT - the amount of time an order spends waiting in the queue (waiting time) WTH - unit load width A - the throughput level demanded from the system.
X
SUMMARY
This study is concerned with the development of an algorithm for constructing a minimum cost automated warehouse.
The cost elements
considered include the land cost, the building cost, the rack cost, the storage/retrieval machine cost and the annual maintenance cost associated with the building and S/R's.
The primary decision variables
are the number of aisles, the number of levels and the number of columns in each aisle. Two important variables that affect the final solution are the total number of openings required and the throughput level demanded from the system. inventory system.
The former has been analyzed for a particular A method is presented for determining the total
storage requirements of the warehouse based on the available inventory records.
The approach is demonstrated within the framework of a hypoÂ
thetical example. The throughput capacity of the system is a function of both the S/R travel time and the storage method.
Hence, the mean and variance
of the S/R travel time for single and dual command have been analyzed for both randomized and dedicated storage methods.
Closed form
expressions to compute S/R travel time are presented for randomized storage.
In addition, single and dual command travel time distributions
are derived for randomized storage. Combining the above findings with the algorithm to minimize total present worth cost offers the user a method to arrive at a final design
xi
that satisfies the throughput requirement while minimizing cost. Lastly, sensitivity analysis has been performed for those variables the user has to provide.
1
CHAPTER I
INTRODUCTION Automated high-rise warehousing systems are having a dramatic effect on the design and operation of large capacity, high volume storage facilities.
High-rise, storage is one of the fastest growing
areas of material handling.
It has the ability to store thousands of
unit loads and retrieve any particular one within a few minutes, when under computer control.
Such systems make it possible to totally
integrate material handling and storage into the manufacturing and dis tribution processes.
The installations that are reported to be successÂ
fully operating have provided considerable savings in terms of space, time, manpower and the physical and inventory control over all materials in process or in storage.
Objective The primary objective of this study is to develop an algorithm to aid in the design of automated warehouses.
The purpose is to give
the user the ability to quickly develop final designs and their associated costs for different parameters.
It is necessary to understand that the
algorithm does not provide the detailed design. With the high number of design options provided by numerous firms, it would have been impossible to consider the detailed design and still maintain the flexibility of the approach.
Hence, the user should view the resulting cost as a
benchmark to compare various design alternatives.
In developing the
2
algorithm, care has been exercised to keep it as general as possible.
Importance of the Study This study examines automated high-rise warehousing systems. Economic justification for such systems is often complex. savings alone are usually not enough.
Labor
Some other factors that are
often included in the justification are:
- Lower building cost - Better space utilization - Improved inventory control - Lower land cost - Increased throughput - Less damage to stored material - Increased automation opportunities
Automated storage and retrieval systems generally require considerable investment in capital and tend to be more inflexible than manually operated conventional storage systems.
Estimating the
system performance and the associated costs becomes extremely impor tant in making investment decisions for such systems. Hence, a quantitative approach is essential for decision making. In the light of the above fact, an attempt was made to learn more about the decision process employed by various firms operating in this field.
At the 1978 Material Handling Show (sponsored by the Material
Handling Institute, Inc.) 34 firms joined the exhibition as automated handling system manufacturers.
Several among these firms were actually
3
AS/RS manufacturers.
To the author's knowledge, none of these firms
had a general purpose design package.
Instead, most of them reported
that they had various design procedures based on judgement and past experience.
Furthermore, these procedures were tailored to the type
of equipment manufactured by the firm.
Foundation Material Various designs are available for high-rise storage systems. This study considers one basic type.
Product(s) is stored in pallet
racks whose height is a decision variable.
Each storage/retrieval machine
(S/R) operates in a single aisle with pallet racks on either side. Figure 1-1 portrays an S/R traveling between two rows of racks. The configuration of one S/R and two rows of racks is defined as one aisle of storage.
The aisle consists of a number of columns (up to L ) ;
height is indicated by the number of levels (up to H).
Thus, the total
number of pallet loads stored in one aisle is equal to 2 x H x L (assuming one load per opening).
Only one S/R accesses a given rack.
It
moves vertically along a support and the support moves horizontally along a track.
Both motions can occur at the same time so that the
S/R moves diagonally.
Thus, travel time to a particular opening is
determined by the maximum of the horizontal and vertical travel times. There are two options available for the S/R itself.
It can either be
a fully automated S/R with no operator or one with a man-on-board. In the first case the S/R receives instructions via the computer; in the latter case, the operator picks up a list of orders from the inputoutput point and operates the S/R accordingly.
The operator or the S/R
Figure 1-1.
One Aisle of Storage , Reprinted from (19) with permission of the author.
5
may access the product in either of two ways.
The shuttle or forks
can move in and out to either side to pick up full or empty pallets. Since the aisle is only a little wider than the S/R, the S/R cannot turn around while traveling back to the origin.
The origin, also called the
input-output (I/O) point, the pick up and delivery (P/D) point, etc. consists of space for a few pallets to queue outside the aisle at a level low enough for either fork trucks or conveyor I/O.
The fork
trucks or conveyors deliver product for storage in the racks and pick up product for shipping. A number of terms remain to be defined. command and dual command.
The first two are single
Single command is used simply to refer to
the storage or retrieval of one pallet load.
The S/R starts from the I/O
point, travels to the required opening, stores or retrieves the pallet load and returns to the origin. in the following manner.
Dual command involves two pallet loads
The S/R leaves the I/O point with a load to
be stored, travels to the particular opening, stores the load, then travels to another opening from which a load is retrieved, and returns to the origin.
As can be seen, the dual command calls for simultaneous
storage and retrieval orders. Two additional terms encountered in the warehousing area are dedicated and randomized storage.
Some use randomized storage to
refer to the situation where the location for a load to be stored is the closest open location and a first in, first out retrieval policy is used.
Such a policy is called "randomized storage" since, at any
point in time, the distribution of pallet loads of various products
6
over the rack looks as if they were randomly allocated to the openings. In this thesis, randomized storage refers to the situation in which both storage positions and retrieval positions are selected on a purely random basis. Dedicated storage is used when each opening is assigned to a particular product and storage/retrieval is performed randomly withr'i a given set of openings assigned to one product. As will be seen later, assumptions relating to single-dual command travel time computations and type of storage method used have important effects on system through put. This study has made the following assumptions: 1.
Each pallet contains only one item type, i.e., pallet assignment (the assignment of multiple items to the same pallet) will not be analyzed.
2.
It is assumed that all openings are the same size. One pallet load is stored in each opening.
3.
The storage orders are performed on the FCFS basis. The retrievals are on a FIFO basis.
4.
Dual cycles will be performed on a percent basis (the percent of S/R trips made on a dual basis is to be provided by the user).
5.
The S/R operates either on a single or on a dual command basis, i.e. multiple stops are not allowed.
6.
Each item is carried in every aisle. Therefore, transfer cars are not considered, i.e. there is one S/R per aisle.
7.
Each aisle is equally likely to be selected when a storage and/or retrieval order arrives.
8.
Decision variables such as unit load dimensions, unit load weight, and_ S/R logic and travel velocities are assumed to be provided by the user. The decision variables considered are the number of aisles, number of levels, and the number of columns.
7
Description of the Problem Assuming that the unit load dimensions, the total number of slots required, S/R travel velocities and values for other parameters to be discussed later are given, the problem is to find the' design that will minimize the total present worth cost, yet satisfy the throughput requirement.
The final design will indicate the number of aisles and
the number of levels and columns in each aisle.
It will also provide the
system throughput capacity with and without including the waiting time associated with each storage/retrieval operation.
In addition to these
variables, the building dimensions for the resulting design are provided. One side-problem encountered has been the computation of the expected travel time of the S/R in filling storage/retrieval orders. As stated before, travel time is crucial in determining the throughput capacity of a given design.
Hence, a separate chapter (Chapter IV)
treats the analysis of travel time. Literature Search Earliest Research Efforts Previous publications on AS/R systems date back to the late 1960's with the subject of interest being the stacker cranes.
Except
for those which describe successful applications, there are very few articles (10, 11, 14, 17, 18) that review the advances made in this area. As indicated in Bafna's (2) literature survey, prior to 1972 there have been several attempts to develop a design methodology for stacker crane systems.
One of the earliest known attempts to be made
was by Thompson and Cnossen (15) at the Ford Motor Company.
They
developed a simulation model for a computer controlled warehouse.
Their
8
study is reported to be a model which shows how the warehouse would perform under alternative operation rules.
The major drawback of their
study is that it did not consider the cost of the system. simulation study appeared in late 1969.
Another
The program was developed by the
Storage Systems Department of Clark Equipment Company; Schwind (14) has only made mention of it.
Bafna (2) reports this simulator to be
insufficient for two reasons.
First, it does not simulate a complete
system but only the aisles associated with one S/R.
Secondly, the
simulator is not tied to the program which yields the cost of the system.
Bafna (2) also presents an analytical model developed by
one of the firms operating in this area.
It consists of a formula to
find the optimum number of aisles, N, given the fraction of total inventory moved in and out per day, the number of openings visited by the crane per trip, the depth of stacking, the horizontal travel speed of the crane, the required fork cycles per trip and the fork cycle time. However, how it takes into account the various cost elements involved in such an analysis is not mentioned.
In addition, the system throughput
is not considered and above all, the height of storage (the principle advantage of these systems) is ignored.
Bafna (2) concludes the survey
by stating that there has not been much work done in this area. Following Bafna (2), research in the automated warehousing area has increased.
It can be grouped under two main categories:
that done
by firms which manufacture AS/R systems and by the Material Handling Institute (MHI); and that done in the universities.
9
Recent Research.:
Commercial Sector
Work done by private firms addresses two areas. The first is the analysis of S/R travel time under various circumstances and the second covers economic justification for these systems. Sand (13) has analyzed the S/R travel time for a rack with fixed picking area and randomized reserve storage. not considered turnover-based storage assignment.
However, she has She assumes a fixed
picking area with two pallet openings per product and random reserve area. First she presents a two part heuristic that minimizes the number of picking lots and then minimizes the travel time for each lot. Then, assuming that the picking area is square in time, the average travel time to the reserve area is found.
However, she does not complete the
analysis to find total travel time for replenishment, stating that it is not straightforward. She concluded her study by indicating that further research is necessary for analyzing storage (putaway) and possibilities for dual cycles (interleaving). Barrett (3) used a simulation model to evaluate the effectiveness of various rules for order picking systems that allow multiple stops for each crane trip into the aisle.
Four heuristic methods are investiÂ
gated for their ability to reduce the number of picking lots. model tests system performance for three factors:
The
storage assignment
(the assignment of pallet loads to storage locations), lot assignment (the assignment of an order to a picking lot) and batch quantity (the number of orders from which picking lots are formed).
System performance
is measured by the number of lots, percent of minimum lots, number of
10
stops and travel time (comparisons are given for inventories with 1
different ABC* characteristics and customer ordering patterns).
For
combined effects of lot assignment and batch quantity it compares the effectiveness of the above mentioned four heuristics in regard to number of lots, percent of minimum lots and number of stops.
From the standÂ
point of travel time it is stated that it can be reduced by 20% - 35% when turnover-based storage is used instead of random.
However, all of
the results are highly empirical. The second area of research for private firms has been the inves tigation of the factors which effect the total cost of an AS/RS.
One
such study is due to Allred (1). It mainly covers the justification of automated storage systems, and the techniques for purchasing and constructing such systems. A simple hypothetical model is presented to compute the cost difference between a conventional and an automated storage system. (How the cost of the automated system has been computed is not stated).
The two systems are compared on three bases:
the
tangible annual costs, less tangible annual costs, and the least tangible annual costs.
Both systems are evaluated for the receiving, storage
and order picking functions on the basis of material queue, paper work queue and average elapsed time.
The study is concluded with a discussion
on methods of system procurement. Another study that develops a basis for estimating AS/RS invest ment cost is presented by Zollinger (1.9) . He has compiled detailed cost information for more than 60 AS/RS installations. Assuming that the data related to the particular design under question is given, cost estimates are developed for the following elements:
11
1.
Rack prices including installation and stacker support rail.
2.
Storage machine price including controls, electrification, guiderail and installation.
3.
Building price for conventional construction including services and sprinklers (construction height between 25
T
and 85'). In addition, a method of computing building dimensions is presented, including the required clearances to be added.
The cost model used
in this thesis is based on that of Zollinger (19). One slight modifica tion has been the addition of land cost and recurring costs to the above list of cost elements. Hence, to avoid duplication, details of this approach are left to Chapter II. The Material Handling Institute, Inc. (MHI) has several publica tions related to the study. nature.
However, they are mostly descriptive in
One of them (.9) deserves mention; it has been developed by the
member-companies of the AS/RS Product Section of the Material Handling Institute.
It is a useful source for quick reference and discusses the
following areas: - system objectives
- system installation
- system performance
- system acceptance procedure
- system engineering
- system warranty
- system control
- system maintenance
- system hardware
- system glossary
- system justification The considerations given in (9) are basically recommendations. They are intended only to provide guidelines for technical procedure
12
and contain information that may be helpful to the purchasers and/or users of AS/R systems. Rygh (12) discusses the common uses, configurations and benefits of four basic types of systems mentioned as unit load S/R systems, orderpicking systems, in-process systems, and combined systems. Also, the effect of new computer technology on such systems is briefly explained. Recent Research:
University Sector
From the standpoint of work done at universities, there are very few published articles.
Hausman, Schwarz and Graves have analyzed
S/R travel times and published their findings in
(5), (6) and (7).
In (5) it is assumed that the rack is square in time, i.e. the vertical and horizontal speeds are such that the time to reach the row most distant from the I/O point equals the time to reach the most distant column.
In addition, interleaving (dual command) is ignored.
Expected one-way travel is computed for the discrete case as follows: let, R = number of rows in each rack, C = number of columns in each rack, N = number of pallet storage locations in one rack (N = RC), y^ = the ranked one-way time for the S/R machine to travel from the I/O point to location i, i = 1, y
y
i " i+l'
N; ranked so that
all i < N.
the turnover of the item on pallet j.
It is the number
of times that the item on pallet j will be stored (and subsequently retrieved) per unit time.
For convenience
the A. are ranked so that A. 2 A for j < N. j+1 J J
13
Under random storage assignment the expected one-way travel time for a pallet, T , is: K
N
1 T
R
=
y
=
N. V i 1=1
With the highest turnover pallet assigned to the closest location, the expected one-way travel time is: N T = T
1 = 1
N I X. Âą-l
1
To switch to the continuous representation of the rack, the travel time is first normalized.
So, the travel time to a location in
th the k
percentile is:
y(k) = k^
0 < k < l
The turnover distribution is derived by making use of the "ABC" phenomenon for inventories and the basic EOQ model.
The "ABC" curve
is represented by the function:
G(i) = i
S
for
0 < s < 1
where demand is measured in full pallet loads. Let:
D(i) = demand rate (pallets per unit time) of item i, Q(i) = the economic order quantity of item i.
14
By definition G(i) = i
S
1 i i = / D(j) dj / / D(j)dj 0
0
s
Letting / D(j)dj = 1, i
s
0
i = / D(j)dj 0
which has the solution: .s-1 D(i) = si
0 < i < 1
Since all items are ordered using the EOQ model, then Q(i) = (2KD(i))
2
where K is the ratio of order cost to holding cost. Hence the average turnover is:
2D(i)/Q(i) = (2T)(1)/1L)^. th
In order to determine the time index, i, of the j
pallet, the
following equation has to be solved: j = /
2
((2KD(k)) /2)dk
0
If j is normalized by letting j = j/L (L = total number of rack locations required for the average inventory of all items), then
i(j)
Hence,
= D K j )
.2(s-1)/(s+1). =
Therefore, the turnover of the j
S J
pallet is
5
! 5
( s
X(j) - (2DJ/K)* = ( 2 s / K ) ( j -
1 ) / ( s + 1 )
)
th
With the above results, it is shown that the one-way travel time for randomized storage is:
E
y(Âą)
\ - [ ]
= 2/3 0
and for full-turnover based storage it is
1 f *(j>y(j)dj 4s (5s + 1)
/ X(j)dj
It is also shown that for typical inventory distributions, the percentage reduction in S/R travel time ranges from 26% to 71% when full-turnover based storage is used instead of random. Another storage assignment rule which is similar to full TOR based storage is class-based turnover (dedicated) storage.
For a two
class system with the Class I region to be used for the higher turnover pallets, and the Class II region for the lower turnover pallets, the one-way travel time is (see Figure 1-2):
T (R) =
=0
2
1
i=R
2
1 / X(j)dj 0
where R = the partitioning value y
= average travel time to region K.
1 1
16
Clou
Clou I
Figure 1-2.
J
Two Class Storage Assignment (Reprinted from (6))
Cc)
i
i i
lb)
(â&#x20AC;˘)
I/O Point
Figure 1-3.
0
ii
j
I
J
Representation of Continuous Rack Locations (Reprinted from (7))
17
Since y^. = yR and
y
3
IZ
2
= | (1-R ) / (1-R )
the above equation will reduce to
1
2
(5s+l)/(s+l)
R
+
( 1
_ 3 R
) ( 1
_ 4s/(s l) R
+
)
/
( 1
_
r 2 )
The percentage inprovement over random storage for this case ranges between 18% and 53%. Of course, the percentage improvement increases as the skewness of the inventory distribution increases, i.e. as s approaches zero.
Similar results are presented for a three
class system. This study is concluded with a comparison and continuous representations of the rack. in the comparison:
between the discrete
Four cases are included
Random, Full Turnover, Two-Class, and Three-Class.
The continuous approximation performs well only for the random storage case.
For a 20%/60% inventory distribution (i.e. 20% of the items
represent 60% of total demand) the continuous model underestimates the S/R travel time by 6%, 3.4%, and 7.2% for the full TOR, Two-Class and Three-Class systems respectively.
Likewise, for a 20%/80% inventory
distribution, the travel time is underestimated by 24.8%, 16%, and 22%. As the skewness increases, the underestimation gets considerably larger. Following the above paper, Graves, Hausman and Schwarz published two complementary papers (6) and (7). The assumptions stated for (5)
18
also hold for their subsequent studies, with the exception that interleaving (dual cycle) is not ignored.
In (6) an expression is
developed to find the expected interleave time between any two percentile sets of rack locations.
Specifically, the expected S/R travel time from
the set of rack locations i units from the I/O point to the set of rack locations j units from the I/O point is (see Figure 1-3):
i
E[d(iJ)j
12ij
3
17i
2
- 9i j + 12ij" ! if 2i < j
3
2 2 3 - 33i j + 24ij - 2j
if 2i > j
(1-1)
where 0 ÂŁ i < j * 1. The expected interleave time for the randomized storage method is determined using the following expression:
L = 2 / / E[d(i,j)1 f(i,j) djdi i=0 j=i L
(1-2)
J
where f(i,j) is the joint density of travel from the set of locations i time units from the I/O point to the set of locations j time units from the I/O point.
Using the above expression for E^d(i,j)J
and by
decomposing the integral boundaries accordingly, L is found to be 7/15. The expected interleave time is also developed for the full TOR based storage.
Given full TOR, the probability that an item is within
i time units of the I/O point, F(i), satisfies:
i
19
2 l
F(i) = |
= i
s + 1
/ A (j) dj 0
A s
Thus,
f ( 1 )
di
or
-1 o < i * 1
=
s+1
f(i) = 2 z i
2z
1
where z = 2s/(s+l)
2 2z - 1 Consequently f(i,j) = 4z (ij) . When f(i,j) is substituted in (1-2), with (1-1) used for E [d(i,j)],
L
=
l48z 3
3
2
+ 3 6 z + 42z -
' ( 4 z + l)(2z
+2)(2z
48 + 9 6 ( 2
2
z
+ l)(2z)(2z -
2
)
v
(
1
_
3
)
1)
is obtained. The expected interleave time in a "square-L" region is also determined.
Assuming that storage/retrieval within the region is
randomized, the expected interleave time E Jd(I,J)J, is
E|d(I,J)] = 2
J J / / E[d(I,J)] f(i,j) djdi i=I j=i
(1-4)
The probability of a store/retrieve within iunits of the rack, F(i), is
.2 _ 2 F(i) = -^-=â&#x20AC;&#x201D;Âą-=- for I < i < J a - r) z
20
2
Differentiating F(i) yields:
1
f(i) = — for I £ i £ J. (J - I ) by the independence assumption: Z
4 i j
f(i,j) = f(i)f(j) =
9
9
Therefore,
Z
I - i, j - J
?
(1-5)
(J - O
Now substitute (1-5) for f(i,j) and (1-1) for E|d(I,J)J in (1-4). Evaluating the resulting expression in pieces gives:
3
2
2
3
E[d(I,J)| = ^-(13J + 36J I - 21JI + 32I ) / (J+I) J
2
if J < 21
J U
= ~(14J
(1-6) 5
3
2
2
3
4
2
2
2
- 40J ! + 30J ! - 5JI ) / (J -I ) if J > 21
The study is also concerned with the determination of the expected interleave time, L, given two-class storage assignment.
Conceptually,
L is:
L = p(I,I)L + p(II,II)L I
II
+ 2p(I,II)L
I z l
(1-7)
Under randomized storage (1-7) can be written as:
7/15 = R* " L
2
r
2
+ (1 - R ) ' L
2
I X
is found from (1-6) by setting 1 = 0 and J = R^. found by setting I = Rj- and J = 1 in (1-6). from (1-8) as:
)L
+ R (l - Ri I,II
( 1
~
8 )
Likewise, L ^ is
Hence, L^ ^ will be found
21
L
I,II
~k
=
( 4
° "
3 0 R
1
+
5 R
1 "
1
4
R
1
} 1
( 1
R
" l
}
i f
R
<
l
h
(1-9) (1-9R + 71R - 39R + 46R ) / (R (l +R )) 2
=
3
4
if R > h
2
±
X
±
Under full TOR, the probability of an item being within R^ units of the I/O point, F(R ), is: 1
4 s
F( ) = R /<
s + 1
2 z
> = R
Rl
Consequently, p(I,I)
= R
4 z
p(II,II) = (1 - R and
2 z
)
2
Z
p(I,H)
Z
= R* (1 - R^ )
Substituting the above probabilities into (1-7) with the corresponding expressions obtained for L ^ , ! ^ and L^.
one obtains the following
expression for L :
L =
[(R
+
2[R
L =
[(R
+
2[R
2 Z
2
) (7R /15)] 1
2
Z
2 Z
(1 -
R
2
Z
)
2
) (7R /15)j 1
2
Z
(1 -
R
2
2
)
+[(1
-
R
' -^(40 -
+
-
Z
)
2
30R
[(1-R
' -^(1
2
2
Z
9R
)
X
1
2
' -^(14 -
40R
2
3
+
5R
-
' 3^(13 +
71R
2
-
+
2
+
30R
14R )
/
(1 -
36R
-
21R
39R
1
3
4
+ 46R )
3
-
4
5R )
2
R )]
2
+ /
32R (R (l 2
/ R
if
3 )
(1 <
±
1
+
if
(
1
+
R
>
R
1
h
Graves, et al, considered the case in which the retrieve portion of the dual cycle is not FCFS, instead the retrieve order is selected
)
2
)
2
J
k
R ^ ) ] R,
2
]
22
from a queue of length K.
Let
that the store is in class X.
RT =
be the expected round-trip time given Then the round-trip time, RT, is:
PCDTj + PUDTjj
(1-10)
2z 2z It was previously shown that p(I) = R^ and p(II) = 1 - R^ . Now consider T^:
the expected time for the store is the expected one-way
time in class I, W^.
The expected time for the retrieve depends on
whether or not an inspection of the first K retrieves yields a retrieve of class I.
Let X represent the number of class I retrieves in the
retrieve queue, 0 - X - K.
Then, given independence between stores and
retrieves, as well as the default policy (if no match is found, select the first retrieve), X is a Markov-Chain whose steady-state probabilities, 7T
x >
are derived and,
after evaluating T^ and T ^ , substituted in (1-10).
The resulting expression for RT is quite complicated.
The approach
becomes much more tedious for three-class systems. In ( 6 ) the analysis concludes with a treatment of the additional rack openings required in order to keep fixed the default probability (the probability of not being able to store an item in a desired loca tion) • It has been shown that the steady-state probability of n in the system is Poisson with parameter m - X/y (assuming a M/G/°° queue with expected arrival rate X and expected service rate y).
Considering a
rack with m + k locations, the problem is to find how large must k be in order to ensure that the probability of n exceeding (m + k) is some acceptably small number.
For large numbers of openings the Poisson
23
distribution with parameter m can be approximated by the Normal distribu tion.
Consequently, the ratio of k to m, representing additional rack
locations required as a fraction of m, is:
5
(k/m) = 2.575 (c/m)*
where c is the number of classes (classes are assumed to be equal size). The companion paper, (7), extends the work done in (6). Using both the continuous and discrete approach, it compares the operating performance of several storage assignment/interleaving policies.
The
expected round-trip time is the expected time for the system to complete one store and one retrive.
In mandatory interleaving (MIL) policies
the expected round-trip time is twice the expected one-way time plus the expected interleave time.
In no interleaving (NIL) policies, the
expected round-trip time is four times the expected one-way travel time. Results are presented for the policies shown in Table 1-1.
24
Table 1-1.
Storage Assignment/Interleaving Policies
1 - RAN/NIL/FCFS:
Random Storage Assignment; No Interleaving FCFS retrieve queue
2 - FULL/NIL/FCFS: Full TOR based Storage Assignment; etc. 3 - C2/NIL/FCFS:
2-Class Storage Assignment; etc.
4 - C3/NIL/FCFS:
3-Class Storage Assignment; etc.
5 - RAN/MIL/FCFS:
Random Storage Assignment; Mandatory Interleaving; FCFS Retrieve Queue
6 - FULL/MIL/FCFS 7 - C2/MIL/FCFS 8 - C3/MIL/FCFS 9 - C2/MIL/Q = K: 10 - C3/MIL/Q = K
Selection queue of K retrieves
25
For a 20/80 inventory distribution, expected round trip time for above listed cases has been found as shown in Table 1-2 (discrete enumeration results are given in parenthesis; the third number denotes percent differences):
Table 1-2.
Expected Round Trip Times
Continuous Rack
Discrete Rack
Percent Difference
1
2.667
(2.664)
-.11
2
1.310
(1.640)
20.12
3
1.709
(1.933)
11.59
4
1.501
(1.774)
15.39
5
1.800
(1.795)
-.28
6
1.041
(1.223)
14.88
7
1.261
(1.390)
9.28
8
1.145
(1.296)
11.65
9
1.119
(1.328)
15.74
10
1.072
(1.247)
14.03
From the above results it can be readily seen that, except for randomized storage, the continuous model always underestimates the true answer. percent underestimation varies between
The
9-28% and 20.12%.
Bafna ( 2 ) used a simulation model to estimate the values of the principle parameters of an AS/RS.
The number of aisles, the rack height,
the number of S/R's, and the horizontal and vertical velocities of the S/R's are the design variables.
From the standpoint of transfer cars, it
is assumed that any S/R having more than one aisle assigned to it must
26
have a transfer car (a conservative assumption) . The entire approach is composed of three stages:
constructing the cost model, developing the
simulation model and combining the cost model with the simulator in order to arrive at a design that minimizes total cost. The cost model includes those costs that change with variations in the warehouse and S/R parameters.
They include cost of floor space
(land), cost of building, cost of racks, cost of S/R's, cost of transfer cars, and cost of fire protection.
The equivalent annual cost of the
system is found by computing and summing the capital recovery for each cost element.
The annual labor cost is added to the above total in order
to find the equivalent annual cost for the entire system.
However,
investment tax credits are ignored and the calculations are based on before tax returns. The simulation model operates as follows:
the initial values
(lower limits) of the number of aisles, rack height, the number of S/R's and the horizontal and vertical travel speeds of the S/R's are provided by the user. variables.
The throughput is simulated with this initial set of
The calculated value of throughput is transferred to the main
program and compared with the last throughput.
If it has increased,
a new series of variable sets are generated and the one giving the lowest annual cost is transferred to GASP to simulate the next throughput. If the throughput has not increased, the next set of variables from the series generated earlier is transferred to GASP and
the throughput is
calculated (the method by which a new series of variable value sets is generated is discussed in detail in (2)). This cycle is repeated until the calculated throughput satisfies the throughput demanded from the
27
system.
Every satisfactory throughput is resimulated with a longer
simulation time to assure steady-state.
If the revised value of throughÂ
put still exceeds the required level, simulation stops and final values of the variables are printed.
Otherwise, the procedure described above
starts repeating itself until the throughput requirement is satisfied. The above described simulation model is analyzed with a sample run for an AS/RS with 5900 openings and 90 operations per hour through put capacity.
The solution is reached after 24 iterations.
Required
computer time is stated to be 243 decimal seconds on a CDC 6500 computer. The final horizontal and vertical velocities of the S/R's are equal to their initial values. Lastly, a sensitivity analysis is presented.
The
sensitivity of the cost to changes in the throughput of the system is analyzed.
It has been found that the incremental cost per unit throughÂ
put is lower for higher throughputs.
Other results reached by Bafna are
subject to discussion and will be presented in Chapter VII. Based on the literature survey, it can be concluded that previous analytical approaches failed to combine their results with an appropriate cost model and ignore any throughput requirement. Hence, this study has been performed to develop a general purpose design package based on analytical findings. Summary The purpose of this study is to develop an analytical approach to design automated warehouses.
The objective is to minimize the total
cost of the system, subject to the throughput requirement imposed over the system.
The literature search does not indicate the existence of
a similar study that has been published.
28
The above mentioned goal is reached in several stages. First, a method is presented for determining the total storage requirements based on inventory records (Chapter III). Chapter IV is concerned with the analysis of S/R travel time in regard to randomized and dedicated storage methods.
The cost functions are analyzed in Chapter V.
Based
on this analysis and previous findings from Chapter IV, an algorithm that will minimize the total cost and satisfy the throughput requirement, is developed.
Chapter VI is concerned with the discussion of sample
run results and sensitivity analysis. conclusions and recommendations.
Lastly, Chapter VII presents the
29
CHAPTER II
DEFINITION OF THE TOTAL PRESENT WORTH COST
Introduction The total cost of an AS/RS is composed of numerous elements. They can be identified as the land cost, the building cost, the rack cost, and the S/R cost.
These cost elements can be viewed as the
initial cost of the system.
Of course, there are other costs that can
be classified as the initial cost.
They include the required hardware,
acquisition of the software, investment required in the service area, etc.
However, such costs remain constant irrespective of the system
design.
Therefore, analyzing only those costs which vary with the design
will serve the purpose of comparing alternative designs. Recurring costs, on the other hand, are composed of those costs associated with maintenence, direct personnel, secondary personnel, utilities, inventory carrying, etc.
Once the total number of unit loads to be accommodated by the
system is determined, such costs tend to show little variation
with
changes in the design, except for the maintenance cost on the building and on the S/R's.
One other exception is the labor cost incurred
if the S/R's are of the man-on-board type.*
*The costs obtained in this chapter are based on 1978 costs, verified by Zollinger in October 1978 and communicated by him in a private conver sation at the Automated Material Handling Systems short course at Georgia Tech.
30
Cost of Land The importance of land cost will vary with the land price. Depending on the location of the warehouse, it may have a considerable effect on the final design.
The total area taken up by the warehouse
will simply be the base area of the building.
Let
BLTH = building length (in feet) BWTH = building width (in feet) LCOST = unit land price ($/square foot)
Then total land cost, say, C
= BLTH x BWTH x LCOST dollars.
A method
of computing BLTH and BWTH will be presented under Building Cost.
Cost of Building The method used to compute the building cost is referred to as the square-foot-floor-area method.
Hence, as for the land cost, the
base area of the building is required to compute the building cost. The length and width of the building is found by using tables provided by Zollinger (19). Let WTH
unit load width (in inches)
HEHT
unit load height (in inches)
DEPTH
Then
unit load depth (in inches)
ICOL
number of columns of storage
N0AI
number of aisles
31
Rack length
HDST
(WTH + 8") x 1/12 x ICOL (in feet)
Rack height
VDST
(HEHT + 10") x 1/12 x NOL (in feet)
Storage width
S = w
3 x (DEPTH + 6")x 1/12 x NOAI (in feet)
Storage width is found assuming that sprinklers are provided. Next, the building dimensions can be found by adding the proper clearances. Building length = BLTH = HDST + IADD* Building height = BHEHT = VDST +4.0* Building width (not rack supported) = BWTH = S + 2.0* w where IADD = an estimate of the space required for P/D stations, S/R extension beyond the end of the rack and conveyor or truck aisle. The amount of allowance depends upon the unit load width. Zollinger (19) provides an estimate as shown in Table 2-1. These allowances are given assuming no transfer cars. For practical and programming purposes, the following regression line was derived from the table. IADD = 12.504243 + 0.447478 WTH The 4.0
1
added to building height is due to the fact that
the lower storage level cannot be at floor level and the ceiling is not put directly on the top of the upper storage level. Thus, an allowance of 18" from the floor level and an allowance of 20" above the upper storage level has to be added.
(Sprinklers are assumed
to be in the flue spacing.) The 2.0' added to building width is the allowance provided for the space between the walls and rack.
This allowance need not be
added if the building is rack supported.
Table 2-1.
Minimum Building Length Addition.*
ADD TO RACK LENGTH Pallet Length
System Without Transfer
System With Transfer
T
30" 36"
26 29'
42' 45
40" 42"
30' 31
47' 48
48" 52"
34 36
Table 2-2.
1
1
1
T
52 54
1
Conversion Factors for Unit Building Cost.*
CLEAR HEIGHT (ft.)
FACTOR NUMBER
25 40 55 70 85
Table 2-3.
1 1
1.00 1.25 1.50 1.90 2.50
Factor Numbers to Determine Rack Cost.*
LOAD VARIABLE
1
FACTOR NUMBER 2 3
Load Size (cu.ft.)
40
80
120
Load Weight (lbs.)
500
3000
6000
4
10
24
Rack Height (loads)
*Taken from (19) with permission of the author.
33
One remaining variable required to compute building cost is BCOST.
That is, the cost per square foot of the building base assuming
conventional construction (not rack supported) and including sprinklers. Zollinger (19) has proposed the following approach.
Let BCOST be the
1
cost incurred per square foot in building a 25 high building. base cost will increase with the height of the building.
The
Thus, if 25'
is taken as the base, Table 2-2 shows the factor numbers to be multi plied by BCOST in order to determine the corresponding cost per square foot.
The factor will subsequently be referred to as CFR (the conversion
factor).
Again for practical and programming purposes the following
regression line will be used to find CFR:
CFR =
0.986508 - 0.005349 BHEHT + 0.0002698 (BHEHT)
2
Hence, the building cost is obtained using the following expression:
Building Cost = Câ&#x20AC;&#x17E; = BLTH x BWTH x BCOST x CFR
Rack supported warehouses have increased in use. various advantages.
They bring
As given in (16)
A rack supported warehouse is supported entirely or at least partially by the actual rack structure. The weight of the warehouse roof and metal sidings or skins to enclose the rack structure normally does not significantly increase the load support characteristics of the racks. As an example, if unit loads weighing from 2,000 to 4,000 pounds are stored in a rack structure to a height of from 8 to 10 tiers, the floor loading due to the rack
34
and its contents can exceed 1,000 pounds per square foot. If the rack must also support the skins and roof, another 70 pounds per square foot must be accommodated by the storage rack. Thus, the rack supported building requires storage racks with the capability of handling a 7% increase in the load handled by conventional storage racks. Cost savings for rack supported buildings include the reduction of redundant steel structures for supporting the roof and siding. Erection cost is usually decreased since the roof and siding are added after the rack structure is complete. One estimate of the savings in fabricating and erecting a rack supported warehouse is that it can be as much as $6 per square foot less than a conventional warehouse with free-standing rack. In addition to the significant savings in the price of the rack supported warehouse, there exists another cost savings which can be even more significant - tax savings! A rack supported warehouse qualifies as equip ment due to its special purpose design. Hence, a writeoff period of 8 - 10 years, rather than 50 - 60 years, can be used. Furthermore, since the rack supported warehouse is treated as equipment, it qualifies for the investment tax credit. Hence, if the building is going to be rack supported, BC0ST x CFR will be decreased by 6.0.
Building Cost = C
Thus,
= BLTH x BWTH X
(BCOST x CFR) - 6.0
Rack supported buildings also provide tax savings, as discussed subsequently.
Cost of Racks The function of the racks is to support multiple levels of unit loads.
For rack supported buildings they also support the skin and
roof of the building. There are various options for designing racks, depending upon the type of unit load and storage module.
Conventional AS/RS racks have
35
load support rails with space between them for shuttle clearance. Conventional pallet racks have load beams parallel to the aisle. Taking all forces into account and performing the stress analysis, will dictate the type of design required.
Including external forces for rack
supported buildings is especially important. Zollinger (19) presents the following approach for estimating the rack cost.
It includes the cost of installation and S/R support rail.
Table 2-3 shows the corresponding factor values based on the load size, the load weight, and the rack height expressed in number of unit loads. The factor values are added together and then multiplied by $14 to find the cost per rack opening.
The figure obtained is then
multiplied by the total number of openings in the system to obtain the total rack cost.
The following example is provided:
the unit load
is 40" x 48" x 48" (depth x width x height) = 53.2 cubic feet. Load weight is 2,000 lbs, rack height is 10 levels. up to 1 + 2 + 2 = 5 .
Thus, the factor adds
Hence, the cost per rack opening is 5 x $14 = $70.
An alternative approach to factor value addition is to use a regression line. White (16) provides the following expression:
2
r
Cost/rack opening = $14 |0.92484 + 0.025x + 0.0004424y - 82,500,000 ^ 2
2
+ 0.23328z - 0.00476z ] where x = number of cubic feet in a unit load y = weight of the unit load in pounds z = height of the rack expressed in unit loads For the above example this line gives $69.27 per opening. Hence,
36
if we let TNOA = total number of openings ALS
= unit load size (cu.ft.)
WEGT = weight of unit load (lbs) then
TNOA = NOL x ICOL x 2.0 x NOAI ALS
= (DEPTH x WTH x HEHT)/1728.0 2
RCOST = 0.92484 + (0.025 x ALS) + (0.0004424 x WEGT) -
(WEGT) 82,500,000 2
Therefore, RCOS = [RCOST + (0.23328 x NOL) - (0.00476 6 xx NOL NOL )] ) x 14.0 and, Total Rack Cost = C
= RCOS x TNOA.
Cost of S/R Machines The basic function of the S/R machine, as the name implies, is to perform the storage/retrieval orders. Apart from other types of material handling equipment it is capable of performing high lifts, fast movements, and accurate positioning. mechanical drives.
It consists of three
The first is the horizontal drive which moves the
machine back and forth along the aisle.
The second is the hoist drive
which raises and lowers the carriage along the height of the rack.
The
third drive is the shuttle drive which moves the shuttle in and out of the rack openings while transferring the load.
It also powers the
shuttle for transferring the load at the P/D station. Based on Zollinger (19), the S/R cost is a function of the height of the AS/RS, the weight of the unit load, and the type and location of the control logic.
If the height of the AS/RS is less
37
than 30' then a base cost of $13,000 per S/R applies; if the AS/RS is in the range from 30' to 42' then an additional cost of $13,000 is added to each S/R; for heights from 42' to 60' an additional cost of $26,000 is added to the base cost for each S/R; and for heights from 60' to 85' an incremental cost of $39,000 is incurred (above the base cost). If the load weighs less than 1000 lbs, then $13,000 is contributed to the cost per S/R; for weight in the interval from 1000 lbs. to 3500 lbs., $26,000 is contributed to the cost per S/R; for loads weighing from 3500 lbs. to 6500 lbs., $39,000 is contributed to the cost per S/R; and for loads above 6500 lbs., $52,000 is added to the overall cost of each S/R. If the control logic is on-board a cost of $26,000 is contribu ted to the cost of each S/R; if the control logic is off the machine, a cost of $39,000 is contributed per S/R; and if a central console is used to control all S/R's, a cost of $52,000 is contributed to each S/R. As will be noticed the range of the AS/RS height and load weight for each interval is quite wide.
Thus, the use of a regression line
for this case will not provide a good approximation.
The Annual Labor Cost The annual labor cost is included in the analysis if man-on board S/R's are used. each S/R.
It is assumed that there is one operator on
Despite the increased labor cost, the cost per S/R will
decrease due to reduction of controls on the machine.
Recall that if
control logic is on-board a cost of $26,000 is contributed to the
38
cost of each S/R.
If man-on-board S/R is used, then instead of
$26,000, $13,000 is added to the cost of each S / R ^ .
Also, if we
let LBCOST = annual labor cost on a per operator basis
then, total annual labor cost, C
TTJ
, will be:
LiJ
C
T
= LBCOST x NOAI
($/yr)
Lb
The Annual Maintenance Cost The annual maintenance cost is composed of the maintenance cost on the building and the S/R's.
The building maintenance cost will be
found as follows, let: BMCOST = the annual maintenance cost per sq. ft. of the building base (to be provided by the user).
Then, total building maintenance
cost, Cj^g, will be:
C ^ = BLTH x BWTH x BMCOST
dollars/yr.
Furthermore, if we let: SRMCOST = the annual maintenance cost per S/R (to be provided by the user), then, the total S/R maintenance cost, C
, will be:
Above approach has been developed through verbal communication with Mr. H.A. Zollinger.
39
= NOAI x SRMCOST dollars/yr. (assuming one S/R per aisle.)
Hence, the total annual maintenance cost, C., will be:
°m - Sffi
+
SMSR
dollars/yr.
Present Worth Cost of the System The present worth cost of the system will be computed in two different ways, depending upon whether the building is rack supported or not.
The following assumptions have been made in determining the
present worth cost of the non-rack supported building: 1 - The planning horizon is ten years 2 - Land has no depreciation 3 - The S/R's, the rack, and the controls have a ten year write-off period, and no salvage value 4 - The building has a forty year write-off period and the salvage value equals the book value at the end of the planning horizon. 5 - The sum-of-years-digits method will be used for depreciation 6 - There is a 10% investment tax credit for the S/R's, the rack and for 50% of the controls (software is not eligible for tax credit).
40
For the rack supported building the following are assumed: 1 - The planning horizon is ten years. 2 - Land has no depreciation 3 - The S/R's, the rack, the controls, and the building have a ten year write-off period, and no salvage value 4 - The sum-of-years-digits method will be used for depreciation 5 - There is a 10% investment tax credit for the S/R's, the rack, the building and for
50% of the controls
It is also assumed that the after tax minimum attractive rate of return (MARR) and the applicable income tax rate will be provided by the user.
The approach can be demonstrated by an example.
Suppose the
following are given: ATMARR = 10% Income Tax Rate = 50% Land Cost = $100,000 Building Cost (non-rack supported) = $300,000 Building Cost (rack supported) = $250,000 Rack Cost = $300,000 S/R Cost = $450,000 (includes $125,000 for controls) Total Recurring Cost = $18,000 Sum-of-years-digits = ^ y t i l ^ for n = 10, SOYD = 55 and for n = 40, SOYD = 820. The annual depreciation figures for the building, rack and S/R's are tabulated in Table 2-4.
(Building 1
denotes non-rack supported construction, likewise 2 denotes rack supported building.)
From the table it is found that:
Table 2-4. The Annual Depreciation on the Building, Rack, and S/R's
Rack
Bldg 2
Bldg 1
Year
S/R
(P/F,i,n)
In
2n
1
14,634.12
45,454.525
54,545.43
81,818.145
0.9091
137,271.99
165,290.83
2
14,268.27
40,909.075
49,090.89
73,636.335
0.8264
113,213.07
135,229.03
, 3
13,902.42
36,363.625
43,636.35
65,454.525
0.7513
92,404.858
4
13,536.57
31,818.175
38,181.81
57,272.715
0.6830
74,440.914
86,927. 254
5
13,170.72
27,272.725
32,727.27
49,090.905
0.6209
58,978.604
67,734.539
6
12,804.87
22,727,25
27,272.7
40,909.05
0.5645
45,716.946
51,318.13
7
12,439.02
18,181.8
21,818.16
32,727.24
0.5132
34,376.404
37,323.599
CO
12,073.17
13^636.35
16,363.62
24,545.43
0.4665
24,716.205
25,445.429
9
11,707.29
9,090.9
10,909.08
16,363.62
0.4241
16,531.413
15,421.802
10
11,341.44
4,545.45
5,454.54
8,181.81
0.3855
9,628.938
7,009.084
10 I n=l
10 P W
ln
=
6 0 7
'
2 7 9
'
3 0
E n=l
PW = 700,979 .61 2n 0
109,279.95
42
10 J PW. = 607,279.30 In n=l L
10 J PW n=l
and
0
= 700,979.61
Hence, the present worth of tax savings due to increased depreciation is: for non-rack supported = (607,279.30)(0.50) = 303,639.65 for rack supported
= (700,979.61)(0.50) = 350,489.80
The total investment tax credit will be:
for non-rack supported =
300,000 + (450,000-125,000) + (125,000x0.50) 10%
= $68,750 for rack supported
= 68,780 + (250,000) 10% = $93,750
The book value of the non-rack supported building at the end of the tenth year will be:
B V = FC 10 i n
10 y D. = 300,000 - 129,877.89 = 170,122.11 l ' ' ' i=l
Therefore the PW cost of the system (non-rack) will be:
PW
M
= 300,000 + 300,000 + 450,000 + 18,000 x 0.50(P/A,10%,10) + 100,000 (1- (P/F,10%,10)) - 303,639.65 - 68,750 - 170,122.11(P/F,10%,10)
p
W
X T O
= $728,774.33
43
The PW cost for the rack-supported case will be:
P W - 250,000 + 300,000 + 450,000 + 18,000 x 0.50(P/A,10%,10) + 100,000 R
(1 - (P/F,10%,10)) - 350,489.80 - 93,750 PWâ&#x20AC;&#x17E; = $672,506.20
It should be noted that the investment tax credit is limited to a certain amount which depends on the present tax liability of the firm as a whole.
Let y denote the tax liability before credit, then
the investment tax credit cannot exceed 12,500 + 0.5y.
Hence, the
investment tax credit, W, will be:
W = Max ^(initial investment cost) 10%; 12,500 + 0.5yj
The computer program computes the 10% of the initial investment cost and finds the value of y which equates the above expression to the computed value.
The value of y is printed on the output as "The tax
liability should be greater than".
If the user finds that the firm's
tax liabilities are smaller than the required amount, the necessary adjustment shall be made on the PW cost of the system. It should be noted that if the user finds that the assumptions are not met, corresponding changes in the program can be made.
In
any case, the initial cost of each cost element appears in the output, hence the user has the freedom of computing the system cost through any desired approach.
44
Summary The initial cost of an AS/RS is composed of the land cost, the building cost, the rack cost, and the S/R cost. Recurring costs include the S/R maintenance cost, the building maintenance cost, and labor cost. The corresponding expressions to compute each of the above cost elements have been presented throughout the chapter.
In addition, an
example demonstrating the calculation of the present worth cost has been presented.
The analysis of the above mentioned cost elements in
terms of the system variables will be treated in Chapter V.
45
CHAPTER III
DETERMINING SPACE REQUIREMENTS
Introduction Determining the appropriate space required to operate economically and efficiently is an important step in the design of storage systems. The entire design is affected by the number of pallets to be stored in the system. _From the standpoint of designing AS/R systems it is especially important to achieve a balance between the storage and through put capacity of the system. This chapter aims to develop a closed form expression to predict the space requirement "relationship" between dedicated and randomized storage given the ordering policies for items carried in stock.
The
particular inventory system at hand operates under the <R,r,T> policy. The results presented in this chapter are not claimed to be theoretical and/or generalized in nature.
The purpose is to demonstrate a unique
approach for determining the space requirements from the inventory records of the company. Definition of the System In this study it is assumed that economic ordering policies for each item group are already determined according to procurement, inventory holding, shortage and system operating costs.
In other words,
the r and R values for each item group are assumed to be pre-determined with T, the review period, being kept fixed at one week.
From (8):
46
It will be recalled that if a periodic review system uses an Rr operating doctrine, then if at a review time the inventory position (in the backorders case) or the quantity on-hand plus on-order (in the lost sales case) is less than or equal to r, a quantity is ordered which is sufficient to bring the inventory position or the quantity on-hand plus on-order up to R. For this study it is assumed that shortages are not backordered, i.e. shortages are considered to be lost sales. Hence, the inventory level to be checked is the on-hand plus on-order; henceforth, this quantity will be referred to as the inventory position.
The R and r
values for each item are explicitly stated in the model according to its demand group.
There are 100 items carried in stock.
The demand
rate of each item is assumed to be independently Poisson distributed. The first 15 items are high demand items with 6 having seasonal demand pattern; the corresponding average demand rate is 20 items/week.
The
following 25 items are classified as medium demand items with 10 having seasonal variations; the average demand rate is 12 items/week.
The
remaining (60 items) are low demand items with no seasonal variation and an average demand rate of 4 items/week.
The seasonal variation in
the demand rate is generated by a sine curve with different amplitudes assigned to every item.
The lead time is assumed to be deterministic
and the same for each item stored in the warehouse. Another important assumption is that all items require equal space.
At first, such an assumption may seem unrealistic but the model
can also be used when it does not hold.
For instance, if item i requires
twice as much space as -that of item j, then item i can be considered as two separate items having equal mean demand rates and space requirements
47
equivalent to that of j. It is also assumed that replenishments are instantaneous.
In other words, when an order arrives the on-hand
inventory is instantly increased by the amount received and no retrievals are allowed before the entire order is entered. One controversial issue in regard to space requirement has been the relation between dedicated and randomized storage methods. For reasons which will become clear in the following paragraph, randomized storage will almost always require less space than dedicated storage. On the other hand, it is possible to achieve higher throughput levels by using dedicated storage.
Thus, the question of whether randomized
or dedicated storage is "best" does not have a simple answer. Let I k be the on-hand inventory level of item k at the beginning of week t, and
denote the maximum value taken by 1..^*
Then
by definition, the storage space required for randomized storage, IGT, will be:
100 I G T
=T Z
I *tk k=l t k
where t denotes the time period.
The storage space required for dedicated storage, IDS, will be:
IDS -
100 I M. k=l k
In words, IGT is equal to the maximum level reached by the "aggregate" on-hand inventory, while IDS is found by adding individually the
48
maximum on-hand inventory level reached by each item.
It can be
readily seen that, unless all the orders for all items arrive exactly at the same point in time, randomized storage will always require less space than dedicated storage. Finding the maximum on-hand inventory level analytically is a simple matter for the basic EOQ model.
In continuous review model with
a probabilisitc demand, it may be possible to set up confidence inter vals for the on-hand inventory.
Also, assigning a certain probability
to the maximum on-hand inventory level that may be reached will be possible.
However, when shortages are not backordered
and when a
periodic review model is used, even when a Poisson process generates the demand, the analytical approach becomes much more undesirable (see (8)). Hence, using the simulation approach seems more suitable for this particular inventory system under question.
The Simulation Model The model proposed to simulate the system described in the previous section will be presented.
The purpose for keeping track of
certain variables of the system is motivated in the following sections. The variables of interest are taken into account after 52 weeks of simulation in order to ensure steady-state inventory levels.
Each week
is one iteration and the review period, T, is kept constant at one week. Definition of the parameters: DEM(i)
=
the mean demand rate of item i (Poisson distributed)
IAMP(i) = amplitude of the sine curve for item i (i = 1,2,...,16) MC(i)
= any integer number between 1 and 12; denotes the cycle iength of the sine curve for item i.
49
RLAM(i) = any real number; denotes the origin of the sine curve for item i. 211
Hence, mean demand rate of item i = (DEM(i) 4- IAMP(i) s i n ^ , ^ (KL + RLAM(i)) . Where, IAMP(i) = 0 if item i does not have a seasonal varia tion. IR(K) = re-order level for item k (r in the <R,r,T> notation). IRT(K) = target inventory level for item k (R in the <R,r,t> notation). ID(K)
= is an array used to store weekly generated demand for each item.
K
= denotes the item (K=l,...,100) (any integer number following k serves the same purpose).
KL
= denotes the weeks (KL = 1,2,...,260)
LT
= Lead time (deterministic)
Definition of the variables: INV(K)
= on-hand inventory level of item K
LS(K)
= total number of lost sales for item k (is increased by the amount of shortage each time it occurs.)
ISAY
= total number of lost sales summed over all items. th
N0R(K,KL) = number of parts on the order placed in the KL for item K. MON(K)
week
= total number of parts standing in the "on-order" status for item K (i.e. level of the on-order inventory of item K)
MCUM(K) = stores the total number of parts carried in inventory for item K (will be divided by number of weeks to find the average on-hand inventory). MAX(K)
= keeps track of the maximum on-hand inventory level reached by item K.
IGIT
= stores the up-to-date maximum "aggregate" on-hand inventory level reached by the system.
ITOT
= used to compute the aggregate on-hand inventory in a given week.
50
CUDF = stores (in a cumulative fashion) the difference between the total number of parts received in two consecutive weeks (will be divided by number of weeks to find the average difference) KMX
= keeps track of the maximum difference between"the total number of parts received in two consecutive weeks.
A computer program was written to simulate the model defined by above parameters and variables. 3-1. 1.
The flow-chart is presented in Figure
The program listing (written in Fortran IV) is given in Appendix (The detailed flow-chart is given in Appendix 14).
Regression Analysis Before the results obtained through regression analysis are presented, the reasons for selecting the independent variables will be motivated.
The first independent variable is the average difference
between the total number of parts received in two consecutive weeks (is equal to CUDF divided by the number of weeks).
The purpose for selecting
this variable can be explained in reference to Figure 3-2a. assumed that only two items are carried in stock.
It is
Item A has a demand
of 6 parts/week and B has a demand of 5 parts/week (both demands are deterministic).
Cycle length for both items equals two weeks. On
the left part of the figure (i through iii), both of the orders are received on the same day.
Thus, the total on-hand inventory level
fluctuates between 0 and 22. At the very beginning of week 1, a total of 12 + 10 - 22 parts are received. 0 parts are received.
At the beginning of week 2 however,
The difference is 22 - 0 = 22.
difference between weeks 2 and 3 is 22, as well. difference for this system will be 22 parts.
Similarly, the
Hence, the average
For this case it will be
51
Initialize the R,r value for each item group in stock
Start iterating over the number of weeks
t
Generate the weekly demand for each item
Consider the next item (1=1,2,...,100)
Store the difference (of the orders re ceived) from those received the previous week.
Reduce its inventory level by that week's demand. Check if any lost sales have occured.
i
Compare the new agg. on-hand inv. level to present max. If it is ^ max, up-date the max.
Increase the inven tory level of the item by the amount of order received. Store the order received.
I
NO
Reduce the on-order inventory level by the amount received.
Take the difference between its inv. YES pos. and on-hand K level and place an order equal to the difference.
3-1.
Find the Up-dated inventory position. Is it less than r.?
YES
Have all the items been considered?
NO
Do not place any order for item i.
Add the on-hand inv. level of i to that weeks agg. on-hand inv. Up-date item i's max. on-hand inventory.
Flow-Chart of the Simulation Program
52
Figure 3-2.
The effect of the difference between total number of parts received (in two consecutive weeks) on the relative space requirement.
53
noticed that both randomized and dedicated storage methods require 22 openings.
For dedicated storage this figure is obtained by adding the
maximum on-hand inventory levels for both items (shown in i -and ii). For randomized storage on the other hand, it is obtained by looking at the maximum of the aggregate on-hand inventory level (shown in iii). Figure 3-2b portrays a different configuration with the cycle of item B shifted by one week. is received.
At the beginning of week 1 an order of 12 parts
In the following week another order of 10 parts is
received, yielding a difference of 2 parts. Likewise, the difference between weeks 2 and 3 is 2 parts as well.
Thus, the average difference
for this system is 2 parts and there is a difference between the space requirements of the two storage methods.
Dedicated storage requires
22 openings, again, while randomized storage requires 17 openings. The same logic will hold for any inventory system where n items are carried.
Hence, the number of openings required by the two storage
methods is affected by the difference between total number of items received in consecutive time periods. The second independent variable is the "total number of lost sales".
The relation between this second variable and storage space
requirements of the two methods is not as obvious as for the first variable.
However, it is true that the required storage space will be
affected by the r and R values that have been chosen for each item.
But
this effect, in fact, is the result of the "combined" effect of the demand pattern and the <r,R> values.
In this model there are only two
variables which reflect this combined effect.
They are the total lost
54
sales and the average on-hand inventory of each item.
Also, note
that lost sales and on-hand inventory are inversely proportional. Predicting the direction of the above mentioned effect on the "relative" space requirements of the two methods is not straightforward. This point will be discussed in the last section of the present chapter. The variable of interest has been selected as the "total" lost sales because all items are assumed to require the same space. The third and last independent variable was taken to be the maximum of the differences defined for the first variable.
There is no
readily observable reason for including this variable, but intuitively it is true that a very sharp change in the difference under question will not affect the average too much (because the system is simulated for 5 years, i.e. 260 weeks); on the other hand, it will have a dramatic effect on storage requirements (the system is assumed to be capable of storing any order size).
Obviously, it can be argued that
this variable may not be significant if such sharp changes are not expected.
Let:
xj^ - average pre-defined difference (first independent variable) x^ - total number of lost sales (second independent variable) x^ - the maximum of the pre-defined differences (third independent variable) y
- the ratio of the space requirement of dedicated storage to that of randomized (the dependent variable)
Eleven simulation runs were made with each run having different <r,R> values and lead-times. The results are shown in Table 3-1. outputs can be found in Appendix 2.
The related
Since we have few independent
55
TABLE 3-1
RESULTS OF THE SIMULATION RUNS
â&#x20AC;¢
XI
X2
X3
X4
X5
Y
62.7260
7.6914
229
52441
26.836
1.4270
61.0144
5.6041
212
44944
27.862
1.4189
65.9183
10.1478
252
53424
24.814
1.4753
62.1010
2.6207
223
49729
30.1053
1.3331
76.2115
4.1793
239
57121
28.3083
1.3472
90.2500
7.0406
359
128881
26.4136
1.3871
92.7644
19.4226
329
108241
19.0091
1.6233
87.6923
1.000
331
109561
34.7995
1.2109
104.4375
2.5713
369
136161
31.6281
1.2460
117.7308
9.9023
443
196249
26.1584
1.2900
126.1971
14.4369
411
168921
22.0214
1.3713
56
variables the "all possible regressions" technique seems to be appropriate.
The result is presented in Table 3-2, where
x
i x = (xp
x
=
l
13/
2
x
=
3
x
3
2 4 3 x,- = average aggregate on-hand inventory x
and
x
p = number of variables in the model including the intercept 2 R = multiple correlation coefficient PSS (p) = sum of squares due to regression K
SS (p) = sum of squares due to error ill
MS (p) = mean square of error £
—2 Rp
= adjusted multiple correlation coefficient
2 R =1 p
n—1
2 (1-R ), where n = number of observations n-p p'
C = Mallows statistic (a measure of bias and variance) P SS (p) C p
=
n + 2p c
2
In general, there are three criteria in selecting the best regression equation: 1 - MS„ should be minimized hi
—2 2 - R should be maximized P 3 - Cp should bd less than or equal to p.
TABLE
NUMBER
3 - 2
A L L P O S S I B L E REGRESSIONS
S S R ( P )
S S E ( P )
MSE(P)
" R ( P ) * * 2
0.0784
0.0513
0.0057
0.5606
56.63
0.6976
0.0905
0 . 0 3 9 0
0.0044
0.6640
41.33
X 1 , X 2
0.9526
0.1237
0.0062
0.0008
0.9408
2.63
X 1 , X 3
<0.50
3
X 1 , X 4
<0.50
3
X l r X 5
0.9297
0.1207
0.0091
0.0011
0.9120
6.302
2
3
X 2 , X 3
0.9237
0.1199
0.0099
0.0012
0.9050
7.269
2
3
X 2 , X 4
0.9365
0.1216
0.0082
0.0010
0.9206
5.212
2
3
X 2 , X 5
0.7077
0.0918
0.0379
0.0047
0.6349
41.97
2
3
X 3 , X 4
<0.50
2
3
X 3 , X 5
0.8959
0.1163
0.0135
0.0017
0.8706
11.73
2
3
X 4 , X 5
0.9139
0.1186
0.0112
0.0014
0.8929
8.880
3
4
X 1 , X 2 , X 3
0.9526
0.1237
0.0062
0.0009
0.9322
4.63
3
4
X 1 , X 2 , X 4
0.9537
4
X 1 , X 2 , X 5
0.9575
0.0060 0.0055
0.0009 0.0008
0.9339
3
0.1238 0.1243
4.440 3.840
3
4
X 1 , X 3 , X 4
<0.50
3
4
X 1 , X 3 , X 5
0.9338
0.1212
0.0086
0.0012
0.9054
7.650
3
4
X l r X 4 , X 5
0.9297
6.1207
0.0091
0.0013
0.8996
8.310
3
4
X 2 r X 3 , X 4
0.9401
0 . 1 2 4 0
0.0078
0.0011
0.9145
6.630
3
4
X 2 , X 3 , X 5
0.9258
0.1202
0.0096
0.0014
0.8944
8.927
P
V A R I A B L E S
R ( P ) * * 2
!
2
X I
<0.50
1
2
X2
0.6045
1
2
X3
<0.50
1
2
X4
<0.50
1
2
X5
2
3
2
3
2 2
O F
V A R I A B L E S
0.9392
C ( P )
3
4
X 2 f X 4 , X 5
0.9403
0.1221
0.0075
0.0011
0.9145
6.330
3
4
X 3 , X 4 , X 5
0.9346
0.1213
0.0085
0.0012
0.9068
7.510
4
5
X l r X 2 , X 3 ,
0.9592
0.1245
0.0053
0.0009
0.9319
5.569
0.9579
0.1246
0.0055
0.0009
0.9298
5.766
0.9579
0.1244
0.0055
0.0009
0.9300
5.766
0.9557
0.1241
0.0058
0.0010
0.9261
6.130
0.9481
0.1231
0 . 0 0 6 7
0.0011
0.9140
7.350
0.9689
0 . 1 2 5 8
0.0040
0.0008
0.9378
6.000
X4 4
5
X l r X 2 , X 3 » X5
4
5
X 1 , X 2 , X 4 , X5
4
5
X l f X 3 , X 4 . X5
4
5
X 2 f X 3 , X 4 » X5
5
6
X 1 . X 2 . X 3 * X 4 . X 5 . X 6
58
Usually, it will not be possible to find one regression line that satisfies all three criteria simultaneously.
However, inspection of
Table 3-2 reveals the fact that the line including x^, x^ is the best among all possible regression lines. Furthermore, note that the second best line which includes x^, x^ and x^ also provides a good fit. The regression line obtained by including x^ and x^ is (see Appendix 3):
3
2
y = 1.487 - (3.153 x 10~ ) x + (2.079 x 10" )x Âą
(3-1)
2
The regression obtained by including x^, x^ and x^ is (see Appendix 3):
3
y = 1. 723 - (2.971 x 10~ ) x. + (1.478 x 1 0 " J-
2)
x
3
0
- (7,589 x 10" )x
I
c
D
(3-2) Note that use of (3-2) requires information on an additional independent variable, namely x^.
Also, from Table 3-2 it is seen that a third best
line does not exist.
(Table 3-2 has been constructed from the computer
printout presented in Appendix 3). Conclusions One remark is concerned with the selection of the first indepen dent variable, i.e. the average difference between the total number of "parts" received in two consecutive weeks. A system which uses a fixed order size policy can well use the total number of "orders" instead of parts. Another remark is on the relation between lost sales and y. Recall that the regression coefficient of a particular independent
59
variable will depend on other independent variables already in the model.
However, it is instructive to note that both in 3-1 and 3-2,
which are the only two lines that provide a good fit, x coefficient.
2
has a positive
That is, the relative space required by dedicated storage
increases as the total number of lost sales increases (the increase in storage space is at a decreasing rate).
The reason underlying the above
relation can be examined as follows; by definition
tj .
rr
tvyd
Turnover Rate = TOR =
average annual sales average on-hand inventory
But (lost sales) + (sales) = (demand). - (lost sales).
Therefore, (sales) = (demand)
Hence:
,j,q _ average demand - average lost sales average on-hand inventory R
Therefore,
average lost sales = average demand - (TOR)(av. on-hand inv.) (3-3)
In (3-3) as TOR decreases, lost sales increases.
But from the
regression line as lost sales increases, the relative space requirement of dedicated storage will increase.
Hence, as TOR decreases the relative
space requirement of dedicated storage increases.
Likewise, as TOR
increases, the so-called space requirement decreases.
Therefore, in
the light of the regression line and the above definition for TOR, it
60
can be concluded that TOR and the relative space required by dedicated storage are inversely proportional, which is true. As stated by (13): This method (dedicated storage) is probably advantageous if each item is close to its maximum inventory on most days or if there is a high turnover rate. The result of this study indicates that the word "probably" in the above statement can be dropped. Note that the regression line only predicts the "relative" space requirements of dedicated and randomized storage methods. Mathe matically, space required under dedicated storage = y x space required under randomized storage
i.e.
IDS = y x IGT
(3-4)
Hence, in order to compute the required storage space under one method, one needs to estimate the required storage space under the other method, Recall, the definitions of IDS and IGT as:
100 k=l
and
IDS =
100 ÂŁ k=l
^
From above expressions it is seen that estimating IDS is a trivial task, as long as the required data is provided.
On the other hand, estimating
IGT requires more computation and extensive data. Hence, in using
61
the regression line for estimating space requirements one would first estimate IDS, and then use y to estimate IGT.
Summary The relative space requirement under randomized and dedicated storage has been a controversial issue.
This chapter has demonstrated
a simulation approach for determining the relative space requirements of the two storage methods.
The approach has been demonstrated within
the framework of a hypothetical example.
Simulation results show that
1.21 < y < 1.62, i.e. dedicated storage may require up to 60% more space than randomized storage.
Also, regression analysis shows that the
relative space requirements of the two storage methods is best explained by the difference in total number of parts received in two consecutive weeks and the total lost sales.
62
CHAPTER IV
ANALYSIS OF THE STORAGE/RETRIEVAL MACHINE TRAVEL TIME
Introduction The S/R travel time is the time required to perform a storage (or retrieval) for single command; and a storage and retrieval for dual command (definitions for single and dual command were given previously). The activities involved in a single command retrieve are: 1 - Leave the I/O point empty 2 - Travel to the pre-determined opening 3 - Retrieve the load 4 - Return to the I/O point and put-down the load The activities involved in a single command storage are: 1 - Pick up the load at the I/O point 2 - Travel to the pre-determined opening 3 - Store the load 4 - Return empty to the I/O point The activities involved in a dual command are: 1 - Pick up from the I/O point the load to be stored 2 - Travel to the first pre-determined opening 3 - Store the load 4 - Travej to the second pre-determined opening 5 - Retrieve the load
63
6 - Return to the I/O point and put down the load Obviously, from a standpoint of increased throughput, it is always better to operate on a dual command; but since it requires both a storage and retrieval order simultaneously, this may not always be possible. There fore, the system is assumed to operate on the dual command basis by a pre-determined percent of total operation numbers provided by the user (this parameter is referred to as "DUAL" in the program listing).
The
rest of the time the system operates on a single command basis. Hence, the expected travel time, ETH, will be:
ETH = (DUAL x DT) + (1 - DUAL)ST
(in minutes)
where
DT = expected travel time for dual command
and
ST = expected travel time for single command
The variance of expected travel time, VTT, will be:
2
2
VTT = [DUAL x E(DT )] +[(1 - DUAL) x E(ST )]
- ETH
Also,
VDT =
variance of travel time for dual command,
and
VST =
variance of travel time for single command.
The travel time is a crucial variable of the system.
2
The through
put capacity is directly determined by the expected travel time. Mathe matically, throughput, TPUT, will be:
TPUT = NOAI/ETH (Operations/nr. )
64
where
1 ETH
(in operations/hr) is the throughput capacity of one aisle.
and
NOAI = number of aisles.
The above expression for TPUT assumes that there will always be a storage and/or retrieval order present at the I/O point.
Such a situation will
occur when the storage/retrieval orders are scheduled in advance, so the arrival rate of the orders is deterministic.
A typical example of
this case is in-process storage where production schedules determine the storage/retrieval rates.
On the other hand, if the orders arrive
according to a probabilistic distribution, the actual throughput of the system has to be computed by:
TPUT = NOAI/(ETH + WT) where WT = the time an order spends waiting in the queue. The waiting time in the queue, WT, is found by assuming a (M/G/l) : (GD/°°/») queue. Hence, WT is given by:
2
WT =
X' V(t) + 1/y 2(l-n)
where
X
1
= arrival rate per aisle = X/NOAI
V(t) = expected variance of travel time = expected mean travel time = X Vu(n < 1)
From the above expression given for TPUT, note that throughput will
65
always decrease when WT is included, as long as WT ^ 0.
The amount of
decrease in throughput will depend on the design under question. Specifically it will depend on the number of aisles because in determining A',
A is divided by NOAI.
The effect of including WT will be further
discussed in Chapter VI. Another remark is concerned with the estimation of A .
Recall
that A is the throughput level demanded from the system, expressed in operations/hr. retrieval.
By definition, an operation is either a storage or a
Hence, A will be determined by adding the expected number
of storages per hr. to the expected number of retrievals per hr. Further more, note that the magnitude of the aforementioned expected values should be equal in the long range. slightly greater than A ,
In addition, in cases where TPUT is only
the user should check the "peak" value of total
operations/hr. against TPUT.
If TPUT does not meet the peak value,
then the program should be reexecuted with a higher value assigned to A . The program prints out both of the above mentioned throughput levels.
Small variations in estimating the expected travel time may
seem unimportant; but for a system with multiple aisles, this is not true. For example, underestimating the throughput of one aisle even by, say, three operations/hr. will lead to an underestimation of total through put by 18 operations/hr. for a system with six aisles.
In other words,
the negative effect of slight variations in travel time estimation will grow seriously with the number of aisles. Hence, travel time estimation gains importance.
66
The Conventional Method of Computing Travel Time: A method of computing S/R travel time has been developed by the AS/RS Product Section of the Material Handling Institute (9). It only holds for the randomized storage method.
For single command travel time, it is
found by adding the duration of the following operations (see Figure 4-1): 1.
Retrieve a load at home station(I/0 point).
2.
Store the load at a storage location 1/2 the number of
storage addresses along the aisle, and up 1/2 the number of vertical storage locations in that aisle.
(In cases of simultaneous travel, use
the longer of the travel or the lift times computed individually.) 3.
Return to home position, empty and stop.
For dual command the total cycle time is found by adding: 1.
Pick up a load at home station.
2.
Store the load at a storage location 1/2 the number of storage
addresses along the aisle, and up 1/2 the number of vertical storage locations in that aisle.
(For simultaneous travel use the longer of the
two times) . 3.
Retrieve a load at a storage location 3/4 of the number of
storage locations in that aisle, and up 3/4 the number of vertical storage locations in that aisle.
(For simultaneous travel use the longer of the
two times.) 4. Note:
Deposit load at home position.
In cases where 1/2 or 3/4 of the addresses equals a fractional
number, round off to the next higher (longer) address.
67
Method of Computing Cycle Time for One Command Operation START AT "HOME" WITH PICKUP. RETURN TO "HOME" EMPTY.
T TOTAL STORAGE LOCATIONS] HIGH
"HOME" AT END OF SYSTEM AND AT LEVEL OF 1 S T STORAGE LOCATION.
Method of Computing Cycle Time for Dual Command Operation START AT "HOME" WITH PICKUP. RETURN TO HOME FOR DEPOSIT. RETRIEVE LOAD DEPOSIT LOAD
s
TOTAL STORAGE %* LOCATIONS HIGH
T
TOTAL STORAGE LOCATIONS LONG "HOME" AT END OF SYSTEM AT LEVEL OF 1 S T STORAGE LOCATION.
Figure 4-1.
The Conventional Method of Computing S/R Travel Time for Single and Dual Command Trips (Reprinted from (9) with permission)
68
There are mainly two drawbacks of the above approach:
First it
is very unlikely that expected single command travel time will correspond to travel time to the opening at the center of gravity of the rack. The above statement is especially true if the rack is not square-in-time, and it can be supported by the following argument.
Suppose in a given
rack, vertical travel time is dominating, i.e. the travel time to the opening at the higher right-hand corner of the rack (across the I/O point) is equal to the vertical travel time to that opening (recall that travel time to an opening is determined by selecting the maximum of vertical and horizontal travel times to that opening).
In this case,
the openings at reasonably high levels will all have high (vertical) travel times compared to low (horizontal) travel times at lower levels of the rack.
Hence, upon taking the average over all openings, we will definÂ
itely expect the center of travel time to be "above" the center of gravity of the rack.
Likewise, for a rack in which horizontal travel
time is dominating, one will expect the center of travel time to fall to the right of the center of gravity.
Hence, based on the rack shape,
the center of travel time will not coincide with the center of gravity. (The proof will be provided on an empirical basis in the next section.) The second drawback is based on the fact that the third element in dual travel time computation will cause the dual command travel time to be over-estimated.
Namely, given that the S/R has
traveled to the first opening for storage (on an expected basis this will be travel time center), there is no reason for assuming that (on an expected basis) the second opening for retrieval will be at .
69
the indicated location ( 3 / 4 of the horizontal and vertical number of locations).
Randomized Storage The expected travel time under randomized storage can be found accurately by complete enumeration.
Let
H = number of levels (height) L = number of columns (length) N = total number of openings = L x H t ^ = travel time from origin to the 1^ opening th th t ^ = travel time from the i opening to the j opening. The expected single command travel time, E(SC), can be found from
ÂąI
4
N
1
E(SC) -
L
N ..
N
2 t
oi
.
=
L
N , . oi
1=1
and its variance,
V(SC),
I t .
( 4 - 1 )
1=1
will be:
2
[e(sc)J
2
V(SC) = E(SC ) " V(SC) - | ,
V(SC) = ~N
J
4
t
2
N I t . oi 1=1
.
-
[E(SO] - \ 2 N
2
N ( I t Y oi 1=1
The expected dual command travel time, E(DC), can be found from
( 4 - 2 )
70
o E ( D C )
and its variance,
=
NOP!)
V(DC),
2
(
D
C
)
" NCifc
?
V(DC) = 2
N -N
.?
.
. ,
+ (t- .+ +t - -t.. + t. ;
( t
O l
n
J=l+1
1J J
J
( 4
J O
"
3 )
2
( t
i=l j=i+l
N
i
1=1
N ^y
will be
V(DC) = E(DC ) - [e(DC)]
V
.
N y ^
+ 1
oi
+
'ij
"
+
J
N
M
9
J I (t ill j-i+1
0 1
A
. + t,, + t. T 1 J
J
°
5â&#x20AC;&#x201D;52
(N -N) N
2
(4-4) 2
N
I
I
1=1
J=l+1
(t . + t.. + t. ) J
Equations (4-1), (4-2), (4-3) and (4-4) have been used in a computer program that prints the true travel time and the travel time based on the conventional method. Appendix 4.)
(The program listing is presented in
The results are shown in Table 4-1.
One of the drawbacks
mentioned earlier for conventional single command travel time computation can be readily seen by observing columns E(SC) and conventional SC (con.sc.) in Table 4-1; it is seen that the conventional method always underestimates the true travel time. When columns E(DC) and con.DC. are observed, it can be seen that in some cases the conventional method under estimates the true travel time.
It will be recalled that in mentioning
Table 4-1.
VV
L
H
WTH
Travel Times Obtained by Complete Enumeration and the Conventional Method
HEHT
E ( S C )
V ( S C )
E ( D C )
V ( D C )
CON.SC
CON.DC
220.0
4 0 . 0
10
5
4 0 . 0
48.0
0.5136
0.0708
0.6877
0.0472
0.5000
0.7500
230.0
5 0 . 0
20
7
42.0
46.0
0.5932
0.0651
0.7975
0.0473
0.5600
0.8400
230.0
5 0 . 0
40
12
42.0
46.0
1.0537
0.1797
1.4185
0.1354
0.9200
1.4000
240.0
6 0 . 0
40
10
40.0
40.0
0.7403
0.0682
0.9995
0.0539
0.5667
0.8417
240.0
4 0 . 0
50
10
40.0
40.0
1.0256
0.1398
1.3834
0.1091
0.8500
1.2500
220.0
4 5 . 0
50
10
40.0
4 0 . 0
0.9984
0.1243
1.3478
0.0983
0.7636
1.1364
260.0
5 0 . 0
53
10
48.0
40.0
0.9966
0.1342
1.3443
0.1047
0.8154
1.2231
200.0
4 5 . 0
48
12
44.0
4 4 . 0
1.2412
0.1965
1.6753
0.1548
0.9778
1.4667
205.0
5 2 . 0
30
7
48.0
4 0 . 0
0.6994
0.0695
0.9432
0.0534
0.5854
0.8780
225.0
5 6 . 0
35
8
44.0
4 4 . 0
0.7301
0.0530
0.5778
0.8622
6 0 . 0
30
7
3 3 . 0
3 2 . 0
0.4184
0.0674 0.0217
0.9856
260.0
0.5650
0.0171
0.3333
0.4872
260.0
6 0 . 0
15
3
3 6 . 0
4 1 . 0
0.2276
0.0062
0.3073
0.0047
0.2000
0.2846
2 6 0 . 0
6 0 . 0
20
4
3 1 . 0
36.-0
0.2647
0.0086
0.3573
0.0066
0.2000
0.3000
250.0
35.0
30
6
51.0
5 1 . 0
1.0594
0.2220
1.4226
0.1603
0.9714
1.4857
295.0
35.0
30
8
5 1 . 0
5 1 . 0
1.0342
0.2390
1.3865
0.1689
0.9714
1.4857
295.0
3 5 . 0
29
8
51.0
5 1 . 0
1.0301
0.2423
1.3806
0.1706
0.9714
1.4857
295.0
35.0
28
8
5 1 . 0
5 1 . 0
1.0261
0.2456
1.3748
0.1723
0.9714
1.4857
2 9 5 . 0
3 5 . 0
27
8
5 1 . 0
5 1 . 0
1.0221
0.2490
1.3692
0.1740
0.9714
1.4857
295.0
3 5 . 0
25
8
5 1 . 0
5 1 . 0
1.0146
0.2558
1.3585
0.1774
0.9714
1.4857
200.0
6 2 . 0
40
9
6 0 . 0
45.0
1.0984
0.2335
1.4762
0.1707
1.0000
1.5000
200.0
6 2 . 0
40
7
6 0 . 0
4 7 . 0
1.0648
0.2564
1.4285
0.1823
1.0000
1.5000
240*0
7 0 . 0
19
4
5 0 . 0
5 2 . 0
0.3908
0.0220
0.5272
0.0166
0.3333
0.5000
240.0
7 0 . 0
18
4
5 0 . 0
5 2 . 0
0.3768
0.0195
0.5086
0.0148
0.3167
0.4750
2 4 0 . 0
8 5 . 0
13
4
5 0 . 0
5 2 . 0
0.2860
0.0103
0.3867
0.0078
0.2333
0.3417
240.0
8 5 . 0
11
3
5 0 . 0
5 1 . 0
0.2290
0.0072
0.3097
0.0053
0.1917
0.2917
72
the drawbacks of the conventional method, however, it was stated that the third element in the dual cycle computation will cause total cycle time to be over-estimated.
The underestimated dual command times are
due to the underestimation in single command. operation
In other words, the second
of a dual command is computed as if it is the second operation
of a single command cycle and travel between two openings is added to this figure.
Hence, if the underestimation in computing the second
element overcomes the over-estimation in computing travel time between two openings, then the total dual cycle time will be underestimated as opposed to what was stated before.
Therefore, in light of the above
fact, it will be more correct to state the following for the conventional method:
it will "always" underestimate single command cycle time and may
underestimate or over-estimate dual cycle time. Equations (4-1) through (4-4) are very useful in terms of providing a reliable answer.
However, to use them in an algorithm, over and over
again, will not be efficient.
This is especially true for the double
summation involved in computing dual cycle time. As N increases (roughly above 400 openings) the computation time for the double summation expression exceeds 60 cpu seconds (the default computation time for jobs executed through the terminal).
Hence, from a computational
viewpoint, it will be preferred to develop closed form expressions for the travel time computation.
Furthermore, the existence of such
expressions will prove to be very useful in establishing several facts concerning throughput (Chapter V ) .
73
One approach would be to represent the rack in a continuous manner instead of discretely spaced openings, and replace previous summations by integrations.
In this case, the expected travel time for single
command, E(SC), will be:
L /
E(SC) =
H /
x=0 y=0 L / ~
E(SC)=
H /
x=0 y=0 n
2 max {-ÂŁ-,â&#x20AC;&#x201D;> f(x) ' f(y) dy dx W V
2 max
w'hv
7- . ^ dy dx L H
where hv = horizontal travel velocity of the S/R w
= vertical travel velocity of the S/R
We cannot integrate the above expression as long as the max operator is present.
Hence, the rack will be partitioned as shown in Figure
4-2.
(It is assumed that horizontal travel time is dominating). Hence,
L' = HR =
H ^ w
Therefore. ir~ ~ â&#x20AC;&#x201D; which means that the rack is square-in-time up to * hv w ?
L .
?
Beyond L , horizontal travel time will be dominating.
Since any
point in region I lies above the diagonal, vertical travel time will be dominating in this region. dominating in region II. Tj, will be:
Likewise, horizontal travel time will be
Subsequently, expected travel time to region I,
74
H
L'=HR
Figure 4-2.
Partitioning of a rack where horizontal Travel time is dominating
H'=L/R -
Figure 4-3.~ Partitioning of a rack where vertical travel time is dominating
75
HR
T »
T_
= ~~
/
1
1
T = T
T
T H
2
=
I
l
l
2
H
-
9
/ W
HR
i
. —
x=0 y=x/R
(w)LH ^
I
H
—
2L. _.__L_ | 2
2
dy
« 7—cttt
dx
(
L'H
" 2 " OMlH i
H
H
X
R
x , 3
v
- ^2
l
Q
w
HR
)
L
H
2
/
I 2
x=0
1
H
,3
" UvTLH >
H
R
dx
x/R 3
H R, "
~I
R
3(w)L
Expected travel time to region II,
r \r \
, HR x/R t
-
—
^I " . L
II
x=0 y=0
will be:
9
— 1 hv ' L'H
3 3 2 H R (hv)HLR * 3
j a y
2 (hv)LH
j
^
HR
2 3 HR x_ , _ 2 ix i R (hv)LHR 3 ' x=0 0 r
J
d X
1
n
2 2 2H R 3(hv)L
If we substitute (hv) = R ( w ) in above expression for T^j., we will obtain
ml
U
=
2
2H R
= T
3(w)L
The above equality is intuitively what one will expect because the rack formed by combining regions I and II is square-in-time. time to region
III, j-jj» t
will be:
Expected travel
76
T
III
1 (hv)LH
=
( L
H
2 "
2
H
R
2
3 3 L H R > = H(hv)L
2
Hence, the expected single command travel time, E(SC), is:
E(SO) = T ,
+
T „
+
T
i
-
n
+
i^jf^
(*-5)
The variance of travel time, V(SC), can be computed by first finding —2 E(SC ) . In region I, it will be found by: , HR s
i-r-
'
H
'
? (
w>
9
i
• Fh
x=0 y=x/R
Upon integration, one obtains the following expression for S j
m
s. (w)
3
L
In region II, we will have:
f
HR x/R x=0 y=0
Upon integration, one obtains:
S
3 3 = H R (hv) L
2
77
If we substitute (hv) = R ( w ) , then
II
3 3 H R 2 2 (w) R L
(w) L
In region III, we will have:
L-L'
(—)
'III
y=0 x=HR
dx dy
L-L'
Upon integration, one obtains
3 3 3 _ _ 4(1/ - HV) III" 2 3(hv) L 1 1 1
Z
Hence 2
E(SC ) - S,
S„
+
S
+
I X I
- »sL
4 ( +
3
; -»
(w) L
3 r 3 )
(4-6a)
3(hv) L
But 2
2
V(SC) = E(SC ) - E (SC)
r n u s
,
-s-c) -
+
v(
2
(W) L
*a
3
- hV) _ I V R 2
3(hv) L
L 3 ( W ) L
+
2
lL=hV] ( h v ) L
J
(4-6)
Similar results can be derived for the case where vertical travel time is dominating. Figure 4-3. &
We have H
The rack is partitioned as shown in 7
= ^ = M? ^ . Therefore, r^ = — which means R (hv) hv w
f
1
that the rack is square-in-time up to H*. time will be dominating.
Beyond H
f
vertical travel
The dominating travel times for regions I
and II are the horizontal and vertical times, respectively. I, the expected travel time, T p is:
_ T
_H'
i "
\
X/
iT '
.
R
2x
1
'
. . d y
• FT
d x
x=0 y=0 Upon integration we obtain
T
I
=
2 J j 2
3(hv)HR
In region II, the expected travel time, , T
n = i-
L
L/R
_
'
'
£
x=0 y=x/R
Upon integration we obtain
II
2
Substituting ( w ) = (hv)/R, then
T-j-j*
is :
. d
d
•ipt * *
In region
79
2
2L 3(hv)HR
II
I
In region III, the expected travel time, T-j-j^. is:
H-H
1
^ x
^
2v
1
1
=0 y=L/R
After integrating we will obtain
T
III
2 2 2 H R -L „, v „2 H(vv) .R
Hence, the expected single command travel time (when vertical travel time is dominating) will be:
T + E(SC) -- T +T T F(SC)
r
n
+ +T T
Z I I
--
++
3 ( h v ) R R
2
H^R - L
2
(4-7)
2
(w)HR
E(SC?
Development of the expression for equation (4-6).
f
h»
2
is very similar to that of
It is obtained from the following expression:
f
4
i
2 X
H
1
x=0 y=0 (hv) H-H* +
—
— H
\ /
* /
^
^
R
4y^
x=0 y=x/R ( w ) 4y
2
— * - j
x=0 y=L/R~ ( w )
1 1 . =^=r . " H
H
L
. , dy dx
1
80
which, upon integration, reduces to
J
4(H R
2
E (
SC ) =
2 3 ( w ) HR
J
- L ) 2
3(w) HR
2L
3
+ 4H R 2
3(w) HR
(4-8a)
3
But V(SC) â&#x20AC;&#x201D; x V(SC)
Thus,
2
2
E(SC ) - E (SC) 3^3 2L" + 4H R 2
3(w) HR
3
3 ( h v ) H R
+
2 2 2' HR - L
" (w)HR
(4-8)
2
The last case to be considered is the one where the rack is square-in-time.
The expressions for this case can be developed by
integration as done for the two previous cases. However, one can utilize the previous results from either one of the cases.
Suppose the
second case, where vertical travel time is dominating, is selected. Let R
f
= â&#x20AC;&#x201D; . Then, since the rack is square-in-time, the following
equations will hold
L hv Therefore,
H w (4-9)
L ( w ) = H(hv) and
Now, consider equation (4-7)
1 . = R, i.e. RR R
f
= 1
(4-10)
81
^ E(SC)
2 2 2 H R - L
41/
=
3 ( h v ) H R
(w)HR
4L 3(hv)R'R
E(SC) =
4L 3(hv)R'R
2
R'HR - L (w)R*R
2
R'HR - L (hv)R'R
+
Substituting equation (4-10) one obtains
E
(
S
C
)
-
^
3(hv)
M +
L
-
L +
(hv)
3HR
3(hv)
and using equation (4-9) we have
E
fsc^ -
E ( S C )
*HR
(4-11)
' 3(hv)
for a rack that is square-in-time (the integration approach gives the same expression).
For variance of travel time, equation (4-8a) will
be modified as follows:
w
â&#x20AC;&#x201D; 2 , _ 2L
Mot, J -
3
3 3 + 4H R x
~
3(w) HR But, from (4-10) RR
1
= 1 , therefore
2 2 3 2IT + 4R'H*R =
9
o
3(w)Vir
82
Substitute (4-9) for L to obtain
2 2 2 2 2H R + AH R 2 2 3(w) R
(4-12a)
Hence, V(SC)
2 2H* (w)
2
4HR I |_3(hv)
2 (4-12)
for a square-in-time rack. Notice that in the f irst two cases the expressions for E(SC) and â&#x20AC;˘2 E(SC ) could have been further simplified by using ( w ) = R(hv).
However,
they were left as they are because the following transformation will be made.
Let the longer side in travel time of the rack be 1.0.
the shorter side in travel time will be b, where 0 < b < 1.
Then
For
example, in a rack where horizontal travel time is dominating, we have
travel time to opening A travel time to opening B
where A is located at the upper left hand corner of the rack and B is located at the lower right hand corner of the rack.
Consider now the
first case where horizontal travel time was assumed to dominate. Mathematically, this means that
=
H/(W)
L/Chv) E(SC) is given by equation (4-5) as
83
4
v(«r\ (
E(
}
3
"
2 r
" <™>
4. L L
2
2
- H R h v
< >
SC) . jllHf
2
2
"
3(vv) L
JL
2
AH (hv)
=
L
H (hv)
+
"
2
h V
(w) L
L
+
2
2
( h v )
3(w) L
Recall that we standardize the rack by setting the larger travel time, in this case L/hv, to 1.0. Hence,
L/hv = 1.0 and b = H/(w) —
Therefore,
h
(4-13)
2
E'(SC)=-y
+1
(4-14a)
Now consider equation (4-6a): E(
Sc2
) m
2RHi^
3
2
2
( S C ) = 2(hv)H 3
(w) L
3
2
(w) L
E
3
4L - 4H R
+
3(hv) L
3 +
4L
2
3
_ 4H (hv)
3(hv)
2
=
3
3(w) L
2(hv)H
3
3
3(w) L
+
_4L
2
3(hv)
2
Using equation (4-13), we obtain
f
2
3
E (SC ) = | b + ^
(4-15a)
Now consider the case where the vertical travel time is dominating. From equation (4-7), we have
84
# i
2
„2„2
4L
— s E(SC) =
,2
,,2, ,
3 ( l w ) f f i
(w)HR
E(SC) =
Therefore,
,
3(hv) H
2
2
.2 L"(w)
H (
™>
H(hv)
2
jh^L +(JwL) 2
3 { h v )
But
„
I T r " - L " _ 41/" ( w )
^
H
H/(w) =1.0 and b = L/(hv)
(4-16)
2
— b E (SC) = -r- + 1 1
Therefore,
(4-14b)
—2 For E(SC ) , equation (4-8a) gives:
E(
Sc2
3
}
=
+ 2
(w) HR Y
3
3
—2.
2L (w) ,
4H R
3
4L
3(w) HR 2
4H
2
,(sc) = — 3 - + (hv) H
3(w)
3(w) HR
3
2
3
4L (w)
2
3
_
3~ • 3(hv) H J
3
2L (w) , 4 3
T 3(hv) H J
3 J
Using equation (4-16) we obtain
E'(SC ) = I b 2
3
+ ^
(4-15b)
It will be readily seen that equations (4-14a) and (4-14b) are identical.
So are equations (4-15a) and (4-15b).
This result is not
surprising because given the corresponding b (henceforth referred to as the "shape factor"), one would expect to obtain the same expressions 2
for E(SC) and E(SC ) in both types of racks. The answer for a square-in-
85
time rack is now obvious.
It can either be obtained from equations
(4-11) and (4-12a) in a fashion similar to previous derivations or it will be obtained by simply setting b=l.
From equation (4-11), we
have:
E<SC> = 3(hv) 4 H R
4 H
3(w)
T
Since the rack is square-in-time the above equation reduces to E (SC)=4/3, which will also be obtained by letting b=l in equations (4-14a) or (4-14b) (henceforth referred to as equation 4-14).
The same value for
E'(SC/b=l) is found by Graves et al. (5 ) , who assumed a rack that is square-in-time.
Furthermore, from equation (4-12a) we have:
9
E(SC ) =
9H 9
but H/(w) = 1.0
(w) â&#x20AC;&#x201D;2 .*. E'(SC ) = 2.0
(again same result is obtained by setting b=l in
equation (4-15a) or (4-15b), henceforth referred to as equation (4-15)). Next consider dual command travel time.
For dual command the
integration approach may not provide a solution because we can have travel "between" regions with different dominating times. Furthermore, even within a region where, say, horizontal travel time is dominating, the two locations may be so located that the vertical travel time will become dominant when travel between the two locations is considered. addition, since we have two points (locations) to
take into account,
single command's double integration will be replaced by a quadruple
In
86
integration.
Hence, using an approach similar to that of single command
does not seem to be promising.
The following approach, however, makes
it possible to-develop an expression for the expected travel time between any two points. As before, let the longer side in travel time of the rack be 1.0 and let b be defined as the shape factor. randomly select two points, x^ and ^ travel time between x^ and x^.
Suppose we
(see Figure 4-4). Let z denote the
Assuming that x^ is fixed, the probability
that travel time between the two points is less than or equal to z will be equal to the probability that x.^ lies anywhere in the square ABCD (we will initially assume that z ÂŁ b).
Define two sets:
set I and set II.
As seen from Figure 4-4, set I is the horizontal strip containing the square ABCD, while set II is the vertical strip containing the same square.
Then, let us partially relax the assumption that x^ is fixed
and assume that it is now allowed to be anywhere on the dashed line labeled L^, i.e. the square ABCD, in a sense, can now move anywhere within set I.
Note that since 0 - x^ - 1, the left half of the square will
fall outside the rack limits when x^ = 0 (labeled A" and D" in the figure). Likewise, when x^ = 1 the right half will fall out (labeled B' and C the figure).
in
Hence, in finding the corresponding probability the above
outliers will be taken into account as follows: P (moving square coincides with ABCD) = area of square/area of set I P(x^ - 0) = (h x area of square)/area of set I = P(x^ = 1). .'. P(moving square coincides with ABCD; adjusted for x x = 1) =P(H) 2 2 4z /2z_. 2.
n
= 0 "or"
1
n
( n
Now let us further relax x
and assume that it can also be anywhere
II
Figure 4 - 4 .
Location of the square ABCD
•
Z=b
LI >
<
1> Y
• i. i
1.0
Figure 4 - , 5 . Location of the rectangle XYWZ
88
on the dashed line labeled L2, i.e. the square ABCD can also move within set II. This time since 0 < x^ ^ b, we need a similar adjustment for x^ = 0 and x^ - b.
Thus,
P(moving square coincides with ABCD) = area of square/area of set II P(x^=0) = ( h x area of square)/area of set II = P(x^=b) P(moving square coincides with ABCD, adjusted for x- = 0 "or" = b) =P(V) X ; L
- ^ 2 _ /2z2 2 _ 2z_ _ z2 2zb 4zb b ^2 z
v ;
b But,
P(travel time between x^ and
- z)= P(Having a square 2z x 2z) = P(H) . P(V) 2
2
2
= (2z-z ) . (2z/b - z /b )
Throughout the above argument it was assumed that z £ b. Con sider now the case where z > b. As before, the two points x^ and randomly selected.
are
However, for this case, since z ^ b the previous
square (denoted by ABCD) will become a rectangle (denoted by XYWZ) in Figure 3-5). With an argument similar to the case z ^ b, it is true that P(moving rectangle coincides with XYWZ) = area of rectangle/area of rack P(x^ = 0 ) =Q$ x area of rectangle)/area of rack = P(x^ = 1) P(moving rectangle coincides with XYWZ; adjusted for x, = 0 "or" x = 1) = P(H') x
89
Note that
the rectangle XYWZ cannot be moved in the vertical direction.
Mathematically, this means P(V') = 1.0, i.e. in the vertical direction the travel time is already less than or equal to z because z > b. Hence,
P(travel time time between x^ and x
f
f
- z) = P(H ) . P(V ) =
2
2
9
2z - z . The above result can be justified by setting z=b in P(V).
2
P(V) = 2z/b - z /b
Recall that
2
if z = b, then P(V) = 2 - 1 = 1,0. In summary, if F(z) denotes the cumulative probability distribution of z, where z is the travel time between any two points, it was shown that
2
2
2
(2z - z )(2z/b - z /b )
if 0 < z ÂŁ b
F(z) = 2z - z'
if b < z
s
1
Therefore: r 2
2
2
2
(2-2z)(2z/b - z /b ) + (2z - z )(2/b - 2z/b ) if 0 < z < b f(z)
=1 if
2 - 2z
< z i
b
Hence, E<z) = fz . f(z) dz = / zf(z)dz + / zf(z)dz = E (z) + E (z) 0 0 1 L
2
2
2
2
3
2
E (z) = / (2z - 2z )(2z/b - z /b ) + (2z - z )(2/b - 2z/b )dz 0 T
b
90
Upon opening the parentheses we obtain
â&#x20AC;&#x17E;
E
i
\
N
(
( z )
=
8z
2
6z
3
6z
3
Az
4
7,2 b
b " - X X 6
^"b U
b
7,3 b
~io .
b
1
2 2 2 2 3 Also, E (z) = / 2z - 2z dz = l - - | - b + - | b b 2
3 Hence, E(z) = ^ - b
2
+ j b
3
+ | b
2
- ^ b
3
2
= -| + | b - i Q b
(4-17a)
The theoretical distribution of a complete dual cycle is difficult to determine.
?
However, its expected value, E (DC), is
E'(DC) = E*(SC) + E(z) 2
_ b . ~ 3 "
.1 .1 2 3 6
n
+
EÂť (DC) = f
+
J
K
1
+
b
2
+
" 3o
b
1 3 " 30
1 ) 3
b
4
17
C " )
It should also be noted that the E(z) when b-1, is:
E(z) = 1/3 + 1/6 - 1/30 = 7/15
The same value, i.e. 7/15 is found by Graves, et al (6 ) who studied a rack that is square-in-time. The variance of dual command, V'(DC), is found by the following analysis.
First, a simulation program was written to estimate E'(DC)
and V(DC) for a given value of b.
The program listing can be seen in
91
Appendix 5. 4-2.
The computer output for 0 ^ b H ,
is presented in Table
From the aforementioned table one can graph the estimated r \, 1
E'(DC)J against b.
2
coefficient of variation |_V'(DC) /
it can be seen that the relation is almost linear.
From Figure 4-6
The least squares
estimates of the parameters produces the following model: Est. Coeff. of variation = 0.3588 - 0.1321 x b 2
(R = 0.9705)
Also, note that the estimated values for E'(DC) in Table 4-2 can be compared to those obtained from equation (4-17).
Taking the percentage
T
difference between the value of E (DC) obtained from (4-17) and the one taken from the table produces an average difference of 0.586% over the 11 runs. In summary, for randomized storage the following are true:
h
â&#x20AC;&#x201D;
2
E* (SC) = ~
+ 1
E (SC ) = I b T
2
3
E' (DC) = -| + -|- b
?
V (DC)^ =
[o.3588
(from 4-14)
+ -| (from 4-15)
2
- ~
b
3
(from 4-17)
- 0.1321 x b] E'(DC)
(4-18)
where b is the shape factor and the rack is normalized (i.e. longer travel time is equal to 1.0). The formulae can be tested for a run taken from Table 4-1.
Let us consider run number 12; we have:
92
Table 4-2. Simulation Results for Different b Values
SHAPE FACTOR" 0.00 1 MEANUC T. TIME SHAPE FACTOR- .10 MEAN DC T.TIML SHAPE FACTOR^ .20 MEAN DC T.TIME SHAPE FACTOR- .30 MEAN DC 7.TIME SHAPE FACTOR- .40 MEAN DC T .TIME SHAPE- FACTORS- .50 MEAN DC T.TIML" SHAPE FACTOR- .60 MEAN DC T•TIME SHAPE FACTORS .70 ML ANDC T.TIME SHAPE F" AC TOR-.30 MEAN DC T.TIML SHAPE FACTOR- .90 MEAN DC T.T1MC SHAPE FACTOR-_____ 1 .00 MEAN DC T.7IME
1.340? VAR. 0! DC 7•TIME 1.3493 VAR. CI- DC I. TIME 1.354 6 VAR. OP DC T.TIME 1.4041 VAR. CP DC T.TIML 1.4143 \ OAR . OF DC T.TIME 1.4416 [ VAR. 01- DC T.TI ML 1.5092 ; OAR. 01 DC T.I IMC 1.5503 i OAR. OF DC T.TIME 1.6303 j VAR. or DC T.TIME 1 . 704;' jOAR < T.TIME 1 . .'955 VAR . Oi DC 7 .TIME i
!
1
OF
DC
|
1
a/E(DC) 0.4000-f
0.2000. 0.1 0.2 0.30.4 0.5 0.6 0.7 0.8 0.9 1.0 Figure 4-6.
b
Scatter Diagram of Simulation Results
. 2
0
o
.2233 . 2234 • 1053 . 1922 . 1001 . 1675 •1/ .1735 • 1663 . 1608
OV
93
hv = 200 fpm
vv = 45 fpm
L = 48
H = 12
slot width = slot
height = 44"
Therefore, horizontal travel time = [(44/12)48] /200 =0.88 min. vertical travel time =
[(44/12)12] /45
= 0.9778 min.
. . b = 0.88/0.9778 = 0.900
Hence,
E ' ( S C ) = b /3 + 1 = 1.27 min. 2
T
2
E ( S C ) = 2b /3 + 4/3 = 1.8193 min. 3
T
E ( D C ) = 4/3 + b /2 - b /30 = 1.7140min. V'(SC) T
V (DC)
2
2
3
[e(SC)] = 1.8193 - (1.27) = 0.2064 2
=
E(SC )
=
[|o.3588 - (0.1321) (0.90^
-
2
2
1.714} = 0.1691
Thus, we have the following results (numbers in parenthesis are the answers from Table 4-1).
E(SC) = 0.9778(1.27)= 1.2418(1.2412) V ( S C ) = (0.9778) (0.2064) = 0.1973 (0.1965) 2
E(DC) = (0.9778)CI.7140) = 1.6759 (1.6753) V ( D C ) = (0.9778) (0.1691) = 0.1653 (0.1548) 2
From the above it can be noted that the expressions developed by assuming a continuous rack provide a good approximation to the true answer.
94
This will always be true if there are sufficiently high number of openings per rack.
In practice, this number hardly falls under 30-40
openings.
Dedicated Storage In dedicated storage particular slots are assigned to particular product classes, and within a set of slots assigned to one product class, storage and/or retrieval is performed randomly.
One question which
immediately arises from the above definition is to find a rule for assigning slots to product classes, so that the advantage gained by using dedicated storage (reduction in expected travel time) is maximized.
One straightforward
answer would be to assign high turnÂ
over product classes (where turn-over is defined as operations/hr) to locations closer to the I/O point.
In fact, throughout their study,
Graves, et al (6,7 ) have used the above rule in assigning product classes to storage locations.
In terms of minimizing expected single
command travel time, such an assignment is not necessarily optimal. This point is proved in Francis and White ( 4 ) , on the basis that it is not only the magnitude of the turnover rate that determines the priority given' to a particular product class, but the number of openings required by that class should also be considered.
Hence, if we let
t\ = number of operations/hr corresponding to product class i, and
= number of openings required by product class i,
then the product class with the highest priority will be the one with the highest C./A. ratio.
(The inverse of this ratio is known as the
95
Cube Per Order, CPO index).
Thus, one has to find the C/A ratio for
every product class and then assign the closest locations to the highest C/A ratio class, the remaining closest locations to the second highest C/A ratio class and so forth, until all product classes are assigned to a particular set of openings.
Suppose we had three product
classes with decreasing C/A ratios. Then according to the value of A^. and A.^, the class boundaries will look either as in Figure 4-7a, 4-7b or 4-7c when the above assignment rule is used.
In terms of
dual command, however, we are not assured that such an assignment rule will minimize travel time because we have "travel between" two points (the fourth element of dual command operations).
For a rack
square-in-time, Graves, et al. (7) states the following:
The exact shapes of the optimal boundaries are quite difficult to specify, although they are symmetric about the line from the I/O point to the opposite corner of the rack (due to square-in-time crane travel). "Square-L" boundaries possess this symmetry property. An alternate boundary configuration, also possessing square class boundaries and the symmetry property, has class I centered in the rack with class II and III boundaries forming concentric squares around it. This boundary pattern can be shown to minimize expected interleave (travel between) time, but not expected one-way (single command) time. The optimal class boundary minimizes expected round-trip time.
The study continues with an investigation of expected travel time sensitivity to class boundary shape.
Their empirical findings
are (1) expected travel between time is fairly insensitive to boundary shape, and (2) expected travel between time contributes "approximately" one-third to expected round-trip time (note that same result is
96
II
I
[
III
1
I/O
1.0
(a)
II
III
I I/O
1.0
(b)
II
I/O
Figure 4-7.
b Cc)
III
1.0
Alternative Configurations of a Rack with Three Product Classes
97
obtained by letting b=l in equation 4-17a).
A lower bound on expected
travel time is twice the minimum expected single command time (achieved by square-L boundaries) plus the minimum expected travel between time (achieved by concentric-square boudaries).
Their findings point out
that square-L boundaries will result (at most) in an expected travel time only 3% above the aforementioned lower bound. Based on the above mentioned results, this study has assumed "square-L" boundaries as those shown in Figure 4-7b; i.e. product classes are allocated to openings starting from those close to the I/O.
Now, let
th denote the travel time to the j opening assigned to product class i.
Suppose the k.. values arek., , k.~, • • •» k. for some product ij ii i2 lm r
class i requiring m slots.
Then, if x denotes single command travel
time, we have:
x
260
2k. 1
1/m
x
2
1/m
2k m
1/m
Hence, I E(x/class i) i
E(x)
EGO where
C
(class i)
(4-19)
"I i c
98
v
9
Also,
A
E(x ) = I E(x /class i) . p(class i) 2
E(x ) =
I (— i i
I k. j=l
m
2
1 3
(4-20)
c
Using (4-19) for E(x) and (4-20) for E(x ) we have: m.
9 2 4 V(x) = E(x^) - E (x) =
r c. "i if 2 Z — _Z_ k
Z
±1
i
m. " j>
i j=l
Next consider the dual command travel time.
m
Li i
j=i
i
The probability of a dual
trip that involves slot i and slot j is:
c. c. P(i,j) = — . -1 m, m. 1
where c. = c./C
J
k
k
Since we are not allowed to visit the same opening twice, i.e. since we cannot have i=j, each P(i,j) for i^j has to be divided by
^2 c. 1 - (probability that i=j) = 1 - £ —|- = 1 - 6 i m.
Hence, if Y denotes dual command travel time, we have c. c . /m.m. E(Y) =
T J i
E
0O
jH
—V-
(t01 . + 13 t.. + 3t. )
c. . c. =[ [ , (t . + t.. + t. ) h . h . m.m.Cl-6) oi in jo 1
3
99
The above expression for E(Y) can be further simplified by noting that t.+t.+t. 01
13
= t . + t . . + t . . Hence,
30
03
31
10
2c.c. E(Y) = T T J i j>i i j 1
m
Also,
m
( 1
"
-rr- (t . + t.. + t. ) ^ J°
6 )
2c c E(Y ) = 1 1 J-A 7T" (t . + t.. + t. ) . .r". m.m.(l - 6 ) oi 13 30 2
1
J > i
1
(4-22)
0 1
2
(4-23)
3
2 Therefore, V(Y) = E(Y ) , which is obtained from (4-22) and (4-23). The double summation involved in randomized storage dual command is also present in the expressions for dedicated storage dual command travel time.
Hence, we once again face the issue of the inefficiency
involved in using (4-22) and (4-23) in a repetitive manner.
Partitioning
the rack in a fashion similar to Figures (4-2) and (4-3) is not feasible for this case because in addition to zones with different dominating velocities, we also have zones that correspond to different product classes.
Furthermore, in terms of programming it is difficult
to determine a general partitioning scheme based on product class zones because any particular configuration of the rack will depend upon the number of product classes and their corresponding iik values, both of which are provided by the user.
In addition, any analytical approach
will be based on that particular configuration obtained at a particular instance of the problem.
Hence, an analytical approach seems to be
100
very inefficient if the number of product classes is not restricted to a certain value.
However, fixing the number of product classes
will lead to loss of generality.
One solution to the above problem
is to simulate the rack after allocating the classes to the openings. Another solution would be to pull four unit loads together when total number of openings exceed a certain value, and then use (4-22) and 2 (4-23) to evaluate E(Y) and E(Y ) . Computational experience shows that, on the average, pulling four loads in to one provides a better approxi mation than simulation.
However, when the the number of openings on
the rack exceeds approximately 1600 openings (a situation which occurs in the Fibonacci search over the lower end of the uncertainty interval where number of aisles is a small number) the computation time again grows seriously.
Hence, simulation has been used for approximating the
the dual cycle travel time. (Single command travel time is automatically found by enumeration while allocating the product classes to the rack openings.)
A sample of 12 runs has been analyzed in terms of checking
the accuracy achieved by simulation. The data related to the sample runs are presented in Table 4-3. Table 4-4 tabulates the results. The first two columns represent the true answers given by enumeration.
The third and fourth columns represent simuÂ
lation results obtained by a simulation time equal to 2*G, where G = total number of slots in the rack.
The last two columns represent the results
obtained by a simulation time equal to 10,000 trips. Table 4-5 tabulates the percent differences between the true answers and those obtained by simulation.
TABLE
4 - 3
SAMPLE RUNS
DATA
NOL
ICOL
2
5
10
5 0 / 2 5
2 5 / 2 5
2
5
10
1 0 0 / 2 0
1 0 / 3 0
3
7
15
5 0 / 3 0
3 0 / 3 0
3 0 / 4 5
3
7
15
9 0 / 3 0
3 0 / 3 0
1 5 / 4 5
3
7
15
3 0 / 3 0
3 0 / 3 0
4 5 / 4 5
4
10
20
1 0 0 / 5 0
9 0 / 5 0
8 0 / 5 0
7 0 / 5 0
4
10
20
1 0 0 / 5 0
8 5 / 5 0
8 4 / 5 0
7 0 / 5 0
4
10
20
1 0 0 / 2 5
7 0 / 3 5
6 0 / 6 0
4 0 / 8 0
4
10
20
1 0 0 / 5 0
9 8 / 5 0
9 6 / 5 0
9 4 / 5 0
5
10
30
3 0 0 / 6 0
2 4 0 / 6 0
1 8 0 / 6 0
1 2 0 / 6 0
6 0 / 6 0
5
10
40
2 0 0 / 4 0
1 8 0 / 6 0
1 5 0 / 8 0
1 1 0 / 1 0 0
6 0 / 1 2 0
5
10
40
1 6 4 / 8 0
1 6 0 / 8 0
1 5 6 / 8 0
1 5 2 / 8 0
NOPR
1.
C/A
2.
C/A
3.
C/A
4.
C/A
5 .
C/A
1 4 8 / 8 0
TABLE
1.
E(DC)
4 - 4 .
2.
ENUMERATION
V<DC)
3.
V£
E<OC>
SIMULATION
4.
V<DC)
RESULTS
5.
E<DC)
6.
V(TIC)
(TRUE)
(TRUE)
L.SIMUL.
L.SIMUL.
2.SIMUL.
2.SIMUL.
0 . 5 2 2 4
0 . 0 3 0 3
0 . 4 6 4 6
0 . 0 2 2 6
0 . 4 7 9 7
0 . 0 2 6 2
0 . 3 7 1 0
0 . 0 1 9 8
0 . 3 4 6 5
0 . 0 1 4 0
0 . 3 4 6 6
0 . 0 1 5 4
0 . 7 1 8 2
0 . 0 6 3 0
0 . 6 9 3 2
0 . 0 4 8 7
0.6767
0 . 0 5 5 8
0 . 5 7 4 8
0 . 0 5 0 7
0 . 5 8 5 2
0 . 0 3 7 3
0 . 5 4 8 1
0 . 0 4 3 3
0 . 8 2 3 9
0 . 0 5 3 7
0 . 7 7 8 8
0 . 0 4 8 6
0 . 7 7 7 5
0 . 0 4 9 6
1 . 1 1 1 2
0 . 1 2 0 4
1.0473
0 . 1 0 8 7
1 . 0 6 6 2
0 . 1 1 1 2
1.1157
0 . 1 1 9 0
1.0517
0 . 1 0 7 4
1.0711
0 . 1 1 0 1
0 . 8 4 9 0
0 . 1 3 6 6
0 . 8 0 7 6
0 . 1 1 4 5
0 . 8 1 2 1
0 . 1 2 2 9
1 . 1 5 5 4
0 . 1 1 5 3
1.0887
0 . 1 0 6 9
1 . 1 0 7 9
0 . 1 0 7 3
1 . 0 7 4 2
0 . 0 9 9 8
1.0466
0 . 0 8 4 0
1 . 0 5 0 6
080918
1.0933
0 . 1 3 2 5
1.0800
0 . 1 2 3 0
1 . 0 6 8 0
0 . 1 2 8 1
1.4101
0 . 1 1 1 6
1.3614
0 . 1 0 5 1
1.3707
0 . 1 0 7 5
103
TABLE
4 - 5
PERCENT AND
PERCENT 1.
VS
3 .
DIFFERENCES
SIMULATION
DEVIATION 2.
MS
4.
BETWEEN
ENUMERATION
RESULTS
PERCENT 1.
MS
5.
DEVIATION 2.
MS
6.
1 1 . 0 6
2 5 . 4 1
8 . 1 7
1 3 . 5 3
6 . 6 0
2 9 . 2 9
6 . 5 8
2 2 . 2 2
3 . 4 8
2 2 . 7 0
5 . 7 8
1 1 . 4 3
- 1 . 8 1
2 6 . 4 3
4 . 6 5
1 4 . 6 0
5 . 4 7
9 . 5 0
5 . 6 3
7 . 6 4
5 . 7 5
9 . 7 2
4 . 0 5
7 . 6 4
5 . 7 4
9 . 7 5
4 . 0 0
7 . 4 8
4 . 8 8
1 6 . 1 8
4 . 3 5
1 0 . 0 3
5 . 7 7
7 . 2 9
4 . 1 1
6 . 9 4
2 . 5 7
1 5 . 8 3
2 . 2 0
8 . 0 2
1 . 2 2
7 . 1 7
2 . 3 1
3 . 3 2
3 . 4 5
5 . 8 2
2 . 7 9
3 . 6 7
4 . 8 2
1 5 . 4 2
4 . 5 5
9.71
104
It is seen that by increasing simulation time, the average percent deviation reduces from 4.82% to 4.55% for mean travel time. For its variance, the average percent deviation reduces from 15.42% to 9.71% (a considerable reduction).
Furthermore, from Table 4-5, note that
high percent deviations are caused by widely spread C/A ratios. For the purpose of this study a simulation time of 10,000 trips is sufficient. However, in cases where the C/A ratios are widely spread and/or number of openings in the rack is a comparatively large number, the corresponding statement in the computer program should be modified.
(The program
listing from which the true answers were obtained is presented in Appendix 6.
The listing of the program written to simulate trips is
presented in Appendix 7.
The computer printout from which Table 4-4
has been prepared, is presented in Appendix 8.) Note that, in computing S/R travel time we have not considered the load handling times. Pick-up and put-down times are assumed to be constant and deterministic, i.e. they do not vary with the shape of the rack.
Therefore, they are simply added to the expected mean travel
time in order to find the time required to complete an entire cycle. - Statistical Distribution of Travel Times Analyzing the travel time distribution is important in terms of highlighting future simulation studies.
First consider randomized storage.
Expressions for the mean and variance of single and dual command travel times have been presented earlier in the chapter.
However, the theoretical
distribution of travel time was only developed for travel between tow points.
The theoretical distribution of single command travel time
can be developed as follows; let,
105
F(v) = the probability that the travel time to a point is less than or equal to v.
If the rack is normalized by setting the larger travel time equal to 1, then 2
2 /b if v < b v
F(v) = 2v
if v > b
where b = the shape factor, and 0 < v - 1. Note that when b=l, f(v) is a triangular distribution, and when b=0, f(y) becomes a uniform distribution.
For 0 < b < 1, f(v) will linearly
increase up to 2b, and then will become an uniform distribution after 2b. Also, there will be a discontinuity at b. Using the above expression for i i F(v) to derive f(v), and evaluating / v . f(v) and / v . f(v) produces 0 _ 0 _____ the same expressions obtained earlier for E (SC) and V'(SC), respectively. 2
1
The program listing presented in Appendix 5 was used to plot the histogram of single command travel time. Figure 4-8.
The result is presented in
Note the shape of f(v) as predicted earlier.
The histogram for dual command travel time was also obtained from the above mentioned program.
The result is presented in Figure 4-8.
Theoretically, dual command travel time will be Beta distributed (it is the sum of the maximum of two uniform random variables).
Plots shown
in Figure 4-9 confirm this fact. The theoretical distribution of single command travel time is difficult to develop when dedicated storage is used.
The same is true
106
Figure 4-8.
The Statistical Distribution of Randomized Storage Single Command Travel Time
REUIND'FBIO
9AMPLE Mt,AN» 1.27392 SAMPLE VARIANCE- .200296 BTP. DEVIATION^ .447544 MEDIAN-' 1.34:'83 Ml IDA! CLASS INTERVAL 19 1.7 TO 1.81 MDHt - 1.710.13 RANI i f . - 1.V:JV74 MIDLAND!- 1.01941 MIDI (UNI *TI>I (IIJCNCY BASED MEAN- 1.27307 M1DIU1NI*! KI-OIJLNCY BASED VARIANCE- .201464 SID. DtV.^ .44111147
READY. RUN 78/11/01. 21.23.14. PROGRAM HISTOMF INPUT DATA FILE MANE T FP10 INPUT ENPPOINTS OF ENTIRE INTERVAL*A.I 9 T 0.05 2.10 INFUT UNIT LENGTH OF CLASS INTERVALS* U T 0.14 INPUT TOTAL NO. OF DATA POINTS* N T 900
0
SAMPLE MEAN- 1.33634 SAMPLE VARIANCE- .212721 STP. DEVIATION- .461216 MEDIAN- 1.41202 MUPAL CLASS INTERVAL IS 1.07 TO 2.01 MODC- 1.00067 RANGE- 1.95974 NIM.'ANDE- 1.01941 MIIil 01NT*PRE0UENCY BASED MEAN- 1.33336 MIITOl NT*PREOUENCY BASED VARIANCE- .213343 STP. DEV.- .461891
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HISTOMP SAMPLE MEAN" 1.17391 SAMPLE VARIANCE* .208242 SI P. DEVI AT IUN-» .454333 Ml. HI AN-- 1.1U4J9 MUDAL CLASS INTERVAL IS 1.09 TO 1.22 MOPE = 1.19065 RANGE" 1.95974 HI liRANGE-- 1.01941 Hll'f••OINT*FRCnurNCY BASED MEAN- 1.17398 MIIiFOINT*( RLUIILNCY BASED VARIANCE- .207176 SID. DEV.- .4^,5166
INPUT DATA FILE NAME T FB08 INPUT ENDPOINTS OF ENTIRE INTERVAL** I B T 0.05 2.2 INPUT UNIT LENGTH OF CLASS INTERVALS' U T 0.13 INPUT TOTAL NO. OF DATA POINTS* N —T 1000 SAMPLE NEAN- 1.22172 SAMPLE VARIANCE" .199719 9TD. DEVIATION' .4469 MEDIAN- I.26B? HOl'AL CLASS INTERVAL IS 1.48 TO 1.61 MOLE- 1.4B848 RANGE" 1.95974 HI ('RANGE* 1.01941 MIPPOINT*FREOUENCr BASED MEAN- 1.2216? HIDFOINTtFREOUENCr BASED VARIANCE- .199042 STB. DEV.- .446141 0
.OS .113 .18 .243 .31 .373 .44 .503 .57 .633 .7 .743 .S3 .893 .96 1 .025 1.09 1.155 1.22 1.265 1.35 1.413 1.48 1.545 1.61 1.473 1.74 1.803 1.87 1.933 2. 2.043
IS
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3.141 UNTS.
RUN COMPLETE.
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SAMPLE MEAN- 1.09387 BAMPLE VARIANCE- .24623? STD. DEVIATION .4V6244 MEDIAN- 1.01946 MODAL CLAOS INTERVAL IS .71 TO .82 MODE- .790 RANtiE^ 1.95974 MIIKANOE- 1.01941 MIDPOINT*! REHUENCY BASED MEAN- 1.09621 MIDPOINT*! R! C1ULNCY BASED VARIANCE- .246662 BTD. DEV.- .49665
SAMPLE MEAN- 1.13212 SAMPLE VARIANCE- .22441 8TD. DEVIATION' .47372 MEDIAN- 1.07276 MODAL CLASS INTERVAL IS .93 TO 1.03 MODE"'I.00508 RANCC= 1.959 74 ' HI1T.'ANGL<- 1.01941 MIHr01NT*rRE0UENCr BASED MEAN" 1.132 MIDPOINMFREQUENCY BASED VARIANCE- .223702 STD. DEV." .475163
3
10 10
20
30
40
30
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70
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BAMPLE MEAN- 1.04013 SAMFLE VARIANCE- .297674 STD. DEVIATION- .545597 MEDIAN' 1.00/14 MODAL CLASS INTERVAL IS .3
SAMPLE MEAN- 1.044*6 SAMPLE VARIANCE- .27142 STD. DEVIATION- .321172 MEDIAN" 1.01577 MODAL CLASS INTERVAL IS .423 TO .74 MODE- .481327 RANGE- 1.93974 MIDKANGE- 1.01941 HIDFOINT.FREQUENCY BASED MEAN- 1.06440 HIDPOINTtFREOUENCY BASED VARIANCE- .27143 STD. DEV.* .5212 0
15
30
.03 .1073 .163 .2223 -»** .28 .3375 -*** .395 .4323 -*** .31 .5675 -»*» .625 .6023 -*«« .74 .7973 .653 .9123 -*** .97 1.0273 -**» 1.095 1.1423 1.2 1.2575 -»** 1.313 1.J725 -*** 1.43 1.4375 -*** 1.545 1.6025 -*** 1.66 1.7175 -*** 1.775 1.0325 -*** 1.69 1.9473 2.003 2.0623 -1 0 2.12
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RANGE= 1.V5974 MIDRANGi:- 1.01741 MIDI OINTtTftttlULNCY BASED MEAN- 1.0407 MlHf 01NT*F KLDI/tNCY BASED VARIANCE- .29488 STD. DEV.-" .5441167 103
120
109 >
-
-
90
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HOHL-- .536
0
10
20
30
40
50
40
70
BAMPLE MEAN- 1.01179 I SAMPLE VARIANCE- .337143 STD. DEVIATION .580639 MCDIAN- 1.00714 MODAL CLASS INTERVAL 18 .1 TO .2 h0Dr_= .172.500 RANOL"- 1.77129 M1DKANCJL = 1.01363 MIMfOlNUI \il (IULNCY BASED MEAN- 1.0110 MIDI'UIMl *fKl IIUI.NCY BASED VARIANCE- .338019 STD. DLV•- .5U1394
SAMPLE MEAN- 1.02321 3AMPLC VARIANCE- .317720 SID. DEVIATION- .545443 MEDIAN" 1.00324 MUDAL CLASS INTERVAL IS .3 TO .43 MJl>L» .370046 KAMtiL* 1.75774 H I I ' R M N U C - 1.01741 MUiHIINT.FKEUUENCy BASED MEAN- 1.02343 HH'T UIHI»rntOUCMCY BASED VARIANCE- .323734 SID. DLV.= .t.40994
3
10 15
0 .073 .13 .223 .3 .375 .45 .525 .6 .6/5 .75 .325 .9 .975 1.05 1.125 1.275 1 .35 1.425 1.5 1.573 1.65 1.723 1.0 1.075 1.93 2.023 2.1
30
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SAMPLE MEAN- 1.00843 SAMPLE VARIANCE" .34341 STD. DEVIATION- .086183 HE MAN" 1.00595 NODAL CLASS INTERVAL IS 1.4 TO 1.8. MODI • » 1.7 r....JOL» 1.97037 (IIDKANOC- 1,00008 midpoint «n.nuun/cr based mean- 1.0073 Ml DPUINI *l RLIIUL/ICY DA'JLD VARIANCE" .339213 51D. LLV.• .5UJ421 0
15
30
43
60
73
70
103
120
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113
Figure 4-9.
The Statistical Distribution of Randomized Storage Dual Command Travel Time
SAMPLE MEAN- 1.695*3 • SAMPLE V A R I A N C E - .139B7 S T D . D E V I A T I O N - .399837 MEDIAN- 1.73532 MODAL CLASS INTERVAL I S 1.71 TO . 1 . 8 4 MODE= 1.77933 RANUL= 2.41914 MHK;.1NGF= 1.52696 MIDI U I N T * r i . r t l U L N C Y BASED MEAN- 1.69109 MIDPOINT*FRCOUENCY BASED V A R I A N C E - .165277 STD. D E V . - .406543
S * W L E MEAN* 1.79342 SAMPLE V A R I A N C E " .17634 S T t i . DEVI A n o w - .420191 MEDIAN* 1.0326 JMDAL CLASS INTERVAL I S 1.88 TO 2 MODL" 1.91407 RA/;Gf= 2.50061 K I : r.'A.'.GE « 1.61126 MIDI O I i H t r f i C f l U E N C Y BASED MEAN- 1.79472 MID! Otl»T«KRtOUENCY BASED V A R I A N C E - .177003 S I D . l ' t V . = .42072 0
IS
30
43
60
73
90
103
0 120
10
133
KUN COMPLETE. Rl U I N D . F B O y
20
30
40
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40
70
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90
100
SAMPLE MEAN- 1.3699 SAMPLE V A R I A N C E - .160932* ' S T D . D E V I A T I O N - .401180 MEDIAN" 1.6042 MODAL CLASS INTERVAL I S - 1 . 3 9 TO 1.73 MODE- 1.47244 R.U'LJt: 2.3553. MIDTiANfJL 1.370B1 MIDPOINT*! liLOIJf.NCY BASED MEAN- 1.37138 MIDI 01NT4I hLMill.NCY DASED V A R I A N C E - .162044 S I D . f i E V . ^ .402572
SAMPLE MEAM" 1,43919 SAMPLE V A R I A N C E - .138439 S I P . D E V I A T I O N - .398049 H L D I A N - 1.65791 MODAL CLASS INTERVAL I S 1.5B TO 1.71 M . l l E ' 1.650 • RriUGE- 2.4498 Ml WV.WX.L" « 1.44259 MI I T U i NT t r k r i ' U E NCV BASED MEAN- 1.4383 #41 !•! OINTtrREOUEHCY BASED V A R I A N C E - .138043 S I D . D L V . « .397572 0
15
30
43
60
73
90
109
0 120
139
130
15
30
45
40
73
90
105
120
133
ISO
SAMPLE MEAN- 1.43818 SAMPLE V A R I A N C E - .177247 STD. D E V I A T I O N - .423376 MEDIAN" 1.47607 MODAL CLASS INTERVAL I S 1.64 TO 1.78 HODL- 1.70657 RANGE" 2.15727 M1UKANGL= 1.^7173 MIDPOINT*! IttOUCNCY BASED MEAN- 1.43758 M I D I 0 I N 1 * ! RLC1KCNCY BASED V A R I A N C E - .181446 S I D . D E V . * .41.'5?63
SAMPLE MEAN" 1.30736 SAI1PLE V A R I A N C E " .169534 O T b . D E V I A T I O N - .410556 MEDIAN* 1.54113 MCD.'.L CLASS INTERVAL 18 1.44 TO 1.77 IIU1'L= 1.70522 I^.10L= 2.25633 MIDr.ANUL- 1.34125 M I D I U l f U t F R C Q U E N C Y BASED MEAN 1.30803 M I D I O l N I t r R E O U E N C Y BASED VARIANCE 170141 S I D . DCV.= .412401
30 30
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120
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SAMPLE MEAN* 1.41348 SAMPLC V A R I A N C E " .191048 S T D . D E V I A T I O N - .43709 MCDIAN" 1.47095 MODAL CLASS INTERVAL 18 1.45 T O 1.78 MUDC- 1.67143 RANGE- 2.05953 MlIiRANGt- 1.24158 HIUi 01MMI f.'CUULNCY BASED MEAN- 1.4102 MlliF 01N1*I RLOUENCY BASED V A R I A N C E - .196392 S I D . DEV.= .443162 0
10
20
30
40
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AO
70
80
90
0
100
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15
30
3.272 UNT8.
RUN COMPLETE.
45
60
73
90
103
120
SAMPLE MEAN- 1.33718 SAMPLE V A R I A N C E - .212172 S 1 U . DEVIATIONMEDIAN =• 1 . 4 3 6 0 3
MODAL CLASS I I U C R V A L I S MUDL«-
SAMPLE MEAN- 1.34336 ' SAMPLE V A R I A N C E - .218342
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15
30
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60
73
90
105
120
MCAN- i.340?2 V A R I A N C E - .220644
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30
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60
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120
120
for dual command.
The particular distribution under question will
depend upon the number of product classes and the spread of their C/A ratios.
Summary The S/R travel time is an important variable in terms of meeting the throughput required from the system.
This chapter has analyzed
mean travel time and its variance both for randomized and dedicated storage.
Closed form expressions were obtained for randomized storage
travel time.
Simulation has been used to approximate dedicated storage
travel time.
In addition, histograms were presented for travel time
under randomized storage.
121
CHAPTER V
THE OPTIMIZATION MODEL
Introduction This chapter will present the mathematical model of the AS/R system.
Following a discussion of the model, the cost elements
introduced in Chapter II, will be analyzed in terms of convexity.
In
addition, the expected throughput level is analyzed as a function of the shape factor.
Based on the above findings, an optimization algorithm
will be developed and presented in full detail. Recall the notation previously developed, where: X
= throughput level demanded from the system
TNOA
= total number of openings
NOAI
= number of storage aisles
NOL
= number of levels in each aisle
ICOL
= number of columns in each aisle
DEPTH
= unit load depth
WTH
= unit load width
HEHT
= unit load height
hv
= horizontal travel velocity of the S/R
w
= vertical travel velocity of the S/R
BHEHT
= building height
BLTH
= building length
BWTH
= building width
122
IADD
= amount cf clearance provided at the end of the aisles. It is added to rack length in order to determine the building length
BCOST
= unit cost (in dollars/sq.ft.) to construct a 25' high free standing building
CFR
= the conversion factor used to compute the unit construc tion cost of a building with a height greater than 25*
LCOST
= unit land cost (in dollars/sq.ft.)
The Mathematical Model The system under question can be modeled as follows:
min TC = Land Cost + Building Cost + Rack Cost + S/R Cost + Total Annual Recurring Costs (P/A, i%, 10)
s. t. (expected throughput level) k X 2 x NOAI x NOL x ICOL > TNOA
The possibility of solving the above model with classical optimization techniques can be discussed with respect to the storage method used.
First, consider dedicated storage.
It was not possible
to develop closed form expressions for the expected travel time (Chapter IV).
Hence, it is not possible to state the first constraint
in a closed form.
The closed form expressions for travel time with
randomized storage were presented in Chapter IV. model variables the shape factor, b, will be:
In terms of the
123
r . /(ICOL x WTH)/hv (NOL x HEHT) / w |(NOL x HEHT)/w * (ICOL x WTH)/hv Recall that the expressions were obtained for a normalized ln
rack (longer travel time being unity).
Hence, an inverse transformation
was required to find the true expected travel time for the rack under question.
As a result, the throughput constraint has a complex form.
Furthermore, when waiting time is included in determining the throughput, the variance of travel time also has to be considered and the expression for expected throughput will get further complicated.
In addition, note
that both the objective function and the constraints are non-linear in NOL, ICOL and NOAI.
Consequently, solving the entire model with
classical optimization techniques does not seem to be promising. However, we can initially fix one of the variables (NOAI) and ignore the throughput constraint in order to see how the cost functions behave in terms of the remaining variables, namely NOL and ICOL.
Analysis of Cost Elements Assuming that the number of aisles is fixed, individual cost elements can be analyzed in terms of convexity.
The following analysis
assumes that the variables NOL and ICOL are continuous. The Land Cost:
Recall the expression for land cost as
c
c
T
T
= BLTH x BWTH x LCOST
= f(ICOL) x g(NOAI) x LCOST
124
let K = g(NOAI) x LCOST, then
c = f(ICOL) x L
K
= [(ICOLxWTH) +
IADDJ
X
K
let G = number of openings per rack = TNOA/2 x NOAI (a fixed number) then NOL x ICOL = G ..
-
-
x WTH x xWTHxK K| +
" [N5T ] D
L n q l
(
N
0
L
)
IADD x K
= \r- |G x WTH x K|
Since NOL > 0 and (G x WTH x K) > 0, dC^/d(NOL) < 0.
= 2(NOL)"
D(NOL)
(5-1)
NOL
3
|G x WTH x KJ
Furthermore,
>
0
2
Therefore, land cost is a convex decreasing function of NOL. The Building Cost:
Recall the expression for building cost as:
c_ = BLTH x BWTH x BCOST x CFR
also, recall the expression for CFR as:
2
CFR = 0.986508 - 0.005349BHEHT + 0.0002698(BHEHT) let
CFR = c c
fi
1
-(c x BHEHT) + c (BHEHT) 2
2
3
= [(ICOL x WTH) + IADDJ X BWTH x BCOST x ^ 2
+ c (BHEHT) ] 2
- (c x BHEHT) 2
125
Substitute ICOL = G/NOL and BHEHT = (NOL x HEHT) + 4.0 in the above expression for c . Upon opening the parentheses one obtains: B
o = c . -r^B 1 NOL n
WTH . BWTH . BCOST + c, . IADD . BWTH . BCOST 1
- c2 ' NOL . WTH . BWTH . BCOST £(N0L . HEHT) + 4~|
"
C
2
. IADD . BWTH . BCOST [(NOL . HEHT) + 4J
+ c„ . t ^ T 3 NOL + c
. WTH . BWTH . BCOST(NOL . HEHT +8 . NOL . HEHT + 1 6 ) 2
. IADD . BWTH . BCOST(NOL . HEHT + 8N0L . HEHT + 1 6 ) 2
3
2
2
Recall that IADD, BWTH and BCOST are constant.
Hence, if we add
the constant terms and let
Y = G . WTH . BWTH . BCOST 1
Y = HEHT . IADD . BWTH . BCOST 2
we have
c = (N0L)" (c Y - 4c Y + " c ^ ) 1
1
fi
1
2
1
+ NOL(c Y (HEHT) - c ^ 2
3
+ NOL
Hence,
2
1
(c . HEHT . y ) 3
2
+
+ 8.2^) constants
( 5
_
2 )
126
- + c
( N 0 L )
"
Y
2
3Yl
(HEHT) - c Y 2
+ 2(NOL)(c
2
1
1 6 C
+ 8C Y 3
Y
3 1
}
2
2
-
3
B
2
+
. HEHT . Y >
3
Hence, d c / d ( N O L ) = 2 ( N O L ) ~ ( C Y 2
4
1 1 " Vl
2 ( C
1
^1 2
1
+ 16 c ^ ) + 2(c . HEHT . y ) 3
2
2
The building cost will be convex if d c /d(NOL) I}
> 0; that is if the
following hold: 3
(N0L)"" (c Y - ^ y 1
1
2
c
+
1 6 c
Y
2 l
>
4 c
> 0
0
We know that NOL > 0. Y
have to verify that c-p'i ~ ^ 2 i Y
±
2
3
Consider the first inequality.
1 1
3
(c . HEHT . Y > >
and
C
+ lSc y )
±
+
c
Y
^ 3 i
>
Therefore we
® holds; i.e. is
Y
2 l'
but
y
1
i.e.
= G . WTH . BWTH . BCOST > 0
is
is
c^ + 16c > 4c 3
2
0.986508 + 16(0.0002698) > 4(0.005349) 0.990324
>
0.021396
£
127
—3 Thus (NOL) ^ i^i c
~ ^°2 1 Y
the second inequality.
Y
2
+
C
"^ 3^1^
i S
str
ct
* - ^ y positive.
Now consider
Recall
= HEHT . IADD . BWTH . BCOST > 0
Also
c
= 0.0002698 and HEHT > 0
3
2
Therefore,
2
d c /d(NOL) > 0 TJ
Jo
was
j^Note that, if
Hence, building cost is a convex function of NOL.
it
not possible to show that the first inequality was true, one would _3 attempt to show that (c . HEHT . Y o ) > (NOL) . (c/f. - 4 c Y - , +
0
0
3
x
2.
2±
i
1 6 c Y ^ ) » where NOL will be assumed to take its maximum valuej. Recall from Chapter II that, total rack cost, cK., is —The — — —Rack — — — —Cost: —— 3
c_. = RCOST x TNOA
K
=
{RCOST +
(0.23328N0L)
-
(0.00476
2
x NOL )]
x 14.0
x TNOA
In the above expression RCOST is fixed because it is a function of ALS (unit load size) and WEGT (unit load weight). ICOL x NOL.
Therefore, c R
= 28 x NOL x
Also, TNOA = 2 x NOAI x
Szr * NOL
N 0 A
T^OST
I
+ 0.23328NOL
L
2
0.00476NOL ] . Since NOAI is also fixed, the variable rack cost, C y ^ * will be
= 28(0.23328)G . NOAI(NOL) - 28(0.00476)G . NOAI(NOL)
2
128
The above function is concave in NOL, due to the second term.
Recall
the original equation (from Chapter II):
2 variable cost/rack opening = 0.23328z - 0.00476z
where z is the height of the rack expressed in unit loads.
The intercept
of the above line equals to 0.142857, which is included in 0.92484 that appears in the cost/rack opening equation in Chapter II. An alternative 2 approach would be to drop the z
term and re-estimate the coefficients.
The least-squares estimate from the data in Table 2-3 gives the equation: variable cost/rack opening = 0.797468 + 0.0949367z
Adjusting the original line for the above modified variable cost, we obtain
cost/rack opening = $14 ÂŁl.579451 + 0.025x + 0.0004424y
2 y 82,500,000
+ 0.0949367zJ
For the example presented in Chapter 2, the above modified line gives $65.73/opening instead of $70/opening obtained from Table 2-3. Hence, in the Fibonacci search, the above convex expression for rack cost was used while the original expression was kept in enumerating over the final interval of uncertainty.
Note that the variable cost/rack
129
opening is now an increasing convex function of z, where z corresponds to NOL.
c
T m
Furthermore, the variable rack cost, c
, will be:
= 28(0.0949367) x NOL x G x NOAI = 2.6582 x NOL x G x NOAI
The S/R Cost:
(5-3)
Recall that the S/R cost is a function of the height of
the AS/RS, the weight of the unit load, and the type and location of the control logic.
In terms of the design variables, the variable
S/R cost is a linear piecewise function of the rack height where the 1
break-points are the number of levels corresponding to 3 0 4 2 , 60
1
and 85'. The Recurring Costs:
The annual maintenance cost on the building was
stated as:
c ^ = BLTH x BWTH x BMCOST BM
Note that the above expression is very similar to land cost, with the exception that it is a recurring cost. Hence, for c
M
we have
(P/A,i%,n)
where
G = NOL x ICOL M = g(NOAI) x BMCOST
(5-4)
130
It was previously shown that land cost is a convex decreasing function of NOL.
Therefore, so is the building maintenance cost.
The Total Cost:
The land cost and the building maintenance cost are
convex decreasing functions of NOL. function of NOL.
The rack cost is a convex increasing
Hence, by adding the above cost elements one obtains
a convex function in NOL. say h^(NOL). convex function of NOL.
The building cost is also a
Hence, adding building cost to h^(NOL) will
produce a convex function in NOL, say h (N0L). o
Recall that S/R cost
is a linear piecewise function of rack height with four break-points, say b^, b^, b^ and b^ (see Figure 5-1). When the S/R cost function is added to t^NOL), the resulting function, TC(NOL), i.e. the total system cost will be a piece-wise convex function of NOL.
In terms of the
minimum point, the total cost function may take different forms depending on the magnitude of the step size at the break-points and the slope of h (x), (see Figures 5-la and 5-lb). 2
Note that, in a general
case there may exist different step sizes at different break-points, and therefore one can develop several different forms of the total cost function.
But, in any case the following rule will be true:
if we
let h' denote the number of levels obtained by minimizing ^^(NOL), then the number of levels that minimizes the total cost function, say NOL^, will either correspond to h' or it will be equal to one of the break points that lie to the "left" of h'.
In terms of Figures (5-la) and
(5-lb) the above rule states that NOL^ equals either to h', b^ or b ^ . Of course, if h' > b^, then all four break-points are candidates for determining NOL.. .
131
S/R
Cost
NOL
Figure 5 - 1 .
The Total Cost Function for Fixed Number of Aisles
132
For a given unit load width (WTH), the b^'s are determined as follows, let:
WTHP = (WTH + 8")/12.0 ft. and [a] = the greatest integer less than or equal to a. Then b
±
= [(y /WTHP) + 0.5] i
where y^ - 30, 32, 60 and 85 for i=l, 2, 3 and 4, respectively. Determination of h' is as follows. Recall that h' minimizes l^CNOL) where ^^(NOL) is the total land, building, rack and maintenance cost, The land cost is given by (5-1) as:
KJ + [lADD x WTH x K| I IADDxxK~] Kl
L = ITT^T L NOL
c
T
where
G = NOL x ICOL K = g(NOAI) x LCOST = BWTH x LCOST
and
Y
Let
= G x WTH x BWTH x LCOST
3
then
C
The maintenance cost on the building is given by (5-4).
Y
then
4
(5-5)
+ [lADD x KJ
L " NOL
Let
= G x WTH x BWTH x BMCOST x (P/A,10%,10)
°BM
=
Y N0~L
+
[
IAD
X
M
X
( P / A > 1 0 %
»
1 0 )
]
( 5
"
6 )
133
Next consider the building cost given by (5-2) and the variable rack cost given by (5-3).
If we delete the constant terms (IADD x K) in
(5-5) and (IADD x M x (P/A,10%,10)) in (5-6) and add (5-5), .(5-2), (5-3) and (5-6), after rearranging the terms, we obtain h (N0L) as 2
follows:
V
N 0 L )
- NOT
( c
Y
l l
+
Y
- 2 l 4 c
l b c
2
Y
+
3
V
+ 8 C Y + 2.6582 x G x NOAI)
2
+ NOL(c^(HEHT) - c ^
+ NOL (c
+
Y
3 l
3
2
5
. HEHT . y >
3
In the above equation, let
4c
"
a
and
0.5a
2
3
=
~ 2'Y
C
HE
3
'Y ( HT)
=
2
-
1
C (HEHT)Y 3
+
1 6 c
1
C Y 2
Y
+
3 i
2
+
Y
3
+
8C Y 3
Y
2
4
+ 2.6582 x G x NOAI
2
Taking the first derivative with respect NOL, one obtains:
dh (N0L) 2
-^(NOLT
1
=
~Z72 NOL
a
l
+
a
2
+
(
N
0
L
)
A
T
Recall that h (NOL) should be minimized at h . 2
1
3
Therefore, we have
a .1 + a .9 + (h )a = 0 v " / ^ o f
9 f
(h )
7
(~)
2
0
134
Hence, h' satisfies
2
3
(h')cc + (h')cx = ÂŤ 2
3
1
Since convexity of t^CNOL) has previously been established, we do not need to check h^ (NOL).
Furthermore, note that the S/R maintenance and
labor cost are not included in determining NOL^ because they are not a function of the number of levels, once the number of aisles is fixed. It has been demonstrated that, given a fixed number of aisles, the number of levels that minimize total system cost, namely NOL^, will either be equal to h' obtained from (5-8) or one of the b_/s to the left of h'.
Mathematically,
(5-9)
Note that the throughput constraint has not been considered in the above approach.
In other words, we are not guaranteed that the required
throughput will be met with the rack obtained by setting NOL - NOL^. The analysis of throughput as a function of the shape factor is presented in the following section.
Analysis of Throughput In analyzing the effect of throughput requirements on the optimi zation, it is instructive to first consider the randomized storage case. It will be shown that the expected single command travel time is
135
minimized when b=l.
Suppose we have a rack for which b=l, i.e. H / w
= L/hv and LH = K (where K - required rack area).
—
h
2
Hence,
T
E(SC/b = 1) = ( V + 1) ~ = 4L/3hv J nv
Now, suppose we increase the length of the rack to L
T
(5-10)
= cL, where c - 1;
the new height H' will be:
L*
CL
cK
c
v _ H'/w H . hv b = . = —= but H . hv - L . w L'/hv 2 c .L . w T
c
.\
2
E(SC/b=c ) =
+ 1) ^ 3c
Now, compare (5-10) and (5-11).
(A-4 + 3c
i . e .
3c
1
Dh^v
We need to show
* 3hv ^
for all c * 1
+ 6 * 1 + -|
for all c * 1
A /
(5-11)
136
The above equation becomes an equality at c=l, and for any c>l, the left-hand side will be greater than (1 +
. The proof is very similar
if we increase the rack height instead of the rack length.
In any case,
the single command expected travel time is minimized at b=l. The variance of single command travel time, V(SC), will also be minimized at b=l. Recall that
â&#x20AC;&#x201D; 2 V(SC) = f
/.
b
3
4 2 â&#x20AC;&#x201D; + f - E (SC) Z
V(SC/b=l) = ^
(^)
2
(5-12)
2
Also,
2
V(SC/b=c" ) = (^g + - ~ + 9c 3c 3c
+ ^)c (^)
2
(5-13)
In terms of (5-12) and (5-13) we need to show that
1 ~~6 9c
x +
2 " T 3c
2 +
~
7 2 > 34 3 "
+
C
J
3c
T
At c=l the left-hand side of above inequality becomes 34/9, for any c>l it is always greater than the right-hand side. Next consider the expected value of dual command travel time, namely E(DC).
From (4-17) we have
E'<DC) = 3 + f b
2
- ^ b
E(DC)/b=l) = 1.80r^-
3
(5-14)
137
E(DC/b=c ) - 4 2
Also
3
+ - V " —^-7" 2c 30c 4
(5-15) 6
h v
In terms of (5-14) and (5-15) we need to show
Gc+-ArJ
) - 1.80
2c
30c
The left-hand side of the above inequality equals 1.80 when c=l; for c>l it will always be greater than 1.80. Now consider V(DC).
It was previously shown that
V'(DC)^ =
Hence,
V(DC) =
[o.3588
{ [o.3588
1
- 0.1321b| E (DC)
t
2
- 0.1321bj E ( D C ) } ( — )
2
if the horizontal travel time is dominating.
Therefore,
V(DC/b=l) =
)
2
and
For
V(DC/b=c
b=l,
anf for L
(5-16) 1
= {[o3.588
gives
= cL,
f
0.2267 x E (DC/b=l)
(0.2267
i.e. b
x
2
(^)
2
2
- °-
1.80)
-2 = c , (5-17)
2
1321
2
] E' (DC/b=c" ) }c
2
C^;)
gives
= 0.1665
2
O^;)
2
.
(^) (5-17) (5-18)
138
V(DC/b=c ) = [4 2
+ ^ 2c
\ ) (0.3588 - ° ' ^ 30c 6 2
1
2 1
)
We have to show:
U~ + ~ 2c
- â&#x20AC;&#x201D; ~ 30c
2
) (0.3588 -
Q , 1
2 1 2
2
) ] c
2
> 0.1665 for c > 1
c
The left-hand side of the above inequality is equal to 0.1665 at c=l. Calculations show that the function on the left-hand side increases as one further increases c. In summary, it has been shown that, under randomized storage the expected single and dual command travel times and their corresponding variances are minimized at b=l. travel time and its variance. are dual cycles.
Now, consider the over-all expected
Recall that, DUAL percent of the trips
Since expectation is a linear operator, one can
conclude that the over-all expected mean travel time will be minimized at b=l.
Variance is not a linear operator; therefore, it is difficult
to demonstrate that the over-all variance is minimized at b-1. However, the following argument will motivate the point. Assuming that DUAL > 0, there are only two events that may occur: and both have positive probabilities.
single cycle or dual cycle;
But we have already shown that
the variance is minimized both for single and dual cycle when b=l. Therefore, no matter which event occurs we know that the corresponding variance will be minimized if b=l. Hence, in the long-range, one would expect the over-all variance to be minimized when b-1.
139
Next consider dedicated storage.
Closed form expressions are
not available for mean travel time and its variance.
However, there
is probably no obvious reason to state that the conclusion reached for randomized storage will not be true for dedicated.
In fact, consider
Figure 5-2, where it is assumed that b-1. Taken individually, we know that the area corresponding to class I will have minimum expected travel time and variance because it is square-in-time. "total" area taken by classes I and II.
Now, consider the
If (C/A) = (C/A> , then we 1
2
know that, once again the expected travel time and variance will be minimized since the aforementioned total area is square-in-time.
If
(C/A)^ ^ (C/A)^, then the only difference is that, comparatively more stops are made in either area I or area II (depending on their relative c. values); but the mean and variance of travel time is still minimized
1
because both area I and area (I + II) are square-in-time.
The argument
for the entire rack is similar to that presented above, because the number of areas that are square-in-time will approach zero, as b approaches zero.
To support the above discussion a computer program was
written, where for given descending (C/A) ratios the program first allocates the product classes to the openings and then computes the mean and variance of the travel time for five different levels. If NOL denotes the number of levels that maximizes the value of b, then s 5
the previously mentioned five levels correspond to the range defined by NOL + 2. s -
NOL
s
is determined from the following expression:
NOL s
h
(5-20)
140
1.0 III
II
I
I/O
Figure 5-2.
1.0
Allocation of three product classes over a rack that is square-in-time
141
where
G = total number of openings on the rack w = slot width h = slot height R = hv/w
Ten different runs have been made with the above program. is seen in Table 5-1.
One of them
The rest are presented in Appendix 9.
from the printouts, the third level, i.e. NOL mean and variance of the travel time.
As seen
always minimizes the
(The program listing of the above
program is presented in Appendix 10. In summary, this section has shown that whichever storage method is used and whether waiting time is included or not, the throughput will be maximized when the rack is square-in-time.
This result will
be used in developing the algorithm which is presented in the following section.
Description of the Solution Procedure First, recall the following: A
= the required throughput level specified by the user,
TPUT( )= the throughput level corresponding to s aisles. S
h'
= the number of levels that minimize t^NOL), where hÂŤ,(N0L) is the cost function corresponding to (total variable cost - S/R cost). Also, h is determined from (5-8). f
N0L
= the number of levels that minimize TC(NOL), where TC(NOL) is the total variable cost. NOL^ is determined from (5-9)
1
NOL
= the number of levels that maximize b (determined from (5-20)).
S
Also, let
* NOL(s)
= the number of levels that minimize TC(NOL) and meet the throughput requirement, given s number of aisles.
Table 5-1.
THE
RACK
Sample Run Showing That Travel Time and Its Variance are Minimized at b-1
IS SQUARE
NUMBER
OF
LEVELS"
SINGLE
COMMAND(MEAN
NUMBER
OF
SINGLE
COMMAND(MEAN
NUMBER
OF
SINGLE
COMMAND<MEAN
NUMBER
OF
SINGLE
COMMAND(MEAN
NUMBER
OF
SINGLE
COMMAND(MEAN
LEVELS'
LEVELS-
LEVELS*
LEVELS=
IN TIME
FOR
7
LEVELS
5
8 VAR.)
.2168
.0150
DUAL
COMMAND(MEAN
& VAR.)
.3114
.0175
& VAR.)
.2094
.0109
DUAL
COMMAND(MEAN
& VAR.)
.2976
.0115
& VAR.)
.2074
.0098
DUAL
COMMAND(MEAN
& VAR.)
.2938
.0100
& VAR.)
.2076
.0100
DUAL
COMMAND(MEAN
& VAR.)
.2942
.0101
& VAR.)
.2103
.0114
DUAL
COMMAND(MEAN
& VAR.)
.2990
.0121
6
7
8
9
143
TC(s)
= total cost of a system with s aisles.
The following algorithm is proposed: 1 - Set the upper limit on the number of aisles, call it M. 2 - Set NOAI = M and determine NOL .
Then compute TPUT(M)
g
for NOL = NOL . g
If TPUT(M) * A, go to step 3. Otherwise,
print an appropriate message and stop. 3 - Initiate the Fibonacci search over NOAI, with the initial interval of uncertainty being [l , M | . 4 - Set NOAI = F, where each F is dictated by the Fibonacci search. 5 - Determine NOL , and compute TPUT(F) for NOL = NOL . If s s TPUT(F) * A, go to step 6.
Otherwise, set TC(F) = ÂŤ> and
go to step 4. T
6 - Determine h .
Then find NOL^
Compute TPUT(F) for NOL =
NOL . If TPUT(F) > A, then set NOL(F)* = NOL TC(F) with NOL = NOL(F)* to step 4.
and return
If TPUT(F) < A , go to
step 7. 7 - Determine the direction d, in which the throughput increases. (This direction will be unique because it was previously established that throughput increases as b approaches 1). If NOL- < NOL , then set d = +1 1 s* If NOL- > NOL , then set d = -1 1 s* Set NOL = NOL and go to step 8. d
1
8 - Let NOL, = NOL, + d. Compute TPUT(F) for NOL = NOL,. d d d If TPUT(F) - A, go to step 9. Otherwise, repeat this step
144
until throughput is satisfied (note: we are guaranteed that throughput will eventually be satisfied or alse we would have not been able to pass step 5). 9 - Set NOL(F)* = NOL, and determine TC(F) with NOL = NOL(F)*. a Return this cost value to step 4. The algorithm stops when the Fibonacci search ends.
Suppose the final
uncertainty interval was ^F^jF^J . The minimum cost design is then found by enumerating over the range F^ to F^. One remark is concerned with the possibility of missing the global minimum through the Fibonacci search.
It is difficult to
establish the convexity of the total variable system cost in terms of NOAI and NOL, where both are allowed to vary simultaneously.
The
difficulty arises from the piecewise nature of the S/R cost, which is a function of both NOAI and NOL.
However, based on empirical findings,
there is very strong evidence that the total variable system cost is a convex function of NOAI and NOL.
In all of the runs, where the program
was forced to enumerate over the entire range J^>M] > there was no indication of local mimina. The flow-chart of the above algorithm is presented in Figure 5-3. The corresponding program listing is given in Appendix 11. The functional definitions of the subroutines are given throughout the program listing. Summary The mathematical model of the system has been presented in terms of the design variables.
The complexity of the throughput constraint
and the non-linearity of the model prevents one from using classical
145
Set the upper limit on the. number of aisles. Call it M.
i.
Set NOAI = M; and determine the throughput for b = 1. Is TPUT > A?
NOJ Print an appropri ate message and stop.
YES Initiate the Fibonacci search with an initial interval of uncertainty equal to [l,M]. By the search procedure determine the next "number of aisles to be checked. Call it F. >
Given F aisles, determine the throughput for b = 1. Is TPUT > A?
NO
Assign an infi nite cost to the design with F aisles.
YES > f
f
T
Determine h (h is the smallest integer that satisfies: 2
3
(h') a + (h') a > a 2
3
!
Find h ( h ) . 2
x
Then check
h (h'-l) for a smaller 2
cost and select the one that minimizes cost. Call it h \
Check the cost at the break points to the left of h' to — determine NOL,
Figure 5-3.
Flow-Chart of the Optimization Algorithm
146
Determine the throughput for F aisles and NOLj_ levels. Is TPUT > A? NO
YES |
Find the PW of total system cost for the design with F aisles and NOL^ levels. Call it TC(F).
Is NOL < NOL x
7
s'
NO
.YES
d = -1
d = +1
Is the Fibonacci search over? NO YES
NOL = NOL a J
>
1
f
NOL = NOL 4â&#x20AC;˘ d d
d
Find the throughput for F aisles and NOL, levels. Is TPUT > A? YES Find the PW of the total system cost for the design with F aisles and NOL, levels. Call it TC(FJ.
Enumerate over the final interval of uncertainty.
Print results for each aisle. Stop after the PW cost has increased,
147
optimization techniques.
The. analysis of cost elements, however,
indicates that when the number of aisles is fixed, the variable cost elements (except S/R cost) are convex functions of the number of levels.
Also, it is established that throughput is maximized when the
rack is square-in-time.
Hence, based on the results of the above two
findings, it was possible to develop an algorithm that finds the minimum cost design with a corresponding feasible throughput level.
148
CHAPTER VI
DISCUSSION OF SAMPLE RUNS AND SENSITIVITY ANALYSIS
Introduction Four sample runs will be presented and discussed in terms of cost minimization.
Furthermore, a sensitivity analysis will be performed on
two model parameters, namely "DUAL" and "X".
Sample Runs The program presented in Appendix 11 was executed with the data set shown in Figure 6-1.
Inspection of the data indicates a high
throughput required from the system (310 operations/hr). to this data set is shown in Figure 6-2.
The solution
Note that, enumeration over
the final interval of uncertainty does not continue when present worth cost increases.
In reference to Figure 6-2, since 611,502.59 > 576,560.58
the program stops after 8 aisles. Also, it can be seen that for 7 aisles, the system achieves a throughput of 330.50 operations/hr, which is greater than 310. At this point, one may argue that, for any data set the optimum number of aisles will always be the smallest number of aisles that meet the throughput constraint and therefore adding additional aisles will always cause the cost to increase while providing additional throughput capacity. but it is not true.
The above argument may look intuitively correct, Consider the data set shown in Figure 6-3.
The
G L. I.) K U.!. A 1 UI i * - li i... G .t. C N U! A u r. L< . /' i G 7 YPE Tl-lE LiEI'M v UIDT'!â&#x20AC;¢! ? IIEIGi 11 GF LiNIT L0AD ( INCIIES ) ? 30. 40. 30. TYPE UNIT EGAD WEIGHT ( LB > - S) ? 800. TYPE DUAL COMMAND PERCENTAGE ? 0.3 TYPE VERTICAL?HORIZONTAL VELOCITY OF S/R(FPM> ? 61. 250. TYPE PICK -UP 9 PUT -DO WN TIMES (MIN.S) ? 0.25 0.25 TYPE UNIT BUILDING*UNIT LAND COST<$/Ff#*2> ? 22.0 0.40 TYPE 0*192 OR 3 FOR S/R LOGIC 0----MAN -ON -BOARD l^GN THE S/R 2=GFF THE G/R 3-CENTRAL CONSOLE ? 1 TYPE ANNUAL MAINTENANCE COST/SR ? 150. TYPE ANNUAL BLDG. MAIN 1". COST/SQ.FT. ? 1.5 TYPE 0 (RACK) OR 1 (NGN-RACK) SUPPORTED BLDG. ?0 TYPE 0 TO INCLUDE WAITING TIME ?OTHERWISE TYPE 1 ?1 TYPE ATMARR?APPLICABLE INCOME TAX RATE ? 0.1 0.5 TYPE 0 FOR RANDOMIZED?1 FOR DEDICATED STORAGE ?0 TYPE TOTAL NO. OF SLOTS REQUIRED(REAL) ? 5000. TYPE THE REQUIRED THROUGHPUT(OP.NS/HR.) ? 310.
Figure 6-1.
Data of Sample Run Number 1
150
HU.
in
A1GLI..G
NO.
0\
I.L'v'Ll. G
7
9.0 0
tilt. bLiK.-. I'LUG. 01
l: L - i; l, . i'LL'L . LAND
CULUMNG
i . . L . r : i
III
u.idiii
HL. J liU'l
COS I
40.00 22;;. 6/ 63. oo
34. 00
5t>6v.G0 COS! 2GG0/..\2i.
RACK CCS! 2LJJV2-T.U6 b, k rtrtUiiNL COST 3640OO.00 1 ML. RLLURR1NG LUG I b ARL HLDG.
MAINTENANCE
COG)
KHLN
S/R OblEM gtgilm
rCNANGL COST 1050.00 IHRUUGHI'UI (0) 1C.0U UP. NG. HR . thruuompurcij jgo.go up.ng/itk.
L\l
.
SINGLE
CUnVANU
i i DUAL
I HL GIIAI 'L I AC! OK lb TOTAL NG. 01 GI'LNlNGb
I'U COG I 01 (TAX
ABUVL
Lintl-m
. 66 504 0.00
DESIGN
SHOULD
NO.
01" A]GLEG Gl
LEVELS
9.00
NG.
Df
COLUMNS
35.00
DLDG.
LENG1H
IV/. 33
U111T II
72.00
DLDG.
ITE1CHI
CObf
b bo0.5J w
DULL
7599 . 26 >
0
DLDG.
LAND
lb
Di_
NO.
:
DULLARG 20GG3.00
34.00
5632.20
HLI.iL. COG I 263732.73 RAG!, LOG I 2Ul";GW.06 MALiiiNL MIL
GObl
KLCUKKING
JfLDG . S/R
416C-00.00
DULLAPG
GOG IS AttL
MAiiJ 1 LNA1 'C. GOG i
MAINTENANCE
COST
2i.il. .v V. 1200.00
SYSTEM
THROUGHPUT CO)
G1.01
SYS)EM
ThROGGIIPUT ( 1->
399.75
EXP. IHL
SINGLE SHAPE
(GTAL
PW
NO.
COST
(TAX
8 DUAL
FACTOR
OP
:
0F
ADOVE
Figure 6-2.
CP
TRAVEL
' I MEG
1 . 2G
5040.00
DESIGN
SHOULD
SECONDS
UP.Nb/lik.
.7b
OPENINGS
LIABILITY
.157
COMMAND
IS
GI'.Nb/MR.
BE
IS >
611502.59
DOLLARS
146535.36)
EXECUTION
TIME
Solution to Sample Run Number 1
1 KftVt.!_
I I MEG
151
data tabulated in Figure 6-3 is identical to the data set previously given in Figure 6-1 with the exception that the required throughput level has been increased to 340 operations/hr. Figure 6-4.
The solution is shown in
Note that, for this case, while the number of aisles has
increased from 7 to 8, the cost has decreased from 629,237.77 to 611,502.59 dollars.
Furthermore, the throughput increases from 340.80
to 399.75 operations/hr.
That is, it was possible to achieve a higher
throughput level with a lower cost. follows:
The result can be explained as
in the solution to the first run (Figure 6-2) note that both
7 and 8 aisle designs have 9 levels, hence by adding an aisle we have not gained much in terms of cost savings; in fact, we have incurred the cost of an additional S/R (note the sharp increase in S/R cost vs. slight changes in the other cost elements).
Now suppose we increase the
throughput requirement to 340 operations/hr.
To meet the new throughput
level, the design with 7 aisles now has to move towards a rack that is square-in-time.
Figure 6-4 shows that for a 7 aisle design, the rack
becomes square at 11 levels (note that the shape factor is 0.98).
Hence,
we are still able to meet the required throughput with 7 aisles, but in doing so, we forced the 7 aisle design to move further away from the constraint-free design.
Now consider 8 aisles. The 8 aisle design has
a throughput capacity of 399.75 operations/hr in Figure 6-2.
Therefore,
increasing the required throughput will not change the answer to the 8 aisle design (note that the 8 aisle design is identical in Figures 6-2 and 6-4). Hence, by adding an aisle, we were able to arrive to a design that comes closer to its constraint-free design and thus provides a
152
GEORGIA TECH.
DESIGN Of AN AG/KG
type ti-ie dep rh »wid rh y i-ieigh7 of unit lgad < i n c h s ) ? 30. 48. 30. TYPE UNI T LOAD WE I GHT(LB.G) ? 800. TYPE DUAL COMMAND PERCENTAGE ? 0.3 TYPE VERT I GAL y HORIZONTAL VELOCITY OF S/R<FPM) ? 61. 250. tyi"e p:i:ckup >.• pu r•••• down i'Ihes < min . s; ? 0.25 0.25 TYPE UNIT BUILDING r UNIT LAND COST ( $/FT##2) ? 22.0 0.40 TYPE Oyly2 OR 3 FOR S/R LOGIC O^MAN-ON-BOARD l^ON THE S/R 2"-• 0FF THE S/R 3--CENTRAL CONSOLE ? 1 TYE ANNUAL MA INTENANGE CGS if/SR ? 150. TYPE ANNUAL BLDG. MAINT. COST/SQ.F). ? 1 .5 TYPE 0 (RACK) OR 1 (NON-RACK) SUPPORTED BLDG. ? 0 TYPE 0 TO INCLUDE WAITING TIME v07 HERWISE TYPE 1 ? 1 TYPE ATMARRyAPPLICABLE INCOME 7 AX RATE ? 0.1 0.5 TYPE 0 FOR RANDOM IZED y1 FOR DEDICATED STORAGE ? 0 TYPE TOTAL NO. OF SLOTS REQUIRED(REAL) ? 5000. T YPE TI-IE REQUIRED THR0UGIIPUT ( 0P . NS/HR . ) ? 340. ;
Figure 6-3.
Data of Sample Run Number 2
153
in.:
,
01
1..LVLLG
NO.
LOLUflNb
Ul
7
A3 „,LL
i'O. 01
3 1.00
33.0 0
DLDG.
LLNGIH
101... 0 0
DLDu,
WIH III
6J.CJ
DLDG.
HLlbIM
LAND
COOT
40.
6/
4/3 7.6C
ULIiG. COST
1455/0.74
KACK COG J 31 74b G . 2 4 S/k MACHINE COG i -355000.UO THL
RECUFv'RlNG
DLDG. S/R
COSTS
MAIN I LNANCL
MAINT LNANCC
ARL
LUG'!
THROUblll'U T(0)
SYSTLM
THROUGHPUT(1)
SHAF'E
1GTAL
NO.
F'U COST (TAX
FACTOR
1050.00 .71
H DUAL
SINGLE
THE
1/,'uu.vO
COG I
SYSTEM
LXi'.
COMMAND
DESIGN
IS
or
aisles
OF
LEVELS
9.00
NO.
OF
COLUMNS
35.00
LENGTH
IV/.3b
DLDG.
UiDfTI
/2.00
DLDG.
HEIGHT
34.00 5683.20
COS I 263752.73 CUS1
2S1924.06
S/R
MAC HI Nl.
THE
RECURRING
BLDG. S/R
DOLLARS
a
DLDG.
RACK
62923/.7/
COG I 41c.000.OC COSTS
MAINTENANCE
MAINTENANCE
DOL:.
Akb
ARE
GOG I
COST
21312.00 1200.00
SYSTEM
T FIROOGHF'UT l 0 ;
53,45
OF-'. NG/HR •
SYSTEM
THROUGHPUT 11>
399.75
OP.NS/HR.
TXP.' THE
SINGLE SHAPE
TOTAL
iU
& DUAL
FACTOR
NO.
COST
(TAX
2.10
i'L .: 1604U? , G<-->
no.
C0S1
1.31
50GV.0O
LI AD XL 3 I'Y GHOUL i'
DLDG.
II MLS
. 9C
NO.
LAND
UP.NG/HR.
TRAVEL
IS
ABOVE
GP.Nb-HR.
340 . SO
OF" OI'LNJNbb
01
DOLLARS
OF
OF
ABOVE
IIMES
1.28
611502.59
DOLLARS
5040.00
DLS1GN
SHOULD
TFfAVLL .75
OPENINGS
LIABILITY
Figure 6-4.
COMMAND
IS
BE
IS >
146535.3&>
Solution to Sample Run Number 2
2.05
154
;i0.
01
LLVLLG"
NO.
01
LULiJflNij
KLHC.
?.OU .51.00
LENGTH
I/O. 6?
fcLHG. UII i r,
81.00
ULI.IG. 11L1G I IT LANli
CCL-.
34.00
G,'GG. Gj
KriLt, CD'.'!' .'BOV2 / ' _ > / I ' 'riALHJ.NLLUG1
46GuOO.^0
T'Ht: KLL'UKK i N'G CGG1G 1>LDG . G/k
AI •.' L
btiJTLM
llO. . / Ll'.K^.Nk. 1/0. 4J,
! I IKOUGHI U I il )
i_xr.
' iN...lI
] ML
GHAI 'L. I M L I OK' lb
iota,, no. i'U cor;: or
i: uual. wGMf:,,;;i.i
or
) KHVlc.
Of . NG/IlK . :if-,c. g
. a;
QPLtiittuc t-v:.:\ .><.•
a hove
LiAblLltl
.1UA
...:. ,'Ot!. oo
UUG I
1 HUiJUGI I! 'U 1 i 0 )
ST r G1 LM
UAA
..,':..<). CO
hAJ NTL.KA.IGC. g.;:, :
MAIN I LNANLL
fcULLAKL
uegign
SUGULt
CI"' bECONuC
VL
j g
\
i4G_'Gv.oi
dogl^kg
1 G:.ii. 1 • J. J . ' , j )
CXLGUTIDK
Tim..
155
smaller cost.
It is also instructive to note the S/R costs for 7 and
8 aisles in Figure 6-4.
It is seen that the S/R cost is 455,000 and
416,000 dollars for 7 and 8 aisles, respectively.
That is, though we
purchased an additional S/R for the 8 aisle design, the "total" S/R cost has decreased.
This reduction in total S/R cost is due to a high cost
per S/R in the 7 aisle design. the number of levels).
(Recall that S/R cost is a function of
Also note that, since the present worth cost
decreased, the program does not stop and carries the search over to 9 aisles. The third sample run demonstrates a man-on-board design. data set is presented in Figure 6-5.
Note that, for the S/R logic
equal to 0, the program asks the annual operator cost. is shown in Figure 6-6.
The
The solution
The additional labor cost is printed under "total
operator cost". The fourth sample run shows the dedicated storage case. The data set is presented in Figure 6-7.
Note that, as opposed to sample
runs presented previously, the number of openings and the throughput level are determined from the A^ and c^ values entered respectively by the user.
Furthermore, the user has the freedom to choose between
simulation and enumeration for dual cycle travel time computation. solution is shown in Figure 6-8.
The
The number of openings for each
product class are given on a per rack basis.
Sensitivity Analysis There are many parameters that stand as candidates for sensitivity analysis, but among them only two require considerable judgement and/or
GEORGIA
T E C I!.
DESIGN
OF
AN
AS/R S
Wlimb
? ? ? ? ? ?
T Y P E T ME D E E T H * H E I G H T OF 4 8 . 4 8 . 4 8 . T Y P E U N I T LOAD W E I G H T ( L B . S ) 2 0 0 0 . T Y P E D U A L COMMAND P E R C E N T A G E 0 . 5
T Y P E ANNUAL 2 0 0 0 0 .
COST/OPERATOR
?
ANNUAL
MAINTENANCE
ANNUAL
DLDG.
r°
TYPE 2 2 0 0 . TYPE 5 • b
?
0
? ? ?
TYPE
0
(RACK)
TYPE
0
TO
INCHES )
S/R(FPM)
OR
INCLUDE
COST/SR
MAIN")". 1
COST/SO . F T .
(NON-RACK) WAITING
SUPPORTED
11MEY O T H E R W I S E
BLDG. TYPE
1 T Y P E A T M A R R V A P P L I C A B L E INCOME TAX R A T E 0 . 1 0 0 . 5 0 T Y P E 0 FOR RANDOMIZED 5 1 FOR DEDICATED STORAGE 0 T Y P E TOTAL NO. OF S L O T S REQUIRED<REAL) 6 0 0 0 . TYPE
?
i.
TYPE UNIT BUILDING?UNIT LAND COST($/FT#*2) 2 5 . 6. T Y P E 0 X 1 * 2 OR 3 F O R S / R LOGIC 0---= M A N - O N - B O A R D 1="0N T H E S/R 2 = O F F THE S / R 3-CENTRAL CONSOLE 0
?
LOAD
T Y P E U E R T I C A L v H 0 E I "1.0 N T A L V E L O C I T Y O F 60. 2 4 0 . T Y E P I C K U P v P U T D OWN TIM E S ( I I I N . S ) 0 . 2 5 0 . 2 5
?
?
UNI T
THE
REQUIRED
::
TL-LR0UGH! 'U T ( 0 P . N S / H R . )
150.
Figure 6-5.
Data of Sample Run Number 3
1
157
NO. 01"' A1 SLL G 5 NO. Ol" LEVELS' 11.00 NO. Ol COLUMNS jj.00 dldg. length 290.6/' dldg. width 67.50 dldg. heigh! 5/.17 land cost 117720.00 dldg. cost 640655.54 kack cost 502036.82 g/r mauunl cijgt 390000.00 dollars the recurring costs are dldg. . maintenance cog i 10?','] 0.00 s/r maint enancl cog! 11000.00 total oi era tor lost 100000.00 dollars/vr. system throughput u)> 31.6s of'.ns/hr. system throughputci> 107.47 op.ns/hr. exp, singlefactor s dualiscommand.33 travel times 1,81 2,77 i he share TOTAL NO. Of UE 'lNINUG 6050.00 PW COST OF ABOVE DESIGN lb 1594757.29 DOLLARS (TAX LIADiLI'l i SHOULD Dl , 2/o63G.4/) NU. 01" AISLES 6 NO. Of" LEVLLG 11.00 ttu. 01 LOLOMNS 4o.00 DLDG. LENGTH 24G.67 DLDG. WD ITH 01.00 DLDG. HEIGHT 57.17 LAND COST 120052.Oo DLDG. COST 6oG9i3.35 RACK COST 503062.41 S/1v- h ACH ', f If." COST 4â&#x20AC;&#x17E;000C . 0C DOEL ARG THE RECURRING COSTS ARE DLDG. MAN ITENANCE COST 1]0731.CO S/R MAN ITENANCE COS I 13200.00 TOTAL OPERATOR COST 120000.00 DOLL ARS/YR. S Y S T E M TH HR RO O U G H PT UT(0(.>1) 79.77 OO P.N GS/H RK .. S Y S T E M T U G H P U 240. 9 4 P . N / H E XEP. SH SIN D TRAVEL TIMES 1.69 2.60 TH AG PELEFAS CTOD RUAL IS COMMAN.99 TOTAL NO. OF" OPENINGS 6072.00 PU COST OF ASOVE DESIGN IS 1727704.90 DOLLARS (TAX LIABILITY SHOULD BE > 294755.15) 182 CP SECONDS EXECUTION TIME Figure 6-6.
Solution to Sample Run Number 3
158
GEORGIA TECH.
- DESIGN OF AN AS/RS
TYPE THE DEPTHvWIDTHyHEIGHT OF UNIT LOAD(INCHES) ? 4 8 . 46. 46. TYPE UNIT LOAD WEIGHT<LB. S) ? 1300.0 TYPE DUAL COMMAND PERCENTAGE ? 0.40 TYPE VERT ICAE?HORIZONTAL VELOCITY OF S/R<FPM) ? 55. 2 3 0 . TYPE PICK -UP y PU"!' -D0WN TIMES (M1N.S) ? 0.20 0.20 TYPE UNIT BUILDING y UNIT LAND COST ( */FT#*2 ) ? 23.0 3.5 TYPE 0yly2 OR 3 FOR S/R LOGIC 0-M AN -â&#x20AC;¢ON-BOARD I-ON THE S/R 2::::0FF THE S/R 3--CENTRAL CONSOLE ? 3 TYPE ANNUAL MAINTENANCE COST/SR ? 1800. TYPE ANNUAL BLDG. MAINT . COST/SO.FT. ? 5.5 TYPE 0 (RACK) OR 1 (NGN-RACK) SUPPORTED BLDG. ? 1 TYPE 0 TO INCLUDE WAITING TIME>OTHERWISE TYPE 1 ? 0 TYPE ATMARRyAPPLICABLE INCOME TAX RATE ? 0.10 0.50 TYPE 0 FOR RANDOMIZED v1 1 OR DEDICATED STORAGE ? 1 TYPE NO. OF PRODUCT CLASSES(INTEGER) ? 2 TYPE OP.N/HRyNO. OF SLOTS P OR EACH PROD. CLASS ? 6 0 . 2000 ? 50. 2100 FOR ENUMERATION TYPE
1yFOR SIMULATION TYPE 0
? 0
Figure 6-7.
Data of Sample Run Number 4
BLDG.
205.00
LENGlli
i:;._Db. Wll.:i;i DLDG. LA Nil
ML i Gl I l LOG i"
I!i_J."G . k'ALL
59552.50
COS I G J ^ l yo. 58
COG "I
294/56.85
S-'R
MACHINE
THE
RLCUKfUNG
HLDG. S/R
GUST
o24<J00.0C
DOLLARS
COG) G AKt.
MAINTENANCE
MAINILNANCL
COG I
93582,50
LOG I
IOJOO.00
STEM
THRDCGHPUT<0)
167.38
GVSlLrt
T HROUOI ll-'U It 1)
2V0 . 9&
LXI'.
SINGLE,
i HI.
PW
SHAPIL
i
DUAL
COMMAND
F AC f OR
OI'CNJNOS
I'OK IkUD.
CLASS
i
0PLN1NGS
I OR
CLASS
2
Of
ABOVE
PROD.
DESIGN
LI A D I E U 7
SHOULD
7
IS
OF
AISLES
NO.
01
LEVELS
9.00
NU.
OF
COLUMNS
33.00
DLDG.
LENGTH
182.50
DLDG.
WIDTH
96.50
DL-DG.
HEIGHT
LAND
COS F
DLDG. Ri.CK
46.00
531174.12 2VS3635.23
G, I; MACHINE
COG 1 /L.'TJUOO . O O
RECORRING
DLUG.
COSTS
MAINTENANCE
MAINTENANCE
SYSTEM CfSfEM
DOLLARS
ARE
COG I
96861.88
COST
12600.00
THRGUGHF'U I < 0 ) THROUGHPUT(1)
232.95 OP.NS/HR. 3G5.9G OP.NS/HR.
EXP. SINGLE H DUAL COMMAND TRAVEL THE SHAPE FACTOR IS .88
143,00 150.00
COST
(TAX
OF
OPENINGS OPENINGS
ABOVE
LIABILITY
33.250
CP
DOLLARS
61639.38
COSf COST
1208991.24
HI". • 1//G51.37;
NO.
PW
I 1MLS
.93
175.00
COST
S/R
IKAVLL
IS
lo/.Ou
, I .-•>'
THE
. NS-'ltK'. OP.NS/HR.
PROD. CLASS PROD. CLASS
DESIGN
SHOULD
SECONDS
Figure 6-8.
FOR FOR
BE
IS >
TIMES
1.31
1,98
1 2
1299366.29
DOLLARS
143927.05)
EXECUTION
TIME
Solution to Sample Run Number 4
160
past data for estimation.
Those are, "DUAL" (indicates the percent
of trips done on a dual command basis) and A (the throughput level demanded from the system). In general it is common practice to perform sensitivity analysis studies on optimization programs that require many and/or difficult parameter estimations.
However, in the context of the optimization
algorithm developed in this study, it is not possible to give general statements about sensitivity because the importance (or effect) of a certain parameter will vary from problem to problem, depending on the relative locations of the constraint-free optimum and that optimum dictated by constraint involving the parameter of interest. "DUAL".
For example, consider
At any particular instance of a problem, if the optimum design
has some "slack." throughput capacity, then, decreasing "DUAL" will not change the answer, and hence one will be tempted to conclude that the minimum cost is not sensitive to "DUAL". However, imagine an optimum design in which there is virtually no slack throughput capacity.
Then,
reducing "DUAL" will force the design and the corresponding cost to change immediately.
This argument is supported by the results tabulated
in Table 6-1.
(The computer printouts of Table 6-1 can be seen in
Appendix 12).
Note that, different conclusions can be made in terms
of sensitivity over different ranges of "DUAL".
Hence, it is strongly
recommended that every user should perform their own sensitivity analysis over a particular data set. Next consider the required throughput, A. given for "DUAL" holds for this parameter also.
The above argument Table 6-2 presents the
161
TABLE
6-1
SENSITIVITY
PU
DUAL
ANALYSIS
NOAI
RESULTS
1 , 3 6 5 , 4 6 4 . 7 2
5
165•36
0.2
1 , 3 6 5 , 4 6 4 . 7 2
5
1 7 3 . 5 5
0 . 4
1 , 3 6 5 * 4 6 4 . 7 2
5
1 8 2 . 5 9
0 . 6
1 , 3 6 5 , 4 6 4 . 7 2
5
1 9 2 . 6 2
0 . 8
1 , 2 8 5 , 2 6 4 . 6 5
4
1 4 6 . 8 1
1 , 2 8 5 , 2 6 4 . 6 5
4
1 5 6 . 4 2
i
TABLE
6-2,
„,,,,,
,.„.
SENSITIVITY REQUIRED
TPUT
PW
"DUAL"
TPUT
0 . 0
1.0
FDR
ANALYSIS
THROUGHPUT
NOAI
100
1 , 2 8 5 , 2 6 4 . 6 5
4
125
1 , 2 8 5 , 2 6 4 . 6 5
4
150
1 , 3 6 5 , 4 6 4 . 7 2
5
200
1 , 4 5 2 , 5 5 3 . 8 3
6
RESULTS LEVEL
FOR
THE
162
the results obtained from four different throughput levels (denoted TPUT in the table).
Note that, even we have mentioned the hazards of making
general conclusions,
Table 6-2 indicates that present worth cost is
fairly insensitive to A (TPUT). Another remark is concerned with the waiting time associated to each operation.
It would be recalled that in computing the system
throughput, the user has the option of including "WT", where "WT" denotes the time spent by an order waiting in the queue.
Hence, if
the user has required that "WT" should be included in the throughput calculation, then in the computer printout, the throughput level given under the heading "system throughput (0)" should be greater than or equal to A.
(Obviously, "system throughput (1)" will always be greater than
"system throughput (0)", because (1) denotes that "WT" is not considered in computing the throughput level of the system).
It should also be
noted that the expression used for WT assumed a (M/G/l): (GD/W°°) queue. From previous sample runs note that in a given run, the difference between "system throughput (1)" and "system throughput (0)" is always larger for the initially printed number of aisles compared to the difference found from the following design which has an additional aisle. This is due to the fact that as the number of aisles decreases, A increases.
1
(Recall from queueing theory that the waiting time will
exponentially increase as the arrival rate increases).
Summary Four sample runs were presented and discussed.
It was demonstrated
that it is possible to increase throughput while decreasing cost.
In
163
addition, in the light of the sensitivity analysis, it can be concluded that total cost seems to be fairly insensitive to the required throughput level.
Also, the degree of importance of each parameter will vary with
the nature of the optimum design.
164
CHAPTER VII
CONCLUSION AND RECOMMENDATIONS
Introduction In analyzing AS/R systems and developing the optimization model for minimum cost AS/RS design, a number of insights were obtained. In this chapter, the conclusions reached during the study are provided and recommendations for further study are presented. Conclusions The conclusions reached in the conduct of the study can be stated as follows: 1)
The relative space requirement for randomized and dedicated
storage methods is a function of two variables:
the average difference
between the total number of parts received in two consecutive time periods and the average number of lost sales.
(Note that, additional
variables may have to be considered if a different inventory model is used).
Simulation results obtained showed that, under the <R,r,T>
policy, dedicated storage may require 20% to 60% more space than randomized. 2)
It is possible to reduce the mean S/R travel time by using
dedicated storage instead of randomized.
Graves, et.al. (7) and Barrett
(3) report various ranges of reduction depending upon the spread of the
165
turnover on each pallet. However, computational experience has shown that the expected variance of travel time is not necessarily reduced when dedicated storage is used.
In fact, there have been some cases
where the expected variance increased when DUAL - 0.80.
Hence, if WT
(waiting time) is included in the throughput calculation, the expected throughput level may not significantly increase when one uses dedicated storage instead of randomized. 3) Whether waiting time is included or not, throughput will be maximized when b=l, i.e. when the rack is square-in-sime.
(A similar
conclusion is also presented by Bafna (2)). However, when design costs are included, the optimum design, will not necessarily be squarein- time. 4)
Under randomized storage the single command mean travel time
follows a theoretical distribution presented in Chapter IV; the dual command mean travel time is Beta distributed. 5) When the number of aisles is fixed, the variable cost elements constitute a piece-wise convex function in the number of levels. The break-points are those number of levels that correspond to the height intervals defined under the S/R cost. 6)
By adding an aisle to a given design, in some cases it may be
possible to increase the throughput while decreasing the total present worth cost of the system. 7)
It is not correct to make general conclusions regarding the
sensitivity of total system cost to "DUAL".
The importance of "DUAL"
166
will depend upon the particular design under analysis. On the other hand, sample runs show that the total system cost is fairly insensitive to the required throughput. 8)
The waiting time grows exponentially when the arrival rate per
aisle is increased.
Hence, systems with a small number of aisles will
always show a larger difference in throughput levels obtained by including and not including waiting time. 9)
The conventional method of computing travel time presented
in Chapter IV will always underestimate single command mean travel time. It may underestimate or overestimate the dual command mean travel time.
Recommendations A number of areas for further study were identified in performing the study.
Areas for further study felt to be significant due to either
their theoretical contribution or their usefulness in designing real world systems include the following: 1)
Consider transfer cars for systems with low throughput
requirements and/or for systems where every item is not carried in every aisle.
One alternative would be to consider the S/R's and the transfer
cars within a network of queues and use simulation to analyze system behavior and draw conclusions. 2)
Develop closed form expressions for the expected mean and
variance of the single command and dual command travel times under dedicated storage for fixed and/or variable number of product classes. 3) Further analyze the effect of square-L boundaries versus
167
concentric squares on "travel between" and round-trip time.
Extend the
work done by Graves, et.al. in (6) and (7). 4)
Develop an analytical approach to establish the proper balance
between storage capacity and system throughput.
Constract a mathematical
relation between the two in terms of system parameters and/or variables. 5)
Extend the present analysis to consider multiple sized,
vertical openings in the rack and multiple part numbers on a pallet. Analyze the effect of multi-level I/O stations on throughput.
Analyze
the effect of the vertical location of the I/O station on throughput and cost. 6)
Extend the analysis of AS/RS configurations to include mini-
load systems.
Develop appropriate cost models and consider multiple
part numbers per storage bin. 7) Develop analytical and/or simulation models for carousel S/R systems.
Analyze the effects on throughput of centralized versus
decentralized storage and retrieval and randomized versus dedicated storage. 8)
Extend the present analysis to include deep lane storage
systems, as well as multi-depth AS/R systems. 9) Analyze the economic impact of increasing S/R travel velocities in order to meet the throughput requirement over a given rack. 10) Analyze the effect of deviating from FCFS on retrievals. Compare the advantage of achieving lower dual command trip times against the cost of a more sophisticated software.
168
11) Analyze the travel time for the "pure random" and "closestopen location" storage methods.
Determine the amount of overestimation
in travel time caused by using the pure random approach. 12) Update the current program to reflect changes in the tax laws for 1979.
APPENDIX 1 Program listing for simulating the storage space requirement under dedicated and randomized storage
170
PROGRAM
M A I N ( I N P U T * OUTPUT , TAPE5=INPUT
DIMENSION DIMENSION
I R T ( I O O ) » M C U M ( I 0 0 > , M O N ( 1 0 0 ) , M A X ( 1 0 0 ) *
DIMENSION COMMON DO
2
tTAPE6=0UTPUT)
N O R ( 1 0 0 , 2 6 0 ) » I N V ( 1 0 0 ) , I R ( 1 0 0 ) » L S ( 1 0 0 > D E M ( 1 0 0 ) » M C ( 1 6 ) , I A M P ( 1 6 ) » R L A M ( 1 6 >
RLAM» DEM»IAMP»MC
K 7 « l » 9
D E M ( K 7 > * 2 0 . 0 I R T ( K 7 ) * 6 2 2
I R ( K 7 > = 4 4 DO
3
K8=10»24
DEM(KB>=12.0 I R T < K 8 > * 4 6 3
I R < K 8 ) « 3 1 DO
4
K9=25r84
DEM(K9>«4.0 I R T ( K 9 ) « 1 7 4
I R < K 9 > * 7 DO
5
5
K 6 = l f l 6
R E A D < 5 r * ) I A M P ( K 6 ) F R L A M ( K 6 ) » M C < K 6 ) DO
33
K14=85»94
D E M < K 1 4 ) « 1 2 . 0 I R ( K 1 4 > » 5 0 33
I R T ( K 1 4 > * 6 0 DO
34
K 1 5 = 9 5 r l 0 0
D E M ( K 1 5 ) = 2 0 . 0 I R ( K 1 5 > « 6 8 34
IRT<K15>«85 LT=1 I G T = 0 DO
10
K l » l » 1 0 0
I N V < K 1 ) « I R T < K 1 > MCUM(K1)«0 M0N(K1>»0 LS<K1>«0 10
M A X < K 1 ) » 0 C U D F « 0 . 0 RMX=0.0 DO
999
KL<=1>260
I T 0 T = 0 D I F F - O . O C A L L DO
D G E N ( I D rKL >
888
K = l » 1 0 0
I N V ( K ) - I N U ( K ) - I D ( K ) I F ( I N V C K > > 1 3 » 1 4 » 1 4 13
L S < K ) - L S ( K ) - I N V < K > I F ( K L . L E . 5 2 > L S < K ) - 0 I N V ( K > * 0
14
I N - K L - L T I F ( I N . L E . O ) B O
T O
15
I N V ( K > « I N V < K > + N O R ( K » I H ) B I F F - D I F F + N O R ( K » I N > H O N ( K ) « M O N < K > - N O R ( K r I N ) 15
IPO«INV<K)+MON<K> I F ( I P 0 - I R « K ) ) 1 9 f 1 9 # 2 0
19
N O R < K » K L > « I R T ( K > - I P O M 0 N ( K ) - M O N ( K ) + N 0 R ( K , K L ) ©0
20 17
TO
17
N O R < K f K L > » 0 _
I F ( K L . L E . 5 2 ) 6 0 _ T D
BB8
-
I D ( 1 0 0 )
171
MCUM(K>«MCUM<K>+INV(K> ITOT=ITOT+INV(K> IF<INV(K).GT.MAX(K)>MAX<K>«INV<K) eee continue IF<ITOT.GT.IGT>IGT=ITOT IF(KL.LE.52)G0 TO 998 RR«ABS(DIFF-TD > CUDF=CUDF+RR IF(RR.GT.RMX)RNX-RR 998 TD=DIFF 999 CONTINUE RINVEN«0.O IDS=0 ISAY=0 DO 21 Jl-l.lOO AV«MCUM<Jl>/208.0 RINVEN=RINVEN+AV IDS=IDS+MAX(J1> ISAY=ISAY+LS(J1> 21 URITE(6»23>AV>LS<Jl>rJl 23 FORMAT(3X » * AV. INT.•»2X»F9.4»2Xr*L. SALES•T2XT15r2X»I3> WRITE(6>24)IGT»IDS 24 FORMAT(/3X,•RANDOMIZED * 12X119, 2X T•DEDIGATED', 2X RI9) AVER=CUDF/208.0 RINVEN=RINVEN/100.0 URITE(6 r 74)AVER TRMX 74 FORMAT</3Xr•AV.DIFF.'r2X»F10.4r2Xr"MAX.DIFF.•r2X»F10.4> WRITE(6»75>ISAY»RINVEN 75 FORMAT(3X»•TOT« LOST SALES"»2X»I9»2X»"AV. AGG. INV."r2X»F10.4> STOP END SUBROUTINE DGEN(IDrKL) DIMENSION ID(IOO) DIMENSION DEM(100)rIAMP(16>»RLAM(16)»MC<16> COMMON RLAM TDEM TIAMPTMC DO 11 Kl«lr84 CALL P0S(IAR»DEM(K1>> 11 IDCKD-IAR DO 12 K2-1.16 IW-B4+K2 A=IAMP < K2 > *SIN <(6.283/MC(K2 > > *(KL+RLAM(K2 >)) ARR*DEM<IW>+A CALL POS(IARrARR) 12 ID<IW)-IAR RETURN END SUBROUTINE POS(NrAL) I«l *«1.0 P-1.0/EXP<AL> 2 ' S-S*RANF(X) : a IF(6.LE.P)G0 TO 5 I-I+l . 00 TO 2 3 -N-I-l i RETURN i END • —v
172
APPENDIX 2 Output of the simulation runs obtained from the program listing in Appendix 1
AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AC, AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV.
f NT. INT. INF . TNT. TNT • INT . TNT . INT. INT. INT. TNT. INT. INT. INT. INT. INT. TNT. INT. INT. INT. INT . INI. TNT. INT. INT. INT. INT. INT . INT. INT. INT. INT. INT. INT. INT. INT. INT. INT < INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. TNT.
37.88V4 34.6442 35.4856 '34.6250 34,4856 34.3990 34.6058 34.5962 20.3173 28.11269 29.8702 28.3317 29.2788 27.9087 29.8029 29.7452 30.4808 29.1394 28.9087 29.2067 28.8846 29.4567 2.8.2452 17.9712 18.1442 18.7692 18.9183 10.0481 18.6394 18.4327 18.4087 18.0721 18.1250 18.5337 18.5913 17.7981 17.8654 18.5577 17.9712 18.1010 17.6298 18.6779 18.0240 18.3942 17.4663 18.5144 17.9904 18.1106 18.9423 18.8221 18.6106
L • SALFS L. SALES L. SALES L. SAl ES L. SALES L. SALFS L. SALES L • SALES I.. SALES L. SALES • L . SALFS L, SALFS L. SALFS L. SALES L , SALES L, SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L, SALFS L. SALES L. SALES L. SALES L. SALFS L. SAl ES L. SALES L, SALES L, SALFS L, SALES L. SALES L. SALFS L. SALES L. SALES L, SALFS L, SALES L. SALES L, SALES L, SALES L. SALES L. SALES L. SALES L, SALES L, SALES «... SAl ES L. SALES L. SALES L • SALES
9 5 21 35 34 12 13 23 21 14 11 0 2 15 1 3 12 3 5 4 7 5 7 2 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 6
0 0 0 0 0
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 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
52
AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV, AV, AV, AV. AV. AV. AV, AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV, AV, AV. AV. AV. AV. AV.
INT. INT. INT. TNT. TNT. INT. INT. INT, INT. INT, INT. INT, INT. INT. INT. INT. INT. INT. INT. INT • INT. INT. INT. INT. INT. INT. INT. INT. INT. INT . INI . INT. INT. INT. INT. INT. INT. INT. INT, INT, INT. INT. INT. INT. INT. INT. INT. INT.
RANDOMIZED
18.5096 19.2212 IB.437/ 18.8365 17.846 2
18.9471 18.3173 17.9712 17.2452 17.7500 17.8654 18.5000 18.1779 18.5385 18.5433 18.2067 18,3942 18.2019 18.0625 18.3654 18.1496 19.0240 10.4808 18.1635 17.9760 19.2019 18.6923 18.7067 17.6298 17.8029 18.8173 18.6538 49.9808 49.7067 49.2548 49,0192 49.6106 50.0817 48.5913 47.8269 48.8317 48.4712 56.7019 57.2596 56.2308 58.5096 56.4904 55.2452
L, SALFS L . SAL ES 1 . SAL FS 1 . SAL ES L.. SAL IS
L. SALES L. SALES L. SALES L. SALES L. SALFS L. SALES L. SALES L. SALES L. SALES L, SALES L. SALES L. SALES L. SALES L. SALES . L. SALES L. SALES L. SALES L, SALES L. SALES L, SALES L. SALES L, SALES L, SALES L. SALES L, SALES L • SALES L. SALES L. SALES L. SALES L. SALES L • SALES L. SALES L. SALES L. SALES L. SALES L. SALES L • SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES
2876
DEDICATED
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 o' 0 0 0 0 0 5 7 0 0 2 0 0 3 3 0 7 43 50 11 26 27
53 54 55 56 5>
58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 89 89 90 91 92 93 94 95 96 97 98 99 100 4104
AV.DIFF. 62.7260 HAX.DIFF. 229.0000 TOT. LOST SALES 455 AV. AGO. INV.
AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV • AV. AV. AV. AV. AV. AV. AV* AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV, AV, AV, AV. AV. AV, AV, AV. AV. AV,
INT. INT. INT. INT. INT. INT, INT. INT. INT. INT. INT. INT. INT. INT. INT, INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT, INT. INT. INT. INT. INT, INT, INT. INT. INT. INT. INT , INT. INT. INT. INT. INT. INT. INT. INT. INT, INT. INT.
32.6683 35.0048 31,9615 32.61175 32.0000 31.0894 31 . ' 6 6 8 3 31.9231 31.8702 24.9327 25.5337 26,3990 25,0769 25.9038 24.6875 26.5577 26.5240 26.9471 25.6587 25.5913 26.0048 25. 7115 26.1442 24.9760 15.9712 16.1587 16,7692 16.9183 16.0481 16.6394 16.4327 16.4375 16,1010 16.1394 16.5337 16.5913 15.8269 15.8654 16.5577 15.9712 16.1010 15.6298 16.6779 16,0240 16.4087 15.5096 16.5529 15.9904 16.1106 16.9423 16.8221 16,6106
L. L, L. L, L , I.. L. L. L, L, L. L. L. 1.» L, U L. L, L. L, L. L. L, L. L. u. L. L, L. L, L, L. L. L. L. L. L, u. L, L. L • L, L. L. L. 1.. L. L. L. L. L. L.
SA! E5 SALES SAl ES 5AI. E S SAl ES SALES SALES SALES SALES SAL ES SAl F S SALES SALES SALES SALES SALES SAl ES SALES SALES SAl ES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SAl ES SALES SALES SALES SALES SALES SALES SALES SALESSALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES
35 13 43 49 60 40 34 45 40 35 23 10 19 31 17 17 36 20 21 IB 19 41 26 31 0 1 0 0 0 0 0 2 3 1 0 0 2 0 0 0 0 0 0 0 1 3 8 0 . 0 0 0 0
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 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV, AV, AV, AV, AV, AV, AV. AV. AV. AV.
INT. JNI . INT. INI . INT. INI . INI . TNT. INT. INT. INT. INT. INT. INT. INT. INI. INI . INT. INT. INT. INT. INT. INT, INT. INT. INT. INT. INT, INT, INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT, INT. INT, INT, INT. INT.
RANDOMIZED AV.DIEF. TOT. LOST
16,3385 17.2212 16.4327 16.8365 15.8750 16.9760 16.3173 15.9712 15.2596 15.7500 15.9038 16.5000 16.1923 16.5385 16.5433 16.2212 16.4087 16.2308 16.0625 16.3654 16.1779 17.0240 16.4808 16.1779 15.9760 17,2019 16.6923 14,7067 15.6298 15,8317 16.8317 16.6538 50,4231 50.3173 49.6202 49.3125 50.1346 50.3798 48.8029 48.3510 49.1298 48.9423 54.0433 55.1106 54.5769 56,2163 54,5240 53.3269 2655 65.9183 SALES
L.
1
t
L. 1.. L. L.. L. 1... 1.. L, L, L. L. L. L. L, L. L. L. L. L. L. L. L. L, L, L, L. L. L. L. L. L. L. L. L. 1.. L. L. L, L, L. L. L. L. L, L. L.
SALES SAL ES SAl ES SAl I'S SALES Sal. e s SA i S SAl ES SALES SALES SALES SALES SALES SALES SAl E S SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES iD E D I C A T E D
*y 0 0 0 'J '•}
0 0 1 0 2
53 54 55 56 5? 511 5V 60 61 62
0 1 0 0 1 1 2 0 0 2 0 0 1 0
63 64 65 66 67 68 69 70 71 72 73 74 75 76 77
0 0 0 0 2 1 0 5 6 0 0 2 0 0 3 0 0 9 57 78 21 39 61
78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 3917
MAX.DIFF. 252 .0000 1045 AV. ABG.INV.
24.8140 t-
1
AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV.AV. AV. AV. AV. AV. AV.
INT. INT. INT. INI . INT. INT. INT. INT. INT. INT. INF. INT. INT. INT. INT. INT. INT. INT. INT. INI . I NT. INI . INT. TNT. INT. INT. INT . INT. INT, INT. INF. INT. INT. INT. INT. INT. INT. INT, INT. INT. INT. INT. INT. INT, INT, INT. INT. INT. INT. INT. INT. INT.
39.1635 41.8173 38.4567 39.2981 38.3365 38.3125 38.2356 38.3894 38.4375 31.5048 32.3125 33.2500 31.8221 32.5865 31.4856 33.3125 33.0817 33.6971 32.4231 32.3750 32.7115 32.2356 32.7981 31.5017 17.9712 18,1442 18,7692 18,9183 18,0481 18,6394 18,4327 18.4087 18,0721 18,1250 18,5337 18,5913 17,7981 17.8654 18,5577 17.9712 18.1010 17.6298 18.6779 18.0240 18.3942 17.4663 18.5144 17.9904 18.1106 18.9423 18.8221 18.6106
L.SALES SAL rs L 1. .SALFS L. SALFS SAL ES SALES SALES SALES SALES SALES SALES SALES L. SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES • SALES L • SALES SALES SALES SALES SALES SALES L, SALES SALFS SALES L, SALES SALES L, SALES SALES SALES SALES SALES SALES SALES SALFS SALES SALES SALES SALES SALES SALES SALES SALES SALES
L, L. L. L, L. L. L, L. L, L, L. L. L. L. L. L. L, L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L, L. L. L, L, L. L. L. L. L. L.
0 0 8 22 14 0 4 8 10 4 4 0 0 7
0 0 7 0 1 0 1 0 0 1 0
0 0 0 0 0 0 0 1 0
0
0 0 0 0 0 0 0 0 0 0 0 6 0 0 0 0 0
1
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 IB
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
AV. AV. AV. AV. AV. AV. AV.
AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV, AV, AV, AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV.
INT . INT. INT. INT . INT. INT. INT .' IN 1 , TNT . INT. INT. INT. INT. INT . INT. INT. INT. INT. INT. INT. INT. INT. INT. INT, TNT. INT. INT. INT. INT. INT. INT. INT. INT, INT, INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT.
RANDOMIZED
18.5096 19.2212 18.4327 18.8365 17.8462
I . SAL IS L. SAL FS L. SALES L. SAl ES L. SAl ES 18.V471 L . SAILS 18 . M M L .SAl I., n 1 7.W 1 L •> . SAI r:s 17.2452 L. SALES 17.7500 L. SALES 17.8654 L. SALES 18.5000 L. SALES 18.1779 L. SALES 18.5385 L. SALES 18.5433 L, SALES 18.2067 L. SALES 18.3942 L. SALES 18.2019 L. SALES 18.0625 L. SALES 18.3654 L. SALES 18.1490 L. SAL IS 19.0240 L. SALES 18.4808 L. SALES 18.1635 L. SALES 17.9760 L. SALES 19.2019 L. SALES 18,6923 L. SALES 18.7067 L, SALES 17.6298 L. SALES 17.8029 L, SALES 18.8173 L. SALES 18.6538 L. SALES 49.9808 L. SALES 49.7067 L. SALES 49.2548 L. SALES 49.0192 L, SALES 49.6106 L. SALES 50.0817 L. SALES 48.5913 L. SALES 47.8269 L. SALES 48.8317 L, SALES 48.4712 L, SALES 59.7740 L, SALES 60.2212 L. SALES 59.2115 L. SALES 61.6635 L. SALES 59.5385 L. SALES 58.0529 L. SALES
0 0 o 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 7 0 0
2 0 0 3 3 0 4 15 15 5 13 6
2967 DEDICATED
61.0144 AV.DIFF. TOT. LOST SALES
2
53 54 55 56 5/ 58 *.<'•> .')>'>
61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 4210
12.0000 MAX.DIFF. 176 AV. AOG. INV.
AV. AV, AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV.
INT. INT. INT. TNT. INT. INT . INI . INT. INT. INT • TNT. INT. INT. INT. INT. INT. TNT. INT. INT . INT. INT. INT. INT. INT. INT. INT. INT. INT . INT. INT. INT. INT, INT, INT, INT, INT, INT. INT. INT. INT. INT. INT, INT. INT. INT. INT. INT. INT. INT. INT. INT. INT.
35.29.13 37.8894 34.6442 35.4856 '34.6250 34.4856 34.3990 34.6058 34.5962 28.3173 28.8269 29.8702 28.3317 29.2788 27.9087 29.8029 29.7452 30.4808 29.1394 28.9087 29.2067 28.8846 29.4567 28.2452 17.9712 18.1442 18.7692 18.9183 18.0481 18.6394 18.4327 18.4087 18.0721 18.1250 18.5337 18.5913 17.7981 17.8654 18.5577 17.9712 18.1010 17.6298 18.6779 18.0240 18.3942 17.4663 18.5144 17.9904 18.1106 18.9423 18.8221 18.6106
L, SALES L. SALES L. SALES L, SAL ES L. SALES L. SALES L. SALES L. SALES L• SALES L. SALES L • SALES L. SALES L, SALES L, SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SAL FS L. SALES L. SALES L. SALES L. SALES L. SAIES L. SALES L. SALES L. SAIES L. SAIES L. SALES L. SALES L. SALES L. SAL ES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. BALES L. SALES L. SALES L. SALES L. SALES L. SALES T. SALES L. SALES I. • SALES L. SALES
9 5 21 35 34 12 13 23 21 14 11 0 2 15 1 3 12 3 5 4 7 5 7 2 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 6 0 0 0 0 0
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 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV, AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV, AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV, AV, AV, AV, AV. AV. AV.
INT. INT. INT. INT. TNT. INT. INT, INT, INT. INT, INT, INT. INT. INT. INT, INT, INT. INT. INT. INT. INI . INT. INT. INT. INT. INT. INT. INT. INT. INT . INT. INT. INT. INT. INT. INT. INI , INT. INT • INT. INT, INT. INT, INT. INT. INT . INT. INT.
RANDOMIZED
18.5096 19.22J2 111. 43..'/ 1H.8.5.,5 17.8462 18.9471 18.3173 17.9712 17.2452 17,7500 17,8654 IB.5000 IB.1779 18.5385 18.5433 18.2067 18.3942 18.2019 18.0625 18.3654 IB.1496 19.0240 18.4808 18.1635 17.9760 19.2019 18.6923 18.7067 17.6298 17.8029 18.8173 18.6538 49.9808 49.7067 49,2548 49.0192 49.6106 50.0817 48.5913 47.8269 48.8317 48.4712 56.7019 57.2596 56.2308 58.5096 56.4904 55.2452 876
AV.DIFF. 62.7260 TUT. LOST SAL ES
L , SAL ES L . SAL ES I . SAL FS I , SAL ES L . SAL IS L. SALES L. SALES L. SALES L. SALES L, SALES L. SALES L. SALES L. SALES L. SALES L. SALES L • SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L, SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES U SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L, SALES L. SALES L. SALES L. SALES L. SALES DEDICATED
0 53 0 54 0 0 0 58 0 59 0 60 0 61 0 62 0 63 0 64 0 65 0 66 0 67 0 68 0 69 0 70 0 71 0 72 0 73 0 74 0 75 0 76 0 77 0 78 0 O' 7 9 80 0 81 0 82 0 83 0 84 0 85 5 86 7 87 0 88 0 89 2 90 0 91 0 92 3 93 3 94 0 95 7 96 43 97 50 98 11 26 99 2 7 100 4104
229,0000 MAX.DIFF. 455 AV, AGFI. INV,
26,6360
1
AV.
INT,
9.6250
L.
SALES
47
53
L.
SALES
9.6058
L.
SALES
35 44
54
3
INT. INT.
9.1)029
16
AV. AV, AV.
INT.
9.8365
L,
SALES
35
56
AV. AV.
INT . INT.
9.5385 9.8365
L.
45
L.
SAl ES SAl L S
49
57 50
INT.
39.0288
L,
SALES
23
AV.
INT,
38.6923
L.
SALES
34
AV.
INT •
38.4471
1 . SAl ES
55
AV.
TNT.
38.4423
L.
SALES
19
4
AV.
INT.
38.6394
L.
SALES
29
5
AV. AV.
INT. INT.
38.2596 38.8077
L. L.
SALES SAl. ES
22 15
6 7
AV,
INT.
9.A058
L.
SALES
65
59
AV.
INT.
38.8365
1.. SALES
19
8
AV.
INT.
9.5048
L,
SALES
20 0
9 10
AV.
INT.
9.5048
L.
SALES
44 58
61
AV.
INT.
9.6202
L.
SALES
57
62
SALES
3
11 12
AV. AV.
INT. INT.
9.5144 9.6875
L.
0
SALES SALES
61 35
63 64
.
AV, AV.
INT. INT.
38.8702 29.3029
L. SALES L . SALES
AV.
INT.
29.1442
L.
60
AV. AV.
INT. INT.
29.6010 28.8510
L . SALES L, SALES
' 8
13
AV,
INT.
9.BB46
L.
SALES
39
65
AV.
INT.
29.0817
L.
SALES
14
14
AV.
INT.
9.6971
L,
SALES
37
66
28.6394 29.4663
L. L.
SALES SALES
1 2
15 16
AV.
INT.
9.6010
L,
SAl ES
33
67
AV.
INT.
9.5529
L,
SALES
58
68
8
17
69
4
AV, AV.
INT. INT.
L.
AV.
INT.
29,4952
1.. SALES
AV.
INT.
9.9038
L,
SALES
45
AV. AV.
INT. INT.
29.6971
L. L,
SALES SALES
INT.
9.7308
L,
SALES
48
70
4
18 19
AV.
29,2708
AV.
INI .
9,6779
L.
SALES
61
71
AV.
INT.
29,1779
L.
SAl ES
0
20
9.7308
INT.
29,7163 29.4519
L. L.
SAL ES SALES
6 8
21
AV. AV,
INT.
AV. AV.
INT.
9.6731
L, L,
SALES SALES
40 46
73
22
AV.
INT.
9.7404
L,
SALES
28
74
AV.
INT.
29.1971
L.
SALES
10
23
AV. AV.
INT. INT.
29.2885
SALES
AV. AV.
INI . INT.
9.6587 9.7212
L, L,
SALES SALES
52 58
75 76
SALES
3 28
24
9.4135
L. L.
25
AV.
INT.
9.6394
L.
SALES
47
77
AV.
INT.
9.8029
L,
SALES
43
26
INT.
9.7837
SALES
41
27
9.8077 9.7115
L.
SALES SALES
25 35
78 79
AV.
INT.
9.7452
L. L.
INT. INT.
L,
AV.
AV. AV.
SALES
42
28
AV.
INT.
9.6971
L,
SALES
25
80
AV,
INT.
9.7212
L.
SAL ES
41
29
INT,
L.
SALES
45
L,
SALES SALES
71 49
81 82
L.
SALES
33
30 31
9.4663 9.5433
INT.
9.7500 9.8221
INT. INT.
L.
AV, AV.
AV. AV. AV.
INT.
9.6779
L.
SALES
50
83
AV.
INT,
9.6875
L.
SALES
45
32
AV.
INT. INT.
9.5577 9.7067
L. L.
SALES SALES
52 47
33 34
AV.
9.8125 46.3413
L, L,
SALES SALES
35
AV, AV.
INT. INT.
0
84 85
AV.
INT.
46.3173
L,
SALES
0
86
AV.
INT.
L.
SALES
51
35
AV.
INT.
46.3413
L.
SALES
0
87
AV. AV.
INT. INT.
9.4904 9.7067
L. L.
SALES SAl ES
36 78
36 37
AV.
INT.
46.2885
L,
SALES
0
88
AV.
INT.
46.3798
L.
SALES
0
89
L.
SALES
0
L,
SALES
0
90 91
INT.
9.5625
72
AV.
INT.
9.5625
L.
SALES
64
38
AV,
AV. AV.
INT.
L. • SALES L. SALES
47 49
39 40
AV.
INT. INT.
46.6683 46.2115
INT.
9.6683 9.5769
AV.
INT.
45.7885
L.
SALES
0
92
AV.
INT.
9.7212
L.
SALES
42
41
AV. AV.
INT.
9.6394
49
L, L.
SALES SALES
0 0
93 94
41
42 43
46.2260 46.1442
9.8029
SALES SALES
INT. INT.
INT.
L. L,
AV. AV. AV.
INT.
61.8846
L,
SALES
0
95
AV.
INT.
9.6202
L.
SALES
36
44
INT. INT.
9.6298
49 50
45 46
61.9471 61.6346
L.
SALES SALES
0 0
96 97
9.5000
L. SALES L.. SALES
INT. INT.
L.
AV. AV.
AV. AV. AV.
INT.
62.7115
L.
SALES
0
98
AV.
INT.
9.5721
L.
SALES
43
47
AV.
62.3702
AV.
9.5673 9.5577
L. L.
SALES
48 49
AV.
SALES SALES
0 7
100
SALES
50 45
L. L.
AV.
INT. INT.
INT. INT.
AV.
INT.
9,8798
L.
SALES
40
50
RANDOMIZED
AV. AV.
INT. INT.
9.6394 9.6587
L. L.
SALES SALES
39 51
51 52
AV.DIFF. TOT.
LOST
61.5192
232 4 126,1971 SALES
3187
DEDICATED MAX,DIFF. 3009
AV,
99
411.0000 AGG.
INV.
AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV.
INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INI . INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT, INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT.
54.5192 55.3413 54.4519 54.7805 54.4279 54.5865 54.4423 54.4519 54.3990 28.7115 28.6779 29.0721 28.3942 28.7981 28.0433 28.8990 29.0769 29.3846 28.5673 28.7933 28.9808 28.8462 28.7019 28.8798 11.9519 11.8365 12.2260 12.1106 11.7644 11.9856 11.9519 12.1635 11.8750 11.8990 11.8221 12.0096 11.8173 11.8894 11.9471 11.7163 12.0433 11.8462 12.0192 11.8798 12.0385 11.6635 11.7933 11.6875 11,8269 12.3654 12.1058 11.9567
L. L, L. L. L. L. L. L. L. L. L. L, L. L. L. L. L. L. L. L. L, L. L. U L, L. L. L, L, L. L. L. L. U L. L. L. L. L. L. L. L. L• L. L. L. L. L. L. L. L. L.
SALES SALES SALFS SALES SAL ES SALES SALES SALES SALES SALES SAL ES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALFS SALES SALFS SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES
0 0 0 0 0 0 0 0 0 . 3 5 4 12 12 1 4 8 6 4 1 6 15 11 3 7 20 8 12 14 20 14 20 25 15 16 18 28 22 11 20 7 17 9 12 17 17 10 24
a
14 5 7
<
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 2*? 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV, AV, AV. AV. AV. AV. AV. AV. AV, AV. AV, AV. AV, AV, AV, AV, AV, AV. AV, AV, AV,
INT. INT. INT. INT. INT. INT. INT. INT. INT • INT. INT. INT. INT. INT. INT. INI . INT. INT. INT. INI. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT • INT. INT. INT. INI. INT. INT. INT. INT. INT. INT. INT. INT. INT.
RANDOMIZED
12.0048 1J.8654 12.1779 12.2356 11.7356 12.158 7 12.0529 11 .8990 11.8750 11 .8510 11.9375 11.8798 12.0192 11.8173 11.9471 12.0625 12.0577 12.0048 11.7692 11.8094 11.8750 12.1683 11.9952 11.8462 11.8077 12.1587 12.0529 12.0962 11.8125 11.7067 12.0240 12.1010 54.7837 54.9423 54.6779 54.5817 54.8990 54.9663 54.4663 54.3125 54.5673 54.5721 71.6106 71.7644 70.9760 72.2740 71.4904 70.6442 2776
AV.DIFF, 117.7308 TOT. LOST SALES
L. L. L, L. L. 1... L. L. L. L, L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L• L. L. L. L. L. L. L. L. L. L. L. L.
SALES SALES SALFS SALES SALES
s
•>AI FS S Al R FS SAL SALES SALES SALES SALES SALFS SALES SALES SALES SALES SALES SALES SALES SALES SALES SALESSALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES
DEDICATED
11 7 6 6 16 4 25 15 24 14 28 9 14 11 8 15 17 14 27 16 13 7 19 14 23 8 11 16 22 11 15 13 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
53 54 55 56 57 58 59 AO 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 3581
443.0000 MAX.DIFF. 971 AV. AGG. INV.
26.1584 On
AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV.
INT. INT. INT . INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT, INT. INT, INT.
60.0577 60.9279 59.7115 59.9567 59.6971 59.7452 59.7163 59,7644 59.7644 34.7644 34.6202 35.0000 34.7212 34.9760 34.3750 34,7260 34.7933 35.2644 34,6FT 75 35.0192 34.9471 34.9H08 34.8094 34.6298 16.7596 16.7596 16.9471 16.9183 16.7260 16.9760 16.9183 16.9856 16,7740 16.8413 16.9856 17.0240 16.8750 16.5721 16.7933 16.4087 16.6779 16.5625 16.9327 16.8173 16.9327 16,7308 16,8510 16.8462 16.5577 17.1683 17.1/79 17.1010
L, SALES L. SALES L. SALES L. SALES L, SALES L. SALES 1.. SAL ES L. SALES L • SALES L. SALES L. SALES L. SALES L, SALES L. SALES L. SALES L. SALES L. SALES L. SALES L . SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L, SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L, SALES L, SALES L. SALES L. SALES L. SALES L, SALES L. SALES 1.. SALES L. SALES L. SALES L. SALES
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 1 0
2
1 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 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV.
INT. INT. INT. INI. TNT. INI . TNF . INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT . INT. INT» INT . INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT . INT. INT, INT. INT. INT . INT. INT. INT. INT. INT. INT. INT» INT.
RANDOMIZED
16,8077 17.1394 16.8462 17.0529 16.6731 17.0V62 16.00 77 16.6298 16.4231 16.6394 16.0510 17.0305 16.8990 17.0096 16.9008 16.0990 16.9712 16.8462 16.6731 16.7708 16.0942 17.1538 16,9375 16.0077 16.0046 17.3125 16.8558 "16.7548 16.5577 16.6154 17.1010 16,9327 61.7837 61.9423 61.6779 61.5817 61,8990 61.9663 61.4663 61.3125 61.5673 61.5721 79.4519 78.9183 78.3750 79.4567 78.6490 77.9615
I , SALES I . SALES L. SALES L. SALES L . SAL. ES L. SAL ES 1 . (•.Ml F S L, SALES L. SALES L. SALES L. SALES L. SALES L, SALES L. SALES L, SALES L, SALES L , SALES L, SALES L. SALES L. SALES L, SALES L. SALES L, SALES L, SALES L, SALES L. SALES L, SALES L. SALES L. SALES L, SALES L. 5ALES L. SALES L. SALES L. SALES L. SALES L. SALES L, SALES L, SALES L. SALES L. SALES L, SALES L, SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES
3273
AV.DIFF. 104,4375 TOT. LOST SALES
DEDICATED
0 0 0 0 0 0 0 0 6 0 0 0 0 0 0 2 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
53 54 55 56
57
50 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 68 89 90 91 92 93 94 95 96 97 98 99 100
4078
MAX.DIFF. 369.0000 17 AV. AGO. INV.
AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV, AV. AV, AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV.
INT. INT. INT. INT. INT. INT, INT. INT, INT, INT, INT. INT, INT. INT, INT. INT. INT. INT, INT, INI, INT. INT. INT, INT. INT. INT, INT, INT. INT. INT. INT. INT. INT. INT. INT, INT. INT. INT, INT. INT, INT, INT. INT. INT. INT. INT. INT, INT. INT, INT. JNT. INT.
64,0577 64,9712 . 63,7540 63.9567 63.6971 63.7452 63.7163 63,7644 63.7644 39.0962 39.1490 39.2981 39.3221 39.5962 38.8750 39.4279 39.5865 39.7452 39,4056 39.6587 39.3510 39.5721 39.5481 39.3846 19.1731 19,3750 19,3510 19.4904 19.3269 19.4519 19.4327 19,2788 19.2692 19.1442 19.4663 19.3125 19.0529 19.1731 19.3462 19.1058 19.0769 19.2452 19.4087 19.3558 19.3846 19,0817 19.3894 19.3654 19.2596 19.4856 19.5240 19.3462
L. L. L, L. L. L. L. L. L. L. L. L. L. L. L.
SALFS SAL. ES SALES SALES SALES SALES SALE'S SALES SALES SALES SALES SALES SAILS SALES SAL ES SALES L. SALES L. SAL ES L, SALES L. SALES L. SALES L. SALES L. SALES L. SALES L, SALES L, SALES L. SALES L, SALES L, SALES L. SALES L. SALES L, SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L, SALES L. SALES L. SALES !_. SALES L. SALES L, SALES L_ , SALES L. SALES L. SALES L, SAL ES L. SALES L. SALES L. SALES L. SALES
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
1 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 •>•)
23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV, AV. AV. AV, AV, AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV.
INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INI , INT. INT, INT, INT, INT. INT, INT, INT, INT, INT, INT, INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT.
RANDOMIZED
19.278B 19.6250 19,2260 19.5000 19.1346 19.4856 19.30 77 19.1346 19.1490 19.206 7 19,1346 19.4808 19.2212 19.4375 19.3894 19.3029 19.4375 19.3317 19.3269 19.3702 19.2596 19.4567 19.3702 19.2500 19.2452 19.5962 19.4567 19.5577 19.1490 19.2115 19,5865 19.5481 65.7837 65.9423 65.6779 65.5817 65.8990 65.9663 65.4663 65.3125 65.5673 65.5721 83.4519 82.91B3 82.3750 83.4567 82.6490 81.9615
L.
1... L. L. L. L. L. L. 1.. L, L• L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L. L, L. L.
3603
AV.DIFF. 87.6923 TOT. LOST SALES
IIAL. T-.Y SAL ES SAL ES SALES SALES SALFS SALES SAL. ES SALES SALES SAL ES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES
DEDICATED
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 4363
331.0000 MAX.DIFF. 1 AV. AGO. INV.
AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV.
INT. INT. IN R. IN R. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT.
26,644? 28.0144 25,7548 25.9904 26.0625 25.7596 25.7452 25.8365 25.5721 18.2740 18.0433 18.7404 18.4567 19.1587 17.8558 18.7692 18.5673 18.9760 18.5000 18.8173 18.6010 18.8221 18.7933 18.4327 11.2596 11.7308 11.9760 12.1106 11.4087 12.1442 11.7788 11.8702 11.5481 11.4760 11.7740 11.7548 11.3365 11.2404 11.7404 11.5192 11.3942 11.3029 11.8894 11.4327 11.7356 11.1442 11.8029 11.4615 11.3654 12.1731 12.0048 11,7837
L. L. L. L, L. L. L. L. L. L. L. L. L. L. L.
L. I. L. L.
SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES SALES
172 83 170 142 203 171 178 176 158 250 223 191 265 277 275 196 203 164 218 248 235 231 207 263 18 31 15 10 17 31 18 33 32 21 18 24 38 22 12 35 21 31 16 28 23 39 29 25 27 11 8 9
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 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV. AV, AV. AV. AV. AV. AV. AV. AV.
INT. INT. INT. INT. INT, INT. INT. TNT. INI . INT. INT. INT. INT. INT. INT. INT. INT . INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT. INT, INT. INT, INT. INT, INT, INT, INT.
RANDOMIZED
11,8413 12,3894 11,7019 12,0673 11,1058 12,3413 11,5913 IT,1075 10,9712 11,2644 11,2596 11,7308 11,8221 11,8029 11,6635 11,7115 11,9760 11.5913 11.5433 11.8029 11.4808 12.2212 11.8317 11.5721 11.3894 12.4856 11.9760 11.9231 11.1346 11.3029 12.1587 11.9663 42.8125 42.7067 41.6202 41.3125 42.4087 42.3798 41.1490 40.7692 41.3606 40.9423 43.2644 45.8558 45.3990 45.4327 44.5288 43,8125 2105
AV.DIFF. 92.7644 TUT. LOST SALES
L, SALES L. SALES L. SALES L. SAL ES L. SAL.F:S L. SALES L, SALES 1 . SAL ES L. SAL ES L. SALES L. SALES L. SALES L. SALES L, SALES L. SALES L, SALES L, SALES L. SALES L, SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L, SALES L. SALES L. SALES L. SALES L. SALE9 L, SALES L. SALES L. SALES L. SALES L, SALES L, SALES L. SALES L. SALES L. SALES L. SALES L. SALES L, SALES " L. SALES L â&#x20AC;¢ SALES L. SALES L. SALES L. SALES L. SALES DEDICATED
23 9 21 11 23 28 29 20 40 34 22 13 36 10 15 37 30 23 25 31 28 14 29 30 31 14 16 8 38 25 10 15 32 33 0 0 21 0 24 32 16 0 38 201 230 58 151 212
53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 3417
MAX.DIFF. 329.0000 7327 AV. AOG. INV,
19.0091 00
INT. 44.1346 L . SALES AV. INT. 45.8798 L. SALES AV. INT. 43.5337 1...SALES AV. INT. 43.9567 L. SALES AV. INT. 43.4135 L . SALE'S AV. INT. 43.5385 L. SALES AV. INT. 43.4615 L •SALES AV. INT. 43.5337 L. SALES AV. INT. 43.5144 L, SALES AV. INT. 26.8990 L. SALES AV. INT. 27.0817 L. SALES AV. INT. 27.6731 L. SALES AV. INT. 27.2885 L. SALES AV. INT. 27.7885 L. SALES AV. INT. 26.5865 L . SALES AV. INT. 27.8365 L . SALES AV. INT. 27.8413 L, SALES AV. INT. 28,2356 L. SALES AV. INT. 27.5192 L. SAl F S AV. INT. 27.5577 L. SALES AV. IN I . 27.4952 L. SALES AV. INT. 27.7019 L. SALES AV. INT. 27.8173 L. SALES AV. INT. 27.4231 L. SALES AV. INI • 15.0962 L. SALES AV. INT. 15.3269 L. SALES AV. INT. 15.5481 L. SALES AV. I NT . 15.7115 L. SALES AV. INT. 15.2788 L. SALES AV. INT. 15.6106 L. SALE'S AV. INT. 15.6154 L. SALES AV. INT. 15.3413 L. SALES AV. INT, 15.2260 L» SATES AV. INT. 15.1106 L. SALES AV. INT. 15.5481 L. SALES AV. INT. 15.3798 L. SALES AV. INT. 14.8798 L. L. SALES AV. INT. 15,0000 SALES AV. INT. 15.4663 L. SALES AV. INT. 15.0385 L. SALES AV. INT, 15.0625 L.SALES AV. INI. 14.9712 L. SALES AV. INT. 15.5192 L. SALES AV. INT. 15.2596 L. SALES AV. INT. 15.4135 L. SALES AV. INT. 14.8558 L. SALES AV. INT. 15.4856 1..SALES AV. INT. 15,2308 L. SALES AV. INT, 15.1202 1 .SALE'S AV. INT. 15.7644 L. 1 .SALES AV. INT. 15.6923 SALES AV. INT . 15.4 760t. SALFS %
0 0 0 3 0 0 0 0 0 13 5 15 20 28 24 14 8 6 17 1 16 18 24 28 1 4 1 0 4 5 6 3 5 1 0 '2 1 'o 0 0 1 0 0 1 4 3 1 0 0 1
i X r
}
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
AV. INT. 15.3702 L. SALES 1 53 AV. INT. 15.9279 1. S .ALES 0 54 AV. INT. 15.3125 L .SALES 1 55 AV. INI. 15.7067 1 . 0 56 AV. INF. 14.9663 1 .S AlLES S 0 57 A AV. INI. L5.7163 1. S S .A l L S 2 59 AV, INI . L •SAl i < : > 1 59 AV. INI . 1 5 , 0 6 '51 . SAl 1 S 60 1 5 . 3 3 1 7 AV. INT . 14.8173 L. SALES 6 61 AV. INT. 14.9904 L. SALES 2 62 AV. INT. 15.0625 L. SALES 63 AV. TNT. 15.5865 L. SALES 0 64 AV. INT. 15.2115 L. SALES 65 AV. INT. 15.5000 L. SALES 0 66 AV. INT. 15.4856 L. SALES 0 67 AV. TNT. 15.3029 L. SALES 4 68 AV, INT. 15.4615 L. SALES 0 69 AV. INT. 15.3077 L. SALES 1 70 AV. INT. 15.3317 L. SALES 3 71 AV. INT. 15.4135 L. SALES 0 72 AV. INT. 15.2019 L. SALES 4 73 AV. INT. 15.7163 L. SALES 0 74 AV. INT . 15.4279 L. SALES 0 75 AV. INT. 15.2019 L. SALES 2 76 AV. INT. 15.2500 L. SALES 3 77 AV. INT. 15.9231 L. SALES 0 78 AV, INT. 15~.58i7 L. SALES 0 79 AV. INT. 15.7163 L. SALES 0 80 AV. INT. 14.9087 L. SALES 0 81 AV. INT. 15.0865 L. SALES 0 82 AV. INT. 15.8125 L. SALES 1 83 AV. INT. 15.6490 L. SALES 0 B4 AV. INT. 54.0673 L. SALES 0 85 AV. INT. 54.1250 L. SALES 0 B6 AV. INT. 53.6106 L. SALES 0 87 AV. INT. 53.4135 L. SALES 0 88 AV. INT. 53.9808 L. SALES 0 89 AV. INT.- 54.1779 L. SALES 0 90 AV. INT. 53.1490 L. SALES 0 91 AV. INT. 52.8029 L. SALES 0 92 AV. INT. 53.3606 L .SALES 0 93 AV. INT. 53.2404 L. SALES 0 94 AV. INT. 63.0962 L. SALES 0 95 AV. INT. 62.9856 L. SALES 0 96 AV. INT. 62.2981 L. SALES 16 97 AV..INT. 64.0048 L. SALES 0 98 AV. INT. 62.4856 L.SALES 0 99 AV. INT. 61.4760 L. SALES 9 100 RANDOMIZED 2816 DEDICATED 3906 AV.DIFF. 90.2500 MAX.DIFF. 359.0000 TOT. LOST SALES 349 AV, AGG, INV.
f
•»
•t
AV.
INT.
16.9279
AV.
INT •
17,4135
L.
SAl IS
0
54
INT.
1
55
INT.
16,9327 17 .290.1
L . SALES
3
AV. AV.
0
56
4
AV.
INT.
16.3462
1... SALTS
0
57
0
5
SALES
0
6
INT. INT .
1"'. 1027 16.9327
L. SAl ES 1 . SAl IS
0 1
58 59
SALES SAl. ES
0 0
7 8
AV. AV. AV .
TNT .
16.A 779
1 . SAl 1 S
0
60
INT.
61
0
9
INT.
L. L.
3
SALES
16.2067 16.3654
SALES
L,
AV. AV.
SALES
0
62
L, L.
SALES
10
AV.
INT.
16,6106
L.
SALES
1
63
SALES
2 1
11
AV.
SALES
0
64
SALES
2
12
AV.
16.8990 16.6490
L.
L,
INT. INT.
L.
SALES
1
65
SALES SALES
5 4
13 14
AV.
INT.
16.9135
L.
SALES
0
66
30.7500
L. L.
AV,
INT,
16.9038
L.
SALES
0
67
INT.
30.3077
L.
SALES
5
15
AV,
INT.
16.8413
L.
SALES
3
68
AV. AV.
INT. INT.
31.2500
0 5
16 17
AV.
INT,
16.8510
L,
SALES
0
69
30.9B56
L.. SALES L. SALES
AV,
INT.
16,7548
I... SALES
0
70
AV.
INT.
31.6683
L.
SALES
0
AV.
INT.
16,4056
L.
SALES
0
71
AV. AV.
INT. INT,
30.8462 30.8317
L. L.
SALES
1
. 18 19
AV,
INT,
16.7837
L.
SALES
0
72
SALES
0
20
AV.
INT ,
30.8798
L • SALES
6
21
AV. AV.
INI . INT.
16.8558 17.1923
L, L,
SALES SALES
0 0
73 74
AV.
INT,
30.7163
L.
SALES
0
22
AV.
INT.
16.9103
L.
SALES
0
75
AV,
INT ,
31.1923
L,
SALES
6
23
AV,
INT,
30.6442
L,
SALES
3
24
AV. AV.
INT. INT.
16.7212 '16.6154
L. L.
SALES SALES
0 0
76 77
AV. AV,
INT,
16.5721 16.6875
L, L,
SALES
INT,
SALES
0 3
25 26
AV,
INT,
16.9135
L,
SALES
0
27
AV. AV,
INT, INT,
17.3269 16,6923
L, L,
SALES SALES
0 0
28 29
AV.
INT,
16,9471
L,
SALES
2
30
AV, AV,
INT, INT.
16.8510 16.8798
L. SALES L, SALES
0 0
31 32
AV,
INT,
16,6971
L.
SALES
0
33
AV, AV,
INT. INT.
16.6010 16.9615
L, L,
SALES SALES
0 0
34 35
AV.
INT.
16,8894
L,
SALES
1
AV.
INT.
16,5529
L,
SALES
AV.
INT.
16,4231
L,
SALES
AV.
INT.
17.0337
L,
AV.
46.1346 47,8790
L. SALES L. SALES
0 0
INT.
45.57;?t
L.
SAl ES
0
INT,
45.9760
L.
SALES
1
AV.
INT,
45.4135
L.
SALES
AV.
INT.
45.5385
L.
AV. AV.
INT. INT,
45.4615 45.5337
L, L.
AV.
INT.
45.5144
AV. AV.
INT. INT.
30.2019 30.7163
AV,
INT.
31.3269
AV.
30.4056
AV.
INT . INT.
AV.
AV. AV.
INT. INT.
AV. AV.
1 2
53
L.
sail
AV.
INT,
17.403B
L.
SALES
0
78
AV. AV.
INT. INT.
16,9008
SALES
17.1250
L. L.
SALES
3 0
80
AV.
INT.
16.3077
L,
SALES
0
81
AV. AV.
INT. INT.
16.5433 17.0962
L. L.
SALES SALES
0 0
82 83
AV.
INT,
17.0288
L.
SALES
0
84
AV. AV.
INT.
56.0673
L.
56.1250
L.
0 0
85
INT,
SALES SALES
AV.
INT.
55.6106
L,
SALES
0
87
INT, INT.
55.4135
L.
36
AV. AV.
55.9808
L.
SALES SALES
0 0
88 89
0
37
AV.
INT,
56.1779
L.
SALES
0
90
0
38
SALES
0
39
AV. AV.
INT. INT.
55,1490 54.8029
L. L.
SALES SALES
0 0
91 92
79
86
INT. INT.
16,6346
L,
SALES
55.3606
L.
SALES
0
93
SALES
40 41
INT.
L,
0 0
AV,
16.8317
AV,
L.
SALES
16,4519
L,
SALES
0
42
AV.
65.2019
L.
SALES
0 0
94
INT.
INT. INT,
55.2404
AV. AV, AV.
INT,
17,1106 16.6923
L. L.
SALES SALES
0 0
43 44
AV,
INT.
65.4183
L.
SALES
0
96
INT.
AV,
INT,
64.4038
L.
SALES
3
97
AV.
INT.
16,7885
L,
SALES
0
45
AV,
INT,
66.4808
L.
SALES
0
98
AV. AV.
INT. INT.
16.2885
SALES
1
INT,
64.8894
L.
SALES
0
99
SALES
0
46 47
AV.
16.8654
L. L.
AV.
INT.
63.6058
L.
SALES
7
100
AV,
INT.
16.6442
L.
SALES
0
48
AV, AV.
INT. INT .
16.6394 17.2404
L. L.
SALES SALES
0 0
49 50
RANDOMIZED
AV.
INT.
17.0240
L . SALES
0
51
AV.DIFF.
AV.
INI .
17,0673
L . SALES
. 0
52
TOT.
AV.
LOST
3004 76.2115 SALES
DEDICATED MAX.DIFF. 73
AV.
95
4047 239.0000
AGO.
INV.
AV. INT. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. AV. INT. AV. INT. AV. INT. INT. AV. INT. AV. INF . AV. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. AV. INT. INT, AV. INT. AV, INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. AV. INT. AV, INT. AV. INT. AV. INT. AV. IN r. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. AV. INT. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV. INT. AV.
40.1346 49.8798 47,3721 47.9663 47.4135 47.5385 47.4615 47.5337 47.5144 32.8/50 33.40B7 33.9663 33.1346 33.5337 33.0048 34.1058 33.8846 34.2212 33.4183 33.4279 33.6538 33.2740 33.6923 32.8125 17.9567 18.1875 18.5385 18.7788 18.1538 18.4760 18.2692 18.4038 18.0913 18.1971 18.4471 18.2212 17.8654 18.1298 18.5577 18.2788 18.2260 17.7356 18.5192 18.1442 18.2163 17.7596 18.3462 18.1202 18.0577 18.7740 18.5385 18.4183
L, SALES L. SALFS 1..SALFS L, SALES L. SALES L, SALES L. SAl ES L .SAILS L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L. SALES L . SALES L, SALES L. SALES L. SALES L. SALES L. SAL ES L. SALES L. SALES L. SALES L . SALES I_. SALES L. SALES L . SALES I_. SALES L. SALES L. SALES L. SALES L. SALES L. SAl ES 1..SALES L. SALES L. SALES L. SALES 1..SAl ES L. SALES L. SALES L. SALES L. SALES L. SALES L. SAl. ES L. SALES L. SALES L. SAl ES
0 0 0 0 0 0 0 0 0 2 0 0 1 0 0 0 5 0 0 0 3 3 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
AV. INT. 18.3221 L. SAl ES 1 53 AV. INF. 18.9038 L. SAl ES 0 54 AV. INT. 18.2212 L. SALES 0 55 AV. INT. 18.6683 L, SALES 0 56 AV. INT. 17.8125 L. SALE'S 0 57 AV. INT. 18.8029 L. SAL E S 0 58 AV. INT. 18.3269 L, SAl i S 0 59 AV. INT. 18.2452 L. SAL ES 0 60 AV. INT. 17.6827 L. SALES 2 61 AV, INT. 17,8 798 L. SALES 0 62 AV. INT. 18,0048 L. SALES 0 63 AV. INT. 18.5385 L. SALES 0 64 AV. INT. 18.2308 L. SALE'S 0 65 AV. INT. 18.3269 L. SALES 0 66 AV, INT, 18.5337 L, SALES 0 67 AV. INT. 18.3077 L. SALES 0 68 AV. INT. 18.3942 L, SALES 0 69 AV. INT. 18.2596 L. SALES 0 70 AV. INT. 18.1442 L, SALES 0 71 AV. INT. 18.3365 L. SALES 0 72 AV. INT. 18.1154 L. SALES 0 73 AV. INT, 18.5962 L. SALES 0 74 AV. INT. 18.432 7L. SALES 0 75 AV. INT. 18.1635 L. SALES 0 76 AV. INT. 18.0337 L. SALES 0 77 AV. INT. 18.7644 L. SALES 0 78 AV. INT. 18.3846 L. SALES 0 79 AV. INT. 18.5673 L. SALES 0 80 AV. INT. 17.8413 L. SALES 0 81 AV. INT. 18.0817 L, SALES 0 82 AV. INT. 18.8846 L. SALES 0 83 AV. INT. 18.5337 L. SALFS 0 84 AV. INT. 58.0673 L. SALES 0 85 AV. INT. 58.1250 L. SALES 0 86 AV. INT. 57.6106 L. SALES 0 87 AV. INT. 57.4135 L. SALES 0 88 AV. INT. 57.9808 L. SALES 0 89 AV. INT. 58,1779 L. SALES 0 90 AV. INT. 57.1490 L. SALES 0 91 AV, INT. 56.8029 L. SALES 0 92 AV. INT. 57.3606 L. SALES 0 93 AV. INT. 57.2404 L. SALES 0 94 AV. INT. 67.2500 L.SALES 0 95 AV. INT. 67.6587 L. SALES 0 96 AV. INT. 66.7019 L, SALES 0 97 AV. INT. 68.7356 L. SALES 0 98 AV. INT. 67.2404 L. SALES 0 99 AV. INT. 65.8413 L. SALES 0 100 RANDOMIZED 3143 D1 EDICATED 4190 AV.DIFF. 62.1010 MAX.DIFF. 2'3.0000 TOT.LOST SALES 18 AV. AGO. INV. 1
185
APPENDIX 3 Output of all possible regressions based on simulation runs
1
INPUT DATA 7.4914 . 7 2 4? '624 11 .. 4C 2 • 144 5.6041 1.418? I AS,9183 10.1478 -"1.4733 2.4207 ) 62.101 '' 1.3331 4.1793 i 7 I 14 .. 32 417123 t « 9 03 .8 27 31 7,0406 • J 1. 9 12 .. 47 24 34 34 • 19.422* i 87.4923 1 • 1.2109 2.3713 1 I 1.1 20 44 4.43-8 '• 9.9023 117.731 14.4349 • 1.29 '• 1 13 T7 12 .4 3. 71 SU MS O3 F VARI8 AB 9 4L .E 4S 17 14 57 .. 10 34 01 ; WEA NS OF VARIABLES < - 6 16 .. 30 79 54 48 4 7.4924S 1
SI 3 4 S r CROSS775P5R 1M 28 70 30 .. 8O9 .O 6D 4U .C -T >S • 7 7 0 3 . 6 4 121.303 95S.773 310478. 27607.4 1 . 0 6 ? 1 7 £ t 9 4 14 .4 36 0. 23 22 7E+6 r9.41781EU 2029.84 1 2 9 3 . 8 20.9407 121.303 35407.
403.701
229 212 223 239 i 339 : 1 329 331 369 443 ; 411 ' 33?7j
52441 44944 33424 49729 37121 128881 100241 109541 136141 194249 168921
24.014 30.1053 28.3083 26.4134 19.0091 34.7993 31.4201 24.1504 22.0214
'- 1103473 £ - •297.942' i
306.018
100514. i 27.0874
310478. - 1.06917E+8- • 25407. 9.U701E+4 2059.84 27607.4 384493957 * 91330.4 1115733 1.39270Etlt 2.75340E+7 18 384493957 2.7D340Cr7 8259.0 91350.6 1.50227E+6 405.701 4446.32
CO S70.M5A5T2 RIX 5R 2R 6E 5C .T 3E ?D S 4 4 54 39 2 47 .0 9. 15 307.862 1 8 0 1 4 . 2 -26.1317 1474.33 1.17243C+7 912468. 0 -414 05 0. 47 03 .4 -232.213 . 1 3 1 0 -6 . 0 1 9 2 7 4.91549 .127025 SI MFLE CORRELA. T3 I6 O7 N503MATRIX 12 -. 60021 . .3 74 79 75 50 13 5 . 7 6 1 2 4 7 .325804 -.2C0G23 .963181 .31001 ' -• . 2 .. 384 0 6 7 3 8 -.743233 34 54 27 2? 3 -. 2 6 0 0 2 1 1 .777513
18014.2 1474.35 44697.4 4.30420E+7 -443.41 , -24.1317
1.17243C+7 -243.934 912468. -232.215 4.J042OE+7 -465.41 2.01402EH0 41C3C0. -415880. 103.774 -18540. -4.1348
.941247 .325804 1 .993313 -.107324 -.200823
.943181 .31001 .993513 1 -.18044 -.304738
-.244479 -.943253 -.187526
HOW MANY VARIABLES (INCLUDES A DEPENDENT VARIABLE) * IF NO MORE» PLEASE TYPE O(ZCRO) ? 2 WHAT ARE THEY. (LIST THE DEPENDENT VARIABLE LAST) T 1,6 VARIABLES ARE 1 4 J
HOU MANY VARIABLES (INCLUDES A DEPENDENT VARIABLE) IT NO MORE. FLEASE TYPC 0<ZCRO) ? 2 UIIAT ARE THEY. (LIST THE DEPENDENT VARIABLE LAST) T 3,4 |, VARIABLES ARE 3 4
NORMAL EQUATION MATRIX 3245.39
NORMAL EOUATION MATRIX , 64497.6
RIGHT HAND SIDE VECTOR -4.81927
RIGHT HAND SIDE VECTOR -26.1317
INVERSE OF THE MATRIX 1 .B9920E-4
INVERSE OF THE MATRIX i 1.49930E-3
T R E O C0EF-1.2f311E-3 INTERCEPT 1.48697
ANALYSIS Of UARIANCCVAfil.MICN DF TOTAL 10 REGRESSION 1 ERROK 9
! 3 REG C0EF-3.91793E-4 1 INTERCEPT 1.47646
... g . SS .129825 B.B3173E-3 .120993
ANALYSIS OF VARIANCE VARIATION I>F •TOTAL 10 REGRESSION 1 ERROR 9
B.B3173E-3 1.34437E-2
MULTIPLE CORRELATION - R .240821 MULTIFLE CORR COEF - R*R * 4.80278E-2
F TEST 1.02382C-2 1.32874E-2
MULTIFLE CORRELATION - R .200023 MULTIPLE CORR COEF » R*R - 7.BB415E-2
hou Many variables (includes a dependent variable) IF NO MORE. PLEASE TYPE O(ZEftO) T 2 WHAT ARC THEY. (LIST THE DEPENDENT VARIABLE LAST) T VARIABLES ARE 2 4
HOU MANY VARIABLES (INCLUDES A DEPENDENT VARIABLE) IT NO MORE. FLEACE TYPE O(ZERO) T 2 UIIAT ARE THEY. (LIST THE DCrCHBENT VARIABLE LAST) T 4,6 VARIABLES ARE 4 6
4.
NORMAL EOUATION MATRIX 307.E62
- NORMAL EOUATION MATRIX 2.01402EU0
H I OUT HANO SIDE VECTOR 4.91349
j • RIGHT HAND 9IDE VECTOR -** -10340.
INVERSE OF THE MATRIX 3.240Z1E-3
—-.-«-'
'INVERSE OF THE MATRIX 3.3G363E-11 1
2 REO COEF 1.59645E-2 INTERCEPT 1.25244
ANALYSIS OF VARIANCE VARIATION IT TQTitL 10 REGRESSION 1 ERROR 9
S3 .129823 1.023C2E-2 .117^37
4 REO C0EF-4.3QB43E-7 INTERCEPT 1.44169
CS .129023 7.04832E-2 .051342
MULTIPLC CORRELATION -ft.777513 MULTIPLE CORA COCF - RtR - .40433
MS 7.B4832E-2 5.70467E-3
'
ANALYSIS OF VARIANCE VARIATION LK TOTAL 10 regression 1 f:rro(; 9
SS .12982S .012213 • 117.il
MULTIFLE CORRELATION = R .306730 MULTIPLE COfcR COCf - R*R « 9.400E1E-2
F TEST .012215 I.3CA70E-
.934741 0 0
HOU MANY VARIABLES
i L
(INCLUDES A DEPENDENT VARIADLE)
IF NO MORE, FLEASE
•4 HOU M A N Y T T
NO
jWHAT - |
VARIABLES
MORE, ARE
PLEASE
THEY.
VARIABLES
HAND
TYFE
(LIST
ARE
5
NCI«HI%L E O U A T I C N ;08.774
RIGHT
(INCLUDES A D E P E N D E N T O(ZERO)
THE
? 2 '
DEPENDENT
VARIABLE)
.
"
i
VARIABLES ARE.
VARIABLE
LAST)
4
i
5265.39
470.55?
470.532
307.842
VECTOR
MATRIX INVERSE OF THE MATRIX 2.19945E-4
OF
3.74200E-3
1.94877
VARIANCE
VARIATION
-
D F
S S
10
.129823
REGRESSION
1
9.05662E-2
9.05662E-2
.03923?
4.36211E-3
-
9
-
1
REG
2
REG COEF
C0EF-3.15242E-3
INTERCEPT
M S X
TOTAL ERROR
-3.34203E-4
-3.36205E-4
COEF-2.19034E-2
ANALYSIS
4
NORHAL EQUATION MATRIX
• RIGHT HAND SIDE VECTOR i-4.01927 4.91349
INVERSE OF THE 3.27733E-J
INTERCEPT
2
MATRIX
SIDE
t
REG
1
T 5.4
-4.1348
3
TYPE O(ZERO) T 3
WHAT ARE THEY. (LIST THE DEPENDENT VARIABLE LAST) f 1.2.6
2.07051E-2 1.487
* ANALYSIS OF VARIANCE VARIATION
CORRELATION
- R
MULTIPLE
CORA
- R*R - .477401
COEF
.835223
S3
PF 1 0
.129025
REGRESSION
2
.123668
ERROR
8
6.15763E-3
TOTAL MULTIPLE
MULTIPLE CORRELATION
R
MULTITLC CORR
RtR *
COEr
4.18338E-2 7.69704E-4
.975997 ,95237
CO CO
W O H M A N Y V A R B I L E S N I C ( L U D E S A D E P N D E N T V A A R I B L E ;J. H O U M A N Y V A R B I L E S N I C ( L U D E S A D E P N D E N T V A A R I B L E ) F I N O M O R E , L E A r G C T Y P E O Z C ( R O ) T 3 iV rW N O M O R E . P A S E T Y P E O Z E ( R O ) T 3 W H A T A R C T H E Y . L ( S I T T H E D E P N D E N T V A R B I L E L A S H A T R E EY.L (S IT 4T H ED E P N D E N TV A R B I L E L A S T ) f » t 3 4 . A A R I P L E 3 R E 1 4 4 ' A R B IL E SA A P ETH 1 3 N R M Q T I7t M A X T R I! 3O 26A 53L .? 1E 7.U 2O 4A 3N E {G N O R M A L E Q U O A N T M A X T I G R H I T H A N D D S I E V E C T O R R I189T H H A N D D S I E V E C T O R C 4 . 1 7 2 7 1 0 5 1 0 . 6 8 . 1 2 7 2 6 1 . 3 1 7 27R 44S 37E 7I3-Or01-9.4T 00S 240.9E3-O 4F6A 7.47-.T N IV E H A X T R I N E R H E M A X T R I 6127..E C 7E 1E4-M ;IV .321E 3 2 0 3 2 3 8 E 4 4 3 5 7 . 1 8 0 1 2 1 . ! 1 R E G C O E F 3 2 7 . 8 4 4 E 3 1 R E G C C E F S 9 6 . 4 1 4 E 4 4 K E G C 0 C 1 . r 4 4 7 7 ? E 4 3 D C E 3 F 2 5 3 . 7 E N T I E R C C T 3 1 3 . 6 5 2 IT E R C E P T 4 1 7 . 4 8 5 A -IN .A 7 5 2 3 8 4 9 1 7 . 3 4 6 4 7 4 1 . 4 E 1 0 -I.07471E-4 A N A L Y S S I O F V A R N I C E N A L S Y S I O * f V A R N I C E V A O R T I N D T S V A O R T I N D T S r T E S T 1 2 . 7 8 2 3 T O T L 1 0 F T E S T T O T L • 1 0 1 2 . 7 0 2 3 ' 0 1 4 3 6 8 R E G R O E S N I 2 R E G R O E S N I 2 1 3 C 0 5 E 2 1 3 9 . 0 2 4 E 3 4 9 . 7 7 7 8 1 . 3 4 5 7 3 4 . 7 4 2 3 E O R 8 E R C R 8 1 . 7 4 4 3 4 7 3 2 R 3O 3.E 2-F67141.0672178.401E3U P L E T IC O RE L O A N IC !M M U L E T IC C O RE L O A N T IO R *T V F L E T IP O R C E F--RRt28.274•7979.374E2- M i-l'J.irE-s 1
J
I-*
C O
I I HOU
MANY
VARIABLES
(INCLUDES
A
DEPENDENT
HO KONCR FLEASE TYPE C ( Z E R O ) T 3 : WHAT ARC T H E Y . < L I S T T H E DEPENDENT I VARIABLES ARE I S A
VARIABLE)
R O W MANY VARIABLES (INCLUDES A D E P E N D E N T VARIABLE) IF NO MORE. PLEASE TYPE O ( Z E R O ) T 3 WHAT ARC THEY. (LIST THE DEPENDENT VARIABLE LAST) T ' VARIABLES ARE 2 3 6
"
! IF
VARIABLE
LAST)
T
2.366
1.5.A
1
NORMAL EQUATION MATRIX 307.062 1476.35
'
; N03.1AL E O U A T I O N MATRIX I 5J6S.39 -245.734 ' -245.934
188.774
-
-
RIGHT HAND SIDE VECTOR 4.915-1? -26.1317 RIGHT
HAND
SIDE
-6.01927 INVERSE
VECTOR
-4.134S
OF
THE
2.02225E-4
2.6345BE-4
2.A3458E-4
5.44056E-3
1 5
REG RED
. INVERSE OF THE MATRIX J 3.43374C-3 ' -0.04J73E-5 -8.04373E-5 1.67735E-5
MATRIX
2 REG COEF 1.77644E-2 3 REG C0ER-8.33700L~-4 INTERCEPT 1.47735
C0EF-2.46037E-3 C0EF-2.51192E-2
INTERCEPT 2.26C3? , ANALYSIS OF ; VARIATION TOTAL ; REGRC3SI0N • ERROR
VARIANCE DF 10 2 8
SS .129823 .120693 9.12984E-3
•
ANALYSIS OF VARIANCE VARIATION W* TOTAL . 10 REGRESSION " 2 ERROR 8 A.03477E-2 1.14123E-3
MULTIPLE
CORRELATION CORR
COEF
R R*R
.964197 -
SS .127025 .117722 9.90340E-3
MS 5.9740?E-2 1.23793C-3
32.8793
MULTIPLE MULTIFLE MULTIPLE
-
CORRELATION
•
CORR
•
CCEF
R .961102 R«R .923717
.929676
t
V0 O
O U M A N Y V A R B I L E S N I C ( L U D E S ft D E P N D E N T V A A R I B L E ) * \•iIF IH N O M O R E . F A S E T Y E r O Z E ( R O ) T 3 H O U M A N Y V A R B I L E S N I C ( L U D E S A D E P N D E N T V A A R I B L E ) W H A T A R E T H E Y . L ( S I T T H E D E P N D E N T V A R B I L E L S T ) T 2 4 r > 6 i r N O M O R E , L F E A D E T Y P E O Z C ( R O ) f 3 V R B I L E S A S E 2 4 * W H A T A R E T H E Y . L ( S I T T H E D E P N D E N T V A R B I L E L A S T ) V A R B I L E S A R E 2 5 6 IJ N C K M A L E O U O A N T I M A X T R I 30976.2 • 912466. N O R M A L E O U O A N T I2.15M A X T R I 3 0 B 7 . 6 2 2 3 *•}] O R I93124T H H A N D D S I E V E C T O R 9 4 1 . 9 1 0 5 4 0 . C R H I T H A N D D S I E V E C T O R 4 6 B . 8 2 1 . 4 0 2 E 1 0 . 9 4 1 . 5 4 9 4 1 . 3 4 8 2 3 2 . 1 5 0 7 3 7 . 4 N IV E R S E O F T H E M A X T R I 39.337E3- 6I.I-3247N IV E R S EC2-O F53.4T H EE2-M A X T R I 5 4 0 . 2 3 B A ]j N R E O F 0.42-43E2< 4C R E O13.3C C OE 30F 1.J6-179E •N '53.04C 2 R E O C 0 E F 6 7 9 . 0 4 5 E 3 I2 T E R E P T 3O 0E 5 R C C 1 3 4 3 6 C 2 I T E R E P T 2 0 . 4 1 8 6 2 3 7 . 2 6 9 E 2 A O R T I N • D F G9 HB A N A L Y 3 S I O F V A R N I C E .\V T A L 1 0 1 2 . 9 8 2 3 V A O R T I N O F S M S R E G R O E S N I 2 1 , 2 1 3 8 A 0 7 . 9 0 1 E 2 1 2 . 9 8 2 5 T O T L 1 0 E O • • 8 • • B 2 4 . 3 0 S 3 0 1 3 . 0 6 4 T R E G R O E S N I 2 9 8 7 9 S E 2 3 4 9 . 3 9 9 E E O R 8 7 3 . 4 S 4 7 1 7 IM U P L E T IC O R R E L O A N T I -ft M U P L E T IC O RE L O A N T IC .4*1261-70.772 O E F- R•8R C O RC O E F - Rt 1
«•
-1.I6324E-?
1
ANALYSIS
O F
3.V3147E-11
VARIANCE
-j-
*
-
«•
. W 7 7 2 5
'
MULTIPLE
•
.736471
• I
i /
'.:..,.L,
V
- 4
i
HOU
HOU HAHY VARIABLES <INCLUDES A DEPENDENT VARIABLE) IF NO MOREf FLEASE TYPE O(ZERO) T 3 WHAT,ARE THEY. (LIST THE DEPENDENT VARIABLE LAST) ? 3.4-4 VARIABLES ARE 3 4 6 NORMAL EQUATION MATRIX 64497.4 4.30420E47 4.30420E47
IT
RIGHT HAND -26.1317
3.47751E-3
2.74809E-?
.
3 REO COEF 2.58130E-3 4 R E O C0EF-4.40709E-4 I N T E R C E P T 1.0414
SS .129825
1.79419E-2 .111863
MULTIPLE CORRELATION -
|
MULTIPLE CORR
L.
COEF
ARE
R .37194
MS 8.98093E-3 1.39829E-2-
(INCLUDES TYPE THE
3
6
5
A DEPENDENT
O(ZERO)
(LIST
VARIABLE)
T 3
DEPENDENT
VARIABLE LAST) T
3,3.4
103.774
SIDE
INVERSE O FTHE 1.55395E-3
INVERSE OF THE MATRIX 1.13944E-3 -1.77343E-A
ANALYSIS OF VARIANCE VARIATION DP TOTAL 1 0 REGRESSION 2 ERROR 8 •
PLEASE
THEY.
-665.41
2.B1402E+10
'
VARIABLES
NORMAL EQUATION MATRIX 66697.6 -665.41
; 1
-18340.
•1.77343E-4
ARE
VARIABLES
RIGHT HAND SIDE VECTOR -24.1317
MANY
N OMORE.
WHAT
F TEST .642278
VECTOR -4.1348
MATRIX ~ ' 5.47751E-3 5.49041E-3
3 REG C0EF-6.32557E-4 5 REG C0CF-2.41331E-2 INTERCEPT 2.22451
ANALYSIS O F VARIANCE VARIATION D F TOTAL 1 0 RECRESSION 2 ERROR 0
MULTIPLE CORRELATION MULTIPLE CORR COEF
SS .129823 .116315 1.35090E-2
F TEST 3.81377E-2
34,4380
1.68073E-3
R .94434 RtR » .895930
- R*R - .130334
. . . ... _ .
;
TO
I
NOW MANY VARIABLES (INCLUDES A DEPENDENT VARIABLE) IR NO MORE. FLEASE TYPE O(ZERO) T 3 UHAT ARE THEY. (LIST THE DEPENDENT VARIABLE LAST) T 4.3.4 VARIABLES ARE 4 3 6 NORMAL EQUATION MATRIX 2.B1402EU0 -415080. -413380.
NORMAL EOUATION MATRIX 5265.39 470.552
I80.774
RIOHT HAND SIDE VECTOR -1B540. -4.1348 INVERSE OF THE MATRIX 3.67323E-11 8.07232C-8 8.07232E-8
HOU MANY VARIABLES (INCLUDES A DCPCNDENT VARIABLE) IF NO MORE. TLEACE TYFC 0<ZCRO) ? 4 WHAT ARE THEY. (LIST THE DEPENDENT VARIABLE LAST) T 1.2.3.6 VARIABLES ARE 1 2 3 6
4 RED COEF-1.01342E-A 3 REO C0EF-2.41409E-2 INTERCEPT 2.13146 ANALYST9 OF VARIANCE VARIATION DF TOTAL 10 REGRESSION 2 F ERRO* 0 MULTIPLE CORRELATION MULTIPLE CORR COEF
307.862
1476.33
10014.2
1476.35
66697.6
RIGHT HAND SIDE VECTOR -6.01927 4.91549
-26.1317
INVERSE OF THE MATRIX , 2.62295E-3 -6.0444BE-4
3.47341E-3
SS .127823 .118647 1.11779E-2 R .753781 R*R - .713701
MS 3.73237E-2 1.37723C-3
18014.2
470.352
-6.7327SE-4
-6.04440E-4
3.01255E-3
1.00470E-4
-6.93275E-4
1.00470E-4
2.00014E-4
1 REO C0EF-3.13457E-3 2 REG COEF 2.07023E-2 3 REG C0EF-5.20448E-6 INTLRCCFT 1.48707 ANALYSIS OF VARIANCE VARIATION DF TOTAL 10 REGRESSION 3 ERROR 7 MULTIFLE CORRELATION MULTIPLE CORR COEF
SS .127823 .123668 6.15750E-3
MS 4.12226E-2 B.79643E-4
R .775777 R»R - .732371
1
T-
MOW MANY VARIABLES (INCLUDES A DEPENDENT VARIABLE) IF NO MORE. PLEASE TYPE O(ZERO) T 4 •*. • WHAT ARC THEY. (LIST THE DEPENDENT VARIABLE LAST) T l»2t4»6 VARIABLES ARE 1 2 4 6 NORMAL EQUATION MATRIX 5265.39 470.352 1.17243E+7 470.352 307.862 912468. 2.81402EM0 1.17243E+7 912468.
HOU MANY VARIABLES (INCLUDES A DEPENDENT VARIABLE) IT NO MORCt FLEACC TYPE 0(ZCRO> ? 4 UHAT ARE THEY. (LIST THE DEPENDENT VARIABLE LAST) T 1.2.5.4 VARIADLCS ARE 1 2 5 4 NORMAL EQUATION MATRIX 5265.3? 470.532 -245.934 470.532 -232.215 307.062 180.774 -243.934 -232.213
RIGHT HAND BIDE VECTOR -4.31927 4.9154?
RIGHT HAND SIDE VECTOR •
INVERSE OF THE MATRIX 2.84707E-3 -9.24717E-4 -7.24717E-. -1.13421E-4
3.89391E-3 2.39010E-7
-6.01927
INVERSE OF THE MATRIX 2.72192E-4 -2.05914E-3
-1.13421E-4 2.59010E-7 3.00B41E-10
-2.03914E-3 -2.1793BE-3
4.04013E-2 7.1B641E-2
-2.17838E-3 7.18641E-2 9.O0607E-2
1 REG C0EF-2.97067E-3 2 REG COEF 1.47020C-2 3 RCG COEF-7.50097C-3 ' INTERCEPT 1.72307
i J REO C0EF-2.32412E-3 2 REO COEF 2.04443E-2 4 REG C0CX-r.74&04E-7 INTERCEPT 1.44170
1
ANALYSI3 OF VARIANCE VARIATION DF TOTAL 10 REGRESSION 3 ERROR 7
4.91549
-10540.
SS .127823 .123818 4.00720E-3
MULTIPLE CORRELATION - R .97439 MULTIPLE CORR COEF - R«R « .953728
F TEST 4.12724E-2 8.50182E-4
ANALYSIS OF VARIANCE VARIATION ur TOTAL 10 REGRESSION 3 ERROR 7
SS .127023 .124301 3.52378E-3
MULTIPLE CORRELATION R .978493 MULTIPLE CORR COEF R»R - .937452
4.14338E-2 7.09111E-4
j HOU MANY VARIABLES (INCLUDES A DEPENDENT VARIABLE) ' IT NO MORE. PLEASE TYPE O(ZERO) T' "4 WHAT ARE THEY. (LIST THE DEPENDENT VARIABLE LAST) T 1.3.4,4 VARIABLES ARE 1 3 4 6 NORMAL EOUATION MATRIX 5263.3? 10014.2 | •\
18014.2 1.17243E+7
66677.4 4.30420E+7
1
HOU MANY VARIABLES <INCLUDES A DEPENDENT VARIABLE) IF NO MORE. PLEASE TYPE O(ZERD) T 4 UHAT ARE THEY. (LIST THE DEPENDENT VARIABLE LAST) T 1.3.3.4 VARIABLES ARE 1 3 3 6
' NORMAL EOUATION MATRIX
l.I7243Ef7 4.30420E+7
,
0263.37
10014.2
-243.734
2.B1402EH0
'
18014.2
64497.6
-465.41
-245.734 RIOHT HAND SIDE VECTOR -4.81727 -26.1317
j INVERSE OP THE MATRIX 2.68133E-3 -2.52474E-4
-18340.
RIGHT HAND SIDE VECTOR -6.01727 -24.1317
1.18322E-3 -1.70440E-4
INVERSE OF THE MATRIX 2.66017E-3 -7.00030E-4 2.04415E-4
7.67105E-4 -2.02720E-4
2.74735E-7 9.67105E-4
-2.02720E-4
,003042
j
') 1 REO COEF 1.84508E-3 3 REO COEF 2.40547E-3 4 REO C0EF-S.I1552E-4 INTERCEFT ,784164
ANALYSIS OF VARIANCE VARIATION DF TOTAL XO REGRESSION 3 ERROR 7
-4.1348
-1.70440E-* , -7.00830E-4
-7.30932E-7
130.774
!
-7.30952F.-7 *
•2.32474E-4
-665.41
'
1 REG C0EF-3.61631E-3 3 REO COEF 3.31044E-4 5 REG C0EF-2.54470E-2 INTERCEPT 2.27309
-• SS .127825 1.72S72E-2 .110544
i MULTIPLE CORRELATION • R .383138 MULTIPLE CORR COEF • R»R - .148347
MS 4.41772E-3 1.57931E-2
F TEST ,406436
! ANALYSIS OF VARIANCE VARIATION Dr TOTAL 10 REGRESSION 3 ERROR 7
SS .129823 .121231 B.57372E-3
M S 4.04103E-2 1.22767C-3
P TEST 32.9163
MULTIPLE CORRELATION - R .966336 MULTIPLE CORR COEr • R*R = » .933C05
\0 L/1
!
i
HO U NM ANYMORVE A. RIABFLLEESASE(INTCYLPUEDES0(2 AEROD) EPEN?DEN4 T VARIABLE) r .i WV HA AR TIAOBA R E T H E Y . ( L I S T T H E D C P E N D C N T V A RIABLE LAS LES ARE 14 3 4 "T) T 1,4,3,4 M7 A2 T4 R3 IE Xt7 5765.3E 9QUATION1.1 -243.934 1.17243E+7 2.81402EMO -415880. 188.774 -243.934" -415BBO. R I G H T H A N D S I D E V E C T O R -4.81727 -1S540. -4.1348 I2.£X344E-3 NVERSE OF T H E M A T R I X -1.14778E-4 1.13133E-3 -1.1479BE-4 S.14480E-10 -3.83781E-7 1.13133E-3 -3.B37B1E-7 3.92373E-3 1R RE EG OC CO OE EF F-2 2. .4 41 10 37 02 4C E-8 3 4 3 R E G C O E F 2 . 5 1 0 1 3 1 2 INTERCCFT 2.24344 A A LA YT SI IO SN OF VA R VN AT RA I D FIANCE S S F TEST , 1 2 9 8 2 T O L 1 0 . 1 22 0S 67 73 4 R E G R E S S I O N 3 31 20 2E E-23 30.8S03 9.1 1E-3 4 ERROR 7 1. .0 32 04 NORMAL
M UL LT TI IP PL LE EC CO OR RR RELA TE IF ON«•R» RR .»944. 29 02 29663 MU CO
H U NO MOVR AE R. IABLP EL SEAS( INCTLYUrDECSO( AZERDO E) PENDTENT4 VARIABLE) IO F ET U H A T T H E Y . ( L I S T H E D E P E N D E N T VARIABLE LAST) T 2.3.4,6 VARIABLES ARE 2 3 4 4 307.862 1476.35 712468. 1476.35 66697.6 4.30420E+7 2.81402E+10 912468. RI G H T H A N D S I D E VECTOR 4.91549 -26.1317 MANY
ARC
NORMAL
EQUATION
MATRIX
4.30420E+7
INVERSE OF T HE -3.15333E-4 MATRIX
.3.69375E-3
4.08742E-7 -3.45533E-4 1.19174E-3 1.81144E-4 4.08742E-7 -1.B1144E-4 2.79332E-9 2 RE OC E F . 9 40 06 77 9E E-4 2 3 CO OE EF r-2.1 7 . 4 7 4INTR REE ERG GCEPC 0 4 3 7 1 ' 3 E 6 T 1.2J071 A N A LA YT SI IO SN OF Vt A RIANCE V A R I i r S S . 1 2 9 0 2 3 T O T A L 10 . 1 27 26 07 48 8E-3 4 R E G R E S S I O N 3 01 60 B9 27 BE E-3 2 7.7 ERROR 7 1. .1 4958 M IP PL LE EC OR RR RELAT MU UL LT TI CO CI OO EN FR R*R.9.7 940098
/ 1.4...
I . . «. ..-V. I •«5 1
MOW MANY VARIABLES (INCLUDES A DEPENDENT VARIABLE) IP NO MOREr PLEASE TYPE C(ZENO) » 4 * WHAT ARE THEY. (LIST THC DEPENDENT VARIABLE LAST) T 2.3,5.6 VARIABLES ARE 2 3 3 6 * NORMAL EOUATION MATRIX .» ! 307.862 1474.33 -232.213 -443.41 44497.4 1476.33 -232.213 188.774 -443.41 RIGHT HAND SIDE VECTOR 4.9154? -24.1317
-4.1348
HOW HANY VARIABLES (INCLUDES A DEPENDENT VARIABLE) ir N O MORE, FLEASE TYPE 0<ZCRO> T 4 WHAT ARC THEY, (LIST THE DEPENDENT VARIABLE LAST) 9 2,4,3.4 VARIABLES ARE 2 4 3 4 ; NORMAL EQUATION MATRIX i 307.042 912460. -232.213 912468. 2.81402E+10 -415000, -415080. -232.215 100.774 RIGHT HAND SIDE VECTOR 4.91549
INVERSE OF THE MATRIX 4.43737E-2 -6.40084E-4 2.22B70E-S -4.40084E-4 -7.33422E-4 7.71044E-2 2 REO COEF 1.S8399E-2 3 RES C0EF-7.94473E-4 S REO C0EF-3.21892E-3 INTERCEPT 1.44033
7.71044E-2 -7.33422E-4 9.73424E-2
INVERSE OF THE MATRIX . 6.13223E-2 -9.02997E-7 -9.02997E-7 7.34443E-2
i '
5.00293E-11 -1.00038E-4
-4.1348 7.34443E-2 -1.00038E-4 9.34383E-2
2 REO COEF 1.44920E-2 4 R E O COEF-1.22903E-4 3 R E O COCF-4.70314E-3 INTERCEPT
ANALYSIS OF VARIANCE VARIATION DF TOTAL 10 REGRESSION 3 ERROR 7
-18540.
SS .129825 .120201 9.62430E-3
MULTIPLE CORRELATION R ,94222 MULTIPLE CORR • COEF R*R - .925047
,040067 • 1.37490E-3
F TEST 29.1417
1.57125
ANALYSIS OF VARIANCE VARIATION DF TOTAL 10 REGRESSION 3 ERROR 7
SS .127023 .122073 7.75244E-3
MULTIPLE CORRELATION - R .949482 , MULTIPLE CORR COEF - R*R - .940284 .1
4.04908E-2 1.10752E-3
• HON MANY VARIABLES CINCLUDES A DEPENDENT VARIABLE) " IF NO MORE, PLEASE TYPE O(ZERO) ? 4 . UHAT ARE THEY. (LIST THE DEPENDENT VARIABLE LAST) T 3.4.3.4 i VARIABLES ARE 3 4 5 4 , NORMAL EOUATION MATRIX j 64497.4 4.30420ET7 -"443.41 4.30420E*7 2.B1402EH0 -415080. -445.41 -415S00, 188.774 RIOHT HAND SIDE VECTOR -24.1317 -18540.INVERSE OF THE MATRIX 1.14579E-3 -1.70038E-4 -1.7803BE-4 2.73370E-? 1.87022E-4 -2.04694E-7
-4.134S
RIGHT HAND SIDE VECTOR .-4.81927 4.91549
1.S7022E-4 -2.04694E-7 5.30541E-3
INVERSE OF THE MATRIX • 2.87461E-3 -8.79431E-4 -B.79431E-4 3.94273E-3 -1.B8395E-4 -2.87845E-4 j -8.81510E-7 4.7829BE-7
3 REO COEF 1.77090E-3 4 REO C0EF-3.72011E-4 3 REG C0EF-2.38568E-2 INTERCEPT 1.04073 ANALYSI9 Of VARIANCE VARIATION DF TOTAL 10 REGRESSION 3 ERROR 7
SS .129823 .121337 8.48777E-3
MULTIPLE CORRELATION - R .964739 MULTIPLE CORR COEF - R»R - .934622
HOW MANY VARIABLES (INCLUDES A DEPENDENT VARIABLE) IF NO MORE. PLEASE TYPE O(ZERO) T 5 UHAT ARE THEY. (LIST THE DEPENDENT VARIABLE LAST) T i 2 3 4 * VARIABLES ARE 1 2 3 4 * NORMAL EQUATION MATRIX 5265.39 470.352 18014.2 1.17243E+7 307.862 1474.33 470.552 912448. 1474.33 44497.4 18014.2 4.30420E47 4.30420E+7 1.17243E+7 912448. 2.81402EU0
F TE8T 4.04438E-2 I.21254E-3
-24.1317 -1.88593E-4 -2.07845E-4 1.20413E-3 -1.753D7E-4
-8.81518E-7 4.78290E-7 —1.75387E-4 3.06343C-7
SS .129825 .124524 5.30127E-3
.031131 8.0354SC-4
1 REO COEF-2.44853E-3 2 REG COEF 2.04237E-2 3 REO COEF 9.22020E-4 4 REC COCF--1.41T57C-6 INTERCEFT 1.32616 ANALYSIS OF VARIANCE VARIATION DF TOTAL 10 REGRESSION 4 ERROR 6
MULTIPLE CORRELATION R .97937 MULTIPLE CORR COEF R*R - .959146
F TEST 33.2342
1
t--
00
| I
H O U MANY VARIABLES (INCLUDES A DEPENDENT VARIABLE) * IF NO MORE. PLEASE TYPE O(ZERO) T 5 • WHAT ARE THEY. (LIST THE DEPENDENT VARIABLE LA3T) T 1 2 3 3 6 VARIABLES ARE 1 2 3 3 A NORMAL EOUATION MATRIX 5245.37 470.552 18014.2 -243.934 470.332-
307.842
1474.33
-232.213
18014.2
1476.33
64497.6
-443.41
'245.734
-232.213
RIOHT HAND SIDE VECTOR -6.81727 4.71547 INVERSE OF THE MATRIX 2.48447E-3 1.23832E-3 1.25B32E-3
4.31434E-*-
-643.41
-24,1317
HOU MANY VARIABLES (INCLUDES A DEPENDENT VARIABLE> IF NO MORE. PLEASE TYPE 0<ZERO> J 5 UHAT ARE THEY. (LIST THE DEPENDENT VARIABLE LAST) T 1 2 4 3 4 VARIABLES ARE 1 2 4 3 4 NORMAL EQUATION MATRIX 5265.37
470.553
188.774
1.17243E+7 -245.934
-4.1348
-243.934
307.062
712468.
-232.213
912448.
2.31402E+10
-415800.
-415880.
-232.213
RICHT HAND SIDE VECTOR -4.81927 4.91549 -7.2814BE-4
2.47848E-3
-1.00141E-3
7.824B4E-2
INVERSE OF THE MATRIX 2.04706C-3
-7.28148E-4
-1.00141E-3
2.I7804E-4
-1.40371E-3
2.47848E-3
7.B2484E-2
-1.40371E-3
' 7.78307E-2
-5.B6309E-4
1 REO C0EF-3.34047E-3 2 REO CCEF .014274 3 KEG C O E F 1.114BBE-4 3 REO COEF-8.30325E-3 INTERCEPT 1.7437
S3 .127823 .124358 3.44703E-3
MULTIPLE CORRELATION - R .776718 MULTIPLE CORR COEF - RTR • .937887
180.774
-4.1348
'
-5.06387E-4
.061443
-1.16164E-6
-6.63910E-7
4.31254E-4
7.33356E-2
-1.14144E-4 -4.639I0E-7 S.23662C-10 -I.17641E-6
' 4.31254E-4 7.33336E-2 -1.176.1C-6 7.3503SE-2
1 REO C0EF-2.55719E-3 2 REG COEF 1.50191E-2 4 REG C0EF-I.86374E-7 3 REO COEF-7.I7023E-3 INTERCEPT 1.67305
1 ANALYSIS OF VARIANCE VARIATION DF . TOTAL 10 REGRESSION . 4 ERROR 4
I.17243E+7
470.552
\
MS
3.10895E-2 7.11171E-4
f TEST 34.1204
ANALYSIS OF VARIANCE VARIATION DF TOTAL 10 REGRESSION 4 ERROR 6
•
8 9
.129823 .124368 3.45743E-3
3.I0919E-2 9.09572C-4
MULTIPLE CORRELATION - R .978756 MULTIPLE CORR COEF * RTR = .757963
\
I
9 TEST
/
< HOU MANY VARIABLES (INCLUDES A DEPENDENT VARIABLE) T IF NO MORE, FLEASE TYPE OCERO) T 3 WHAT ARE THEY. (LIST THE DEPENDENT VARIABLE LAST) T 1 3 4 3 6 VARIABLES ARE 1 3 4 3 6
HOU MANY VARIABLES (INCLUDES A DEPENDENT VARIABLE) IF NO MRE, PLEAGE TYPE O(ZERO) ? 3 WHAT A R E THEY. (LIST THE DEPENDENT VARIABLE LAST) T 2 3 4 5 4 VARIABLES ARE 2 3 4 5 4
NORMAL EQUATION MATRIX 18014.2 I 3263.3?
NORMAL EQUATION MATRIX 307.062 1476.33
} 18014.2 1 J 1.17243E+7 j• -243.?34
1.17243E+7
-245.934
66697.4
4.30420E+7
-443.41
4.30420E47
2.81402E+10
-415880.
-663.41
-4158B0.
RIGHT HAND SIDE VECTOR -4.81927 -24.1317 INVERSE OF THE MATRIX 2.BS982E-3 -2.34344E-4 -2.34344E-4
188.774
-232.213
66697.6
4.30420E47
-665.41
912468.
4.30420E+7
2.81402EF10
-415880.
-232.215
-415880.
-665.41
1B0.774
RIOHT HAND 8IDE VECTOR 4.91549 -26.1317
;
-4.1348 ¥
-8.29130E-7
' 1.1121BE-3
1.1B480E-3- -1.71314E-4
9.48321E-3
-8.29130E-7
-1.71314E-4
2.99339E-?
-3.23794E-7
1.11218E-3
9.48321E-3
-3.23794E-7
5.93363E-3
• INVERSE OF THE MATRIX 7.20253E-2 -3.8282J!E-3 ! -3.02822E-3 1.3692TE-3
1 REO COEF-2.B0919E-3 3 REG COEF L.??871E~3 4 REO CCEF-2.91412E-4 3 REO C0EF-2.4937?E-2 INTERCEPT 1.9603 ANALYSIS OF VARIANCE VARIATION DF TOTAL 10 ' REGRESSION 4 * ERROR 6
912468.
1474.35
4.70577E-6
8.S6489E-2
-2.03473E-6
-4.36530E-3
4.70579E-6
-2.03475E-6
3.07370E-7
5.48633E-6
8.56439E-2
-4.36530E-3
S.48633E-6
.107355
SS .129825 .123081 6.74396E-3
3.07703E-2 1.12399E-3
2 REO COEF 1.12071E-2 3 REG COEF 1.17524E-3 4 REG C0CF-2.97545E-6 5 REO C0EF-1.05279E-2 INTERCEPT 1.51063 SS .129825 .124068 3.73696E-3
MULTIPLE CORRELATION - R ,977377 MULTIPLE CORR COEF - RTR « .955654
F TEST 3.10171E-2 9.39494E-4
ANALYSIS OF VARIANCE VARIATION NR TOTAL 10 - RECRE3SI0N 4 I ERROR 6 MULTIPLE CORRELATION MULTIPLE CORR COEF
R .97368 RTR « .948054
F TEST
'I
/
I
J-NON MANY VARIABLES (INCLUDES A DEPENDENT VARIABLE) j IF NO MORE. FLEASE TYPE O(ZERO) T 6 WHAT ARE THEY. (LIST THE DEPENDENT VARIABLE LAST) T • VARIABLES ARE 1 2 3 4 S 6 NORMAL EQUATION MATRIX 3263.37 470.352
• 1,2,3,4,5,6
18014.2
1.17243E+7
-243.934
470.352
307.862
1476.33
912468.
-232.213
10014.2
1474.33
44497.4
4.30420E+7
-443.41
1.17243Ef7
912460.
4.30420E47
2.81402E+10
-415B80.
-243.734
-232.213
-443.41
-415880.
188.774
-2.3S121E-4
-0.24435E-7
1.19438E-3
I
RIGHT HAND SIDE VECTOR -4.81927 4.91549
INVERSE OF THE MATRIX 2.BB989E-3 7.07498E-3 7.09498E-3 -2.3B121E-4
.072027 -3.93404E-3
-3.83404E-3
4.74335E-*
8.347S3E-2
1.388B9E-3
-l.?4482E-4
-4.46390E-3
-0.24433E-7
4.7453SE-4
-1.94482E-4
3.30B89E-9
3.I4497E-4
1.1943BE-3
8.S4783E-2
-4.44390E-3
3.14497E-4
.107831
1 REO C0EF-2.7982IE-3 2 REG COEF 1.11384E-2 3 REG COEF 1.40380E-3 4 REG C0EF-2.17717E-6 5 REO C0EF-1.148B5E-2 INTERCEPT 1.43201
ANALYSIS OF VARIANCE VARIATION DF TOTAL 10 REGRESSION S ERROR 5
, MULTIPLE CORRELATION • MULTIPLE CORR COEF
!
I
i
SS ' .129823 .125791 4.034S1E-3
R .984339 R1R > .948924
MS 2.31581E-2 8.04902E-4
F TEST 31.1787
202
APPENDIX 4 Program listing for computing the expected S/R travel time by complete enumeration and the conventional method
203
PROGRAM MAIN(INPUTTOUTPUTtTAPES-INPUT.TAPE6-0UTPUT) DIMENSION TIME(50»350) C C C
INITIALIZE THE SYSTEM
PARAMETERS
HEHT=50.0 WTH=50.0 HEHT«=HEHT/I2.0 WTH=UTH/12.0 HVEL=240.0 VVEL-60.0 ICOL-34 NOL-3 C C C C C C C
INITIALIZE PROGRAM VARIABLES CUMT=CUMULATIVE TOTAL TRAVEL TIME FOR S C CUMS=CUMULATIVE TOTAL OF SQUARE OF TRAVEL TIME FOR S C TOT«=CUMULATIVE TOTAL TRAVEL TIME FOR DC TOS-CUMULATIVE TOTAL OF SQUARE OF TRAVEL TIME FOR DC CUMT«=0.0 CUMS=0.0 TOT»=0.0 TOS=0.0 NSL»=ICOL*NOL PHET—HEHT/2.0
C C C C C
13 C C C C
12
32
DETERMINE THE TRAVEL TIME TO EACH OPENING STORE THEM IN T I M E U F J ) ADD THE TRAVEL TIME AND ITS SQUARE TO CUMT AND CUMSRRESPECTIVELY DO 1 3 K 1 = 1 F N O L PHET=PHET+HEHT FWT=UTH/2.0 DO 13 K2=1»IC0L TH=FWT/HVEL TV»PHET/VVEL TIME(KLRK2)*TH IF(TV.GT.TH)TIME(KLFK2)«TV FWT«=FWT+WTH CUMT=CUMT+TIME(K1»K2> CUMS=CUMS+(TIME(KLFK2)**2) NS1-NSL-1 DETERMINE TOT X TOS BY ENUMERATION OVER TOTAL NUMBER OF OPENINGS DO 1 2 LM*LRNSL — IX1«( ( L M - O . O O O O D / I C O D + L.O IYL«LH-<(IX1-1)*IC0L) LT-LM+1 DO 1 2 KK=LTF NSL IX2=<(KK-O.OOOOL)/ICOL)+L.0 IY2«=KK-( (IX2-1)*IC0L> SLT»ABS< <IY1-IY2)*WTH>/HVEL VTM=ABS((IX1-IX2)*HEHT)/VVEL IF(VTM.GT.SLT)SLT»VTM ALLT=TIME(IXLFIYL)+TIME(IX2»IY2)+SLT TOT=TOT+ALLT T0S»T0S+(ALLT*»2) CONTINUE ETS=(2.0/NSL)*CUMT VTS«= < (4,0/NSL > *CUMS )-ETS**2 DEN=2,0/<NSL**2-NSL) ETD=DEN*TOT VTD=(DEN*T0S)-ETD**2 WRITE(6>32)ETSRVTSTETDTVTD FORMAT< 3X R•SINGLE"R 2X»2F?.4 F 2X »•DUAL•12X F2F9.4 >
»,
S C A N D DC T R A V E L T I M E AND T H E I R CORRESPONDING V A R I A N C E HAS BEEN COMPUTED. FOLLOWING I S FOR THE C O N V E N T I O N A L METHOD OF C O M P U T I N G E X P . SC t DC TRAVEL TIMES. HDT-WTH*ICOL VDT-HEHT4NOL IHI-((HDT/2.0>-0,OOOOOl)+1.0 IH2*(<0.75*HDT)-0.000001>+1,0 IV1-C<VDT/2,0)-0.OOOOOl)+1.0 IV2=(<0.75*VDT)-0.000001)+1.0 SIT-IH1/HVEL CT-IV1/VVEL IF<CT.GT,SIT>SIT«CT T8T«CIH2-IH1>/HV/EL CTB»(IV2-IV1)/VV£L IF(CTB.GT.TBT)TBT-CTB SC-SIT*2.0 DC-<SIT+TBT>*2,0 WRITE<A»15)SC»DC FORMAT(3X»'CONVENTIONAL SINGLE 6T0P END
I
DUAL
COMM.•rIX?2F9.
205
APPENDIX 5 Program listing for simulating dual and single command trips over a continuous normalized rack
206
PROGRAM C c C C C C C C
c C C C C C C
2 C C C
MAIN<INPUT.OUTPUTtTAPE5=INPUTRTAPE6=0UTPUTRTAPE?)
FILE NAME: SIMB THIS PROGRAM SIMULATES A RACK WITH A GIVEN S H A P E F A C T O R » B IN O R D E R T O F I N D T H E E X P E C T E D MEAN AND VARIANCE OF A DUAL C Y C L E . IT IS A L S O U S E D TO PLOT HISTOGRAPHS OF SINGLE AND DUAL C Y C L E MEAN TRAVEL TIMES. DIMENSION BB=1.000 DCT=0.O DCS=0.0
SC<1000)RDC(LOOO)
START SYSTEM SIMULATION. NO. OF T R I P S = 1 0 0 0 L O C A T I O N OF THE F I R S T SLOT= (RLFR2> L O C A T I O N OF THE S E C O N D S L O T = <R3FR4> •DCT' STORES CUMULATIVE DUAL CYCLE TRAVEL TIMES "DCS" S T O R E S CUMULATIVE SQUARE OF DUAL C Y C L E TRAVEL DO 2 K2=L»1000 R1=RANF(X) R2=BB*RANF<X> R3=RANF(X> R4=BB*RANF(X> SC(K2)=R1 IF(R2.GT.R1)SC(K2)=R2 TBT=ABS<R3-R1) T1=ABS<R4-R2) IF(T1.GT.TBT)TBT=T1 TBK=R3 IF(R4.GT.R3)TBK=R4 DC < K2)=SC < K2)+TBT+TBK SC<K2)=SC<K2)*2.0 DCS=DCS+<DC<K2)**2> DCT=DCT+DC<K2> CONTINUE END OF TRAVEL
SIMULATION. STORE T I M E S O N T A P E 9.
BOTH
SINGLE
AND
DUAL
CYCLE
C 3 4
DO 3 K3=LRL000 WRITE(9F*)SC<K3) DO 4 K4=L,1000 WRITE(9R*)DC(K4> RMEAN=DCT/1000.0 VAR=(DCS-<1000. 0*RMEAN**2))/999.0
C C C
PRINT SHAPE FACTOR? SIMULATED DUAL CYCLE TRAVEL TIME.
MEAN
AND
VARIANCE
OF
C
WRITE(617)BB ? R M E A N
7
R VAR FORMAT</7XR"SHAPE FACTOR •»F5.2»2X»"MEAN *F8.4R2X»"VAR. OF DC T.TIME"FF7.4) STOP . END 0
DC
T.TIME"»
TIMES
207
APPENDIX 6 Program listing for finding the expected mean and variance of travel time by enumeration under dedicated storage
208
R P O G R A M M A N I N < I P U T O r U T P U T A t P E S N * I P U T A t P E 6 = 0 U T P U T ) C cC fT iH lH eESInaN m e : d e d a i R P O G R A M C O M P U T E S T H E M A E N A N D V A A R I N C E S I G L E A N D D U A L C O M M A N D T R V E L M T I E S F O C A G V E N R C K . T H E S U B R O U N T I E T I U S E S I S T H T H E A S M E S U B R O U N T I E U S E D B Y < E M P > . CD M I E N O S I N A I O I ( C , ) C 1 0 > ) T N C < 1 0 ) C O M N H V E V L r V E N L f O P R H E H T = 5 0 8 1 / . 0 2 . W T H 6 H V E L = 2 0 4 . V V E L = 0 6 . N 0 P R = 5 L = 1 0 C I 0 L = 4 0 H O 7 7 K 7 N = r l 0 P R 77 N R E A D < C * 5 ) r < K 7 A » I ) < K 7 ) O A = 0 C A P C = 0 0 . D E L T O 2 K N 2 f = l 0 P R N 0 A = N 0 A A + I < ) • 23 D C A P C = C A P C + C K ( 2 ) O 3 K 3 = 1 » N 0 P R T N C K ( = 3 ) C K ( C / 3 A ) P C D O 4 4 = 1 » N 0 P R 4D D E L T = D E L T + T ( N C K ( 4 * ) * A I 2 / K ( 4 * ) * 2 * A ) I K ( 4 ) E L T = < 0 D 1 . E L T 0 > 2 . / C A L T R A V E < H E H T » W T H A I » C C r A P C T r N C D t E L N T t O L tI * E S C V , S C E r D C V r D C ) W R T I E < E 6 » 7 ) S C » V S E C D , C » V D N C , 0 A 7 *X F O M A 4 X T / / R » < ' / U E N S I G L E V ( A R ) X D U A V L ( A R ) T M I E S * t lE » 4 F 8 4 . 3 , X r I 5 ) S T O P N D S U B R O U N T I E T R A V E L < H E H W T t T H r M A I » C r C A P C T r N C D r E L T N r O *E T S rVTS f E D T rVDT) D M I E N O S I N M A I < 1 C 0 , ) < 1 0 > T , N C < 1 0 T r ) M I E 5 ( 3 0 0 , P 0 I f ) R D < 5 0 3 r 0 0 ) C O M N H V E V L r V E N L r O P R T O T M 0 = 0 . S = P H E T H E H 0 T 2 . / D O 1 2 M J N f = O l L P H E T = P » H E T + H E H T F W T = W T 0 2 . / D O 1 2 C J « 1 » C I 0 L T H = F W H T V / E L V P H E V L T M I M E J ( » O J T H F I < T V G . T . H T 1 ) r E I < M J C J r « ) T V W W F = T « T W + T 12 PIRD<1rJ» JO-0
209
D O 1 6 N N P , = O l P R C U T M 0 = 0 . S = . L ID G = M A I N ( P > 0 T 9 1 r N = l t L I G T M N I = I 1 0 0 0 0 . L X = 1 Y D O 1 7 L I E N = r l O L C O C I = r O l L FT I<M PT IR D < L I E » C I O > N . E . ) G O T O 1 7 M I E < L I E » C I O G . ) E T . M N I ) I G O T O 1 7 N I = I T M I < L E C » I O ) L X = L I E Y C O 17PC C O N N T I U E IR D L ( X L f Y = ) N P U T M = C U T M M + T I E < L X L » Y > S = S + < T L X » L Y > * M T I L E X ( » L Y > 19 T C O N N T I U E O T M = O T M < C + ( < N L G I * C P > ) / U T M > T O T S = T O T S C + ( N ( L I P G / ) * C ) U T S ) 16 E C O N N I U E T S = T 0 T M C * 0 < / A 2 . P C > V T S = < ( O 4 . C * ) T O T S < ) < C 4 / A P C * C A P C > ) T * O T M T * O T M ) S T A M 0 = 0 . R N S L C = I O L * N O L 1 N S L 1 D O 8 8 L 6 * N r l S l JJL P R = ( L ( 6 0 O . O 0 0 D C I / 0 D + 1 0 . P = L 6 J ( P R * ) l C I O L ) 7 6 + 1 D O 8 8 L 8 = L N 7 r S L K G = < < L O 0 8 . O O O C > I l / O L > + 1 0 . L M = L < 8 < K G C * 1 I ) 0 L ) S L T = < A E S < < K L M * P W J T H ) H > / V E L « • V T T M = < A B S < K G P R J * H ) E H T V > / V E L FM I(A T M G . T S L . T ) S L T = V T M T Y M M = T I E < P J R rJP)+TIME < K G A 1 = P I R D < P J R P J r ) P R 2 = I D K ( G K r L M > F I < M A * M l A 2 4 ) ? » 8 4 f 9 49 R B 0 L = A M I ( A * M 1 ) A M I ( A * D 2 ) E L T T A M R * « T A M T A + Y ( M T * N C < M A > I T * N C < M A B > 2 ) / O L ) S T A M = S T A M + < < T A Y M T * A Y M T * N M C T ( * N A ) C I < M A 2 > B > / O L > 88 E C O N N T I U E D T = R « T A M . V D = S T A M < E D E T D t T > R E T U R N E N D
210
APPENDIX 7 Program listing for simulating dual command trips over a discrete rack under dedicated storage
211
• C c C C C C C C C C
77
2 3 4
7
12 C
PROGRAM
MAIN<INPUT,0UTPUT,TAPE5«INPUT,TAPE6=0UTPUT)
FILE NAME: DEDISIM T H I S PROGRAM IS SIMILAR TO <DEDIA>. THE S U B R O U T I N E IT USES, HOWEVER, HOES NOT ENUMERATE OVER ALL P O S S I B L E DUAL C Y C L E S INSTEAD, IT U S E S S I M U L A T I O N TO C O M P U T E THE E X P E C T E D MEAN AND VARIANCE OF DUAL C Y C L E S . THE E X P E C T E D MEAN AND V A R I A N C E OF S I N G L E C Y C L E S ARE C O M P U T E D WHILE ALLOCATING THE PRODUCT C L A S S E S TO THE O P E N I N G S . H E N C E , SINGLE CYCLE IS FOUND A C C U R A T E L Y . DIMENSION IA(10),C(10),TNC<10) COMMON H V E L ,VVEL R NOPR HEHT=58.0/12.0 WTH=56.0/12.0 HVEL=240.0 VVEL=40.0 N0PR=5 N0L=10 IC0L=40 DO 77 K77*1,N0PR READ(5,*)C(K77),IA(K77) NOA=0 CAPC=0.0 DELT=0.0 DO 2 K2=1,N0PR N0A=N0A+IA(K2) CAPC=CAPC+C(K2) DO 3 K3=1,N0PR TNC(K3)=C(K3)/CAPC DO 4 K4=1,N0PR DELT=DELT+((TNC(K4)**2/IA(K4)**2)*IA(K4)> C A L L T R A V E L ( H E H T R W T H , I A , C , C A P C , T N C , D E L T t N O L tICOL t . *ESC,VSC,EDC,VDC> WRITE<6,7>ESC,VSC,EDC,VDC,N0A F0RMAT<///4X,"SIMULATED SINGLE(VAR) S DUAL<VAR) TIMES'* *1X»4F8.4»3X»I5) STOP END SUBROUTINE TRAVEL<HEHT,WTH,MIAtC*CAPC,TNCRDELT,NOLtICOLt •ETSRVTS,EDTRVDT) DIMENSION MIA(LO)RCDO),TNC<10),TIME<50,300),IPRD<50R300> C O M M O N H V E L R V V E L rNOPR TOTM=0.0 TOTS-=0.0 PHET=-HEHT/2.0 DO 12 JM=1,N0L PHET=PHET+HEHT FWT*WTH/2.0 DO 12 JC=1,IC0L TH=FWT/HVEL TV=PHET/VVEL" TIME<JM,JC>«TH IF(TV.GT.TH)TIME<JM,JC)*TV FWT«FWT+WTH IPRD(JM,JC)*0 DO 16 NP«LRNOPR CUTM=0.0 CUTS=0.0 ILG=MIA(NP)
212
D O 1 9 M N L I G r = l T M N I = I 1 0 0 0 0 . L X « = 1 Y » D O 1 7 L I E » = N l O L C O C » = I l O L F IT< P I R D L I ( E C I r O N . ) E O . ) 6 0 T O 1 7 T M E < » G E T . M N I > I G O T O 1 7 M N I = T I M I E L < I E C » I O > L X = L E Y C I O 17PR C O N N T I U E ICU D < L X » L Y = N ) P T M = C U T M + M T I L E X ( L Y r ) S = C U S + < T M I < X » L Y * T M ) I E < L X L Y , ) ) 19 T C O N N T I U E O T M = T 0 T M + C < L N ( I G / # P C ) U T M ) Q S = T O T S + < < L I G / * C ) U T S > 16 E C O N N I U E T S = T 0 T M * C < 0 / 2 A . P C ) V T S = < < 4 . O C / A P C * T ) C O ( 4 A / ( T ( S P ) C * C A P C * T ) ) O T M * T O T M ) S T A M 0 = 0 . R T A M 0 = 0 . C W 'O E T I ' I S U S E D T O K E E P T R A C K O F T H E P R C A S O C A T E D W T H I N I D V D I I U A L D U A L T R P I S W C H I F R M E D B Y C H O O S G W T O O P E N I G S R A N D O M L Y CN W E T I = 0 . S L C = I O L * N O L N = 1 0 0 .S F IT (A S L G . E 1 0 . 0 0 0 N ) N = 1 5 0 0 0 C R T M S I U L A O T I N N I O R D E R T O N F I D D U A L C C A N D V A R A I N C E . N O T E T H A T M S I U L A O T I N M T I E = N ' N CP D O 2 0 K 2 0 = 1 » N N JK R = < R A N F < X * N ) O L + 4 ) 9 0 . 9 = < R X ( * C ) I 0 L > + ? G = < R A N F < X * N ) 0 L 4 9 ) 0 . L M < R A N F < X * ) C I O L + ) 0 4 . ? ? S T = < A B S < < K L M * W P ) J T H H > / V E L = < A B S < < K G P J R * H ) E H T > V > / V E L FV IM (TTM T T M G . T S L . T ) S L T = V T M A Y M = M I P E J R ( + P » T J M ) I E < K G K r L M + S ) L T A 1 = P I R D < J , P R = 2 I D < K K G L , M > F I M ( A * M l A 2 4 ) ? f 0 » 4 ? 4? PR B 0 L = A I < M A * M 1 ) A I < M A * D 2 ) E L T 0 B = < T N C < M A > 1 T * N C < M B A / O ) 2 ) L T A M = R T A M + < T T A Y M P R f O B ) S S T A M < T A Y M T * A Y M P R * O B > W T E = I W T E + I P R O B 2C 0 C O N N U E -E C O M U PT ET T H EM W G E H IA T E D A V E R A G EO FT H ET C ED G E N R A T E D H R O U G S I U L O T I N R T T A M W / T E I V D S T ( A M W / E T I D ) T * 2 R E T U R N E N D
APPENDIX 8 Outputs obtained from those programs listed in Appendices 6 and 7
TRUE SINGLE<VAR) X DUAL(VAR) TIMES .158 CP SECONDS EXECUTION TIME
.3783
.0353
.5224
.0303
.2659
.016?
.3710
.0198
.5174
.0708
.7182
.0630
.4096
.0473
.5748
.0307
.6131
.0751
.8239
.0537
TRUE SINGLE(VAR) 3 DUAL(VAR) TIMES 2.818 CP SECONDS EXECUTION_TIME 100.»50 85..50 84.,50 70.»50
.8172
.1550
1.1112
.1204
TRUE SINGLE(VAR) I DUAL(VAR) TIMES 2.1B4 CP SECONDS EXECUTION TIME
.8216
.1557
1.1157
..1190
?100.R20 ? 10.»30
TRUE SINGLE(VAR) 8 DUAL<VAR) T.IMES .193 CP SECONDS EXECUTION TIME ? 50.R30 ? 30.»30 ? 3 0 . R45
TRUE SINGLE(VAR) T DUAL(VAR) TIMES .676 CP SECONDS EXECUTION TIME ? 90.,30 T 30.,30 ? 15.,45
TRUE SINGLE(VAR) i DUAL(VAR) TIMES .626 CP SECONDS EXECUTION TIME /LGO ? 30.,30 T 30.,30 ? 45.»45
TRUE SINGLE(VAR) I DUAL(VAR) TIMES .704 CP SECONDS EXECUTION TIME ? W0.»50 ? ?0.,50 ? 80.T50 ? 70.,50
? ? ? ?
100.,2,5 70.,35 60.fAO
t BO
40;
TRUE
SINGLE(VAR)
3.201
CP
S DUAL(VAR)
SECONDS
EXECUTION
TIMES
.5922
.1306
.8490
.0586
.1589
1.1554
.7803
.1120
1.0742
.7775
.1413
1.0933
1.0423
.1403
1.4101
TIME
100..50 98.»50 96.»50 94..50
TRUE
SINGLE(VAR)
3.023
CP
S DUAL(VAR)
SECONDS
EXECUTION
TIMES TIME
300.,60 240..60 100.,60 120.,60 60.
f60
TRUE
SINGLE(VAR)
4.944
CP
SECONDS
t
DUAL(VAR) EXECUTION
TIMES TIME
200.»40 180.
f60
liiO. .80 110..100 60.»120
TRUE
SINGLE(VAR)
8,003
CP
SECONDS
& DUAL(VAR) EXECUTION
TIMES TIME
164..80 160. .CO 106.tBO 15.0 52.
*DEL*
t CO
148..80
TRUE
SINGLE(VAR)
13.948
CP
SECONDS
8 DUAL(VAR) EXECUTION
TIMES TIME
? 50.R25 ? 25.,25
SIMULATED SINGLE(VAR) X DUAL<VAR) TIMES .067 CP SECONDS EXECUTION TIME
.4646
.3783
.0353
.2659 „
.0169
.3465
.5174
.0708
.6932
.4096
.0473
.5852
.6131
.0751
.7788
.8172
.1550
1.0473
? 100.»20 ? 10.»30
SIMULATED SINGLE(VAR) & DUAL(VAR) TIMES .064 CP SECONDS EXECUTION TIME 50.»30 30.130 30.,45
SIMULATED SINGLE(VAR) I DUAL(VAR) TIMES .192 C P SECONDS EXECUTION TIME
? 90..30 ? 30..30 ? 15.,45
SIMULATED SINGLE(VAR) X DUAL(VAR) TIMES .202 CP SECONDS EXECUTION TIME /LGO ? 30.»30 ? 30..30 ? 45..4S
SIMULATED SINGLE(VAR) X DUAL(VAR) TIMES .193 CP SECONDS EXECUTION TIME ? ? ? ?
100.,50 90.»50 80.,50 70.,50
SIMULATED SINGLE(VAR) X DUAL(VAR) TIMES .639 CP SECONDS EXECUTION TIME
r ioo.»so ? 85..30 ? 84..50 'r 70..50 SIMULATED SINGLE(VAR) & DUAL(VAR) TIMES .62? CP SECONDS.EXECUTION TIME ? 100..25 T 70..35 ? 60.160 ? 40..80
.8216 .1557 1.0517 .1074
SIMULATED SINGLE(VAR) & DUAL(VAR) TIMES .641 CP SECONDS EXECUTION TIME 100.f50 98.,50 96..50 94..50
.5922 .1306 .8076 .1145
.B586 .1539 1.0387 .1069
? ? ? ? ?
SIMULATED SINGLE(VAR) & DUAL(VAR) TIMES .655 CP SECONDS EXECUTION TIME 300..60 240..60 180..60 120..60 60..60 SIMULATED SINGLE(VAR) 8 DUAL(VAR) TIMES 1.250 CP SECONDS EXECUTION TIME k'UO.flO 180..60 150.,80 110..100 60..120
.7303 .1120 1.0466 .0840
Y ? ? ? ?
SIMULATED SINGLE(VAR) & DUAL(VAR) TIMES 2.205 CP SECONDS EXECUTION TIME 164..80 160..00 156..80 152.ȣ0 148..80
.7775 .1413 1.0800 .1230
SIMULATED SINGLE(VAR) S DUAL(VAR) TIME1 S.0423 .1403 1.3614 .1051 2.242 CP SECONDS EXECUTION TIME
? 50.F25 T 25.,205 *DEL* 25..25
SIMULATED SINGLE(VAR) X DUAL(VAR) TIMES 1.660 CP SECONDS EXECUTION TIME
.3783
.0353
.4797
.265?
.016?
.3466
? 100.,20 T 10.,30
SIMULATED SINGLE(VAR) X DUAL(VAR) TIMES 1.662 CP SECONDS EXECUTION TIME ? 50.,30 ? 30.,30 ? 30. ,45
SIMULATED SINGLE(VAR) X DUAL(VAR) TIMES 1.51? CP SECONDS EXECUTION TIME
.5174
.0708
.6767
? 90.,30 ? 30.,30 T 15.»45
SIMULATED SINGLE(VAR) X DUAL(VAR) TIMES 1.819 CP SECONDS EXECUTION TIME
.4096
.0473
.5481
.6131
.0751
.7775
.8172
.1550
1.0662
T 30.,30 T 30.,30 T 45.,45
SIMULATED SINGLE(VAR) X DUAL(VAR) TIMES 1.603 CP SECONDS EXECUTION TIME
? 100.,50 ? 90.,50 T 80.,50 T 70.,50
SIMULATED SINGLE(VAR), X DUAL(VAR) TIMES 2.074 CP SECONDS EXECUTION TIME
219 ? ? ? ?
100.»50 8 5 . ,50 81.,50 70.,50
SIMULATED SINGLE(VAR) X DUAL<VAR) TIMES 2.045 CP SECONDS EXECUTION TIME
.8216
.1557
1.0711
.1101
.5922
.1306
.8121
.1229
.3586
.158?
1.1079
.1073
SIMULATCD SINGLE(VAR) X DUAL(VAR) TIMES 3.276 CP SECONDS EXECUTION TIME 200.,40 180.,60 150.,80 110.,100 60.,120
.7803
.1120
1.0506
.0918
SIMULATED SINGLE(VAR) X DUAL(VAR) TIMES 3.O87 CP SECONDS EXECUTION TIME
.7775
.1413
1.0680
.1281
1.J.23
.1403
1.3707
.1075
7 100.,23 ? 70.,35 ? 60.,60 ? 10.,GO
SIMULATED SINGLE(VAR) X DUAL(VAR) TIMES 1.950 CP SECONDS EXECUTION TIME ? ? ? ?
100.,50 98.,50 96.,50 94.,50
SIMULATED SINGLE(VAR) & DUAL(VAR) TIMES 1.846 CP SECONDS EXECUTION TIME ? ? ? T ?
? ? ? ? ?
300.,60 240.,60 100.,60 120.,60 60.,60
164,,80 160.,30 156.>30 152.,80 143.,80
SIMULATED SINGLE(VAR) X DUAL(VAR; TIMES 4.300 CP SECONDS EXECUTION I:..L
220
APPENDIX 9 Sample runs to demonstrate that the expected mean and variance of travel time under dedicated storage are minimized when b is maximized
T H ER A C K IS S Q U A R EN I M TIE F O R 2 LEVELS N U M B E RO F LEVEL *S 1 N SIGLE C O M M A M N D (E A N S VAR ).D 3A .5L35C 0.7M 20A U O M M N D ( E A N & V A R ) . 5 . 1 4 7 0 . 9 3 4 N U M B E RO F LEVELS- 2 N SIGLE C O M M A M N D (E A N & VAR ).D 2L .94?C 0.M 20A 7M U A O M N D (E A N 4.*118VA 0.R 2)3.6 N U M B E RO F LEVELS- 3 D O M N D (E A N 4.& )1. 497V 0.A 45R N SIGLE C O M M A M N D (E A N S VAR ). U 3A .1L71C 0.3M 7A 3M N U M B E RO F LEVELS- 4 D U A L C O M M A M N D ( E A N & V A R ) . 5 . 1 9 6 0 . 9 1 0 N SIGLE C O M M A M N D (E A N & VAR ). 3.583 0.722 N U M B E RO F LEVELS- 5 D 311V1A .5R 6)0. U A LC O M M A M N D (E A N 6.& N SIGLE C O M M A M N D (E A N S VAR ). 4.267 1.273 8
T H ER A C K IS S Q U A R EN I M TIE F O R 3 LEVELS N U M B E RO F LEVELS- 1 L2.0C M A M N D (E A NftVAR ). 7.704 2.471 N SIGLE C O M M A M N D (E A NftVAR ). 5.1D 5U ?A 6O 7M N U M B E RO F LEVELS" 2 N SIGLE C O M M A M N D (E A NftVAR ). 3.146 0.398 D U A LC O M M A M N D (E A NftVAR ). 4.562 0. N U M B E RO F LEVELS" 3 N SIGLE C O M M A M N D (E A NftVAR ). 2.875 0.18? D U A LC O M M A M N D (E A NftVAR ). 4.060 0. N U M B E RO F LEVELS- 4 N SIGLE C O M M A M N D (E A NftVAR ). 2.?12 0.214 D U A LC O M M A M N D (E A NftVAR ). 4.126 0. N U M B E RO F LEVELS- 5 N SIGLE C O M M A M N D (E A NftVAR ). 3.076 0.332 D U A LC O M M A M N D (E A NftVAR ). 4.404 0. N5 N3
THE RACK IS SQUARE IN TIME FOR 4 LEVELS NUMBER OF LEVELS- 2 SINGLE COMMAND(MEAN t VAR.)
.1750 .0159
DUAL COMMAND(MEAN 8 VAR.)
.2547 .0167
.1372 .0056
DUAL COMMAND(MEAN 8 VAR.)
.1951 .0057
.1283 .0033
DUAL COMMAND(MEAN S VAR.)
.1804 .0031
SINGLE COMMAND(MEAN S VAR.)
.1295 .0036
DUAL COMMAND(MEAN S VAR.)
.1823 .0034
NUMBER OF LEVELS- 6 SINGLE COMMAND(MEAN I VAR.)
.1328 .0045
DUAL COMMAND(MEAN I VAR.)
.1881 .0045
NUMBER OF LEVELS- 3 SINGLE COMMAND(MEAN 1 VAR.) NUMBER OF LEVELS- 4 SINGLE COMMAND(MEAN & VAR.) NUMBER OF LEVELS- 5
to
T H ER A C K IS S Q U A R EN I M TIE F O R 4 LEVELS N U M B E RO F LEVELS- 2 N SIGLE C O M M A M N D (E A N & VAR ). 2.411 0.28? D U A LC O M M A N D C M A E N * VAR ). 3. N U M B E RO F LEVELS- 3 N SIGLE C O M M A M N D (E A N & VAR ). 1.922 0.102 D U A LC O M M A M N D (E A N & VAR ). N U M B E RO F LEVELS4 N SIGLE C O M M A M N D (E A N 4 VA> R. 1.825 0.066 D U A LC O M M A M N D (E A N * VAR ). 2.55 N U M B E RO F LEVELS" 5 N SIGLE C O M M A M N D (E A N * VA> R. 1.861 0.079 D U A LC O M M A M N D (E A N S VAR ). 2.6 N U M B E RO F LEVELS- 6 N SIGLE C O M M A M N D (E A N S VA> R. 1.924 0.102 D U A LC O M M A M N D (E A N & VAR ). 2 58
u THE
RACK
IS
SQUARE
NUMBER
OF
LEVELS-
SINGLE
COMMAND(MEAN
NUMBER
OF
SINGLE
IN
TIME
FOR
5
LEVELS
3
&
VAR.)
.3569
.0321
DUAL
COMMAND(MEAN
ft
VAR.)
.4803
.0226
COMMAND(MEAN
ft
VAR.)
.2951
.0140
DUAL
COMMAND(MEAN
S
VAR.)
.3988
.0106
NUMBER
OF
5
SINGLE
COMMAND(MEAN
ft
VAR.)
.2814
.0100
DUAL
COMMAND(MEAN
ft
VAR.)
.3811
.0078
NUMBER
OF
6
SINGLE
COMMAND(MEAN
ft
VAR.)
.2909
.0128
DUAL
COMMAND(MEAN
ft
VAR.)
.3935
.0098
NUMBER
OF
7
SINGLE
COMMAND(MEAN
VAR.)
.3232
.0220
DUAL
COMMAND(MEAN
ft
VAR.)
.4357
.0158
LEVELS-
LEVELS-
LEVELS-
LEVELS-
4
ft
THE
RACK
IS
SQUARE
NUMBER
OF
LEVELS*
SINGLE
COMMAND(MEAN
NUMBER
OF
SINGLE
COMMAND(MEAN
NUMBER
OF
LEVELS*
LEVELS*
SINGLE
COMMAND(MEAN
NUMBER
OF
SINGLE
COMMAND(MEAN
NUMBER
OF
SINGLE
COMMAND(MEAN
LEVELS-
LEVELS-
IN
TIME
FOR
6
LEVELS
4
*
VAR.)
.8750
.2577
S
VAR.)
.7875
.1456
DUAL
COMMAND(MEAN
S
VAR.)
1.2300
.2442
DUAL
COMMAND(MEAN
&
VAR.)
1.0984
.1357
DUAL
COMMAND(MEAN
%
VAR.)
1.0544
.0992
DUAL
COMMAND(MEAN
&
VAR.)
1.0679
.1097
DUAL
COMMAND(MEAN
*
VAR.)
1.1210
.1531
5
6
&
VAR.)
.7595
.1107
7
S
VAR.)
.7687
.1219
8 &
VAR.)
.8026
.1640
t-o ON
THE RACK IS SQUARE IN TIME FOR 7 LEVELS NUMBER OF LEVELS- 5 SINGLE COMMAND(MEAN i VAR.)
.2187 .0154
NUMBER OF LEVELS- 6 SINGLE COMMAND(MEANftVAR.)
.3141 .0178
DUAL COMMAND(MEANftVAR.)
.2998 .0117
.2110 .0111 DUAL COMMAND(MEAN & VAR.) .2958 .0101
NUMBER OF LEVELS- 7 SINGLE COMMAND(MEAN & VAR.)
DUAL COMMAND(MEANftVAR.)
.2088 .0100
NUMBER OF LEVELS- 8 SINGLE COMMAND(MEANftVAR.)
.2090 .0101
NUMBER OF LEVELS- 9 SINGLE COMMAND(MEANftVAR.)
.2119 .0116
DUAL COMMAND(MEANftVAR.)
.2963 .0102
DUAL COMMAND(MEANftVAR.)
.3013 .0123
T H ER A C K IS S Q U A R EN I M TIE F O R 7 LEVELS N U .M B E RO F LEVELS- 5 N SIGLE C O M M A M N D (E A NftVAR ). 2.202 0.156 D U A LC O M M A M N D (E A NftVAR ). 3.163 0 N U M B E RO F LEVELS- 6 N SIGLE C O M M A M N D (E A NftVAR ). 2.123 0.113 D U A LC O M M A M N D (E A NftVAR ). 3.016 N U M B E RO F LEVELS- 7 N SIGLE C O M M A M N D (E A NftVAR ). 2.101 0.101 D U A LC O M M A M N D (E A NftVAR ). 2.976 N U M B E RO F LEVELS- 8 N SIGLE C O M M A M N D (E A NftVAR ). 2.103 0.102 D U A LC O M M A M N D (E A NftVAR ). 2.980 N U M B E RO F LEVELS- 9 N SIGLE C O M M A M N D (E A NftVAR ). 2.132 0.117 D U A LC O M M A M N D (E A NftVAR ). 3.032 0 OO
T H ER A C K IS S Q U A R EN I M TIE F O R 8 LEVELS N U M B E RO F LEVELS- 6 D U A L C O M M A M N D ( E A N ft V A R ) . 4 . 1 3 0 0 . 3 3 8 N SIGLE C O M M A N D M <A E NftVAR ). 2.861 0.293 N U M B E RO F LEVELS- 7 D U A LC O M M A M N D (E A NftVAR.> 4.007 0.262 N SIGLE C O M M A N D C M A E NftVAR ). 2.7D 9U 2A 0 . 2 3 8 O M M A M N D (E A NftVAR ). 3.997 0.255 N U M B E RO F LEVELS- 8 L C N SIGLE C O M M A M N D (E A NftVAR ). 2.7D 8U 6A 0.2C 3M 3M L O A M N D ( E A N ft V A R ) . 4 . 0 5 ? 0 . 2 9 3 N U M B E RO F LEVELS- 9 N SIGLE C O M M A M N D (E A NftVAR.> D 2A .8L20C 0.2M 60A U O M M N D (E A NftVAR ). 4.120 0.332 N U M B E RO F LEVELS- 10 N SIGLE C O M M A M N D (E A N S VAR ). 2.854 0.288
230
APPENDIX 10 Program listing from which sample runs presented in Appendix 9 were obtained
231
PROGRAM
MAIN<INPUT*OUTPUTrTAPE5-INPUT.TAPE6-0UTPUT>
FILE NAME. TEMP THI5 PROGRAM WAS WRITTEN WITH THE PURPOSE OF SHOWING TH^T UNDER DEDICATED STORAGE THE EXPECTED MEAN AND VARIANCE OF DUAL CYCLES WILL DE MINIMIZED AT B»l. DIMENSION lA<10)rC(10> rMIAt10),TNC<10) COMMON/BLK1 / HEHT r WTH f HVEL t VVEL t CAPC t DELT t TNC,C»"RR» *NOL»ICOL »NOPR> MIA INITIALIZE SYSTEM PARAMETERS NOPR-NUMBER OF PRODUCT CLASSES HEHT-42.0 WTW>44.0 HVEL*=220.0 VVEL=80.0 N0AI»1 •NOAI" HAS NOT BEEN EXCLUDED IN ORDER TO FACILITATE POSSIBLE MODIFICATIONS OF THE PROGRAM NOPR-3 INITIALIZE THE <C/A) VALUES FOR EACH PRODUCT CLASS IA(3)-150 IA<2>«100 IA(1)=50 > N C<3)-1.0 C<2)«2.0 C<1)=5.0 HEHT=HEHT/12.0 WTH=WTH/I2.0 RAT=HVEL/VVEL TA»0.O CAPC-O.O DELT=0.0 DO 2 K1=>1»N0PR TA=TA+IA(K1> CAPC=CAPC+C<K1) PA=TA/<N0AI*2> ICM-0 DO 3 K2-1.N0PR MIA(K2)=<(IA(K2)/TA)*PA>+1.0 ICM=ICM+MIA(K2) TNC<K2)=C(K2)/CAPC DO 4 K3=lrN0PR • DELT=DELT+(<TNC(K3)**2/MIA(K3)**2>*MIA<K3)> DELT> (1.0-DELT > /2.0 B
INITIALLY INOL IS THE NUMBER OF LEVELS THAT MAXIMIZE B. IN0L=<(SORT((ICM*WTH*HEHT)/RAT)>/HEHT)+0»5 URITE<6rl4)IN0L FORMAT<///3Xf"THE RACK IS SQUARE IN TIME FOR'rlX»I2»IX.'LEVELS
*v>
—
IN0L=INDL-2 IF<IN0L.LE.0)IN0L-1 JNOL-INOL+4 START ITERATING OVER THE RANGE INOL-2 TO INOL+2. DO 5 NOL»INOLrJNOL IC0L=(PA/N0L)+1.0 NOAS=ICOL*NOL RR=(ICOL*WTH)/(NOL*HEHT> CALL TRAVEL(ETSfVTS,EDTrVDT> WRITE(&f8)N0L
232
6 7 5
FORMAT(/3X T * NUMBER OF LEVELS-*TIX,12) WRITE<6»7)ETSFVTS»EUT»VDT FORMAT(/3X?"SINGLE COMMAND(MEAN ( VAR.)"RLXR2F7•4»3X»'DUAL • (MEAN I VAR . > * • 1X»2F7.4/) CONTINUE STOP END SUBROUTINE
C C C C C
COMMAND
TRAVEL(ETSRVTS,EDT»VDT)
T H I S SUBROUTINE DETERMINES THE EXPECTED MEAN AND VARIANCE OF SINGLE AND DUAL CYCLES BY U S I N 3 THE COMPLETE ENUMERATION SCHEME DIMENSION TIME(50,300)RIPRD(50.300> »MIA(10>,C(10),TNC<10) DIMENSION SC(10> »RH(10)RRV(10) »ASC( 10) F RFC( 1 0 ) » R M ( 1 0 ) R D T ( 1 0 ) COMMON/BLK1/HEHT ?WTHR HVELRVVEL »CAPC »DELT, TNC,C»RR, *NOL •ICOL RNOPR»MIA TOTM-0.0 TOTS-0-0
C C C C C C C C C C C C C C C
12 C C C C C C
17
19
FIND THE TRAVEL TIME TO E A C H OPENING A N D STORE IN 'TIME'. CUTM - STORES SINGLE CYCLE TRAVEL TIME CUMULATIVELY C U T S - STORES THE SQUARE OF SINGLE CYCLE TRAVEL TIME CUMULATIVELY E T S - EXFECTED MEAN TRAVEL TIME OF SINGLE CYCLE V T S - EXPECTED VARIANCE OF SINGLE CYCLE STAM - STORES THE ADJUSTED DUAL CYCLE TRAVEL TIME CUMULATIVELY RTAM - STORES THE SQUARE OF THE ADJUSTED DUAL CYCLE TRAVEL TIME CUMULATIVELY EDT - EXPECTED MEAN TRAVEL TIME OF DUAL CYCLE VDT - EXPECTED VARIANCE OF DUAL CYCLE PHET—HEHT/2.0 DO 12 JM=1»N0L PHET=FHET+HEHT FUT=WTH/2.0 D O J2 JC=1»IC0L TH=FUT/HVEL TV=PHET/VVEL TIME( JM» J O - T H IF (TV. G T . T H ) TIME (JM, J O - T V FUT=FMT+WTH IPRDUMR J O - 0 A L L O C A T E THE PRODUCT CLASSES T O THE RACK OPENINGS STARTING FROM THE F I R S T PRODUCT CLASS* W H I L E DOING T H I S A L S O STORE THE TRAVEL TIME AND ITS SQUARE IN 'CUTM' AND 'CUTS'»RESPECTIVELY D O 1 6 NP-LRNOPR CUTH«0.0 CUTS-0.0 ILG=MIA(NP> DO 19 MN»LFILO TMINI-10000.0 LX-1 LY-1 D O 1 7 ILE*1,N0L D O 1 7 ICO=LFICOL IF(IPRD(ILE»ICO).NE.O> GO T O 1 7 IF<TIME(ILERICO>.GE.TMINI) G O T O 1 7 TMINI=TIME(ILE»ICO) LX»ILE LY-ICO CONTINUE IPRD(LX,LY)«NP CUTM=CUTM+T-IME(LX»LY) CUTS=CUTS+(TIME(LX»LY)*TIME<LX»LY)> CONTINUE
*
•nrrn=Tuin+(
»L-<NP>/ILG)*CUTH>
TOTS=TOTS+<<C<NP)/ILG)*CUTS> CONTINUE ETS=TOTM*< 2.O/CAPC) VTS= < <4.O/CAFC)*TOTS)-< <4/(CAPC*CAPC))*TOTM*TOTM) COMPUTE CYCLE
EXPECTED TRAVEL
MEAN
AND
VARIANCE
OF
DUAL
TIME
STAM-0.0 RTAM=0.0 NSL=ICOL*NOL NS1=NSL-1 DO
88
L6=1»NS1
JPR=((L6-0,OOOOl)/IC0L>+1.0 JP=L6-(<JPR-1)*IC0L> L7»L6+1 DO
88
L8=L7,NSL
K6=(<L8-0.00001)/IC0L>+1.0 KLM=L8-< <KG-1)*ICOL) SLT=<AKS<<KLM-JF>*UTH> >/HVEL VTTM=<AfcS<<KG-JPR)*HEHT))/VVEL IF(VTTM.GT.SLT)
SLT^VTTM
'
TTAYM=TIME(JPR p JP)+TIME < KG»KLM>+SLT MA1 = I F'RD < JPR » JP ) MA2=IPRD<KG»KLM) IF<MA1*MA2>4?,88»49 BOL = MI A (MA1)*MIA<MA2)*DELT RTAM=RTAM+(<TTAYM*TNC(MAI>*TNC(MA2>
>/B0L>
STAM=STAM+<<TTAYM*TTAYM*TNC<MAI>*TNC<MA2)>/B0L> CONTINUE EDT=RTAM .VDT-STAM-(EDTtEDT) RETURN END
APPENDIX 11 Program listing of the algorithm developed to find an optimum design
235
c c c 60 61
62
63 64
66 67 505 506
502 501 503 304 68 69
76
70
71
PROGRAM MAIN<INPUT,OUTPUT,TAPE5=INPUT,TAPE6=0UTPUT> DIMENSION LFIB(20),ANS(30>,C(10),IA(10>,IBPT(4>,FACT(10) COMMON ANOP tWTH,HEHT , DEPTH -WTHP,HEHTP,DEPTHP,DUAL , ALAMDA, *VVEL»HVEL»RATIO.0PERN1,0PERN2,BCOSTrRLCOST,SC0ST1.SC0ST2.RCOST, *C , ATMAR fRUkt FACT , PGIVA » RLBCOST , SRMCOST , BMCOST , *IPULL.IADD.NOPR.IA»IBPT,ISTM,IBLDG,ITPUT,INFES.IRACK INPUT OF DATA WRITE<6.60) F0RMAT(//3X,'GEORGIA TECH. - DESIGN OF AN AS/RS") WRITEC6.61) FORMAT(//3X,"TYPE THE DEPTH,WIDTH,HEIGHT OF UNIT LOAD(INCHES)') READ(5,*)DEPTH,WTH,HEHT WTHP=(WTH+S.0)/12.0 HEHTP=(HEHT+10.0>/12.0 DEPTHP=(DEPTH+6.0>/12.0 WRITE(6,62) F0RMAT(3X,"TYPE UNIT LOAD WEIGHT(LB.S)") READ(5.*)WEGT WRITE(6>63> FORMAT<3X,"TYPE DUAL COMMAND PERCENTAGE*) READ(5,*)DUAL WRITE<6.64> FORMAT(3X,"TYPE VERTICAL.HORIZONTAL VELOCITY OF S/R(FPM)»> READ(5.*)VVEL.HVEL RATID=HVEL/VVEL WRITE(6,65) FORMAT<3X,"TYPE PICK-UP,PUT-DOWN TIMES (MIN.S)') READ(5»*)0PERN1,0PERN2 WRITE(6,66) F0RMAT(3X,"TYPE UNIT BUILDING.UNIT LAND COST(*/FT**2)") READ(5,*)BCOST,RLCOST WRITE(6,67) FORMAT(3X,"TYPE 0,1,2 OR 3 FOR S/R LOGIC") WRITE(6,505) F 0 R M A K 5 X , "0=MAN-ON-BOARD 1=0N THE S/R") WRITE(6,506> F0RMAT(5X,"2=0FF THE S/R 3=CENTRAL CONSOLE") READ(5,*)L0GC RLBCOST=0.0 IF(LOGC.NE.O)GO TO 501 WRITE(6,502) FORMAT<3X,"TYPE ANNUAL COST/OPERATOR") READ(5.*)RLBCOST WRITE(6,503) FORMAT(3X,"TYPE ANNUAL MAINTENANCE COST/SR") READ(5,*)SRMCOST WRITE(6,504) FORMAT(3X»"TYPE ANNUAL BLDG. MAINT. COST/SO.FT.•) READ(5.*)BMCOST WRITE(6,63) F 0 R M A K 3 X , "TYPE 0 (RACK) OR 1 (NON-RACK) SUPPORTED BLDG. *) READ(5,*)IBLDG WRITE(6,69) FORMAT(3X."TYPE 0 TO INCLUDE WAITING TIME.OTHERWISE TYPE 1") READ(5,*)ITPUT WRITE<6,76) FORMAT(3X,"TYPE ATMARR.APPLICABLE INCOME TAX RATE') READ(5,*)ATMAR»RITR WRITE(6,70) FORMAT(3X,"TYPE 0 FOR RANDOMIZED,1 FOR DEDICATED STORAGE") READ(5.*)ISTM IF(ISTM.EQ.0)G0 TO 2 WRITE(6,71) FORMAT(3X."TYPE NO. OF PRODUCT CLASSES(INTEGER)") READ(5,*)N0PR WRITE(6,72) FORMAT(3X."TYPE 0P.N/HR,N0. OF SLOTS FOR EACH PROD. CLASS") ALAMDA=0.0 ANOP=0.0 DO 3 K1=1.N0PR READ<5,*)C(K1)»IA<K1) ALAMDA=ALAMDA+C(K1> CONTINUE
236
102
2 73
74 C C C C 4
21
5
6
7
8
9
WRITE(6.102) FORMATI3X,'FOR ENUMERATION TYPE LRFOR SIMULATION TYPE 0") READ(5 R *)1PULL GO TO 4 URITE(6,73) F0RMAT(3X»"TYPE TOTAL NC. OF SLOTS REQUIRED(REAL)•) READ(5F*)AN0P WRITE(6,74) FORMAT(3XF"TYPE THE REQUIRED THROUGHPUT<OP.NS/HR.)'> READ < 5 » # ) ALAMIIA END-OF-DATA-INPUT COMPUTE CONSTANT COSTS AND PARAMETERS IBPT(1>=(30.0/HEHTF)+0.OOOOOL IPPT < 2 > = (42.O/HEH TP)+0.OOOOOL IPPT<3>=(60.O/HEHTP)+0.000001 IBPT(4)=(85.O/HEHTP)+0.000001 PGIVA=((1.0+ATMAR)**10-L.0)/CATMAR*((1.O+ATMAR>**10)) DO 21 K21*LTL0 FACT(K21)=1.0/((1.O+ATMAR)**K21) IADD=12.5042434(0.447478*UTH)+1.0 ALS=(WTH*HEHT*DEPTH)/1728.0 RCOST=1,579451+(0. 025*ALS > + (0.0004424*WEGT) *-(<UEGT*UEGT)/82500000.0) IF(UEGT.LT.1000.0) SC0ST1=13000.0 IF(UEGT.GE.1000.0.AND.UEGT.LT.3500.0) SC0ST1=26000.0 IF(UEGT.GE.3500.0.AND.UEGT.LT.6500.0) SC0ST1=39000.0 IF(UEGT.GE.6500.0) SCOST1=52000.0 IF(LOGC.EQ.0)SC0ST2=13000. 0 IF(LOGC.EQ.L)SC0ST2=26000.0 IF(LOGC.EQ.2)SCC)ST2 = 39000.0 IF(L0GC.EQ.3)SC0ST2=52000.0 IRACK=0 NFE=4 LFIB<1)=0 LFIB<2)=1 DO 5 K=3>20 LFIB(K)=LFIB(K-L)+LFIB(K-2) MLL=0 MUL=40 CALL DOLRS(MULFSYSC.ANS) I F D N F E S . E Q . D G O TO 12 KK=1 MXL = (MLL) + ((LFIB(NFE-KK+1)*(MUL-MLL))/LFIB(NFE-KK+3) ) CALL DOLRS(MXLFAKRANS) MX2=(MLL) + << LFIB < NFE-KK+2)*(MUL-MLL))/LFIB(NFE-KK+3 ) ) CALL D0LRS(MX2»BK»ANS) DO 7 K6=2»2 IF<AK.GT.BK)GO TO 6 MUL=MX2 MX2=MX1 BK=AK MXL=(MLL)+<(LFIB(NFE-K6+1)*(MUL-MLL))/LFIB(NFE-K6+3)) CALL D0LRSCMX1RAKIANS) GO TO 7 MLL=MX1 MX1*MX2 AK=BK MX2= < MLL) + ((LFIB(NFE-K6+2)*(MUL-MLL))/LFIB< NFE-K6+3)) CALL D0LRS(MX2»BK»ANS) CONTINUE IF(AK.GT.BK)GO TO 8 MUL=MX2 MX2=MX1+1 CALL D0LRS(MX2»BKFANS> IF(AK.GT.BK)MLL=MX1 IF(AK.LE.BK)MUL=MX2 GO TO 9 MLL=MX1 AK=BK _MX2=MX2+1 CALL D0LRS(MX2»BKFANS) IF(AK. GT.BK)MLL=MX2-1 IF(AK.LE.BK)MUL=MX2 IF(MLL,EQ.0)MLL«1
237
c C C C
END OF FIBONACCI SEARCH FINAL INTERVAL OF UNCERTAINITY
IS
<MLL,MUL>
JJ=0 IRACK=1 TAMIN=999999999.0 DO 11 MM=MLLrMUL CALL DOLRS(MMfAMINrANS) IF(ANS( 1 ) . E G . O . O G O TO 11 IF<AMIN.GT.TAMIN)JJ=1 WRITE(6»10)MM
10 30 31
FORMAT(///3X»"NO. OF AISLES"•IXr14) WRITE(6r30)ANS<1> F 0 R M A K 3 X , "NO. OF LEVELS".IXrF9.2) URITE<6,31)ANS<2> FORMAT(3X•"NO. OF COLUMNS"»lXrF9.2> WRITE<6F32)ANS<3>
32 33 34
F0RMAT(/3X,"BLDG, LENGTH"•IXTF9.2) URITE(6.33)ANS(4) F0RMAT(3X»"BLDG. WIDTH"r1X»F9.2> WRITE(6,34)ANS<5> FORMAT(3X R"BLDG. HEIGHT",IXTF9.2) WRITE<6.35)ANS<6)
35 36 37 38 600 601 602
604 603 39 40 41 42
43 13 44 15 14 45 48 11 12 46 16
C C C C C C C
F 0 R M A T ( / 3 X » " L A N D COST"TIXTF9,2) WRITE(6»36)ANS<7) FORMAT(3Xf"BLDG, COST"T1X»F9.2> WRITE<6»37)ANS<8> F O R M A T < 3 X » " R A C K COST"TIXrF9.2) WRITE(6»38)ANS(9) FORMAT(3X»"S/R MACHINE COST*rIXrF9.2.3Xr"DOLLARS") WRITE(6f600) F0RMAT(/3X»"THE RECURRING COSTS ARE"/) WRITE(6f601)ANS(16) FORMAT(3X»"BLDG. MAINTENANCE COST*T1X.F11.2) WRITE(6f602)ANS(17) FORMAT(3X»"S/R MAINTENANCE C O S T " T I X T F l 1 . 2 ) I F ( L 0 G C N E . 0 ) G 0 TO 603 WRITE(6,604)ANS(18) F0RMAT(3X»"TOTAL OPERATOR C O S T " T I X T F l 1 . 2 T 1 X T " D O L L A R S / Y R . " ) WRITE(6»39)ANS<11) F0RMAT(/3Xr"SYSTEM THROUGHPUT(0)* rlX»F9.2rlXr"OP.NS/HR."> WRITE<6f40)ANS<12) F0RMAT(3X»"SYSTEM THROUGHPUT(1)"TIX,F9.2rIX»"OP.NS/HR.") WRITE(6,41)ANS<13)TANS<14) F 0 R M A K / 3 X , "EXP. SINGLE & DUAL COMMAND TRAVEL TIMES" 11X.2F7.2) WRITE(6.42)ANS(15) FORMAT(3X,"THE SHAPE FACTOR IS"rIXTF9,2/) IFCISTM.EQ.l)G0 TO 13 WRITE(6,43)ANS(19) FORMAT(3Xf"TOTAL NO. OF OPENINGS"»IX.F9.2) GO TO 14 DO 15 K7=1.N0PR WRITE<6.44>AN3<18+K7> »K7 F0RMAT(3X»F9.2»lXr"OPENINGS FOR PROD. C L A S S " T I X , 1 3 ) CONTINUE WRlTE(6f45)AMIN F0RMAT(//3Xf"PW COST OF ABOVE DESIGN IS"r1X,F12.2TlXf"DOLLARS"> WRITE(6»48)ANS(10) FORMAT(3X»"(TAX LIABILITY SHOULD BE >"•IX,F9.2»")"///) IF( JJ.EQ. D G O TO 16 TAMIN=AMIN GO TO 16 WRITEC6.46) FORMAT<//3X»"WITH THE GIVEN INPUT 40 AISLES IS NOT ENOUGH TO MEET •THE REQUIRED THROUGHPUT,REVIEW YOUR INPUT DATA") STOP END SUBROUTINE DOLRStNOAI'AMINrANS) THIS IS THE KEY SUBROUTINE USED IN MINIMIZING THE SYSTEM COST WITHOUT VIOLATING THE THROUGHPUT REQUIREMENT. IT FINDS THE CONSTRAINT-FREE OPTIMUM NUMBER OF LEVELS FOR THE FIXED NUMBER OF AISLES TRANSFERRED FROM THE MAIN PROGRAM. THEN IT WORKS TOWARDS A RACK THAT SATISFIES THE THROUGHPUT IN CASE THE ABOVE FOUND OPTIMUM NUMBER OF LEVELS DO NOT
238
C
HEET
C
APPROPRIATE
THE
THROUGHPUT
C
CORRESPONDING
CONSTRAINT -
OPTIMUM
NUMBER
COST
OF
THE
A F T E R
O FL E V E L S > DESIGN
FINDING
THE
I T RETURNS
T D THE
MAIN
THE
PROGRAM.
C DIMENSION
A N S ( 3 0 ) p I A (1 0 )pC<10)pMlA(10)pAC(10)rTNC(10)t
*IBPT(4)fS0LN(30)pFACTUO)
ANDPtUTH
C O M M O N
tHEHT pDEPTH tUTHP»HEHTP,DEPTHP tDUAL r ALAMDA p
* W E L f H V E L f RATIO f OPERN1,0PERN2pBCOST,RLCOSTtSCOST1pSC0ST2p
RCOST,
* C f A T M A R F RITRF F A C T F PGIVA F RLBCOSTF SRMCOSTF BMCOST F • I P U L L F I A D D F N O P R F I A F I B P T F I S T M F I B L D G F I T P U T F I N F E S F I R A C K INFES=0 N O A = <AN0P/(N0AI*2.0>)+0.99999 MN0L=IBPT(4> BLA=ALAMDA/NOAI IF< I S T M . E Q . O G O
TO
2
CAPC=0.0 NOA=0 DO
3
K1=1FN0PR
MIA(K1)=((IA(Kl)*1.0)/(NOAI*2.0>)+0.99999 N0A=N0A+MIA(K1) AC(K1>=C(K1)/NOAI 3
CAPC=CAPC+AC(Kl) IF<N0A.GE.400.AND.IPULL.E0.0)G0 DO
4
TO
6
4 K2=1FN0PR
TNC(K2)=AC(K2)/CAPC DELT=0.0 DO
24
24
K24=1FN0PR
DELT=DELT+((TNC(K24>**2/MIA(K24>**2>*MIA(K24>) DELT=(1.O-DELTJ/2.0
2
NOL=((SORT((NOA*WTHP#HEHTP)/RATIO))/HEHTP)+0.5 ICOL=<(N0A*1.0)/NOL)+l.0 IF(ISTM.EQ.0)G0 C A L L
TO
5
DTRAV(NOLFICOLFCOMPFOTHERFSCFDCFBBF
* C A P C F D E L T F A C F M I A F T N C ) IF(COMP.LT.CAPC)GO GO 5
TO
CALL
6
RTRAV<NOL FICOL F BLA F COMPF O T H E R r SC tDC F B B )
IF(COMP.LT.BLA)GO 7
TO
7
TO
6
BWTH=DEPTHP*3*N0AI IF(IBLDG.EG.1>BWTH=BWTH+2.0 GM1=N0A*UTHP*BUTH*BC0ST GM2=HEHTP*IADD*BUTH*BC0ST GM3=N0A*WTHP*BWTH*RLC0ST GM4=NQA*UTHP*BUTH*BMC0ST*PGIVA ALP1=GM3+(0.9694288*GM1)+GM4 ALP1=GM3+(0.9694288*GM1) ALP2=(0,0002698*HEHTP**2*GM1)-(0.003190&*GM2)+(2.6582*N0A*N0AI) ALP3=0.0005396*HEHTP*GM2 VAR=ALP2+ALF3 IF(VAR.GT.ALP1)G0 DO
9
T O8
K9=1FMN0L
DIFR=(K9**2*ALP2 ) + ( K9*K9*K9*ALP3) -ALP1 IF(DIFR.GE,0.0)GO 9
T O
1 0
CONTINUE I O N O L = M N O L GO
1 0
T O 11
I0N0L=K9 TCK9=<ALFl/I0NOL>+CIONOL*ALP2)+(I0N0L**2*(ALP3/2.O)> KB=I0NDL-1 TCK8=(ALP1/K8)+(KB*ALP2) + ( K 8 * * 2 * ( A L P 3 / 2 . 0 ) ) IF(TCK8.LT.TCK9)I0N0L=K8 GO
8 11
CALL DO
1 3
T O
1 1
I 0 N 0 L = 1
13
COSTT<NOAI»IONOL»NOA»TKOST»SOLN) K13=1F30
ANS(K13)=S0LN(K13) AMIN-TKOST L E V E L = I O N O L IF(I0N0L.LT.IBPT(1))G0 DO
14
T O
IF(IBPT(K14> .GE.IONODGO CALL
2 3
T O
1 2
COSTT(NOAIFIBPT(K14)fNOAFTKOSTFSOLN)
I"F( TKOST . G E . A M I N ) G 0 DO
1 2
K14=1F4
2 3 K23=1F30
A N S ( K 2 3 ) = S 0 L N < K 2 3 )
T O
1 4
239
14 12
18 19
17
20 21
6 94
99
97
96 95 98
C C C C C C
2
AMIN=TKOST LEVEL=IBPT(K14> CONTINUE RRR=BLA IF<ISTM.EQ.l)RRR=CAPC JC0L=ANS(2> IF(ISTM.EG.O)CALL RTRAV<LEVEL.JCOL,BLA.COMP.OTHER.SC.DC.BB) IF(ISTM.EQ.1>CALL DTRAVC LEVEL »JCOL »COMP tOTHER ,SC »DC»BB » *CAFC.DELT.AC.MIA,TNC) IF(COMP.GE.RRR)GO TO 99 IF(LE"VEL.GT. NOL ) GO TO 17 LEV2=LEVEL+1 DO 18 K18=LEV2»MNOL ICOL=((NOA*l.0)/K18)+0.99999 IF <ISTM.EQ.0)CALL RTRAV( K1 8.1 COL .BLA,COMP,OTHER .SC»DC.BE) IF(ISTM.EO.1)CALL DTRAV<Kl8.ICOL,COMP,OTHER,SC,DC.BE. *CAPC,BELT.AC,MIA,TNC) IF(COMP.GE.RRR)GO TO 19 CONTINUE IFN0L=K18 CALL C O S T K N O A I ,IFNOL,NOA.TKOST.ANS) AMIN=TKOST GO TO 99 DO 20 K20=l.LEVEL II=LEVEL-K20 IC0L=(<N0A*1.0)/II)+0.99999 IF(ISTM,EQ.O)CALL RTRAV(II.ICOL.BLA.COMP.OTHER.SC.DC.BB) IF(ISTM.EQ.l)CALL DTRAV(II.ICOL.COMP,OTHER.SC.DC.BB. *CAPC,HELT,AC,MIA,TNC> IF(COMP.GE.RRR)GO TO 21 CONTINUE IFNOL=II CALL C O S T K N O A I . IFNOL »NO A»TKOST»ANS ) AMIN=TKOST GO TO 99 INFES=1 DO 94 K94=1.30 ANS(K94)=0.0 AMIN=999999999.0 GO TO 98 ANS(11)=C0MP*N0AI ANS(12)=0THER*NOAI IF(ANS(12).GT.ANS(ll))G0 TO 97 ANS(11)=OTHER*NOAI ANS(12)=C0MP*N0AI ANS(13)=SC ANS(14)=DC ANS(15)=BB IF(ISTM.EO.1)G0 TO 96 ANS(19)=ANS(1>*ANS(2)*2*N0AI GO TO 98 DO 95 K95=1.N0PR ANS(18+K95)=MIA(K95> RETURN END SUBROUTINE COSTT(NOAI.NOLL»NOA.TKOST.SOLN) GIVEN THE DESIGN VARIABLES.THIS SUBROUTINE WILL RETURN THE VALUES OF EACH VARIABLE COST ELEMENTS. IT USES SUBROUTINE <PWORTH> TO FIND THE PRESENT WORTH OF ABOVE COMPUTED VARIABLE COST ELEMENTS. DIMENSION SOLN(30).IA(10).C(10)tFACT(10).IBPT(4) COMMON ANOP.WTH.HEHT.DEPTH.WTHP.HEHTP.DEPTHP.DUAL.ALAMDAr JKVVEL.HVEL.RATIO.0PERN1.0PERN2.BCOST.RLCOST.SC0ST1.SC0ST2.RCOSTr •C.ATMAR.RITR.FACT,PGIVA.RLBCOST.SRMCOST.BMCOST. *IPULL.IADD.NOPR.IA.IBPT.ISTM.IBLDG.ITPUT,INFES.IRACK DO 2 K2=l,30 SOLN(K2)=0.0 ICOL=((N0A*1.0)/N0LL)+0.99999 S0LN(1)=N0LL S0LN(2)=ICGL S0LN(3)=(WTHP*IC0L)+IADD SOLN(4)=(DEPTHP*3.0)*NOAI IF(IBLDG.EQ.l)S0LN(4)=S0LN(4)+2.0
240
S0LN(5>=(HEHTP*N0LL >+4 .0 S0LN(6)=S0LN(3)*S0LN(4)*RLC0ST SOLN(16)-SOLN(3)*SOLN < A>*BMCOST SOLN(17)=N0AI*SRMC0ST SOLN<18>=N0AI*RLBC0ST RECUR=SOLN(16)+S0LN(17)+S0LN(IB) CFR-0.986508-(0.005349*S0LN<5))4 <0.0002698*S0LN(5)**2) IF(SOLN < 5).LE.25.0)CFR=1.0 S0LN(7)=S0LN(3 >*S0LN(4)*BC0ST*CFR IF<IBLDG.E0.0)S0LN(7)=S0LN(3)*SGLN(4)*((BCOST*CFR)-6.0) S0LN(8)=(RC0ST+(0.0949367*N0LL))*SOLN(1)*SOLN(2)*2*N0AI*14.0 IF (I RACK . EQ. 1 )S0LN(8) = <RC0ST+-( < 0 . 23329*NC)LL )-( 0 .00476* *N0L**2)-0.654611))*SOLN(1)*S0LN(2)*NOA1*2*14.0 TSR=SC0ST14SC0ST2 IF(NOLL.LE.IBPT(1 ) ) SOLN ( 9 ) =TSR+13000. 0 IFiNOLL.GT.IBFT(1).AND.NOLL.LE.IBPT(2)>S0LN<9)= •TSR+26000.0 IF(NOLL.GT. IBF'T(2) • AND. NOLL »LE.IBPT(3) )S0LN(9)= *TSR+39000.0 IF(NOLL.GT.IBPT(3).AND.NOLL.LE•IBPT(4))SOLN(9)= *TSR+52000.0 IF(NOLL.GT.IBPT(4))SOLN(9)=999999999,0 SOLN(9)=SOLN(9)*NOAI CALL PWDRTH(NOAI»S0LN(6)tS0LN(7)rSOLN(8)fS0LN(9)tRECUR,YYtPWC) SOLN(10)=YY TKGST=PWC RETURN END SUBROUTINE RTRAV(NOLI,IC0L1,FLOW.COMPtOTHER,STtDT,BB) C C C C C C C C
2
5 3
GIVEN THE RACK SHAPE AND TRAVEL VELOCITIES• THIS SUBROUTINE RETURNS THE MEAN AND VARIANCE OF SINGLE AND DUAL COMMAND TRAVEL TIMES UNDER * RANDOMIZED * STORAGE. IT ALSO RETURNS THE OVER-ALL MEAN AND VARIANCE OF S/R TRAVEL TIME AND THE VALUE OF THE SHAPE FACTOR. DIMENSION IA(10),C(10).IBPT(4),FACT(10) COMMON ANOP tWTH r HEHT r DEPTH,WTHP tHEHTPf DEPTHP,DUAL tALAMDA» • VVEL»HVEL»RATI 0,OPERN1.0PERN2,BCOST>RLCOST,SC0ST1>SC0ST21RCOST, *C tATMAR,RITR,FACT,PGIVA tRLBCOSTiSRMCOST.BMCOST• * I PULL , I ADD r NOPR f IA . IBPT f ISTM f I BLDG r I TF'UT r INFES»IRACK HTT=(WTHP*IC0L1)/HVEL VTTT=(HEHTP*N0L1)/VVEL BB=VTTT/HTT BASE=HTT IF(BB.LE.1.00)G0 TO 2 BB=HTT/VTTT BASE=VTTT RST=(BB**2/3,0)+l.0 ST=(RST*BASE)+0PERN1+0PERN2 RXSO=((BB*BB*BB*2.0)/3,0)+l.3333 RVST=RXS0-(RST**2) VST=RVST*BASE**2 EXSSQ=VST+ST**2 RDT=(0.5*BB*BB)-(BB*BB*BB/30.0)4-1.3333 DT=(RDT*BASE)+(2*0PERN1)+(2*0PERN2) RVDT=( (0.3588-0.1321*BB)*R'DT )**2 VDT=RVDT*BASE**2 EXDS0=VDT+DT**2 ETT=( (DUAL*(I(T/2.0) )4-( (1. O-DUAL )*ST) )/60.0 VTT=((DUAL*(EXDSQ/2.0))4((1.O-DUAL)*EXSSQ)*ETT**2)/3600.0 CHK=FLOW*ETT WTIME=(FL0W*(VTT4-(ETT**2) ) >/ (2.0* (1.0-CHK) ) IF(CHK.GE.1.0)WTIME=99999.0 COMF-1.O/ETT 0THER=1.0/(ETT4WTIME) I F d T P U T . E O . D G O TO 3 COMP=OTHER 0THER=1.O/ETT FORMAT(3X»"YAVUZ'?21712X19F10.2?"BOZER*) RETURN END SUBROUTINE PWORTH(NOAItQLANDtQBLDGrQRACK,QSRtRECURtYY?PWC)
241
c C
G I V E N
C
WILL
C
THE
C
D E P E N D S
THE
V A L U E S
C O M P U T E
P A R T I C U L A R ON
OF
THE
THE
C O S T
P R E S E N T
D E S I G N
W H E T H E R
U N D E R
THE
E L E M E N T S ,
WORTH
COST
Q U E S T I O N .
B U I L D I N G
T H I S
OF
THE THE
I S R A C K
S U B R O U T I N E S Y S T E M
FOR
A P P R O A C H
S U P P O R T E D
OR
NOT.
C D I M E N S I O N
I A d O ) r C ( l O ) rIBPT(4) - FACT (10)
?UTHPrHEHTP,DEPTHP» D U A L
ANOPtWTH,HEHT,DEPTH
C O M M O N
*VVEL>HVEL *C r A T M A R
r
tALAMDA
t
rRATIOrOPERNl»0PERN2»BCOSTrRLCOST*SC0ST1rSCDST2»RC0ST»
. RI T R
r FACT , P G I V A . RL.BCGS I » S R M C O S T
B M C O S T ,
• I P U L L F I A D D P N O P R P I A r I B P T P I S T M P I B L D G , I T P U T » I N F E S F I R A C K T 1 = R R A C K + Q S R IF(IBLDG.EQ.1)GO
T O 3
T1=T1+QBLDG SUM=0.0 DO 2
2
K2=l»10
SUM=SUM+(((10-(K2-1>)/55.0)*T1*FACT(K2)) RITC=(T1-(0.5*SC0ST2*N0AI))*0,10 YY=(2.0*RITC)-25000.0 PUC=T1+(QLAND*<1-FACT(10)))-(SUM*RITR)-RITC+(RECUR*PGIVA* *(1.O-RITR)) GO
3
T O 4
S U M = 0 . 0 TDEP=0,0 DO
5
K5=lrl0
Rl=((10-(K5-1))/55.0)*Tl DEP=(<40-(K5-l))/820.0)*QBLDG T D E P = T D E P + D E P 5
SUM=SUM+((DEP+R1)*FACT(K5>) RITC=(T1-(0.5*SC0ST2*N0AI))*0.10 Y Y = (2,0*RI T O - 2 5 0 0 0 . 0 PUC=T1+QBLDG+(QLAND*(1-FACT(10))>-(SUM*RITR)-RITC *-<(QBLDG-TDEP)*FACT(10))+(RECUR*PG1VA*(1.O-RITR))
4
C
R E T U R N E N D
DTRAV(N0L2tIC0L2,COMPtOTHERtSC,DC;BB,
S U B R O U T I N E
C.
G I V E N
C
S U B R O U T I N E
T H ER A C K
C
AND
C
S T O R A G E .
C
ENTERED
C
S U B R O U T I N E
D U A L
S H A P E
R E T U R N S
C O M M A N D IT
IS
IN
M E A N
T R A V E L
A S S U M E D
UILL
T H EC O M P L E T E
C
OVER-ALL
THAT
C
THE
OF
S I M U L A T E
AND THE
T H I S
V A R I A N C E
S I N G L E
U N D E R
P R O D U C T
THE
R A C K
S C H E M E .
V A R I A N C E S H A P E
V E L O C I T I E S *
C / A R A T I O S .
E N U M E R A T I O N
M E A N
AND
T I M E S
" D E S C E N D I N G "
C
V A L U E
A N DT R A V E L
THE
OF
O F
* D E D I C A T E D " C L A S S E S IF
A R E
D E S I R E D ,
I N S T E A D I T A L S O
S / RT R A V E L
OF
T H I S
F O L L O W I N G
R E T U R N S T H E T I M E A N D
F A C T O R .
C * C A P C P D E L T p A C r MIA r T N C ) D I M E N S I O N
T I M E ( 5 0 » 3 0 0 ) » I P R D ( 5 0 , 3 0 0 ) f I A ( 1 0 > F C ( 1 0 > F M I A < 1 0 > F
*AC(10)FIBPT(4)FFACT(10)FTNC(10) C O M M O N
A N O P , W T H >HEHT » D E P T H
tUTHP» H E H T P
tDEPTHP
rDUAL rALAMDA F
*VVELFHVEL»RATIOFOPERNiF0PERN2FBCDST,RLCOSTFSC0ST1FSC0ST2FRC0STF * C F A T M A R F R I T R F F A C T F P G I V A F R L B C O S T F S R M C O S T F B M C O S T F * I P U L L F I ADD F N O P R F I A F I B P T F I S T M F I B L D G F I T P U T TOTM=0,0 T O T S = 0 . 0 P H E T — H E H T P / 2 . 0 DO
12
JM=1FN0L2
P H E T = P H E T + H E H T P F U T - U T H P / 2 . 0 D O
12
J C = 1 F I C 0 L 2
T H = F U T / H V E L T V = P H E T / V V E L TIME(JM p J C ) = T W I F ( T V
. GT . T H ) T I M E ( J M F J C ) = T V
FUT=FUT+UTHP 1 2
I P R D ( J M F J C ) = 0
C C D O
1 6
N P = 1 F N 0 P R
C U T M = 0 , 0 C U T S = 0 . 0 I L G = M I A ( N P ) D O
1 9
M N = 1 F I L G
T M I N I = 1 0 0 0 0 . 0
F I N F E S » I R A C K
242
17
19 16
49
88 2 C C C C C
LX = 1 LY=1 DO 17 ILE=1»N0L2 DO 17 IC0=lrIC0L2 IF<IPRD<ILE,ICO).NE.O) GO TO 17 IF(TIME(ILE,ICO>.GE.ThlNI) GO TO 17 TMINI=TIME(ILE,ICO> LX=ILE LY=ICO CONTINUE IFRD(LX»LY>=NP CUTM=CLITM+TIME(LXf LY) CUTS=CUTS+(TIME(LX»LY)*TIME<LX»LY)) CONTINUE ,TOTM=TOTM+ ( <AC<NP)/ILG>*CUTM) TOTS=TOTS+<(AC(NP)/ILG)*CUTS> CONTINUE SC=(TOTM*(2 . O/CAPC))+OPERN1+0PERN2 VTS=(<4.0/CAPC)*TGTS>-((4/(CAPC*CAPC)>*TGTM*TOTM) EXSSQ=VTS+SC**2 STAM=0.0 RTAM=0.0 NSL=ICOL2*NDL2 IF(IPULL,EQ.O)GO TO 2 NS1=NSL-1 DO 88 L6=lfNSl JPR=< <L6-0.00001)/IC0L2)+1.0 JP=L6-<(JPR-1)*IC0L2) L7=L6+1 DO 88 L8=L7,NSL KG=((L8-0.OOOOl)/IC0L2)+1.0 KLM=L8-<<KG-1)*IC0L2> BLT=(APS(<KLM-JP)*WTHP>)/HVEL VTTM=(ABS< <KG-JPR)*HEHTP> >/VVEL IF(VTTM.GT.SLT) SLT=VTTM TTAYM=TIME<JPR»JP>+TIME<KG»KLM)+SLT MAI=IPRD(JPR tJP) MA2 = IPRD(KG tKLM) IF(MAl*MA2)49r88r49 B0L=MIA(MA1)*MIA(MA2)*DELT RTAM=RTAM+((TTAYM*TNC(MAI)*TNC(MA2))/BOL) STAM = STAM+ <<TTAYM*TTAYM*TNC(MAI)*TNC<MA2))/BOL) CONTINUE GO TO 99 WEIT=0.0 "UEIT" IS USED TO KEEP TRACK OF THE PROBABILITIES ASSOCIATED WITH INDIVIDUAL DUAL TRIPS WHICH ARE FORMED BY CHOOSING TWO OPENINGS RANDOMLY NN=10000 IF(NSL.GE.10000>NN=15000
C C C
START SIMULATION IN ORDER TO FIND DUAL CYCLE MEAN AND VARIANCE. NOTE THAT SIMULATION TIME* »NN» DO 20 K20=1»NN JPR=(RANF(X)*N0L2)+0.4999
243
79
20 C C C C
99
5
JP=(RANF<X)*IC0L2) +0.4999 KG=< R ANF(X)*NDL2)+0.4999 KLM=<RANF(X)*1C0L2>+0.4999 SLT=<ABS<<KLM-JP>*UTHP>>/HVEL VTTri = <ABS(<KG-JPR)*HEHTP)>/VVEL IF(VTTM.GT•SLT) SLT=VTTM TTAYM=TIME(JPR•JP)+TIME(KG.KLM)+SLT MA1=IPRD<JPR.JP) MA2= I PREi ( KG t KLM ) IF(MA1*MA2>79,20.79 BQL=MIA(MA1 >*MIA(MA2)*DEL.T PROD-<TNC(MAI)*TNC(MA2))/BUL RTAM=R TAM+<TTAYM*PR0B> SIAM=STAM+(TTAYM*TTAYM*PROB> UEIT=UEIT+PR0B CONTINUE COMPUTE THE WEIGHTED AVERAGE OF THE TRIP TIMES GENERATED THROUGH SIMULATION RTAM=RTAM/WEIT STAM=STAM/WEIT HC=RTAM+(2*0PERN1>+(2*0PERN2) VDT=STAM-<RTAM*RTAM> EXIiSQ^VDT+DC**2 BB=TIME(1fIC0L2)/TIME(N0L2.1) IFCBB.GE.1.0)BP=1.O/BB ETT=((DUAL*(DC/2.0))+((1.O-DUAL)*SC))/60.0 VTT=< <DUAL*(EXD5Q/2.0>)+((1.O-DUAL>*EXSSQ>*ETT**2)/3600.0 CHK=CAPC*ETT WTIME=(CAPC*(VTT+(ETT**2))>/(2.0*(1.O-CHK)) IF(CHK.GE.1.0)WTIME=99999.0 C0MP=1.O/ETT 0THER=1.O/CETT+WTIME) IFCITPUT.EO.l)G0 TO 5 COMF-OTHER 0THER=1.O/ETT RETURN END
APPENDIX 12 Outputs related to the sensitivity analysis on "DUAL"
245
G E O R G I A
T Y P E
THE
?
4 8 .
4 8 ,
?
2 0 0 0 .
?
0.0
TYPE
T Y P E
T Y P E ?
6 0 .
?
0.25
?
2 5 .
TECH.
D E S I G N
GF
AN
A G / E S
DEPiliv W I D T H H-IE1GHT
OF
UNIT
L O A D ("INCHES )
4 8 .
U N I T
LGAD
W E I G H T <L B «G )
DUAL
0 U M M A N D
P E R C E N T A G E
V E R T I C A L - H O R I Z O N T A L
V E L O C I T Y
OF
S/R(FPM>
2 4 0 .
TYPE
P I C K - U P y P U I - D O W N
TYPE
UNIT 6.
T Y P E
B U I L D I N G , U N I T
0*1>2
OR
0 = HAN-
0N -D0ARD
2::::QFF
THE
?
2
?
2 2 0 0 .
?
5*5
T Y P E
TYPE
S/R
(MIN.S)
BLDG.
T Y P E
0
TO
0
?
1
?
0.10
?
0
?
6 0 0 0 .
?
1 4 5 .
T Y P E
OR
I N C L U D E
LOGIC S/R C O N S O L E
C O S T / S R
M A I N T .
1
C O S T ( * / F T # # 2 )
S/R THE
3"-CENTRAL
A N N U A L
(RACK)
LAND
FOR
1~0N
M A I N T E N A N C E
0
?
3
A N N U A L
TYPE
C O S T / S Q
( N O N - R A C K )
W A I T I N G
A T M A R R y A P P L I C A B L E
.FT.
S U P P O R T E D
BLDG.
T I M E y O T H E R W I S E
INCOME
TAX
TYPE
RATE
0.50
TYPE
0
T Y P E
T O T A L
TYPE
TIMES
0.25
FOR
THE
R A N D O M I Z E D ?I
NO.
OF
R E Q U I R E D
S L O T S
FOR
D E D I C A T E D
S T O R A G E
R E Q U I R E D ( R E A L )
T H R O U G H P U T ( O P . N S / H R . )
1
246
NO. OF AISLES NO* OF LFUELS NO. OF COLUMNS
11 .00 55.00
BLDG. LENGTH BLDG. WIDTH BLDG. HEIGHT LAND COST 117720,00 BLDG * COST 648655.54 RACK COST 502036.82 S/R MACHINE COST 520000*00
DOLLARS
THE RECURRING COS IS ARE BLDG. MAINTENANCE C0ST 10 7910.00 S/R MAINTENANCE COST 11000.00 SYSTEM 7"I-IR0UGHPUT < 0 ) S YSTEh T1-1R00GHPUT ( 1) e: x p. s x n g l e
X
19.76 0P . NS/i IU * 165.36 0P . N3/ HR .
dual command trameL t i mes
THE SHAPE FACTOR IS TOTAL NO. OF OPENINGS
.03 6050.00
P U! COST OF ABO ME DESIGN IS 1365464.72 DOLLARS (TAX LIABILITY SHOULD BE > 209630.47)
i . e j.
G E O R G I A
T YPE 40.
TECH.
THE
4 8 .
'TYPE
•••• D E S I G N
OF
AN
AS/RS
D E P T H * W 1 D T I-I I-IEIGIIT
0F
V
UN![T
LOAD (INCHES )
4 8 .
U N I T
LOAD
W E I G H T (L B .S )
DUAL
COMMAND
2 0 0 0 . TYPE
PERCENTAGE
0.2 TYPE 60.
V E R T I C A L Y II0 R1Z
0 N T A L
VELOCITY
OF
S/R(FPM)
2 4 0 .
TYPE 0*25
P I C K - U P ? P U T - D O W N
TYPE 25 .
TIMES
(M1N.S)
0.25 U N I T
BUILDING
VU N I T
LAND
COS"! ( $ / F T # > K 2 )
6 •
TYPE
0»l-2
OR
3
FOR
O ^ M A N - O N - B O A R D 2--=0FF
THE
S/R
I=ON
S/R
LOGIC THE
S/R
3-CENTRAL
CONSOLE
2 TYPE
ANNUAL
MAINTENANCE
ANNUAL
BLDG.
COST/SR
2 2 0 0 . TYPE C I::; VJ + :J
MAINT.
COST/SQ.FT.
-
TYPE
0
(RACK;
TYPE
0
TO
OR
1
(NON-RACK)
SUPPORTED
BLDG.
0 INCLUDE
WAITING
T I M E vOTHERWISE
TYPE
J. TYPE 0.10
A T M A R R V A P P L I C A B L E
INCOME
TAX
RATE
0.50
TYPE
0
FOR
TYPE
TOTAL
R A N D O M I Z E D v1
FOR
DEDICATED
STORAGE
0 NO.
OF
SLOTS
R E Q U I R E D ( R E A L )
6 0 0 0 . TYPE 145.
THE
R E Q U I R E D
THROUGHPUT
(O P .NS/TIR » )
I
248
N O .. O F A SV ILE ELSS N O O F L E NO. OF C O L U M N S 1 1 . 0 0 B E D S . L E N G T H BL LD DG G.. H W D H B E G IIT H T L A N D C O S T 1 1 7 7 2 0 0 . 0 5 5 0 , 0 D L D G . C O S T 6 4 0 6 5 5 5 . R A C K COST 5020360.24 SR / M A C H N IE COST 5200000.0 D O L L A R S T H E R E C U R R N I G C O S T S A R E B LDG.M M A N IT E N A N C ECOS C OST 1110007090.1000.0 /SY S R A N I T E N A N C E T S T E M T H R 0 U P I U 0 I T ( 0 ) 2 6 2 . 8 0 P + N H S / R , S Y S T E M T H R 0 U G 1 P U T E X P . S N I G L E & D U A L C O M M A N D T R A V E L T M I E S T H E S H A P E F A C T O R I S 8 . 3 W P C O S T O F A B O V E D E G S I N I S 1 3 6 5 4 6 4 7 . 2 D O L L A T (T A X L A I B L I T I Y S H O U L D B E > 2 8 9 6 3 8 4 . 7 ) O T A L NO. O F OPEN N IGS 60500.0
249
G EO R G A I48P.TE C I. T *N -D G SINDO F•A R /U SD Y EE ZI 'N "f 11A yS I ?T 4 3 . 4 8 TY PE UN T IL O A D WEG IHTL (B S.) ?T 2Y0P 0 0 . D U A LC O M M A N DP E R C E N T A G E ? TY 04P .E ER T C IAL H ?OR ZIONTAL VELOC TIY O F SR /F (PM ? TY 60PE .E2V 4 0 . I2.K U -P vU D T -O W NT M IES M (N IS.) ? T02.5YP 0C 5 P E U NII" B UL ID N I G*U TY PEN-•0yv l2N O R 31~0N F O RT SR / SR LOG C I M A 0 O B O A R D H E / O -FF T H E SR / CE3NT R A LC O N S O L E ? TY 22P A N N U A L .M A N IT E N A N C E COS/SR 'I ? TY 22PE 0 0 . A N N U A L BLDG. MA N IT. COSTS/QF.T. ? TY 5P 5 .E E 0R (ACK) O R 1N (GN R -ACK) SUPORTED ? TY 0PE 0T ON ICLUDE W A N F T IG T M IExOTHERW SE I E 0A -rAPPL C IABLEN IC O M ET A X RATE ? TY 0P.1 5 .TMARR TYPE 0 F O RR A N D O M Z E D I ?1 F O R DED C IATED ? TY 0PE T O T A L NO. O F SLOTS REQU R IEB R (EAL) ? 7"600Y0. P E T H E R E Q U R I E D ? 145. ::::
?
1
::::
250
N O .. O F A SV ILE ELSS 5110.0 N O O F L E NO. OF C O L U M N S 550.0 B L D G .. W L E N G T H 269750.60.7 B L D G D I T H BLDG. HE G IHT 571.7 L A N D*COST 11767429006.0555.4 B L D G C O S T R A C K COST 5020368.2 S/R M A C H N IE C O STC5O 2S 0"!0000.010791D O L L A R S B L D G . M A N I T E N A N C E 0 0 . 0 T H E/ RRECU RR N IGA COT STSEA R E AN0£ C0S S M N I N S Y STIR T IR T I PU SYSTEM 0UGE HPU "M I"( 1T)T 1825.09U0pG. US/IIR E PAPE. SN IF G T H EXSH A C T O R LISE 8 .& 3 D U A I. C
T O T A L N O . O F O P E N N I G S 6 0 5 0 0 . 0 W P C O S T O F A B O V E D E G S I N I S O L L T(AX LA IBL IT IY S H O U L D BE > 2896318346.75)4647.2 D
251
G E O R G I A
T Y P E ?
4 8 .
THE
4 8 .
T Y P E ?
2 0 0 0 .
?
0.6
T Y P E
TYPE ?
6 0 .
?
0.25
?
2 5 .
T E C H .
•••
D E S I G N
OF
AN
A S / R S
DEP'i H y W I I T H y H E I G H T
OF
UNIT
LOAD (INCHES )
4 8 .
U N I T
L O A D
W E I G H T ( L B . B )
D U A L
C O M M A N D
P E R C E N T A G E
0 E R T I C A L yH 0 R I Z 0 N T A L
V E L O C I T Y
OF
S / R ( F P M )
2 4 0 . P I C K - U P * PUT•-DOWN
TY1"'E
0.25
T Y P E
U NIT
6.
T Y P E
B UIL DIN G
0vlv2
OR
3
0"M AN --ON-BOARD 2~0FF ?
2
?
2 2 0 0 .
?
5.5
?
0
?
1
?
0.10 TYPE
?
0
?
6 0 0 0 .
?
1 4 5 .
TYPE
TYPE
THE
S/R
3
: ;
BLDG.
TYPE
0
TO
OR
I N C L U D E
T O T A L
TITE
OF
R E Q U I R E D
L O G I C S/R C O N S O L E
C O S T / S R
C O S T / S O . F T .
W A I T I N G
R A N D O M I Z E D
NO.
S/R
(NON-RACK)
A T M A P R y A P P L I C A B L E 0.50 0 FOR
S L O T S
S U P P O R T E D
B L D G .
T I M E y O T H E R W I S E
INCOME ?1
C 0 S T ( * / F T # * 2;
THE
M A I N T .
1
(M I N . S )
L A N D
"CENTRAL
A N N U A L
(RACK)
T YPE
FOR 1~QN
M A I N T E N A N C E
0
T Y P E
U N1T
A N N U A L
TYPE
TYPE
v
TIMES
FOR
TAX
T Y P E
RATE
D E D I C A T E D
S T O R A G E
R E Q U I R E D ( R E A L )
T I T R 0 U G H P U T (0 P .NS/ITR , )
1
:U iN .O ,. O FFA SIE L EE SLS 5110,0 O L V N O . O F C O L U M N S 5 5 * 0 0 D L D GG.*W L E N G T H B L D D I T H BLDG• HE G IHT6 75.0 L A N D COST 117C 7200.00S I 6 4 8 6 5 5 * 5 B L L < 0< • R C C 2L 0S3 6 * S SR / AM A C H N IEKCO ST05S 2000T00,05 0D O L A R B DG.M M A N IT E A N C ECC C T H RE C U R R N IN G O ST SOSTA R E1110007090.1000.0 /SLE S R A N I T E N A N C E O S T YS ST TE EM MT T iH G H SY -i RR00UU GP HUPH U E X S N IA G L & U A O M N D TRAVEL T M IES T H EP. S H P EE F A C T O RD IL SC 0.M 3A T O T A L N O . O F O P E N N I G S 6 0 5 0 0 . 0 W P C O S T O F A B O V E D E G S I N I S O L {TAX LA IBL IT IY S H O U L D BE > 2896313864.574)64*72 D
R G A I48I.:!TE C L E O IND F N G EG /T T4EO Y "f8IH E.* IE•I'T H W yG IO TI- vA E IG IA H ?G 8 . 4 Y PE UN T IL O A D UElGML TB ( . G) ?T 2 0 0 0 . Y PE0 D U A LC O M M A N DP E R C E N T A G E ?? T 0 8 . T Y P E V E R C T I A L v H O R Z O I N T A L V E L O C T I Y O F R 3 / C F P M ) 6 0 . 2 4 0 . Y PE5 C P IK U -PvPUT D -OWN T M IES M (N IS.) ?T 0 2 . 0 2 . 5 Y PE. 6U N T I BU LID N IG ?UN T IL A N D COST$(F/T**2) ?T 2 5 . Y PE 0*1,2 O R 3~U F O RT SR / SR L OG C I MTA 0 N • 0 N • B 0 A R D . J N H E / OFF T H E SR /CEN 3TRAL C O N S O L E ? TY 22PE=A N N U A L M A N I T E N A N C E C O S T S R / ? TY 22PE 00.A N U A L B L D G . M A N I T . C O S T S / O F . T . ? TY 5P 5 .E 0N R (ACK) O R 1N (ON R -ACK) SUPORTED ?0 TYPE 0 T ON ICLUDE WA N TIG T M IE yOTHERW SE I ? TY 1 E A T M A R R y A P C L I A B L E N I C O M E T A X R A T E ? TY 01.P 0 0 5 . 0 E0F O RR A N D O M Z E D I y1 F O R DED C IATED ? TY 0P O T A L NO. O F SLOTS REQU R IED R (EAL) ?? T61"04PE05Y 0.. T P E TT I E REQU R I ED : : : :
; : ; ;
254
NO. OF AISLES NO. OF LEVELS NO. OF COLUMNS BLDG. LENGTH BLDG. WIDTH BLDG. HEIGHr
I6O .9l0.O 0 5640.-0_)' 6 . 0 u 4
'57.17
LAND COST 115344.00 BLDG. COST 635563.41 RACK COST 503S62.41 S/R MACHINE COS7 416000.00
DOLLARS
THE RECURRING COSTS ARE BLDG. MAINTENANCE COST S/R MAINTENANCE COS I"
105732.00 8000.00
::
TI-IR0UGI-I!U 'T
S YSTEM ( 0) SYSTEM THROUGHPUT(1)
1.30 0P. NO/I-iR . 146. 81 OP . NS/HR .
EXP. SINGLE % DUAL COMMAND TRAVEL TIMES THE SHAPE FACTOR IS .66 TOTAL NO. OF OPENINGS
2.04
6072.00
W P COST OF ABOVE DESIGN IS 1285264.65 DOLLARS C TAX LIABILITY SHOULD BE > 270485.16)
3.07
255
GEORG A I TEC I-I. - DE G SIN O FA N AS R /S Y P*E 4T H E48D E P T Hy W D IT H * HG EIHT Of UN T I ?T 4 8 8 * * Y PE00.UN T IL O A D WEG IHTL (B S.) ?T 2 0 TY PE*D U A LC O M M A N DP E R C E N T A G E ?T 1 0 T C IA »H L.OR ZIONTAL VELOC TIY O F SR /F (PM) ? TY 60PE.YPE24V 0E *R PC IK•- U P9P U D T -0 W N "IM IES ( M N 1 ? TY 02.PE5 U 02.N 5 I BU L ID N IGyUN T IL A N D COST$(F/T*#2) ? TY 25P*E 60*.1T *A 2R O R1 3NF O R SR /SR LOG C I 0 ~ M A N O B N O D ~ G T H E / 2FF T H E SR / C 3-ENTRAL C O N S O L E ?O 2 TY N N U A LM A N IT E N A N C E COST SR / ?T 22PE 0Y 0.A EA N N U AL BLDG . M A N I ? TY 5*PE 5 0PR (ACK) O R IN (ON R -ACK) SUPORTED ? TY 0PE 0 T ON ICLUDE WA N TIG T M IE O -THERW SE I Y PE A TM ARR*APC LIABLEN IC O M ET A X RATE ?T 0 1 . 0 0 5 . 0 T0YPE 0 F O RR A N D O M I ZED?1 F O R DED C IATED ?T Y PE T O T A L NO. O F SLOTS REQU R IED R (EAL) ? TY 60P 0 0 . H E REQU R IED THROUGHPUT O (N P.H S/R ). ? 14E5. T ;:::
?
1
256
N O .. O F A SV ILE E S1 1 . 0 . 0 N O O F L E L S 6 9 0 . 0 N G . O F C O L U M N S 4 BL LD DG G.. L E N G T H535406.00.0 B W D I T H DLDG.HE G IHT L A N D C O ST 1613553546430.4.01 D L D G . C O S T R A C K C O S T 5 0 3 8 6 2 4 . 1 S/R MACH N IE'COST 4160000.0 D O L L A R S T "BH E . R E C U R R N I G C O S T S A R E M A N IT E N A N C ECOS C SR /LDG.M A N IT E N A N C E TOST 818005070.3020.0 SY YS ST TE EM MT T H R O U G H P U TPUT P .R SH /.R. S H R O U G H 1)C (O1)564.276.4OP N .O SH /N E X S N IA G L & U A O M N D TRAVEL T M IES T H EP. S H P EE F A C T O RD IL SC 6.M 6A T O T A L N G . O F O P E N N I G S 6 0 7 2 0 . 0 W P O OFS A B O V EB D SIN> I2S704815218.65)2646.5 D O L T (AX C L A ISTBL IT IY H O U L D EEG
257
APPENDIX 13 Outputs reXated to
sensitive, ^
°n XCTPUX)
258
G EO G A I48.T I4E C DE G SW IN A N G / Y PR E FU El > D E P 7 H- * D IT HOFv F E lG I FIlGG 0E F ?T 4 8 . 8 . T2Y0PE UN T IE G A D WE G IHT E(D . 8) ?T 0 0 . P ED U A LC O M M A N DP E R C E N T A G E ? TY 0 5 . C IALyHOR ZION A !L VELOC TIY O FG E/CFPM) ? TY 6P 0E.Y2V 4P0ER .T E PC I K •- U P y F" U T •? TY 02.PE5 U 02.N 5 I . BU L ID N IGyUN T IL A N D COST$(F/T##2) VTYP2E5 .0vy 6lT 2N O R 3^O F O RT SE R / SL OG C I O M ^ A N O B O A R D l N H R / T H E SR / C --3 E -NTRAL C O N S O L E ? TY 22^OFF E00.A N N U A LM A N IT E N A N C E COST SR / ? TY 22P A N N U A L BLDG. MA N IT. COSTS/QF.T. ? TY 5P .PE 5 E 0R (ACK) O R 1N (GN R -ACK) SUPORTED Y PE 0 T ON ICLUDE WA N TIG T M IEyOTHERW SE I ?T 1 Y PE A TM ARRyAPC LIABLEN IC O M ET A X RATE ?T 0 1 . 0 0 5 . 0 TYPE 0 F O R RANDOM ZE ID 1? F O R DED C IATED S T O Y PE T O T A L NO. O F SLOTS REQU R IED R (EAL) ?T 6 0 0 0 . H E REQU R IED T H R O U G H P U T(O P . NS FR /I ? TY 10PE0. T ?
0
?
0
259
N O .O O F A SV ILE E SS0.0 N O . F L E L 11 N O . O F C O L U M N S 6 9 . 0 0 D L D G . L E N G T H D L D G . W D I T H DLDG. HE G IHT" L A N D C O ST 1613533543430.4.01 D L D G . C O S T R KM C O STIE5C 03O 0S 6T 2'4.14160000.0 D SA R /C A C H N O L L A R S T H E R E C U R R N I G C O S T S A R E D GM .AM A N IT E N A N C EA C0S1C0S1T0"373200.0000 . 00 S /SLD R N I T E N N C E Y S T E M T i i R 0 U G H U SXPY S)G "' LE E M8TD I-A R 0C U GI-T l-'RAV U TT (ES1 E . S N I U L O M M A N D E L M I T H E SHAPE F A C T O R IS 6.6 T O T A L N O . O F O P E N N I G S 6 0 / 2 0 . 0 W P O O FS A B O V EB D G SIN> 2IS70401251S O T (AX C L A IST BL IT IY H O U L D EE .65)2646.5 D :
260
E O R G A I40P. 1 .:iT H E OLI O N EG /Ii T Y E0.C E I, D D IN -'O l vO JltDA "f I A IO rG 4 8 . 4 T200Y l' E U NI"I I. 0 A B li EG IT I( ?T 0 . D U A L C0hM Ai D PR P.CEN I A 0E ? TY 05.PE0YPEVER A E >• i I O RZU INTAL VELOC TIY ? T60.Y24l'0E .T1l'C C I K •- U l' >•p U T D ? TY 02P .E5 U 02.N 5 I BU L ID N IGyUN T IL A N D COST$(F/7**2) ? TY 2P 5E. 60. T y1v2 O R 3 !O R SR / LO 2O -FF T M E SR / SC ^ENTRAL C O N S O L E ?I 2 "20Y E ANNUAL MA 'i 2T 0Y .PEPA N N U A L BLDG, M Ai N T . 0;iS 1S /'0 ? TY 5P 5 .E 0 R (ACK) O R 1N (ON R -ACK) SUPORTED ? TY 0PE 0 T ON ICLUDE WA N TIG T M IE G ?THERW S ? TY 1 E0 A TM ARRyAPC LIABLEN IC O M ET A X RATE ? TY 01.P 0 5 . 0 PE 0 F O RR A N D O M Z E D I v1 F O R DED C IATED ? TY 0P E00.T O T A L NO. O F SLOTS REQU R IED R (EAL) ? TY 60PE H E REQU R IED THROUGHPUT<O N PH .S/R.) ? 125. T :
:
:
:
:
0 ••••••• M A N -
:
0 N -
B 0 A R D
1 ~
0
261
r:uO .. O FFA SIE L EE S N O L V L S II 0 . 0 NO. O FC O L U M N S 6 9 . 0 0 4 DL LD DG G.. W L E N G T H D D I T H BLDG. HE G IHT L A N D C O S T 1613533546430.4.01 B L D G . C O S T R KM C O S TIE3C 0O 38ST 624.14160000.0 D SA E /C A C H N O L L A R S T H E R E C U R R N I G C O S T S A R E D GM .A M A N IT E N A N C ESCSC OST 88100050.70320.0 /SLD S R N I T E N A N C E I Y S T E M T T II R 00 U G T II U P U S Y S T E M T T R U G T l E X P . S N I G L E & D U A L C O M M A N D T R A V E L T M I E S T H E SHAPE F A C T O R IS 6.6 T O T A L N O . O F O P E N N I G S 6 0 7 2 0 . 0 W P O O FS A D O V EB D G SIN O T (AX C L A IST BL IT IY H O U L D EE > 2IS704812518.632;648.3 D 1:
262
GEORG A I TECH. • DE G SIN O FA N AS R /S YI-'4E T8.H E D E E T1 v W D IT ?? T 4 8 . 8 . 4 TY PE00.UN T IE G A D WE G IHT E(D . S) 2 0 Y P ED U A LC O M M A N DP E R C E N T A G E ?T 0 5 . I60Y. 24E0. 0 E E1 J: C A L y I0 RIZ ?? T Y02E I"'C I K - U l' P U T •- D 0 2 . 5 . 5 Y PE. 6U N T I BU L ID N IGyUN T IL A N D COST ($T/l#*2 ?T 2 5 . TY 0A *O 1N *N 2-O O R 31G F O RNSR / I' H LO G C ISR 02P^EO M B A R D E / F F T H E S R / C 3 E N T R A L C O N S O L E ? TY 2PE A N N U A LM A N IT E N A N C E COST SR / ? TY 22PE 00.A N U A L BLDG. M A INT. COSTS/QF.T. ? TY 5P 5 .E 0N (ACK) O R 1N (ON R -ACK) SUPORTE ? TY 0PE 0 R T ON ICLUDE WA N TIG T M IE vOT TTY1 E A TM ARRyAPC LIABLEN IC O M ET A X RATE ? TY 01.P 0 0 5 . 0 E0F O RR A N D O M Z E D I y1 F O R DED C IATED ? TY 0P PE O T A L NO. O F SLOTS REQU R IED R (EAL) ?]'"60Y 00.PT R I ED ? 150. E T F! E R E Q U :
P
P
i
:
::
263
NO O.. O F A SV ILE ELSS 5110.0 N O F L E N O . O F C O L U M N S 5 3 0 . 0 B L D G .. L E N G T H 269750.60.? D L D G W D I T H DLDG. HE G IHT 571.7 L A N DG.C O ST 1614707625050.5.04 D L D C O S T R A C K COST 5020360.2 SR / M A C H N IE COST 3200000.0 D O L L A R S T H E R E C U R R N I G C O S T S A R E D L D G . M A N I T E N A N C E C O S T 1 0 7 9 1 0 0 . 0 S R / YM A N IT E N A N C E COSTT I-11R 0000.00U G H P S S T E M SY STEM T H R O U G H P U T > C lU 1074.C 7O O P N .SH /T R .VEL T E X P . S N I G L E S D A L M M A N D R A M IES T H E S H A P E F A C T O R I S 8 . 3 W P COSTLA O F A B O V E DEG SIN IS> 123869564368447..72) D O C T A X I B L I T I Y S H O U L D D E T O T A L NO. OF OPEN N IGS 60500.0
G E O R G I A
T Y P E ?
4 0 .
?
2 0 0 0 . TYPE
6 0 .
AN
A S / R S
01-
U NI T
L 0 A D (IN C H E S
4 0 .
UNIT
L O A D
WE I G H T ( L D . S )
D U A L
C O N N A N D
P E R C E N T A G E
OERTI C A E y H O R I Z O N T A L
0.25 TYPE
i
V E L O C I T Y
OF
S/R(FPM)
2 4 0 .
"I" Y p E ?
OF
0.50 T Y P E
r
•••• H E S I G N
D E P T IT ? W I D T IT v H E I G H T
T IT E
4 0 .
TYPE
?
T E C H .
*
P I C K - U P •> 0.25 UNIT
U T •- D CJ W N
B U I L D I N G * U N I T
T I fl E S
( MIN , S )
LAND
C O S T ( $ / F T * * 2 >
£> .
TYPE
0*1*2
OR
0-~M A N - O N 2:;::OFF ?
2
?
2 2 0 0 .
?
5.5
?
0
?
1
?
0.10
TYPE
T Y !•' E
THE
A N N U A L
A N N U A L
TYPE
0
TO
T O T A L
2 0 0 .
T YPE
1
C O S T / S R
M A I N "I".
(NON
C 0 S I'/S0
RACK)
W A I T I N G
A T M A R R y A P P L I C A B L E
TYPE
?
S/R C O N S O L E
v
FT .
S U P P O R T E D
BLDG.
T I M E y O T H E R W I S E
INCOME
TAX
TYPE
RATE
0.50 0
6 0 0 0 .
OR
I N C L U D E
TYPE
?
LOGIC THE
3-CENT RAL
BLDG .
(RACK)
0
S/R
1~GN
M A I N T E N A N C E
0
?
FOR
S/R
TYPE
T Y P E
3
"BO A R B
FOR
THE
R A N D O M I Z E D
NO.
OF
R E Q U I R E D
?1
S L O T S
FOR
D E D I C A T E D
S T O R A G E
R E Q U I R E D ( R E A L )
THR0UGIII""UT ( 0 P .N S / H R . )
1
265
NO. OF AISLES NO. OF LEVELS NO. OF COLUNMS
i i. oo 4600
BLDG. LENGTH BLDG. WIDTH BLDG. HEIGHT LAND COS7 120052.00 BLDG. COST 665913". 35 RACK COST 503062.41 S/R MACHINE COST 624000.00
DOLLARS
THE RECURRING COSTS ARE BLDG. MAINTENANCE COG7" 110701.00 S/R MAINTENANCE COST 13200.00 SY S TE M T H R 0UGHPUT(0) 34.50 0 P. N S/HR. S Y S I E M T H R 0 U G IT P U 7 (1) 240.94 0 f'. N S / IT R . :
EXP. SINGLE S DUAL COMMAND TRAVEL TIMES THE SHAPE FACTOR IS .99 TOTAL NO. OF OPENINGS
1.69
6072.00
PW COST OF ABOVE DESIGN IS 1452553.83 DOLLARS (TAX LIABILITY SHOULD BE > 310355.15)
2.60
APPENDIX 14 Detailed flow-chart of the program listing presented in Appendix 1
267
START
Initialize the DEM, IRT, IR values for each item }
Initialize LT IGT = 0
<^ D0
Kl = 1, 100
^>
INV(Kl) = IRT(Kl) MCUM(Kl) = 0 MON(Kl) = 0 LS(K1) = 0 MAX(Kl) = 0
f
CUDF = 0.0 RMX == 0.0
<^ D0
v1,260
KL = ITOT = 0 DIFF := 0.0 >
CALL DGEN(ID.KL)
^Cy^<^
D0
K -1,100
^ >
268
INV(K)
= INV(K)
-
ID(K)
(-) INV(K) LS(K)=LS(K)-INV(K)
(0) or (+)
YES LS(K)
INV(K)
=
0
IN = KL -
= 0
YES
LT
— <
NO >
INV(K) = INV(K) + NOR(K,IN) DIPT = DIFF + NOR(K,IN) MON(K)
—>-
N0R(K,KL)
=
0
«= MON(K)
IPO = I N V ( K )
(+)
-
+
NOR(K,IN)
MON(K)
IPO = I R ( K ]
(0) or (-) NOR(K,KL) MON(K)
= IRT(K)
= MON(K)
+
-
IPO
N0R(K,KL)
269
DO Jl = 1, 100
I
>
AV = MCUM(Jl)/208.0 RINVEN * RINVEN + AV IDS = IDS + MAX(Jl) ISAY = ISAY + LS(J1) V
AV,LS(J1),J1
IGT,IDS
t
AVER = CUDF/208.0
AVER,RMX, ISAY
271
BIBLIOGRAPHY
1.
Allred, James K. , "Large Automated Warehousing Systems - New Methods of Justifying, Buying and Applying Them," Presented at the MHI Material Handling Seminar, The 1975 National Material Handling Show, Cincinnati, Ohio, June 3, 1975.
2.
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3.
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4.
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6.
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7.
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8.
Hadley, G. and Whitin T.M., Analysis of Inventory Systems, PrenticeHall, Inc., New Jersey, 1963.
9.
Material Handling Institute, Inc., "Considerations for Planning and Installing an Automated Storage/Retrieval System," AS/RS Document 100 7M, 1977.
10.
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11.
Perlman, Justin A., "Materials Handling: New Market For Computer Control," Datamation, May 1970, pp. 132-137.
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272
13.
Sand, Gene M., "Stacker Crane Product Handling System," Eastman Kodak Co., September 1976.
14.
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16.
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18.
Wilkie, George S., "Hands-Off Warehousing System," Industrial Engineering, Vol. 5, No. 5, May 1973, pp. 12-19.
19.
Zollinger, Howard A., "Planning, Evaluating and Estimating Storage Systems," Presented at the Advanced Material Handling Technology Seminar, Purdue University, Indiana, April 29-30, 1975.
20.* Apple, J.M., Inc., 1972.
Material Handling Systems Design, Ronald Press Co.,
21.
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22.
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23.
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24.
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* The following were not cited, but were useful in the conduct of the research.