INSTRUCTOR’S MANUAL TO ACCOMPANY
FACILITIES
PLANNING
THIRD EDITION
JAMES A. TOMPKINS
Tompkins Associates, Inc.
JOHN A. WHITE
Universi ty of Arkansas
YAVUZ A. BOZER
University of Michigan
J. M. A. TANCH OCO
Pu rdu e U niv e rsi ty
PREFACE
The Instructor's Manual provides answers to the questions and solutions to the
problems at the end of the chapters in the Third Edition of Facilities Planning.
When a question or problem is open-ended, either no answer is provided or
guidance is provided relative to the response intended.
Due to human error, it is possible that mistakes exist in the responses provided.
Several changes were made to end-of-chapter questions and problems during the
production process.
possibletothat
arethat
provided
in anticipation itofischanges
theresponses
manuscript
did notin the
Instructor's
Manual Consequently,
occur. If you encounter an error in the Instructor's Manual, the authors request
that you bring it to the attention of Dr. James A. Tompkins, who was coordinating
author for this edition of Facilities Planning. Correspondence should be sent to
Dr. James A. Tompkins, President and CEO, Tompkins Associates, Inc.,8970
Southall Road, Raleigh, NC 27616. If more convenient, you can communicate
errors via email using the address: jtompkins@tompkinsinc.com.
This electronic version of the Instructor's Manual is available to faculty who
adopt Facilities Planning. To prevent widespread dissemination of answers and
solutions to end-of-chapter questions and problems, the authors and publisher
require that you use password-protected access to course websites containing
solutions to problems for students. Further, passwords must be changed after each
offering of the course. Many instructors do not want students to have access to the
Instructor's Manual. Hence, it is important for access to be limited.
To prepare the Third Edition of Facilities Planning, coordinating authors were
assigned for each chapter. Dr. Tompkins had coordinating responsibility for
Chapters 1, 7, 9, and 12; Dr. White had coordinating responsibility for Chapters 2,
3, and 11; Dr. Bozer had coordinating responsibility for Chapters 4 and 6; Dr.
Tanchoco had coordinating responsibility for Chapters 5 and 8; and Drs. Bozer
and White shared coordinating responsibility for Chapter 10. Hence, depending on
the nature of questions you have regarding material in a chapter, you might wish
to contact directly the coordinating author. Contact information for each author is
provided on the following page.
To assist you in finding material for a particular chapter, the page numbering in
the Instructor’s Manual incorporates the chapter number. In this way, we believe
you will be able to turn more quickly to the response of interest to you.
We are pleased to acknowledge the assistance of a number of individuals in
developing the material for the Instructor’s Manual. In particular, we express our
appreciation to David Ciemnoczolowski (University of Michigan), Ronald
Gallagher (Tompkins Associates), E. Harjanto (Purdue University), Mark
Rukamathu (University of Arkansas), and S. Sugiarta (Purdue University).
Thank you for adopting Facilities Planning. We trust it measures up to your
expectations and you view it as an improvement over the previous editions. We
have endeavored to respond to all feedback we received from instructors who
used the previous editions of the text. And, as in the past, we welcome your
suggestions for ways to improve the book.
James A. Tompkins
John A. White
Yavuz A. Bozer
J. M. A. Tanchoco
Authors’ Addresses
James A. Tompkins
President and CEO
Tompkins Associates, Inc.
8970 Southall Road
Raleigh, NC 27616
919-876-3667 (B)
919-872-9666 (F)
jtompkins@tompkinsinc.com
John A. White
Chancellor and Distinguished
Professor of Industrial Engineering
University of Arkansas
425 Administration Building
Fayetteville, AR 72701
479-575-4148 (B)
479-575-2361 (F)
jawhite@uark.edu
Yavuz A. Bozer
Professor
Department of Industrial and
Operations Engineering
University of Michigan
Ann Arbor, MI 48109
734-936-2177 (B)
734-764-3451 (F)
Yavuz.Bozer@umich.edu
J. M. A. Tanchoco
Professor
School of Industrial Engineering
Purdue University
Grissom Hall
West Lafayette, IN 47907
317-494-4379 (B)
317-494-5448 (F)
tanchoco@ecn.purdue.edu
Table of Contents
Chapter
Ti t l e
Part One
Defining Requirements
1
2
3
4
Part Two
5
6
Part Three
7
8
9
Part Four
10
Part Five
11
12
Introduction
Product,Process,andScheduleDesign
Flow,Space,andActivityRelationships
Personnel
Requirements
No. of
P a g es
16
44
9
8
Developing Alternatives: Concepts and Techniques
Material
Handling
LayoutPlanningModelsandDesignAlgorithms
7
28
Facility Design for Various Warehouse Functions
Warehouse
Operations
Manufacturing
Systems
Facilities
Systems
14
4
9
Developing Alternatives: Quantitative Approaches
QuantiativeFacilitiesPlanningModels
80
Evaluating, Selecting, Preparing, Presenting, Implementing,
and Maintaining
EvaluatingandSelectingtheFacilitiesPlan
Preparing, Presenting, Implementing, and Maintaining
Facilities
Planning
9
4
PART ONE
DEFINING REQUIREMENTS
Chapter 1
Introduction
1-1
Answers to Questions and Problems at the End of Chapter 1
1.1a
To plan for a football game, a visiting head football coach must:
Make sure that his players and coaches understand the game plan
Prepare his players physically and mentally for the game
Study opposing team game films
Understand strengths and weaknesses of the opponent
Maintain the discipline of his players
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•
To operate the team during a football game, a visiting head coach must:
Continuously update his game plan based upon what the opposing
team does
Communicate these updates with his coaches and players
Keep players physically ready to play
Replace injured players with other quality players if at all possible
Call the best plays that will score the most points and make victory
most likely
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•
1.1b
To plan for a football game, the home team quarterback must:
Practice drills and fundamentals
Become familiar with pass receivers skills and abilities
Understand role in game plan
Assume leadership role of team
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The operating activities that a home team quarterback must do are:
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1.1c
Operate
coaches’
plan on the field
Lead
team
againstgame
opposition
Use skills learned in practice
Minimize mistakes
Relay plans from sidelines to all players in the huddle
Win the game
To plan for a football game, the manager of refreshment vending must:
Make sure all food, beverages, condiments, and paper supplies are
ordered and stocked
Hire employees to work the vending locations
Pay suppliers
Understand what refreshments field personnel will need during the
game
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The operating activities that a manager of refreshment vending must do
are:
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•
Serve customers as efficiently and professionally as possible
Restock vending locations whenever necessary
1-2
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•
1.1d
Keep fans updated on game activities with televisions at vending
locations
Solve customer problems
To plan for a football game, the ground crew manager must ensure that:
Lines are painted on the field
Grass is mowed
Damaged turf is replaced
Necessary equipment for the teams is in its proper place
Locker rooms are prepared for the teams
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The operating activities that a ground crew manager must do are:
Replace damaged turf, if possible, during the game if weather is bad
Be ready for any grounds problems that may arise during the game
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•
1.1e
To plan for a football game, the stadium maintenance manager must:
Examine plumbing fixtures to make sure they are operational
Clean the stadium and pick up trash from the previous event
Repair any seats or bleachers that are damaged
Clean bathrooms
Repair and replace any necessary audio or video equipment for the
stadium
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The operating activities that a stadium maintenance manager must do are:
Respond to any equipment breakdown
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1.2
Respond to
breakdown
Respond
to any
any plumbing
power outage
Ten components of a football facility are:
The stadium structure
Parking lots around the stadium
Vendor selling areas
Maintenance components
Grounds-keeping equipment and personnel
Security
Customers
Athletic team personnel
Locker rooms
Vend ing equipment and supplies
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1.3a Activities that would be involved in planning the location of an athletic
stadium are:
Marketing analysis: Is there a fan base to support the teams or
events that will occupy the facility?
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1-3
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•
Determining a suitable plot of land for the facility and its associated
parking lots in terms of size and levelness
Determine by what routes suppliers will supply the facility
Determine if there are other facilities that will interact with the
stadium and how the location of the facility will affect those
interactions
Determine if existing structures will need to be demolished to
accommodate the new stadium and how those people or businesses
will be compensated
Determine the stability of the land that will be used to hold this
structure
1.3b Activities that would be involved in planning the design of an athletic
stadium are to determine the facility layout and systems and the material
handling systems that are necessary to operate the facility. Items that need
to be determined withinthe facility layout are as follows:
Number of people it will need to hold
Equipment is necessary to run this facility.
How the facility can host multiple types of sports—the design must
accommodate all of them
Whether the stadium be an outdoor stadium or a domed stadium
Artificial or natural turf
Materials of the track surface be
How and where upper-deck access will be
Vending locations in the facility
Where administrative offices of the facility will be
How many restrooms there will be and their locations
Where the locker rooms will be
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Facilities systems that need to be examined are:
Structural and enclosure elements.
Power and natural gas requirements
Lighting requirements
Heating, ventilation and air conditioning requirements
Water and sewage needs
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•
Handling systems that need to be examined are:
Material handling system
Personnel required to operate the stadium
Information systems required to operate the stadium
Equipment needed to support the stadium
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•
1.3c Activities that would be involved infacilities planning for an athletic
stadium are determining the facility location and design, explained in
1-4
greater detail in 1.3a and 1.3b, plus the ongoing maintenance and
improvement of the facility.
1.4
Customers in the transportation, communication, and service sectors do
have a need for facilities planners. Service facilities such as hospitals,
restaurants, athletic facilities, and retail shopping establishments all can and
do use facilities planners to optimize the facility layout, handling systems,
and facility systems on a continuous basis.
The communications industry uses the world as its facility. Communication
networks can
of as changing
the handling
systems offound
a communications
customer.
Duebetothought
the rapidly
technologies
in the
communications industry, these networks have to be continuously updated
and improved, or new ones have to be created. Also, the equipment and
personnel required to run multiple communication networks have to be
considered as well as where the communication points within the network
will be located. Finally, the facilities required to house a communication
hub or hubs must be located and designed, and a facilities planner is the
optimal person to do this job.
Just like the communications industry, the transportation industry uses the
world as its facility. Facilities planners can play an integral role in
determining where airports, train terminals, bus depots, truck depots, and
shipping docks are located and in designing them to accommodate the
traffic that travels through them. Also, facilities planners can assist in
determining the equipment and layout of those facilities as well.
Service industries such as retail shopping establishments, restaurants,
hospitals, and athletic facilities all use facilities planners to lay out their
buildings. Also, retail establishments and restaurants generally have
warehouses in which product is stored before it comes to its point of use.
Facilities planners play a large role in the location and design of warehouses
of all types.
1.5
Use the following criteria for determining the optimal facilities plan:
Does the facility provide for the company’s future storage
requirements?
Is there cost justification for the facility?
Is the facility centrally located for suppliers or customers if it is a
manufacturing facility; for manufacturing centers if it is a shipping
warehouse; for fans if it is an athletic facility, etc.?
Does it minimize receiving and putaway times while providing
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•
enough spaces for those functions?
Does the product from the manufacturing operations flow smoothly
through the facility?
Are there enough dock doors for shipping and receiving functions?
1-5
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•
1.6
Is the material handling equipment proper for the product that is
being moved? Is there enough space for the equipment to maneuver
around the facility?
Is the lease cost, property cost, or building cost cheaper than other
alternatives?
Can the manufacturing or warehouse facility accommodate sales
forecasts?
Does the plan minimize the cost of operation in terms of labor and
equipment?
Will scrap be minimized with this plan?
Can the facility be expanded easily to accommodate growth?
Is space being utilized to the highest extent?
This answer will vary from campus to campus, but some things of which
the student should be aware are:
Traffic patterns on campus
Whether administration offices are easily accessible
Handicapped accessibility to all buildings
Whether areas mix high vehicle and pedestrian traffic
What it would take to remove dilapidated buildings
The locations of certain buildings or functions if they are related to
one another, e.g. financial aid should be close to the cashier’s office
Maintenance facilities should be hidden from the main traffic areas
Whether there are adequate parking facilities for students, faculty,
and staff
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Whether adequate
there are adequate
establishments,
etc.
Whether
computereating
facilities
are accessiblerestrooms,
to all students
Whether there are adequate recreational facilities
If there is room to expand in the future
If there is room on or near campus for all students to live
Whether there adequate security on campus
1-6
1.7a
Facilities Planning Process for a Bank
Phase 1
1. Define (or
redefine) the
objectives of the
bank
Phase 2
Phase 3
2. Specify the
processes required to
3. Determine the
interrelationships
4. Determine the
space requirements
accomplish the bank’s
objectives
among departments
for all activities
5. Generate
alternative facilities
plans
6. Evaluate the
alternative facilities
plans
7. Select a facilities
8. Implement the
9. Maintain and
adapt the facilities
plan
facilities plan
plan
1.7b
Facilities Planning Process for a University
Phase 1
1. Define (or redefine) the
objectives of the university, the
majors it wishes to offer, and the
students it wishes to admit
Phase 2
2. Specify facilities and
personnel needed to
accomplish the
university’s objectives
Phase 3
among facilities
5. Generate
alternative facilities
6. Evaluate the
alternative facilities
plans
plans
8. Implement the
9. Maintain and
adapt the facilities
facilities plan
4. Determine the
space requirements
for all activities
3. Determine the
interrelationships
plan
7. Select a facilities
plan
1-7
1.7c
Facilities Planning Process for a Warehouse
Phase 1
Phase 2
Phase 3
1. Define (or
redefine) the
objectives of the
warehouse
2. Specify facilities
and personnel to
accomplish the
warehouse’s
objectives
3. Determine the
interrelationships
among activities
4. Determine the
space requirements
for all activities
5. Generate
alternative facilities
6. Evalu ate the
alternative facilities
7. Select a facilities
plans
plans
8. Implement the
9. Maintain and
adapt the facilities
facilities plan
plan
plan
1.7d
Facilities Planning Process for a Consulting Office
Phase 1
Phase 2
Phase 3
1. Define (or
redefine) the
objectives of the
office
2. Specify facilities
and personnel
required to accomplish
the office objectives
3. Determine the
interrelationships
4. Determine the
space requirements
among departments
for all activities
5. Generate
alternative facilities
6. Evaluate the
alternative facilities
7. Select a facilities
plans
plans
8. Implement the
9. Maintain and
adapt the facilities
facilities plan
plan
plan
1-8
1.8
The first task that needs to be accomplished is to determine the site of the
library on the campus. Questions such as do existing structures need to be
demolished, do roads need to be rerouted, do possible sites meet federal,
state, and local codes need to be asked. Next, a determination of office
space, book storage space, storage rack layout, and student study areas must
be made so that the structural design of the facility can be created. Also, the
power, gas, heat, ventilation, security, plumbing fixtures, and maintenance
issues must be settled. Finally, material handling issues such as how to
transport
supplies
and books,
howrequired
to reshelve
books,
how
from floor
to
floor, the
information
systems
to keep
track
of to
theget
locations
of
all of the books, and the personnel who will be required to maintain the
library must be resolved.
1.9
Facilities planning is never completed for an enterprise. If facilities planners
believed this, then numerous enterprises would be doomed to failure. New
technologies, different enterprises would be doome d to failure. New
technologies, different packaging methods, better storage methods, and
more advanced material handling equipment make the process of updating
and continuously improving facilities a must, or else the company will be
left behind with antiquated structures, information systems, and storage and
material handling mechanisms.
1.10 Certain items to look for from this paper are that an architect is more
involved with the design of the structure of the facility, the materials that
will go into
the structure,
andelectrical
the location
of items
as plumbing,
lighting,
HVAC
systems, and
systems.
Thesuch
facilities
planner will
be consulted on the above issues, but is more likely to plan the material
flow within the facility, how materials will be handled within the facility,
how materials will be stores, how materials will be manufactured, the
location of storage areas, the location of shipping and receiving docks, the
location and types of manufacturing equipment, the types of material
handling equipment required in the facility.
1.11 The IIE description of Industrial Engineering is:
Industrial engineering (IE) is about choices. Other engineering disciplines
apply skills to very specific areas. IE gives an opportunity to work in a
variety of businesses. The most distinctive aspect of industrial engineering
is the flexibility that it offers. Industrial engineers figure out how to do
things better. They engineer processes and systems that improve quality and
productivity. They work to eliminate waste of time, money, materials,
energy, and other commodities. Most important of all, IEs save companies
money. Industrial Engineering draws upon specialized knowledge and skills
in the mathematical, physical and social sciences together with the
principles and methods of engineering analysis and design to specify,
1-9
predict, and evaluate the results to be obtained from such systems.
“Facilities planning” could easily be substituted for the worlds “industrial
engineering.” Clearly, facilities planning is concerned with the design,
improvement, and installation of tangible fixed assets to achieve an
activity’s objectives. While designing, improving, and installing these
assets, the impacts on people, material, information, equipment, and energy
are very important. It is also clear that facilities planning draws upon
sciences, together with the principles and methods of engineering analysis
and design to specify, predict, and evaluate the results on an activity’s
tangible fixed assets. Therefore, it is clear that the profession most able to
perform facilities planning is the profession of industrial engineering.
1.12 The articles read should relate to the strategic planning process and should
draw parallels to the facilities planning process. An awareness of the true
meaning of strategy should be demonstrated.
1.13a Strategic uses for an airport that must be addressed in facilities planning
include:
Location
Types of planes that will be flying into the airport
Type of air traffic control system
Location of air traffic control tower
Location of baggage carousels
Location of ticket counters
Number of hangars required for servicing airplanes and for safety
inspections
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Employee skill levels for servicing airplanes between arrivals and
departures and for maintaining airpla nes
Number of employees required to run vending services, ticketing,
baggage claim, etc.
Location of parking lots
Many other items, including security control
1.13b Strategic issues for a community college that must be addressed include:
Location of classroom facilities
Location of administrative facilities
Skill level of faculty in each major
Majors that will be offered
Location of roads and sidewalks through campus
Skill level of students that will be admitted
Financial aid distribution to students
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Services
students to be provided to the students and the fees charged to the
Level of athletic focus, if any
Level of maintenance and the location of facilities
1-10
1.13c Strategic issues for a bank that must be addressed include:
Types of services provided
Skill levels of different types of employees
Security system and personnel
Interest rates for loans
Types of loans granted
Minimum borrower qualifications
Interest rates for savings and money market accounts
ATM machine on premises?
How many drive-thru windows?
Accounting procedures
Vault size
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1.13d Strategic issues for a grocery store chain that must be addressed include:
Types of products to sell
Arrangement of shelves within the store
Locations of items on those shelves
Parking area required
Accounting procedures
Skill level of employees
Freshness qualifications for open-air food such as produce
Amount of warehouse space where customers do not shop
Location of grocery stores throughout the country
Should stores have the same layout or should they vary?
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Number of shopping carts required
Number of docks required for receiving purposes
Unpacking methods
Predicted sales volume
Product pricing considerations
1.13e Strategic issues for a soft drink bottler and distributor that must be
addressed include:
Location of bottling facilities
Distribution routes
Number of trucks required for distribution
Types of equipment required for bottling
Number of bottles shipped per day
Number of SKUs that have to be bottled
Methods of conveyance through the facility
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Methods
of loading and
unloading trucks
Material composition
of bottles
Location and type of liquid components of beverage to be bottled
Skill level of employees
1-11
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Automation level
Inventory control methods
Storage requirements
1.13f Strategic issues for a library that must be addressed include:
Location of the library
Number of books that the library can contain
Number and location of shelves to hold the books
Skill level of employees
Types of magazine and newspaper subscriptions
Manual or automated system for finding items
Types of reference materials
From where will funding come?
Amount of study area required
Number of computers needed for public use
Size of children’s section within the library
Types of annual special events
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1.13g Strategic issues for an automobile dealership that must be addressed
include:
Parking area required
Level of knowledge of automobiles that employees must have
Number of salespeople
Sales tactics, professionalism of salespeople
Types of automobiles to sell
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Sales trends in the automobile industry
Relationship with automobile manufacturer
Service levels offered at the dealership
Parts for cars offered at the dealership
Number of service bays
Storage area required for service parts
Elegance of automobile showroom
Relationships with lending institutions
1.13h Strategic issues for a shopping center that must be addressed include:
Types of shops to be included in the center
Number of shops and their sizes
Location of shopping center
Lease rates for tenants
Public facilities for customers
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Parking
Securityarea for customers, should it be decked parking, etc.
Shopping center design, how elegant should it be?
Will it be a strip mall or an enclosed shopping center?
1-12
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Finding possible investors in the shopping center
1.13i Strategic issues for a public warehousing firm that must be addressed
include:
Size of the warehouse
Location of the warehouse
Level of automation within the warehouse
Number of warehouses
Types of products that can be stored in the warehouse
Material handling mechanisms within the warehouse
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Rental rates
Policies for unclaimed product
Number of racks required for the storage of the product
Number of docks required for shipping and receiving of the product
Will there be crossdocking capability?
Skill level of employees
1.13j Strategic issues for a professional sports franchise that must be addressed
include:
How does the ownership manage the franchise?
Location of a facility in which to play
Rent a facility or own one
Team colors, name, logo
Advertising methods
Determining the makeup of the fan base
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Ticket prices
Revenues from luxury suites
Training facilities
Selection criteria for choosing players and coaches
Salaries for players and coaches
Revenues from licensed sportswear sales
Player and coach discipline if they break the rules
Management’s role in determining the direction of the team
1.14a When there are many critical short-term problems that a company has to
solve at once, it is generally due to a lack of good strategic planning. Many
times, critical problems such as not enough storage space, too few material
handlers, improper storage techniques, and too much work-in-process
inventory all result from poor strategic facilities planning. Strategic
facilities planning is important when there are many “critical” problems to
solve because planning itself will generally solve many of the problems.
1.14b Everyone needs to have a say in the strategic planning process. However, if
critical individuals are too busy to do some sound strategic facilities
planning, they will always remain too busy because the problems that result
1-13
from poor strategic planning will keep these individuals busy attacking
short-term problems. Furthermore, there are consulting firms that specialize
in this type of work, so not as much time would have to be taken by these
individuals who are too busy, but their input would still be required while
the consultants are working on the problem.
1.14c A good facilities plan will be flexible enough to change with future events.
For example, a company that supplies the automobile industry builds a
warehouse during a downturn in automotive sales. If good strategic
planning has occurred, the company will not build the warehouse based on
the storage
requirement
of some
the sales
volume
duringofthe
the
company
will build on
large
percentage
thedownturn;
maximuminstead,
sales
volume they had during an upswing in the economy and on the forecast of
automotive sales over some time frame. Otherwise, the supplier could run
out of storage space in the new facility and have to add on or build another
warehouse before it has the financial capacity to do so.
1.14d This is a response of laziness. It takes much work and research to determine
what alternatives are available besides the one the company is using. Data
collection is needed to determine product throughput, production flow,
storage capacities, and inventory control procedures so that the proper
facility size and alternative control systems can be determined.
Development of labor standards and evaluation of material handling
methods is needed to look at different alternatives. If this information is
gathered and certain types of vendors are brought in to look at the problem,
numerous alternatives for a facility can be generated. Those that apply must
meet
scientific and financial criteria that are determined at the beginning of
the
project.
1.14e Strategic facilities planning is an ongoing process. Technologies come and
go, and a good facilities plan will enable a company to adapt to rapidly
changing technologies as well as to discard those that will not help the
company achieve its goals. A strategic plan is a plan for the future, not the
present. Do not incorporate technologies into a facility for the present
product mix—incorporate those technologies that can be used to produce
the future product mix. However, a facilities plan cannot take into account
technologies that have not yet been created; if a company creates a strategic
facility plan that incorporates technologies that have not been created, there
is no guarantee that the technology will be available when the company is
ready to implement the facilities plan and it will be doomed to failure.
1.14f One cannot get an exact dollar figure on the cost of a strategic plan
implementation and the savings it will generate. However, examination of
past trends and the prices of different types of technologies can give the
strategic planner a good estimate of the costs and the savings. The only way
to get exact costs is to get quotes from vendors, and the only way to
1-14
determine the exact savings is to implement the facility plan. This should
not stop a company from doing strategic planning, however. Sometimes the
plan may be infeasible and may not be implemented, but a company that
does not engage in strategic planning will not get a return on an investment
because the investment will never be made, or there will never have been a
determination on how much money the investment would make or lose.
1.15 Doing facilities planning for a manufacturing facility is a positive exercise
for a company in terms of its competitiveness. The production flow can be
examined by simulation, and potential bottlenecks can be smoothed. Also, a
facilities plan
for a manufacturing
facility outlines
skill levels floor,
of
employees
required
to operate the equipment
on the the
manufacturing
and it discovers the latest and best equipment to make the product.
Companies that do not engage in facilities planning usually end up having
uneven production flows, improper labor skills, too much or too little labor
on the floor, and outdated equipment that must be heavily maintained.
These problems as well as many others that could be eliminated by facilities
planning cause companies to lose their competitive edge.
1.16 Using strategic planning to assist with your career allows you to evaluate
where you are compared with the goals you set for yourself. If you have
fallen short of your goals, it allows you to easily evaluate why and
determine how you intend to correct the shortcoming, if it is possible.
Furthermore, it can show you if your career is at a dead end and where you
need to go to make a change in your career. Strategic planning gives you a
path to follow once the plan is in place. However, this plan should be
continuouslyedge
updated
likeonany
plan or you will lose your
competitive
overjust
others
theother
samestrategic
career path.
1.17 Automation, if planned properly, can have a positive impact upon facilities
planning. If there is a large product throughput, automation can reduce costs
by reducing labor requirements, improving quality, and perhaps improving
product throughput. However, many things could go wrong, and a facilities
planner needs to be aware of them. First, the automation may not justify
itself. If the throughput required to meet sales is 1,000 units per day and
automation equipment is purchased that can produce 20,000 units per day,
cost justification probably will not occur. Also, is the automated equipment
flexible enough to handle changes in product design or production
methods? While the first example showed the automation that could
produce 20,000 units per day had too much excess capacity and therefore
was not cost-justifiable, when a process is automated, there needs to be
extra capacity built into it so that it is not obsolete when the manufacturing
requirement increases to meet future sales. Also, the automated process
needs to be able to make product of higher quality than if it was made
manually or with a cheaper automated process. It does no good to have a
machine that can produce 5,000 units per hour to meet production
1-15
requirements of 4,000 units per hour when only half of the pieces pass a
quality inspection. Finally, it is necessary to examine how the automated
process will fit within the existing facility and how much rearrangement
will have to occur so that production will continue to flow smoothly.
Automation is a wonderful way to improve the quality and throughput of a
product, but only if it is done properly. Automation that does not do what it
is supposed to do ends up giving manufacturing personnel more problems
than the process it replaced.
1.18 Issues that should be addressed during strategic planning for
warehousing/distribution
include:
Number of shipping
and receiving docks
How the product will be shipped from the warehouse to the
customer
Product that will be stored
Number of SKUs
Size of the product that will be stored
Storage cube requirement
Pallet rack, flow rack, bulk storage, conveyor requirements
Inventory turns per a specific time
Warehouse staffing levels
Inventory investment levels
Material handling procedures and equipment
Building size
Will refrigeration be necessary?
Inventory control methods
Inventory control equipment
Building and rack layout
Power requirements
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
The primary customer service consideration for a warehouse/distribution
strategic plan is that the faster the turnover from receipt of an order to
shipment to the customer, the better. Also, proper product storage and
product shipment will reduce the amount of damaged product that a
customer receives. Finally, the better the inventory control system, the
easier it is to determine when there are stockouts and where all of the
inventory is, thereby making it easier to fill a customer’s order in less time.
Cost implications are that if excess inventory has to be carried due to
inaccurate inventory control data, then a larger facility has to be built, more
rack has to be installed, more labor has to be used to find the inventory in a
larger
building,
capitalproductive
is tied up way,
in inventory
rather
than earning
money
being used
in a more
and more
inventory
will have
to beor
thrown away because it is outdated. Facilities planning can reduce or
eliminate all of these problems, thereby reducing a company’s warehousing
1-16
costs, which can result in product being shipped at a cheaper price to
customers, which will result in more sales and revenues.
1.19 All of the requirements for success in Supply Chain Synthesis are directly
linked to the facilities planning process. A proper facilities plan will enable
a company to achieve synthesis much more quickly than a competitor who
does not use facilities planning. Strategic facilities issues in manufacturing
such as smoothing the production flow, eliminating bottlenecks, using the
proper amount of labor that has the appropriate skill level, using the proper
equipment that can meet or exceed throughput at a high level of quality will
help reduce
manufacturing
costs,
enable
manufacturing
marketing
to
work
together
to achieve sales
goals,
reduce
lead-times, and
reduce
setup times
and production lot sizes, reduce work-in-process inventories, simplify
process, balance the production flow, adapt to changing product and
technologies, reduce uncertainty, increase quality, and reduce process
failures. Also, reducing problems in the manufacturing process will increase
the number of happy employees and encourage them to become team
players. This will allow every part of the manufacturing process to become
integrated and facilitate not only a greater understanding by the emp loyees
of the entire manufacturing process and how to achieve synthesis, but also
an understanding of the company’s goals and directions.
1.20 The main difference between strategic planning and contingency planning is
that strategic planning is aproactive process in which problems are being
eliminated before they occur, while contingency planning is a reactive
process in which plans are made to eliminate problems as or after they
occur.
1.21 Facilities planning is not a thing that can be rushed or done halfway because
the personnel have “more important things to do.” If facilities planning is
done that way, the cost of implementation will increase because all
alternatives may not have been examined, or errors will have been made in
calculations. In order for facilities planning to be done properly, sufficient
lead-time in the implementation project must be granted. The amount of
lead-time is never the same for two different projects because each project
has a different level of complexity.
Chapter 2
Product, Process, and
Schedule Design
2-1
Answers to Questions and Problems at the End of Chapter 2
2.1 Identify the banking functions and departments to be incorporated in the
facilities plan. Identify the key entities for which flow requirements will be
needed, e.g., people, coin and currency, paperwork, vehicular movement.
Identify the various criteria that will be used to evaluate the alternative
facilities plans generated, e.g., customer service, security, ease of expansion,
cost. Determine the time period over which the facilities requirements will be
estimated. Estimate space and flow requirements and determine activity
relationships. Generate alternative facilities plans.
2.2 It is important for the various design decisions to be integrated so that all
critical issues have been considered before product and process designs are
finalized. Using a linear or series approach can result in multiple re-starts of
the design process because of “down stream” consequences of “up stream”
design decisions that are made. Overall optimization is the goal, rather than
piecewise optimization.
Knowledgeable representatives from each of the activities or functions need
to be involved in the design process.
Several concurrent engineering techniques can be used to improve the design
process. Quality Function Deployment is one technique that can prove
extremely beneficial. However, all of the approaches described in Section 2.5
should receive serious consideration. Since the text is devoted to facilities
planning, every technique presented in the text is a candidate for use in a
specific application.
2.3 Research question. Depending on the comprehensiveness of the collection of
periodicals and manuscripts in the university library, it might be more helpful
to the students to modify the assignment and encourage them to use the
Internet in performing the assigned search.
2.4 Research question. Depending on the comprehensiveness of the collection of
periodicals and manuscripts in the university library, it might be more helpful
to the students to modify the assignment and encourage them to use the
Internet in performing the assigned search.
2.5 Research
question.
Depending
comprehensiveness
of thebecollection
of
periodicals
and manuscripts
in on
thethe
university
library, it might
more helpful
to the students to modify the assignment and encourage them to use the
Internet in performing the assigned search.
2-2
2.6 Research question. Depending on the comprehensiveness of the collection of
periodicals and manuscripts in the university library, it might be more helpful
to the students to modify the assignment and encourage them to use the
Internet in performing the assigned search.
2.7 The assigned chart submitted for a cheeseburger will vary depending on the
assumptions regarding ingredients, e.g., mayonnaise, mustard, catsup,
onions, pickles, lettuce, tomato, multiple beef patties, multiple slices of
cheese, intermediate layer of bread. Likewise, the chart submitted for a taco
will vary depending on the ingredients included. It is important to verify that
the student follows the steps described in Section 2.3.
2.8 The assembly chart shows only the operations and inspections associated
with the assembly of the product. The operation process chart includes all
operations and inspections, fabrication and assembly operations and
processing times, and purchased materials.
2.9 OB = 3,000
3
3
IB = lOB/(1-PB)m = l3,000/(1-0.05) m = OA = QB
IA = l3,158/(1-0.02) m = 3,223 = QA
F=
lSA(QA)/[(EA)(H)(RA)] + SB(QB)/[(EB)(H)(RB)] +
30(QA)/[(500)(H)]m
H = 5 days/wk(18 hrs/day)(60 min/hr) = 5,400 min/wk
F=
l3(3,223)/[0.95(5,400)(0.95)] + 5(3,158)/[0.95(5,400)(0.90)] +
30(3,223)/[500(5,400)] = 5.4398 = 6 milling machines
m l
m
2.10 Follow the instructions provided in Section 2.5 for each of the 7 M&P tools.
2.11 The number of units of each product to be produced by each machine is
determined as follows (note:lxxm means the smallest integer$ xx).
OX = 111,000 3 IXC = lOX/(1-PXC )m = l111,000/(1-0.03) m = 114,433
IXB = lIXC/(1-PXB)m = l114,433/(1-0.04) m = 119,202
IXA = lIXB/(1-PXA)m = l119,202/(1-0.05) m = 125,476
OY = 250,000
3
IYC = lOY/(1-PYC)m = l250,000/(1-0.03) m = 257,732
IYA = lIYC/(1-PYA)m = l257,732/(1-0.05) m = 271,297
IYB = lIYB/(1-PYB)m = l271,297/(1-0.04) m = 282,602
Setup times are identical for machines A, B, and C for a particular product.
The setup time for product X, regardless of the machine, is 20 mins; the
setup time for product Y is 40 mins., regardless of the machine. A critical
piece of information needed to determine the number of machines required is
the length of production runs between setups. If a single setup is needed to
2-3
produce the annual requirement of a product on a machine, then the number
of machines required is determined as follows:
FA = l0.15(125,443)/[0.90(1600)(0.90)] + 20/[60(1600)] +
0.10(271,297)/[0.90(1600)(0.90)] + 40/[60(1600)]m
FA = l35.4529m = 36 machines of type A
FB =
l0.25(119,170)/[0.90(1600)(0.90)] + 20/[60(1600)] +
0.10(282,602)/[0.90(1600)(0.90)] + 40/[60(1600)]m
FB =
l44.7944m = 45 machines of type B
FC =
l0.10(114,403)/[0.95(1600)(0.95)] + 20/[60(1600)] +
0.15(257,732)/[0.95(1600)(0.95)] + 40/[60(1600)]m
l34.6960m = 35 machines of type C
FC =
If setups occur more frequently, then additional machines might be required
due to the lost production time consumed by setups.
2.12 S = 15, S/E = 20, P = 0.2, R = 7/8 = 0.875, H = 8(60) = 480
O = 750
3
I = lO/(1-P)m= l750/(1-0.2)m = l937.5m = 938 parts = Q
F = l(S/E)(Q/HR)m = l20(938)/[480(0.875)] m = l44.67m = 45 machines
2.13 I1 = lO6/[(1-P1)(1-P2)(1-P3)(1-P4)(1-P5)(1-P6)m
2.14 Shown below is the probability mass function for number of good castings
produced (x), based on Q castings scheduled for production.
2-4
Shown below is the matrix of net income for each combination of Q and x.
For a given value of Q, multiplying the net income in the column by the
probability of its occurrence and summing over all values of x yields the
following expected profits for each value of Q.
Based on the results obtained, scheduling 29 castings for production yields
the maximum expected profit of $11,050.
2-5
2.15 Shown below is the probability mass function for number of good castings
produced (x), based on Q castings scheduled for production.
Shown below is the matrix of net income for each combination of Q and x.
For a given value of Q, multiplying the net income in the column by the
probability of its occurrence and summing over all values of x yields the
following expected profits for each value of Q.
2-6
Based on the results obtained, scheduling 25 castings for production yields
the maximum expected profit of $8,338.
2.16 Shown below is the probability mass function for number of good castings
produced (x), based on Q castings scheduled for production.
Net incomes for feasible combinations of Q and x are shown below.
For a given value of Q, multiplying the net income in the column by the
probability of its occurrence and summing over all values of x yields the
2-7
following expected profits for each value of Q. Based on the results obtained,
scheduling 21 castings for production yields the maximum expected profit of
$22,663.
2.17 Shown below is the probability mass function for the number of good
custom-designed castings produced (x), based on Q castings scheduled for
production.
Net incomes for feasible combinations of Q and x are shown below.
a. For a given value of Q, multiplying the net income in the column by the
probability of its occurrence and summing over all values of x yields the
following expected profits for each value of Q. As shown, the optimum
number to schedule is 5, with an expected net profit of $35,335.
2-8
b. The probability of losing money on the transaction is the probability of
the net income being negative when Q equals 5. From above, a negative
net cash flow occurs if less than 4 good castings are produced. The
probability of producing less than 4 good castings equals 0.0005 +
0.0081 + 0.0729, or 0.0815.
c. Using Excel it is easy to perform the sensitivity analysis. By varying the
cost parameter, it is found that a cost of $22,044.96 yields an expected
profit of zero for Q = 5. Likewise, when the cost of producing a casting
is reduced to $7,873.19, the optimum number to schedule increases to 6.
We could not find a cost that would reduce the optimum production
batch to 4 and still have a positive expected profit.
d. Again, using Excel it is easy to perform sensitivity analyses. For example,
if the probability of a good casting is reduced to 0.84248 then the
optimum production batch increases to 6; likewise, if the probability of a
good casting increases to 0.963781, then the optimum batch production
quantity decreases to 4.
2.18 Shown below is the probability mass function for number of good high-
precision formed parts (x), based on Q parts scheduled for production.
Shown below is the matrix of net income for batch sizes of 10, 11, and 12.
Also shown below is the expected profit based on batch sizes of 10, 11, and
12, as well as the probability of losing money. A batch size of 12 yields the
smallest expected profit. Based on the probability of losing money, the least
attractive alternative is a batch size of 10.
2-9
2.19 Shown below is the probability mass function for the number of good wafers
(x) resulting from a production batch size of Q.
Shown below is the matrix of net profit resulting from combinations of Q and
x.
2-10
Shown below are the expected profits and probabilities of losing money for
various batch sizes. The optimum batch size is 7, with a 0.0129 probability of
losing money.
2.20 Shown below is the probability mass function for the number of good die
castings (x) in a production batch of size Q.
Shown below is the matrix of net profits resulting from various combinations
of Q and x.
2-11
Shown below are the expected profits and probabilities of losing money for
various values of Q, the batch size. From the results obtained, the optimum
batch size is 28. The probability of losing money, which is the probability of
less than 25 die cast parts being acceptable, equals 0.0022.
2.21 a. Shown on page 2-12 are the prob ability mass function for the number of
good castings (x) produced when Q castings are produced, based on a
probability of 0.85 that an individual casting is good. Also shown on page
2-12 is a matrix of net profits resulting from the combination of Q and x.
Finally, the expected profit is shown for various values of Q. Based on
the results obtained, the optimum lot size is 65, with an expected profit
of $55,925.
b. Shown on page 2-13 are the prob ability mass function for the number of
good castings (x) produced when Q castings are produced, based on a
probability of 0.98 that an individual casting is good. Also on page 2-13
is a matrix of net profits resulting from the combination of Q and x.
Finally, the expected profit is shown for various values of Q. Based on
the results obtained, the optimum lot size is 56, with an expected profit
of $64,315.
2-12
2 .21 a .
2-13
2 .21 b .
2.22 In Example 2.5, a = 2 min, b = 1 min, t = 6 min, n’ = 2.67, C
o = $15/hr, and
Cm = $50/hr. Without other constraints the optimum number of machines to
assign to an operator was shown to equal 2. Since 2 < n’ < 3, the economic
choice was between 2 and 3 machines. Hence, two groups of 2 would be less
costly, on a cost per part produced basis, than one group of 4 machines.
Here, 11 machines are required to meet the production requirements. How
should they be assigned? One of the alternatives being considered is to assign
2 machines to each of 4 operators and then assign 3 machines to one
operator; the alternative assignment being considered is to assign 2 machines
to each of 5 operators and then assign 1 machine to one operator.
2-14
To calculate the cost per unit produced for each scenario, it is useful to
evaluate each alternative using a length of time equal to the least common
multiple of the cycle times for each machine-operator assignment in the
scenario. For example, with the {2, 2, 2, 2, 3} scenario, T c = 8 min for m = 2
and Tc = 9 min for m = 3; therefore, a time period equal to 72 minutes will be
used. During a period of 72 mins. each 2-machine combination will perform
9 cycles and produce 18 parts; likewise, over the same time period, the 3machine assignment will perform 8 cycles and produce 24 parts. Hence, over
a 72 min. period, a total of 4(18) + 24, or 96 parts will be produced. The
total cost per unit produced over a 72 min. period equals [(5 op)($15/hr-op)
+ (11 mach)($50/hr-mach)](72 min/60 min/hr.)(1/96) parts), or $7.81/part.
For the {2, 2, 2, 2, 2, 1} scenario, T c = 8 min for both m = 1 and m = 2.
During an 8 min. time period a total of 11 parts are produced. The total cost
per unit produced over an 8 min. period equals [6(15)+11(50)](8/60)(1/11),
or $7.76/part. Hence, the least cost alternative, in terms of cost per unit
produced, is the {2, 2, 2, 2, 2, 1} scenario.
Are there other scenarios that are less costly than the two considered? No!
From Example 2.5, n* = 2. Hence, any scenario involving multiple
assignments of single machines will be more costly than assignments of 2
machines per operator. Likewise, from the analysis performed above, any
scenario involving a 3-machine assignment will be more expensive than one
with a 2- machine and a 1-machine assignment. Further, any scenario having
a 4-machine assignment will be more costly than one that substitutes two 2machine assignments for the 4-machine assignment. From the analysis
performed above, a 5-machine assignment will be more expensive than a {2,
2, 1} assignment. By similar analyses, there are no other scenarios that need
to be considered for the assignment of 11 machines.
2.23 For the optimum assignment in Example 2. 5 to remain unchanged,M < 1.
Recall, in Example 2.5, n’ = 2.67, Co = $15/hr, Cm = $50/hr, and , = Co/Cm
= 15/C m. Therefore, for M < 1, (, + n)(n’) < (, + n + 1)(n), or
(, + 2)(2.67) < (, + 3)(2), or (15 + 2C m)(2.67) < (15 + 3Cm)(2), or
(15)(2.67 - 2) < (6 - 5.33)C m, or C m > $15/hr. Hence, for a machine cost of
$15 or more per machine-hour, the optimum assignment will be 2 machines
per operator.
2-15
2.24
2-16
2.24 (continued)
2-17
2.24 (continued)
2-18
2.24 (continued)
2-19
2.24 (continued)
2-20
2.24 (continued)
2-21
2.24 (continued)
2-22
2.24 (continued)
2-23
2.24 During 7 hours of work for the operator between 8:00 a.m. and 4:00 p.m.,
136 units were produced by the 3 machines. In steady state conditions, the
repeating cycle is 9 minutes. Hence, in steady state conditions a total of 140
units are produced. Transient conditions due to start-up and shut-down for
breaks, lunch, and beginning/ending of the shift diminish the production by
only 4 units. If replacement labor is provided to keep the machines working
during the entire 8-hour shift and 3 shifts operate per day, then steady state
production will result in 160 units being produced per 8-hour shift.
2.25 The problem statement was overly simplified, assuming sufficient demand
exists to keep 5 machines busy and sales prices of the products are such that
the calculation of cost per unit produced can be performed by summing the
units produced for both products. Also, it is assumed that a machine is
dedicated to producing either product 1 or product 2 and cannot be assigned
to produce a combination of the two products due to changeover times.
From Example 2.5, we know that two machines producing product 1 should
be assigned an operator to minimize the cost per unit produced. For product
2, the optimum number of machines to assign an operator is obtained as
follows for producing product 2: n 2' = (2.5 + 8)/(2.5 + 1.5) = 2.625; , = 0.3;
and M = (2.3/3.3)(2.625/2) = 0.9148 < 1; therefore, n2* = 2. Hence, it
appears that 2 machines should be assigned to pro duce product 1 and 2
machines should be assigned to produce product 2; however, that leaves 1
machine unassigned. From the solution to Problem 2.22, we know that {2,
1} is less costly than {3} for product 1.
The alternatives to be evaluated are as follows: assign 2 machines producing
product 1 to an operator and 3 machines producing product 2 to an operator;
assign 2 machines producing product 1 to an operator, 2 machines producing
product 2 to an operator, and 1 machine producing product 2 to an operator;
assign 2 machines producing product 1 to an operator, 1 machine producing
product 1 to an operator, and 2 machines producing product 2 to an
operator.
Assign 2 machines producing product 1 to an operator and 3 machines
producing product 2 to an operator. From Example 2.5, for product 1 the
repeating cycle is 8 minutes. The repeating cycle is 3(2.5 + 1.5), or 12 min.,
for product 2. Hence, in 24 minutes, the 2 machines producing product 1
perform 3 repeating cycles and produce 6 parts, and 6 parts are produced by
the 3 machines making product 2 while performing 2 repeating cycles. The
total hourly cost of 2 operators and 5 machines is $280. During a period of
24 minutes, the cost will be $112.00; hence, the cost per unit to produce 12
parts in 8 minutes is $9.33/part.
2-24
Assign 2 machines producing product 1 to an operator, 2 machines
producing product 2 to an operator, and 1 machine producing product 2 to
an operator. The repeating cycle for product 2 is 10.5 minutes. Hence, in
168 minutes there will be 21 repeating cycles for machines producing
product 1 and 16 repeating cycles of machines producing product 2. The
total hourly cost of 3 operators and 5 machines is $295. During a period of
168 minutes, the cost will be $826; hence, the cost per unit to produce 90
parts (42 of product 1 and 48 of product 2) in 168 minutes is $9.18/part.
Assign 2 machines producing product 1 to an operator, 1 machine
producing product 1 to an operator, and 2 machines producing product 2 to
an operator. As in the previous case, the repeating cycles are 8 and 10.5
minutes. Hence, over a 168 minute time frame, there will be 21 repeating
cycles of the 3 machines producing product 1 and 16 repeating cycles of the
2 machines producing product 2. The cost per unit to produce 63 units of
product 1 and 32 units of product 2 is ($295)(168)/(60)(95), or $8.69/part.
Since this is the least cost o ption, it would be recommended.
As noted, a simplified approach was used to arrive at a preference in the
assignment of the 5 machines to the 2 products. With more information
regarding sales prices, demands, changeover times, etc., a more informed
decision could be made. The underlying objective in presenting the machineassignment problem was to provide students with experience in using simple
mathematical models in making decisions regarding the assignment of
machines to operators.
2.26 a. a = 6 + 4 = 10; b = 6; and t = 30. n’ = (10 + 30)/(10 + 6) = 2.5. No more
than 2 mixers can be assigned without idle mixer time.
b. Co = $12/hr, Cm = $25/hr, and , = Co/Cm = 0.48.
Therefore, M = ( , + n)(n’)/[(, + n + 1)(n)] = 2.5(2.48)/[3.48(2)], or
M = 0.89 < 1. Hence, 2 mixers should be assigned an operator.
2.27 The multiple activity chart is provided on the following page. The length of
the repeating cycle is given by the maximum of the following values:
(aA+bA+aB+bB+aC+bC); (aA+tA); (aB+tB); (aC+tC), or max (11, 9, 10.5, 12),or
12. The repeating cycle is determined by machine C. As shown, the operator
will have 1 minute of idle time during a repeating cycle, machine A will have
3 minutes of idle time, machine B will have 1.5 minutes of idle time, and
machine C will have no idle time during a repeating cycle.
2-25
2.28 a = 5 min; b = 1 min; t = 20 min; Co = $12/hr, and Cm = $30/hr.
a. n’ = (5 + 20)/(5 + 1) = 4.167 4 is the maximum number of machines that
can be assigned an operator without creating machine idle time during a
repeating cycle.
b. , = Co/Cm = 0.4. Therefore, M = ( , + n)(n’)/[(, + n + 1)(n)] =
4.4(4.167)/[5.4(4)], or M = 0.8488 < 1. Hence, 4 machines should be
assigned an operator.
c. TC(m = 4) = [12 + 4(30)](5 + 20)/[ (60)(4)] = $13.75/unit.
d. For n* = 4, either M < 1 when n = 4 or M > 1 when n = 3.
I) M < 1 case: M = (0.4 + 4)(n’)/[(0.4 +5)(4)] < 1 or M = [4.4(a + 20)]/[(a
+ 1)(5.4)(4)]| (4.4a + 88) < (21.6a + 21.6). Hence, 17.2a > 66.4, or a >
3.86 min.
ii) M > 1 case: M = [3.4(a + 20)]/[(a + 1)(4.4)(3)]. Thus, (3.4a + 68) >
(13.2a + 13.2). Hence, 9.8a < 54.8, or a < 5.59 min.
Hence, for 3.86 min < a < 5.59 min.
e. Consider the alternatives: {5, 5, 5}, {4, 4, 4, 3}, and {4, 5, 6}.
i) {5, 5, 5} case: Tc = 30 min.
TC{5, 5, 5} = [3($12) + 15($30)](30/60)(1/15) = $16.20/unit.
ii) {4, 4, 4, 3} case: Tc = 25 min.
à
2-26
TC{4, 4, 4, 3} = [4($12) + 15($30)](25/60)(1/15) = $13.83/unit.
iii) {4, 5, 6} case: Tc(4) = 25 min., Tc(5) = 30 min., Tc(6) = 36 min.In
2,700 minutes, 1,332 units will be produced: 4(108), o r 432, by the 4machine assignment; 5(90), or 450, by the 5-machine assignment; and
6(75), or 450, by the 6-machine assignment.
TC{4, 5, 6} = [3($12) + 15($30)](2700/60)(1/1332) = $16.42/unit.
The least costly assignment of 15 machines is {4, 4, 4, 3}. We do not need
to consider {3, 3, 3, 3, 3} since it has the same repeating cycle as {4, 4, 4,
3} and requires an additional operator. Likewise, there is no reason to
consider an alternative involving a {3, 5} combination since we know
from part a) that {4, 4} is less costly.
2.29 a = 4 min; b = 5 min + 3 min = 8 min; and t = 40 min n’ = (4 + 40)/(4 + 8)
à
or n’ = 3.67 and 3 is the maximum number of automatic palletizers on
operator can be assigned without creating idle time for the palletizers.
2.30 The problem statement does not mention retrievals by the S/R; hence, no
travel between the P/D station and the outbound conveyor is required. It is
assumed that the travel between the P/D station of one S/R aisle and the
inbound conveyor for another S/R aisle also requires 0.5 minute.
For the problem, there is no concurrent activity. As long as a load is at the
P/D station, the S/R machine will retrieve it automatically. Hence, for the
problem, b = 0.5 min to travel to the P/D station + 0.5 min to travel from the
P/D station = 1.0 and t = 4 min to store a load and return to the end-of-theaisle. (In Chapter 10, we define this as a single command cycle for the S/R
machine.) With b = 1 and t = 4, n’ = 4. Hence, as shown on page 2.-27, one
lift truck operator can service 4 S/R machines; the operator and the S/R
machines will be busy 100% of the time. For 5 S/R machines, 2 lift truck
operators will be required.
2.31 a = 0.25 min; b = 0; and t = 1.0. Therefore, n’ = 1.25/0.25 = 5.0. An
operator can tend 5 carousels without creating idle time for the co nveyors;
the operator will also be 100% occupied.
2.32 The solution depends on the recipe chosen.
2.33 The solution depends on the recipe chosen.
2.34 a.
shortages
b. Board
Board for
for part
back-orders
c. Feedback from material handlers when part has low physical inventory
d. Feedback from operator handlers when part has low physical inventory
2-27
2.35
2-28
2.36
2.37 - 2.40
The answers to these questions depend on choices made and course
specifics. See Section 2.5 for details on each of the 7 M&P to ols.
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Chapter 3
Flow, Space, and Activity
Relationships
3-1
Answers to Questions and Problems at the End of Chapter 3
3.1 A material management system for a bank refers to the flow processinto the
bank. The subjects of this system might include, but are not limited to, coin,
currency, checks, other monetary instruments, deposit and withdrawal forms,
loan documents, customers, suppliers, employees, banking supplies, and
equipment required to operate the bank. The material flow system for a bank
refers to the movement of materials, supplies, equipment, and personnel
within the bank. The physical distribution system at a bank describes the flow
of money and information, including loan and other documents,out of the
bank.
3.2 A group technology layout is a candidate for consideration when a medium
volume of a medium variety of products are t o be produced. It is used when
families of products can be grouped according to physical characteristics or
production sequence information.
3.3
3-2
“Flow between,” expressed in equivalent loads/day, is depicted below. As
noted, 90 equivalent loads move weekly between Marriage Licences and the
Lab (60 + 30). Given the differences in numerical values a closeness rating
was assigned, as shown in the activity relationship chart.
Flow Between Chart
Activity Relationship Chart
3-3
The basic layout of the hospital will depend on “who moves” and “who
remains fixed” in location. Actually, all subjects are candidates for
movement, except for the most expensive diagnostic equipment; it cannot
always be taken to the patient. However, advances in medical technology are
such that more and more the technology is being brought to the patient via
distance medicine.
If it is desired that patients not be transported around the hospital to the
various x-ray, cat-scan, and MRI facilities, then either remote sensing
methods must be provided or the diagnostic equipment must be duplicated.
Likewise, if the movement of medical personnel is to be minimized, then
patients can be assigned to rooms/beds that are designated for particular
doctors and nurses; the down-side of such an assignment is the likelihood of
empty beds existing, since not all doctors will always have the maximum
number of patients in the hospital at all times.
Hospitals are often designed using group technology layouts, where the
grouping is based on the medical services provided the patients.
3.5 A fixed position layout would be recommended when the product is very
large, awkward, or expensive to move. It is also a candidate when low or
sporadic volume of low variety production occurs.
3.6 Group technology layout using manufacturing cells is popular in JIT facilities
because cellular layout encourages small lots, kanbans, simple material
handling systems, short setup times, cross trained personnel, teamwork, and
quality at the source.
3.7 Process layouts can result in complex flows, excessive handling, large
inventories, and long production lead-times.
3.8
3-4
3.9
3.10
3.11
3-5
3.11 (continued)
3.12 Research Question: answer depends on the local fast-food restaurant chosen
for study.
3.13 Research Question: answer depends on t he components in the kitchen and
the eating habits of the person for whom the kitchen is designed.
3.14 Backtracking results in excessive flow or travel, longer lead-times, and
complications in scheduling. Backtracking may be avoided by duplicating
machines, redefining the process plans to complete the machining in
consecutive steps, specifying the use of another machine not requiring
backtracking, or redesigning the product to eliminate the processing
requiring backtracking.
3.15 Pros:
Cons:
simpler material handling equipment, shorter travel distances, less
complex flow
duplication of equipment, more discipline to monitor inventory and
quality, more complex logistics to distinguish deliveries and
shipments at different docks
Considerations should include evaluation of aisle requirements, outside
accessibility for trucks, requirements for receiving inspection, delivery lot
sizes, delivery/shipment unit load sizes, frequency of deliveries/shipments,
and required paperwork and communications.
3-6
3.16 Top Management: Alternative issues and strategies to consider in the analysis
such as layout classification, storage-handling strategies, organizational
structure, and environmental policy.
Product Designer: components and their dimensions and quantities per unit.
Process Designer: machines and their specifications and product routings.
Schedule Designer: lot sizes, production control method (e.g., kanbans), and
production volume.
Modern Manufacturing Approaches: kanbans, standardized containers, total
productive maintenance, short set-ups, small lots, manufacturing cells,
quality at the source, delivery to points of use, visual management, total
quality management, multi-functional employees, decentralized storage, and
team-based environments.
3.17 Benefits: reduction of inventories, space, machine breakdowns, rework,
paperwork, warranty claims, storage and handling equipment, employee
turnover and absenteeism, production lead-times, cost, and stockouts;
simplification of communication, handling, and production scheduling; and
improvement of productivity, flexibility, inventory turnover, quality,
customer satisfaction, and employee morale.
3.18 Research Question: answer depends on problem statement.
3.19 A kanban is a signal, typically a card, that indicates to the supplying
workstation that the consuming workstation requests more parts. There are
two types of kanbans: a production card that is used to authorize production
of more parts and a withdrawal card that is used to authorize delivery of
more parts or components. The benefits of kanbans include simplified
production control, reduced inventories, increased visibility, improved
quality, reduced material handling, and reduced lead-times. Kanbans are a
reminder that each employee has a customer, i.e., someone who depends on
the employee doing her or his job well; in this case, kanbans come from
internal “customers” and serve a similar function to a customer’s order for
material. Kanbans are important in cellular manufacturing because of the
limited space and need for better production control to prevent excessive
work-in-process (WIP).
3.20 Benefits of U-shaped flow in manufacturing cells include, but are not limited
to, enhanced visibility, improved communication, improved teamwork,
simplified
flow,reduced
reducedhandling,
travel distance,
improved
quality,
WIP,to
reduced space,
and improved
control
overreduced
input/output
the cell.
3-7
3.21 The impact includes delivery of small lots, the elimination of paperwork due
to electronic data interchange (EDI), receiving material in decentralized
storage areas, by-passing incoming inspection since suppliers have been
certified, reduced movement of material, and simpler material handling
equipment alternatives. The overall impact to the logistics systems includes
shorter lead times, lower cost, and better quality.
3.22 a) Process improvement and quality activities
b)
c)
d)
e)
Training for improved cross-functionality
Production coo rdination
Performance measures tracking
Material handling
The new roles impact resources (less people, less inventories, less space) and
change the structure of traditional activity relationships (storage at points of
use, training and continuous improvement).
3.23 Logistics encompasses the arrival and departure of parts and products,
including the quantities of each. Material handling and storage are extremely
important for the facilities planner. In fact, some define manufacturing as
production processes located strategically in a logistics system.
3.24 Research Question: answer depends on the restaurant visited.
3.25 Parallel parking advantages: space utilization and flow in one direction; okay
if last-in, first-out arrival and departure is used
Parallel parking disadvantages: excessive vehicular movement required; poor
if first-in, first-out arrival and departure is used
3-8
Perpendicular parking advantages: none come to mind
Perpendicular parking disadvantages: flow in two directions, excessive
movement, and requires more open spaces
Diagonal parking advantages: requires less vehicular movement
Diagonal parking disadvantages: requires more space
3-9
3.26 Research Question: answer depends on the individual.
3.27 a. 100' + 75' + 75' + 25' + 25' = 300'
b. 175' + 75' + 150' + 25' + 25' = 450'
c. 300' + 25' + 25' + 75' + 75' + 100' + 300' = 900'
d. 175' + 100' + 175' + 150' + 50' = 650'
3.28 When streets are not at right angles, turning of vehicles can be more
problematic. In such instances, wider turning lanes might be required,
particularly for delivery trucks.
3.29 Research Question: answer depends on assumptions made.
3 .30 A 6 C
8
9
D7 B
3.31 Research Question: answer depends on student and professor.
3.32 Research Question: answer depends on services provided by and the
organization of the bank.
3.33 Research Question: answer depends on student’s perspective and the degree
to which the student interacts with the various activity areas listed.
3.34 Research Question: answer depends on student’s perspective of the professor
and the various roles played by the professor in question.
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Chapter 5
Material Handling
5-1
Answers to Questions at the End of Chapter 5
5.1
Work principle. The measure of work is material flow (volume, weight,
or count per unit of time) multiplied by the distance moved. The work
principle promotes minimizing work whenever possible. Materials
movement should be in as practically large amount as possible and
material-handling devices should be fully utilized by carrying loads at its
capacity.
Ergonomic principle. Ergonomics is the science that seeks to adapt work
or working conditions
to suit
the be
abilities
of the worker.
Human
capabilities
and limitations
must
incorporated
into material
handling
equipment and procedures designed for effective interaction with the
people using the system. As an example, for humans picking in a
warehouse, do not store any items above the top of the head. The
optimum picking zone in order to minimize stooping and stretching is
from the hips to the shoulder.
5.2
Unit Load principle. A unit load is one that can be stored or moved as a
single entity at one time, such as pallet, container, etc, regardless of the
number of individual items that make up the load. In this principle,
product should be handled in as large a unit load as practical. As an
example, instead of moving individual products, a number of products can
be palletized and move as a single load. However, the size of the load
must be considered as excessive unit load can reduce the visibility of the
forklift driver or may not be practical in utilizing truck space. Moving as
large delivery.
a unit loadThere
as practical
mustinbemoving
exercised
in the if
context
just-intime
is no point
materials
it willof
not
be used
for a long duration.
Space Utilization principle. Space in material handling is three
dimensional and therefore is counted as cubic space and effective
utilization of all cubic space is the essence of this principle. If products
are stored in a container, the product should fill up the container. In
storage application, the storage dimension should be designed to fit a unit
load with some tolerance for storage and retrieval.
5.3
An illustrative example is given below.
Environmental principle. Environmental consciousness stems from a
desire not to waste natural resources and to predict and eliminate the
possible negative effects of our daily action to the environment. Practical
application to this principle is to replace combustion engine forklift to
electric forklift.
5-2
Standardization principle. Standardization means less variety and
customization in the methods and equipment employed. With unit load
standardization, a common material handling device can be employed to
move different products. Higher space utilization can also be attributed to
unit load standardization. Storage space is the same across all products;
therefore there is no need for differentiating the storage area. By doing
this, space wasted from excess storage space dedicated to each product can
be eliminated due to the pooling effect.
5.4
An illustrative example is given below.
Bridge Crane. A bridge crane can be classified into two basic types: top
running and under running, both with single or multiple girders design.
The girders which functions as a bridge beam may support one or multiple
trolley and hoist. The bridge travels along the runway on wheeled
carriages called end trucks, which are mounted to each bridge end. The
hoist body is either mounted on top of the bridge or suspended from it.
The hoist moves along the bridge on a trolley. Bridge crane operates on
three axes of movement and a swivel that rotates the load around the
vertical axis is regularly featured. An operator standing on the floor using
a pendant, remote radio, or infrared control unit can do the controlling.
The operator may also be in a self-contained control cab in larger and
faster cranes. Finally, automation is the last control option.
Jib Crane. The main configuration of a jib crane is a horizontal beam that
pivots along a vertical axis. A trolley and hoist may be perched from the
beam. Jib
degreestoof
freedomwall
- vertical,
radial
and
rotary.
Jibcranes
craneshave
may three
be attached
a vertical
member,
or mounted
to a mast attached to the floor, and some may be constructed to move
along a wall. Jib crane is economical; uses little floor space, and pivots in
a pie-shaped area. However, it cannot reach into corners and it doesn't lift
objects outside the circular area. Another disadvantage is the lack of
portability. Typical application includes localized activity like in a
machine shop.
Gantry Crane. A gantry crane is a bridge crane using a horizontal beam
that is supported at each end. There are two versions of gantry crane:
single leg and double leg both with provisions for the legs to travel along
rails. Gantry cranes are used when overhead runway are not practical. It
can be built on wheels for portability, is economical, and is able to lift
heavy loads. Its major disadvantage is the limited reach since it cannot
expand much beyond the inside span.
5.5
Attributes for comparing sorting conveyor:
!
Maximum number of sorting capacity per minute
!
Load range
5-3
!
!
!
!
!
!
5.6
Attributes for comparing automated guided vehicles systems:
!
Weight capacity
!
Safety
!
!
!
!
!
5.7
Load size
Minimum distance between spurs
Diverter impact on load
Safety
Initial cost
Maintenance cost.
Travel speedcapacity
Throughput
Vehicle Cost
System cost per vehicle
Operating costs
Attributes for comparing unit load storage system:
!
Cost per position
!
Potential storage density
!
Load Access: the ease to load/unload the materials stored
!
Throughput capacity - the rate of which materials can flow through
!
Inventory and location control - visibility permitting for easier
inventory control/counting
!
FIFO maintenance - storage arrangement so that the first material
that is stored will also be the first one that is retrieved
!
Ability to house variable load sizes
!
5.8
5.9
Ease of installation
Attributes for comparing unit load retrieval technologies:
!
Vehicle cost
!
Lift height capacity
!
Aisle width - aisle width required for vehicle to operate
!
Weight capacity
!
Lift speed - the speed in the vertical axis
!
Travel speed - speed of the vehicle
Attributes for comparing small part storage alternatives:
!
Gross system cost - initial cost/purchased ft3
!
Net system cost - initial cost/available ft 3
!
Floor space requirements - ft 3 of inventory housed per ft 2 of floor
space
!
Human factors - ease of retrieval
!
Maintenance requirements
!
Items security
!
Flexibility - ease of reconfigure
!
Pick rate - order lines per person-hour
5-4
!
Key
5.10
Attributes for comparing automated data collection systems:
!
Real time - data collected will be straight recorded in the database
!
Hands free
!
Eyes free
!
Cost
5.11
Attributes for comparing bar code readers:
!
Range - the range where the bar code can still be read
!
!
!
!
Depthrate
of field - orthogonal distance to read the bar code
Scan
Resolution
Price
5.12
Attributes for comparing bar code printers:
!
Print Quality
!
Throughput (in 2 / sec)
!
Typical price range
!
Relative ownership cost
5.13
a. A pallet truck is used to lift and transport pallet loads of material for
distance preclude walking. Platform truck is a version of industrial
truck. A platform truck uses a platform for supporting the load. It does not
have lifting capabilities and used for transporting. The main difference is
that a platform truck needs another device to load and unload the material
while a pallet truck doesn’t need one.
b. A reach truck is mostly used on double deep rack. A reach truck has
the capability to extend its forks for easier storage/retrieval on deep racks.
A turret truck is mostly used on narrow aisle. A turret truck has a pivot
point on the fork and a constant mask for more maneuverability along
narrow aisles.
5.14
a. Drive-in rack extends the reduction of aisle space that begun with the
double deep rack. Drive-in rack allows a lift truck to drive in to the rack
several positions and store or retrieve a unit load. A drive-thru rack is a
drive-in rack that is accessible from both sides of the rack. It is staged for
a flow-thru fashion where load is loaded at one end and retrieve at the
other end. Both racks have the same design considerations.
b. Push back rack is a carrier with rail guide for each pallet load. Push
back rack provides a last-in-first-out storage system. A mobile rack is a
single deep selective rack on wheels/tracks that permits the entire row of
racks to move on adjacent rows.
5-5
5.15
An illustration is given below.
Bridge Crane vs. Stacker Crane. Bridge Crane is a bridge that spans a
work area. The bridge is mounted on tracks. The bridge crane and hoist
can provide 3 dimensional coverage of the department. A stacker crane is
similar to a bridge crane, but instead of using a hoist, a mast is supported
by the bridge. The mast is equipped with forks or a platform, which are
used to lift unit loads.
5.16
The
first
pallet sizecorrespond
correspondtotothe
thelength
lengthof
ofthe
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stringer
board
anddimension
the secondofdimension
board.
48
40
48
40
40x48
48 x 40
5-6
5.17
Two-way pallet - fork entry can be only on 2 opposite sites of the pallet
and is parallel to the stringer board. Four-way pallet - fork entry can be on
any side of the pallet.
A two-way pallet:
A four-way pallet:
5.18
This is an independent study question.
5.19
This is an independent study question.
5.20
This is an independent study question.
5-7
5.21
A-F-E-D-C-B-A-F
There are several different answers depending on the assumption. Two
different cases will be presented in these examples: The vehicle is
assumed to stop at point F.
Case 1:
Unit load size = 50, so assuming for every 100 loads the vehicle will make
the following trip:
A-F-A-F-E-F-E-D-E-D-C-B-A-F
A
F
E
D
C
B
A
F
Number of trips = 13
There are 13 trips for every 100 loads. Since we have 2000 pieces, then
the total number of trip is 2000/100 * 13 = 260 trips
Case 2:
Unit load size 50, so assuming for every 100 loads the vehicle will make
the following trip:
A-F-E-D-A-F-E-D-C-B-A-F
A
F
E
D
C
B
A
F
Number of trips = 11
There are 11 trips for every 100 loads. Since we have 2000 pieces, then
the total number of trips is 2000/100 * 11 = 220 trips
These two examples are just two possible answers available. There are
many other possibilities depending on the assumption.
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Chapter 7
Warehouse Operations
7-1
Answers to Questions and Problems at the End of Chapter 7
7.1
A project to evaluate existing dock area, including receiving and shipping, will
allow the determination of those aspects that can potentially prove economically
advantageous. The following is a partial list of such aspects that warrant attention.
a. Eliminate the receiving area. For some materials, e.g., large and bulky ones,
drop shipping (having the vendor ship to the customer directly) can save the
time and labor associated with receiving and shipping.
b. Reduce or eliminate staging in the receiving and shipping area. Determine the
c.
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are stored
as soon identification
as they arrive,prior
reducing
the need
large so
staging area.
Employ the fastest and most productive receiving process possible, i.e.,
crossdocking or shipping directly from the receiving dock. Palletized materials
with a single SKU per pallet, floor-stacked loose cases, and backordered
merchandise are excellent candidates for crossdocking.
Bypass receiving staging and put materials away directly to primary picking
locations, essentially replenishing those locations from receiving. In direct
putaway systems, the staging and inspection activities are eliminated, saving the
time, space, and labor associated with those operations.
Minimize the floor space required for staging by providing storage locations for
receiving staging.
Prepare shipping from the time the materials are received, thus reducing the area
for shipping.
Ship materials in larger quantities and preferably in unit loads, i.e., pallets with
one SKU
per from
pallet.storage and without the need for staging, having prepared the
h. Ship
directly
shipping information prior to picking.
i. Reduce the number of docks, if possible. For example, can receiving and
shipping be modified so that less frequent visits to docks are necessary? This
will save more space in and around the receiving and shipping areas.
j. Modify docks to 90-degree docks. Finger docks should be eliminated, if
possible. Otherwise, can the largest angle finger docks be used? This will allow
shrinking of maneuvering and staging areas inside the receiving and shipping
areas.
k. Eliminate dock levelers by requiring uniform- height carriers for loading and
unloading, thus reducing or eliminating dock maintenance costs.
l. Remove dock shelters in favor of a more streamlined dock, if dock shelters
exist. Again, maintenance costs will be reduced or eliminated.
7.2 This problem should not be assigned if the students have not been exposed to either
Monte Carlo simulation or queueing theory. In Chapter 10, we provide queueing
models to address problems of this type. The instructor might want to assign this
problem before covering the material in Chapter 10 to motivate the students to learn
the material on queueing theory.
7-2
Based on the material in Chapter 10, we provide the following solution to the dock
design problem. Let ? denote the arrival rate of trucks and µ denote the service rate
at the docks. Since arrivals occur at a Poisson rate and service times are
exponentially distributed, the problem is of the class (M|M|c):(GD| !| ! ). The
problem reduces to determining the smallest value of c (number of docks) such that
the average time trucks spend in the system (W) is less than 50 minutes.
The relationship between W and the expected number of trucks in the system (L) is
given by L = " W. Hence, sinte it is desired that W < 50/60 hr., then it is desired
that L/" < 50/60. Since "= 20/8, or 2.5 trucks per hour, then the problem reduces to
determining the smallest value of c such athat L < 2.0833.
From figure 10.28, we can determine the value of c. The utilization factor, ?, is
given by ? = ?/(cµ). With µ= 60/40 = 1.5 trucks per hour, we find that for c equal to
3, ? = 2.5/[3(1.5)] = 0.5556; from Figure 10.28, L < 2.0833. For c = 2, ? = 0.8333
and, form Figure 10.28, L > 2.0833. Therefore, 3 docks are required to ensure that
the average amount of time spent waiting and being loaded or unloaded is less than
50 minutes.
7-3
7.3
7-4
7.4 The impact on space requirement of this proposal is that 90-degree docks require
greater apron depth but less bay width, both impacting outside space requirements,
as well as requiring greater outside turning area. Finger docks, however, require
greater inside maneuvering area. Because inside space costs considerably more to
construct and maintain than outside space, finger docks should be used as little as
possible. If outside space is sparse, then finger docks should be used, although
keeping the angle closer to 90 degrees.
If 45-degree finger docks are utilized, the bay width is increased from the width of
the truck to about 45 feet. Furthermore, the dock width will increase from about the
width of the
to about
percent
more10than
width
thedock
truck.
Thatwould
is,
assuming
thattruck
the old
dock 40
width
is about
feet,the
then
the of
new
width
have to be around 14 feet to accommodate a 45-degree finger dock.
7.5 Permanent, adjustable dockboardsare fastened to the dock and are not moved form
position to position. Therefore, the dock board may be longer and wider than
portable dockboards. The extra length results in a smaller incline between the dock
and the carrier. This allows easier and safer handling of hand carts, reduced power
drain on electrically powered trucks, and less of a problem with fork and
undercarriage fouling on the dockboard. The greater width allows for safer and
more efficient carrier loading and unloading. Permanent, adjustable dockboards also
eliminate the safety, pilferage, and alignment problems associated with dockboards.
For these reasons, permanent adjustable dockboards should always be given serious
consideration, despite their high installation costs.
7.6
A survey of the receiving areas on a typical campus shows that most dock and
carrier height
differences
are handled
by portable
In accommodate
many areas, the
however,
the problem
is solved
by walking
up or dockboards.
down a step to
difference in carrier and dock height.
Because surveys indicate that the campus has scattered receiving docks with
minimal carrier visits, it is not justified to invest in expensive dock shelters. The
bookstore docking area, however, may benefit from a dock shelter because more
frequent stops are made at this location in delivering and shipping more valuable
products.
7.7 Increases in the length of trucks imply that more space will be needed to maneuver
trucks around shipping and receiving docks. Furthermore, certain access roads about
the plant/warehouse may need to be modified to better accommodate longer
trucks/trailers.
Increases in both width and height of trucks and trailers imply that docks may have
to be modified to accept taller and, in many cases, wider trucks and trailers. Some of
the changes can be handled by using permanent adjustable dockboards whose
dynamic ranges can accept taller or wider trucks. Redesigning the dock area as well
as the inside of the shipping and receiving areas may nevertheless be required
7-5
should the width and height of the new vehicles fall outside of the current
equipment’s dynamic ranges.
7.8
About 55% of all operating costs in a typical warehouse can be attributed to order
picking
7.9
a. Strict order picking, i.e., each batch is an order, or single-order-pick (SOP),
b. Batch picking two or more orders, or sort-while-pick (SWP),
c. Batch picking two or more (partial) orders with progressive assembly across
zones,picking
and
d. Batch
two or more orders with downstream sorting, or pick-and-sort
(PAS)
7.10
a. Traveling to, from, and between pick locations,
b. Searching for pick locations,
c. Documenting picking transactions,
d. Reaching (and bending) to access pick locations, and
e. Extracting items from storage locations
7.11
a. Simple ranking of SKUs based on the ratio of pick frequency (the number of
times a SKU is requested) to shipped cube (the product of unit demand and unit
cube) establishes a baseline for an intelligent stock assignment planning. SKUs
with high rankings should be assigned to the most accessible locations.
b. picking
A distribution
illustrating
SKUsisranked
by popularity
and the
of total stock
activity
they represent
the ABC
or Pareto plot
usedportion
in intelligent
assignment planning. More popular SKUs with the highest picking activity are
distributed closer to the input/output point.
c. Correlated stock assignment planning ranks SKUs by popularity as well as
considering the correlations between various items in storage. A small picking
zone dedicated to high-density, high-throughput order picking is used for SKUs
with high correlations.
7.12
a. Popularity of items or SKUs
b. Similarity of items or SKUs
c. Size of items or SKUs
d. Characteristics of SKUs, such as perishability, and
e. Space utilization of SKUs, i.e., maximizing space used in the layout to store
SKUs.
7.13 travel time per line itempicked … sorting
7-6
7.14 By reducing the amount of stock in the forward picking area, forward picking costs
(b) decrease and the cost to replenish the forward picking area (c) increases.
7.15
a. Activity
b.Quantity, and
c.Size
7.16
a. Pick frequency (the number of times the item is requested) and
b. Shipped cube (the product of unit demand and unit cube).
7.17 travel times … productivity
7.18 a. Introduction of Just-in-time operating program
b. Introduction of cycle time reduction operation program
c. Introduction of quick response operating program, and
d. Introduction of micro-marketing and megabrand strategies
7.19
a. Wrong SKU is picked, or
b. Wrong quantity of the SKU is picked.
7.20 the same or nearby locations
7.21 Visiting a typical warehouse may bring about the following discoveries:
a. The
Bothmain
dedicated
randomized storage
schemes
b.
modeand
of transporting
materials
within are
the used.
warehouse is
counterbalanced lift trucks, also known as fork lift truck.
c. In some areas of the warehouse pallet jacks are also used.
d. During order picking materials are transported using lift trucks, pallet jacks, or
order picking carts.
e. Although the warehouse is rather new, the condition of some ofthe equipments
is not good because not all equipment was purchased new when the warehouse
was constructed.
f. The bar code symbols used are the familiar “three-of-nine” code, though some
parts of the warehouse have abandoned the use of bar codes.
g. Hand-held optical bar code readers are used in areas where the technology is still
being exploited.
h. Receiving and shipping are on the same side of the warehouse, though in
opposite corners.
i. Both receiving and shipping areas employ three 90-degree docks with adjustable
permanent dock levelers, but without any shelters.
j. The pallets used are all 42 in. x 42 in. double- faced, non-reversible type.
k. The warehouse is mostly clear of clutter, but the item pick area could benefit
from a better empty case (carton) removal system.
7-7
l. Most of the warehouse, e.g., 75%, is dedicated to storing the materials in pallet
racks, giving ample space to aisles for order picking using counterbalanced lift
trucks and walk-behind order picking carts.
m. Receiving and shipping areas consist of about 10-15% of the warehouse, while
the remaining 10-15% of the warehouse is used for other functions, such as
office space.
n. The aisles represent 25-30% of the total floor space of the warehouse.
o. More than 35% of the pallet rack openings are empty.
p. Since most of the loads are partial loads, stored product represents less than 2025% of the rack face.
q. 20%.
The storage space (cube) utilization factor within a storage opening is about 15r. The overall cube space utilizedin the warehouse is less than 15%.
7.22
a. Total cube demand for item A
= ( 100 units ) x ( 1 ft3 / unit A )
= 100 ft3
Total cube demand for item B
= ( 50 units ) x ( 0.5 ft3 / unit B )
= 25 ft3
Total cube demand for item C
= ( 200 units ) x ( 2 ft3 / unit C )
= 400 ft3
b. ( 400 ft3 / Year ) / ( 20 ft3 / Replenishments )
c. Item A:
Item B:
Item C:
d. Item A:
Item C:
Item B:
= 20
Replenishments
/ Year.
[ ( 100 units / 50
weeks
) x ( 1 ft3 / unit
A)]
= 2 cubic ft / week.
( 2 / 10.5 = 19.0% of forward pick area.)
[ ( 50 units / 50 weeks ) x ( 0.5 ft3 / unit B ) ]
= 0.5 ft3 / week.
( 0.5 / 10.5 = 4.8% of forward pick area. )
[ ( 200 units / 50 weeks ) x ( 2 ft3 / unit C ) ]
= 8 ft3 / week.
( 8 / 10.5 = 76.2 % of forward pick area. )
( 25 Requests / 50 weeks )
= 0.5 requests / week = Rank 1
( 16 Requests / 50 weeks )
= 0.32 requests / week = Rank 2
( 5 Requests / 50 weeks )
= 0.1 requests / week = Rank 3.
7.23
a. When product volume is low and/or product is to be stored for short periods of
time, i.e., work in process, storing the product directly on the floor is practiced
7-8
frequently. For example, liquid products in barrels (55 gallon drums) are often
stored directly on the floor.
b. When product volume is high and/or product mix is low, e.g., large production
runs of palletized and uniform products in cases (cartons), storing the product in
block storage or block stacking is quite advantageous. In particular, tight-block
stacking wastes the least amount of storage space and thus has a high storage
space utilization.
c. Pallet racks are mostly used when product mix is high and accessibility to a
particular product must be quick for faster order-processing.
d. Pallet flow racks provide good storage space (cube) utilization and allow
products
be processed
in first-in,
for
dated to
products
with high
productfirst-out
mix, but(FIFO)
lower fashion.
volume. They are very good
e. Drive-in racks allow large pallet counts (product volume) of high product mixes
to be stored and still obtain good storage space (cube) utilization. Drive-in racks
are best used along the walls of plants or warehouses.
f. Drive-through racks can be used the same as drive-in racks, but they also allow
accessing the product from both sides of the rack. They are thus used for areas in
the middle of the plant or the warehouse.
g. Placing cases directly on the flow rack is used when the warehouse operation is
geared toward full or partial broken case (carton) order picking. It also allows for
faster replenishment, but it does lower the storage space (cube) utilization
(factor).
h. Storing in cantilever racks provides long, unobstructed storage spans. They are
mostly helpful in sorting self-supporting long stock, such as bar stocks, pipes,
and lumber.
i. Portable racks are mostly goodfor loads (open stock), suchas pipes, that need to
be protected
against
and otherofdamage.
Portable
result
in for
flexibility
and
good crushing
space utilization
bulk storage,
andracks
are also
useful
crushable materials and allow access to materials on all levels.
j. Smaller SKUs and/or SKUS that require various protections from damage and
loss are usually stored in bin shelving. Orders that require picking items, rather
than cartons or pallets, use bin racks or shelving systems.
7.24
A detailed approach to storage space planning for the 1200 different SKUs this
warehouse would have to receive, store, and ship is not possible with the data
given. However, the information from the shipping and receiving analysis chart
and the storage analysis chart can be used to determine the space requirement of
the 1200 SKUs.
As part of the storage space planning, certain inventory parameters would have to
be determined. For example, determining the safety stock and knowing the order
quantity are crucial in determining theaverage quantity to be stored for each of
the 1200 SKUs. The average quantity to be stored will then help in choosing
among dedicated storage, randomized storage, and class-based dedicated storage.
7-9
The main difference between dedicated storage and randomized storage is it
implication for what happens when a storage location becomes empty or available.
In randomized storage, the closest-available-slot is designated as the storage
location; retrievals are performed on a first-in, first-out (FIFO) basis, which
provides a uniform stock rotation policy. The closest-available-slot storage policy
can yield results that are similar to the “pure” randomized storage policy if the
storage level remains fairly constant and the level of utilization is high.
Dedicated storage locations (and class-based dedicated storage locations) remain
active even after stock has been removed from that location and the location is
empty.must
Thisaccommodate
is partly due its
to the
fact thatinventory
the number
of openings
assigned
to an
SKU
maximum
level.
The planned
quantity
of unit
loads to be stored for dedicated storage is thus equal to the sum of the openings
required for each SKU. With randomized storage, however, the planned quantity of
unit loads to be stored in the system is the number of openings required to store all
SKUs. Since typically all SKUs will not be at their maximum inventory levels at the
same time, randomized storage will generally require fewer openings than dedicated
storage.
As part of the storage space planning, consideration must be given to throughput as
well. For example, when using dedicated storage, SKUs should be assigned to
storage locations based on the ratio of their activity to the number of openings or
slots assigned to the SKU. The SKU having the highest ranking is assigned to the
preferred opening, with the lowest-ranking SKU assigned to the least-preferred
openings. Because fast movers are up front and slow movers are in back,
throughput is maximized.
Finally, in ranking SKUs, it is important to define activity as the number of
storage/retrieval actions per unit time, not the quantity of materials moved. Also, it
is important to think of “part families” as well. “Items that are ordered together
should be stored together” is a maxim of activity-based storage.
7.25 The receiving/shipping ratios (the ratio of the trips to receive and the trips to ship a
material) for each of the products A-N are calculated and given in the following
table. A receiving/shipping ratio of 1.0 indicates the same number of trips are
required to receive as to ship. Therefore, for products B, G, I, and L, the travel
distance will be the same no matter where along the main aisle the products are
stored. A receiving/shipping ratio less than 1.0 indicates that fewer trips are
required to receive the product than to ship it. Therefore, products having ratios less
than 1.0 should be located closer to shipping. Products A, C, D, J, and M should be
closer to shipping.
7-10
Product
A
B
C
D
E
F
G
H
I
J
K
L
M
N
Monthly
Throughput
Quantity per
Receipt
Trips to
Receive
High
Low
Low
High
High
Low
High
High
Low
High
Low
High
High
Low
300 pallets
200 cartons
10 pallets
400 pallets
6000 cartons
40 cartons
200 cartons
9000 cartons
50 pallets
500 pallets
80 pallets
400 pallets
7000 cartons
700 cartons
30
50
10
400
1000
40
200
2250
50
500
80
400
1167
140
Average
Customer
Order Size
2.0 pallets
4.0 pallets
0.2 pallets
0.5 pallets
10.0 cartons
2.0 pallets
1.0 pallets
5.0 cartons
1.0 pallets
0.7 pallets
2.0 pallets
1.0 pallets
3.0 cartons
7.0 cartons
Trips
to
Ship
150
50
50
800
600
20
200
1800
50
715
40
400
2334
100
Receiving/
Shipping
Ratio
0.2
1.0
0.2
0.5
1.67
2.0
1.0
1.25
1.0
0.69
2.0
1.0
0.5
1.4
Similarly, a receiving/shipping ratio greater than 1.0 indicates that fewer trips are
required to ship than to receive. Therefore, products having ratios greater than 1.0 (i.e., E,
F, H, K, and N) should be located closer to receiving.
The layout shown is typical of an aisle-based system. Products are distributed according
to throughput,
quantity for
receipt,
and the ratio
receiving to shipping trips. In Chapter 10,
we
provide guidelines
minimizing
travelofdistances.
7-11
7.26
Minimum Total Cube Requirement :
[(4 ft x 4 ft x 4.5 ft)/load]x[300 loads/product]x[30 products] = 648,000 3ft
Max Usable Height Requirement:
[(4.5 ft/load)x(4 loads)] = 18 ft
Minimum Total Floor Space Requirement:
[648,000 cubic ft]/[18 ft] = 36,000 ft2
Approximate Dimensions of Block Stacking Area:
[36,000 ft2]0.5 = 190 ft (width or depth of stacking area)
Minimum Aisle Width and Maximum No. of Lanes per Product :
Minimum aisle width = 13 ft.
Maximum no. of lanes per product = 2
Practical Dimensions of Block Stacking Area:
[30 products]x[3 load widths/product]x[4 ft/load width] = 360 ft (width)
{[300 loads]/[4 loads high]/[3 loads wide]}x[4 ft/load depth] = 100 ft (depth)
Add [30 products]x[13 ft –3 x 4 ft] = 30 ft to 260 ft
Therefore, the practical dimensions for the block stacking storage area are 390 ft
wide by 300 ft deep.
Layout for the Warehouse:
A rectangular block stacking area of 390 feet wide by 100 feet deep, minimum
dimensions. Each product will be stacked in three lanes, each containing 25 stacks
of four loads high, for a total of [ 25 + 25 + 25 ] x 4 = 300 loads per product. The 30
products will be stacked in 90 side-by-side lanes (30 aisles), with one foot
separating each product aisle.
7-12
This scheme tries to reduce the pick face of the stacks, which increases congestion
and the time it takes to empty a (deep) lane. Addressing those issues requires a more
rectangular block stacking area at the expense of cube utilization. A typical layout is
depicted below.
7.27
Cube utilization increases as products are more densely stored in a warehouse.
Thus, cube utilization and product accessibility are inversely proportional. For
example, in block stacking storage scheme, cube utilization increases as lane depth
increases. This makes accessing (deep) product(s) more difficult.
7.28
The slightest advantage that layout (a) has over layouts (b) and (c) of Figure 7.36 is
that certain conditions may exist to make the overall travel distance somewhat less.
In general, however, layouts (b) and (c) are preferred over (a) because they allow
more stock accessibility. Layout (c) is preferred over layout (b) because it allows
the most accessibility of all layouts. The disadvantage of layout (c) is that it uses
more aisle space and its overall storage space utilization (factor) will be slightly
less. The overall preference is thus (c) over (b), which are both preferred over (a).
7.29
Storage space utilization
Load cubic volume
Storage area cubic volume
Load height
Load width
Load depth
= (load cubic volume)/(storage area cubic volume)
= (load height)x(load width)x(load depth)
= (area height)x(area width)x(area depth)
= 5x(pallet height + product height)
= 5x(5”+46”) = 255”
= 2x(pallet width)
= 2x(48”) = 96”
=
full pallet depth
+ half- full pallet depth
= (41”)+(20.5”)
= 61.5”
7-13
Storage area (rack) height
Storage area (rack) width
Storage area depth
Storage Space Utilization
= bottom support height + 5x(pallet height +
product height + product height clearance + top
support height)
= 4 + 5 x (4”+5”+46”+4”+4”)
= 319”
= half left beam width + left product clearance +
left pallet width + pallet-to-pallet clearance + right
pallet width + right product clearance + half right
beam width
= (4/2)”+5”+48”+5”+58”+5”+(4/2)”
=
= 115”
half flue spacing + second pallet depth + first
pallet depth + half aisle width
= (12”/2)+41”+41”+(8.5’x12ipf/2)
= 139”
= [255”x96”x61.5”]/[319”x115”x139”]
= 0.2952 = 29.52%
7.30
Storage space utilization
Load cubic volume
Storage area cubic volume
Load height
Load width
Load depth
Storage area (rack) height
Storage area (rack) width
Storage area depth
Storage Space Utilization
= (load cubic volume)/(storage area cubic volume)
= (load height)x(load width)x(load depth)
= (area height)x(area width)x(area depth)
= 4x(pallet height + product height)
= 4x(6”+48”) = 216”
= 3x(pallet width)
= 3x(42”) = 126”
= 1x(36”)
full pallet
depth
=
= 36”
= 4x(pallet height + product height + product height
clearance + top support height)
= 4x(6”+48”+3”+4”) = 244”
= half left beam width + left product clearance +
left pallet width + pallet-pallet clearance + middle
pallet width + pallet-pallet clearance + right pallet
width + right product clearance + half right beam
width
= (3/2)”+4”+42”+4”+42”+4”+42”+4”+(3/2)”
= 145”
= half flue spacing + single pallet depth + half aisle
width
= (15/2)”+36”+(8.5’x12ipf/2)
= 94.5”
= [216”x126”x36”]/[244”x145”x94.5”]
= 0.2930 = 29.30%
7-14
7.31
Storage space utilization
Load cubic volume
Storage area cubic volume
Load height
Load width
Load depth
Storage area (rack) height
Storage area (rack) width
Storage area depth
Storage Space Utilization
= (load cubic volume)/(storage area cubic volume)
= (load height)x(load width)x(load depth)
= (area height)x(area width)x(area depth)
= 6x(pallet height+product height)
= 6x(6”+40”) = 276”
= 2x(pallet width)
= 2x(42”) = 84”
=
full pallet= depth
= 1x(36”)
36”
= 6x(pallet height + product height + product height
clearance + top support height)
= 6x(6”+40”+4”+4”) = 324”
= half left beam width + left product clearance +
left pallet width + pallet-to-pallet clearance +
middle pallet width + pallet-to-pallet clearance +
right pallet width + right product clearance + half
right beam width
= (3/2)”+4”+42”+4”+42”+4”+(3/2)”
= 99”
= half flue spacing + single first pallet depth + half
aisle width
= (12/2)”+36”+(10’x12ipf/2)
= 102”
= [276”x84”x36”]/[324”x99”x102”]
= 0.2551 = 25.51%
Chapter 8
Manufacturing Systems
8-1
Answers to Questions at the End of Chapter 8
8.1
To some extent, the automatic factory is still valid today. There are three
components in an automatic factory – manufacturing, material handling,
and the information system. In terms of manufacturing, some decisive
factors to justify automation are:
!
Volume of production. Economics of scale can be achieved by mass
production and the financial benefit can compensate the high capital
cost of an automatic factory
!
Expensive machinery. Some industries, such as semiconductor,
!
require
expensive
machinery.
By automating,
these machines can be
fully
utilized
to reduce
production
cost.
Variability reduction. Manual machining, while still within tolerance,
often produce parts with high variability. This variability can be
reduced significantly by automation.
From a material handling perspective, automation is desirable to reduce
cost in time due to savings in labor cost. In addition to cost saving, some
product may require careful handling; therefore automation is an
alternative to prevent product damage. In addition, the declining costs of
computing and data storage continue to fuel the desire to invest in
automation.
8.2
The semiconductor industry would be a good target for the automatic
factory. Machinery for semiconductor production cost dearly and should
be fully utilized. Product value is also very high; material-handling
automation is needed to avert damage.
Another sector would be continuous flow manufacturing such as chemical
products.
8.3
Advantages of automatic warehouse:
!
High throughput
!
Reduced labor cost
!
Elimination of human error
!
Reduced material damage
!
Greater security
Disadvantages of automatic warehouse:
!
High initial capital cost
!
Downtime or reliability of equipment
!
Software related problems
!
Backup is needed to cover the risk of complete reliance on automation
if a disaster should strike.
!
User interface and training
!
Obsolescence
8-2
!
!
8.4
A cross docking facility can be totally automated. A list that is required for
a fully automate cross docking facility with respect to the material
handling aspect:
!
Software for warehouse management. Software must be provided for
a fully automated cross-docking facility since there is no man-labor
!
!
8.5
Lack of flexibility
Risk of having all eggs in one basket if a disaster should strike the
warehouse
available tomaterial
make thetransport
arrangement
to unload/load.
Automatic
equipment
for moving the materials. For
example: conveyors, racks that is designed to accommodate cross
docking facility, i.e. multi-lane directional conveyor lines from
receiving dock to the shipping dock.
Industrial vehicles for transporting from the dock to storage or storage
to dock. For example: automatic crane/hoist to load/unload material
to/from trucks/containers. This device will allow automated retrieval
of loads from truck.
Criteria for comparing transfer line and FMS:
!
Volume of production
!
Number of products manufactured
!
Production of families of work parts
!
Reduced flexibility in processing orders because of long production
line
!
!
!
!
Manufacturing
lead time
Machine
utilization
Direct and indirect labor
Management control
8.6
Criteria for comparing FMS and SSMS:
!
Part transportation requirement
!
Number of setup
!
Tool allocation and management
!
Flexibility
!
Capital cost
8.7
Tool management problem in:
a. FMS
In a FMS environment, tool sharing among machines is common. This
trend is perpetuated by limited tool magazine size and more importantly;
keeping a complete set of tools in a magazine may not be economically
feasible since tools are generally expensive. Tools can be categorized into
8-3
two types: resident, which reside in a machine permanently and transient,
which is shared among machines and kept in central tool storage.
Determining how many resident and transient tools is the problem. This
also translates to how to allocate the transient tools among machines and
how many transport devices is needed. Given the machining schedule,
usually based on order priority, the required tool sequence can be known.
Simulation or integer programming can be used to find the optimal
solution. In practice, keeping active inventory of all tools in the cell with
their size, type, number and location, improving tool forecasts and
warnings
of toolsreliability
changes, reducing
delays
themanagement
system, and system.
improving
tool
information
are the core
of a in
tool
b. SSMS
SSMS essentially faced the same tools management problems as in FMS.
Tools are still separated into resident tools and transient tools; however the
proportion of resident tools is considerably higher than in a FMS setting.
This can be attributed to the nature of SSMS where part only visit a
machine once; thus more resident tools are needed. This arrangement is
more costly; in return it offers more versatility, more machine utilization,
easier part scheduling and higher throughput.
8.8
Alternatives for moving part as shown:
!
Spurs can be removed so that the system will have an on-track
upload/offload. An on-track unload/offload system will allow
queueing on tracks. If the system has many AGVs and the process of
!
uploading/offloading
takes
a significant
of time,
then having
spurs
is more favorable
since
it allows anamount
off-track
upload/offload
system.
Instead of using AGVs, conveyors can be used for transporting
materials. When using conveyors, a spine layout can also be
implemented. When using a spine layout, there is no more loop around
the system.
8.9
This is a research question.
8.10
This is a research question.
8.11
JIT and lean manufacturing have essentially the same objectives. Both
strive for eliminating or minimize waste, produce only what is demanded,
minimize the use of time and space resources, and manufacture in the
shortest cycle time possible. Different companies have different name for
JIT, at Hewlett-Packard it is called Stockless Production, Material as
Needed at Harley-Davidson, Continuous Flow Manufacturing at IBM.
8-4
8.12
It is not applicable to all types of manufacturing systems. Mass
production is still the best process to use for high volume, repetitive
products. JIT may also be difficult to implement to very low volume or
unique products such as in a job shop environment unless there is
flexibility in reordering the machine.
8.13
This is a research question.
8.14
This is a research question.
8.15
In
straight
line-balancing
problem,
setassigned.
of assignable
limited
to athat
task whose
predecessors
havethe
been
In a tasks
U lineisbalancing
problem, the set of assignable task is enlarge by those tasks whose
successors have been assigned, therefore U line balancing problem is more
complex since now task grouping not only move in forward or backward
direction as in a straight line, but it can also move in both directions at the
same time.
In practice, rebalancing of the line is done quite often following demand
changes. Rebalancing involves adding or removing machine from on the
line or changing the standard time bases on new layout configuration; it
also involves determining the number of operators required and assigning
the machines that each operator tends.
Chapter 9
Facilities Systems
9-1
Answers to Questions and Problems at the End of Chapter 9
9.1
Step 1: Determine the Level of Illumination
From Table 9.5, for a regular office environment, the minimum level of
illumination is 100 foot-candles
Step 2: Determine the Room Cavity Ratio
Using Equation 9.8:
RCR = ( 5 ) ( 10 – 3 ) ( 250 + 150 ) = 0.373
(250) (150)
Step 3: Determine the Ceiling Cavity Ratio
Since the lights are ceiling mounted, the reflective property of the ceiling
will not be impacted by the luminaries’ mounting height.
Step 4: Determine the Wall Reflection (WR) and the Effective Ceiling
Reflectance (ECR)
From Table 9.6, since the walls and ceiling are painted white, the WR and
the base ceiling reflectance (BCR) are 80%. Since the lights are ceiling
mounted, BCR = ECR = 80%.
Step 5: Determine the Coefficient of Utilization (CU)
From Table 9.8, since the ECR and the WR are 80%, one can extrapolate
from the table for an RCR value of 0.373. The CU is approximately 0.68
and will be the value utilized.
Step 6:Table
Determine
thefluorescent
Light Losslamps
Factorin(LLF)
From
9.10, for
prismatic lens fixtures and six
months between cleanings, the LLF = 0.92
Step 7: Calculate the Number of the Laps and Luminaries
From Table 9.9, Lamp output at 70% of rated life for an 85-watt lamp is
5,400 lumens. Using Equation 9.10:
Number of Lamps = (100) (250) (150)
= 1,110 Lamps
(0.68) (0.92) (5,400)
If there are 2 lamps per luminary, then 555 luminaries are required for the
facility.
Step 8: Determine the Location of the Luminaries
From Table 9.8, Column 2, the spacing between banks of luminaries is
equal to 1.2 times the mounting height, or 12 feet between banks. In a
250’X150’ office, 20 banks of luminaries space 12 feet apart can be
mounted along the length of the facility. Since each fluorescent lamp is 4’
long, 37 luminaries per bank can be placed along the width of the facility
9-2
for a total of 740 luminaries, exceeding the required amount of 555
luminaries determined in step 7.
9.2
The changes in the facility would be that the minimum level of
illumination would be 150 foot-candles. This would make the number of
lamps and luminaries change as well, as shown in the following equation:
Number of Lamps = (150) (250) (150)
= 1,665 Lamps
(0.68) (0.92) (5,400)
Number of Luminaries = 1,665/2 = 833 luminaries
The spacing between banks of luminaries must be reduced from 12 feet to
10.5 feet to get the required 833 luminaries within the facility. This will
allow 851 luminaries to be placed within the facility.
9.3
From equation 9.8, the RCR changes from 0.373 to 0.64. This result
changes the coefficient of utilization in Table 9.8 slightly to 0.67. This
changes the number of lamps, as shown in the following equation:
Number of Lamps = (100) (250) (150) = 1,127 Lamps
(0.67) (0.92) (5,400)
Number of Luminaries = 1,127/2 = 564 luminaries
The spacing between banks of luminaries can remain 12 feet to get the
required
564 This
luminaries
within750
the luminaries
facility, andthat
must
exceed within
the 12foot
spacing.
result yields
cannot
be placed
the facility.
9.4
The Ceiling Cavity Ratio (CCR) would have to be determined because the
luminaries are no longer mounted on recessed into the ceiling. From
Equation 9.8:
CCR = (5) (0.373)
(7)
= 0.266
Because the CCR has to be considered in this problem the ECR is no
longer equivalent to the BCR. The ECR is determined by examining Table
9.7 and by interpolation is found to be 0.78. Since the ECR changed, this
affects the coefficient of utilization slightly. The CU decreases to 0.67 by
extrapolation using Table 9.8. As in Problem 9.3, the number of
luminaries required is 564, and 12-foot spacing between banks with 37
luminaries per bank will allow 750 luminaries to be placed within the
facility.
9.5
If the walls and ceiling were painted a medium color, then the WR and
ECR would change to 50% instead of 80%. This changes the value of the
CU to 0.62, using extrapolation from Table 9.8. This changes the number
of lamps required to:
Number of Lamps = (100) (250) (150)
(0.62) (0.92) (5,400)
= 1,217 Lamps
Number of Luminaries = 1,217/2 = 609
The
number
of luminaries
luminaries per
required
609,allow
and 12spacing between
banks
with 37
banksiswill
750foot
luminaries
to be placed
within the facility.
9.6
If the fixtures were cleaned once every 36 months instead of once every
six months, the LLF value changes from 0.92 to 0.78. This changes the
number of lamps to:
Number of Lamps = (100) (250) (150)
(0.62) (0.78) (5,400)
= 1,436 Lamps
Number of Luminaries = 1,436/2 = 718
The number of luminaries required is 718, and 12- foot spacing between
banks with 37 luminaries per banks will allow 750 luminaries to be placed
within the facility.
9.7
QF = (120,000 ft2)[0.81 Btu/(hr)(ft2)(ºF)](72ºF - 12ºF)
QF = 5.83 x 106 Btu/hr
QR = (120,000 ft2)[0.20 Btu/(hr)(ft2)(ºF)](72ºF - 12ºF)
QR = 1.44 x 106 Btu/hr
QD = (6) (24 ft2)[1.13 Btu/(hr)(ft2)(ºF)](72ºF - 12ºF)
QD = 0.0098 x 106 Btu/hr
QW = [(2)(8,000 ft2)+(2)(6,000 ft2)-(6)(24 ft2)][0.39 Btu/(hr)(ft2 )(ºF)]
(72ºF - 12ºF)
QW = 0.65 x 106 Btu/hr
QI = [(120,000 ft2)+(28,000 ft2)][0.02 Btu/(hr)(ft2)(ºF)](72ºF - 12ºF)
QI = 0.18 x 106 Btu/hr
Using equation 9.1, the total heat loss may be calculated as follows:
QH = 5.83 x 106 Btu/hr + 1.44 x 106 Btu/hr + 0.0098 x 106 Btu/hr + 0.65 x
106 Btu/hr + 0.18 x 106 Btu/hr
QH = 8.11 x 106 Btu/hr
9.8
QF = (120,000 ft2)[0.81 Btu/(hr)(ft2)(ºF)](72ºF - 12ºF)
9-4
QF = 5.83 x 106 Btu/hr
QR = (120,000 ft2)[0.20 Btu/(hr)(ft2)(oF)](72oF – 12oF)
QR = 1.44 x 106 Btu/hr
QD = (6) (24 ft2)[1.13 Btu/(hr)(ft2)(oF)](72oF - 12 oF)
QD = 0.0098 x 106 Btu/hr
QG = (20)(32 ft2)[0.63 Btu/(hr)(ft2 )(o F)](72oF - 12oF)
QG = 0.024 x 106 Btu/hr
QW = [(2)(8,000 ft2)+(2)(6,000 ft2 )-(6)(24 ft2)-(20)(32 ft2)]
[0.39 Btu/(hr)(ft2)(oF)](72ºF - 12ºF)
QW = 0.64 x 106 Btu/hr
2
2
2 o
o
o
I
Q
[(120,000
)+(28,000 ft )][0.02 Btu/(hr)(ft )( F)](72 F – 12 F)
QI =
= 0.18
x 106 ft
Btu/hr
Using Equation 9.1, the total heat loss may be calculated as follows:
QH = 5.83 x 106 Btu/hr + 1.44 x 106 Btu/hr + 0.0098 x Btu/hr + 0.024 x
106+ 0.64 x 106 Btu/hr + 0.18 x 106 Btu/hr
QH = 8.12 x 106 Btu/hr
9.9
QF = (120,000 ft2)[0.81 Btu/(hr)(ft2)(ºF)](72ºF - 12ºF)
QF = 5.83 x 106 Btu/hr
QR = (120,000 ft2)[0.13 Btu/(hr)(ft2)(oF)](72oF – 12oF)
QR = 0.94 x 106 Btu/hr
QD = (6)(24 ft2)[1.13 Btu/(hr)(ft2)(oF)](72oF – 12oF)
QD = 0.0098 x 106 Btu/hr
QW = [(2)(8,000 ft2)+(2)(6,000 ft2 )-(6)(24 ft2)]
2 o
[0.39xBtu/(hr)(ft
QW = 0.65
106 Btu/hr )( F)](72ºF - 12ºF)
QI = [(120,000 ft2)+(28,000 ft2)][0.02 Btu/(hr)(ft2)(o F)](72oF – 12oF)
QI = 0.18 x 106 Btu/hr
Using Equation 9.1, the total heat loss maybe calculated as follows:
QH = 5.83 x 106 Btu/hr + 0.94 x 106 Btu/hr + 0.0098 x Btu/hr + 0.65 x 106
Btu/hr + 0.18 x 106 Btu/hr
QH = 7.61 x 106 Btu/hr
9.10
QF = (120,000 ft2)[0.55 Btu/(hr)(ft2)(ºF)](72ºF - 12ºF)
QF = 3.96 x 106 Btu/hr
QR = (120,000 ft2)[0.20 Btu/(hr)(ft2)(oF)](72oF – 12oF)
QR = 1.44 x 106 Btu/hr
QD = (6)(24 ft2)[1.13 Btu/(hr)(ft2)(oF)](72oF – 12oF)
Q = 0.0098 x 106 Btu/hr
2
2
2
QD
W = [(2)(8,000 ft )+(2)(6,000 ft )-(6)(24 ft )][0.39
2 o
Btu/(hr)(ft )( F)](72ºF - 12ºF)
QW = 0.65 x 106 Btu/hr
QI = [(120,000 ft2)+(28,000 ft2)][0.02 Btu/(hr)(ft2)(o F)](72oF – 12oF)
QI = 0.18 x 106 Btu/hr
Using Equation 9.1, the total heat loss maybe calculated as follows:
QH = 3.96 x 106 Btu/hr + 1.44 x 106 Btu/hr + 0.0098 x Btu/hr + 0.65 x 106
Btu/hr + 0.18 x 106 Btu/hr
QH = 6.24 x 106 Btu/hr
9.11
QF = (120,000 ft2)[0.81 Btu/(hr)(ft2)(ºF)](98ºF - 78ºF)
5
F
2
Q
x 10 ftBtu/hr
QR =
= 19.4
(120,000
)[0.20 Btu/(hr)(ft2)(oF)](98oF – 78oF)
5
QR = 4.80 x 10 Btu/hr
QD = (6)(24 ft2)[1.13 Btu/(hr)(ft2)(oF)](98oF – 78oF)
QD = 0.033 x 105 Btu/hr
QW = [(2)(8,000 ft2)+(2)(6,000 ft2 )-(6)(24 ft2)][0.39
Btu/(hr)(ft2)(oF)](98ºF - 78ºF)
QW = 2.17 x 105 Btu/hr
QI = [(120,000 ft2)+(28,000 ft2)][0.02 Btu/(hr)(ft2)(oF)](98oF – 78oF)
QI = 0.59 x 105 Btu/hr
QS = {[(3)(24 ft2 )][110 Btu/(hr)(ft2)(oF)] + [(3)(24 ft2)]
[55 Btu/(hr)(ft2)(oF)]}(98oF – 78o F)
QS = 2.37 x 105 Btu/hr
QL = (700)(60 watt)[4.25 Btu/(hr)(watt)]
QL = 1.79 x 105 Btu/hr
QP = (50)(1,500 Btu/hr)+(60)(700 Btu/hr)
5
QP = 1.17 x 10 Btu/hr
Using Equation 9.3, the total cooling load may be calculated as follows:
QC = 19.4 x 105 Btu/hr + 4.80 x 105 Btu/hr + 0.033 x 105 Btu/hr
+ 2.17 x 105 Btu/hr + 0.59 x 105 Btu/hr + 2.37 x 105 Btu/hr
+ 1.79 x 105 Btu/hr + 1.17 x 105 Btu/hr
QC = 32.32 x 105 Btu/hr
9.12
QF = (120,000 ft2)[0.81 Btu/(hr)(ft2)(ºF)](98ºF - 78ºF)
QF = 19.4 x 105 Btu/hr
QR = (120,000 ft2)[0.20 Btu/(hr)(ft2)(oF)](98oF – 78oF)
QR = 4.80 x 105 Btu/hr
QD = (6)(24 ft2)[1.13 Btu/(hr)(ft2)(oF)](98oF – 78oF)
QD = 0.033 x 105 Btu/hr
QG = (10)(32 ft2)[0.63 Btu/(hr)(ft2 )(oF)](98oF – 78 oF)
Q = 0.040 x 105 Btu/hr
2
2
2
2
QG
W = [(2)(8,000 ft )+(2)(6,000 ft )-(6)(24 ft )-(10)(32 ft )]
2 o
[0.39 Btu/(hr)(ft )( F)](98ºF - 78ºF)
QW = 2.15 x 105 Btu/hr
9-6
QV = [(120,000 ft2)+(28,000 ft2 )][0.02 Btu/(hr)(ft2)(oF)](98o F – 78oF)
QV = 0.59 x 105 Btu/hr
QS = {[(3)(24 ft2 )+(5)(32 ft2 )][110 Btu/(hr)(ft2 )(oF)] + [(3)(24 ft2)
+(5)(32 ft2)][55 Btu/(hr)(ft2)(oF)]}(98o F – 78oF)
QS = 7.65 x 105 Btu/hr
QL = (700)(60 watt)[4.25 Btu/(hr)(watt)]
QL = 1.79 x 105 Btu/hr
QP = (50)(1,500 Btu/hr)+(60)(700 Btu/hr)
QP = 1.17 x 105 Btu/hr
Using Equation 9.3, the total cooling load may be calculated as follows:
QC = 19.4 x 105 Btu/hr + 4.80 x 105 Btu/hr + 0.033 x 105 Btu/hr
+ 0.040 x 105 Btu/hr + 2.15 x 105 Btu/hr + 0.59 x 105 Btu/hr
+ 7.65 x 105 Btu/hr + 1.79 x 105 Btu/hr + 1.17 x 105 Btu/hr
QC = 37.62 x 105 Btu/hr
9.13
QF = (120,000 ft2)[0.81 Btu/(hr)(ft2)(ºF)](98ºF - 78ºF)
QF = 19.4 x 105 Btu/hr
QR = (120,000 ft2)[0.20 Btu/(hr)(ft2)(oF)](98oF – 78oF)
QR = 4.80 x 105 Btu/hr
QD = (6)(24 ft2)[1.13 Btu/(hr)(ft2)(oF)](98oF – 78oF)
QD = 0.033 x 105 Btu/hr
QG = (10)(32 ft2)[0.63 Btu/(hr)(ft2 )(oF)](98oF – 78 oF)
QG = 0.040 x 105 Btu/hr
QW = [(2)(8,000 ft2)+(2)(6,000 ft2 )-(6)(24 ft2)-(10)(32 ft2)]
2 o
[0.39xBtu/(hr)(ft
QW = 2.15
105 Btu/hr )( F)]*(98ºF - 78ºF)
QV = [(120,000 ft2)+(28,000 ft2 )][0.02 Btu/(hr)(ft2)(oF)](98o F – 78oF)
QV = 0.59 x 105 Btu/hr
QS = {[(3)(24 ft2 )+(5)(32 ft2 )][110 Btu/(hr)(ft2 )(oF)] + [(3)(24 ft2)
+(5)(32 ft2)][55 Btu/(hr)(ft2)(oF)]}(0.3)(98oF – 78oF)
QS = 2.30 x 105 Btu/hr
QL = (700)(60 watt)[4.25 Btu/(hr)(watt)]
QL = 1.79 x 105 Btu/hr
QP = (50)(1,500 Btu/hr)+(60)(700 Btu/hr)
QP = 1.17 x 105 Btu/hr
Using Equation 9.3, the total cooling load may be calculated as follows:
QC = 19.4 x 105 Btu/hr + 4.80 x 105 Btu/hr + 0.033 x 105 Btu/hr
+ 0.040 x 105 Btu/hr + 2.15 x 105 Btu/hr + 0.59 x 105 Btu/hr
+ 2.30 x 105 Btu/hr + 1.79 x 105 Btu/hr + 1.17 x 105 Btu/hr
QC = 37.27 x 105 Btu/hr
If blinds are used, the cooling load required would be QC = 35.26 Btu/hr
9.14
The required fire protection in a building, which is governed by the
Uniform Building Code (UBC). There are 20 classifications or groups of
buildings governed by the UBC.
9.15a Maximum Population = 120 ft x 200 ft = 1200 people
20 sf/person
9.15b Minimum Number of Exits (from Table 9.12) = 4
2
2 0.5
9.15c Minimum Distance Between Exits = ((200) + (120) )
9.16
/ 2 = 117 feet
The major benefit of using a sprinkler system is that it is a very effective
means of extinguishing or controlling a fire in its early stages and it can
provide protection when the building is not occupied.
9.17
1. Safety of regular building occupants
2. Safety of firefighters
3. Salvage of the building
4. The goods and equipment in the building
9.18
The main purpose of face sprinklers in an in-rack sprinkler system is to
oppose vertical development of fire on the external face of the rack.
9.19
The purpose of a trap in a sewage system is to prevent odor from passing
back into the facility.
9.20
Vent stacks can be used to improve water-seal integrity and to provide
assurance that smells are effectively removed from the facility and vented
to the atmosphere.
9.21
The statement, “The grid structure should defer to the function of the
facility” is true. Some reasons why this is so are:
In a warehouse facility, the grid spacing should be dictated by the rack
dimensions and access aisles between racks
In a parking garage, the grid element should be a multiple of the car
bay width and length and the circulation needs
As long as the structural integrity is not compromised, the dimensions
of the building grid are secondary to the optimal layout of the facility
Specific structural members other than 20’ and 40’ lengths can be
obtained without much of a penalty cost
•
•
•
•
9.22
The
thermal
performance
a facility
can the
be improved
making better
use of
solar gain.
This willofgreatly
reduce
building’sby
dependence
on
artificial atmospheric systems. Using double-paned windows with a
9-8
reflective glass member can reduce the solar transmission through
windows from 90% to 25%
A good roof design will also help thermal performance. A membrane layer
to prevent water penetration, an nsulation
i
layer to assist with thermal
comfort, and a vapor check to stop vapor migration are all requirements of
a good roof. The floor should have integral water-proofing and an applied
membrane to seal the floor against water migration.
The primary purpose of an enclosure system for a manufacturing facility is
9.23
to keep out undesirables.
Allocate the mechanical equipment room space
Total Gross Area = 120 ft x 180 ft = 21,600 ft2
Assume the system will be roof-mounted
Determine the air-handling requirements
Air supply = 21,600 ft2 x 1.0 ft3 /min-ft2 = 21,600 ft3/min (CFM)
Air exhaust = 21,600 ft2 x 0.3 ft2/min-ft2 = 6,480 ft3/min (CFM)
Determine Louver Supply and Exhaust sizes
Louver Supply = 21,600 CFM / 1,000 ft/min = 21.6 ft2
2
Louver Exhaust = 6,480 CFM / 2,000 ft/min = 3.24 ft
Determine Main Duct Sizing
2 main ducts, each covering a half of the building’s area, times the air
supply
rate: ft2 x 1.0 ft3/min-ft2 = 10,800 ft3/min (CFM)
0.5
x 21,600
10,800 CFM / 1,800 ft/min = 6 ft2
PART FOUR
DEVELOPING ALTERNATIVES:
QUANTITATIVE APPROACHES
Chapter 10
Quantitative Facilities
Planning Models
10-1
Answers to Questions and Problems at the End of Chapter 10
10.1 Using the half-sum method: half of the sum of the weights equals 35
x-coord
0
5
wi
15
25
3 wi
15
40>35
y-coord
wi
5
10
30
35
10
30
70
15
5
(x*,y*) = (5,10), which is the location of existing facility 2
3 wi
30
65>35
70
f(10,10) = 15(|10-0|+|10-10|) + 20(|10-5|+|10-10|) + 5(|10-5|+|10-15|) +
30(|10-10|+|10-5|)
f(10,10) = 450
10-2
10.2 Using the half-sum method: half of the sum of the weights equals 15
x - co o rd
wi
-5
-3
1
2
3
8
3
4+3=7
7
5
3 wi
8
11
18>15
25
30
y-coord
wi
-3
-1
0
5
6
3
8
4
7
3
3 wi
3
11
15=15
22
25
8
5
30
x* = 1, y* = [0,5]
Any point on the defined line segment is an optimum location.
10.3 Using the half-sum method: half of the sum of the weights equals 3
wi
3 wi
y-coord
wi
0
1
1
0
1
20
1
2
20
1
30
1
3=3
40
1
60
1
4
70
1
70
1
5
80
1
90
1
6
90
1
x* = [30,60], y* = [40,70]
Any point in the defined square is an optimum location.
x - co o rd
3 wi
1
2
3=3
4
5
6
10.4 It is desired to determine the unweighted minimax location, i.e., it is
important that the farthest house be as close as possible, regardless of the
monetary value of the house.
a
b
a+b
-a+b
20
25
13
25
4
18
15
25
32
14
21
8
35
50
45
39
25
26
-5
0
19
-11
17
-10
c1 = 25, c2 = 50, c3 = -11, c4 = 19
c5 = max (c2-c1, c4-c3) = max (25, 30) = 30
The minimax locations are all the points on the line segment joining the
points:
(x1*, y1*) = !(c1-c3, c1+c3+c5) = !(36,44) = (18,22)
(x2*, y2*) = !(c2-c4, c2+c4-c5) = !(31,39) = (15.5,19.5)
10-3
10.5 a.
Since the weights are uniformly distributed over the indicated regions,
the optimum x-coordinate location will be the point at which no more
than half the weight is to the left of the point and no more than half the
weight is to the right of the point; likewise, the optimum y-coordinate
is the point at which no more than half the weight is below the point
and no more than half is above the point. Show below are plots of the
cumulative weights along each axis.
10-4
10.5 b.
Since the minimax location is to be determined with all points equally
weighted, the rectangular regions can be represented by their corners.
Hence,
a
b
a+ b
-a+b
a
b
a+b
-a+b
2
2462
4
2
4
0
41
6
10
2
8
88
10
10
2
16
20
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c1 = 4, c2 = 28, c3 = -12, c4 = 8
c5 = max (c2-c1, c4-c3) = max (24, 20) = 24
The minimax locations are all the points on the line segment joining
the points:
(x1*, y1*) = !(c1-c3, c1+c3+c5) = !(16,16) = (8,8)
(x2*, y2*) = !(c2-c4, c2+c4-c5) = !(20,12) = (10,6)
Hence, the optimum location is inside region A4.
10.6 Since there are only 3 sites to consider, enumeration is the quickest means of
solving the problem.
f(50,50) = 600(|50-20|+|50-25|) + 400(|50-36|+|50-18|) + 500(|50-62|+|5037|) + 300(|50-50|+|50-56|) + 200(|50-25|+|50-0|) = 80,700
f(30,45) = 600(|30-20|+|45-25|) + 400(|30-36|+|45-18|) + 500(|30-62|+|4537|) + 300(|30-50|+|45-56|) + 200(|30-25|+|40-0|) = 70,500
f(65,28) = 600(|65-20|+|28-25|) + 400(|65-36|+|28-18|) + 500(|65-62|+|2837|) + 300(|65-50|+|28-56|) + 200(|65-25|+|28-0|) = 76,900
(x*,y*) = (30,45)
10.7 a.
Using the half-sum method: half of the sum of the weights equals 1200
x-coord
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35
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(x*, y*) = (30,20)
2000
2400
25
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1800
2400
10-5
b.
Shown below is a partial contour line which passes through the
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line and Building 3 is located outside the contour line, the least cost
location is Building 1 at (20,20).
10-6
10.8 The contour line passing through (1,5) is shown below.
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10-18
10.20 a. Average distance item i travels between dock k and its storage region,
Ri = 3 dk j / Ai
j 0 Ri
b. Average cost of trans porting item i between dock k and its storage region,
Ri = wi k3 dk j / Ai
j 0 Ri
c. Integer programming formulation
m p
Minimize 3
3 wi k (3 dk j / Ai)
i = 1 k = 1 j 0 Ri
subject to:
m
3 Ai
= n
i=1
n
3 xij
= Ai ú i
j=1
m
3 xij
= 1 ú j
i=1
xij
= 0,1
10-19
10 .21
P ro d u c t
A rea
# B ays
(j)
(ft 2)
(Sj)
A
4,375
7
B
7,500
12
C
1,500
3
D
6,250
10
Product Ranking: C > A > B > D
(multiple optimum layouts exist)
LoadRate
(Tj )
500
600
700
200
T j / Sj
500/7=71.43
600/12=50.00
700/3=233.33
200/10=20.00
10-20
10 .22
P ro d u c t
A rea
# B ays
LoadRate
(j)
1
2
3
4
5
6
(ft 2)
2,400
3,200
2,000
2,800
4,000
1,600
(Sj)
6
8
5
7
10
4
(Tj )
600
400
800
400
400
800
Product Ranking: 6 > 3 > 1 > 4 > 2 > 5
(multiple optimum layouts exist)
T j / Sj
100
50
160
57.14
40
200
10-21
10 .23
P ro d u c t
# B ays
LoadRate
(j)
(Sj)
(Tj )
A
15
1
B
5
1
C
16
1
Product Ranking: B > A > C
T j / Sj
0.0667
0.20
0.0625
10-22
(multiple optimum layouts exist)
10.24 Distances to/from storage bays are determined by summing the distance
traveled from the receiving door (A) to t he center of the storage bay and
from the center of the storage bay to the shipping door (B). Consider the
shaded storage bay, shown below. Travel from the receiving door to the midpoint of the cross-aisle is 92'; travel up the cross-aisle to the center of the
two-way aisle is 52'; travel along the two-way aisle to the center point of the
storage bay is 54'; travel from the mid-point of the aisle to the mid-point of
the storage bay is 14'. Hence, a total of 212' is required to storage a load in
the storage bay. Travel from the storage bay to the shipping door requires 14'
from bay to aisle, 202' along the aisle to the mid-point of the farthest oneway aisle, 52' down the aisle to the mid-point of the center aisle, and 4' to the
shipping door for a total distance of 272'. The total distance traveled to/from
the storage bay is 488', as shown. We do not compute the distance traveled
by the lift truck after storage and before retrieval since we do not know its
destination and srcin points, respectively. Once the distance is known for
one storage bay, the other distances can be easily calculated.
10-23
The distances to/from storage bays located along the center aisle are
determined as follows: the distance from from door A to door B is 244'; the
distance from the mid-point of the center aisle to the mid-point of a storage
bay is 18'. Hence, the total distance equals 244' + 18' + 18', or 280'.
P ro d u c t
A rea
#B ays
(j)
(ft 2)
(Sj)
X
6,400
16
Y
8,800
22
Z
2,400
6
Product Ranking: Z > X > Y
LoadRate
(Tj )
300
400
500
T j / Sj
300/16=18.75
400/22=18.18
500/6=83.33
10-24
10.25
Create Distance Matrix:
10-25
10.26 a. Shown below is an Excel spreadsheet solution of the problem. To
compute the number of storage rows required to accommodate the
designated inventory level the ROUNDUP Function is used, e.g., if the
inventory level is shown in cell D6 and the row depth is given in cell F5,
then if the product is stacked 4 levels high, the following entry would be
entered: =ROUNDUP(D6/(4*F5),0).
10-26
10.26 (continued)
To determine the average number of storage rows required over the life of
a storage lot (0), the value obtained using the SUM function over the
range of storage rows required is divided by the number of entries, e.g.,
=SUM(F6:F65)/60.
Likewise, to determine the average amount of floor space required during
the life of the storage lot (S), the following was used:
=F69*(K1+M1)*(F5*I1+0.5*O1), where F69 is the value of 0, K1 is the
value of W, M1 is the value of c, F5 is the value of x, I1 is the value of L,
and O1 is the value of A.
Alternately, the following equation can be used to compute the value of
SBS: SBS = y(W + c)(xL + 0.5A)[2Q - xyz + xz]/2Q, where y is obtained
10-27
using the ROUNDUP function. Shown below is an Excel solution to the
problem. In enumerating over x, SBS is minimized with x equal to 5.
b. Using a continuous approximation,
xcBS = [AQ/2Lz] = [144(60)/2(48)(4)] = 4.74
Therefore, rounding off the value to the nearest integer gives xcBS = 5.
!
!
10.27 L = 42", W = 48", c = 4", r = 6", f = 12", A = 96", z = 7
a. Q = 300
xcDLSS = [(A + f)(Q)/2L] = [(96 + 12)(300)/2(42)] = 19.64
xcDLSS = 20
Using Excel to solve by enumerating over x gives the same result, as
shown below.
!
!
b. Q = 35, s = 0
Using Excel to solve by enumerating over x gives the following result.
10-28
10.28 L = 40", W = 48", c = 8", A = 156", z = 5
a. Q = 300
xcBS = [AQ/2Lz] = [156(300)/2(40)(5)] = 10.82
Therefore, rounding off the value to the nearest integer gives xcBS = 11.
Using Excel to solve by enumeration gives x*BS = 10, as shown below.
!
!
Excel’s SUMPRODUCT function was used to calculate the value of0; the
values in the percentage column were multiplied by the number of storage
rows required to accommodate the inventory level; the sum of the products
is displayed in the 0 row. Knowing the value of0, S is easily determined by
multiplying by the size of the foot-print of a row, (W + c)(xL + 0.5A).
Developing a spreadsheet allows analyses of changes in inventory profiles
over lifetimes of production lots to be performed easily.
10.29 Q = 30, L = 48", W = 40", c = 4", r = 4", f = 12", A = 96", z = 7
a. SSD = (W + 1.5c + 0.5r)[L + 0.5(A + f)](Q + 1)/2z
SSD = (40 + 6 + 2)[48 + 0.5(96 + 12)](31)/2(7)
SSD = 10,841.14 sq. in.
b. SDD = (W + 1.5c + 0.5r)[2L + 0.5(A + f)](Q + 2)/4z
SDD = (40 + 6 + 2)[96 + 0.5(96 + 12)](32)/4(7)
SDD = 8,228.57 sq. in.
c. STD = <(W + 1.5c + 0.5r)[3L + 0.5(A + f)](2Q - 3 < + 3)/2Qz
where < = jQ/3k = j30/3k = 10
(note: jxxk denotes the largest integer # xx).
STD = 10(40 + 6 + 2)[144 + 0.5(96 + 12)](60 - 30 + 3)/60(7)
STD = 7,467.43 sq. in.
10-29
10.30 a. Q = 30, L = 48", W = 50", c = 10" A = 156", z = 3
b. Using a continuous approximation, with Q = 300,
c
!
!
x BS = [AQ/2Lz] = [156(300)/2(48)(3)] = 12.75
Therefore, rounding off the value to the nearest integer gives xcBS = 13.
Enumerating over values of x, as shown below the optimum value is
actually 14.
c. Using a continuous approximation, with Q = 300 and s = 30
xcBSSS = [A(Q + 2s)/2Lz] = [156(300 + 60)/2(48)(3)] = 13.96
Therefore, rounding off the value to the nearest integer gives xcBSSS = 14.
Enumerating over values of x, as shown below the optimum value is
actually 14.
!
!
d. Using enumeration for Q =30 and s = 5 gives x*BSSS = 5, as shown below.
10-30
10.31 a. Q = 30, L = 36", W = 60", c = 12", A = 156", z = 3
b. Q = 300, L = 36", W = 60", c = 12", A = 156", z = 3
c. Q = 300, L = 36", W = 60", c = 12", A = 156", z = 3 , s = 30
d. Q = 30, L = 60", W = 36", c = 24", A = 156", z = 3, s = 0
The best configuration is 36" x 60" x 48".
10-31
10.32 Q = 30, L = 3', W = 4', c = 1', A = 8', z = 3
The problem solution is shown below.
a. For x = 2, the average amount of floor space required in the freezer is
194.4 sq. ft.
b. For x = 3, the average amount of floor space required in the freezer is
195.0 sq. ft.
(Notice that the optimum storage depth is 5, with an average floor space
requirement of 158.3 sq. ft.)
10.33 a. Q = 30, L = 60", W = 38", c = 10", A = 156", z = 4
b. Q = 30, L = 60", W = 38", c = 3", f = 12", r = 4 ", A = 72", z = 8
xcDL = [(A + f)(Q)/2L] = [(72 + 12)(30)/2(60)] = 4.58
xcDL = 5 (Enumeration yields the same result, as shown below.)
!
!
10-32
c. Q = 30, L = 60", W = 38", c = 5", f = 12", r = 5 ", A = 96", z = 6
SSD = (W + 1.5c + 0.5r)[L + 0.5(A + f)](Q + 1)/2z
SSD = (38 + 7.5 + 2.5)[60 + 0.5(96 + 12)](31)/2(6)(144)
SSD = 98.17 sq. ft.
d. SDD = (W + 1.5c + 0.5r)[2L + 0.5(A + f)](Q + 2)/4z
SDD = (38 + 7.5 + 2.5)[120 + 0.5(96 + 12)](32)/4(6)(144)
SDD = 77.33 sq. ft.
e. STD = <(W + 1.5c + 0.5r)[3L + 0.5(A + f)](2Q - 3 < + 3)/2Qz
where < = jQ/3k = j30/3k = 10
STD = 10(38 + 7.5 + 2.5)[180 + 0.5(96 + 12)](60 - 30 + 3)/60(6)(144)
STD = 71.50 sq. ft.
10.34 a. Q = 250, L = 36", W = 48", c = 4", r = 4", f = 12 ", A = 60", z = 15.
Computing the value of SDL yields the following results. The lane depth of
5 is the worst of the three.
b. As shown below, 15 is the best lane depth of the three choices.
c. Q = 250, s = 50, L = 36", W = 48", c = 4", f = 12", r = 4 ", A = 96", z = 6
SDD = (Q/2)(W+1.5c+0.5r)[2L+0.5(A+f)][2(Q+s)-Q+2]/2(Q+s)z
SDD = 125(48 + 6 + 2)[72 + 0.5(96 + 12)](352)/2(300)(6)(144)
SDD = 598.89 sq. ft.
10-33
10.35 a. Q = 200, L = 5', W = 4', A = 10', z = 5, c = not stated
(Note: it is not necessary to know the values of W and c in order to
determine the optimum row depth.)
Using a continuous approximation, with Q = 200,
xcBS = [AQ/2Lz] = [10(200)/2(5)(5)] = 6.32
Therefore, rounding off the value to the nearest integer gives xcBS = 6.
Enumerating over values of x, demonstrates this is not a convex function,
!
!
as shown below. There are two local optima at x = 6 and x = 8. If only
values of x = 5, 6, and 7 were considered, the alternate optimum at x = 8
would be missed.
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10-50
10.63 a. The least common multiple of the individual periods (5, 2, 4, 3) is 60.
b. k = 543; p = 60; r = 543mod60 = 3; r/p = 3/60 = 1/20. Since r/p is not a
proper fraction, the answer is no.
c. k/p = 543/60 = 9.05. Since k/p is not an integer, the answer is yes.
d. p is not a prime number. Therefore, the answer is no.
Note: even though r/p is not a proper fraction, the solution procedure
presented in the chapter can be used for this problem. Applying the solution
procedure and indexing r carriers yields the following sequence of values for
H*1(n): H*1(1), H*1(4), H* 1(7), H*1(10), H*1(13), H*1(16), H*1(19), H*1(2),
H*1(5), H*1(8), H*1(11), H*1(14), H*1(17), H*1(20), H*1(3), H*1(6), H*1(9),
H*1(12), H*1(15), and H*1(18). This is not always the case, as shown in
Problem 10.65(e).
10.64 a. k = 32; p = 3, r = 32mod3 = 2
{f1(n)} = (-5, 0, 0), {f 2(n)} = (0, -3, 0), {f 3(n)} = (0, 0, 4), and
{f4(n)} = (4, 0, 0). Therefore, {F 1(n)} = (-1, -3, 4).
H*1(1) = 0 à H*1(3) = H*1(1) + F1(3) = 0 + 4 = 4 and H*1(2) = H*1(3) +
F1(2) = 1. Hence, {H* 1(n)} = (0, 1, 4).
Thus, {H*2(n)} = {H*1(n)} - {f1(n)} = (0, 1, 4) - (-5, 0, 0) = (5, 1, 4).
{H*3(n)} = {H*2(n)} - {f2(n)} = (5, 1, 4) - (0, -3, 0) = (5, 4, 4).
{H*4(n)} = {H*3(n)} - {f3(n)} = (5, 4, 4) - (0, 0, 4) = (5, 4, 0).
Since c = 0, {Hi(n)} = {H*i(n)} ú i, n
B=5
b. k =33 is not feasible since p = 3; k = 31; p = 3, r = 31mod3 = 1
As before, {F1(n)} = (-1, -3, 4).
H*1(1) = 0 à H*1(2) = H*1(1) + F1(2) = 0 - 3 = -3 and H* 1(3) = H*1(2) +
F1(3) = -3 + 4 = 1. Hence, {H* 1(n)} = (0, -3, 1).
Thus, {H*2(n)} = {H*1(n)} - {f1(n)} = (0, -3, 1) - (-5, 0, 0) = (5, -3, 1).
{H*3(n)} = {H*2(n)} - {f2(n)} = (5, -3, 1) - (0, -3, 0) = (5, 0, 1).
{H*4(n)} = {H*3(n)} - {f3(n)} = (5, 0, 1) - (0, 0, 4) = (5, 0, -3).
Since c = -3, Hi(n) = H*i(n) + 3 ú i, n
{H1(n)} = (3, 0, 4), {H 2(n)} = (8, 0, 4), {H 3(n)} = (8, 3, 4), and
{H4(n)} = (8, 3, 0)
B=8
i, n
i
Inventory
= 3From
+ 0 +part
4 +a),
8 +for
0 +k 4=+32,
8 +inventory
3 + 4 + 8is+ given
3 + 0 by
Inventory =
for3k ={H
31(n)}
is 45.
Inventory = 0 + 1 + 4 + 5 + 1 + 4 + 5 + 4 + 4 + 5 + 4 + 0 = 37
k = 32 minimizes inventory an d minimizes B.
10-51
c. 1) {f1(n)} = (0, 0, -4), {f 2(n)} = (-4, 0, 0), {f 3(n)} = (0, 0, 4), and
{f4(n)} = (4, 0, 0). {F 1(n)} = (0, 0, 0) à {H*1(n)} = (0, 0, 0).
{H*2(n)} = (0, 0, 0) - (0, 0, -4) = (0, 0, 4)
{H*3(n)} = (0, 0, 4) - (-4, 0, 0) = (4, 0, 4)
{H*4(n)} = (4, 0, 4) - (0, 0, 4) = (4, 0, 0).
c = 0 à {Hi(n)} = {H*i(n)} ú i, n
Inventory = 16; B = 4
2) {f1(n)} = (-4, 0, 0), {f 2(n)} = (0, -4, 0), {f 3(n)} = (0, 0, 4), and
{f4(n)} = (4, 0, 0). {F 1(n)} = (0, -4, 4) à H*1(1) = 0, H* 1(3) = H*1(1)+
F1(3) = 0 + 4 = 4, H* 1(2) = H*1(3) + F1(2) = 4 - 4 = 0.
{H*1(n)} = (0, 0, 4). {H* 2(n)} = (0, 0, 4) - (-4, 0, 0) = (4, 0, 4).
{H*3(n)} = (4, 0, 4) - (0, -4, 0) = (4, 4, 4)
{H*4(n)} = (4, 4, 4) - (0, 0, 4) = (4, 4, 0)
c = 0 à {Hi(n)} = {H*i(n)} ú i, n
Inventory = 32; B = 4
Of the two considered {f1(n)} = (0, 0, -4) and {f2(n)} = (-4, 0, 0) are
the preferred loading sequences based on inventory on the conveyor.
10.65 k = 33, p = 4, r = 33mod4 = 1
a. Since r Ö 0, the answer is yes.
b. Since r/p = 1/4 is a proper fraction, the answer is yes.
c. {f1(n)} = (-2, 0, -2, 0), {f2(n)} = (0, -2, 0, -2), {f3(n)} = (0, 0, 4, 0), and
{f4(n)} = (4, 0, 0, 0). {F1(n)} = (2, -2, 2, -2) à H*1(1) = 0.
H*1(2) = H*1(1) + F1(2) = -2. H* 1(3) = H*1(2) + F1(3) = 0.
H*1(4) = H*1(3) + F1(4) = -2 à {H*1(n)} = (0, -2, 0, -2).
{H*2(n)} = (0, -2, 0, -2) - (-2, 0, -2, 0) = (2, -2, 2, -2)
{H*3(n)} = (2, -2, 2, -2) - (0, -2, 0, -2) = (2, 0, 2, 0)
{H*4(n)} = (2, 0, 2, 0) - (0, 0, 4, 0) = (2, 0, -2, 0)
c = -2 à Hi(n) = H*i(n) + 2 ú i, n
{H1(n)} = (2, 0, 2, 0), {H2(n)} = (4, 0, 4, 0), {H3(n)} = (4, 2, 4, 2), and
{H4(n)} = (4, 2, 0, 2)
H3(2) = 2
d. B = 4 (Although not asked for, the inventory totals 4 + 8 + 12 + 8, or 32.)
e. k = 34, r = 34mod4 = 2. Since r/p is not a proper fraction, it is not clear if
the
sequences
are feasible.
Tofollowing
determinesequence:
the values
of
H*1general
(n), notice
they willposed
be generated
in the
H*
1(1),
H*1(3), H*1(5) = H*1(1). Therefore, values for H* 1(2) and H*1(4) cannot
be generated using the solution procedure. Hence, we are unable to solve
10-52
the problem for k = 34. Likewise, we cannot solve it for k = 32, since r =
0. However, a solution is possible for k = 35 since r = 3 and r/p is a proper
fraction.
With r = 3 and {F1(n)} = (2, -2, 2, -2), the following values of H*1(n) are
obtained: H*1(1) = 0, H* 1(4) = -2, H* 1(3) = 0, H* 1(2) = -2, or
{H*1(n)} = (0, -2, 0, -2), which is the same sequence obtained for k = 33.
Therefore, the same results exist for both k = 33 and k = 35. Hence, the
inventory on the conveyor and the carrier capacity will be the same
for any feasible value of k.
(This problem illustrates the value of Muth’s third result, i.e., having p be
a prime number. Conveyor compatibility results for all values of k when p
is a prime number and r Ö 0.)
f. k = 33, p = 4, r = 1
1) {f1(n)} = (-2, 0, -2, 0), {f2(n)} = (0, -2, 0, -2), {f3(n)} = (0, 0, 2, 2),
{f4(n)} = (2, 2, 0, 0). {F1(n)} = (0, 0, 0, 0) à {H*1(n)} = (0, 0, 0. 0).
{H*2(n)} = (0, 0, 0, 0) - (-2, 0, -2, 0) = (2, 0, 2, 0)
{H*3(n)} = (2, 0, 2, 0) - (0, -2, 0, -2) = (2, 2, 2, 2)
{H*4(n)} = (2, 2, 2, 2) - (0, 0, 2, 2) = (2, 2, 0, 0)
c = 0 à Hi(n) = H*i(n) ú i, n
{H1(n)} = (0, 0, 0, 0), {H2(n)} = (2, 0, 2, 0), {H3(n)} = (2, 2, 2, 2), and
{H4(n)} = (2, 2, 0, 0). Inventory totals 16 and B = 2.
2) {f1(n)} = (-2, 0, -2, 0), {f2(n)} = (0, -2, 0, -2), {f3(n)} = (2, 0, 2, 0),
{f4(n)} = (0, 2, 0, 2). {F1(n)} = (0, 0, 0, 0) à {H*1(n)} = (0, 0, 0. 0).
{H*2(n)} = (0, 0, 0, 0) - (-2, 0, -2, 0) = (2, 0, 2, 0)
{H*3(n)} = (2, 0, 2, 0) - (0, -2, 0, -2) = (2, 2, 2, 2)
{H*4(n)} = (2, 2, 2, 2) - (2, 0, 2, 0) = (0, 2, 0, 2)
c = 0 à Hi(n) = H*i(n) ú i, n
Inventory totals 16 and B = 2.
The two alternate sequences have identical results, with an inventory of 16
on the conveyor. With the srcinal loading sequences, the inventory
totaled 32. Hence, either of the alternate loading sequences would be
preferred to the srcinal sequences .
10.66 a. k = 62, p = 3, r = 62mod3 = 2, {f 1(n)} = (0, -4, 0), {f 2(n)} = (3, 3, 3), and
3
{f
= (0, -5, 0) à {F1(n)} = (3, -6, 3). H* 1(1) = 0, H* 1(3) = 3,
H*(n)}
1(2) = -3. {H* 1(n)} = (0, -3, 3). {H* 2(n)} = {H*1(n)} - {f1(n)} =
(0, -3, 3) - (0, -4, 0) = (0, 1, 3). {H* 3(n)} = {H*2(n)} - {f2(n)} = (0, 1, 3)
- (3, 3, 3) = (-3, -2, 0).
10-53
Since c = -3, Hi(n) = H*i(n) + 3 ú i, n
{H1(n)} = (3, 0, 6), {H 2(n)} = (3, 4, 6), {H 3(n)} = (0, 1, 3)
B = 6 (Inventory totals 26.)
b. 1) {f1(n)} = (0, -4, 0), {f 2(n)} = (3, 3, 3), {f 3(n)} = (-5, 0, 0)
à {F1(n)} = (-2, -1, 3). H*1(1) = 0, H* 1(3) = 3, H* 1(2) = 2.
{H*1(n)} = (0, 2, 3), {H* 2(n)} = {H*1(n)} - {f1(n)} = (0, 6, 3).
{H*3(n)} = {H*2(n)} - {f2(n)} = (-3, 3, 0). c = -3 à
{H1(n)} = (3, 5, 6), {H 2(n)} = (3, 9, 6), {H 3(n)} = (0, 6, 3)
B=9
(Inventory totals 41.)
2) {f1(n)} = (0, -4, 0), {f 2(n)} = (3, 3, 3), {f 3(n)} = (0, 0, -5)
à {F1(n)} = (3, -1, -2). H*1(1) = 0, H* 1(3) = -2, H* 1(2) = -3.
{H*1(n)} = (0, -3, -2), {H*2(n)} = {H*1(n)} - {f1(n)} = (0, 1, -2).
{H*3(n)} = {H*2(n)} - {f2(n)} = (-3, -2, -5). c = -5 à
{H1(n)} = (5, 2, 3), {H 2(n)} = (5, 6, 3), {H 3(n)} = (2, 3, 0)
B=6
(Inventory totals 29.)
To minimize B, use either{f3(n)} = (0, -5, 0) or {f3(n)} = (0, 0, -5) for a
value of 6.
c. To minimize inventory on the conveyor, use {f3(n)} = (0, -5, 0) for a value
of 26.
10.67 a. k = 83, p = 3, r = 83mod3 = 2, {f 1(n)} = (1, 0, 0), {f 2(n)} = (0, 2, 0),
{f3(n)} = (0, 0, 1), {f 4(n)} = (0, -4, 0). à {F1(n)} = (1, -2, 1). H* 1(1) = 0,
H*1(3) = H*1(1) + F1(3) = 1, H* 1(2) = H*1(3) + F1(2) = -1.
{H*1(n)} = (0, -1, 1). {H* 2(n)} = (-1, -1, 1), {H*3(n)} = (-1, -3, 1),
{H*4(n)} = (-1, 1, 0). c = -3 à Hi(n) = H*i(n) + 3 ú i, n
{H1(n)} = (3, 2, 4), {H 2(n)} = (2, 2, 4), {H 3(n)} = (2, 0, 4), and
{H4(n)} = (2, 4, 3) à H3(2) = 0.
b. B = 4, inventory totals 9 + 8 + 6 + 9, or 32.
c. {f1(n)} = (1, 0, 0), {f 2(n)} = (0, 2, 0), {f 3(n)} = (0, 0, 1), and
{f4(n)} = (0, -a, -b). Therefore, {F1(n)} = (1, 2-a, 1-b).
H*1(1) = 0 à H*1(3) = H*1(1) + F1(3) = 1-b
H*1(2) = H*1(3) + F1(2) = 1-b + 2-a, or 3 - (a+b). Since a+b = 4,
then H*1(2) = -1. Hence, {H* 1(n)} = (0, -1, 1-b) and
{H*2(n)} = {H*1(n)} - {f1(n)} = (0, -1, 1-b) - (1, 0, 0) = (-1, -1, 1-b).
{H*3(n)} = {H*2(n)} - {f2(n)} = (-1, -1, 1-b) - (0, 2, 0) = (-1, -3, 1-b).
4
3
{H*
= {H*
- {fH3(n)}
= (-1, -3, 1-b) - (0, 0, 1) = (-1, -3, -b).
If b =(n)}
4, then
c =(n)}
-4 and
i(n) = H*i(n) + 4 ú i, n
{H1(n)} = (4, 3, 1), {H 2(n)} = (3, 3, 1), {H 3(n)} = (3, 1, 1), and
{H4(n)} = (3, 1, 0) à B = 4 if b = 4.
10-54
If b # 3, then c = -3 and Hi(n) = H*i(n) + 3 ú i, n
{H1(n)} = (3, 2, 4-b), {H 2(n)} = (2, 2, 4-b), {H 3(n)} = (2, 0, 4-b), and
{H4(n)} = (2, 0, 3-b) à B = 3 if 1# b # 3.
To minimize B, 1# b # 3.
Notice that the inventory totals 8 + 7 + 5 + 4, or 24 when b = 4. It totals
9 + 8 + 6 + 5 - 4b, or 28 - 4b when 1 # b # 3. To minimize inventory, let
b = 3 for a total of 16.
(Note: for any k that results in r = 1, B is minimized when 2 = b = 2.)
d. {f1(n)} = (1, 0, 1), {f 2(n)} = (0, 2, 0), {f 3(n)} = (0, -4, 0).
{F1(n)} = (1, -2, 1). H* 1(1) = 0, H* 1(3) = H*1(1) + F1(3) = 1,
H*1(2) = H*1(3) + F1(2) = 1 - 2 = -1. Hence, {H* 1(n)} = (0, -1, 1) and
{H*2(n)} = {H*1(n)} - {f1(n)} = (-1, -1, 0).
{H*3(n)} = {H*2(n)} - {f2(n)} = (-1, -3, 0).
Since c = -3, then Hi(n) = H*i(n) + 3 ú i, n
{H1(n)} = (3, 2, 4), {H 2(n)} = (2, 2, 3), {H 3(n)} = (2, 0, 3), and
{H4(n)} = (2, 0, 3) for B = 4. Inventory totals 9 + 7 + 5 + 5, or 26.
Although inventory is reduced, there is no effect on B.
10.68 a. flat belt driven roller conveyor, TL = 300', WBR = 33", RC = 9", and S =
200 fpm
Based on the WBR and RC combination,LF = 0.9.
BV = 4.60 + 0.445(WBR) = 4.60 + 0.445(33) = 19.29
LS = load segment length = 24" + 4" = 28"
TW = tote weight, loaded = 25 lbs.
# of full load segments on conveyor =j300(12)/28k = 128
(note: jxxk denotes the largest integer # xx).
length of partial load segment = 300(12) - 128(28) = 16"
weight of partial load segment = 25(16/24) = 16.67 lbs.
L = 128(25) + 16.67 = 3,216.67 lbs.
b. roller supported belt conveyor, TL = 100', WBR = 33", RC = 9", and S =
300 fpm
Based on the WBR and RC combination,LF = 0.62.
BV = b (WBR) = 22
For the following calculation, fpm denotes feet per minute, ipf denotes
inches per foot, ipt denotes inches per tote, and tpm denotes totes per
minute. The rate at which totes exit the roller conveyor is
250
fpm x 12
28 ipt
= 185.71429
tpm of a load segment) will be
Therefore,
foripf
the÷belt
conveyor,
LS (length
300 fpm x 12 ipf ÷ 85.71429 tpm = 42 ipt
Since a tote is 24" In length, the clearance between totes on the belt
10-55
conveyor will be 18"
# of full load segments on conveyor =j100(12)/42k = 28
length of partial load segment = 100(12) - 28(42) = 24", which is the
length of a tote. At its maximum loaded condition there will be 29 totes on
the conveyor at a weight of 29(25) = 725.
à L = 725 lbs.
10.69 a. chain driven roller conveyor, TL = 150', WBR = 39", RC = 9",
S = 90 fpm, length of pallet = 48", clearance between pallets = 24", and
weight of loaded pallet = 1,200 lbs. FF = 0.05 and LF = 1.0.
BV = 4.60 + 0.445(WBR) = 4.60 + 0.445(39) = 21.955
LS = 48" + 24" = 72"
# of full load segments on conveyor =j150(12)/72k = 25
length of partial load segment = 150(12) - 25(72) = 0
L = 25(1200) = 30,000 lbs.
b. For this calculation, ipf denotes inches per foot and ipls denotes inches per
load segment. For the chain belt driven roller conveyor, the rate at which
pallets are discharged is 90 fpm x 12 pf
i ÷ 72 ipls = 15 pallets per minute.
For the roller supported belt conveyor to process 15 pallet loads per
minute, its speed must beS = 15(48" + 36")/12 = 105 fpm.
10.70 a. V-belt driven roller conveyor, TL = 250', WBR = 30", and RC = 7.5"
Using linear interpolation with the data in the table,
for WBR = 27" and RC = 6", LF = 1.0
for WBR = 27" and RC = 9", LF = 0.8
à for WBR = 27" and RC = 7.5", LF. 0.9
for WBR = 33" and RC = 6", LF = 1.2
for WBR = 33" and RC = 9", LF = 0.9
à for WBR = 33" and RC = 7.5", LF. 1.05
à for WBR = 30" and RC = 7.5", LF . 0.975
b. From the weight and length combinations for the cartons shown on page
10-16, notice that the average carton length is 24" and the average carton
weight is 15 lbs. Also, notice that the average weight per inch of carton,
including the 6" clearance between cartons, is 0.5464 lbs/in. Further,
notice that the weight per inch is not the same as the average weight of a
carton divided by the average length of a load segment, since the average
of a ratio is not equal to the average numerator divided by the average
denominator.
An approximation of the average load on the conveyor is the product of
the length of the conveyor and the weight per inch of load segment, i.e.,
10-56
250 x 12 x 0.5464, or 1,639.20 lbs. This is only an approximation because
the weight of partial load segments is not uniformly distributed across the
length of the partial load segment. (Alternately, by simulating the loading
of the conveyor a more accurate estimate can be determined.)
10-57
c. The “worst case” condition occurs when the conveyor is loaded with the
shortest and heaviest cartons, e.g., having the conveyor full of 12" cartons
weighing 25 lbs., each. With a 6" clearance, the maximum number of 12"
cartons that can be placed on the 250' long conveyor can be determined
by, first, determining the number of full load segments on the conveyor:
j(250 x 12)/(12"+6")k = 166
The length of the partial load segment is (250 x 12) - (166 x 18) = 12",
which is the length of the shortest carton. Hence, there will be 167 cartons
on the conveyor during the “worst case” condition and the load on the
conveyor will total (167 x 25), or 4,175 lbs.
10.71
Flat belt driven roller conveyor, WBR = 27", RC = 6", tote box length =
24", tote box clearance on first conveyor = 6", average weight of loaded
tote box = 30 lbs.
a. S1 = 200 fpm, TL1 = 200', TL2 = 100', spacing between tote boxes on
2nd conveyor = 8", S2 = 200 fpm x (24"+8") ÷ (24"+6") = 213.33 fpm
b. # full load segments = j(200 x 12)/(24"+6")k = 80
Exactly 80 load segments will be on the conveyor; the average weight
of a load segment is 30 lbs. Therefore, the average load on the
conveyor is 2,400 lbs.
It is not recommended that the horsepower calculation be based on
average loading conditions. Instead, we recommend using either “worst
case” or 99 percentile conditions. Since there are exactly 70 load
segments on the conveyor at all times during loading conditions, the
load on the conveyor is equal to the sum of 70 load segments.
If the weight in a tote box is statistically independent, based on the
Central Limit Theorem the weight of 70 load segments will be normally
distributed with an average weight of 2,100 lbs. Since the variance of
the sum of independent random variables equals the sum of their
variances, the variance of the weight of 70 tote boxes is given by
Var(wt of loaded tote box) = 0.2(100+400+900+1600+2500)-(30) 2 =
200 lbs2 à Var(wt of 70 loaded tote boxes) = 70(200) = 14,000 lbs 2
The weight on the conveyor is normally distributed with a mean of
2400 lbs and a standard deviation of 118.32 lbs. Using the NORMINV
function from Excel, the following weights should be used in the
horsepowerthat
calculation
in order istoadequate
provide 95%,
99%,
and condition:
99.9%
confidence
the horsepower
for the
loading
1,694.62 lbs, 1,775.26 lbs, and 1,865.64 lbs, respectively.
10-58
10.72 RC = 6", WBR = 33", carton weight = 35 lbs., carton length = 30", spacing
between cartons on first conveyor = 24"
a. roller supported belt conveyor: TL 1 = 100' and S1 = 300 fpm
belt driven roller conveyor: S2 = 300(30 + 12)/(30 + 24) = 233.33 fpm
b. HP = [BV + LF(TL) + FF(L)](S)/14,000
BV1 = b (WBR) = b (33) = 22
LF1 = 0.72, TL1 = 100', FF1 = 0.05, S1 = 300 fpm
The number of full load segments:j(100 x 12)/(30 + 24)k = 22; the length
of the partial load segment is 1200 - 22(54) = 12"; the weight of the
partial segment is 35(12/30) = 14 lbs. Thus, L1 = 22(35) + 14 = 782 lbs.
HP1 = [22 + 0.72(100) + 0.05(782)](300)/14,000 = 2.85 hp
c. HP = [BV + LF(TL) + FF(L)](S)/14,000
BV2 = 4.6 + 0.445(WBR) = 4.6 + 0.445(33) = 19.285
LF2 = 1.2, TL 2 = 50', FF2 = 0.10, S 2 = 233.33 fpm
The number of full load segments:j(50 x 12)/(30 + 12) k = 14; the length
of the partial load segment is 50(12) - 14(42) = 12"; the weight of the
partial segment is 35(12/30) = 14 lbs. Thus, L2 = 14(35) + 14 = 504 lbs.
HP2 = [19.285 + 1.2(50) + 0.10(504)](233.33)/14,000 = 2.16 hp
10.73 roller supported belt conveyors, TL1 = 150',TL2 = 100', WBR = 27", RC =
10", S1 = 300 fpm, clearance between tote boxes on 1st conveyor = 12",
lightest wt. = 10 lb., avg. wt. = 30 lb., heaviest wt. = 50 lb.
a. S2 = 300 fpm x (24" + 6") ÷ (24" + 12") = 250 fpm
b. number of full load segments on 1st conveyor = j(150)(12)/(24+12) k = 50
there is no partial segment on 1st co nveyor
lightest load on 1st conveyor = 50(10) = 500 lbs
average load on 1st conveyor = 50(30) = 1500 lbs
heaviest load on 1st conveyor = 50(50) = 2500 lbs
Although the problem did not ask for it, by computing the variance of the
weight per tote box and utilizing the Central Limit Theorem, one can
determine upper limits for conveyor loading for specified confidence
levels. A calculation establishes that the variance of the weight of a tote
box equals
[100(0.1)+400(0.2)+900(0.4)+1600(0.2)+2500(0.1)] - (30) 2 = 120 lbs2
The sum of 50 independently distributed random variables, each having a
mean of 30 lbs and a variance of 120 lbs2, will be normally distributed with
a mean of 1500 lbs and a variance of 6000 lbs2. Using Excel’s NORMINV
10-59
function, the following weights should be used in the horsepower
calculation in order to provide 95%, 99%, and 99.9% confidence that the
horsepower is adequate for the loading condition: 1627.41 lbs, 1680.20
lbs, and 1739.37, respectively.
10.74
TL1 = TL2 = 100'., S1 = 150 fpm, S2 = 250 fpm, WBR = 33", and
RC = 6"
a. # full load segments on 1 st conveyor = j(100 x 12)/(20 + 6)k = 46; the
length of the partial load segment is 1200 - 46(26) = 4"; the weight of
the partial segment is 40(4/20) = 8 lbs. Thus, L1 = 46(40) + 8 = 1848
lbs.
b. For the totes to be processed at the same rates by both conveyors, the
clearance between tote boxes on the 2nd conveyor, c2, must be such that
S1/(20+c1) = S2/(20+c2) = 150/26 = 250/(20+c2) à c2 = [250(26)/150] 20 = 23.33"
# full load segments on 2nd conveyor = j(100 x 12)/(20 + 23.33) k = 27
length of the partial load segment is 1200 - 27(43.33) = 30.09", which
is longer than a tote box. Hence, there will be 28 tote boxes on the 2nd
conveyor. Thus, L2 = 28(40) = 1120 lbs.
c. maximum RC for load stability
Recall, at least 3 rollers should be under a tote box at all times for
stability of the load. Therefore, the maximum RC value is the length of
a tote box divided by 3, or 20/3 = 6.67"
10.75
roller-supported belt conveyor: TL = 200', WBR = 27", RC = 6", S = 100
fpm, " = 15°, c = 6"
A sequence of 11 consecutive cartons will be defined as a “train segment.”
Based on the load distribution, a train segment will consist of the
following sequence of cartons: A, A, B, B, C, C, C, D, D, D, E. Including
the clearance between cartons, each train segment is 282" long and weighs
310 lbs. There are j(200 x 12)/(282) k = 8 “train segments” on the
conveyor, plus a partial segment of length (200 x 12) - (8 x 282) = 144"
a. To determine the heaviest loading condition on the conveyor, it is
necessary to determine the weight of the heaviest 144" section of a
train segment, since there will be 8 train segments and a 144" partial
segment on the conveyor at all times during loading conditions.
The
tote40,
weights
pounds),clearances
in sequence,
are 15,cartons,
15, 20,the
20,weight
30, 30,is
30, 40,
40, 30.(inIncluding
between
distributed (in pounds per foot) as follows over a train segment: 10, 10,
10, 10, 15, 15, 15, 16, 16, 16, 10. The heaviest condition will occur
10-60
when, in addition to the 8 train segments, a partial segment is on the
conveyor, with the tot es in the partial segment weighing 15, 15, 16, 16,
and 16 pounds per foot. The length of a partial segment of C, C, D, D,
D, including clearances is 24 + 24 + 30 + 30 + 30, or 138"; it weighs
180 lbs.
Another 6" must be accounted for in the partial segment. The next
heaviest 6" over the partial segment will be 6" of C, at a weight of
(6/18)(30), or 10 lbs. (You may wish to verify that if the 6" had come
from E, it would have weighed only 6 lbs.) Hence, the weight on the
conveyor during the heaviest loading condition will total 8(310) + 180
+ 10, or 2,670 lbs.
b. To determine the heaviest loading condition on the conveyor, it is
necessary to determine the weight of the lightest 144" section of a train
segment. Based on the weight distribution shown above, the lightest
section will be E, A, A, B, B, C, with a length, including clearances, of
144" and a weight of 130 lbs. Therefore, the weight on the conveyor
during the lightest loading condition will be 8(310) + 130, or2,610 lbs.
10.76
Grain is to be conveyed horizontally 150' Smax = 200 + 16.667BW, grain
density = 50 lb/ft3, design capacity = 250 tph
BW = 18", Smax = 200 + 16.667(18) = 500 fpm.
BW = 24", Smax = 200 + 16.667(24) = 600 fpm
BW = 30", Smax = 200 + 16.667(30) = 700 fpm
BW = 36", Smax = 200 + 16.667(36) = 800 fpm.
@ 50 lb/ft3 density and BW = 18", the required speed to deliver 250 tph
equals (250/26.5)(100), or 943 fpm > 500 fpm. Thus, BW = 18" is not
feasible.
@50 lb/ft3 density and BW = 24", the required speed to deliver 250 tph
equals (250/52)(100), or 480.77 fpm < 600 fpm. Thus, BW = 24" is
feasible.
HP (BW=24") = (0.7)(480.77/100) t o drive empty conveyor = 3.37 hp
@50 lb/ft3 density and BW = 30", the required speed to deliver 250 tph
equals (250/81)(100), or 308.64 fpm < 700 fpm. Thus, BW = 30" is
feasible.
HP(BW=30") = (0.9)(3.0864) to drive empty conveyor = 2.78 hp
@50 lb/ft3 density and BW = 36", the required speed to deliver 250 tph
equals (250/113.5)(100), or 220.26 fpm < 800 fpm. Thus, BW = 36" is
10-61
feasible.
HP(BW=36") = (1.1)(2.2026) to drive empty conveyor = 2.42 hp.
Thus, the lowest horepower requirement (2.42 hp) to drive the empty
conveyor occurs with the 36" width conveyor belt traveling at a speed of
220.26 fpm . (ignoring the safety factor of 1.2).
b. f(BW,HP) = $300BW + $5,000HP
Here, we assume the HP value is for the total horsepower requirement,
including safety factor, rather than just the horsepower required to drive
the empty conveyor. The horsepower requirements for feasible belt widths
of 24", 30", and 36" are, respectively,
HP(BW=24") = (3.37 + 2.3)(1.2) = 6.80 hp
HP(BW=30") = (2.78 + 2.3)(1.2) = 6.10 hp
HP(BW=36") = (2.42 + 2.3)(1.2) = 5.66 hp
f(BW=24,HP=6.80) = $300(24) + $5,000(6.80) = $41,200
f(BW=30,HP=6.10) = $300(30) + $5,000(6.10) = $39,500
f(BW=36,HP=5.66) = $300(36) + $5,000(5.66) = $39,100
The least cost belt width is36"
(Note: if the conveyor cost $366.67 per inch of width, the optimum is BW
= 30"; if it cost $583.34 per inch of width, the optimum is BW = 36")
10.77 Dry sand; conveyor length = 75'; elevation = 24'; BW = 30"; S = 600 fpm.
a. Maximum angle of incline = 15°
Since 75 sin(15°) = 19.41 < 24', the angle of incline is not feasible.
Likewise, since sin-1(24/75) = 18.66° > 15°, the angle of incline is not
feasible.
b. Smax = 200 + 16.667BW = 200 + 16.667(30) = 700 fpm > 600 fpm.
Therefore, the speed is feasible.
c. 50 lb/ft 3, BW = 30", and S = 600 fpm yields a delivered capacity of
81.0(600/100), or 486 tph.
10.78 Lumpy, abrasive steel trimmings; conveyor length = 50'; elevation = 20'; BW
= 30"; S = 400 fpm
a. Maximum angle of incline = 18°
Since 50 sin(18°) = 15.45 < 20', the angle of incline is not feasible.
10-62
b. Smax = 100 + 8.333BW = 100 + 8.333(30) = 350 fpm < 400 fpm.
Therefore, the speed is not feasible.
c. 100 lb/ft 3, BW = 30", and S = 400 fpm yields a delivered capacity of
162.0(400/100), or 648 tph.
10.79 Unsized, washed gravel; horizontal distance = 232.55'; elevation = 59.33';
BW = 24"; S = 400 fpm; density = 75 lb/ft3
a. Maximum angle of incline = 12°
Since 232.55tan(12°) = 49.43 < 59.33', the angle of incline is feasible.
b. Smax = 200 + 8.333BW = 200 + 8.333(24) = 400 fpm = 400 fpm.
Therefore, the speed is feasible.
c. 75 lb/ft 3, BW = 24", and S = 400 fpm yields a delivered capacity of [(52.0
+ 104.0)/2](400/100), or 312 tph.
d. conveyor length = [(232.55) 2 + (59.33) 2]0.5 = 240'
hp to drive empty conveyor = [0.8 + (4/5)(0.9 - 0.8)](400/100) = 3.52 hp .
10.80 horizontal distance = 200'; angle of incline = 10°; rate = 100 tph; density =
60 lb/ft3
a. BW = 24"; delivered capacity/100 fpm = 52.0 + (1/5)(104.0 - 52.0) = 62.4
tph/100 fpm. Hence, S = (100 tph)/(0.624 tph/fpm) = 160.26 fpm .
b. Conv. length = 200cos(10°) = 196.96', S = 160.26 fpm, BW = 24".
HP to drive empty conveyor = [0.7 + (10.26/50)(0.8 - 0.7](160.26/100) =
1.155 hp
c. 100 tph; conv. length = 196.96'
) HP to convey material horizontally = [0.9 + (10.26/50)(1.1 - 0.9)] =
0.94 hp.
d. Elevation = 200tan(10°) = 35.27'
) HP to lift material 35.27' = [3.0 + (5.27/10)(3.0 - 4.0)] = 3.527 hp.
e. From a), the delivered capacity of material of this density is 62.4 tph/100
fpm of belt speed. If the material is classified as unsized material, then
Smax = 200 + 8.333BW = 200 + 8.333(24) = 400 fpm.
At
fpm, ismaterial
be ore,
delivered
rate of
62.4(4),
or249.6 tph . If
the 400
material
crushedwill
stone,
grain,atorasand,
then
the delivered
capacity will be greater.
10-63
3
10.81 coffee beans; angle of incline = 10°; elevation = 30'; density = 30 lb/ft
a. BW = 24"; S max = 200 + 16.667BW = 200 + 16.667(24) = 600 fpm
Maximum delivered capacity = 31.0(600/100) =186 tph
b. To achieve 200 tph, speed must be 200 tph(0.31 tph/fpm) = 645.16 fpm .
c. Conveyor length = 30/sin(10°) = 172.763'; S = 500 fpm
HP to drive empty conveyor = [0.7 + (22.763/50)(0.8 - 0.7)](500/100) =
3.73 hp
d.
)HP
to convey material horiz = 1.8 + (22.763/50)(2.1 - 1.8) = 1.94 hp
e. 200 tph; 30' elevation
)HP to elevate material =6.1 hp
3
10.82 oats; angle of incline = 15°; elevation = 50'; density = 30 lb/ft
a. BW = 30"; S = 400 fpm
Conveyor length = 50/sin(15°) = 193.185'
HP to drive empty conveyor = [0.9 + (43.185/50)(1.0 - 0.9)](400/100) =
3.945 hp
b. Delivered capacity = 48.5(400/100) = 194 tph
@ 150 tph, )HP to convey material horiz = 1.4 + (43.185/50)(1.6 - 1.4) =
1.573 hp
@ 200 tph, )HP to convey material horiz = 1.8 + (43.185/50)(2.1 - 1.8) =
2.059 hp
Thus, @ 194 tph, )HP to convey material horiz = 1.573 +(44/50)(2.059 1.573) = 2.00 hp
To elevate 50' @ 194 tph
)HP to elevate material = 8.8 + (19/25)(10.1 - 8.8) = 9.788 hp.
To drive empty conveyor, convey material 193.185' horizontally, and to
elevate material 50' requires 3.945 hp + 2.000 hp + 9.788 hp = 15.733 hp,
not including the 20% safety factor.
c. BW = 18" delivers oats at rate of 16.0 (4) = 64 tph < 200 tph; BW = 24"
has rate of 31.0(4) = 124 tph < 200 tph; BW = 30" has rate of 48.5(4) =
194 tph < 200 tph; BW = 31" has rate of [48.5 + (1/6)(68.0 - 48.5)](4) =
207 tph. > 200 tph. 31" belt width required.
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10-66
10.84 (M|M|c):(GD|N|4)
8 = 0.5/min, : = 1.25/min, D = 0.5/1.25 = 0.4
Determine the smallest value of N such that Pr(n=N)# 0.05
P0 = (1-D)/(1-DN+1) = 0.6/[1-(0.6)N+1] and
PN = (0.4)N(0.6)/[1-(0.6) N+1 ] # 0.05 which reduces to
(0.4) N - (0.4) N+1 # 0.05 - 0.05(0.4) N+1 or
(0.4) N[1-0.95(0.4)] # 0.05
(0.4) N $ 0.080645
Nlog(0.4) $ log(0.080645)
N $ 2.75
Therefore, N = 3 and the accumulation line must have the capacity for
two cartons waiting to be processed by the workstation.
10.85 (D|M|1):(GD|4|4)
8 = 3/min, : = 4/min, D = 3/4 = 0.75
Lq = 22/(1-2), where 2 = exp-(1-2)/D
As shown on the following page, using the iterative solution procedure until
|2k - 2k+1| # 0.00000001 yields an estimate for2 of 0.54560503, which yields
an estimate for Lq of 0.6551.
10.86 (M|G|c):(GD|c|4)
8 = 0.5/min, : = 0.333/min, F = 1 min. Find the smallest value of c such that
Pc # 0.05. From the figure in the text, at least 4 packaging stations should be
provided in order to have a loss no greater than 5%. Using Excel to compute
the vlaue of Pc yields the results shown below, verifying that 4 is the answer.
10.87 (M|G|1):(GD|4|4)
8 = 10/hr, : = 15/hr, F = 1/60 hr, and D = 0.6667.
Wq = 8(F2 + :-2)/(2(1-D) = 10[(1/3600) + (1/125)]/2(1/3), or
Wq = 0.124167 hr or 7.45 min
4 4
10.88 (M|M|1):(GD|
)
8 = 10/min, : =| 12/min,
and D = 5/6
.
Lq = D2/(1-D) = (25/36)/(1/6), or Lq = 4.1667 cartons
10-67
10-68
10.89 (M|M|2):(GD|4|4)
8 = 18/hr, : = 10/hr, c = 2, and D = 0.9
a. Pr(both inspectors are idle) = P 0 = 1/[1 + 1.8 + (1.8) 2/(2)(0.1)] = 0.0526
b. Lq = D(cD)P0/c!(1-D)2 = (0.9)(1.8)(0.0526)/2(.01), or Lq = 4.26 unit loads
10.90 (M|M|1):(GD|N|4)
The solution procedure is identical to that for Problem 10.84.
8 = 1/min, : = 1.25/min, D = 0.8
Determine the smallest value of N such that Pr(n=N)# 0.02
P0 = (1-D)/(1-DN+1) = (0.2)/[1-(0.8) N+1] and
PN = (0.2)N(0.8)/[1-(0.8) N+1 ] # 0.02 which reduces to
(0.8) N - (0.8) N+1 # 0.02 - 0.02(0.8) N+1 or
(0.8) N[1-0.98(0.8)] # 0.02
(0.8) N $ 0.0925925
Nlog(0.8) $ log(0.0925925)
N $ 10.66
Therefore, N = 11 and the accumulation conveyor must have the capacity
for ten pa rts waiting to be processed by the workstation.
10.91 (Mb|M|1):(GD|4|4)
E(b) = [1+2+3+4]/4 = 2.5], 8 = 5/min, : = 24/min, D = 5(2.5)/24 = 0.52083
E(b2) = [1+4+9+16]/4 = 7.5 à V(b) = 7.5 - (2.5) 2 = 1.25
W = [1.25 + (2.5)2 + 2.5]/{2(2.5)[24 - 5(2.5)]} = 0.1739 min.
Wq = W - (1/:) = 0.1739 - 0.0417 = 0.1322 min
10.92 (Mb|M|1):(GD|4|4)
E(b) = [1+2+3+4]/4 = 2.5], 8 = 6/min, : = 24/min, D = 6(2.5)/24 = 0.625
E(b2) = [1+4+9+16]/4 = 7.5 à V(b) = 7.5 - (2.5) 2 = 1.25
L = 6[1.25 + (2.5) 2 + 2.5]/{2[24 - 6(2.5)]} = 3.3333
Lq = L - D = 3.3333 - 0.625 = 2.70833
10.93 c = 1, : = 12/hr
a. (M|M|1):(GD|4|4)
What is the largest value of8 such that L # 4?
L = D/(1-D) = ( 8/12)/[1 - (8/12)] # 4
8/[12 - 8]# 4 à 8 # 9.6/hr
b. (D|M|1):(GD|4|4)
L
2) = 4.8Solving
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8 ) 2 = 0.8
0.8= =2/(1exp(-0.2/(
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gives ln(0.8) = -2.4/8 = -0.2231425. Thus, 8 = 10.755/hr
10-69
c. The arrival and service rates behave like an (M|M|1):(GD|3|3) queue with
K = 3, 8 = 5/hr, : = 12/hr, and D = 5/12 = 0.41667.
The results for P0 can be verified by using the equations in the book,
P0 = 1/[1 + 3(5/12) + 3(2)(5/12) 2 + 6(5/12) 3] = 1728/6438 = 0.2684
10.94 (M|M|1):(GD|5|4) variation
a. Pr(idle server) = P0 = 0.0861
b. If P 0 = 1, what is the value of P1? Since P1 = 2P0, then P1 = 0.2
10-70
10.95 (M|M|2):(GD|8|8) 8 = 0.1/hr, : = 0.5/hr
a. Pr(at least one idle recharging machine) = P0 + P1 = 0.52936
b. Lq = 0.39776
10.96 (M|M|1):(GD|6|6) 8 = 0.5/hr, : = 2.5/hr
a. Pr(idle attendant) = P0 = 0.19185
b. Lq = 1.1511
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0'10,+0 ,/ 6<6*'08 @_'+'/1**< 0.'1?)+=D ,+' 07,:*> !"# 1I'/1=' -7' >1-1
6,**'6-'> ,I'/ 0'I'/1* 0'10,+0 ,/ 6<6*'08H
!"#$X
!"#!"E 97' ./)41/< 671**'+=' )+ >'I'*,.)+= 0:67 1 0)4:*1-),+ 4,>'* )0 -, ?''.
-/16? ,3 -7' .,0)-),+ ,3 '167 6/1+' 0, -71- 2J*,6?)+=5 @)8'8D ,+' 6/1+'
61++,- 4,I' J'61:0' )- )0 J*,6?'> J< -7' +')=7J,/)+= 6/1+'H )0 166:/1-'*<
61.-:/'>8 R*0,D ,+' (,:*> 71I' -, >'I'*,. @1+> )+6,/.,/1-' )+-, -7'
0)4:*1-),+ 4,>'*H 1 6/1+' >)0.1-67)+= *,=)6` )8'8D >'6)>)+= (7)67 6/1+' )0
>)0.1-67'> -, 4,I' (7)67 *,1> 1+> (7'+8 &)-7,:- 1 /'10,+1J*< =,,>
6/1+' >)0.1-67)+= *,=)6D -7' 6/1+'0 41< J*,6? '167 ,-7'/ -,, ,3-'+8 K+
-'/40 ,3 >1-1D ,+' (,:*> 1- *'10- +''> -, )>'+-)3< '167 .)6?:.a>'.,0).,)+- )+ -7' *1<,:-D -7' /1-' 1- (7)67 4,I' /';:'0-0 1//)I' 3/,4 '167
.)6?:. .,)+-D 1+> -7' >'0-)+1-),+ ,3 -7' *,1>0 .)6?'> :.8 \10-*<D ,+' +''>0
-7' 6/1+' -/1I'* -)4'0 1+> -7' *,1> .)6?:.a>'.,0)- -)4'08
!"#!"F A)4:*1-),+ 710 'O-'+0)I' 1..*)61-),+0 )+ 316)*)-< *,61-),+ 1+> *1<,:-8 R
3:** -/'1-4'+- ,3 -7' -,.)6 )0 J'<,+> -7' 06,.' ,3 -7)0 J,,?8 97' ./)41/<
;:'0-),+ )0 ,+' ,3 .:/.,0' 1+> 06,.'8 971- )0D (71- )0 -7' .:/.,0' ,3 -7'
0)4:*1-),+ 'O.'/)4'+- @(71- 1+0('/0 1/' (' -/<)+= -, ,J-1)+ 3/,4 -7'
4,>'*H 1+> 7,( 4:67 >'-1)* 07,:*> -7' 4,>'* 61.-:/'8 9<.)61* 0,:/6'0 ,3
/1+>,4+'00 )+ 0:67 0)4:*1-),+ 'O.'/)4'+-0 1/' -7,0' 100,6)1-'> ()-7
>'41+>D .1/-0 3*,(D 1+> /'0,:/6' 1I1)*1J)*)-<8
!"#!"G 97' ./)41/< 0,:/6' ,3 I1/)1-),+ )0 -7' 1//)I1* .1--'/+ ,3 -7' 0-,/1=' 1+>
/'-/)'I1* /';:'0-08 A)4:*1-),+ 41< J' :0'> -, 0-:>< -7' ><+14)6
)+-'/16-),+0 )+ -7' 0<0-'4 10 ('** 10 -7' $%&#'%()#%"! ,3 6'/-1)+ I1/)1J*'0
@10 ,..,0'> -, L:0- -7')/ 1I'/1=' I1*:'0H8 M,/ 'O14.*'D ,+' 41< >'-'/4)+'
-7' >)0-/)J:-),+ ,3 -7' 0.16' :-)*)b1-),+ @)+6*:>)+= 4)+)4:4 1+> 41O)4:4
I1*:'0H 10 ('** 10 -7' 41-'/)1* 71+>*)+= 61.16)-< +''>'> -, .'/3,/4 1** -7'
+'6'001/< 0-,/1=' 1+> /'-/)'I1* ,.'/1-),+08
!"#!"H A)4:*1-),+ 41< .*1< 1+ )4.,/-1+- /,*' )+ -7' >'0)=+ ,3 RAaF 0<0-'40 1+>
,/>'/ .)6?)+= 0<0-'408 M,/ 'O14.*'D -7' /'0:*-0 07,(+ )+ !"8Y 1+> !"8$
1/' J10'> ,+ 6'/-1)+ 0)4.*)3<)+= 100:4.-),+0 0:67 10 /1+>,4)b'> 0-,/1='
1+> 3)/0-#6,4'#3)/0-#0'/I'> @MPMAH 0-,/1=' 1+> /'-/)'I1* /';:'0-08 E+'
41< :0' 0)4:*1-),+ -, ?''. -/16? ,3 -7' *,61-),+ ,3 -7' )-'40 )+ -7' /16?
1+> -7:0 :0' 16-:1* AaF 4167)+' -/1I'* -)4'0 @/1-7'/ -71+ 100:4)+= '167
,.'+)+= )0 ';:1**< *)?'*< -, J' I)0)-'>H8 E+' 41< 1*0, 4,>'* AaF 4167)+'
-/1I'* 4,/' 166:/1-'*<D )+6*:>)+= 166'*'/1-),+a>'6'*'/1-),+ '33'6-0 1+>
4:*-).*' -/1I'* 0.''>0 -71- 1/' >)0-1+6'#>'.'+>'+-8
!"#$V
!"#!"I A)4:*1-),+ 41< J' :0'> -, 4,>'* 610'0 (7'/' -7' *,1> 1--'4.-0 @,/ -7'
+:4J'/ ,3 .1/-0 *,1>'> .'/ 1--'4.-H )0 /1+>,4*< >)0-/)J:-'> 1+> +,- 1**
.1/-0 1/' 0:66'003:**< :+*,1>'> 1- -7' :+*,1>)+= 0-1-),+0` )8'8D 0,4' )-'40
41< /'6)/6:*1-'8 ^0)+= 0)4:*1-),+D ,+' 41< 1*0, ,J-1)+ 4,/' >'-1)*'>
/'0:*-0 ,+ 7,( *,+= .1/-)6:*1/ .1/-0 6)/6:*1-' ,+ -7' 6,+I'<,/ J'3,/' -7'<
1/' :+*,1>'>8 K+ 0,4' 41+:316-:/)+= 1..*)61-),+0D -7' -)4' 1 .1/- 0.'+>0
,+ -7' 6,+I'<,/ 41< 71I' 1 *,('/ *)4)- @01<D 3,/ 6,,*)+= -7' .1/- >,(+H ,/
1+ :..'/ *)4)- @)3 -7' +'O- ./,6'00 ,+ -7' .1/- 4:0- J' .'/3,/4'> ()-7)+ 1
6'/-1)+ -)4' .'/),>H8 K3 -7' 6,+I'<,/ )0 0:JL'6- -, /1+>,4 0-,..1='0 @>:'
-, J/'1?>,(+0 ,/ -7' (,/?'/0 (7, +''> 1>>)-),+1* -)4'HD 1=1)+ ,+' 41<
+''> 0)4:*1-),+ -, 0-:>< )-0 .'/3,/41+6'8
!"#!"% \'3- 10 1+ 'O'/6)0' 3,/ -7' /'1>'/D (7, 41< ()07 -, -/< -7' 3,**,()+=
/'3'/'+6' 10 1 0-1/-)+= .,)+-Q
F1L1+D A8D C'/*'-7D C8D 1+> 9,40)6'?D ]8D 2],>'*)+= 41+:316-:/)+=
0<0-'40 :0)+= ]RG^B\RG 1+> AK]A9RF9NF # R -:-,/)1*D5 !VV"
&)+-'/ A)4:*1-),+ P,+3'/'+6' B/,6''>)+=0D G'( E/*'1+0D \RD ^AR8
!"#!"' 97'/' 1/' 41+< .1.'/0 ,+ -7)0 -,.)68 97' 3,**,()+= /'./'0'+-0 ,+*< 1
014.*' *)0- )>'+-)3)'> -7/,:=7 1 6,4.:-'/#J10'> ?'<(,/> 0'1/67Q
P7'(D N8B8 1+> c,7+0,+D \8R8D 2A'/I)6' *'I'*0 )+ >)0-/)J:-),+ 0<0-'40 ()-7
/1+>,4 6:0-,4'/ ,/>'/ 0)b'D5 *+,+- /0&0+'12 3"4%&#%1&5 d,*8 WSD G,8 !D
..8 [V#TYD !VVT8
_/1<D
R8 N8D e1/41/?1/D ^8
A8D 1+> A')>41++D
R8D 1..*)61-),+D5
2C'0)=+ 1+> ,.'/1-),+
,3
1+ ,/>'/#6,+0,*)>1-),+
(1/'7,:0'Q
4,>'*0 1+>
6)'"70+!
8")'!+- "9 :70'+#%"!+- /0&0+125 d,*8 TXD G,8 !D ..8!W#[YD !VVS8
\)++D F8c8 1+> f)'D f8D 2A)4:*1-),+ 1+1*<0)0 ,3 0';:'+6)+= /:*'0 3,/ RAFA
)+ 1 .:**#J10'> 100'4J*< 316)*)-<D5 ;!#0'!+#%"!+- 8")'!+- "9 <'"$)1#%"!
/0&0+'125 d,*8 [!D G,8 !"D ..8S[TT#S[Y$D !VV[8
F1+>71(1D A8 ^8 1+> A7/,33D F8D 2A)4:*1-),+#J10'> >'0)=+ 'I1*:1-),+ ,3
:+)- *,1> 1:-,41-'> 0-,/1='a/'-/)'I1* 0<0-'40D5 =">7)#0'& ? ;!$)&#'%+6!4%!00'%!45 d,*8 SXD G,8!D ..8 $!#$VD !VVT8
91?1?:(1D A8D 2B/'6)0' 4,>'*)+= 1+> 1+1*<0)0 ,3 *1/='#061*' RAaFAD5
B/,6''>)+=0 ,3 -7' !VVW &)+-'/ A)4:*1-),+ P,+3'/'+6'D g:'+1 d)0-1D M\D
^AR8
!"#X"
!"#!"$ 97'/' 1/' 41+< .1.'/0 ,+ -7)0 -,.)68 97' 3,**,()+= /'./'0'+-0 ,+*< 1
014.*' *)0- )>'+-)3)'> -7/,:=7 1 6,4.:-'/#J10'> ?'<(,/> 0'1/67Q
P1/>1/'**)D _8 1+> B'*1=1=='D B8 ]8D 2A)4:*1-),+ -,,* 3,/ >'0)=+ 1+>
41+1='4'+- ,.-)4)b1-),+ ,3 1:-,41-'> )+-'/J1< 41-'/)1* 71+>*)+= 1+>
0-,/1=' 0<0-'40 3,/ *1/=' (13'/ 31JD5 ;666 @'+!&+1#%"!& "!
A0>%1"!$)1#"' B+!)9+1#)'%!45 d,*8 XD G,8 !D ..8 WW#WVD !VVT8
P71+D M8 98A8D c1<1./1?107D g8D 1+> 91+=D G8 e8h8D 2C'0)=+ ,3 1:-,41-'>
6'**:*1/ 41+:316-:/)+= 0<0-'40 ()-7 0)4:*1-),+ 4,>'**)+=Q 1 610' 0-:><D5
;!#0'!+#%"!+8")'!+"9 =">7)#0' C77-%1+#%"!& %! @012!"-"4D5 d,*8 XD
G,8 !#SD ..8 !#!!D
!VVT8
h1>>,6?D c8D 2R:-,41-'> 0)4:*1-),+ 4,>'**)+= 1+> 1+1*<0)0 3,/
41+:316-:/)+= 0<0-'40D5 <'"$)1#%"! <-+!!%!4 +!$ ="!#'"-5 d,*8 YD G,8 WD
..8 [TS#[T$D !VVT8
h:;D i8D 2B/,6'00 ,/)'+-'> 41+:316-:/)+= 0<0-'4 0)4:*1-),+ -, 4'10:/'
-7' '33'6- ,3 07,. 6,+-/,* 316-,/0D5 A%>)-+#%"!5 d,*8 YSD G,8 WD ..8 S!X#
SSXD !VVW8
B/1?107D R8 1+> P7'+D ]8D 2A)4:*1-),+ 0-:>< ,3 3*'O)J*' 41+:316-:/)+=
0<0-'40D5 =">7)#0'& ? ;!$)&#'%+- 6!4%!00'%!45 d,*8 SXD G,8!D ..8 !V!#
!VVD !VVT8
PART FIVE
EVALUATING, SELECTING,
PREPARING, PRESENTING,
IMPLEMENTING, AND
MAINTAINING
Chapter 11
Evaluating and Selecting the
Facilities Plan
11-1
Answers to Questions and Problems at the End of Chapter 11
11.1 If equipment need change sooner than anticipated, leasing can provide more
flexibility than purchasing equipment. Depending on the lease conditions,
lease payments can be treated as annual expenses and, thus, yield tax
advantages over purchased equipment. (Lease payments can be expensed;
depreciation deductions are allowed with purchased equipment.) Lease
payments might be “affordable” in a manager’s annual operating budget;
capital investments might require higher level approval and, thus, introduce
more uncertainty in the selection process. Leases minimize the uncertainty in
maintenance cash flows resulting from the use of capital equipment.
When equipment is purchased, risks of escalating lease charges in the future
are avoided. Purchasing has the advantage of ownership of the asset; if
owned, the equipment can be modified. Because of different types of
flexibility, purchasing the equipment can provide certain types of flexibility
not provided by leasing; for example, if the equipment is purchased it can be
relocated to another plant location outside the service region of the leasing
company) if changes in equipment needs occur.
11.2 Manufacturing space is usually more expensive than storage space. As
manufacturing needs expand, storage space is often converted to
manufacturing space. If sufficient land is not available for expansion of
manufacturing and storage, then storage space is often sought off-site. Using
either leased space, public warehousing, or another option, off-site storage is
provided.
On-site storage of materials and supplies has the advantage of proximity. As
a result, the time required to respond to demands for materials and supplies is
minimized; likewise, the amount of inventory required is generally less than
that required with off-site storage. On-site storage has the advantage of
ensuring that storage personnel are made to feel a part of the overall team.
Often when storage is assigned to off-site space, morale problems arise; if
storage personnel are split between on-site and off-site locations, the quality
of management often suffers. When off-site storage is used, delivery trucks
are required; likewise, material must be loaded and unloaded several times in
the process of satisfying the ultimate needs of manufacturing.
Off-site storage has the advantage of being dedicated to storage activities.
When located on-site, manufacturing pressures can result in a reactive mode
of operation, instead of a planned, responsive mode that can be provided by
off-site storage. Since on-site storage is close to manufacturing, the discipline
11-2
needed to ensure a carefully managed activity is often displaced by a brushfire, crisis management style. Because material is stored off-site, it
necessitates that manufacturing managers plan daily activities.
11.3 High-rise warehouses can be significantly less expensive to construct in terms
of cubic footage of storage space than conventional storage facilities;
likewise, when land is expensive, less land is needed for high-rise storage.
Conventional storage has the advantage of flexibility. In a conventional
facility, conventional and automated storage and retrieval equipment can be
used; whereas, in a high-rise facility it is less likely that conventional
equipment can be used. Likewise, if the need for the storage facility declines
over time, alternative uses of a conventional facility might arise; such is not
likely with high-rise storage. If the facility is to be sold, fewer buyers are
likely for a high-rise facility than a conventional facility.
11.4 Depending on the application, many factors should be considered in
evaluating horizontal handling alternatives. Among the various factors to be
considered are the following: discounted present worth; installation time;
discounted payback period; maintenance requirements; flexibility in coping
with changes in throughput requirements; flexibility in accommodating
changes in load sizes and weights; labor skills required; in-process inventory
buffer sizes required; attitude of plant floor personnel toward the
alternatives; previous experience in using the equipment alternatives.
11.5 Depending on the application, many factors should be considered in
evaluating alternate site locations for corporate headquarters. Among the
various factors to be considered are the following: proximity to major
customers; proximity to major operating locations; proximity to major capital
sources; ease of access by customers, suppliers, members of the Board of
Directors; location preferences of upper management; tax incentives and
economic consequences; potential for expansion; access to major airport;
compatibility with image to be projected by the firm; quality of educational
facilities in the city or area; availability of accommodations for visitors;
quality and magnitude labor supply; attractiveness in recruiting mid to upper
management; and quality of life for employees.
11.6 Depending on the application, many factors should be considered in
evaluating alternate site locations for a distribution center. Among the
various factors
be considered
are the
following:
proximity
to to
major
supply
locations
(time to
required
to replenish
from
suppliers);
proximity
major
customer locations (time required to ship to customers); access to major
transportation facilities (rail, air freight, interstate highways, river and ocean
11-3
terminals); number of distribution centers and service regions of each;
availability of existing facilities; cost to have a leased facility built to
specifications; labor availability, labor cost, and attitude of organized labor;
locations of competitors (ability to meet or exceed competition in terms of
service to customers); and overall economic impact of the location decision,
including taxes.
11.7 From the Excel results shown above, Alternative B is preferred.
11.8 This question is intended to ascertain the value of annual operating expenses
for B that yields a net present value identical to A. Shown on the following
page is a spreadsheet that makes the NPV values nearly equal (they differ by
$0.04). The break-even value for annual operating expenses for alternative B
is $369,803.69. Hence, the annual operating cost for B can increase by 23.3
percent before A will have a larger net present value.
11.9 The Excel spreadsheet for this problem is shown on the following page. The
net present value for the problem, as stated originally, is $3,042,149.90.
Hence, if the annual savings and salvage value remain as projected, the initial
investment can increase by $3,042,149.90 and still be economically viable.
Likewise, if the initial investment is as projected then the annual savings can
be reduced to $ 1,254,904.11 without making the investment unattractive
economically. Finally, if the initial investment can increases by 16.22 percent
and the annual savings can decrease by a like percentage, the investment will
be attractive economically.
11-4
11.10 True
11.11 False; the alternative preferred when using the payback period method is
seldom the one that is preferred when using discounted cash flow methods.
11.12 True
11.13 False; it is not always the lowest cost alternative
11.14 Net Investment $3,842
Cost of Capital
Cost of Capital
NOPAT
EVA
11.15
Cost of Capital
NOPAT
EVA
10%
0.1($3,842) = $384.20
$364
$364.00 - $384.20 = -$20.20
FirmC
FirmD
(0.1)($100) = $12
(0.1)($200) = $20
$25-$10=$15
$40-$16=$24
$15-$12=$3
$24-$20=$4
Firm C has the greatest EVA.
Note: a number of different approaches are used by firms in computing EVA
values.
Fortraining
example,
somecalculations
firms incorporate
amortized
values
fornot
R&D
employee
in their
of capital
invested.
So, do
be and
disturbed if your calculations differ from those given in annual reports to
stockholders.
11-5
11.16
WEIGHTED FACTOR COMPARISON FORM
A
Factor
Wt.
Rt
B
Sc
Rt
C
Sc
Rt
Sc
1.NetPresentValue
40
10
400
8
320
6
240
2. TimetoFillCustomer’sOrder
35
7
245
10
350
8
280
3.Flexibility
Totals
25
100
5
125
770
7
175
845
10
250
770
Alternative B would be preferred. Notice that Alternatives A and C are tied
in preference.
11.17 a.
WEIGHTED FACTOR COMPARISON FORM
A
Factor
B
C
Wt.
Rt
Sc
Rt
Sc
Rt
Sc
1.NetPresentValue
50
10
500
8
400
6
300
2. TimetoFillCustomer’sOrder
10
7
70
10
100
8
80
3.Flexibility
40
Totals
5
200
100
7
280
770
10
400
780
780
Alternatives B and C are equally preferred. However, Alternatives A has
almost the same weighted factor total as B and C. This looks like a “coin
toss” will be needed for a decision to be made!
b.
WEIGHTED FACTOR COMPARISON FORM
A
Factor
Wt.
Rt
B
Sc
Rt
C
Sc
Rt
Sc
1.NetPresentValue
40
10
400
8
320
6
240
2. TimetoFillCustomer’sOrder
35
5
175
7
245
10
350
3.Flexibility
Totals
Alternative C is preferred.
25
100
5
125
700
7
175
740
10
250
840
11-6
c.
WEIGHTED FACTOR COMPARISON FORM
A
Factor
Wt.
Rt
B
Sc
Rt
C
Sc
Rt
Sc
1.NetPresentValue
30
10
300
8
240
6
180
2. TimetoFillCustomer’sOrder
20
7
140
10
200
8
160
3.
Flexibility
Safety
4.
10
40
Totals
5
10
50
400
100
7
8
70
320
890
10
9
100
360
830
800
Alternative A would be preferred.
11.18
WEIGHTED FACTOR COMPARISON FORM
A
Factor
Wt.
1. Annual material handling
cost
2.Constructioncost
3.Easeofexpansion
4.Employee’spreference
5.Railsidingaccess
Totals
Rt
20
8
7
20
20
Sc
10
25
15
B
Rt
200
6
10
100
Alternative A would be preferred.
4
8
120
200
Rt
140
100
120
8
6
825
Sc
7
200
105
C
10
120
120
Rt
100
9
9
140
100
10
7
740
Sc
2
250
150
7
5
600
Sc
5
10
D
40
225
135
200
140
740
11-7
11.19
WEIGHTED FACTOR COMPARISON FORM
A
Factor
Wt.
1.Annualmaterialhandlingcost
2.Constructioncost
15
10
25
3.Easeofexpansion
4.Employee’spreference
15
30
5.Railsiding
access
15
Sc
150
Rt
7
8
200
105
180
8
6
150
6
7
10
Totals
BC
Rt
100
D
Sc
105
4
8
100
Rt
Sc
5
75
10
250
2
120
10
150
240
7
210
90
785
5
75
655
Rt
Sc
30
9
225
9
135
10
300
7
105
740
740
Alternative A would be preferred.
11.20 a.
WEIGHTED FACTOR COMPARISON FORM
A
Factor
Rt
1.Annualmaterialhandlingcost
2.Constructioncost
3.Easeofexpansion
4.Employee’spreference
5.Railsidingaccess
Totals
BC
D
Wt.
20
8
25
8
15
7
20
20
10
10
100
Alternative A would be preferred.
Sc
160
Rt
10
200
105
4
8
200
200
865
Rt
2
100
120
8
6
Sc
200
9
10
160
120
700
Rt
5
225
150
7
5
Sc
40
10
9
140
100
630
250
135
6
8
Sc
100
120
160
765
11-8
b.
WEIGHTED FACTOR COMPARISON FORM
A
Factor
Wt.
1.Annualmaterialhandlingcost
2.Constructioncost
BC
Rt
15
8
Sc
120
10
25
8
200
3.Easeofexpansion
4.Employee’spreference
15
30
7
10
105
300
8
5.Railsiding
access
15
150
6
10
Totals
100
D
Rt
Sc
150
4
8
Rt
2
100
9
Sc
90
5
Sc
30
5
75
225
10
250
120
10
150
240
7
210
875
Rt
75
9
8
700
135
180
6
120
690
760
Alternative A would be preferred.
11.21 a.
WEIGHTED FACTOR COMPARISON FORM
A
Factor
1.Annualmaterialhandlingcost
2.Constructioncost
3.Easeofexpansion
4.Employee’spreference
5.Railsidingaccess
Totals
Wt.
Rt
20
7
25
8
15
7
20
20
10
10
100
Alternative A would be preferred.
BC
Sc
140
Rt
10
200
105
200
7
Rt
6
175
120
8
6
845
Sc
200
8
200
D
9
10
160
120
7
Rt
5
225
150
7
775
Sc
120
10
9
140
100
250
135
9
8
735
Sc
100
180
160
825
11-9
b.
WEIGHTED FACTOR COMPARISON FORM
A
Factor
1.Annualmaterialhandlingcost
2.Constructioncost
Wt.
15
7
25
3.Easeofexpansion
4.Employee’spreference
15
30
5.Railsiding
access
15
Totals
BC
Rt
Sc
105
8
10
200
7
105
10
300
10
100
150
Sc
150
7
8
8
6
860
D
Rt
Rt
6
175
9
Sc
5
775
Sc
90
5
75
225
10
250
120
10
150
240
7
210
90
Rt
75
9
9
8
750
135
180
120
850
Alternative A would be preferred.
11.22 The weighted factor comparison method is easy to use. Sensitivity analyses
can be performed quite easily, particularly if a computer spreadsheet is used
to perform the calculations. However, its qualitative and subjective aspects
are a source of concern. Scaling issues exist with the method. Ordinal data
cannot be treated as ratio data without the potential for errors. Hence,
caution should be used when applying the method.
Chapter 12
Preparing, Presenting
Implementing and
Maintaining
Facilities Plans
12-1
Answers to Questions at the End of Chapter 12
Due to the nature of the questions at the end of Chapter 12, answers will not be
provided for all.In those cases where responses are provided, it is likely that
other, equally appropriate, answers can be devised. The purpose in providing
responses is not to providethe answer, but to provide an answer that will
stimulate class discussion.
12.1
This is an open-ended question; response depends on course location.
12.2
Aesthetics
are greatly
important
facilities
planning,
users’
attitudes
will
be influenced
by in
aesthetic
aspects
of thesince
facility.
Just
as it would
be inappropriate to plan a facility solely on the basis of aesthetics, so is it
inappropriate to ignore aesthetics in facilities planning.
12.3
Computer-aided layout models can be used in regional and city planning.
Computer-aided layout models can be used to determine the locations of
the facility relative to the locations of the airport, rail station,
transportation hubs, distribution centers, manufacturing plans, and
customers. Although some computer-aided layout models are more
amenable than others to performing this type analysis, they are not
restricted to use within a facility.
12.4
2-D layouts are superior to 3-D layouts when considering horizontal
handling alternatives, since the vertical dimension is not a factor. 3-D
layouts can be difficult to evaluate when mezzanines and other vertical
obstructions are in place.
In assigning bins along an aisle, only the length of the aisle and the height
of the storage face need to be represented; the depth of the storage faces
will not be a factor, since they will be uniform.
12.5
Three alternatives will be considered: sketches, templates, and computer
graphics. Sketches can be generated spontaneously on-site or in meetings
with users, planners, and managers; they are easy to generate, requiring no
special equipment.In general, sketches are best used in preliminary
discussions and lend themselves to rough-cut analyses.
Templates are useful in representing draft or final layout plans.
They can
be reproduced and circulated to appropriate parties for feedback; they can
be marked up by operating and other personnel who might not have access
to or familiarity with computer graphics.
Computer-generated layouts require both hardware and software
capability. In general, computer graphics would be the preferred method of
generating draft and final layout plans, since changes can be made to
12-2
12.5 (continued )
the layout easily, solutions can be transferred electronically to other
locations, and there is no need to store large Mylar or paper
representations of the layout.
12.6
3-D approaches are useful when the third dimension plays a role in the
facility plan. For example, when comparing an overhead monorail with an
AGV, it is important to visualize the vertical dimension.Likewise, in
comparing
conventional
storage
withcan
high-rise
storage, having
a 3-D
representation
of the storage
facility
prove beneficial
in
communicating the benefits (and limitations) of high-rise storage.
In terms of the various options that can be used to generate 3-D
representations of the facility plan (block models, scale models, and
computer models), block models are useful in preliminary design,
particularly when no decision has been made as to the specific model of a
machine or even type of machine that will be assigned to a particular
location; block models, in this case, might be used to indicate the general
location of a machine.
Scale models would be used at the point that machine selections have been
made and final locations are being determined. The use of scale models
has tended to decline in the past decade, due to the cost of acquiring and
maintaining the 3-D scale model.
Computer models are gaining in popularity because of the ease of
modifications of both the layout and the 3-D representation of the
individual machines.Of course, 3-D computer models are still being seen
on a 2-D monitor.The power of the 3-D computer model is that it can be
rotated and viewed at any angle of interest to the facility planner.
Although
currently the cost of entering the computer model arena can be prohibitive
for relatively small and infrequent facilities planning activities, it is
anticipated that the power of 3-D computer models will increase and
prices will decrease in the future to the point that any serious facilities
planning will occur using computer models.
12.7
Currently, the integration of computer graphics with computer-aided
layout techniques is accomplished by a human interface. It would be
useful to have a software package that combines the best features of
descriptive and prescriptive computer layout software. As noted in Chapter
6, efforts are under way to develop software packages that incorporate
expert models with computer graphics capabilities.
12.7 (continued )
Until such time as the integration occurs in a software package, the human
designer will be required to take the output from a computer-aided layout
technique and input it into a computer graphics package.The CAL model
can stimulate the human designer to represent via computer graphics
various alternative layouts.
12.8
Block templates or models should be used when the time and expense of
generating contour templates and models is not justified.Regardless of
which
approachDon’t
is taken,
it is important
the facilities
to bethe
viewed
as professional.
be penny-wise
andfor
dollar-foolish
in plan
choosing
method of depicting the facilities plan. At the same time, the investment
required to “sell” the facilities plan should be justified in the same way
other investments are justified.If the perceived benefits of using more
refined models do not off-set the costs of using them, then “keep it
simple!” The choice depends on the application and the audience.
12.9
Five additional reasons for people resisting change.
a. previous changes did not produce the promised benefits
b. a history of management by the “fad of the month,” causing skepticism
regarding any changes
c. WIIFM factor (“what’s in it for me?”) has not been addressed
d. those being asked to change were the authors of the current methods
e. a short-term reward system is being used, and the proposed change
will not produce measurable benefits for several years
12.10 Five additional ways to overcome resistance to change.
a. show video tapes and films of applications of the proposed change in
other locations
b. take key employees on tours of facilities where the changes have been
implemented successfully
c. implement the change ina pilot plant setting to gain experience in a
limited setting
d. begin selling change at the start of the project, rather than at the end
e. involve at the outset of the planning process those whom you believe
will be most resistant to the change; it is particularly important to
involve hourly personnel and first- and second-level supervisors in the
early stages of the planning process.
Also, if an environment of trust is created such that employees are not
afraid they will lose their jobs or employment, then resistance will be
much less. If at all possible, ensure that the changes will be made to
improve the quality of life of the personnel affected by the change.
12.11 The response depends on the manufacturing facility chosen.
12-4
12.12 The response depends on the non- manufacturing facility chosen.
12.13 The response depends on the university.
12.14 A multimedia presentation can embody animation with simulation, as well
as film clips of the various equipment that will be used in the facility.
Depending on the particular application, alternative paths can be provided
to the viewer in addressing question regarding the facilities plan.We can
visualize the use of multimedia in incorporating computer-aided layout
techniques with computer graphics models of the facility.
12.15 Improvements in virtual reality and holographic technologies are expected
to transform the process of facilities planning. Improvements in computing
power and speeds will be such that massive optimization problems will be
solved in the facilities planning process.
In the future, rather than plan a large, complex facility, it is likely that the
plan will be for a large, complex virtual facility.With improved
communication, proximity is likely to be less of a factor in facilities
planning; functions will be distributed throughout the world.