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Department of Mechanical Engineering, New Horizon college
of Engineering, Bengaluru
MEE64: Automation
Engineering
Module 1A: Computer Integrated Manufacturing
Compiled by: Somashekar S M, Assistant Professor
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 1
Module1A: Computer Integrated Manufacturing Systems:
๏ƒ˜ Introduction,
๏ƒ˜ Automation definition,
๏ƒ˜ Types of Automation,
๏ƒ˜ Reasons or Advantages of Automation,
๏ƒ˜ Disadvantages of Automation,
๏ƒ˜ Automation Strategies,
๏ƒ˜ Types of Production Systems,
๏ƒ˜ CIM,
๏ƒ˜ Information Processing Cycle in Manufacturing,
๏ƒ˜ Production concepts and Mathematical Models-Manufacturing Lead Time,
Production Rate, Components of Operation Production Time, Production Capacity,
Utilization and Availability, Work-In-Process, WIP Ratio, TIP Ratio, Problems Using
Mathematical Model Equations.
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 2
1.1 INTRODUCTION:
The systems aspects of manufacturing are more important than ever today. The word
manufacturing was originally derived from two Latin words, manus (hand) and factus
(make), so that the combination means made by hand. This way the manufacturing was
accomplished when the word first appeared in the English language around 1567.
Commercial goods of those times were made by hand. The methods were handicraft
accomplished in small shops, and the goods were relatively simple, at least by todayโ€™s
standards. As many years passed, factories were developed, with many workers at a single
site, and the work had to be organized using machines rather than handicraft techniques.
The products become more complex, and so did the processes. Workers had to specialize
in their tasks. Rather than overseeing the fabrication of the entire product, they were
responsible for only a small part of the total work. More up-front planning was required,
and more coordination of the operations was needed to keep track of progress in the
factories. The systems of production, which rely on many separate but interacting
functions, were evolving.
Today the system of production is indispensable in manufacturing. Modern
manufacturing systems must cope with the economic realities of the modern world. These
realities include the following:
๏ƒ˜ Globalization: Once underdeveloped countries like China, India, and Mexico are
becoming major players in manufacturing, due largely to their high population and
low labor costs. Other regions of the world with low labor costs include Latin
America, Eastern Europe, and Southeast Asia, and the countries in these regions
have also become important suppliers of manufactured goods.
๏ƒ˜ International Outsourcing: Parts and products once made in the United States by
American companies are now being made offshore (overseas, so that cargo ships are
required to deliver the items) or near-shore (in Mexico or Central America, so that
rail and truck deliveries are possible). In general, international outsourcing means
loss of job in US.
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6th Semester BE, Mechanical, NHCE Page 3
๏ƒ˜ Local outsourcing: Companies can also outsource by using suppliers within the
US, reason why companies elect local outsourcing include: 1) benefits from using
suppliers that specialize in certain production technologies, 2) lower labor rates in
small companies, and 3) limitations of available in-house manufacturing
capabilities.
๏ƒ˜ Contract Manufacturing: This refers to companies that specialize in
manufacturing entire products not just parts, under contract to other companies.
Contract manufacturers specialize in efficient production techniques freeing their
customers to specialize in the design and marketing of the products.
๏ƒ˜ Trend towards the service sector in the US economy: There has been gradual
erosion of direct labor jobs in manufacturing while jobs in service industries (e.g.,
healthcare, food services, retail) have increased in numbers.
๏ƒ˜ Quality expectations: Customers, both consumer and corporate, demand that the
products they purchase are of the highest quality. Perfect quality is the expectation.
๏ƒ˜ The need for operational efficiency: To be successful, US manufacturers must
be efficient in their operations to overcome the labor cost advantage enjoyed by their
international competitors.
In order to improve efficiency, increase productivity and quality, the following approaches
and technologies are in use.
๏ƒ˜ Automation: The use of automated equipment compensates for the labor cost
disadvantage relative to international competitors. Automation reduces labor costs,
decrease production cycle times, and increase product quality and consistency.
๏ƒ˜ Material handling technologies: Manufacturing usually involves a sequence of
activities performed at different locations in the plant. The work must be
transported, stored and tracked as it moves through the plant.
๏ƒ˜ Manufacturing systems: These involve the integration and coordination of multiple
automated and/or manual workstations through the use of material handling
technologies to achieve a synergistic effect compared to the independent operations
of individual workstations. Ex: production lines, manufacturing cells and automated
assembly systems.
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 4
๏ƒ˜ Flexible manufacturing systems: Much of the outsourcing to international
competitors, such as China and Mexico, has involved high-volume goods production.
Flexibility in manufacturing allows US manufacturers to compete effectively in the
low volume/high-mix product categories.
๏ƒ˜ Computer integrated manufacturing (CIM): The technologies include computer
aided design (CAD), computer aided manufacturing (CAM), and computer networks
to integrate manufacturing and logistics operations.
๏ƒ˜ Lean production: Accomplishing more work with fewer resources is the general goal
of lean production. which involves techniques to increase labor productivity and
operational efficiency.
1.2 AUTOMATION: Automation is a technology concerned with the application of
mechanical, electronic, and computer-based systems to operate land control production.
This technology includes:
๏‚ฎ Automatic machine tools to process parts
๏‚ฎ Automatic assembly machines
๏‚ฎ Industrial robots
๏‚ฎ Automatic material handling and storage systems
๏‚ฎ Automatic inspection systems for quality control
๏‚ฎ Feedback control and computer process control
๏‚ฎ Computer systems for planning, data collection and decision making to support
manufacturing activities.
1.3 TYPES OF AUTOMATION: Automated manufacturing systems can be classified into
three basic types: 1. Fixed automation 2. Programmable automation 3.
Flexible automation
1.3.1. Fixed automation:
๏ƒ˜ It is a system in which the sequence of processing (or assembly) operations is fixed
by the equipment configuration.
๏ƒ˜ The operations in the sequence are usually simple.
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6th Semester BE, Mechanical, NHCE Page 5
๏ƒ˜ It is the integration and coordination of many such operations into one piece of
equipment that makes the system complex.
๏ƒ˜ The typical features of fixed automation are:
๏‚ฎ High initial investment for custom-engineered equipment
๏‚ฎ High production rates
๏‚ฎ Relatively inflexible in accommodating product changes
๏ƒ˜ The economic justification for fixed automation is found in products with very high
demand rates and volume.
๏ƒ˜ The high initial cost of the equipment is spread over a very large number of units,
thus making the unit cost attractive compared to alternative methods of production.
Examples: Mechanized assembly lines โ€“ product moved along mechanized
conveyors
1.3.2. Programmable automation:
๏ƒ˜ The production equipment is designed with the capability to change the sequence of
operations to accommodate different product configurations.
๏ƒ˜ The operation sequence is controlled by a program, which is a set of instructions
coded so that the system can read and interpret them.
๏ƒ˜ The features that characterize programmable automation include:
๏‚ฎ High investment in general-purpose equipment
๏‚ฎ Low production rates relative to fixed automation
๏‚ฎ Flexibility to deal with changes in product configuration
๏‚ฎ Most suitable for batch production
๏ƒ˜ Automated production systems which are programmable are used in low and
medium volume production.
๏ƒ˜ The parts or products are typically made in batches.
๏ƒ˜ To produce each new batch of different product, the system must be reprogrammed
correspond to new product.
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6th Semester BE, Mechanical, NHCE Page 6
๏ƒ˜ The physical setup of the machine must be changed over: Tools must be loaded,
fixture must be attached to the machine table, and the required machine settings
must be entered. This changeover takes time.
๏ƒ˜ Thus, the typical cycle for a given product includes a period for the setup and
reprogramming, followed by a period in which the batch is produced.
๏ƒ˜ Example: Numerically controlled machine tools and industrial robots,
Machining transfer lines.
1.3.3. Flexible automation:
๏ƒ˜ It is the extension of programmable automation.
๏ƒ˜ A flexible automated system is one that is capable of producing a variety of products
(or parts) with virtually no time lost for changeovers form one product to the next.
๏ƒ˜ There is no production time lost while reprogramming the system and altering the
physical setup (tooling, fixtures, machine settings).
๏ƒ˜ Thus, the system can produce various combinations and schedules of products,
instead of requiring that they be made in separate batches.
๏ƒ˜ The features of flexible automation include:
๏‚ฎ High investment for a custom-engineered system
๏‚ฎ Continuous production of variable mixtures of products
๏‚ฎ Medium production rates
๏‚ฎ Flexibility to deal with product design variations.
๏ƒ˜ The essential features that distinguish flexible automation form programmable
automation are
1. The capacity to change part programs with no loss of production time (the part
programs are prepared off-line on a computer system and electronically transmitted
to automated production system).
2. The capability to change over the physical system to continue production without
the downtime between batches (the physical setup is changeover off-line and then
moving it into place simultaneously as the next part comes into position for
processing, the use of pallet fixtures).
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 7
1.4 REASONS OR ADAVANTAGES OF AUTOMATION:
Companies undertake projects in manufacturing automation and computer
integrated manufacturing for a variety of good reasons. Some of the reasons used to justify
automation are the following:
๏ƒ˜ To increase labor productivity: Automating a manufacturing operation usually
increases production rate and labor productivity. This means greater output per
hour of labor input.
๏ƒ˜ To reduce labor cost: Ever-increasing labor cost continues to be the trend in the
worldโ€™s industrialized societies. Consequently, higher investment in automation has
become commercially justifiable to replace manual operations. Machines are
increasingly being substituted for human labor to introduce unit product cost.
๏ƒ˜ To mitigate the effects of labor shortages: There is general shortage if labor in
many advanced nations, and this has stimulated the development of automated
operations as substitute for labor.
๏ƒ˜ To improve worker safety: Automating a given operation and transferring the
worker from active participation in the process to a monitoring role, or removing the
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6th Semester BE, Mechanical, NHCE Page 8
worker from the operation altogether, makes the work safer. The safety and physical
well being of the worker has become a national objective with the enactment of the
Occupational Safety and Healthy Act (OSHA) in 1970. This has provided an impetus
for automation.
๏ƒ˜ To improve product quality: Automation is not only results in higher production
rates than manual operation, it also performs the manufacturing process with a
greater uniformity and conformity to quality specifications.
๏ƒ˜ To reduce manufacturing lead time: Automation helps reduce the elapsed time
between customer order and product delivery, providing a competitive advantage to
the manufacturer for future orders. By reducing manufacturing lead time, the
manufacturer also reduces work-in process (semi-finished parts) inventory.
๏ƒ˜ To accomplish processes that cannot be done manually: Certain operations
cannot be accomplished without the aid of a machine. These processes require
precision, miniaturization, or complexity of geometry that cannot be achieved
manually. Ex: IC (Integrated Circuit) chips fabrication operations, Rapid prototyping
processes based on computer graphics (CAD) models, and the machining of complex,
mathematically defined surfaces using computer numerical control. These processes
can only be realized by computer controlled systems.
๏ƒ˜ To avoid the high cost of not automating: There is a significant competitive
advantage gained in automating a manufacturing plant. The advantage cannot easily
be demonstrated on a companyโ€™s project authorization form. The benefits of
automation often shown up in unexpected and intangible ways, such as improved
quality, higher sales, better labor relations, and better company image. Companies
that do not automate are likely to find themselves at a competitive disadvantage with
their customers, their employees, and the general public.
1.5 DISADAVANTAGES OF AUTOMATION:
๏ƒ˜ High initial investment
๏ƒ˜ Retrenchment or unfavorable for employees
๏ƒ˜ Difficult to implement in old production systems.
๏ƒ˜ Not economically justifiable for small scale industries (SSI).
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6th Semester BE, Mechanical, NHCE Page 9
1.6 STRATEGIES FOR AUTOMATION AND PRODUCTION SYSTEMS
If automation seems a feasible solution to improving productivity, quality, or other
measure of performance, then the following ten strategies provide a road map to search for
these improvements.
๏ƒ˜ Specialization of operations
๏ƒ˜ Combined operations
๏ƒ˜ Simultaneous operations
๏ƒ˜ Integrations of operations
๏ƒ˜ Increased flexibility
๏ƒ˜ Improved material handling and storage
๏ƒ˜ Online inspection
๏ƒ˜ Process control and optimization
๏ƒ˜ Plant operations and control
๏ƒ˜ Computer integrated manufacturing (CIM)
๏ƒ˜ Specialized Operations:
๏‚ฎ Some special purpose machines or equipmentโ€™s are used to perform particular
operation with the highest possible efficiency.
๏‚ฎ This is employed to utilize the specialization of labour and improve labour
productivity.
๏ƒ˜ Combined Operations:
๏‚ฎ Production of parts is obtained in a sequence of operations. Several processing steps
are required for complex parts.
๏‚ฎ Combined operations eliminate or reduce the processing steps.
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6th Semester BE, Mechanical, NHCE Page 10
๏‚ฎ This is accomplished by performing more than one operation at a given machine
thereby reducing the number of separate machines needed.
๏ƒ˜ Simultaneous Operations:
๏‚ฎ The operations combined in one particular workstation and are carried out
simultaneously to reduce the total processing time.
๏‚ฎ This will result in greater increase in productivity.
๏ƒ˜ Integration of Operations:
๏‚ฎ Integration means linking of several workstations together into a single integrated
mechanism to transfer parts between them by automatic material handling system.
๏‚ฎ The products are easily scheduled and several parts are processed at a time, thereby
resulting in greater increase in overall output of the system.
๏ƒ˜ Increased Flexibility:
๏‚ฎ This strategy attempts to achieve maximum utilization of equipment for producing
variety of parts or products.
๏‚ฎ Prime objectives are to reduce setup time and programming time for production
machine.
๏ƒ˜ Improved Material Handling and Storage
๏‚ฎ Automated material handling is adopted for reducing non-productive time.
๏‚ฎ Benefits include reduced work in process & shorter manufacturing lead time
๏ƒ˜ Online Inspection:
๏‚ฎ Online inspection means that poor quality product is inspected regularly to eliminate
scrap & overall quality can be increased.
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6th Semester BE, Mechanical, NHCE Page 11
๏ƒ˜ Process Control and Optimization:
๏‚ฎ This includes a wide range of control schemes intended to operate the individual
processes and associated equipment more efficiently.
๏‚ฎ Individual processing time is reduced and Product quality can also be improved.
๏ƒ˜ Plant Operations Control:
๏‚ฎ This strategy attempts to manage and coordinate the aggregate operations in the
plant.
๏‚ฎ A high-level computer networking may be incorporated to do this task.
๏ƒ˜ Computer Integrated Manufacturing (CIM):
๏‚ฎ This is the logical extension of one level higher than the earlier.
๏‚ฎ Engineering design and business step of the firm are integrated in this strategy.
๏‚ฎ CIM involves extensive use of computer applications, databases and networking
throughout the enterprise.
1.7 PRODUCTION SYSTEM:
The production system of an organization is that part, which produces products of
an organization. It is that activity whereby resources, flowing within a defined system, are
combined and transformed in a controlled manner to add value in accordance with the
policies communicated by management.
The production system has the following characteristics:
๏‚ฎ Production is an organized activity, so every production system has an objective.
๏‚ฎ The system transforms the various inputs to useful outputs.
๏‚ฎ It does not operate in isolation from the other organization system.
๏‚ฎ There exists a feedback about the activities, which is essential to control and improve
system performance.
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6th Semester BE, Mechanical, NHCE Page 12
1.7.1 Classification/Types of Production systems: Production systems can be classified
as Job Shop, Batch, and Mass Production systems.
๏‚ฎ Job shop Production:
๏‚ท Job shop productions are characterized by manufacturing of one or few quantities
of products designed and produced as per the specification of customers
(customized) within prefixed time and cost.
๏‚ท The distinguishing feature of this is low volume and high variety of products.
Mass Production
Job Shop
Batch Production
Production Quantity
Production Rate
Labor Skill level
Special Tooling
Process Plant Layout Product
flow
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6th Semester BE, Mechanical, NHCE Page 13
Characteristics
๏‚ท High variety of products and low volume.
๏‚ท Use of general purpose machines and facilities.
๏‚ท Highly skilled operators who can take up each job as a challenge because of
uniqueness.
๏‚ท Large inventory of materials, tools, parts.
๏‚ท Detailed planning is essential for sequencing the requirements of each product,
capacities for each work center and order priorities.
Advantages
๏‚ท Because of general purpose machines and facilities variety of products can be
produced.
๏‚ท Operators will become more skilled and competent, as each job gives them learning
opportunities.
๏‚ท Full potential of operators can be utilized.
๏‚ท Opportunity exists for creative methods and innovative ideas.
Limitations
๏‚ท Higher cost due to frequent set up changes.
๏‚ท Higher level of inventory at all levels and hence higher inventory cost.
๏‚ท Production planning is complicated.
๏‚ท Larger space requirements.
Example: space vehicles, aircrafts, machine tools, special tools and equipment and
prototypes of future products.
๏‚ฎ Batch Production
๏‚ท Batch production is defined by American Production and Inventory Control Society
(APICS) โ€œas a form of manufacturing in which the job passes through the
functional departments in lots or batches and each lot may have a different
routing*.โ€
๏‚ท It is characterized by the manufacture of limited number of products produced at
regular intervals and stocked awaiting sales.
๏‚ท As in job order production, the batch order production can be of the following three
types,
o A batch produced only once
o A batch produced repeatedly at irregular intervals when need arises.
o A batch produced periodically at known intervals to satisfy continuous demands.
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6th Semester BE, Mechanical, NHCE Page 14
*routing: Routing means determination of the path to be followed by each part/component
being transformed from input /raw material into final product.
Characteristics
Batch production system is used under the following circumstances:
๏‚ท When there are shorter production runs.
๏‚ท When plant and machinery are flexible.
๏‚ท When plant and machinery set up is used for the production of item in a batch and
change of set up is required for processing the next batch.
๏‚ท When manufacturing lead time and cost are lower as compared to job order
production.
Advantages
๏‚ท Better utilization of plant and machinery.
๏‚ท Promotes functional specialization.
๏‚ท Cost per unit is lower as compared to job order production.
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6th Semester BE, Mechanical, NHCE Page 15
๏‚ท Lower investment in plant and machinery.
๏‚ท Flexibility to accommodate and process number of products.
๏‚ท Job satisfaction exists for operators.
Limitations
๏‚ท Material handling is complex because of irregular and longer flows.
๏‚ท Production planning and control is complex.
๏‚ท Work in process inventory is higher compared to continuous production.
๏‚ท Higher set up costs due to frequent changes in set up.
Examples:
๏‚ท Turret lathes capable of holding several cutting tools are used rather than engine
lathe.
๏‚ท The machine tools used are combined with specially designed jigs and fixtures which
increase the output rate.
๏‚ท Batch production plants include machine shops, casing foundries, plastic molding
factories, press working shops and some types of chemical plants.
๏‚ท Industrial equipment, furniture, textbooks, house hold appliances, lawn movers and
components parts for many assembled consumer products.
๏‚ฎ Mass Production
๏‚ท Manufacture of discrete parts or assemblies using a continuous process are called mass
production.
๏‚ท This production system is justified by very large volume of production. The machines
are arranged in a line or product layout. Product and process standardization exists
and all outputs follow the same path.
๏‚ท The mass production can be distinguished into two categories:
1. Quantity production
2. Flow production
Characteristics
๏‚ท Standardization of product and process sequence.
๏‚ท Dedicated special purpose machines having higher production capacities and output
rates.
๏‚ท Large volume of products.
๏‚ท Shorter cycle time of production.
๏‚ท Lower in process inventory.
๏‚ท Perfectly balanced production lines.
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6th Semester BE, Mechanical, NHCE Page 16
๏‚ท Flow of materials, components and parts is continuous and without any back
tracking.
๏‚ท Production planning and control is easy.
๏‚ท Material handling can be completely automatic.
Advantages
๏‚ท Higher rate of production with reduced cycle time.
๏‚ท Higher capacity utilization due to line balancing.
๏‚ท Less skilled operators are required.
๏‚ท Low process inventory.
๏‚ท Manufacturing cost per unit is low.
Limitations
๏‚ท Breakdown of one machine will stop an entire production line.
๏‚ท Line layout needs major change with the changes in the product design.
๏‚ท High investment in production facilities.
๏‚ท The cycle time is determined by the slowest operation.
๏ถ Quantity production:
๏‚ท It involves the mass production of single parts on fairly standard machine tools such as
punch presses, injection molding machines and automatic screw machines.
๏‚ท The standard machines have been adopted to the production of the particular part by
means of special tools โ€“ die sets, molds and form cutting tools, respectively designed for
the part.
๏‚ท The production equipment is devoted full time to satisfy a very large demand rate for the
item.
Examples: Components for assembled products that have high demand rates
(automobiles, household appliances, light bulbs, etc.), hardware items (screws, nuts &
nails) and many plastic molded products.
๏ถ Flow production:
๏‚ท It suggests the physical flow of the product in oil refineries, continuous chemical
process plants and food processing.
๏‚ท The term also applies to assembled products. In these cases, the items are made to
โ€œflowโ€ through a sequence of operations by material handling devices (conveyors,
moving belts, transfer devices, etc.)
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6th Semester BE, Mechanical, NHCE Page 17
Examples: Automated transfer machines for the production of complex discrete parts and
manual assembly lines for the assembly of complex products.
1.8 COMPUTER INTEGRATED MANUFACTURING (CIM)
Definition: CIM is the integration of total manufacturing enterprise through the use of
integrated system and data communication mixed with new managerial philosophies which
results in the improvement of personnel or organizational efficiencies.
๏‚ฎ Computer-integrated manufacturing (CIM) is the manufacturing approach of using
computers to control the entire production process.
๏‚ฎ This integration allows individual processes to exchange information with each other
and initiate actions.
๏‚ฎ Through the integration of computers, manufacturing can be faster and less error-
prone, although the main advantage is the ability to create automated
manufacturing processes.
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6th Semester BE, Mechanical, NHCE Page 18
๏‚ฎ Computer-integrated manufacturing is used in automotive, aviation, space, and ship
building industries.
๏‚ฎ The term "computer-integrated manufacturing" is both a method of manufacturing
and the name of a computer-automated system in which individual engineering,
production, marketing, and support functions of a manufacturing enterprise are
organized.
In a CIM system functional areas such as;
design, analysis, planning, purchasing, cost accounting, inventory control, and
distribution are linked through the computer with factory floor functions such as
materials handling and management, providing direct control and monitoring of
all the operations.
1.8.1 NATURE & ROLE OF THE ELEMENTS OF CIM SYSTEM
Nine major elements of a CIM system are in figure they are,
i. Marketing
ii. Product Design
iii. Planning
iv. Purchase
v. Manufacturing Engineering
vi. Factory Automation Hardware
vii. Warehousing
viii. Finance
ix. Information Management
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6th Semester BE, Mechanical, NHCE Page 19
i. Marketing:
๏‚ฎ The need for a product is identified by the marketing division.
๏‚ฎ The specifications of the product, the projection of manufacturing quantities and the
strategy for marketing the product are also decided by the marketing department.
๏‚ฎ Marketing also works out the manufacturing costs to assess the economic viability
of the product.
ii. Product Design:
๏‚ฎ The design department of the company establishes initial database for production of
a proposed product.
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6th Semester BE, Mechanical, NHCE Page 20
๏‚ฎ In the CIM system, this is accomplished through activities such as geometric
modelling and computer aided design by taking into account the product
requirements and concepts generated by the creativity of the design engineer.
๏‚ฎ The design process should be constrained by the costs involved in the actual
production and by the capabilities of the existing production equipment and process,
in order to generate an optimized plan.
iii. Planning:
๏‚ฎ The planning department enriches the design database with production data and
information to produce a plan for the production of the product.
๏‚ฎ Planning involves several subsystems dealing with many functional groups in the
industry.
๏‚ฎ In a CIM system, this planning process should be constrained by the production
costs and by the production equipment and process capability, in order to generate
an optimized plan.
iv. Purchase:
๏‚ฎ The purchase department is responsible for placing the purchase orders and follow
up, ensure quality of the received items to the stores for eventual supply to
manufacture and assembly.
v. Manufacturing Engineering
๏‚ฎ It is the activity of carrying out production of the product, involves further
enrichment of database with performance data and information about the
production equipment.
๏‚ฎ In CIM this requires activities such as NC programming, simulation and computer
aided scheduling of the production activity.
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vi. Factory automation hardware:
๏‚ฎ Factory automation equipment further enriches the database with equipment and
process data to carry out the production process.
๏‚ฎ In CIM system this consists of CNC machine tools, flexible manufacturing systems
(FMS), Computer controlled robots, material handling systems, computer controlled
assembly systems, flexibly automated inspection systems and so on.
vii. Warehousing:
๏‚ฎ Warehousing is the function involving storage and retrieval of raw materials,
components, finished goods as well as shipment of items.
๏‚ฎ In todayโ€™s complex outsourcing scenario and the need for just-in-time supply of
components and subsystems, logistics and supply chain management assume great
importance.
viii. Finance:
๏‚ฎ Finance deals with the resources pertaining to money.
๏‚ฎ Planning of investment, working capital, and cash flow control, realization of
receipts, accounting and allocation of funds are the major tasks of the finance
departments
ix. Information Management:
๏‚ฎ Information Management is perhaps one of the crucial tasks in CIM.
๏‚ฎ This involves master production scheduling, database management,
communication, manufacturing systems integration and management information
systems.
1.9 ORGANIZATION AND INFORMATION PROCESSING IN MANUFACTURING
Introduction: For any of three type production (Job shop, Batch shop, Mass production), there
are certain basic functions that must be carried out to convert raw materials into finished product.
For a firm engaged in making discrete products, the functions are:
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1. Processing: Operations transform the product from one state to completion into a more
advanced state of completion.
2. Assembly: In assembly, the two or more separate components are joined together through
mechanical fastening operations.
3. Material handling and storage: A means of moving and storing materials between processing
and assembly operations provided in manufacturing plants.
4. Inspection and test: The purpose of inspection is to determine whether the manufactured
products meet the established design standards and specifications.
5. Control: The control function in manufacturing includes both regulation of individual
processing and assembly operations, and the management of plant-level activities.
Manufacturing firms must organize themselves to accomplish the five functions
described as above. At present, we consider the organizational functions within
manufacturing firm. Many companies make hundreds of different products, each product
consisting of individual components perhaps numbering in thousands. The task of
coordinating all of the individual activities required to make the parts, assemble them, and
deliver the product to the customer is complex indeed. It is a problem in information
processing. The manufacturing model to resort to overcome from the problem of
information processing is as follows:
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Above fig., illustrates the cycle of information-processing activities that typically
occur in manufacturing firm which produces discrete parts and assembles them into final
products for sale to its customers. The information processing cycle, represented by the
outer ring, can be described as consisting of four functions.
๏ƒ˜ Business functions
๏ƒ˜ Product design
๏ƒ˜ Manufacturing planning
๏ƒ˜ Manufacturing control
๏ƒ˜ Business functions:
๏‚ฎ The business functions are the principal means of communicating with the
customer.
๏‚ฎ They are therefore the beginning and end of the information-processing cycle.
๏‚ฎ Included within this category are sales and marketing, sales forecasting, order
entry, cost accounting, customer billing and others.
๏‚ฎ An order to produce a product will typically originate from the sales and
marketing department of the firm.
๏‚ฎ The production order will be one of the following forms: (1) an order to
manufacture an item to the customerโ€™s specifications, (2) a customer order to
buy one or more of the manufacturerโ€™s proprietary products, or (3) an order based
forecast of future demand for proprietary product.
๏ƒ˜ Product Design:
๏‚ฎ If the product is to be manufactured to customer specifications, the design will
have been provided by the customer. Here, the manufacturerโ€™s product design
department will not be involved.
๏‚ฎ If the product is proprietary, the manufacturing firm is responsible for its design
and development.
๏‚ฎ The cycle of events that initiates new product design often originates in the sales
and marketing department.
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๏‚ฎ The departments of the firm that are organized to perform the product design
function might include: research and development (R&D), design engineering,
drafting, and perhaps a prototype shop.
๏‚ฎ The product design is documented by means of component drawings,
specifications, and bill of materials (BOM) that defines how many of each
component go into the product.
๏‚ฎ A prototype is often built for testing and demonstration purposes.
๏‚ฎ The manufacturing engineering department is sometimes consulted to lend
advice on matters of producibility.
๏‚ฎ Upon the completion of the design and fabrication of the prototype, the top
company management is invited for a โ€œshow-and-tellโ€ presentation given by
design engineer, so that management can make decision about manufacturing
of the designed item.
๏‚ฎ The decision is often a two-step procedure: First step is a decision by
engineering management that the design is approved, second step is a decision
by corporate management as to the general suitability of the product.
๏‚ฎ This second decision represents an authorization to produce the item.
๏ƒ˜ Manufacturing Plant:
๏‚ฎ The information and documentation that constitute the design of the product
flow into the manufacturing planning function.
๏‚ฎ The departments in the organization that perform manufacturing planning
include manufacturing engineering, industrial engineering, and production
planning and control.
๏‚ฎ The information processing activities in the manufacturing planning include
๏‚ท Process planning: consists of determining the sequence of the individual
processing and assembly operations needed to produce the part. The
document used to specify the process sequence us called a โ€œroute sheetโ€.
๏‚ท Master scheduling: is a listing of the products to be made, when they are
to be delivered, and in what quantities. Based on this schedule, the
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individual components and subassemblies that make up each product
must be planned.
๏‚ท Material requirements planning (MRP): It is a material control system
that attempts to keep adequate inventory levels to assure that required
materials are available when needed.
๏‚ท Capacity planning: is concerned with planning the manpower and
machine resource of the firm.
๏ƒ˜ Manufacturing Control:
๏‚ฎ Manufacturing Control is concerned with managing and controlling the physical
operations in the factory to implement the manufacturing plans.
๏‚ฎ Included with the control function are;
๏‚ท Shop floor control: is concerned with the problem of monitoring the
progress of the product as it is being processed, assembled, moved, and
inspected in the factory. The sections of a traditional production planning
and control department that are involved in shop floor control include;
scheduling (parts be scheduled one by one through the various production
machines listed on the route sheet for each part), dispatching (involves
issuing the individual work orders to the machine operators to accomplish
the processing of parts) and expediting (comparing the actual progress of
the production order against the schedule).
๏‚ท Inventory control: It attempts to strike a proper balance between the
danger of too little inventory (with possible stock-outs of materials) and the
expense of having too much inventory (called work in progress).
๏‚ท Quality control: is to assure that the quality of the product and its
components to meet the standards specified by the product designer.
1.10 PRODUCTION CONCPETS AND MATHEMATICAL MODELS:
One of the most important factors in the organization is โ€œtime of completion of the
jobโ€. Time is the most important factor which influences the cost of the production,
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scheduling and delivery time. Number of production concepts are quantitative and required
qualitative approach to measure them. The various production concepts are:
1.10.1 Manufacturing Lead Time (MLT)
Definition: Manufacturing lead time is the total time required to process a
given part/product.
MLT comprises of
a) Operation Time (To) b) Non-operation Time (Tno) c) Set-up Time (Tsu)
a) Operation Time (To): The operation time is the time an individual part
spends on a machine. But not all the time is productive. Operation time for
a given manufacturing operation is composed of three elements. They are,
i) Actual machining time (Tm)
ii) Machine handling time (Twh)
iii) Tool handling time per workpiece (Tth)
Therefore, the operation time,
The tool handling time includes
Time spent in changing tools
Changing from one tool to the next for successive operations as in
the case of turret lathe.
Changing between the drill bit and tap in a drill and tap
sequencing performed in the case of turret lathe.
โˆด Tth is the average time per workpiece for any and all of these total
handling activities.
b) Non-operation Time (Tno): It is the total time spend during non-machining
operations.
๐‘‡๐‘‚ = ๐‘‡ ๐‘š + ๐‘‡ ๐‘คโ„Ž + ๐‘‡๐‘กโ„Ž
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๏‚ท Material handling between stations.
๏‚ท Storage between machine or workstation
๏‚ท Inspection
๏‚ท Other sources of delay
c) Setup Time (Tsu): The preparation time spent on the current machine for a
different operation when the product comes from the previous machine is
called setup time.
i) Arrangement of workspace ii) Installation of the tools iii) Installation of
the fixtures
For a Batch Production, it can be expressed as follows;
------------------------- Equn 1.1
Where, โ€˜iโ€™ indicates the operation sequence in processing, ๐‘– = 1,2, โ€ฆ . ๐‘› ๐‘š where
โ€˜nmโ€™ is the total number of machines, and
Q is the number of units produced in a batch
Assuming Tsu, Tno, & To are equal for all โ€˜nmโ€™ machines, then Equn 1.1 reduces
to;
For a Job shop production (only one unit) Q = 1, then Equn 1.2 reduces to
For a Quantity type production, large number of products are made on a single
machine. Then MLT becomes only the operation time for the machines as the
๐‘ด๐‘ณ๐‘ป = เท ๐‘ป ๐’”๐’– ๐’Š
+ ๐‘ธ ๐‘ป ๐’ ๐’Š
๐’ ๐’Ž
๐’Š=๐Ÿ
+ ๐‘ป ๐’๐’ ๐’Š
๐‘ด = ๐’ ๐’Ž(๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’ + ๐‘ป ๐’๐’) ---------------------- Equn 1.2
๐‘ด = ๐’ ๐’Ž(๐‘ป ๐’”๐’– + ๐‘ป ๐’ + ๐‘ป ๐’๐’) ------------------ Equn 1.3
๐‘ด = ๐‘ธ๐‘ป ๐’ -------------------- Equn 1.4
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setup time has already been completed before production commences. In this
situation Equn 1.2 reduces to
Since there is only one machine involved i.e., nm=1, and Tsu =0, Tno=0 during
processing.
For flow type mass production , the entire production line is setup in advance.
Also, the non-operation time between the processing steps consists only the
time to transfer the product from one work station to the next.
When the workstations are integrated to so that, the parts being processed
simultaneously at each station, the station with the longest operation time will
determine the MLT value.
If the operation time of any one of the machines is higher than the remaining,
then it will become bottleneck station which determines the MLT. Similarly,
the transfer time between any of the machines is higher than other transfer
times, then it also determines the MLT.
Hence, MLT will be
1.10.2 Production Rate (Rp): Production rate may be defined as the number of parts
produced per hour. Production rate Rp for an individual process or assembly
operations is usually expressed as an hourly rate i.e., parts produced per hour.
Let us consider To as the operation time and Tsu as the setup time for any given
machine.
For the batch production, the total batch time for the machine is equal to
๐‘ด๐‘ณ๐‘ป = ๐’ ๐’Ž (๐‘ป๐’“๐’‚๐’๐’”๐’‡๐’†๐’“ ๐’•๐’Š๐’Ž๐’† + ๐’๐’๐’๐’ˆ๐’†๐’”๐’• ๐‘ป ๐’) ------- Equn1.5
๐‘ฉ๐’‚๐’•๐’„๐’‰ ๐’•๐’Š๐’Ž๐’†
๐‘ด๐’‚๐’„๐’‰๐’Š๐’๐’†
= ๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’ -----------------------Equn 1.6
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If โ€˜Qโ€™ represents the desired number of parts to be produced and if there is a scrap
rate of q% then the quantity of parts produced in the beginning must
Then the Equn 1.6 becomes
Batch time for a machine is equal to
Dividing Equn 1.7 by Q yields, Average Production Time (Tp) per unit of product
for the given machine.
Then the average production rate Rp for the machine is the reciprocal of average
production time Tp
i.e.,
For a Job shop production situation Q = 1
Then average production time per unit
` โˆด Rp for Job shop production will be
For Quantity type mass production, then the average production time equals to
operation time of the machine. i.e., Tp = To
Then the average production rate Rp is
๐‘ธ
๐Ÿ โˆ’ ๐’’
๐‘ป ๐’”๐’– + เตฌ
๐‘ธ
๐Ÿ โˆ’ ๐’’
เตฐ ๐‘ป ๐’ ----------------------- Equn 1.7
๐‘ป ๐’‘ =
๐‘ฉ๐’‚๐’•๐’„๐’‰ ๐’•๐’Š๐’Ž๐’† ๐’‡๐’๐’“ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†
๐‘ธ
=
๐‘ป ๐’”๐’– + แ‰€
๐‘ธ
๐Ÿ โˆ’ ๐’’
แ‰ ๐‘ป ๐’
๐‘ธ
--------Equn 1.8
๐‘น ๐’‘ =
๐Ÿ
๐‘ป ๐’‘
----------------------------------------- Equn 1.9
๐‘ป ๐’‘ = ๐‘ป ๐’”๐’– + ๐‘ป ๐’ ---------- Equn 1.10
๐‘น ๐’‘ =
๐Ÿ
๐‘ป ๐’”๐’– + ๐‘ป ๐’
--------- Equn 1.11
๐‘น ๐’‘ =
๐Ÿ
๐‘ป ๐’
------------ Equn 1.12
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For Flow line mass production, the production time is the sum of the longest
operation time and transfer time i.e. Tp = Longest operation time (T0max) + transfer
time.
Therefore,
1.10.13 Plant Capacity (Pc): Plant capacity is defined as the maximum rate of
output that a plant is able to produce under a given set of assumed operating
conditions. It is closely related to production rate. The assumed operating
conditions are
i) The number of shifts per day (one, two or three)
ii) The number of days in the week (or months) that the plant operates.
iii) Employment level
iv) Whether or not overtime is included
Plant capacity for a production plant is usually measured in terms of the types of
output produced by the plant.
E.g., Tons of steel in a steel plant, Number of cars produced in an automobile
industry, Barrels of oil in an oil refinery.
Let Pc be the plant capacity of a given work center or group of work-centers.
Capacity will be measured as the number of good units produced per week.
W = No. of work centers. A work center is a production system in the plant typically
consists of one worker and one machine.
Rp = Average production rate produced by โ€˜Wโ€™ work centers in units/hr
H = Total number of hrs/shift each work center operates.
Sw = No. of shifts/week
๐‘น ๐’‘ =
๐Ÿ
๐‘ป ๐’๐’Ž๐’‚๐’™ + ๐‘ป๐’“๐’‚๐’๐’”๐’‡๐’†๐’“ ๐’•๐’Š๐’Ž๐’†
-------------------- Equn 1.13
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The above parameters can be used to calculate the plant capacity i.e.,
The unit of Pc is units/week
If each work unit is routed through no operations requiring a new setup on either a
same or other machine, then the plant capacity equation becomes,
Where, no = number of distinct operations (machines) through which the work units
are routed
Equn 1.15 indicates the operating parameters that affect the plant capacity. The
plant capacity can be appropriately adjusted as per the following plans.
a) Short term plans: To increase or decrease the plant capacity, the following changes
may be adopted
i) Changing the number of shifts per week Sw i.e., making Saturday shifts
authorized to temporarily increase the capacity.
ii) Changing the number of hours per shift (H) i.e., overtime on each regular shift
might be authorized to increase the capacity.
b) Intermediate or long-term plans:
i) Increasing the number of work centers โ€˜Wโ€™ in the shop which might be done by
using equipment that was formerly not in use and hiring new workers or over
the long next term new machines may be acquired.
ii) Increasing the production rate Rp by making improvements in methods or
process methodology.
๐‘ท ๐’„ = (๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ ร— ๐‘น ๐’‘ ------------------ Equn 1.14
๐‘ท ๐’„ =
(๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ ร— ๐‘น ๐’‘
๐’ ๐’
------------------ Equn 1.15
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iii) Reducing the number of operations (no) required per work unit by using
combined operations or simultaneous operations.
1.10.14 Utilization: Utilization refers to the amount of a production facility relative
to its capacity referred as โ€˜Uโ€™. Therefore, we have
The term also can be applied to an entire plant, or a single machine or any other
productive resources.
It is also defined as the proportion of time that the facility is operating relative to
time available under the definition of plant capacity. It is usually expressed in %.
1.10.15 Availability: The term availability is used as a measure of reliability of
equipment. Availability may be defined as the proportion of time the machine is
available for use. They are Mean Time Between Failure (MTBF) and Mean Time
To Repair (MTTR).
Here the MTBF indicates the average length of time between breakdown of the
equipment MTTR indicates the average time required to service the equipment and
place it back into operation when a breakdown occurs.
It is also expressed as a percentage.
1.10.16. Work in Progress(WIP): Work in progress is the amount of product
currently located in the factory that is either being processed or is between
processing operations.
WIP is an inventory that is in the state of being transformed from raw material to
finished product.
๐‘ผ =
๐‘ถ๐’–๐’•๐’‘๐’–๐’•
๐‘ช๐’‚๐’‘๐’‚๐’„๐’Š๐’•๐’š
๐‘จ๐’—๐’‚๐’Š๐’๐’‚๐’ƒ๐’Š๐’๐’Š๐’•๐’š (๐‘จ) =
๐‘ด๐‘ป๐‘ฉ๐‘ญ โˆ’ ๐‘ด๐‘ป๐‘ป๐‘น
๐‘ด๐‘ป๐‘ฉ๐‘ญ
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Measure of WIP is given by
WIP represents the number of units in the process.
Equn 1.16 indicates that the level of WIP will be equal to the rate at which the parts
flows through the factory multiplied by the length of time the parts spent in the
factory.
WIP represents an investment by the firm but that cannot turn into profit until the
processing is completed. Many manufacturing companies sustain major costs
because work remain in process for long time. The units for
๐‘ท ๐’„ร—๐‘ผ
๐‘บ ๐’˜ร—๐‘ฏ
(e.g., parts per
week) must be consistent with the units of MLT (e.g., weeks)
Two measures that can be used to determine the magnitude of the WIP are, WIP ratio
and TIP ratio.
1.10.16.1 WIP (Work in process) ratio: The WIP ratio provides an indication of the
amount of inventory in process relative to the work actually being processed.
Definition: It is the ratio of total quantity of a given part to the quantity of the same
part that is being processed.
WIP ratio obtained by dividing the WIP level by the number of machines currently
engaged in processing the parts.
The divisor that is the number of machines processing can be calculated using the
formula,
๐‘พ๐‘ฐ๐‘ท =
๐‘ท ๐’„ ร— ๐‘ผ
๐‘บ ๐’˜ ร— ๐‘ฏ
ร— ๐‘ด๐‘ณ๐‘ป ---------------- Equn 1.16
๐‘พ๐‘ฐ๐‘ท ๐’“๐’‚๐’•๐’Š๐’ =
๐‘พ๐‘ฐ๐‘ท
๐‘ต๐’–๐’Ž๐’ƒ๐’†๐’“ ๐’๐’‡ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†๐’” ๐’‘๐’“๐’๐’„๐’†๐’”๐’”๐’Š๐’๐’ˆ
๐‘ต๐’ ๐’๐’‡ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†๐’” ๐’‘๐’“๐’๐’„๐’†๐’”๐’”๐’Š๐’๐’ˆ = ๐‘พ ร— ๐‘ผ เตค
๐‘ธ๐‘ป ๐’
๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’
เตจ -- Equn 1.17
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6th Semester BE, Mechanical, NHCE Page 34
Where, W = Number of available work centers
U = Plant Utilization
Q = Batch quantity
Tsu = Setup time
The ideal WIP ration is 1:1 which implies that all parts in the plant are being
processed.
1.10.16.1 TIP (Time in process) ratio: The TIP ratio measures the time that the
product spends in the plant relative to its actual processing time. It is obtained as
the total manufacturing lead time for a part divided by the sum of the individual
operation times for the part.
The ideal TIP ratio is 1:1.
Comments on the Production Concepts
Manufacturing lead time determines how long it will take to deliver a product to the
customer. Here the ability of the firm to deliver the product to the customer in the
shortest possible time is important.
High production rates are important objective in automation. these objectives can be
achieved by reducing workpiece handling time (Twh), processing time (Tm), tool
handling time (Tth) and setup time (Tsu).
Another objective of automation is to increase the plant capacity without the need
for drastic change in employment levels.
Utilization provides a measure of how well the production resources are being used
given that they are available. If the utilization is low, the facility is not being operated
๐‘ป๐‘ฐ๐‘ท ๐’“๐’‚๐’•๐’Š๐’ = เตค
๐‘ด๐‘ณ๐‘ป
๐’ ๐’Ž ๐‘ป ๐’
เตจ --------------------------- Equn 1.18
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nearly to its capacity. If the utilization is higher, it may mean that the facility is being
used fully.
Availability gives an indication of how well the maintenance personnel are servicing
and maintaining the equipment in the plant. If it is 100% it means that the
equipment is reliable and maintenance personnel are doing good job.
WIP is an important issue in manufacturing many firms are attempting to reduce
the high cost of WIP and one of the approaches that is being used is to automate the
operation.
Finally, WIP ratio and TIP ratio should be kept as low as possible and the ideal ratioโ€™s
being 1:1
PROBLEMS
1. A certain part is produced in a batch size of 60 units and requires a sequence of
6 operations in the plant. The average setup time is 3 hours and the average
operation time per machine is 10 min. The average non-operation time due to
handling, delays, inspection is 6 hours. Compute how many days it will take it to
produce a batch, assuming that the plant operates on the 8 hours-shift per day.
Data: Q=60 units, nm=6 operation or machines, Tsu = 3 hours, Tno=6 hours,
To=10 min=(10/60) hours. Plant operates 8 hours shift/day
Solution:
๐‘ด๐‘ณ๐‘ป = ๐’ ๐’Ž(๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’ + ๐‘ป ๐’๐’)
๐‘ด๐‘ณ๐‘ป = ๐Ÿ” แ‰€๐Ÿ‘ + ๐Ÿ”๐ŸŽ ร— แ‰€
๐Ÿ๐ŸŽ
๐Ÿ”๐ŸŽ
แ‰ + ๐Ÿ”แ‰ = ๐Ÿ๐Ÿ๐ŸŽ ๐’‰๐’๐’–๐’“๐’”
โˆด ๐‘ด๐‘ณ๐‘ป =
๐Ÿ๐Ÿ”๐ŸŽ
๐‘ต๐’ ๐’๐’‡ ๐’˜๐’๐’“๐’Œ๐’Š๐’๐’ˆ ๐’‰๐’๐’–๐’“๐’” ๐’Š๐’ ๐’†๐’‚๐’„๐’‰ ๐’”๐’‰๐’Š๐’‡๐’• ๐’‘๐’†๐’“ ๐’…๐’‚๐’š
=
๐Ÿ๐Ÿ๐ŸŽ
๐Ÿ–
= ๐Ÿ๐Ÿ“ ๐’…๐’‚๐’š๐’”
2. Turret lathe section has eight machines all devoted to produce the same part. The
section operates 12 shifts/week. The number of hours per shift averages 8. The
average production rate is 16 units/hour. Determine the production capacity.
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6th Semester BE, Mechanical, NHCE Page 36
Data: W = 8 machines, Sw= 12 shifts/week, Rp=16 unit/hour
Solution: We have, Pc = WSwHRp= (8 x 12 x 8 x 16) = 12,288 unit/week
3. A certain operation is routed through six machines in batch production plant. The
setup and operation times for each machine are given below. The batch size is 100
and the average non-operation time per machine is 10 hours. Determine
i) Manufacturing lead time (MLT)
ii) Production rate for operation number 3.
Solution: i) MLT (Manufacturing Lead time)
We have,
๐‘ด๐‘ณ๐‘ป = เท ๐‘ป ๐’”๐’– ๐’Š
+ ๐‘ธ ๐‘ป ๐’ ๐’Š
๐’ ๐’Ž
๐’Š=๐Ÿ
+ ๐‘ป ๐’๐’ ๐’Š
Total setup time ๐‘‡๐‘ ๐‘ข = เท ๐‘‡๐‘ ๐‘ข ๐‘–
= 5 + 3 + 8 + 4 + 4 + 3 = 27 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ 
6
๐‘–=1
Total operational time ๐‘‡๐‘œ = ๐‘„ โˆ‘ ๐‘‡๐‘œ ๐‘–
= ๐‘„ ( ๐‘‡๐‘œ1
+ ๐‘‡๐‘œ2
+ ๐‘‡๐‘œ3
+ ๐‘‡๐‘œ4
+ ๐‘‡๐‘œ5
+ ๐‘‡๐‘œ6
)
= 100[6 + 3.8 + 12 + 2 + 3.4 + 2.8]
= 100[30] ๐‘š๐‘–๐‘›๐‘  = 3000 ๐‘š๐‘–๐‘›๐‘  = ๐Ÿ“๐ŸŽ ๐’‰๐’๐’–๐’“๐’”
Machineโ€™s Setup time Tsu, (hours) Operation time To (mins)
1 5 6.0
2 3 3.8
3 8 12.0
4 4 2.0
5 4 3.4
6 3 2.8
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 37
Total non-operational time, Tno = nmTo = 6(10) =60 hours
โˆด ๐‘€๐ฟ๐‘‡ = 27 + 50 + 60 = ๐Ÿ๐Ÿ‘๐Ÿ• ๐’‰๐’๐’–๐’“๐’”
ii) Production rate for operation number 3:
We have, Rp =
๐Ÿ
๐‘ป ๐’‘
Where Tp=
๐‘ฉ๐’‚๐’•๐’„๐’‰ ๐’•๐’Š๐’Ž๐’† ๐’‘๐’†๐’“ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†
๐‘ฉ๐’‚๐’•๐’„๐’‰ ๐’’๐’–๐’‚๐’๐’Š๐’•๐’š (๐‘ธ)
๐‘ฉ๐’‚๐’•๐’„๐’‰ ๐’•๐’Š๐’Ž๐’† ๐’‘๐’†๐’“ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†
๐‘ฉ๐’‚๐’•๐’„๐’‰ ๐’’๐’–๐’‚๐’๐’Š๐’•๐’š (๐‘ธ)
= ๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’ = 8 + 100 ร—
12
60
= ๐Ÿ๐Ÿ– ๐’‰๐’๐’–๐’“๐’”
๐‘‡๐‘ =
๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’ ๐‘๐‘’๐‘Ÿ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ž๐‘ข๐‘Ž๐‘›๐‘–๐‘ก๐‘ฆ (๐‘„)
=
28
100
= ๐ŸŽ. ๐Ÿ๐Ÿ– ๐’‰๐’“๐’”/๐’–๐’๐’Š๐’•
โˆด Production rate, Rp =
๐Ÿ
๐‘ป ๐’‘
=
๐Ÿ
๐ŸŽ.๐Ÿ๐Ÿ–
= ๐Ÿ‘. ๐Ÿ“๐Ÿ• ๐’–๐’๐’Š๐’•๐’”/๐’‰๐’“
4. Three products are to be processed through a certain type of work center pertinent data
are given in the following table.
Product Weekly Demand Dw parts (units) Production rate Rp (units/hr)
1 600 10
2 1000 20
3 2200 40
Determine the number of work centers required to satisfy this demand given that the plant
works 10 shifts/week and there are 6.5 hours available for production on each work center
for each shift. The value nm = 1.
Data: Number of products =3, H=6.5 hours/shift, Sw = 10 shifts/week, nm=1
To find: Number of work centers โ€˜Wโ€™
Machineโ€™s Setup time Tsu, (hours) Operation time To (mins)
3 8 12.0
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 38
Solution: We have from Equn 1.15,
๐‘ท ๐’„ =
(๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ ร— ๐‘น ๐’‘
๐’ ๐’Ž
Rearranging this equation, we have,
(๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ =
๐’ ๐’Ž ๐‘ท ๐’„
๐‘น ๐’‘
Replacing Pc by Dw where Dw is the weekly demand similar to plant capacity.
(๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ =
๐’ ๐’Ž ๐‘ซ ๐’˜
๐‘น ๐’‘
for single product
โˆด For the three products as given in the problem, we have,
(๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ = ๐’ ๐’Ž [
๐‘ซ ๐’˜ ๐Ÿ
๐‘น ๐’‘ ๐Ÿ
+
๐‘ซ ๐’˜ ๐Ÿ
๐‘น ๐’‘ ๐Ÿ
+
๐‘ซ ๐’˜ ๐Ÿ‘
๐‘น ๐’‘ ๐Ÿ‘
] = ๐’ ๐’Ž เท
๐‘ซ ๐’˜๐’Š
๐‘น ๐’‘ ๐’Š
๐Ÿ‘
๐’Š=๐Ÿ
(๐‘พ ร— ๐Ÿ๐ŸŽ) ร— ๐Ÿ”. ๐Ÿ“ = ๐Ÿ เตค
๐Ÿ”๐ŸŽ๐ŸŽ
๐Ÿ๐ŸŽ
+
๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ
๐Ÿ๐ŸŽ
+
๐Ÿ๐Ÿ๐ŸŽ๐ŸŽ
๐Ÿ’๐ŸŽ
เตจ
W = 2.54 โ‰ˆ 3 Work centers
5. A production machine is operated 65 hours/week at full capacity and its production
rate is 20 units/hour. During a certain week, the machine produced 1000 good parts and
was idle for the remaining time.
a) Determine the production capacity of the machine.
b) What was the utilization of the machine during the week?
Data: Machine operates = 65 hours/week at full capacity
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 39
Production rate = 20 units/hour
Number of parts produced = 1000
Solution: a) Plant capacity, Pc = (65)(20)=1300 units/week
b) Utilization: It is the ratio of the number of parts produced during productive
use of machine relative to its capacity.
๐‘ˆ =
๐‘‚๐‘ข๐‘ก๐‘๐‘ข๐‘ก
๐ถ๐‘Ž๐‘๐‘Ž๐‘๐‘–๐‘ก๐‘ฆ
=
1000
1300
= 76.92% โ‰ˆ 77%
6. The average parts produced in a certain batch manufacturing plant must be processed
through an average of six machines. There are 20 new batches of parts launched each
week other pertinent data are as follows.
Average operation time = 6 min
Average setup time = 5 hours
Average batch size = 25 parts
Average non-operation time per batch = 10 hours
There are 18 machines in the plant and the plant operates an average of 70 production
hours per week. Scrap rate is negligible.
a) Determine the MLT for an average part?
b) Determine the plant capacity?
c) Determine the plant utilization.
d) How would you expect the non-operation time to be affected by the plant utilization
Data: Batch Production plant
Number of machines, nm = 6 (processed through)
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 40
Operation time, To = 6 min = 6/60 = 0.1 hour
Setup time = Tsu = 5 hours
Batch size, Q = 25 parts
Non-operation time/batch, Tno = 10 hour
Total number of batches = 20
Number of work centers W = 18
Production hours/week = 70
Solution: i) Manufacturing Lead Time (MLT) = ๐’ ๐’Ž(๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’ + ๐‘ป ๐’๐’)
= 6 (5 + 25 ร— 0.1 + 10) = 105 hours
ii) Plant Capacity:
๐‘ท ๐’„ =
(๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ ร— ๐‘น ๐’‘
๐’ ๐’
Sw = Shifts/week
H = Hours/shift
Swร—H = Hours/week
= Production hours/week
= 70
We have,
Rp = Production rate =
1
๐‘‡๐‘
Where, Tp =
๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’/๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
๐‘„
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 41
๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’
๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
= ๐‘‡๐‘ ๐‘ข + ๐‘„๐‘‡๐‘œ = 5 + 25 ร— 0.1 = 7.5
โˆดTp =
7.5
25
= 0.3 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘ข๐‘›๐‘–๐‘ก โˆด Rp = 3.33 units/hour
Substituting all values, we get
Pc =
18ร—70ร—3.33
6
= 699.3 ๐‘ข๐‘›๐‘–๐‘ก๐‘ /๐‘ค๐‘’๐‘’๐‘˜
Plant Utilization(U) =
๐‘‚๐‘ข๐‘ก๐‘๐‘ข๐‘ก
๐ถ๐‘Ž๐‘๐‘Ž๐‘๐‘–๐‘ก๐‘ฆ
Output = There were 20 new batches of parts launched each week and their average
batch size in 25 parts
= 20 ร— 25
= 500 parts
โˆด U =
500
699.3
= 0.715 = 71.5%
iv) We have, ๐‘ด๐‘ณ๐‘ป = ๐’ ๐’Ž(๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’ + ๐‘ป ๐’๐’) and
๐‘พ๐‘ฐ๐‘ท =
๐‘ท ๐’„ ร— ๐‘ผ
๐‘บ ๐’˜ ร— ๐‘ฏ
ร— ๐‘ด๐‘ณ๐‘ป
as it can be observed from the above equation that higher the value of Tno, higher will
be the MLT and also higher the value of MLT, higher will be the WIP.
โˆด The productivity will be reduced as the product spends more time (MLT) in the
factory.
7. Based on the data provided in the problem No. 6 and answers to that problem.
Determine i) the average level of WIP, ii) The WIP ratio, iii) TIP ratio
Solution: i)
๐‘พ๐‘ฐ๐‘ท =
๐‘ท ๐’„ ร— ๐‘ผ
๐‘บ ๐’˜ ร— ๐‘ฏ
ร— ๐‘ด๐‘ณ๐‘ป =
699.3 ร— 0.715
70
ร— 105 = 750 ๐‘ข๐‘›๐‘–๐‘ก๐‘ 
ii)
๐‘พ๐‘ฐ๐‘ท
๐‘ต๐’–๐’Ž๐’ƒ๐’†๐’“ ๐’๐’‡ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†๐’” ๐’‘๐’“๐’๐’„๐’†๐’”๐’”๐’Š๐’๐’ˆ
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 42
๐‘ต๐’ ๐’๐’‡ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†๐’” ๐’‘๐’“๐’๐’„๐’†๐’”๐’”๐’Š๐’๐’ˆ = ๐‘พ ร— ๐‘ผ เตค
๐‘ธ๐‘ป ๐’
๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’
เตจ
๐‘ต๐’ ๐’๐’‡ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†๐’” ๐’‘๐’“๐’๐’„๐’†๐’”๐’”๐’Š๐’๐’ˆ = 18 ร— 0.715 เตค
25 ร— 0.1
5 + 25 ร— 0.1
เตจ = 4.29
โˆด ๐‘Š๐ผ๐‘ƒ ๐‘Ÿ๐‘Ž๐‘ก๐‘–๐‘œ =
750
4.29
= 175 ๐‘š๐‘’๐‘Ž๐‘›๐‘  175: 1
This means for every 175 parts present in the plant there is only one part that is
actually being processed on a machine. This higher value of WIP ratio us due to the higher
value of setup time and the non-operation time per batch which accounts to 15 hours in
total.
iii) TIP ratio:
๐‘ป๐‘ฐ๐‘ท ๐’“๐’‚๐’•๐’Š๐’ = เตค
๐‘ด๐‘ณ๐‘ป
๐’ ๐’Ž ๐‘ป ๐’
เตจ
= เตค
๐Ÿ๐ŸŽ๐Ÿ“
๐Ÿ” ร— ๐ŸŽ. ๐Ÿ
เตจ = ๐Ÿ๐Ÿ•๐Ÿ“
This means if a part has MLT value of 175 hours, then only 1 hour is the actual
operation time for that product.
8. Four products are to be manufactured in a department and it is desired to
determine how to allocate the resources in that department to meet the required demand
for these products for a certain week. The demand and other data for the products are
given as follows:
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 43
Product Dw
(Weekly
Demand)
Tsu Setup
time (hr)
Operation
time, To
(min)
Scrap rate
1 750 6 4.0 0.02
2 900 5 3.0 0.04
3 650 7 2.0 0.05
4 500 6 3.0 0.03
The plant normally operates one shift, 6.75 hours per shift and 6 days per week. Assume
nm=1. Calculate the number of workcenters requires:
Solution: We have, ๐‘Š ร— ๐‘† ๐‘ค ร— ๐ป = โˆ‘
๐ท ๐‘คร—๐‘› ๐‘š
๐‘… ๐‘
Total hours available = 1 shift ร— 6.75 hr/shift ร— 6 days/week = 40.5 hours /week
First calculate Rp for all the product
We have, Rp =
1
๐‘‡๐‘
and Tp =
๐ต๐‘Ž๐‘ก๐‘โ„Ž๐‘ก๐‘–๐‘š๐‘’/๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
๐‘„
Rp for product 1:
๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’
๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
= ๐‘ป ๐’”๐’– +
๐‘ธ
๐Ÿโˆ’๐’’
๐‘ป ๐’
= 6 +
750ร—แ‰€
4
60
แ‰
1โˆ’0.02
= 57.020 โ„Ž๐‘œ๐‘ข๐‘Ÿ
โˆด Tp =
๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’/๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
๐‘„
=
57.020
750
= 0.076 โ„Ž๐‘Ÿ/๐‘ข๐‘›๐‘–๐‘ก
Rp =
1
๐‘‡๐‘
=
1
0.076
= 13.15 ๐‘ข๐‘›๐‘–๐‘ก/โ„Ž๐‘Ÿ
Rp for product 2:
๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’
๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
= ๐‘ป ๐’”๐’– +
๐‘ธ
๐Ÿโˆ’๐’’
๐‘ป ๐’
= 5 +
900ร—แ‰€
3
60
แ‰
1โˆ’0.04
= 51.875โ„Ž๐‘œ๐‘ข๐‘Ÿ
โˆด Tp =
๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’/๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
๐‘„
=
51.875
900
= 0.0576 โ„Ž๐‘Ÿ/๐‘ข๐‘›๐‘–๐‘ก
Rp =
1
๐‘‡๐‘
=
1
0.0576
= 17.35 ๐‘ข๐‘›๐‘–๐‘ก/โ„Ž๐‘Ÿ
Rp for product 3:
๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’
๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
= ๐‘ป ๐’”๐’– +
๐‘ธ
๐Ÿโˆ’๐’’
๐‘ป ๐’
= 7 +
650ร—แ‰€
2
60
แ‰
1โˆ’0.05
= 29.81 โ„Ž๐‘œ๐‘ข๐‘Ÿ
โˆด Tp =
๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’/๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
๐‘„
=
29.81
650
= 0.046 โ„Ž๐‘Ÿ/๐‘ข๐‘›๐‘–๐‘ก
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 44
Rp =
1
๐‘‡๐‘
=
1
0.046
= 21.81 ๐‘ข๐‘›๐‘–๐‘ก/โ„Ž๐‘Ÿ
Rp for product 4:
๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’
๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
= ๐‘ป ๐’”๐’– +
๐‘ธ
๐Ÿโˆ’๐’’
๐‘ป ๐’
= 6 +
500ร—แ‰€
3
60
แ‰
1โˆ’0.03
= 31.77โ„Ž๐‘œ๐‘ข๐‘Ÿ
โˆด Tp =
๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’/๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
๐‘„
=
31.77
500
= 0.064 โ„Ž๐‘Ÿ/๐‘ข๐‘›๐‘–๐‘ก
Rp =
1
๐‘‡๐‘
=
1
0.064
= 15.74 ๐‘ข๐‘›๐‘–๐‘ก๐‘ /โ„Ž๐‘Ÿ
Calculate RHS of the equation
๐‘Š ร— ๐‘† ๐‘ค ร— ๐ป = โˆ‘
๐ท ๐‘คร—๐‘› ๐‘š
๐‘… ๐‘
For product (1)
๐ท ๐‘คร—๐‘› ๐‘š
๐‘… ๐‘
=
750ร—1
13.15
= 57.03 โ„Ž๐‘Ÿ๐‘ 
(2)
๐ท ๐‘คร—๐‘› ๐‘š
๐‘… ๐‘
=
900ร—1
17.35
= 51.87 โ„Ž๐‘Ÿ๐‘ 
(3)
๐ท ๐‘คร—๐‘› ๐‘š
๐‘… ๐‘
=
650ร—1
21.81
= 29.80 โ„Ž๐‘Ÿ๐‘ 
(4)
๐ท ๐‘คร—๐‘› ๐‘š
๐‘… ๐‘
=
500ร—1
15.73
= 31.77 โ„Ž๐‘Ÿ๐‘ 
Substituting into the equation, W ร— 40.5 = [57.03 + 51.87 + 29.80 + 31.77]
W ร— 40.5 = 170.47
W = 4.209
W โ‰ˆ 4 Number of workstation
9. The mean time between failures for a certain production machine is 250 hours and the
mean time to repair is 6 hours. Determine the availability of the machine.
Data: MTBF = 250 hour
MTTR = 6 hours
Solution:
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 45
๐‘จ๐’—๐’‚๐’Š๐’๐’‚๐’ƒ๐’Š๐’๐’Š๐’•๐’š (๐‘จ) =
๐‘ด๐‘ป๐‘ฉ๐‘ญ โˆ’ ๐‘ด๐‘ป๐‘ป๐‘น
๐‘ด๐‘ป๐‘ฉ๐‘ญ
=
๐Ÿ๐Ÿ“๐ŸŽ โˆ’ ๐Ÿ”
๐Ÿ๐Ÿ“๐ŸŽ
= ๐Ÿ—๐Ÿ•. ๐Ÿ”%
10. The mean time between failures and mean time to repair in a certain department of
the factory are 400 hours and 8 hours respectively. The department operates 25 machines
during one 8-hour shift per day, 5 days per week, 52 weeks per year. Each time a machine
breaks down, it costs the company Rs 200 per hour (per machine) in lost revenue. A
proposal has been submitted to install a preventive maintenance program in this
department. The effect of this is the average MTBF will double a repair time will be reduced
to half. The cost of maintenance crew will be Rs 1500/week. However, a reduction of
maintenance personnel on the day shift will result in a savings of Rs 700 per week.
a) Compute the availability of machines before and after the preventive maintenance
program.
b) Determine how many total hours/year the 25 machines are under repair both before
and after preventive maintenance.
c) Will the preventive maintenance program pay for itself for in terms of saving the cost of
lost revenues?
Solution: a) Determination of availability
Case-1: Before preventive maintenance programme.
We have,
๐‘จ๐’—๐’‚๐’Š๐’๐’‚๐’ƒ๐’Š๐’๐’Š๐’•๐’š (๐‘จ) =
๐‘ด๐‘ป๐‘ฉ๐‘ญ โˆ’ ๐‘ด๐‘ป๐‘ป๐‘น
๐‘ด๐‘ป๐‘ฉ๐‘ญ
=
๐Ÿ’๐ŸŽ๐ŸŽ โˆ’ ๐Ÿ–
๐Ÿ’๐ŸŽ๐ŸŽ
= ๐Ÿ—๐Ÿ–%
Case-2: After preventive maintenance programme;
We have,
๐‘จ๐’—๐’‚๐’Š๐’๐’‚๐’ƒ๐’Š๐’๐’Š๐’•๐’š (๐‘จ) =
๐‘ด๐‘ป๐‘ฉ๐‘ญ โˆ’ ๐‘ด๐‘ป๐‘ป๐‘น
๐‘ด๐‘ป๐‘ฉ๐‘ญ
MTBF will be doubled and MTTR will be reduced to half
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 46
=
๐Ÿ–๐ŸŽ๐ŸŽ โˆ’ ๐Ÿ’
๐Ÿ–๐ŸŽ๐ŸŽ
= ๐Ÿ—๐Ÿ—. ๐Ÿ“%
b) How many total hours/year the 25 machines in the department are under repair.
Total hours available= 8 ร— 5 ร— 52 = 2080 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘ฆ๐‘’๐‘Ž๐‘Ÿ ๐‘๐‘’๐‘Ÿ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
Case-1: Before preventive maintenance
Total number of breakdowns =
2080
400
= 5.2
Repair time = 5.2 ร— 8 = 41.6 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
Total repair time for 25 machines = 41.6ร— 25 = 1040 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ 
Case-2: After preventive maintenance
Total number of breakdowns =
2080
800
= 2.6
Repair time = 2.6 ร— 4 = 10.4 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
โˆด Total repair time for 25 machines = 10.4 ร— 25 = 260 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ 
c) Will the preventive maintenance program pay for itself?
Case-1: Before preventive maintenance
Expenditures = Rs 1500/week cost of maintenance crew
Each breakdown costs Rs 200/hour per machine
But for 25 machines, lost in revenue = 200 ร— 1040= Rs 2,08,000/-
โˆด Total cost involved per annum = 1500 ร— 52 + 2,08,000 = Rs 2,86,000/-
Case-2: After preventive maintenance
Expenditures = Rs (1500 โ€“ 700) = Rs 800/week
Loss in revenue = 200ร— 260= Rs 52,000/-
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 47
โˆด Total cost involved per annum = 800 ร— 52 + 52,000 = Rs 93,600/-
โˆด By implementing a preventive maintenance program, the amount saved is
Rs 2,86,000 โ€“ Rs 93,600 = Rs 1, 92, 000/-
11. A certain job specializes in one-of-a-kind orders dealing with parts of medium-to-high
complexity. A typical part is processed through 10 machines in batch sizes of 1. The shop
contains eight conventional machine tools and operates 35-hour per week of production
time pertinent data are given below.
Average machining time per machine = 0.5 hour
Average work handling time per machine = 0.3 hour
Average tool changing time per machine = 0.2 hour
Average setup time per machine = 6 hour
Average non-operation time per machine = 12 hour
A new programmable machine has been purchased by shop which is capable of performing
all 10 operations in a single setup. The programing is done off line. The setup time will be
10-hour. The total machining time will be reduced to 80% of its previous value, the work
handling time will be same as for one machine and total tool change time will be reduced
by 50%. For one machine, non-operation time is expected to be 12 hours.
a) Determine the MLT for the traditional method and for the new method.
b) Compute the plant capacity for the following alternatives
i) A job shop containing the eight traditional machines
ii) A job shop containing two programmable machines.
Data:
Manufacturing type = Job shop
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 48
Number of machines (nm) = 10
Conventional machine tools, w = 8
Production time = 35 hours/week
Tm = 0.5 hour, Twh = 0.3 hour, Tth = 0.2 hour, Tsu = 6 hour, Tno = 12 hour
Solution:
a) We have operation time To = Tm + Twh + Tth
= 0.5 + 0.3 + 0.2
To = 1 hour
i) MLT for traditional method (MLT) = nm (Tsu + QTo + Tno)
= 10 (6 + 1 + 12)
MLT = 190 hours
ii) MLT for new method: performs all operations in single setup
Tsu = 10 hours, Tm = 0.8 (0.5) = 0.4 hours, Twh= 0.3 hours, Tth = 0.1 hours
Tno = 12 hours
โˆด To = Tm + Twh + Tth = 0.4 + 0.3 + 0.1 = 0.8 hours
MLT = 10 + 0.8 + 12 = 22.8 hours
b) Plant Capacity
i) Job shop containing eight traditional machines
๐‘ท ๐’„ =
(๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ ร— ๐‘น ๐’‘
๐’ ๐’
๐‘Š = 8; ๐‘† ๐‘ค ๐ป = 35 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘ค๐‘’๐‘’๐‘˜; ๐‘› ๐‘œ = 10
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 49
Rp =
1
๐‘‡๐‘
=
1
๐‘‡๐‘ ๐‘ข+๐‘‡๐‘œ
=
1
6+1
=
1
7 โ„Ž๐‘œ๐‘ข๐‘Ÿ/๐‘ข๐‘›๐‘–๐‘ก
= 0.1428 unit/hour
๐‘ƒ๐‘ =
(๐‘Šร—๐‘† ๐‘ค)ร—๐ปร—๐‘… ๐‘
๐‘› ๐‘œ
=
8ร—35ร—0.1428
10
= 4 ๐‘ข๐‘›๐‘–๐‘ก๐‘ /๐‘ค๐‘’๐‘’๐‘˜
ii) A job shop containing 2 new programmable machines
i.e., W =2, Tsu = 10 hours, To = 0.8 hours
โˆด Tp = Tsu + QTo = 10 + 1ร—0.8 = 10.8 hours
Rp =
1
๐‘‡๐‘
=
1
10.8
= 0.0926 ๐‘ข๐‘›๐‘–๐‘ก๐‘ /โ„Ž๐‘œ๐‘ข๐‘Ÿ
๐‘ƒ๐‘ =
(๐‘Šร—๐‘† ๐‘ค)ร—๐ปร—๐‘… ๐‘
๐‘› ๐‘œ
=
2ร—35ร—0.0926
1
= 6.48 ๐‘ข๐‘›๐‘–๐‘ก๐‘ /๐‘ค๐‘’๐‘’๐‘˜
c) Determine average level of WIP for the alternatives in part (b) if the alternative
shops operates at capacity.
๐‘พ๐‘ฐ๐‘ท =
๐‘ท ๐’„ ร— ๐‘ผ
๐‘บ ๐’˜ ร— ๐‘ฏ
ร— ๐‘ด๐‘ณ๐‘ป ๐‘ผ = ๐Ÿ๐ŸŽ๐ŸŽ% , ๐’‘๐’๐’‚๐’๐’• ๐’๐’‘๐’†๐’“๐’‚๐’•๐’†๐’” ๐’‚๐’• ๐’‡๐’–๐’๐’ ๐’„๐’‚๐’‘๐’‚๐’„๐’Š๐’•๐’š
๐ฝ๐‘œ๐‘ ๐‘ โ„Ž๐‘œ๐‘ ๐‘๐‘œ๐‘›๐‘ก๐‘Ž๐‘–๐‘›๐‘–๐‘›๐‘” ๐‘’๐‘–๐‘”โ„Ž๐‘ก ๐‘ก๐‘Ÿ๐‘Ž๐‘‘๐‘–๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’๐‘ 
๐‘Š๐ผ๐‘ƒ =
4 ร— 1
35
ร— 190 = 21.744 ๐‘๐‘Ž๐‘Ÿ๐‘ก๐‘  = ๐Ÿ๐Ÿ ๐’‘๐’‚๐’“๐’•๐’”
๐ฝ๐‘œ๐‘ ๐‘ โ„Ž๐‘œ๐‘ ๐‘๐‘œ๐‘›๐‘ก๐‘Ž๐‘–๐‘›๐‘–๐‘›๐‘” ๐‘ก๐‘ค๐‘œ ๐‘๐‘Ÿ๐‘œ๐‘”๐‘Ÿ๐‘Ž๐‘š๐‘š๐‘Ž๐‘๐‘™๐‘’ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’๐‘ 
๐‘Š๐ผ๐‘ƒ =
6.48 ร— 1
35
ร— 22.8 = 4.22 ๐‘๐‘Ž๐‘Ÿ๐‘ก๐‘  = ๐Ÿ’ ๐’‘๐’‚๐’“๐’•๐’”
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 50
12. A factory produces cardboard boxes. The production sequence consists of three
operations i) cutting ii) indenting iiii) printing. There are three machines in the factory,
one for each operator. The machines are 100% reliable and operates as follows when
operating at 100% utilization i) In cutting, large rolls of cardboards fed to cut into the
blanks. Each large roll contains enough material for 4000 blanks. Production cycle time =
0.03 min/blank during production run, but it takes 35 min to change rolls between runs
ii) In indenting, indentation lines are pressed into the blanks to allow the blanks to later
be bent into boxes. The blanks from the previous cutting operation are divided and
consolidated into batches whose starting quantity = 2000 blanks. Indenting is performed
at 4.5 min per 100 blanks. Time to change dies on the indentation machine = 30 min, iiii)
In printing, the indented blanks are printed with labels. The blanks from the previous
indenting operation are divided and consolidates into batches, whose starting quantity
=1000 blanks. Printing cycle rate =30 blanks /min. Between the batches, changes over the
printing plates is required which takes 20 min. In process inventory is allowed to build up
between machines 1 & 2 and between 2 & 3. Based on this data and information determine
the maximum possible output of this factory during a 40-hour week, in completed
blanks/week. Completed blanks have been cut, indented and printed.
Data:
(1) Cutting: 4000 blanks/batch
Cycle time: 0.03 min/blank
Setup time: 35 min
(2) Indenting: 2000 blanks/batch
Cycle time:4.5 min/100 blank
Setup time: 30 min
(3) Printing: 1000 blanks/batch
Printing cycle rate: 30 blanks/min
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 51
Setup time: 20 min
Solution:
(1) Cutting Operation
Calculate the batch processing time (Tb)
Tb = Tsu + QTc
Here, setup time to prepare for the batch is Tsu, batch quantity is Q, and cycle time per
work unit is To Determine the maximum production rate Rp for each operation:
Cutting: Time per batch = 35 min + 4000 blank ร—
0.03 ๐‘š๐‘–๐‘›
๐‘๐‘™๐‘Ž๐‘›๐‘˜
= 155 min/batch
= 2.58 hour/batch
โˆด Production rate =
4000 ๐‘๐‘™๐‘Ž๐‘›๐‘˜๐‘ /๐‘๐‘Ž๐‘ก๐‘โ„Ž
2.58 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘๐‘Ž๐‘ก๐‘โ„Ž
= 1548.4 blank/hour
Indenting: Time per batch = 30 +
4.5
100
ร— 2000
= 120 min/batch
= 2 hours/batch
โˆด Production rate =
2000 ๐‘๐‘™๐‘Ž๐‘›๐‘˜๐‘ /๐‘๐‘Ž๐‘ก๐‘โ„Ž
2 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘๐‘Ž๐‘ก๐‘โ„Ž
= 1000 blank/hour
Printing: Time per batch = 20 +
1000
30 ๐‘๐‘™๐‘Ž๐‘›๐‘˜๐‘ /๐‘š๐‘–๐‘›
= 53.33 min/batch
= 0.89 hour/batch
โˆด Production rate =
100 ๐‘๐‘™๐‘Ž๐‘›๐‘˜๐‘ /๐‘๐‘Ž๐‘ก๐‘โ„Ž
0.89 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘๐‘Ž๐‘ก๐‘โ„Ž
= 1125 blanks/hour
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 52
Out of the three operations, bottle neck operation is indenting, because it requires
more time to complete the operation.
โˆด Weekly output = 40 hour/work ร— 1000 blank/hour
= 40,000 blanks/hour
15. One million units of a certain product are to be manufactured annually on dedicated
machines that run 24 hour/day, 5 days/week, 50 week/year: a) if the cycle time of a
machine to produce one part in 1.0 min, how many of the dedicated machines will be
required to keep up with demand. Assume that availability, utilization and worker
efficiency = 100% and that no setup time will be lost b) Solve part a) except that availability
=0.9.
Solution: a) Hours available = 24 hour/day ร— 5 day/week ร— 50 week/year
= 6000 hour/year per machine
Number of units produced = 1,000,000/year
Cycle time of a machine to produce one part = 1min =
1
60
โ„Ž๐‘œ๐‘ข๐‘Ÿ
โˆด Total work load = 1,00,000/year ร—
1
60
โ„Ž๐‘œ๐‘ข๐‘Ÿ = 16,66,667 hour/year
โˆด Number of machines =
16,66,667 โ„Ž๐‘œ๐‘ข๐‘Ÿ/๐‘ฆ๐‘’๐‘Ž๐‘Ÿ
6000 โ„Ž๐‘œ๐‘ข๐‘Ÿ/๐‘ฆ๐‘’๐‘Ž๐‘Ÿ ๐‘๐‘’๐‘Ÿ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’
= 2.78 โ‰ˆ 3 machines
b) For an availability of 90% =
2.78
0.9
= 3.09 โ‰ˆ 4 ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’๐‘ 
MEE64: Automation Engineering
6th Semester BE, Mechanical, NHCE Page 53

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  • 1. Department of Mechanical Engineering, New Horizon college of Engineering, Bengaluru MEE64: Automation Engineering Module 1A: Computer Integrated Manufacturing Compiled by: Somashekar S M, Assistant Professor
  • 2. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 1 Module1A: Computer Integrated Manufacturing Systems: ๏ƒ˜ Introduction, ๏ƒ˜ Automation definition, ๏ƒ˜ Types of Automation, ๏ƒ˜ Reasons or Advantages of Automation, ๏ƒ˜ Disadvantages of Automation, ๏ƒ˜ Automation Strategies, ๏ƒ˜ Types of Production Systems, ๏ƒ˜ CIM, ๏ƒ˜ Information Processing Cycle in Manufacturing, ๏ƒ˜ Production concepts and Mathematical Models-Manufacturing Lead Time, Production Rate, Components of Operation Production Time, Production Capacity, Utilization and Availability, Work-In-Process, WIP Ratio, TIP Ratio, Problems Using Mathematical Model Equations.
  • 3. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 2 1.1 INTRODUCTION: The systems aspects of manufacturing are more important than ever today. The word manufacturing was originally derived from two Latin words, manus (hand) and factus (make), so that the combination means made by hand. This way the manufacturing was accomplished when the word first appeared in the English language around 1567. Commercial goods of those times were made by hand. The methods were handicraft accomplished in small shops, and the goods were relatively simple, at least by todayโ€™s standards. As many years passed, factories were developed, with many workers at a single site, and the work had to be organized using machines rather than handicraft techniques. The products become more complex, and so did the processes. Workers had to specialize in their tasks. Rather than overseeing the fabrication of the entire product, they were responsible for only a small part of the total work. More up-front planning was required, and more coordination of the operations was needed to keep track of progress in the factories. The systems of production, which rely on many separate but interacting functions, were evolving. Today the system of production is indispensable in manufacturing. Modern manufacturing systems must cope with the economic realities of the modern world. These realities include the following: ๏ƒ˜ Globalization: Once underdeveloped countries like China, India, and Mexico are becoming major players in manufacturing, due largely to their high population and low labor costs. Other regions of the world with low labor costs include Latin America, Eastern Europe, and Southeast Asia, and the countries in these regions have also become important suppliers of manufactured goods. ๏ƒ˜ International Outsourcing: Parts and products once made in the United States by American companies are now being made offshore (overseas, so that cargo ships are required to deliver the items) or near-shore (in Mexico or Central America, so that rail and truck deliveries are possible). In general, international outsourcing means loss of job in US.
  • 4. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 3 ๏ƒ˜ Local outsourcing: Companies can also outsource by using suppliers within the US, reason why companies elect local outsourcing include: 1) benefits from using suppliers that specialize in certain production technologies, 2) lower labor rates in small companies, and 3) limitations of available in-house manufacturing capabilities. ๏ƒ˜ Contract Manufacturing: This refers to companies that specialize in manufacturing entire products not just parts, under contract to other companies. Contract manufacturers specialize in efficient production techniques freeing their customers to specialize in the design and marketing of the products. ๏ƒ˜ Trend towards the service sector in the US economy: There has been gradual erosion of direct labor jobs in manufacturing while jobs in service industries (e.g., healthcare, food services, retail) have increased in numbers. ๏ƒ˜ Quality expectations: Customers, both consumer and corporate, demand that the products they purchase are of the highest quality. Perfect quality is the expectation. ๏ƒ˜ The need for operational efficiency: To be successful, US manufacturers must be efficient in their operations to overcome the labor cost advantage enjoyed by their international competitors. In order to improve efficiency, increase productivity and quality, the following approaches and technologies are in use. ๏ƒ˜ Automation: The use of automated equipment compensates for the labor cost disadvantage relative to international competitors. Automation reduces labor costs, decrease production cycle times, and increase product quality and consistency. ๏ƒ˜ Material handling technologies: Manufacturing usually involves a sequence of activities performed at different locations in the plant. The work must be transported, stored and tracked as it moves through the plant. ๏ƒ˜ Manufacturing systems: These involve the integration and coordination of multiple automated and/or manual workstations through the use of material handling technologies to achieve a synergistic effect compared to the independent operations of individual workstations. Ex: production lines, manufacturing cells and automated assembly systems.
  • 5. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 4 ๏ƒ˜ Flexible manufacturing systems: Much of the outsourcing to international competitors, such as China and Mexico, has involved high-volume goods production. Flexibility in manufacturing allows US manufacturers to compete effectively in the low volume/high-mix product categories. ๏ƒ˜ Computer integrated manufacturing (CIM): The technologies include computer aided design (CAD), computer aided manufacturing (CAM), and computer networks to integrate manufacturing and logistics operations. ๏ƒ˜ Lean production: Accomplishing more work with fewer resources is the general goal of lean production. which involves techniques to increase labor productivity and operational efficiency. 1.2 AUTOMATION: Automation is a technology concerned with the application of mechanical, electronic, and computer-based systems to operate land control production. This technology includes: ๏‚ฎ Automatic machine tools to process parts ๏‚ฎ Automatic assembly machines ๏‚ฎ Industrial robots ๏‚ฎ Automatic material handling and storage systems ๏‚ฎ Automatic inspection systems for quality control ๏‚ฎ Feedback control and computer process control ๏‚ฎ Computer systems for planning, data collection and decision making to support manufacturing activities. 1.3 TYPES OF AUTOMATION: Automated manufacturing systems can be classified into three basic types: 1. Fixed automation 2. Programmable automation 3. Flexible automation 1.3.1. Fixed automation: ๏ƒ˜ It is a system in which the sequence of processing (or assembly) operations is fixed by the equipment configuration. ๏ƒ˜ The operations in the sequence are usually simple.
  • 6. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 5 ๏ƒ˜ It is the integration and coordination of many such operations into one piece of equipment that makes the system complex. ๏ƒ˜ The typical features of fixed automation are: ๏‚ฎ High initial investment for custom-engineered equipment ๏‚ฎ High production rates ๏‚ฎ Relatively inflexible in accommodating product changes ๏ƒ˜ The economic justification for fixed automation is found in products with very high demand rates and volume. ๏ƒ˜ The high initial cost of the equipment is spread over a very large number of units, thus making the unit cost attractive compared to alternative methods of production. Examples: Mechanized assembly lines โ€“ product moved along mechanized conveyors 1.3.2. Programmable automation: ๏ƒ˜ The production equipment is designed with the capability to change the sequence of operations to accommodate different product configurations. ๏ƒ˜ The operation sequence is controlled by a program, which is a set of instructions coded so that the system can read and interpret them. ๏ƒ˜ The features that characterize programmable automation include: ๏‚ฎ High investment in general-purpose equipment ๏‚ฎ Low production rates relative to fixed automation ๏‚ฎ Flexibility to deal with changes in product configuration ๏‚ฎ Most suitable for batch production ๏ƒ˜ Automated production systems which are programmable are used in low and medium volume production. ๏ƒ˜ The parts or products are typically made in batches. ๏ƒ˜ To produce each new batch of different product, the system must be reprogrammed correspond to new product.
  • 7. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 6 ๏ƒ˜ The physical setup of the machine must be changed over: Tools must be loaded, fixture must be attached to the machine table, and the required machine settings must be entered. This changeover takes time. ๏ƒ˜ Thus, the typical cycle for a given product includes a period for the setup and reprogramming, followed by a period in which the batch is produced. ๏ƒ˜ Example: Numerically controlled machine tools and industrial robots, Machining transfer lines. 1.3.3. Flexible automation: ๏ƒ˜ It is the extension of programmable automation. ๏ƒ˜ A flexible automated system is one that is capable of producing a variety of products (or parts) with virtually no time lost for changeovers form one product to the next. ๏ƒ˜ There is no production time lost while reprogramming the system and altering the physical setup (tooling, fixtures, machine settings). ๏ƒ˜ Thus, the system can produce various combinations and schedules of products, instead of requiring that they be made in separate batches. ๏ƒ˜ The features of flexible automation include: ๏‚ฎ High investment for a custom-engineered system ๏‚ฎ Continuous production of variable mixtures of products ๏‚ฎ Medium production rates ๏‚ฎ Flexibility to deal with product design variations. ๏ƒ˜ The essential features that distinguish flexible automation form programmable automation are 1. The capacity to change part programs with no loss of production time (the part programs are prepared off-line on a computer system and electronically transmitted to automated production system). 2. The capability to change over the physical system to continue production without the downtime between batches (the physical setup is changeover off-line and then moving it into place simultaneously as the next part comes into position for processing, the use of pallet fixtures).
  • 8. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 7 1.4 REASONS OR ADAVANTAGES OF AUTOMATION: Companies undertake projects in manufacturing automation and computer integrated manufacturing for a variety of good reasons. Some of the reasons used to justify automation are the following: ๏ƒ˜ To increase labor productivity: Automating a manufacturing operation usually increases production rate and labor productivity. This means greater output per hour of labor input. ๏ƒ˜ To reduce labor cost: Ever-increasing labor cost continues to be the trend in the worldโ€™s industrialized societies. Consequently, higher investment in automation has become commercially justifiable to replace manual operations. Machines are increasingly being substituted for human labor to introduce unit product cost. ๏ƒ˜ To mitigate the effects of labor shortages: There is general shortage if labor in many advanced nations, and this has stimulated the development of automated operations as substitute for labor. ๏ƒ˜ To improve worker safety: Automating a given operation and transferring the worker from active participation in the process to a monitoring role, or removing the
  • 9. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 8 worker from the operation altogether, makes the work safer. The safety and physical well being of the worker has become a national objective with the enactment of the Occupational Safety and Healthy Act (OSHA) in 1970. This has provided an impetus for automation. ๏ƒ˜ To improve product quality: Automation is not only results in higher production rates than manual operation, it also performs the manufacturing process with a greater uniformity and conformity to quality specifications. ๏ƒ˜ To reduce manufacturing lead time: Automation helps reduce the elapsed time between customer order and product delivery, providing a competitive advantage to the manufacturer for future orders. By reducing manufacturing lead time, the manufacturer also reduces work-in process (semi-finished parts) inventory. ๏ƒ˜ To accomplish processes that cannot be done manually: Certain operations cannot be accomplished without the aid of a machine. These processes require precision, miniaturization, or complexity of geometry that cannot be achieved manually. Ex: IC (Integrated Circuit) chips fabrication operations, Rapid prototyping processes based on computer graphics (CAD) models, and the machining of complex, mathematically defined surfaces using computer numerical control. These processes can only be realized by computer controlled systems. ๏ƒ˜ To avoid the high cost of not automating: There is a significant competitive advantage gained in automating a manufacturing plant. The advantage cannot easily be demonstrated on a companyโ€™s project authorization form. The benefits of automation often shown up in unexpected and intangible ways, such as improved quality, higher sales, better labor relations, and better company image. Companies that do not automate are likely to find themselves at a competitive disadvantage with their customers, their employees, and the general public. 1.5 DISADAVANTAGES OF AUTOMATION: ๏ƒ˜ High initial investment ๏ƒ˜ Retrenchment or unfavorable for employees ๏ƒ˜ Difficult to implement in old production systems. ๏ƒ˜ Not economically justifiable for small scale industries (SSI).
  • 10. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 9 1.6 STRATEGIES FOR AUTOMATION AND PRODUCTION SYSTEMS If automation seems a feasible solution to improving productivity, quality, or other measure of performance, then the following ten strategies provide a road map to search for these improvements. ๏ƒ˜ Specialization of operations ๏ƒ˜ Combined operations ๏ƒ˜ Simultaneous operations ๏ƒ˜ Integrations of operations ๏ƒ˜ Increased flexibility ๏ƒ˜ Improved material handling and storage ๏ƒ˜ Online inspection ๏ƒ˜ Process control and optimization ๏ƒ˜ Plant operations and control ๏ƒ˜ Computer integrated manufacturing (CIM) ๏ƒ˜ Specialized Operations: ๏‚ฎ Some special purpose machines or equipmentโ€™s are used to perform particular operation with the highest possible efficiency. ๏‚ฎ This is employed to utilize the specialization of labour and improve labour productivity. ๏ƒ˜ Combined Operations: ๏‚ฎ Production of parts is obtained in a sequence of operations. Several processing steps are required for complex parts. ๏‚ฎ Combined operations eliminate or reduce the processing steps.
  • 11. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 10 ๏‚ฎ This is accomplished by performing more than one operation at a given machine thereby reducing the number of separate machines needed. ๏ƒ˜ Simultaneous Operations: ๏‚ฎ The operations combined in one particular workstation and are carried out simultaneously to reduce the total processing time. ๏‚ฎ This will result in greater increase in productivity. ๏ƒ˜ Integration of Operations: ๏‚ฎ Integration means linking of several workstations together into a single integrated mechanism to transfer parts between them by automatic material handling system. ๏‚ฎ The products are easily scheduled and several parts are processed at a time, thereby resulting in greater increase in overall output of the system. ๏ƒ˜ Increased Flexibility: ๏‚ฎ This strategy attempts to achieve maximum utilization of equipment for producing variety of parts or products. ๏‚ฎ Prime objectives are to reduce setup time and programming time for production machine. ๏ƒ˜ Improved Material Handling and Storage ๏‚ฎ Automated material handling is adopted for reducing non-productive time. ๏‚ฎ Benefits include reduced work in process & shorter manufacturing lead time ๏ƒ˜ Online Inspection: ๏‚ฎ Online inspection means that poor quality product is inspected regularly to eliminate scrap & overall quality can be increased.
  • 12. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 11 ๏ƒ˜ Process Control and Optimization: ๏‚ฎ This includes a wide range of control schemes intended to operate the individual processes and associated equipment more efficiently. ๏‚ฎ Individual processing time is reduced and Product quality can also be improved. ๏ƒ˜ Plant Operations Control: ๏‚ฎ This strategy attempts to manage and coordinate the aggregate operations in the plant. ๏‚ฎ A high-level computer networking may be incorporated to do this task. ๏ƒ˜ Computer Integrated Manufacturing (CIM): ๏‚ฎ This is the logical extension of one level higher than the earlier. ๏‚ฎ Engineering design and business step of the firm are integrated in this strategy. ๏‚ฎ CIM involves extensive use of computer applications, databases and networking throughout the enterprise. 1.7 PRODUCTION SYSTEM: The production system of an organization is that part, which produces products of an organization. It is that activity whereby resources, flowing within a defined system, are combined and transformed in a controlled manner to add value in accordance with the policies communicated by management. The production system has the following characteristics: ๏‚ฎ Production is an organized activity, so every production system has an objective. ๏‚ฎ The system transforms the various inputs to useful outputs. ๏‚ฎ It does not operate in isolation from the other organization system. ๏‚ฎ There exists a feedback about the activities, which is essential to control and improve system performance.
  • 13. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 12 1.7.1 Classification/Types of Production systems: Production systems can be classified as Job Shop, Batch, and Mass Production systems. ๏‚ฎ Job shop Production: ๏‚ท Job shop productions are characterized by manufacturing of one or few quantities of products designed and produced as per the specification of customers (customized) within prefixed time and cost. ๏‚ท The distinguishing feature of this is low volume and high variety of products. Mass Production Job Shop Batch Production Production Quantity Production Rate Labor Skill level Special Tooling Process Plant Layout Product flow
  • 14. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 13 Characteristics ๏‚ท High variety of products and low volume. ๏‚ท Use of general purpose machines and facilities. ๏‚ท Highly skilled operators who can take up each job as a challenge because of uniqueness. ๏‚ท Large inventory of materials, tools, parts. ๏‚ท Detailed planning is essential for sequencing the requirements of each product, capacities for each work center and order priorities. Advantages ๏‚ท Because of general purpose machines and facilities variety of products can be produced. ๏‚ท Operators will become more skilled and competent, as each job gives them learning opportunities. ๏‚ท Full potential of operators can be utilized. ๏‚ท Opportunity exists for creative methods and innovative ideas. Limitations ๏‚ท Higher cost due to frequent set up changes. ๏‚ท Higher level of inventory at all levels and hence higher inventory cost. ๏‚ท Production planning is complicated. ๏‚ท Larger space requirements. Example: space vehicles, aircrafts, machine tools, special tools and equipment and prototypes of future products. ๏‚ฎ Batch Production ๏‚ท Batch production is defined by American Production and Inventory Control Society (APICS) โ€œas a form of manufacturing in which the job passes through the functional departments in lots or batches and each lot may have a different routing*.โ€ ๏‚ท It is characterized by the manufacture of limited number of products produced at regular intervals and stocked awaiting sales. ๏‚ท As in job order production, the batch order production can be of the following three types, o A batch produced only once o A batch produced repeatedly at irregular intervals when need arises. o A batch produced periodically at known intervals to satisfy continuous demands.
  • 15. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 14 *routing: Routing means determination of the path to be followed by each part/component being transformed from input /raw material into final product. Characteristics Batch production system is used under the following circumstances: ๏‚ท When there are shorter production runs. ๏‚ท When plant and machinery are flexible. ๏‚ท When plant and machinery set up is used for the production of item in a batch and change of set up is required for processing the next batch. ๏‚ท When manufacturing lead time and cost are lower as compared to job order production. Advantages ๏‚ท Better utilization of plant and machinery. ๏‚ท Promotes functional specialization. ๏‚ท Cost per unit is lower as compared to job order production.
  • 16. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 15 ๏‚ท Lower investment in plant and machinery. ๏‚ท Flexibility to accommodate and process number of products. ๏‚ท Job satisfaction exists for operators. Limitations ๏‚ท Material handling is complex because of irregular and longer flows. ๏‚ท Production planning and control is complex. ๏‚ท Work in process inventory is higher compared to continuous production. ๏‚ท Higher set up costs due to frequent changes in set up. Examples: ๏‚ท Turret lathes capable of holding several cutting tools are used rather than engine lathe. ๏‚ท The machine tools used are combined with specially designed jigs and fixtures which increase the output rate. ๏‚ท Batch production plants include machine shops, casing foundries, plastic molding factories, press working shops and some types of chemical plants. ๏‚ท Industrial equipment, furniture, textbooks, house hold appliances, lawn movers and components parts for many assembled consumer products. ๏‚ฎ Mass Production ๏‚ท Manufacture of discrete parts or assemblies using a continuous process are called mass production. ๏‚ท This production system is justified by very large volume of production. The machines are arranged in a line or product layout. Product and process standardization exists and all outputs follow the same path. ๏‚ท The mass production can be distinguished into two categories: 1. Quantity production 2. Flow production Characteristics ๏‚ท Standardization of product and process sequence. ๏‚ท Dedicated special purpose machines having higher production capacities and output rates. ๏‚ท Large volume of products. ๏‚ท Shorter cycle time of production. ๏‚ท Lower in process inventory. ๏‚ท Perfectly balanced production lines.
  • 17. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 16 ๏‚ท Flow of materials, components and parts is continuous and without any back tracking. ๏‚ท Production planning and control is easy. ๏‚ท Material handling can be completely automatic. Advantages ๏‚ท Higher rate of production with reduced cycle time. ๏‚ท Higher capacity utilization due to line balancing. ๏‚ท Less skilled operators are required. ๏‚ท Low process inventory. ๏‚ท Manufacturing cost per unit is low. Limitations ๏‚ท Breakdown of one machine will stop an entire production line. ๏‚ท Line layout needs major change with the changes in the product design. ๏‚ท High investment in production facilities. ๏‚ท The cycle time is determined by the slowest operation. ๏ถ Quantity production: ๏‚ท It involves the mass production of single parts on fairly standard machine tools such as punch presses, injection molding machines and automatic screw machines. ๏‚ท The standard machines have been adopted to the production of the particular part by means of special tools โ€“ die sets, molds and form cutting tools, respectively designed for the part. ๏‚ท The production equipment is devoted full time to satisfy a very large demand rate for the item. Examples: Components for assembled products that have high demand rates (automobiles, household appliances, light bulbs, etc.), hardware items (screws, nuts & nails) and many plastic molded products. ๏ถ Flow production: ๏‚ท It suggests the physical flow of the product in oil refineries, continuous chemical process plants and food processing. ๏‚ท The term also applies to assembled products. In these cases, the items are made to โ€œflowโ€ through a sequence of operations by material handling devices (conveyors, moving belts, transfer devices, etc.)
  • 18. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 17 Examples: Automated transfer machines for the production of complex discrete parts and manual assembly lines for the assembly of complex products. 1.8 COMPUTER INTEGRATED MANUFACTURING (CIM) Definition: CIM is the integration of total manufacturing enterprise through the use of integrated system and data communication mixed with new managerial philosophies which results in the improvement of personnel or organizational efficiencies. ๏‚ฎ Computer-integrated manufacturing (CIM) is the manufacturing approach of using computers to control the entire production process. ๏‚ฎ This integration allows individual processes to exchange information with each other and initiate actions. ๏‚ฎ Through the integration of computers, manufacturing can be faster and less error- prone, although the main advantage is the ability to create automated manufacturing processes.
  • 19. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 18 ๏‚ฎ Computer-integrated manufacturing is used in automotive, aviation, space, and ship building industries. ๏‚ฎ The term "computer-integrated manufacturing" is both a method of manufacturing and the name of a computer-automated system in which individual engineering, production, marketing, and support functions of a manufacturing enterprise are organized. In a CIM system functional areas such as; design, analysis, planning, purchasing, cost accounting, inventory control, and distribution are linked through the computer with factory floor functions such as materials handling and management, providing direct control and monitoring of all the operations. 1.8.1 NATURE & ROLE OF THE ELEMENTS OF CIM SYSTEM Nine major elements of a CIM system are in figure they are, i. Marketing ii. Product Design iii. Planning iv. Purchase v. Manufacturing Engineering vi. Factory Automation Hardware vii. Warehousing viii. Finance ix. Information Management
  • 20. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 19 i. Marketing: ๏‚ฎ The need for a product is identified by the marketing division. ๏‚ฎ The specifications of the product, the projection of manufacturing quantities and the strategy for marketing the product are also decided by the marketing department. ๏‚ฎ Marketing also works out the manufacturing costs to assess the economic viability of the product. ii. Product Design: ๏‚ฎ The design department of the company establishes initial database for production of a proposed product.
  • 21. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 20 ๏‚ฎ In the CIM system, this is accomplished through activities such as geometric modelling and computer aided design by taking into account the product requirements and concepts generated by the creativity of the design engineer. ๏‚ฎ The design process should be constrained by the costs involved in the actual production and by the capabilities of the existing production equipment and process, in order to generate an optimized plan. iii. Planning: ๏‚ฎ The planning department enriches the design database with production data and information to produce a plan for the production of the product. ๏‚ฎ Planning involves several subsystems dealing with many functional groups in the industry. ๏‚ฎ In a CIM system, this planning process should be constrained by the production costs and by the production equipment and process capability, in order to generate an optimized plan. iv. Purchase: ๏‚ฎ The purchase department is responsible for placing the purchase orders and follow up, ensure quality of the received items to the stores for eventual supply to manufacture and assembly. v. Manufacturing Engineering ๏‚ฎ It is the activity of carrying out production of the product, involves further enrichment of database with performance data and information about the production equipment. ๏‚ฎ In CIM this requires activities such as NC programming, simulation and computer aided scheduling of the production activity.
  • 22. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 21 vi. Factory automation hardware: ๏‚ฎ Factory automation equipment further enriches the database with equipment and process data to carry out the production process. ๏‚ฎ In CIM system this consists of CNC machine tools, flexible manufacturing systems (FMS), Computer controlled robots, material handling systems, computer controlled assembly systems, flexibly automated inspection systems and so on. vii. Warehousing: ๏‚ฎ Warehousing is the function involving storage and retrieval of raw materials, components, finished goods as well as shipment of items. ๏‚ฎ In todayโ€™s complex outsourcing scenario and the need for just-in-time supply of components and subsystems, logistics and supply chain management assume great importance. viii. Finance: ๏‚ฎ Finance deals with the resources pertaining to money. ๏‚ฎ Planning of investment, working capital, and cash flow control, realization of receipts, accounting and allocation of funds are the major tasks of the finance departments ix. Information Management: ๏‚ฎ Information Management is perhaps one of the crucial tasks in CIM. ๏‚ฎ This involves master production scheduling, database management, communication, manufacturing systems integration and management information systems. 1.9 ORGANIZATION AND INFORMATION PROCESSING IN MANUFACTURING Introduction: For any of three type production (Job shop, Batch shop, Mass production), there are certain basic functions that must be carried out to convert raw materials into finished product. For a firm engaged in making discrete products, the functions are:
  • 23. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 22 1. Processing: Operations transform the product from one state to completion into a more advanced state of completion. 2. Assembly: In assembly, the two or more separate components are joined together through mechanical fastening operations. 3. Material handling and storage: A means of moving and storing materials between processing and assembly operations provided in manufacturing plants. 4. Inspection and test: The purpose of inspection is to determine whether the manufactured products meet the established design standards and specifications. 5. Control: The control function in manufacturing includes both regulation of individual processing and assembly operations, and the management of plant-level activities. Manufacturing firms must organize themselves to accomplish the five functions described as above. At present, we consider the organizational functions within manufacturing firm. Many companies make hundreds of different products, each product consisting of individual components perhaps numbering in thousands. The task of coordinating all of the individual activities required to make the parts, assemble them, and deliver the product to the customer is complex indeed. It is a problem in information processing. The manufacturing model to resort to overcome from the problem of information processing is as follows:
  • 24. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 23 Above fig., illustrates the cycle of information-processing activities that typically occur in manufacturing firm which produces discrete parts and assembles them into final products for sale to its customers. The information processing cycle, represented by the outer ring, can be described as consisting of four functions. ๏ƒ˜ Business functions ๏ƒ˜ Product design ๏ƒ˜ Manufacturing planning ๏ƒ˜ Manufacturing control ๏ƒ˜ Business functions: ๏‚ฎ The business functions are the principal means of communicating with the customer. ๏‚ฎ They are therefore the beginning and end of the information-processing cycle. ๏‚ฎ Included within this category are sales and marketing, sales forecasting, order entry, cost accounting, customer billing and others. ๏‚ฎ An order to produce a product will typically originate from the sales and marketing department of the firm. ๏‚ฎ The production order will be one of the following forms: (1) an order to manufacture an item to the customerโ€™s specifications, (2) a customer order to buy one or more of the manufacturerโ€™s proprietary products, or (3) an order based forecast of future demand for proprietary product. ๏ƒ˜ Product Design: ๏‚ฎ If the product is to be manufactured to customer specifications, the design will have been provided by the customer. Here, the manufacturerโ€™s product design department will not be involved. ๏‚ฎ If the product is proprietary, the manufacturing firm is responsible for its design and development. ๏‚ฎ The cycle of events that initiates new product design often originates in the sales and marketing department.
  • 25. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 24 ๏‚ฎ The departments of the firm that are organized to perform the product design function might include: research and development (R&D), design engineering, drafting, and perhaps a prototype shop. ๏‚ฎ The product design is documented by means of component drawings, specifications, and bill of materials (BOM) that defines how many of each component go into the product. ๏‚ฎ A prototype is often built for testing and demonstration purposes. ๏‚ฎ The manufacturing engineering department is sometimes consulted to lend advice on matters of producibility. ๏‚ฎ Upon the completion of the design and fabrication of the prototype, the top company management is invited for a โ€œshow-and-tellโ€ presentation given by design engineer, so that management can make decision about manufacturing of the designed item. ๏‚ฎ The decision is often a two-step procedure: First step is a decision by engineering management that the design is approved, second step is a decision by corporate management as to the general suitability of the product. ๏‚ฎ This second decision represents an authorization to produce the item. ๏ƒ˜ Manufacturing Plant: ๏‚ฎ The information and documentation that constitute the design of the product flow into the manufacturing planning function. ๏‚ฎ The departments in the organization that perform manufacturing planning include manufacturing engineering, industrial engineering, and production planning and control. ๏‚ฎ The information processing activities in the manufacturing planning include ๏‚ท Process planning: consists of determining the sequence of the individual processing and assembly operations needed to produce the part. The document used to specify the process sequence us called a โ€œroute sheetโ€. ๏‚ท Master scheduling: is a listing of the products to be made, when they are to be delivered, and in what quantities. Based on this schedule, the
  • 26. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 25 individual components and subassemblies that make up each product must be planned. ๏‚ท Material requirements planning (MRP): It is a material control system that attempts to keep adequate inventory levels to assure that required materials are available when needed. ๏‚ท Capacity planning: is concerned with planning the manpower and machine resource of the firm. ๏ƒ˜ Manufacturing Control: ๏‚ฎ Manufacturing Control is concerned with managing and controlling the physical operations in the factory to implement the manufacturing plans. ๏‚ฎ Included with the control function are; ๏‚ท Shop floor control: is concerned with the problem of monitoring the progress of the product as it is being processed, assembled, moved, and inspected in the factory. The sections of a traditional production planning and control department that are involved in shop floor control include; scheduling (parts be scheduled one by one through the various production machines listed on the route sheet for each part), dispatching (involves issuing the individual work orders to the machine operators to accomplish the processing of parts) and expediting (comparing the actual progress of the production order against the schedule). ๏‚ท Inventory control: It attempts to strike a proper balance between the danger of too little inventory (with possible stock-outs of materials) and the expense of having too much inventory (called work in progress). ๏‚ท Quality control: is to assure that the quality of the product and its components to meet the standards specified by the product designer. 1.10 PRODUCTION CONCPETS AND MATHEMATICAL MODELS: One of the most important factors in the organization is โ€œtime of completion of the jobโ€. Time is the most important factor which influences the cost of the production,
  • 27. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 26 scheduling and delivery time. Number of production concepts are quantitative and required qualitative approach to measure them. The various production concepts are: 1.10.1 Manufacturing Lead Time (MLT) Definition: Manufacturing lead time is the total time required to process a given part/product. MLT comprises of a) Operation Time (To) b) Non-operation Time (Tno) c) Set-up Time (Tsu) a) Operation Time (To): The operation time is the time an individual part spends on a machine. But not all the time is productive. Operation time for a given manufacturing operation is composed of three elements. They are, i) Actual machining time (Tm) ii) Machine handling time (Twh) iii) Tool handling time per workpiece (Tth) Therefore, the operation time, The tool handling time includes Time spent in changing tools Changing from one tool to the next for successive operations as in the case of turret lathe. Changing between the drill bit and tap in a drill and tap sequencing performed in the case of turret lathe. โˆด Tth is the average time per workpiece for any and all of these total handling activities. b) Non-operation Time (Tno): It is the total time spend during non-machining operations. ๐‘‡๐‘‚ = ๐‘‡ ๐‘š + ๐‘‡ ๐‘คโ„Ž + ๐‘‡๐‘กโ„Ž
  • 28. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 27 ๏‚ท Material handling between stations. ๏‚ท Storage between machine or workstation ๏‚ท Inspection ๏‚ท Other sources of delay c) Setup Time (Tsu): The preparation time spent on the current machine for a different operation when the product comes from the previous machine is called setup time. i) Arrangement of workspace ii) Installation of the tools iii) Installation of the fixtures For a Batch Production, it can be expressed as follows; ------------------------- Equn 1.1 Where, โ€˜iโ€™ indicates the operation sequence in processing, ๐‘– = 1,2, โ€ฆ . ๐‘› ๐‘š where โ€˜nmโ€™ is the total number of machines, and Q is the number of units produced in a batch Assuming Tsu, Tno, & To are equal for all โ€˜nmโ€™ machines, then Equn 1.1 reduces to; For a Job shop production (only one unit) Q = 1, then Equn 1.2 reduces to For a Quantity type production, large number of products are made on a single machine. Then MLT becomes only the operation time for the machines as the ๐‘ด๐‘ณ๐‘ป = เท ๐‘ป ๐’”๐’– ๐’Š + ๐‘ธ ๐‘ป ๐’ ๐’Š ๐’ ๐’Ž ๐’Š=๐Ÿ + ๐‘ป ๐’๐’ ๐’Š ๐‘ด = ๐’ ๐’Ž(๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’ + ๐‘ป ๐’๐’) ---------------------- Equn 1.2 ๐‘ด = ๐’ ๐’Ž(๐‘ป ๐’”๐’– + ๐‘ป ๐’ + ๐‘ป ๐’๐’) ------------------ Equn 1.3 ๐‘ด = ๐‘ธ๐‘ป ๐’ -------------------- Equn 1.4
  • 29. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 28 setup time has already been completed before production commences. In this situation Equn 1.2 reduces to Since there is only one machine involved i.e., nm=1, and Tsu =0, Tno=0 during processing. For flow type mass production , the entire production line is setup in advance. Also, the non-operation time between the processing steps consists only the time to transfer the product from one work station to the next. When the workstations are integrated to so that, the parts being processed simultaneously at each station, the station with the longest operation time will determine the MLT value. If the operation time of any one of the machines is higher than the remaining, then it will become bottleneck station which determines the MLT. Similarly, the transfer time between any of the machines is higher than other transfer times, then it also determines the MLT. Hence, MLT will be 1.10.2 Production Rate (Rp): Production rate may be defined as the number of parts produced per hour. Production rate Rp for an individual process or assembly operations is usually expressed as an hourly rate i.e., parts produced per hour. Let us consider To as the operation time and Tsu as the setup time for any given machine. For the batch production, the total batch time for the machine is equal to ๐‘ด๐‘ณ๐‘ป = ๐’ ๐’Ž (๐‘ป๐’“๐’‚๐’๐’”๐’‡๐’†๐’“ ๐’•๐’Š๐’Ž๐’† + ๐’๐’๐’๐’ˆ๐’†๐’”๐’• ๐‘ป ๐’) ------- Equn1.5 ๐‘ฉ๐’‚๐’•๐’„๐’‰ ๐’•๐’Š๐’Ž๐’† ๐‘ด๐’‚๐’„๐’‰๐’Š๐’๐’† = ๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’ -----------------------Equn 1.6
  • 30. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 29 If โ€˜Qโ€™ represents the desired number of parts to be produced and if there is a scrap rate of q% then the quantity of parts produced in the beginning must Then the Equn 1.6 becomes Batch time for a machine is equal to Dividing Equn 1.7 by Q yields, Average Production Time (Tp) per unit of product for the given machine. Then the average production rate Rp for the machine is the reciprocal of average production time Tp i.e., For a Job shop production situation Q = 1 Then average production time per unit ` โˆด Rp for Job shop production will be For Quantity type mass production, then the average production time equals to operation time of the machine. i.e., Tp = To Then the average production rate Rp is ๐‘ธ ๐Ÿ โˆ’ ๐’’ ๐‘ป ๐’”๐’– + เตฌ ๐‘ธ ๐Ÿ โˆ’ ๐’’ เตฐ ๐‘ป ๐’ ----------------------- Equn 1.7 ๐‘ป ๐’‘ = ๐‘ฉ๐’‚๐’•๐’„๐’‰ ๐’•๐’Š๐’Ž๐’† ๐’‡๐’๐’“ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’† ๐‘ธ = ๐‘ป ๐’”๐’– + แ‰€ ๐‘ธ ๐Ÿ โˆ’ ๐’’ แ‰ ๐‘ป ๐’ ๐‘ธ --------Equn 1.8 ๐‘น ๐’‘ = ๐Ÿ ๐‘ป ๐’‘ ----------------------------------------- Equn 1.9 ๐‘ป ๐’‘ = ๐‘ป ๐’”๐’– + ๐‘ป ๐’ ---------- Equn 1.10 ๐‘น ๐’‘ = ๐Ÿ ๐‘ป ๐’”๐’– + ๐‘ป ๐’ --------- Equn 1.11 ๐‘น ๐’‘ = ๐Ÿ ๐‘ป ๐’ ------------ Equn 1.12
  • 31. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 30 For Flow line mass production, the production time is the sum of the longest operation time and transfer time i.e. Tp = Longest operation time (T0max) + transfer time. Therefore, 1.10.13 Plant Capacity (Pc): Plant capacity is defined as the maximum rate of output that a plant is able to produce under a given set of assumed operating conditions. It is closely related to production rate. The assumed operating conditions are i) The number of shifts per day (one, two or three) ii) The number of days in the week (or months) that the plant operates. iii) Employment level iv) Whether or not overtime is included Plant capacity for a production plant is usually measured in terms of the types of output produced by the plant. E.g., Tons of steel in a steel plant, Number of cars produced in an automobile industry, Barrels of oil in an oil refinery. Let Pc be the plant capacity of a given work center or group of work-centers. Capacity will be measured as the number of good units produced per week. W = No. of work centers. A work center is a production system in the plant typically consists of one worker and one machine. Rp = Average production rate produced by โ€˜Wโ€™ work centers in units/hr H = Total number of hrs/shift each work center operates. Sw = No. of shifts/week ๐‘น ๐’‘ = ๐Ÿ ๐‘ป ๐’๐’Ž๐’‚๐’™ + ๐‘ป๐’“๐’‚๐’๐’”๐’‡๐’†๐’“ ๐’•๐’Š๐’Ž๐’† -------------------- Equn 1.13
  • 32. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 31 The above parameters can be used to calculate the plant capacity i.e., The unit of Pc is units/week If each work unit is routed through no operations requiring a new setup on either a same or other machine, then the plant capacity equation becomes, Where, no = number of distinct operations (machines) through which the work units are routed Equn 1.15 indicates the operating parameters that affect the plant capacity. The plant capacity can be appropriately adjusted as per the following plans. a) Short term plans: To increase or decrease the plant capacity, the following changes may be adopted i) Changing the number of shifts per week Sw i.e., making Saturday shifts authorized to temporarily increase the capacity. ii) Changing the number of hours per shift (H) i.e., overtime on each regular shift might be authorized to increase the capacity. b) Intermediate or long-term plans: i) Increasing the number of work centers โ€˜Wโ€™ in the shop which might be done by using equipment that was formerly not in use and hiring new workers or over the long next term new machines may be acquired. ii) Increasing the production rate Rp by making improvements in methods or process methodology. ๐‘ท ๐’„ = (๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ ร— ๐‘น ๐’‘ ------------------ Equn 1.14 ๐‘ท ๐’„ = (๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ ร— ๐‘น ๐’‘ ๐’ ๐’ ------------------ Equn 1.15
  • 33. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 32 iii) Reducing the number of operations (no) required per work unit by using combined operations or simultaneous operations. 1.10.14 Utilization: Utilization refers to the amount of a production facility relative to its capacity referred as โ€˜Uโ€™. Therefore, we have The term also can be applied to an entire plant, or a single machine or any other productive resources. It is also defined as the proportion of time that the facility is operating relative to time available under the definition of plant capacity. It is usually expressed in %. 1.10.15 Availability: The term availability is used as a measure of reliability of equipment. Availability may be defined as the proportion of time the machine is available for use. They are Mean Time Between Failure (MTBF) and Mean Time To Repair (MTTR). Here the MTBF indicates the average length of time between breakdown of the equipment MTTR indicates the average time required to service the equipment and place it back into operation when a breakdown occurs. It is also expressed as a percentage. 1.10.16. Work in Progress(WIP): Work in progress is the amount of product currently located in the factory that is either being processed or is between processing operations. WIP is an inventory that is in the state of being transformed from raw material to finished product. ๐‘ผ = ๐‘ถ๐’–๐’•๐’‘๐’–๐’• ๐‘ช๐’‚๐’‘๐’‚๐’„๐’Š๐’•๐’š ๐‘จ๐’—๐’‚๐’Š๐’๐’‚๐’ƒ๐’Š๐’๐’Š๐’•๐’š (๐‘จ) = ๐‘ด๐‘ป๐‘ฉ๐‘ญ โˆ’ ๐‘ด๐‘ป๐‘ป๐‘น ๐‘ด๐‘ป๐‘ฉ๐‘ญ
  • 34. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 33 Measure of WIP is given by WIP represents the number of units in the process. Equn 1.16 indicates that the level of WIP will be equal to the rate at which the parts flows through the factory multiplied by the length of time the parts spent in the factory. WIP represents an investment by the firm but that cannot turn into profit until the processing is completed. Many manufacturing companies sustain major costs because work remain in process for long time. The units for ๐‘ท ๐’„ร—๐‘ผ ๐‘บ ๐’˜ร—๐‘ฏ (e.g., parts per week) must be consistent with the units of MLT (e.g., weeks) Two measures that can be used to determine the magnitude of the WIP are, WIP ratio and TIP ratio. 1.10.16.1 WIP (Work in process) ratio: The WIP ratio provides an indication of the amount of inventory in process relative to the work actually being processed. Definition: It is the ratio of total quantity of a given part to the quantity of the same part that is being processed. WIP ratio obtained by dividing the WIP level by the number of machines currently engaged in processing the parts. The divisor that is the number of machines processing can be calculated using the formula, ๐‘พ๐‘ฐ๐‘ท = ๐‘ท ๐’„ ร— ๐‘ผ ๐‘บ ๐’˜ ร— ๐‘ฏ ร— ๐‘ด๐‘ณ๐‘ป ---------------- Equn 1.16 ๐‘พ๐‘ฐ๐‘ท ๐’“๐’‚๐’•๐’Š๐’ = ๐‘พ๐‘ฐ๐‘ท ๐‘ต๐’–๐’Ž๐’ƒ๐’†๐’“ ๐’๐’‡ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†๐’” ๐’‘๐’“๐’๐’„๐’†๐’”๐’”๐’Š๐’๐’ˆ ๐‘ต๐’ ๐’๐’‡ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†๐’” ๐’‘๐’“๐’๐’„๐’†๐’”๐’”๐’Š๐’๐’ˆ = ๐‘พ ร— ๐‘ผ เตค ๐‘ธ๐‘ป ๐’ ๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’ เตจ -- Equn 1.17
  • 35. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 34 Where, W = Number of available work centers U = Plant Utilization Q = Batch quantity Tsu = Setup time The ideal WIP ration is 1:1 which implies that all parts in the plant are being processed. 1.10.16.1 TIP (Time in process) ratio: The TIP ratio measures the time that the product spends in the plant relative to its actual processing time. It is obtained as the total manufacturing lead time for a part divided by the sum of the individual operation times for the part. The ideal TIP ratio is 1:1. Comments on the Production Concepts Manufacturing lead time determines how long it will take to deliver a product to the customer. Here the ability of the firm to deliver the product to the customer in the shortest possible time is important. High production rates are important objective in automation. these objectives can be achieved by reducing workpiece handling time (Twh), processing time (Tm), tool handling time (Tth) and setup time (Tsu). Another objective of automation is to increase the plant capacity without the need for drastic change in employment levels. Utilization provides a measure of how well the production resources are being used given that they are available. If the utilization is low, the facility is not being operated ๐‘ป๐‘ฐ๐‘ท ๐’“๐’‚๐’•๐’Š๐’ = เตค ๐‘ด๐‘ณ๐‘ป ๐’ ๐’Ž ๐‘ป ๐’ เตจ --------------------------- Equn 1.18
  • 36. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 35 nearly to its capacity. If the utilization is higher, it may mean that the facility is being used fully. Availability gives an indication of how well the maintenance personnel are servicing and maintaining the equipment in the plant. If it is 100% it means that the equipment is reliable and maintenance personnel are doing good job. WIP is an important issue in manufacturing many firms are attempting to reduce the high cost of WIP and one of the approaches that is being used is to automate the operation. Finally, WIP ratio and TIP ratio should be kept as low as possible and the ideal ratioโ€™s being 1:1 PROBLEMS 1. A certain part is produced in a batch size of 60 units and requires a sequence of 6 operations in the plant. The average setup time is 3 hours and the average operation time per machine is 10 min. The average non-operation time due to handling, delays, inspection is 6 hours. Compute how many days it will take it to produce a batch, assuming that the plant operates on the 8 hours-shift per day. Data: Q=60 units, nm=6 operation or machines, Tsu = 3 hours, Tno=6 hours, To=10 min=(10/60) hours. Plant operates 8 hours shift/day Solution: ๐‘ด๐‘ณ๐‘ป = ๐’ ๐’Ž(๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’ + ๐‘ป ๐’๐’) ๐‘ด๐‘ณ๐‘ป = ๐Ÿ” แ‰€๐Ÿ‘ + ๐Ÿ”๐ŸŽ ร— แ‰€ ๐Ÿ๐ŸŽ ๐Ÿ”๐ŸŽ แ‰ + ๐Ÿ”แ‰ = ๐Ÿ๐Ÿ๐ŸŽ ๐’‰๐’๐’–๐’“๐’” โˆด ๐‘ด๐‘ณ๐‘ป = ๐Ÿ๐Ÿ”๐ŸŽ ๐‘ต๐’ ๐’๐’‡ ๐’˜๐’๐’“๐’Œ๐’Š๐’๐’ˆ ๐’‰๐’๐’–๐’“๐’” ๐’Š๐’ ๐’†๐’‚๐’„๐’‰ ๐’”๐’‰๐’Š๐’‡๐’• ๐’‘๐’†๐’“ ๐’…๐’‚๐’š = ๐Ÿ๐Ÿ๐ŸŽ ๐Ÿ– = ๐Ÿ๐Ÿ“ ๐’…๐’‚๐’š๐’” 2. Turret lathe section has eight machines all devoted to produce the same part. The section operates 12 shifts/week. The number of hours per shift averages 8. The average production rate is 16 units/hour. Determine the production capacity.
  • 37. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 36 Data: W = 8 machines, Sw= 12 shifts/week, Rp=16 unit/hour Solution: We have, Pc = WSwHRp= (8 x 12 x 8 x 16) = 12,288 unit/week 3. A certain operation is routed through six machines in batch production plant. The setup and operation times for each machine are given below. The batch size is 100 and the average non-operation time per machine is 10 hours. Determine i) Manufacturing lead time (MLT) ii) Production rate for operation number 3. Solution: i) MLT (Manufacturing Lead time) We have, ๐‘ด๐‘ณ๐‘ป = เท ๐‘ป ๐’”๐’– ๐’Š + ๐‘ธ ๐‘ป ๐’ ๐’Š ๐’ ๐’Ž ๐’Š=๐Ÿ + ๐‘ป ๐’๐’ ๐’Š Total setup time ๐‘‡๐‘ ๐‘ข = เท ๐‘‡๐‘ ๐‘ข ๐‘– = 5 + 3 + 8 + 4 + 4 + 3 = 27 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘  6 ๐‘–=1 Total operational time ๐‘‡๐‘œ = ๐‘„ โˆ‘ ๐‘‡๐‘œ ๐‘– = ๐‘„ ( ๐‘‡๐‘œ1 + ๐‘‡๐‘œ2 + ๐‘‡๐‘œ3 + ๐‘‡๐‘œ4 + ๐‘‡๐‘œ5 + ๐‘‡๐‘œ6 ) = 100[6 + 3.8 + 12 + 2 + 3.4 + 2.8] = 100[30] ๐‘š๐‘–๐‘›๐‘  = 3000 ๐‘š๐‘–๐‘›๐‘  = ๐Ÿ“๐ŸŽ ๐’‰๐’๐’–๐’“๐’” Machineโ€™s Setup time Tsu, (hours) Operation time To (mins) 1 5 6.0 2 3 3.8 3 8 12.0 4 4 2.0 5 4 3.4 6 3 2.8
  • 38. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 37 Total non-operational time, Tno = nmTo = 6(10) =60 hours โˆด ๐‘€๐ฟ๐‘‡ = 27 + 50 + 60 = ๐Ÿ๐Ÿ‘๐Ÿ• ๐’‰๐’๐’–๐’“๐’” ii) Production rate for operation number 3: We have, Rp = ๐Ÿ ๐‘ป ๐’‘ Where Tp= ๐‘ฉ๐’‚๐’•๐’„๐’‰ ๐’•๐’Š๐’Ž๐’† ๐’‘๐’†๐’“ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’† ๐‘ฉ๐’‚๐’•๐’„๐’‰ ๐’’๐’–๐’‚๐’๐’Š๐’•๐’š (๐‘ธ) ๐‘ฉ๐’‚๐’•๐’„๐’‰ ๐’•๐’Š๐’Ž๐’† ๐’‘๐’†๐’“ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’† ๐‘ฉ๐’‚๐’•๐’„๐’‰ ๐’’๐’–๐’‚๐’๐’Š๐’•๐’š (๐‘ธ) = ๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’ = 8 + 100 ร— 12 60 = ๐Ÿ๐Ÿ– ๐’‰๐’๐’–๐’“๐’” ๐‘‡๐‘ = ๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’ ๐‘๐‘’๐‘Ÿ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ ๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ž๐‘ข๐‘Ž๐‘›๐‘–๐‘ก๐‘ฆ (๐‘„) = 28 100 = ๐ŸŽ. ๐Ÿ๐Ÿ– ๐’‰๐’“๐’”/๐’–๐’๐’Š๐’• โˆด Production rate, Rp = ๐Ÿ ๐‘ป ๐’‘ = ๐Ÿ ๐ŸŽ.๐Ÿ๐Ÿ– = ๐Ÿ‘. ๐Ÿ“๐Ÿ• ๐’–๐’๐’Š๐’•๐’”/๐’‰๐’“ 4. Three products are to be processed through a certain type of work center pertinent data are given in the following table. Product Weekly Demand Dw parts (units) Production rate Rp (units/hr) 1 600 10 2 1000 20 3 2200 40 Determine the number of work centers required to satisfy this demand given that the plant works 10 shifts/week and there are 6.5 hours available for production on each work center for each shift. The value nm = 1. Data: Number of products =3, H=6.5 hours/shift, Sw = 10 shifts/week, nm=1 To find: Number of work centers โ€˜Wโ€™ Machineโ€™s Setup time Tsu, (hours) Operation time To (mins) 3 8 12.0
  • 39. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 38 Solution: We have from Equn 1.15, ๐‘ท ๐’„ = (๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ ร— ๐‘น ๐’‘ ๐’ ๐’Ž Rearranging this equation, we have, (๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ = ๐’ ๐’Ž ๐‘ท ๐’„ ๐‘น ๐’‘ Replacing Pc by Dw where Dw is the weekly demand similar to plant capacity. (๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ = ๐’ ๐’Ž ๐‘ซ ๐’˜ ๐‘น ๐’‘ for single product โˆด For the three products as given in the problem, we have, (๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ = ๐’ ๐’Ž [ ๐‘ซ ๐’˜ ๐Ÿ ๐‘น ๐’‘ ๐Ÿ + ๐‘ซ ๐’˜ ๐Ÿ ๐‘น ๐’‘ ๐Ÿ + ๐‘ซ ๐’˜ ๐Ÿ‘ ๐‘น ๐’‘ ๐Ÿ‘ ] = ๐’ ๐’Ž เท ๐‘ซ ๐’˜๐’Š ๐‘น ๐’‘ ๐’Š ๐Ÿ‘ ๐’Š=๐Ÿ (๐‘พ ร— ๐Ÿ๐ŸŽ) ร— ๐Ÿ”. ๐Ÿ“ = ๐Ÿ เตค ๐Ÿ”๐ŸŽ๐ŸŽ ๐Ÿ๐ŸŽ + ๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ ๐Ÿ๐ŸŽ + ๐Ÿ๐Ÿ๐ŸŽ๐ŸŽ ๐Ÿ’๐ŸŽ เตจ W = 2.54 โ‰ˆ 3 Work centers 5. A production machine is operated 65 hours/week at full capacity and its production rate is 20 units/hour. During a certain week, the machine produced 1000 good parts and was idle for the remaining time. a) Determine the production capacity of the machine. b) What was the utilization of the machine during the week? Data: Machine operates = 65 hours/week at full capacity
  • 40. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 39 Production rate = 20 units/hour Number of parts produced = 1000 Solution: a) Plant capacity, Pc = (65)(20)=1300 units/week b) Utilization: It is the ratio of the number of parts produced during productive use of machine relative to its capacity. ๐‘ˆ = ๐‘‚๐‘ข๐‘ก๐‘๐‘ข๐‘ก ๐ถ๐‘Ž๐‘๐‘Ž๐‘๐‘–๐‘ก๐‘ฆ = 1000 1300 = 76.92% โ‰ˆ 77% 6. The average parts produced in a certain batch manufacturing plant must be processed through an average of six machines. There are 20 new batches of parts launched each week other pertinent data are as follows. Average operation time = 6 min Average setup time = 5 hours Average batch size = 25 parts Average non-operation time per batch = 10 hours There are 18 machines in the plant and the plant operates an average of 70 production hours per week. Scrap rate is negligible. a) Determine the MLT for an average part? b) Determine the plant capacity? c) Determine the plant utilization. d) How would you expect the non-operation time to be affected by the plant utilization Data: Batch Production plant Number of machines, nm = 6 (processed through)
  • 41. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 40 Operation time, To = 6 min = 6/60 = 0.1 hour Setup time = Tsu = 5 hours Batch size, Q = 25 parts Non-operation time/batch, Tno = 10 hour Total number of batches = 20 Number of work centers W = 18 Production hours/week = 70 Solution: i) Manufacturing Lead Time (MLT) = ๐’ ๐’Ž(๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’ + ๐‘ป ๐’๐’) = 6 (5 + 25 ร— 0.1 + 10) = 105 hours ii) Plant Capacity: ๐‘ท ๐’„ = (๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ ร— ๐‘น ๐’‘ ๐’ ๐’ Sw = Shifts/week H = Hours/shift Swร—H = Hours/week = Production hours/week = 70 We have, Rp = Production rate = 1 ๐‘‡๐‘ Where, Tp = ๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’/๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ ๐‘„
  • 42. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 41 ๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ = ๐‘‡๐‘ ๐‘ข + ๐‘„๐‘‡๐‘œ = 5 + 25 ร— 0.1 = 7.5 โˆดTp = 7.5 25 = 0.3 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘ข๐‘›๐‘–๐‘ก โˆด Rp = 3.33 units/hour Substituting all values, we get Pc = 18ร—70ร—3.33 6 = 699.3 ๐‘ข๐‘›๐‘–๐‘ก๐‘ /๐‘ค๐‘’๐‘’๐‘˜ Plant Utilization(U) = ๐‘‚๐‘ข๐‘ก๐‘๐‘ข๐‘ก ๐ถ๐‘Ž๐‘๐‘Ž๐‘๐‘–๐‘ก๐‘ฆ Output = There were 20 new batches of parts launched each week and their average batch size in 25 parts = 20 ร— 25 = 500 parts โˆด U = 500 699.3 = 0.715 = 71.5% iv) We have, ๐‘ด๐‘ณ๐‘ป = ๐’ ๐’Ž(๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’ + ๐‘ป ๐’๐’) and ๐‘พ๐‘ฐ๐‘ท = ๐‘ท ๐’„ ร— ๐‘ผ ๐‘บ ๐’˜ ร— ๐‘ฏ ร— ๐‘ด๐‘ณ๐‘ป as it can be observed from the above equation that higher the value of Tno, higher will be the MLT and also higher the value of MLT, higher will be the WIP. โˆด The productivity will be reduced as the product spends more time (MLT) in the factory. 7. Based on the data provided in the problem No. 6 and answers to that problem. Determine i) the average level of WIP, ii) The WIP ratio, iii) TIP ratio Solution: i) ๐‘พ๐‘ฐ๐‘ท = ๐‘ท ๐’„ ร— ๐‘ผ ๐‘บ ๐’˜ ร— ๐‘ฏ ร— ๐‘ด๐‘ณ๐‘ป = 699.3 ร— 0.715 70 ร— 105 = 750 ๐‘ข๐‘›๐‘–๐‘ก๐‘  ii) ๐‘พ๐‘ฐ๐‘ท ๐‘ต๐’–๐’Ž๐’ƒ๐’†๐’“ ๐’๐’‡ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†๐’” ๐’‘๐’“๐’๐’„๐’†๐’”๐’”๐’Š๐’๐’ˆ
  • 43. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 42 ๐‘ต๐’ ๐’๐’‡ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†๐’” ๐’‘๐’“๐’๐’„๐’†๐’”๐’”๐’Š๐’๐’ˆ = ๐‘พ ร— ๐‘ผ เตค ๐‘ธ๐‘ป ๐’ ๐‘ป ๐’”๐’– + ๐‘ธ๐‘ป ๐’ เตจ ๐‘ต๐’ ๐’๐’‡ ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†๐’” ๐’‘๐’“๐’๐’„๐’†๐’”๐’”๐’Š๐’๐’ˆ = 18 ร— 0.715 เตค 25 ร— 0.1 5 + 25 ร— 0.1 เตจ = 4.29 โˆด ๐‘Š๐ผ๐‘ƒ ๐‘Ÿ๐‘Ž๐‘ก๐‘–๐‘œ = 750 4.29 = 175 ๐‘š๐‘’๐‘Ž๐‘›๐‘  175: 1 This means for every 175 parts present in the plant there is only one part that is actually being processed on a machine. This higher value of WIP ratio us due to the higher value of setup time and the non-operation time per batch which accounts to 15 hours in total. iii) TIP ratio: ๐‘ป๐‘ฐ๐‘ท ๐’“๐’‚๐’•๐’Š๐’ = เตค ๐‘ด๐‘ณ๐‘ป ๐’ ๐’Ž ๐‘ป ๐’ เตจ = เตค ๐Ÿ๐ŸŽ๐Ÿ“ ๐Ÿ” ร— ๐ŸŽ. ๐Ÿ เตจ = ๐Ÿ๐Ÿ•๐Ÿ“ This means if a part has MLT value of 175 hours, then only 1 hour is the actual operation time for that product. 8. Four products are to be manufactured in a department and it is desired to determine how to allocate the resources in that department to meet the required demand for these products for a certain week. The demand and other data for the products are given as follows:
  • 44. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 43 Product Dw (Weekly Demand) Tsu Setup time (hr) Operation time, To (min) Scrap rate 1 750 6 4.0 0.02 2 900 5 3.0 0.04 3 650 7 2.0 0.05 4 500 6 3.0 0.03 The plant normally operates one shift, 6.75 hours per shift and 6 days per week. Assume nm=1. Calculate the number of workcenters requires: Solution: We have, ๐‘Š ร— ๐‘† ๐‘ค ร— ๐ป = โˆ‘ ๐ท ๐‘คร—๐‘› ๐‘š ๐‘… ๐‘ Total hours available = 1 shift ร— 6.75 hr/shift ร— 6 days/week = 40.5 hours /week First calculate Rp for all the product We have, Rp = 1 ๐‘‡๐‘ and Tp = ๐ต๐‘Ž๐‘ก๐‘โ„Ž๐‘ก๐‘–๐‘š๐‘’/๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ ๐‘„ Rp for product 1: ๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ = ๐‘ป ๐’”๐’– + ๐‘ธ ๐Ÿโˆ’๐’’ ๐‘ป ๐’ = 6 + 750ร—แ‰€ 4 60 แ‰ 1โˆ’0.02 = 57.020 โ„Ž๐‘œ๐‘ข๐‘Ÿ โˆด Tp = ๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’/๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ ๐‘„ = 57.020 750 = 0.076 โ„Ž๐‘Ÿ/๐‘ข๐‘›๐‘–๐‘ก Rp = 1 ๐‘‡๐‘ = 1 0.076 = 13.15 ๐‘ข๐‘›๐‘–๐‘ก/โ„Ž๐‘Ÿ Rp for product 2: ๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ = ๐‘ป ๐’”๐’– + ๐‘ธ ๐Ÿโˆ’๐’’ ๐‘ป ๐’ = 5 + 900ร—แ‰€ 3 60 แ‰ 1โˆ’0.04 = 51.875โ„Ž๐‘œ๐‘ข๐‘Ÿ โˆด Tp = ๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’/๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ ๐‘„ = 51.875 900 = 0.0576 โ„Ž๐‘Ÿ/๐‘ข๐‘›๐‘–๐‘ก Rp = 1 ๐‘‡๐‘ = 1 0.0576 = 17.35 ๐‘ข๐‘›๐‘–๐‘ก/โ„Ž๐‘Ÿ Rp for product 3: ๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ = ๐‘ป ๐’”๐’– + ๐‘ธ ๐Ÿโˆ’๐’’ ๐‘ป ๐’ = 7 + 650ร—แ‰€ 2 60 แ‰ 1โˆ’0.05 = 29.81 โ„Ž๐‘œ๐‘ข๐‘Ÿ โˆด Tp = ๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’/๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ ๐‘„ = 29.81 650 = 0.046 โ„Ž๐‘Ÿ/๐‘ข๐‘›๐‘–๐‘ก
  • 45. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 44 Rp = 1 ๐‘‡๐‘ = 1 0.046 = 21.81 ๐‘ข๐‘›๐‘–๐‘ก/โ„Ž๐‘Ÿ Rp for product 4: ๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ = ๐‘ป ๐’”๐’– + ๐‘ธ ๐Ÿโˆ’๐’’ ๐‘ป ๐’ = 6 + 500ร—แ‰€ 3 60 แ‰ 1โˆ’0.03 = 31.77โ„Ž๐‘œ๐‘ข๐‘Ÿ โˆด Tp = ๐ต๐‘Ž๐‘ก๐‘โ„Ž ๐‘ก๐‘–๐‘š๐‘’/๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ ๐‘„ = 31.77 500 = 0.064 โ„Ž๐‘Ÿ/๐‘ข๐‘›๐‘–๐‘ก Rp = 1 ๐‘‡๐‘ = 1 0.064 = 15.74 ๐‘ข๐‘›๐‘–๐‘ก๐‘ /โ„Ž๐‘Ÿ Calculate RHS of the equation ๐‘Š ร— ๐‘† ๐‘ค ร— ๐ป = โˆ‘ ๐ท ๐‘คร—๐‘› ๐‘š ๐‘… ๐‘ For product (1) ๐ท ๐‘คร—๐‘› ๐‘š ๐‘… ๐‘ = 750ร—1 13.15 = 57.03 โ„Ž๐‘Ÿ๐‘  (2) ๐ท ๐‘คร—๐‘› ๐‘š ๐‘… ๐‘ = 900ร—1 17.35 = 51.87 โ„Ž๐‘Ÿ๐‘  (3) ๐ท ๐‘คร—๐‘› ๐‘š ๐‘… ๐‘ = 650ร—1 21.81 = 29.80 โ„Ž๐‘Ÿ๐‘  (4) ๐ท ๐‘คร—๐‘› ๐‘š ๐‘… ๐‘ = 500ร—1 15.73 = 31.77 โ„Ž๐‘Ÿ๐‘  Substituting into the equation, W ร— 40.5 = [57.03 + 51.87 + 29.80 + 31.77] W ร— 40.5 = 170.47 W = 4.209 W โ‰ˆ 4 Number of workstation 9. The mean time between failures for a certain production machine is 250 hours and the mean time to repair is 6 hours. Determine the availability of the machine. Data: MTBF = 250 hour MTTR = 6 hours Solution:
  • 46. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 45 ๐‘จ๐’—๐’‚๐’Š๐’๐’‚๐’ƒ๐’Š๐’๐’Š๐’•๐’š (๐‘จ) = ๐‘ด๐‘ป๐‘ฉ๐‘ญ โˆ’ ๐‘ด๐‘ป๐‘ป๐‘น ๐‘ด๐‘ป๐‘ฉ๐‘ญ = ๐Ÿ๐Ÿ“๐ŸŽ โˆ’ ๐Ÿ” ๐Ÿ๐Ÿ“๐ŸŽ = ๐Ÿ—๐Ÿ•. ๐Ÿ”% 10. The mean time between failures and mean time to repair in a certain department of the factory are 400 hours and 8 hours respectively. The department operates 25 machines during one 8-hour shift per day, 5 days per week, 52 weeks per year. Each time a machine breaks down, it costs the company Rs 200 per hour (per machine) in lost revenue. A proposal has been submitted to install a preventive maintenance program in this department. The effect of this is the average MTBF will double a repair time will be reduced to half. The cost of maintenance crew will be Rs 1500/week. However, a reduction of maintenance personnel on the day shift will result in a savings of Rs 700 per week. a) Compute the availability of machines before and after the preventive maintenance program. b) Determine how many total hours/year the 25 machines are under repair both before and after preventive maintenance. c) Will the preventive maintenance program pay for itself for in terms of saving the cost of lost revenues? Solution: a) Determination of availability Case-1: Before preventive maintenance programme. We have, ๐‘จ๐’—๐’‚๐’Š๐’๐’‚๐’ƒ๐’Š๐’๐’Š๐’•๐’š (๐‘จ) = ๐‘ด๐‘ป๐‘ฉ๐‘ญ โˆ’ ๐‘ด๐‘ป๐‘ป๐‘น ๐‘ด๐‘ป๐‘ฉ๐‘ญ = ๐Ÿ’๐ŸŽ๐ŸŽ โˆ’ ๐Ÿ– ๐Ÿ’๐ŸŽ๐ŸŽ = ๐Ÿ—๐Ÿ–% Case-2: After preventive maintenance programme; We have, ๐‘จ๐’—๐’‚๐’Š๐’๐’‚๐’ƒ๐’Š๐’๐’Š๐’•๐’š (๐‘จ) = ๐‘ด๐‘ป๐‘ฉ๐‘ญ โˆ’ ๐‘ด๐‘ป๐‘ป๐‘น ๐‘ด๐‘ป๐‘ฉ๐‘ญ MTBF will be doubled and MTTR will be reduced to half
  • 47. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 46 = ๐Ÿ–๐ŸŽ๐ŸŽ โˆ’ ๐Ÿ’ ๐Ÿ–๐ŸŽ๐ŸŽ = ๐Ÿ—๐Ÿ—. ๐Ÿ“% b) How many total hours/year the 25 machines in the department are under repair. Total hours available= 8 ร— 5 ร— 52 = 2080 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘ฆ๐‘’๐‘Ž๐‘Ÿ ๐‘๐‘’๐‘Ÿ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ Case-1: Before preventive maintenance Total number of breakdowns = 2080 400 = 5.2 Repair time = 5.2 ร— 8 = 41.6 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ Total repair time for 25 machines = 41.6ร— 25 = 1040 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘  Case-2: After preventive maintenance Total number of breakdowns = 2080 800 = 2.6 Repair time = 2.6 ร— 4 = 10.4 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ โˆด Total repair time for 25 machines = 10.4 ร— 25 = 260 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘  c) Will the preventive maintenance program pay for itself? Case-1: Before preventive maintenance Expenditures = Rs 1500/week cost of maintenance crew Each breakdown costs Rs 200/hour per machine But for 25 machines, lost in revenue = 200 ร— 1040= Rs 2,08,000/- โˆด Total cost involved per annum = 1500 ร— 52 + 2,08,000 = Rs 2,86,000/- Case-2: After preventive maintenance Expenditures = Rs (1500 โ€“ 700) = Rs 800/week Loss in revenue = 200ร— 260= Rs 52,000/-
  • 48. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 47 โˆด Total cost involved per annum = 800 ร— 52 + 52,000 = Rs 93,600/- โˆด By implementing a preventive maintenance program, the amount saved is Rs 2,86,000 โ€“ Rs 93,600 = Rs 1, 92, 000/- 11. A certain job specializes in one-of-a-kind orders dealing with parts of medium-to-high complexity. A typical part is processed through 10 machines in batch sizes of 1. The shop contains eight conventional machine tools and operates 35-hour per week of production time pertinent data are given below. Average machining time per machine = 0.5 hour Average work handling time per machine = 0.3 hour Average tool changing time per machine = 0.2 hour Average setup time per machine = 6 hour Average non-operation time per machine = 12 hour A new programmable machine has been purchased by shop which is capable of performing all 10 operations in a single setup. The programing is done off line. The setup time will be 10-hour. The total machining time will be reduced to 80% of its previous value, the work handling time will be same as for one machine and total tool change time will be reduced by 50%. For one machine, non-operation time is expected to be 12 hours. a) Determine the MLT for the traditional method and for the new method. b) Compute the plant capacity for the following alternatives i) A job shop containing the eight traditional machines ii) A job shop containing two programmable machines. Data: Manufacturing type = Job shop
  • 49. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 48 Number of machines (nm) = 10 Conventional machine tools, w = 8 Production time = 35 hours/week Tm = 0.5 hour, Twh = 0.3 hour, Tth = 0.2 hour, Tsu = 6 hour, Tno = 12 hour Solution: a) We have operation time To = Tm + Twh + Tth = 0.5 + 0.3 + 0.2 To = 1 hour i) MLT for traditional method (MLT) = nm (Tsu + QTo + Tno) = 10 (6 + 1 + 12) MLT = 190 hours ii) MLT for new method: performs all operations in single setup Tsu = 10 hours, Tm = 0.8 (0.5) = 0.4 hours, Twh= 0.3 hours, Tth = 0.1 hours Tno = 12 hours โˆด To = Tm + Twh + Tth = 0.4 + 0.3 + 0.1 = 0.8 hours MLT = 10 + 0.8 + 12 = 22.8 hours b) Plant Capacity i) Job shop containing eight traditional machines ๐‘ท ๐’„ = (๐‘พ ร— ๐‘บ ๐’˜) ร— ๐‘ฏ ร— ๐‘น ๐’‘ ๐’ ๐’ ๐‘Š = 8; ๐‘† ๐‘ค ๐ป = 35 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘ค๐‘’๐‘’๐‘˜; ๐‘› ๐‘œ = 10
  • 50. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 49 Rp = 1 ๐‘‡๐‘ = 1 ๐‘‡๐‘ ๐‘ข+๐‘‡๐‘œ = 1 6+1 = 1 7 โ„Ž๐‘œ๐‘ข๐‘Ÿ/๐‘ข๐‘›๐‘–๐‘ก = 0.1428 unit/hour ๐‘ƒ๐‘ = (๐‘Šร—๐‘† ๐‘ค)ร—๐ปร—๐‘… ๐‘ ๐‘› ๐‘œ = 8ร—35ร—0.1428 10 = 4 ๐‘ข๐‘›๐‘–๐‘ก๐‘ /๐‘ค๐‘’๐‘’๐‘˜ ii) A job shop containing 2 new programmable machines i.e., W =2, Tsu = 10 hours, To = 0.8 hours โˆด Tp = Tsu + QTo = 10 + 1ร—0.8 = 10.8 hours Rp = 1 ๐‘‡๐‘ = 1 10.8 = 0.0926 ๐‘ข๐‘›๐‘–๐‘ก๐‘ /โ„Ž๐‘œ๐‘ข๐‘Ÿ ๐‘ƒ๐‘ = (๐‘Šร—๐‘† ๐‘ค)ร—๐ปร—๐‘… ๐‘ ๐‘› ๐‘œ = 2ร—35ร—0.0926 1 = 6.48 ๐‘ข๐‘›๐‘–๐‘ก๐‘ /๐‘ค๐‘’๐‘’๐‘˜ c) Determine average level of WIP for the alternatives in part (b) if the alternative shops operates at capacity. ๐‘พ๐‘ฐ๐‘ท = ๐‘ท ๐’„ ร— ๐‘ผ ๐‘บ ๐’˜ ร— ๐‘ฏ ร— ๐‘ด๐‘ณ๐‘ป ๐‘ผ = ๐Ÿ๐ŸŽ๐ŸŽ% , ๐’‘๐’๐’‚๐’๐’• ๐’๐’‘๐’†๐’“๐’‚๐’•๐’†๐’” ๐’‚๐’• ๐’‡๐’–๐’๐’ ๐’„๐’‚๐’‘๐’‚๐’„๐’Š๐’•๐’š ๐ฝ๐‘œ๐‘ ๐‘ โ„Ž๐‘œ๐‘ ๐‘๐‘œ๐‘›๐‘ก๐‘Ž๐‘–๐‘›๐‘–๐‘›๐‘” ๐‘’๐‘–๐‘”โ„Ž๐‘ก ๐‘ก๐‘Ÿ๐‘Ž๐‘‘๐‘–๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’๐‘  ๐‘Š๐ผ๐‘ƒ = 4 ร— 1 35 ร— 190 = 21.744 ๐‘๐‘Ž๐‘Ÿ๐‘ก๐‘  = ๐Ÿ๐Ÿ ๐’‘๐’‚๐’“๐’•๐’” ๐ฝ๐‘œ๐‘ ๐‘ โ„Ž๐‘œ๐‘ ๐‘๐‘œ๐‘›๐‘ก๐‘Ž๐‘–๐‘›๐‘–๐‘›๐‘” ๐‘ก๐‘ค๐‘œ ๐‘๐‘Ÿ๐‘œ๐‘”๐‘Ÿ๐‘Ž๐‘š๐‘š๐‘Ž๐‘๐‘™๐‘’ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’๐‘  ๐‘Š๐ผ๐‘ƒ = 6.48 ร— 1 35 ร— 22.8 = 4.22 ๐‘๐‘Ž๐‘Ÿ๐‘ก๐‘  = ๐Ÿ’ ๐’‘๐’‚๐’“๐’•๐’”
  • 51. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 50 12. A factory produces cardboard boxes. The production sequence consists of three operations i) cutting ii) indenting iiii) printing. There are three machines in the factory, one for each operator. The machines are 100% reliable and operates as follows when operating at 100% utilization i) In cutting, large rolls of cardboards fed to cut into the blanks. Each large roll contains enough material for 4000 blanks. Production cycle time = 0.03 min/blank during production run, but it takes 35 min to change rolls between runs ii) In indenting, indentation lines are pressed into the blanks to allow the blanks to later be bent into boxes. The blanks from the previous cutting operation are divided and consolidated into batches whose starting quantity = 2000 blanks. Indenting is performed at 4.5 min per 100 blanks. Time to change dies on the indentation machine = 30 min, iiii) In printing, the indented blanks are printed with labels. The blanks from the previous indenting operation are divided and consolidates into batches, whose starting quantity =1000 blanks. Printing cycle rate =30 blanks /min. Between the batches, changes over the printing plates is required which takes 20 min. In process inventory is allowed to build up between machines 1 & 2 and between 2 & 3. Based on this data and information determine the maximum possible output of this factory during a 40-hour week, in completed blanks/week. Completed blanks have been cut, indented and printed. Data: (1) Cutting: 4000 blanks/batch Cycle time: 0.03 min/blank Setup time: 35 min (2) Indenting: 2000 blanks/batch Cycle time:4.5 min/100 blank Setup time: 30 min (3) Printing: 1000 blanks/batch Printing cycle rate: 30 blanks/min
  • 52. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 51 Setup time: 20 min Solution: (1) Cutting Operation Calculate the batch processing time (Tb) Tb = Tsu + QTc Here, setup time to prepare for the batch is Tsu, batch quantity is Q, and cycle time per work unit is To Determine the maximum production rate Rp for each operation: Cutting: Time per batch = 35 min + 4000 blank ร— 0.03 ๐‘š๐‘–๐‘› ๐‘๐‘™๐‘Ž๐‘›๐‘˜ = 155 min/batch = 2.58 hour/batch โˆด Production rate = 4000 ๐‘๐‘™๐‘Ž๐‘›๐‘˜๐‘ /๐‘๐‘Ž๐‘ก๐‘โ„Ž 2.58 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘๐‘Ž๐‘ก๐‘โ„Ž = 1548.4 blank/hour Indenting: Time per batch = 30 + 4.5 100 ร— 2000 = 120 min/batch = 2 hours/batch โˆด Production rate = 2000 ๐‘๐‘™๐‘Ž๐‘›๐‘˜๐‘ /๐‘๐‘Ž๐‘ก๐‘โ„Ž 2 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘๐‘Ž๐‘ก๐‘โ„Ž = 1000 blank/hour Printing: Time per batch = 20 + 1000 30 ๐‘๐‘™๐‘Ž๐‘›๐‘˜๐‘ /๐‘š๐‘–๐‘› = 53.33 min/batch = 0.89 hour/batch โˆด Production rate = 100 ๐‘๐‘™๐‘Ž๐‘›๐‘˜๐‘ /๐‘๐‘Ž๐‘ก๐‘โ„Ž 0.89 โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ /๐‘๐‘Ž๐‘ก๐‘โ„Ž = 1125 blanks/hour
  • 53. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 52 Out of the three operations, bottle neck operation is indenting, because it requires more time to complete the operation. โˆด Weekly output = 40 hour/work ร— 1000 blank/hour = 40,000 blanks/hour 15. One million units of a certain product are to be manufactured annually on dedicated machines that run 24 hour/day, 5 days/week, 50 week/year: a) if the cycle time of a machine to produce one part in 1.0 min, how many of the dedicated machines will be required to keep up with demand. Assume that availability, utilization and worker efficiency = 100% and that no setup time will be lost b) Solve part a) except that availability =0.9. Solution: a) Hours available = 24 hour/day ร— 5 day/week ร— 50 week/year = 6000 hour/year per machine Number of units produced = 1,000,000/year Cycle time of a machine to produce one part = 1min = 1 60 โ„Ž๐‘œ๐‘ข๐‘Ÿ โˆด Total work load = 1,00,000/year ร— 1 60 โ„Ž๐‘œ๐‘ข๐‘Ÿ = 16,66,667 hour/year โˆด Number of machines = 16,66,667 โ„Ž๐‘œ๐‘ข๐‘Ÿ/๐‘ฆ๐‘’๐‘Ž๐‘Ÿ 6000 โ„Ž๐‘œ๐‘ข๐‘Ÿ/๐‘ฆ๐‘’๐‘Ž๐‘Ÿ ๐‘๐‘’๐‘Ÿ ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’ = 2.78 โ‰ˆ 3 machines b) For an availability of 90% = 2.78 0.9 = 3.09 โ‰ˆ 4 ๐‘š๐‘Ž๐‘โ„Ž๐‘–๐‘›๐‘’๐‘ 
  • 54. MEE64: Automation Engineering 6th Semester BE, Mechanical, NHCE Page 53