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1. Industrial Engineering
Introduction
As per American Institute of Industrial Engineers (AIIE),
industrial engineering is defined as follows:
‘It is a branch of engineering concerned with the design,
improvement and installation of integrated systems of
people, materials, equipment and energy. ‘
Industrial engineering is an engineering approach to the
detailed analysis of the use and cost of the resources of
an organisation. The main target for an industrial
engineer is to achieve productivity improvement.
Productivity improvement implies:
More efficient use of resources
Less waste per unit of input applied
Higher levels of output for a fixed level of input
Production Management
Production management focuses on two major areas;
1. Design of the production system which includes
product, process, plant, equipment, and
2. Development of the control systems to manage
inventories, product quality, production
schedules and productivity
Factors related to design in PM cycle are:
1. Product design
2. Job and process design
3. Labour skills and training programs
4. Equipment selection
5. Material selection input
6. Plant selection and layout
7. Scheduling steps of the plan
8. Implementing and controlling the schedule
9. Operating the production system
Factors related to control systems in PM cycle are:
1. Inventory control policies
2. Quality control policies
3. Production schedule control policies
4. Productivity and cost control policies
5. Constructing control systems
6. Implementing and operating control systems
7. Modifying policies and designs
Production Management v/s Industrial Engineering
Production management familiarizes a person
with concepts and techniques specific to the
analysis and management of a production
activity
Industrial engineering deals with the analysis,
design and control of productive systems, i.e.
the system produces either a product or a
service
E.g. The training of an aircraft pilot is analogous to
management education, whereas the designing of the
aircraft is analogous to industrial engineering education.
It is assumed that industrial engineers do not
operate the systems they design.
Managerial Economics
Principles:-
There are four economic principles that managers should
keep in mind;
1. The incremental principle – the decision is
considered good if it increases revenue as
compared to costs
2. The principle of time perspective – the decision
should take into account the long term and
short term effects on costs and revenue
3. The Opportunity cost principle – decision
making should carefully measure the sacrifices
required by the various alternatives
4. The discounting principle – if a decision affects
the costs and revenues of a future date, it is
necessary to discount these costs and revenues
to present values
Concept of Production
Production is any process developed to transform a set of
input elements like men, materials, capitals, information
and energy into a specified set of output elements like
finished products and services in proper quantity and
quality. Factors of production are as follows:
1. Nature (land and other natural resources)
2. Labour (human efforts)
3. Capital (factory building, machinery, tools, raw
materials etc.)
4. Enterprise (activity that organises the factors of
production into an operating unit)
Productivity
It is analogous to the efficiency of a machine. Like we
desire to increase the efficiency of a machine, it is desired
to increase the productivity within the available
resources. Mathematically it is defined as the ratio of
output and input.
Example:
Plant A Plant B
No. of workers 200 300
No. of items
produced per unit
time
10 20
Productivity 10
200
=
1
20
20
300
=
1
15
Factors affecting productivity:
1. Human resources
2. Technology and Capital Investment
3. Product or system design
4. Machinery and equipment

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5. Skill and effectiveness of the worker
6. Production volume
Increasing the productivity of resources:
This implies, producing more number of goods from the
same amount of input (or resources)
Plant location
A plant is a place, where men, materials, money,
equipment, machinery are brought together for
manufacturing products.
Factors governing plant location:
Nearness to Raw material (reduces transportation
cost)
Transport facilities (like road, rail etc.)
Nearness to markets (reduces transportation cost
and the danger of damage to the finished product
before reaching the customer)
Availability of Labour (supply of stable and trained
work force)
Availability of water resources (for industries such as
paper and chemical industries)
Climatic conditions
Financial and other aids
Land (topography, area, shape, cost, drainage etc.
influence the plant location choice)
Plant Layout
Plant layout means the disposition of facilities like
equipments, material, manpower etc. and the services of
the plant within the area of the site selected for plant
setup. It begins with the design of the factory building and
extends up to the location and movement of a work table.
All the equipments and resources are given a proper
place.
Objectives of a good plant layout:
1. Material handling and transportation is
minimized and efficiently controlled
2. Bottlenecks and points of congestion are
eliminated (to make the raw material and semi
finished goods move fast between consecutive
work stations)
3. Suitable and adequate spaces are allocated to
production centres and service centres
4. Minimizing the workers movement
5. Enhancing safety of the working conditions for
all employees
6. Increased flexibility in design changes for future
changes and expansion
7. Reduced plant maintenance cost
Principles of Plant layout:
a) Integration of production facilities in an
efficient manner
b) Minimum movements and material handling
c) Smooth and continuous flow, by implementing
proper line balancing techniques
d) Cubic space utilization by saving the floor space
for storage and making use of ceiling
e) Safe and improved environments in shape of
safe and efficient work places
f) Flexibility for accommodating changing
product designs and production process
Process Layout
It is also known as functional layout and is characterized
by keeping similar machines or similar operations at one
location. In simpler words, all lathes will be in one place,
all milling machines at another and so on. This type of
layout is used in industries engaged in non repetitive type
of maintenance or manufacturing activities.
Advantages of process layout:
1. Greater flexibility in regards to allotment of
work to workers and equipment
2. Better and efficient utilization of available
equipment
3. Reduced capital investment, as the number of
machines or equipments required are less
4. Better product quality obtained
5. Workers in one section is not affected by the
operation carried out in another section
Disadvantages of Process Layout
1. More space is needed
2. Automatic material handling is a difficult task
3. More material-in-process remains in queue for
subsequent operations
4. Production control is difficult
5. Requires more inspections and constant
efficient co-ordination
Product Layout
It is also known as line layout, which means that various
operations on the raw material is performed in a
sequence and the machines are placed along the product
flow line. This type of layout is preferred for industries
where continuous production is performed.
Advantages of product layout:
1. Less space requirements
2. Automatic material handling is easier
3. Material movement and handling, time and
costs are less
4. Less in-process inventory
5. Product completion in less time

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6. Smooth and continuous work flow
7. Less skilled workers may serve the purpose
Disadvantages of product layout:
1. Layout flexibility is considerably reduced
2. The pace of the process depends upon the
output rate of the slowest machine. This
increases the idle time
3. More number of machines of a particular have
to be purchased in order to create adequate
number of standbys in case of any failure. This
increases the capital investment
4. It is very difficult to increase the capacity of the
production beyond the layout capacity
Combination Layout
This type of layout combines the advantages of both,
process and product layout. These kinds of layouts are
very rare. This kind of a layout is possible where an item
is being made in different types and sizes. In these kind of
cases, the machinery is arranged in a process layout but
the process grouping is then arranged in a sequence to
manufacture various size and types of products.
No matter the product varies in size and type,
the operation sequence remain the same.
Fixed Position Layout
This kind of layout is inherent in ship building, aircraft
building and big pressure vessel fabrication. In this type
of layout, the men, materials and equipment move past
the stationary product.
Advantages of Position layout:
1. One or more skilled workers can be employed
from the beginning till the end of the job to
ensure continuity of the process
2. It involves least movement of materials
3. Maximum flexibility available for products and
process
4. Different projects can be taken up for the same
layout
Disadvantages of Position layout:
1. Low content of work-in progress
2. Low utilization of labour and equipment
3. Involves high equipment handling costs
Flow Pattern
Achieving an optimum efficient flow of materials is one of
the most important phases in plant layout. The principle
for optimum effective flow is minimum movement. The
principle of minimum movement reduces material
handling costs, in-process inventory and space for
processing.
A flow pattern must be simple in order to have
an easy supervision and control
Various flow patterns along with their characteristics are
given below in the form of a table;
Line flow It is the simplest, the material
enters at one end (X) and
leaves at the other end (Y).
Used in buildings having long
lengths and smaller widths.
L type flow It resembles Line flow, but
is used in buildings where
width is more as compared
to line flow type buildings
Circular flow It is preferred for rotary
handling systems. Different
work stations are located
along the circular path. Raw
material enters at X and
finished product leaves at Y.
U type flow In this the supervision is
simpler as compared to Line
flow and L type flow. The raw
material and finished
product from the same side.
Preferred in square shaped
buildings.
Combination of line
flow and circular type
As compared to line flow,
this system needs smaller
building lengths.
Processing upwards In this the material is
processed while moving
upwards or downwards in a
multi storeyed

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Work Station Design
The work station design affects the production rates,
efficiency and the accuracy with which an operation can
be performed.
Line Balancing & Process Planning
Line Balancing
Line balancing means balancing the line, i.e. balancing a
production line or an assembly line. Let us consider three
machines, A, B and C, which can process 5, 10 and 15
pieces per unit time respectively. There is a precedence
constraint of A to B to C.
Since machine A has the minimum capacity, this will make
machine B idle for 50% of the time, and machine C idle
for 66.66% of the time. This indicates that the line is
unbalanced. One method to balance this line is to have 3
machines of type A, 2 machines of type B and one
machine of type C. Another method to make sure that the
machines B and C do not remain idle is to give some
additional work to them.
The main task of line balancing is to ensure that the
tasks are evenly distributed among men, machinery
and thereby ensuring minimum idle time.
Line balancing aims at grouping the facilities and
tasks and workers in an efficient pattern in order to
obtain an optimum balance of the capacities of the
processes.
The tasks are grouped in such a way so that their
total time is preferably equal to or a little lesser than
the time available at each work station.
Methods for Line Balancing
1. Heuristic method
2. Linear Programming Model
3. Dynamic Programming
4. Comsoal (a computer method for assembly line
sequencing)
For intermittent flow pattern, Heuristic method is
used as they are simple and involve less time and
money.
For continuous flow pattern involving high volume
production we choose between linear programming
and dynamic programming.
Heuristic Method:
In this method a precedence diagram is drawn in a
particular way which indicates the flexibility available for
transferring tasks laterally from one column to another.
The following procedural steps are taken in this method:
1. Identify the work
2. Break down the work into elemental tasks or steps
3. List the various steps as shown below;
Steps or
elemental
tasks
Immediate
Predecessor
Duration of the
task (minutes)
1 - 3
2 - 4
3 1 2
4 2 5
5 3 4
6 4 8
7 5 2
8 7 4
9 8 6
(Total time=38
minutes)
4. Sketch the precedence diagram and mark the task
duration
5. Assume that the maximum time available for this
problem at any work station is 10 minutes. Or in
other words, cycle time is 10 minutes. The total
duration of tasks is 38 minutes which means that the
minimum number of work stations required are
38/10 = 4. The maximum number of stations may be
equal to as many as the number of tasks or steps, i.e.
nine.
6. Two basic concepts of assigning tasks to stations:
a) Permutability of tasks: It means that any
number of tasks of a column can be combined
to make up their total time closer to cycle time,
provided their total time does not exceed the
cycle time. Analysis is carried out column by
column and one can move to next column only
after the tasks in the previous column have
been assigned to a station.
b) Lateral transferability of the tasks: For making
total time of tasks equal to cycle time, tasks or
steps may be shifted laterally provided the
precedence relationships are maintained.
Keeping these two concepts in mind, the above given
figure (precedence diagram) is modified to the figure
given below. In the below given figure, tasks 1, 2 and 3
are grouped together to station A. Task 4 has been
laterally shifted from column II to column III and has been
grouped with task 5, occupying station B. similarly tasks 6
and 7, 8 and 9 have been grouped and placed at stations
C and D, respectively. This way all the nine steps have
distributed to four stations.

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Linear Programming method of Line Balancing:
Assume that a job is broken down into 6 elemental tasks
and the total duration of all such tasks is 28 minutes. The
cycle time at each work station is 10 minutes. Thus the
minimum number of work stations required are 28/10=3
and the maximum number of work stations may be 6, i.e.
equal to the number of tasks involved.
The problem now reduces to find out the exact number
of work stations needed and which tasks will be assigned
to which station, as shown in the precedence diagram
shown below.
Process Planning
A process is defined as any group of actions performed to
achieve some output from an operation in accordance
with a specified measure of effectiveness. During
designing a product, some specifications are established;
like physical dimensions, tolerances, standards and
quality. The decision of specific details on achieving the
desired outcome is a part of Process Planning.
Process Planning is the systematic determination of the
methods by which a product is to be manufactured,
economically and competitively. It is the intermediate
step between designing a product and manufacturing it.
Process planning takes as its inputs the drawings or
other specifications which indicate what is to be
made and also the forecasts, orders or contracts
which indicate how many are to be made.
Information required to do Process Planning
1. Quantity of work to be done along with
specifications
2. Quality of work to be completed
3. Availability of equipments, tools and personnel
4. Sequence of operations on raw material
5. Names of equipments on which operations will be
performed
6. Standard time for each operation
7. When the operations will be performed
Process Planning Procedure
1. Preparation of working drawings
2. Deciding to make or buy
3. Selection of manufacturing process
4. Machine capacity and equipment selection
5. Material selection and bill of material
6. Selection of jigs, fixtures and other attachments
7. Operation planning and tooling requirements
8. Preparation of documents such as operation and
route sheets
Example:
ABC Company manufactures and sells gas stoves. It
makes some of the parts for the gas stoves and purchases
the rest. The engineering department believes that it
might be possible to cut costs by manufacturing one of
the parts currently being purchased for Rs 8.50 each. The
firm uses 1,00,000 of these parts every year, and the
accounting department compiles the following list of
costs based on engineering estimates.
Fixed costs will increase by Rs 50,000
Labour costs will increase by Rs 1,25,000
Factory overheads, currently running Rs 5,00,000 p.a.
may increase by 12%
Raw materials used to make the part will cost Rs
6,00,000
Given the above estimates, should ABC Company make
the part or buy it?
Solution:
Calculate the total part incurred if the part was
manufactured:
Additional fixed costs Rs 50,000
Additional labour costs Rs 1,25,000
Raw materials costs Rs 6,00,000
Additional overheads
costs
Rs 60,000 ( 0.12 X
5,00,000)
Total cost to manufacture Rs 8,35,000
Cost to manufacture one
part
Rs 8,35,000/ 1,00,000 =
Rs 8.35
As compared to buying cost per piece, the cost of
manufacturing one piece is less, therefore, the company
should manufacture the part instead of buying it.

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Process Analysis
It means the study of the overall process in a plant. It
analyses each step of the manufacturing process and
aims at improving the industrial operations. It helps in
finding better methods of doing a job, and this is achieved
by eliminating the unproductive and unnecessary
elements of the process or through modified layout of
facilities.
A process is analysed with the help of Process Charts and
Flow diagrams.
The steps involved in Process Analysis are:
1. Select the process for analysis
2. Break down the process into operations and
sub-operations
3. Construct a process chart and a flow diagram
4. Analyse the process chart and flow diagram by
subjecting each and every step to detailed
questioning
5. Reconstruct the process chart and flow diagram
6. Test the proposed method for all advantages
claimed
7. Explain the new method to the workers before
putting it into place
Process Chart
It is a diagram which gives an overall view of the situation,
in this case a process. It helps visualising various
possibilities of improvement.
A chart representing a process may be called a Process
Chart. The chart illustrates the process with the help of a
set of symbols and helps in better understanding.
Process Chart Symbols
Event Symbol
Operation
Storage
Temporary Storage
Transport
Inspection
Operation cum transportation
Inspection cum operation
Manual Process Planning
This type of planning is called man-variant
process planning and is the commonest type of
planning used for production today
The planner selects the combination of
processes required to produce a finished part.
In selecting this combination of processes, a
number of criteria are employed
Production cost or time are the dominant
criteria in process selection; but machine
utilization and routing often affects the plans
chosen
Automated Process Planning
Man variant planning is often a boring and
tedious job, with errors in human judgement
To eliminate this automated process planning
has come to aid of industrial process planning
Figure (a) below shows a completely
automated process planning system
Figure (b) shows a fully automated system with
human assistance to code the engineering
drawing data

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Production planning and control (PPC)
Introduction
Production is done my manufacturing different things
with various processes. Planning looks ahead, anticipates
possible difficulties and decides in advance about the
production. The control phase makes sure that the
programmed production is constantly maintained.
A production planning and control (PPC) system has many
functions to perform like:-
Planning phase:- Forecasting, order writing,
product, product design, material control, tool
control, loading etc.
Action phase:- Dispatching
Control phase:- Data processing, expediting and
replanning
Continuous and Intermittent Production
In continuous production, there is a continuous flow of
material, which is achieved by using special machines and
produces standardized items in large quantities. A
continuous production system can be divided into two
categories:-
1. Mass and flow line production
2. Continuous or process production
In an intermittent production, there is an intermittent or
interrupted flow of material. In this system we make use
of general purpose machines and produce different
components of different nature in small quantities.
Intermittent production systems can be classified as:-
1. Batch production
2. Job production
Job Shop, Open Job Shop and Closed Job Shop
In a job shop, there is involvement of intermittent
production. It consists of a number of machine centres,
but each with a different activity to perform.
In a job shop the material in-process follows
different processing patterns in batches through
batch facilities
The material does not flow in a serial fashion
A job shop makes to order and are not open to
orders from just any source
A closed job shop is one which is closed to job orders from
outside the organization, e.g. machine shop of a big
concern making automobile parts. In this standardized
parts are made which have a certain market demand.
An open job shop produces to order with a non-repetitive
trend. In this kind of setup, the products are made to
order as per the requirements of the customer.
Forecasting
It means estimation of type, quantity and quality of
future work, e.g. sales etc. Forecasting plays an important
role in planning for the future. The purpose of forecasting
is:-
To determine the production volume and rate
To prepare production budget for production
processes etc.
Basic elements of forecasting are:-
Trends
Cycles
Seasonal variations
Irregular variations
Forecasting techniques:-
Historic estimate
Sales force estimate
Market survey
Delphi method
Judgemental techniques
Prior knowledge
Forecasting by past average
Forecasting from last period sales
Moving average method
Weighted moving average method
Exponential smoothening method
Correlation analysis
Regression analysis
Forecasting by past average
It is used when our aim is to forecast or predict the sales
of an item for the next sales period.
𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑠𝑎𝑙𝑒𝑠 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑛𝑒𝑥𝑡 𝑝𝑒𝑟𝑖𝑜𝑑
= 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑎𝑙𝑒𝑠 𝑓𝑜𝑟 𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑝𝑒𝑟𝑖𝑜𝑑
Example:
Period No. Sales
1 7
2 5
3 9
4 8
5 5
6 8
Forecasted sales for period number 7 =
7+5=9+8+5+8
6
= 7
Forecasting by last period’s sales
This method eliminates the influence of old data and
bases the forecast only upon the sales of the previous
period.
Example:
Period
No.
Actual
Sales
Forecast
sales
Errors in
forecast
1 5
2 4 5 +1
3 8 4 -4
4 7 8 +1
5 4 7 +3
4
Forecasting by Moving Average
This is a compromise between the two methods
explained above, in which the forecast is neither

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influenced by very old data nor does it solely reflect the
figure of the previous period.
Consider the sales figures shown in the table below,
which can be used to construct a sales forecast for the
next year.
Year Period Sales
Four-period
moving
average
forecast
1987 1 50
2 60
3 50
4 40
1988 1 50 50
2 55 50
3 40 48.75
4 30 46.25
1989 1 35 43.75
2 45 40
3 35 37.5
4 25 36.25
1990 1 35 35
2 45 35
3 35 35
4 30 36.25
Weighted Moving Average (WMA)
In simple moving average, we have equal effects to each
component of the moving average data base, in a
weighted moving average we can place any weights on
each element, provided that the sum of all the weights is
equal to one.
Suppose that in a four month period the best forecast is
derived by using 40% of the actual sales for the most
recent month, 30% of the two month ago, 20% of three
months ago and 10% of four months ago. If actual sales
were as follows;
Month
1
Month
2
Month
3
Month
4
Month
5
100 90 105 95 ?
The forecast for month 5 can be found out by;
𝐹𝑠 = 0.40(95) + 0.30(105) + 0.20(90) + 0.10(100)
= 97.5
Let the actual sales for month 5 came out to be 110, then
the month 6 forecast will be
𝐹𝑠 = 0.40(110) + 0.30(95) + 0.20(105) + 0.10(90)
= 102.5
Forecasting by exponential smoothening
With the help of this technique, we just need to retain the
previous forecast figure and the latest actual sales figure.
𝑛𝑒𝑤 𝑓𝑜𝑟𝑒𝑐𝑎𝑠𝑡 = 𝛼(𝑙𝑎𝑡𝑒𝑠𝑡 𝑠𝑎𝑙𝑒𝑠 𝑓𝑖𝑔𝑢𝑟𝑒)
+ (1 − 𝛼)(𝑜𝑙𝑑 𝑓𝑜𝑟𝑒𝑐𝑎𝑠𝑡)
The term ‘α’ is known as smoothing constant
The use of this technique permits to respond to recent
actual events, but at the same time maintain certain
amount of stability.
The smoothing constant indicates the amount
by which the new forecast responds to the
latest sales figure, and its value lies between
0.1 to 0.3
To find out the smoothing constant, that gives the
equivalent of an N-period moving average, use the
relation,
𝛼 =
2
𝑁 + 1
Process planning
It means the preparation of work detail plan, by
determining the most economical method of performing
an operation of activity. The information needed to do
process planning is:
Quantity of work to be done
Quality of work to be completed
Availability of tools, equipments and personnel
Sequence of operations to be performed
Standard time for each operation
Procedure for process planning:-
1. Selection of process
2. Selection of material
3. Selection of jigs, fixtures and other special
attachments
4. Selection of cutting tools and inspection gauges
Operations Research & Inventory Control
Introduction
Let an industrialist has two industries (A and B) at
different locations. He wants to send the finished goods
to five different locations. To do this task, there are
several ways. The point of discussion here is which way,
out of the several ways, is the best alternative to send the
goods to the five locations in sight. For the industrialist in
question, the best alternative would be the one in which
he has to pay minimum transportation charges, which
becomes an Optimum condition.
By the term optimum condition means, a point where all
the conditions are favourable.
The approach to optimisation involves the following;-
The criteria which judges the best of the several
alternatives
Characteristics of the various alternatives being
judged
Methods available to judge the best
performance for the selected criteria
Operations Research
It signifies research on operations by taking into
consideration a particular view of operations and a
particular kind of research. The purpose of this subject

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and techniques used is to provide the management with
explicit quantitative understanding and assessment of
complex situations, helping them make better decisions.
Linear Programming
It is defined as the optimisation of a linear function of
variables subject to constraints of linear inequalities. A
LPP consists of three components:
1. Decision variables (activities)
2. The objective function (Goal)
3. The constraints (restrictions)
Objective function: It is a clearly identifiable and
measureable quantity
Constraints: These are limited resources within which we
have to obtain optimised solution for the objective
function.
Three different types of solution:
1. Infinite Solution: The objective function slope
equals to one of the constraints which forms
the boundary
2. No solution: These is no solution possible for
the given LPP
3. Unbounded Solution: The greatest value of
objective function occurs at infinity and it
simply means that the common feasible region
is not bounded by limits on constraints
Simplex method
Procedure
RHS of each constraint should be non negative
Each decision variable of the problem should be
non negative
Inequalities in constraints should be converted
to equalities
Set m = no. of equality constraints and n= no. of
variables
Put (n-m) variable equal to zero
(n-m) = non basic variable
m= basic variable
Special case:
Infinite solution: When a non basic variable in
an optimal solution has a zero value for Δj row
then the solution is not unique
Unbounded solution: When all replacement
ratios are either infinite (or) negative then the
solution terminates. This indicates the problem
has unbounded solution
Infeasible solution: When in the final solution
an artificial variable is in the basis then there is
no feasible solution to the problem
Duality in LP
For every LP problem there exists a related unique LP
problem involving the same data which also describes
and solves original problem
Primal Dual
Maximum Minimum
No. of variables No. of constraints
No. of constraints No. of variables
type of constraints Non negative variables
= type constraints Unrestricted variables
Unrestricted variable = type constraints
Big M method: In those situations where an identity
matrix is not obtained initially another form of simplex
method called Big M method is applied. In this method,
artificial variable are put into the model to obtain an
initial solution.
Transportation Problem
These problems are used for meeting the supply and
demand requirements under given conditions in the best
optimal effective manner.
Cij= cost of transportation of one unit from the ith source
to the jth destination
Xij= Quantity to be transported from ith source to the jth
destination
𝑡𝑜𝑡𝑎𝑙 𝑡𝑟𝑎𝑛𝑠𝑝𝑜𝑟𝑡𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑠𝑡 = ∑ ∑ 𝐶𝑖𝑗 𝑋𝑖𝑗
𝑚
𝑗=1
𝑛
𝑖=1
Feasible Solution: A set of non negative individual
allocations which also satisfy the given constraints
Basic Feasible Solution: A basic feasible solution of mXn
TP is basic feasible if the total number of allocations is
exactly the equal to (m+n-1)
Optimal Solution: A feasible solution is said to be
optimal, if it minimizes the total transportation cost.
Non degenerate Basic Feasible Solution: A feasible
solution of mXn TP is non degenerate is
Total number of allocations is exactly equal to
(m+n-1)
These allocations are in independent positions
Unbalanced TP
If total supply from all sources equals total
demand in all destinations, then the TP is
balanced, otherwise unbalanced
If given TP is unbalanced, then make the
problem balanced by using a dummy source or
destination
Degeneracy
When the number of allocations are less then (m+n-1)
then optimality test cannot be performed and such a
solution is called degenerate solution.
Assignment problem
These problems are special cases of TP where
Matrix must be a square matrix
The optimal solution to the problem would
always be such that there would only be one
assignment in the given row or column
Hungarian method is used to solve an AP
Steps for applying Hungarian method:
Subtract the smallest element of each row to
the every element of corresponding row

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Subtracting the smallest element of each
column to every element in corresponding
columns
Now all zeroes are to be covered with minimum
number of lines
If number of lines is equal to the number of
rows or columns then optimal solution is
obtained
Queuing Theory
Queue means the number of customers waiting to be
serviced. The queue does not include the customer being
serviced. The process which serves the customer is called
service facility.
Elements of a Queuing system:
1. Input or arrival process:
Size of queue
Pattern of arrivals
Customer’s behaviour
2. Queue discipline
3. Service mechanism
Single queue one server
Single queue several server
Several queue one server
Several queue several server
4. Capacity of the system
Operating characteristics of a Queuing system:
Expected number of customers in the system is
denoted by [E(n)] or L. it is the average number
of customers in the system, both waiting and
being serviced
Expected number of customers in the queue
[E(m)] or Lq. it is the average number of
customers waiting in queue
Here, m=n-1, i.e. excluding the customer being
serviced, or Lq = L-1
Expected waiting time in the system E(v) or w,
is the average total time spent by a customer in
the system. It is generally taken to be equal to
waiting time + service time
Expected waiting time in queue denoted by
E(w) or wq. it is the average time spent by a
customer in the queue before the
commencement of the service
The server utilization factor, 𝑃 = 𝜆/𝜇. It is the
proportion of time that a service actually stands
with a customer,
Here 𝜆 = average number of customers
arriving per unit time and 𝜇 = average number
of customers completing service per unit time.
The value P is also known as traffic intensity or
the clearing ratio
Deterministic queuing system
A queuing system wherein the customers arrive at regular
intervals and service time for each customer is known
and constant.
1. if > 𝜇 , the waiting line shall be formed and will
increase indefinitely. The service facility would
always be busy and service system will
eventually fail
2. if 𝜆 ≤ 𝜇, there shall be no queue and hence no
waiting time. The proportion of time the service
facility would be idle is 1 −
𝜆
𝜇
.
Probabilistic Queuing system
It is assumed that customers joining the queuing system
arrive in a random manner (variable) and follow a
Poisson’s distribution.
Queuing Model
N(t) = no. of customers in queuing system at time t
S = no. of servers in queuing system
Pn(t) = Probability of n units in queuing system
𝜆 𝑛= mean arrival rate (units/ unit time)
Lq = average number of customers in queue system
n= mean number of units in the queuing system including
the one being served
ws= average waiting time in the queue
wq= average time the queue system
M/M/1 : (∞/ FIFO)
Single service channel, Poisson’s input, exponential
service, no limit on the system capacity. First In First Out.
𝑃𝑛 = 𝜌 𝑛(1 − 𝜌), 𝑤ℎ𝑒𝑟𝑒 𝜌 =
𝜆
𝜇
< 1 𝑎𝑛𝑑 𝑛 ≥ 0
inter arrival time = 1/λ
traffic intensity facto, 𝜌 = 𝜆/𝜇
average number of customers in system, 𝐿 =
𝜌/(1 − 𝜌)
average number of customer of queue or
average queue length, 𝐿 𝑞 = 𝐿 − 𝜌
Waiting time in system, 𝑤 =
1
𝜇−𝜆
Average or expected waiting time in queue,
𝑤𝑞 =
𝐿 𝑞
𝜆
Probability that the service facility is idle,
𝑃0(𝑡) = (1 − 𝜌)
Probability that the service facility has n
customers at time t, 𝑃𝑛(𝑡) = 𝜌 𝑛
𝑃0(𝑡)
Average length of non empty queue, 𝐿 𝑛 =
1
1−𝜌
The fluctuation of queue length, 𝑣(𝑛) =
𝜌
(1−𝜌)2
Probability of n arrivals in time t, 𝑃(𝑛𝑡) =
𝑒−𝜆(𝜆𝑡) 𝑛
𝑛!
Probability that the waiting time in the queue
is greater than or equal to t, 𝑃(𝑤𝑞 ≥ 𝑡) =
𝜆
𝑒−(𝜇−𝜆)𝑡
𝜇
Probability that waiting time in system is
greater than or equal to t, 𝑃 = 𝑒−(𝜇−𝜆)𝑡
Probability that waiting time in system is less
than or equal to t, 𝑃 = 1 − 𝑒−(𝜇−𝜆)𝑡
Inventory Control
Inventory is defined as the list of movable goods which
helps directly or indirectly in production of goods for
sale. We can also defined inventory as a comprehensive
of goods for sale. We can also defined inventory as a
comprehensive list of movable items which are required

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for manufacturing the products and to maintain the
plant facilities in working conditions.
It can be divided in two parts.
Direct Inventories
The inventories which play a direct role in
manufacturing of a product and become an integral part
of the finished product are called direct inventories. e.g.,
raw materials, purchases part and finished goods.
Indirect inventories
The inventories which helps the raw material to get
converted into products but not integral part of finished
product is called indirect inventories, e.g., tool and
supplies (material used in running the plant but do not
go into the product) are indirect inventories.
Inventory Control
It means making the desired item of required quality
and in required quantity available to various
departments when needed.
Determining Inventory Control
The amount of inventory, a company should carry is
determined by five basic variables.
(a) Order quantity
(b) Reorder point
(c) Lead time
(d) Safety stock
(e) Butter stock
Order Quantity
It is the volume of stock at which order is placed or total
quantity of buy or sell order.
Reorder Point
It is time between initiating the order and receiving the
required quantity. Reorder point = Minimum inventory +
Procurement time Consumption rate.
Lead Time
The time gap between placing of an order an its actual
arrival in the inventory is known as lead time. It consist
of requisition time and procurement time.
It has two components.
Administrative Lead Time
From initiation of procurement action until the placing
of an order.
Delivery Lead Time
From placing of an order until the delivery of the odered
material.
Safety Stock
If the maximum inventory would be equal to the order
quantity Q and minimum inventory would be zero.
Average inventory in this case =
Q
2
Safety stock = k
Average Consumption duringlead time
k = A factor abased on acceptable frequency of
stock out in a given number of years.
Buffer Stock
For an average demand during average lead time the
additional stock termed as buffer stock.
Buffer stock = Average demand Average lead time
When no stock outs are desired
Buffer stock = Maximum demand during lead time
(DDLT) Average Demand During Lead Time (DDLT)
When demand rate varies about the average
demand during a constant lead time (LT) period
Reorder Level (ROL) = Average (DDLT) LT + BS
Inventory Cost
The costs that are affected by firm’s decision to maintain
particular level of inventory are called cost associated
with inventories or relevant inventory cost.
Total Inventory Costs (TIC)
TIC = Purchase cost + Total Variable Cost (TVC) of
managing the inventory
TIC = Purchase cost + Inventory cost + Ordering cost +
Shortage cost
Purchase Cost
It is defined as the cost of purchasing a unit of an item.
Purchase cost = Price per unit Demand per unit time
where, Cu = Unit cost
D = Annual demand
Ordering Cost
It is defined as the cost of placing an order from a
vector. This represents the expenses involved in placing
an order with the outside supplier. This includes the
costs involved in processing and ordering for purchase.
expediting over the orders, receiving the consignment
and inspection.

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Annual ordering cost=
oC
Q
D
where,
Q = Produced purchased or supplied throughout the
entire time period (one year) or order quantity.
Co = Cost of placing an order
D = Annual demand
Carrying Cost
Carrying or holding costs are the costs incurred
maintaining the stores in the firm. It is proportional to
the amount of inventory and the time over which it is
held.
Annual carrying cost Cui = Cu i
a
2
where, Cu = Unit purchase cost,
i = Interest rate
Shortage Cost
When an item cannot be supplied on consumer’s
demand, the penalty cost for running out of stock is
called shortage cost or stock out cost.
Shortage cost = Cost of being short one unit in inventory
Average number of unit short in the inventory.
Economic Order Quantity (EOQ)
Economic order quantity is the order quantity that
minimizes total inventory holding costs and ordering
costs. It is one of the oldest classical production
scheduling models.
Assumptions of EOQ Model
The ordering cost is constant
The rate of demand is known
Lead time is fixed Purchase price
of the item is constant
Replenishment is made Only one
product is involved.
instantaneously
EOQ When Stock Replenishment is Instantaneous
Economic Order Quantity (EOQ) is the order quantity
that minimizes total inventory holding costs and
ordering costs. It is one of the oldest classical production
scheduling models.
It is defined as the quantity which will minimise the
total variable cost of managing the inventory.
TVC =
oC Q
D
Q 2
Cu i
EOQ (Q ) =
0
u
2C D
C i
where, Co = Cost of placing an order, Cu = Unit purchase
cost
i = interest rate, D= Annual consumption of the product.
Optimum number of order placed per year
no= o
D
Q
=
u
o
DC i
2C
where, Qo = Economic order quantity
Inventory Models
Broadly categorising, inventory models are of two type:
1. Static Inventory Model
Only one order can be placed to meet the
demand, as repeated orders are deemed too
expensive
2. Dynamic Inventory Model
In this model, the order can be repeated again
and again to replenish the stock. Further, the
dynamic inventory model can be classified into:
a) Deterministic Model
Both the lead time and demand for an item are
pre determined
b) Probabilistic Model
In this, both the demand and lead time are not
known as both keep on varying
Questions:
1. Fifty observations of a production operation
revealed a mean cycle time of 10 min . The worker
was evaluated to be performing at 90% efficiency.
Assuming the allowances to be 10% of the normal
time, the standard time (in second) for thejob is
(A) 0.198 (B) 7.3
(C) 9.0 (D) 9.9
2. When using a simple moving average to forecast
demand, one would
(A) give equal weight to all demand data
(B) assign more weight to the recent demand data
(C) include new demand data in the average without
discarding the earlier data
(D) include new demand data in the average after
discarding some of the earlier demand data
3. Production flow analysis (PFA) is a method of
identifying part families that uses data from
(A) engineering drawings
(B) production schedule
(C) bill of materials
(D) route sheets
4. A project consists ofthree parallel paths with mean
durations and variances of (10, 4) , (12, 4) and
(12, 9) respectively. According to the standard PERT
assumptions, the distribution of the project duration
is
(A) beta with mean 10 and standard deviation 2
(B) beta with mean 12 and standard deviation 2
(C) normal with mean 10 and standard deviation 3
(D) normal with mean 12 and standard deviation 3
5. The supplies at three sources are 50, 40 and 60 unit
respectively whilst the demands at the four

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destinations are 20, 30, 10 and 50 unit. In solving this
transportation problem
(A) a dummy source of capacity 40 unit is needed
(B) a dummy destination of capacity 40 unit is
needed
(C) no solution exists as the problem is infeasible
(D) no solution exists as the problem is degenerate
6. Arrivals at a telephone booth are considered to be
Poisson, with an average time of 10 minutes
between successive arrivals. The length of a phone
call is distributed exponentially with mean 3
minutes. The probability that an arrival does not
have to wait before service is
(A) 0.3 (B) 0.5
(C) 0.7 (D) 0.9
7. An item can be purchased for Rs. 100. The ordering
cost is Rs. 200 and the inventory carrying cost is 10%
of the item cost 𝑝 er annum. If the annual demand is
4000 unit, the economic order quantity (in unit) is
(A) 50 (B) 100
(C) 200 (D) 400
8. In carrying out a work sampling study in a machine
shop, it was found that a particular lathe was down
for 20% of the time. What would be the 95%
confidence interval of this estimate, if 100
observations were made?
(A) (0.16, 0.24) (B) (0.12, 0.28)
(C) (0.08, 0.32) (D) None of these
9. The standard time of an operation while conducting
a time study is
(A) mean observed time + allowances
(B) normal time + allowances
(C) mean observed time × rating factor +
allowances
(D) normal time × rating factor + allowances
10. The principles of motion economy are mostly used
while conducting
(A) a method study on an operation
(B) a time study on an operation
(C) a financial appraisal of an operation
(D) a feasibility study of the proposed manufacturing
plant
11. A project consists of activities 𝐴 to 𝑀 shown in the
net in the following figure with the duration of the
activities marked in days
The project can be completed
(A) between 18, 19 days
(B) between 20, 22 days
(C) between 24, 26 days
(D) between 60, 70 days
12. A manufacturer produces two types ofproducts, 1
and 2, at production levels of 𝑥1 and 𝑥2 respectively.
The profit is given is 2𝑥1 + 5𝑥2. The production
constraints are
𝑥1 + 3𝑥2 ≤ 40
3𝑥1 + 𝑥2 ≤ 24
𝑥1 + 𝑥2 ≤ 10
𝑥1 > 0, 𝑥2 > 0
The maximum profit which can meet the constraints
is
(A) 29 (B) 38
(C) 44 (D) 75
13. Market demand for springs is 8,00,000 per annum.
A company purchases these springs in lots and sells
them. The cost of making a purchase order is Rs.
1200. The cost of storage of springs is Rs. 120 per
stored piece per annum. The economic order
quantity is
(A) 400 (B) 2,828
(C) 4,000 (D) 8,000
14. The sale of cycles in a shop in four consecutive
months are given as 70, 68, 82, 95. Exponentially
smoothing average method with a smoothing factor
of 0.4 is used in forecasting. The expected number
of sales in the next month is
(A) 59 (B) 72
(C) 86 (D) 136
15. A residential school stipulates the study hours as
8.00 pm to 10.30 𝑝𝑚. Warden makes random checks
on a certain student 11 occasions a day during the
study hours over a period of10 days and observes
that he is studying on 71 occasions. Using 95%
confidence interval, the estimated minimum hours
of his study during that 10 day period is
(A) 8.5 hours (B) 13.9 hours
(C) 16.1 hours (D) 18.4 hours
16. Two machines of the same production rate are
available for use. On machine 1, the fixed cost is Rs.
100 and the variable cost is Rs. 2 per piece produced.
The corresponding numbers for the machine 2 are
Rs. 200 and 𝑅𝑒. 1 respectively. For certain strategic
reasons both the machines are to be used
concurrently. The sales price of the first 800 units is
Rs. 3.50 per unit and subsequently it is only Rs. 3.00.
The breakeven production rate for each machine is
(A) 75 (B) 100
(C) 150 (D) 600
17. The symbol used for Transport in work study is
(A)⇒ (B) 𝑇
(C) (D) 𝛻
18. A company produces two types of toys: 𝑃 and 𝑄.
Production time of 𝑄 is twice that of 𝑃 and the
company has a maximum of2000 time units per day.
The supply of raw material is just sufficient to
produce 1500 toys (of any type) per day. Toy type 𝑄
requires an electric switch which is available @ 600
pieces per day only. The company makes a profit of

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Rs. 3 and Rs. 5 on type 𝑃 and 𝑄 respectively. For
maximization ofprofits, the daily production
quantities of 𝑃 and 𝑄 toys should respectively be
(A) 1000, 500 (B) 500, 1000
(C) 800, 600 (D) 1000, 1000
19. A company has an annual demand of 1000 units,
ordering cost of Rs. 100 / order and carrying cost of
Rs. 100/ unit/year. If the stock‐out cost are
estimated to be nearly Rs. 400 each 𝑡 ime the
company runs out‐of‐stock, then safety stock
justified by the carrying cost will be
(A) 4 (B) 20
(C) 40 (D) 100
20. A maintenance service facility has Poisson arrival
rates, negative exponential service time and
operates on a ‘first come first served’ queue
discipline. Breakdowns occur on an average of 3 per
day with a range of zero to eight. The maintenance
crew can service an average of 6 machines per day
with a range of zero to seven. The mean waiting time
for an item to be serviced would be
(A)
1
6
day (B)
1
3
day
(C) 1 day (D) 3 day
21. An electronic equipment manufacturer has decided
to add a component sub‐ assembly operation that
can produce 80 units during a regular 8‐hours shift.
This operation consist of three activities as below
Activity Standard time ( min )
M. Mechanical assembly 12
E. Electric wiring 16
T. Test 3
For line balancing the number of work stations
required for the activities 𝑀, 𝐸 and 𝑇 would
respectively be
(A) 2, 3, 1 (B) 3, 2, 1
(C) 2, 4, 2 (D) 2, 1, 3
22. A soldering operation was work‐sampled over two
days (16 hours) during which an employee soldered
108 joints. Actual working time was 90% of the total
time and the performance rating was estimated to
be 120 per cent. If the contract provides allowance
of 20 percent of the time available, the standard
time for the operation would be
(A) 8 min (B) 8.9 min
(C) 10 min (D) 12 min
23. A standard machine tool and an automatic machine
tool are being compared for the production of
component. Following data refers to the two
machines.
Standard
Machine Tool
Automatic
Machine Tool
Setup time 30 min 2 hours
Machining time per piece 22 min 5 min
Machine rate Rs. 200 per hour Rs. 800 per
hour
The break even production batch size above which the
automatic machine tool will be economical to use, will be
(A) 4 (B) 5
(C) 24 (D) 225
24. There are two products 𝑃 and 𝑄 with the following
characteristics
Product Deman
d
(Units)
Order
cost
( 𝑅𝑠/
order)
Holding Cost
(Rs. /unit/year)
𝑃 100 50 4
𝑄 400 50 1
The economic order quantity (EOQ) of products 𝑃 and 𝑄
will be in the ratio
(A) 1 : 1 (B) 1 : 2
(C) 1 : 4 (D) 1 : 8
25. For a product, the forecast and the actual sales for
December 2002 were 25 and 20 respectively. If the
exponential smoothing constant (𝛼) is taken as 0.2,
then forecast sales for January 2003 would be
(A) 21 (B) 23
(C) 24 (D) 27
26. In PERT analysis a critical activity has
(A) maximum Float
(B) zero Float
(C) maximum Cost
(D) minimum Cost
Common Data for 𝑄.27 and 𝑄.28
Consider a linear programming problem with two
variables and two constraints. The objective function is:
Maximize𝑋1 + 𝑋2. The corner points of the feasible
region are (0,0), (0,2), (2,0) and (4/3, 4/3)
27. If an additional constraint 𝑋1 + 𝑋2 ≤ 5 is added, the
optimal solution is
(A) (
55
3’3
) (𝐵) (
4
3’
4
3
)
(C) (
55
2’2
) (D) (5, 0)
28. Let 𝑌1 and 𝑌2 be the decision variables of the dual
and 𝑣1 and 𝑣2 be the slack variables of the dual of
the given linear programming problem. The
optimum dual variables are
(A) 𝑌1 and 𝑌2 (B) 𝑌1 and 𝑣1
(C) 𝑌1 and 𝑣2 (D) 𝑣1 and 𝑣2
29. A company has two factories 1, 𝑆2, and two
warehouses 𝐷1, 𝐷2. The supplies from 𝑆1 and 𝑆2
are 50 and 40 units respectively. Warehouse 𝐷1
requires a minimum of 20 units and a maximum of
40 units. Warehouse 𝐷2 requires a minimum of 20
units and, over and above, it can take as much as can
be supplied. A balanced transportation problem is to
be formulated for the above situation. The number
of supply points, the number of demand points, and
the total supply (or total demand) in the balanced
transportation problem respectively are
(A)2,4,90 (𝐵)2,4,110

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(C)3,4,90 (𝐷)3,4,110
30. A project has six activities (A to 𝐹) with respective
activity duration 7, 5, 6, 6, 8, 4 days. The network has
three paths A‐B, C‐D and E‐F. All the activities can be
crashed with the same crash cost per day. The
number of activities that need to be crashed to
reduce the project duration by 1 day is
(A) 1 (B) 2
(C) 3 (D) 6
31. The distribution of lead time demand for an item is
as follows:
Lead time
demand
𝑃 robability
80 0.20
100 0.25
120 0.30
140 0.25
The reorder level is 1.25 times the expected value of the
lead time demand. The service level is
(A) 25% (B) 50%
(C) 75% (D) 100%
32. A welding operation is time‐studied during which an
operator was pace‐rated as 120%. The operator
took, on an average, 8 minutes for producing the
weld‐ joint. If a total of10% allowances are allowed
for this operation. The expected standard
production rate of the weld‐joint (in units per 8 hour
day) is
(A) 45 (B) 50
(C) 55 (D) 60
33. A component can be produced by any of the four
processes I, II, III and IV. Process I has a fixed cost of
Rs. 20 and variable cost of Rs. 3 per piece. Process II
has a fixed cost Rs. 50 and variable cost of Rs. 1 per
piece. Process III has a fixed cost of Rs. 40 and
variable cost of Rs. 2 per piece. Process IV has a fixed
cost of Rs. 10 and variable cost of Rs. 4 per piece. If
the company wishes to produce 100 pieces of the
component, form economic point of view it should
choose
(A) Process I (B) Process II
(C) Process III (D) Process IV
34. Consider a single server queuing model with Poisson
arrivals (𝜆 = 4/hour) and exponential service (𝜇 =
4/hour) . The number in the system is restricted to
a maximum of 10. The probability that a person who
comes in leaves without joining the queue is
(A)
1
11
(B)
1
10
(C)
1
9
(D)
1
2
35. The sales of a product during the last four years were
860, 880, 870 and 890 units. The forecast for the
fourth year was 876 units. If the forecast for the fifth
year, using simple exponential smoothing, is equal
to the forecast using a three period moving average,
the value ofthe exponential smoothing constant 𝛼 is
(A)
1
7
(B)
1
5
(C)
2
7
(D)
2
5
36. An assembly activity is represented on an Operation
Process Chart by the symbol
(A) □ (B) 𝐴
(C) 𝐷 (D) 𝑂
37. The standard deviation of the critical path of the
project is
(A)√151 days (𝐵)√155 days
(C)√200 days (𝐷)√238 days
38. The expected completion time of the project is
(A) 238 days (B) 224 days
(C) 171 days (D) 155 days
39. The table gives details of an assembly line.
Work station I II III IV V VI
Totaltask time at the
workstation
(in minutes)
7 9 7 10 9 6
What is the line efficiency of the assembly line?
(A) 70% (B) 75%
(C) 80% (D) 85%
40. A stockist wishes to optimize the number of
perishable items he needs to stock in any month in
his store. The demand distribution for this
perishable item is
Demand (in unit s) 2 3 4 5
𝑃 robability 0.10 0.35 0.35 0.20
The stockist pays Rs. 70 for each item and he sells each at
Rs. 90. If the stock is left unsold in any month, he can sell
the item at Rs. 50 each. There is no penalty for unffilfilled
demand. To maximize the expected profit, the optimal
stock level is
(A) 5 units (B) 4 units
(C) 3 unit 𝑠 (D) 2 unit 𝑠
41. Consider the following data for an item.
Annual demand : 2500 units per year,
Ordering cost : Rs. 100 𝑝 er order,
Inventory holding rate : 25% of unit price
Price quoted by a supplier
Order quantity (units) Unit price (Rs.)
< 500 10
≥ 500 9
The optimum order quantity (in units) is
(A) 447 (B) 471
(C) 500 (D) ≥ 600
42. An manufacturing shop processes sheet metal jobs,
wherein each job must pass through two machines
(𝑀𝑙 𝑎𝑛𝑑 𝑀2, 𝑖𝑛 that order). The processing time (in
hours) for these jobs is

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Machine Jobs
P Q R S T U
M1 15 32 8 27 11 16
M2 6 19 13 20 14 7
The optimal make‐span (in‐hours) of the shop
is
(A) 120 (B) 115
(C) 109 (D) 79
43. A firm is required to procure three items
(𝑃, 𝑄, 𝑎𝑛𝑑 𝑅) . The prices quoted for these items (in
Rs.) by suppliers 𝑆1, 𝑆2 and 𝑆3 are given in table. The
management policy requires that each item has to
be supplied by only one supplier and one supplier
supply only one item. The minimum total cost (in
Rs.) of procurement to the firm is
Item Suppliers
S1 S2 S3
P 110 120 130
Q 115 140 140
R 125 145 165
(A) 350 (C) 385
(B) 360 (D) 395
44. In an MRP system, component demand is
(A) forecasted
(B) established by the master production
schedule
(C) calculated by the MRP system from the
master production schedule
(D) ignored
45. The number of customers arriving at a railway
reservation counter is Poisson distributed with an
arrival rate of eight customers per hour. The
reservation clerk at this counter takes six minutes
per customer on an average with an exponentially
distributed service time. The average number of the
customers in the queue will be
(A) 3 (B) 3.2
(C) 4 (D) 4.2
46. The net requirements of an item over 5 consecutive
weeks are 50‐0‐15‐20‐20. The inventory carrying
cost and ordering cost are Rs. 1 per item per week
and Rs. 100 per order respectively. Starting
inventory is zero. Use “Least Unit Cost Technique”
for developing the plan. The cost of the plan (in Rs.)
is
(A) 200 (B) 250
(C) 225 (D) 260
47. In a machine shop, pins of 15 mm diameter are
produced at a rate of 1000 per month and the same
is consumed at a rate of 500 per month. The
production and consumption continue
simultaneously till the maximum inventory is
reached. Then inventory is allowed to reduced to
zero due to consumption The lot size of production
is 1000. If backlog is not allowed, the maximum
inventory level is
(A) 400 (B) 500
(C) 600 (D) 700
48. The maximum level of inventory of an item is 100
and it is achieved with infinite replenishment rate.
The inventory becomes zero over one and half
month due to consumption at a uniform rate. This
cycle continues throughout the year. Ordering cost
is Rs. 100 per order and inventory carrying cost is Rs.
10 per item per month. Annual cost (in Rs.) of the
plan, neglecting material cost, is
(A) 800 (B) 2800
(C) 4800 (D) 6800
49. Capacities of production of an item over 3
consecutive months in regular time are 100, 100 and
80 and in overtime are 20, 20 and 40. The demands
over those 3 months are 90, 130 and 110. The cost
of production in regular time and overtime are
respectively Rs. 20 𝑝 er it em and Rs. 24 𝑝er item.
Inventory carrying cost is Rs. 2 per item per month.
The levels of starting and final inventory are nil.
Backorder is not permitted. or minimum cost of
plan, the level of planned production in overtime in
the third month is
(A) 40 (B) 30
(C) 20 (D) 0
Common Data For 𝑄.50 and 𝑄.51
Consider the Linear Programme (𝐿𝑃) Max 4𝑥 + 6𝑦
Subject to
3𝑥 + 2𝑦 ≤ 6
2𝑥 + 3𝑦 ≤ 6
𝑥, 𝑦 ≥ 0
50. After introducing slack variables 𝑠 and 𝑡, the initial
basic feasible solution is represented by the table
below (basic variables are 𝑠 = 6 and 𝑡 = 6, and the
objective function value is 0)
−4 −6 0 0 0
𝑠 3 2 1 0 6
𝑡 2 3 0 1 6
𝑥 𝑦 𝑠 𝑡 RHS
After some simplex iterations, the following
table is obtained
0 0 0 2 12
𝑠 5/3 0 1 −1/3 2
𝑦 2/3 1 0 1/3 2
𝑥 𝑦 𝑠 𝑡 RHS
From this, one can conclude that
(A) the 𝐿𝑃 has a unique optimal solution
(B) the 𝐿𝑃 has an optimal solution that is not
unique
(C) the 𝐿𝑃 is infeasible
(D) the 𝐿𝑃 is unbounded
51. The dual for the 𝐿𝑃 in 𝑄. 50 is
(A) Min 6𝑢 + 6𝑣 (B) Max 6𝑢 + 6𝑣
subject to
3𝑢 + 2𝑣 ≥ 4

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2𝑢 + 3𝑣 ≥ 6
𝑢, 𝑣 ≥ 0
(C) Max 4𝑢 + 6𝑣 subject to
3𝑢 + 2𝑣 ≥ 6
2𝑢 + 3𝑣 ≥ 6
𝑢, 𝑣 ≥ 0
subject to 3𝑢 + 2𝑣 ≤ 4 2𝑢 + 3𝑣 ≤ 6 𝑢, 𝑣 ≥ 0
(D) Min 4𝑢 + 6𝑣 subject to 3𝑢 + 2𝑣 ≤ 6 2𝑢 +
3𝑣 ≤ 6 𝑢, 𝑣 ≥ 0
52. The product structure of an assembly 𝑃 is shown in
the figure.
Estimated demand for end product 𝑃 is as follows
Week 1 2 3 4 5 6
Demand 1000 1000 1000 1000 1200 1200
ignore lead times for assembly and sub‐assembly.
Production capacity (per week) for component 𝑅 is the
bottleneck operation. Starting with zero inventory, the
smallest capacity that will ensure a feasible production
plan up to week 6 is
(A) 1000 (B) 1200
(C) 2200 (D) 2400
53. For the network below, the objective is to find the
length of the shortest path from node 𝑃 to node 𝐺.
Let 𝑑𝑖 𝑗
. be the length of directed arc from node 𝑖 to
node 𝑗.
Let 𝑆𝑗 be the length of the shortest path ffom 𝑃 to
node 𝑗. Which of the following equations can be
used to find 𝑆 𝐺 ?
(A) 𝑆 𝐺 = Min {𝑆 𝑄, 𝑆 𝑅}
(B) (𝐵) 𝑆 𝐺 = Min {𝑆 𝑄 − 𝑑 𝑄𝐺, 𝑆 𝑅 − 𝑑 𝑅𝐺}
(C) 𝑆 𝐺 = Min {𝑆 𝑄 + 𝑑 𝑄𝐺, 𝑆 𝑅 + 𝑑 𝑅𝐺}
(D) (𝐷) 𝑆 𝐺 = Min {𝑑 𝑄𝐺, 𝑑 𝑅𝐺}
54. A moving average system is used for forecasting
weekly demand 𝐹1(𝑡) and 𝐹2(𝑡) are sequences of
forecasts with parameters 𝑚1 and 𝑚2, respectively,
where 𝑚1 and 𝑚2(𝑚1 > 𝑚2) denote the numbers
of weeks over which the moving averages are taken.
The actual demand shows a step increase from 𝑑1 to
𝑑2 at a certain time. Subsequently,
(A) neither 𝐹1(𝑡) nor 𝐹2(𝑡) will catch up with the
value 𝑑2
(B) both sequences 𝐹1(𝑡) and 𝐹2(𝑡) will reach 𝑑2 in
the same period
(C) 𝐹1(𝑡) will attain the value 𝑑2 before 𝐹2(𝑡)
(D) 𝐹2(𝑡) will attain the value 𝑑2 before 𝐹1(𝑡)
55. For the standard transportation linear programme
with 𝑚 source and 𝑛 destinations and total supply
equaling total demand, an optimal solution (lowest
cost) with the smallest number of non‐zero 𝑥𝑖𝑗
values (amounts ffom source 𝑖 to destination j) is
desired. The best upper bound for this number is
(A) 𝑚𝑛 (B) 2 (𝑚 + 𝑛)
(C) 𝑚 + 𝑛 (D) 𝑚 + 𝑛 − 1
56. A set of 5 jobs is to be processed on a single machine.
The processing time (in days) is given in the table
below. The holding cost for each job is Rs. 𝐾 per day.
Job Processing time
𝑃 5
𝑄 2
𝑅 3
𝑆 2
𝑇 1
A schedule that minimizes the total inventory
cost is
(A) T‐S‐Q‐R‐P
(B) P‐R‐S‐Q‐T
(C) T‐R‐S‐Q‐P
(D) P‐Q‐R‐S‐T
57. In an 𝑀/𝑀/1 queuing system, the number ofarrivals
in an interval oflength 𝑇 is a Poisson random variable
(i.e. the probability of there being arrivals in an
interval of length 𝑇 is
𝑒−𝜆𝑇(𝜆𝑇) 𝑛
𝑛!
). The probability
density function 𝑓(𝑡) of the inter‐arrival time is
(A) 𝜆2
(𝑒−𝜆2 𝑡
) (B)
𝑒−𝜆2 𝑡
𝜆2
(C) 𝜆𝑒−𝜆𝑡
(𝐷)
𝑒−𝜆𝑡
𝜆
Common Data For 𝑄.58 and 𝑄.59
Consider the following PERT network:
The optimistic time, most likely time and pessimistic time
ofall the activities are given in the table below:
Activity Optimisti
c time
(days)
Most
likely
time
(days)
Pessimisti
c time
(days)
1 − 2 1 2 3
1 − 3 5 6 7
1 − 4 3 5 7
2 − 5 5 7 9
3 − 5 2 4 6
5 − 6 4 5 6
4 − 7 4 6 8
6 − 7 2 3 4

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58. The critical path duration of the network (in days) is
(A) 11 (B) 14
(C) 17 (D) 18
59. The standard deviation of the critical path is
(A) 0.33 (B) 0.55
(C) 0.77 (D) 1.66
60. Six jobs arrived in a sequence as given below:
Jobs Processing Time (days)
I 4
II 9
III 5
IV 10
V 6
VI 8
Average flow time (in days) for the above jobs using
Shortest Processing time rule is
(A) 20.83 (B) 23.16
(C) 125.00 (D) 139.00
61. Consider the following Linear Programming Problem
(LPP):
Maximize 𝑍 = 3𝑥1 + 2𝑥2
Subject to 𝑥1 ≤ 4
𝑥2 ≤ 6
3𝑥1 + 2𝑥2 ≤ 18
𝑥1 ≥ 0, 𝑥2 ≥ 0
(A) The LPP has a unique optimal solution
(B) The 𝐿𝑃𝑃 is infeasible.
(C) The 𝐿𝑃𝑃 is unbounded.
(D) The LPP has multiple optimal solutions.
62. A company uses 2555 units of an item annually.
Delivery lead time is 8 days. The reorder point (in
number of units) to achieve optimum inventory is
(A) 7
(B) 8
(C) 56
(D) 60
63. Which ofthe following forecasting methods takes a
fraction of forecast error into account for the next
period forecast?
(A) simple average method
(B) moving average method
(C) weighted moving average method
(D) exponential smoothening method
64. The expected time (𝑡 𝑒) of a PERT activity in terms of
optimistic time 𝑡0, pessimistic time (𝑡 𝑝) and most
likely time (𝑡𝑙) is given by
(A) 𝑡 𝑒 =
𝑡 𝑜+4𝑡 𝑇+𝑡 𝑝
6
(C) 𝑡 𝑒 =
𝑡 𝑜+4𝑡 𝑇+𝑡 𝑝
3
(B) 𝑡 𝑒 =
𝑡 𝑜+4𝑡 𝑝+𝑡 𝑇
6
(D) 𝑡 𝑒 =
𝑡 𝑜+4𝑡 𝑝+𝑡 𝑇
3
Common Data For 𝑄.65 and 𝑄.66
Four jobs are to be processed on a machine as per data
listed in the table.
Job Processing time
(in days)
Due date
1 4 6
2 7 9
3 2 19
4 8 17
65. If the Earliest Due Date (EDD) rule is used to
sequence the jobs, the number of jobs delayed is
(A) 1 (B) 2
(C) 3 (D) 4
66. Using the Shortest Processing Time (𝑆𝑃𝑇) rule, total
tardiness is
(A) 0 (B) 2
(C) 6 (D) 8
67. The project activities, precedence relationships and
durations are described in the table. The critical path
of the project is
Activity Precedence Duration (in
days)
𝑃 - 3
𝑄 - 4
𝑅 𝑃 5
𝑆 𝑄 5
𝑇 𝑅, 𝑆 7
𝑈 𝑅, 𝑆 5
𝑉 𝑇 2
𝑊 𝑈 10
(A) P‐R‐T‐V (C) P‐R‐U‐W
(B) Q‐S‐T‐V (D) Q‐S‐U‐W
68. Annual demand for window frames is 10000. Each
frame cost Rs. 200 and ordering cost is Rs. 300 per
order. Inventory holding cost is Rs. 40 per frame per
year. The supplier is willing of offer 2% discount if
the order quantity is 1000 or more, and 4% if order
quantity is 2000 or more. If the total cost is to be
minimized, the retailer should
(A) order 200 frames every time
(B) accept 2% discount
(C) accept 4% discount
(D) order Economic Order Quantity
69. Simplex method of solving linear programming
problem uses
(A) all the points in the feasible region
(B) only the corner points of the feasible region
(C) intermediate points within the infeasible region
(D) only the interior points in the feasible region
70. Vehicle manufacturing assembly line is an example
of
(A) product layout (B) process layout
(C) manual layout (D) fixed layout
71. Little’s law is a relationship between
(A) stock level and lead time in an inventory system
(B) waiting time and length of the queue in a queuing
system

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(C) number of machines and job due dates in a
scheduling problem
(D) uncertainty in the activity time and project
completion time
72. The demand and forecast for February are 12000
and 10275, respectively. Using single exponential
smoothening method (smoothening coefficient =
0.25) , forecast for the month of March is
(A) 431 (B) 9587
(C) 10706 (D) 11000
Common Data For 𝑄.73 and 𝑄.74
One unit of product 𝑃1 requires 3 kg ofresources 𝑅1
and 1 kg ofresources 𝑅2 One unit of product 𝑃2
requires 2 kg of resources 𝑅1 and 2 kg of
resources𝑅2. The profits per unit by selling product
𝑃1 and 𝑃2 are Rs. 2000 and Rs. 3000 respectively. The
manufacturer has 90 kg of resources 𝑅1 and 100 kg
of resources 𝑅2.
73. The unit worth of resources 𝑅2, i.e., dual price of
resources 𝑅2 in Rs. per kg is
(A) 0 (B) 1350
(C) 1500 (D) 2000
74. The manufacturer can make a maximum profit of Rs.
(A) 60000 (B) 135000
(C) 150000 (D) 200000
75. The word ‘kanban’ is most appropriately associated
with
(A) economic order quantity
(B) just‐in‐time production
(C) capacity planning
(D) product design
76. Cars arrive at a service station according to Poisson’s
distribution with a mean rate of 5 per hour. The
service time per car is exponential with a mean of 10
minutes. At steady state, the average waiting time in
the queue is
(A) 10 minutes (B) 20 minutes
(C) 25 minutes (D) 50 minutes
Common Data For 𝑄.77 and 𝑄.78
For a particular project, eight activities are to be carried
out. Their relationships with other activities and expected
durations are mentioned in the table below.
Activit
y
Predecess
ors
Durations (days)
𝑎 3
𝑏 𝑎 4
𝑐 𝑎 5
𝑑 𝑎 4
𝑒 𝑏 2
𝑓 𝑑 9
𝑔 𝑐, 𝑒 6
ℎ 𝑓, 𝑔 2
77. The critical path for the project is
(A) 𝑎 − 𝑏 − 𝑒 − 𝑔 − ℎ
(B) 𝑎 − 𝑐 − 𝑔 − ℎ
(C) 𝑎 − 𝑑 − 𝑓 − ℎ
(D) 𝑎 − 𝑏 − 𝑐 − 𝑓 − ℎ
78. If the duration of activity 𝑓 alone is changed from 9
to 10 days, then the
(A) critical path remains the same and the total
duration to complete the project changes to 19 days.
(B) critical path and the total duration to complete
the project remains the same.
(C) critical path changes but the total duration to
complete the project remains the same.
(D) critical path changes and the total duration to
complete the project changes to 17 days.
79. Which one of the following is NOT a decision taken
during the aggregate production planning stage?
(A) Scheduling of machines
(B) Amount oflabour to be committed
(C) Rate at which production should happen
(D) Inventory to be carried forward
ANSWERS:
1 D 31 D 61 D
2 D 32 A 62 C
3 D 33 B 63 D
4 B 34 A 64 A
5 B 35 C 65 C
6 C 36 D 66 D
7 D 37 A 67 D
8 B 38 D 68 C
9 B 39 C 69 D
10 A 40 A 70 A
11 C 41 C 71 B
12 A 42 B 72 C
13 C 43 C 73 A
14 B 44 C 74 B
15 C 45 B 75 B
16 A 46 B 76 D
17 A 47 B 77 C
18 A 48 D 78 A
19 C 49 B 79 A
20 A 50 B
21 A 51 A
22 D 52 C
23 D 53 C
24 C 54 D
25 C 55 D
26 B 56 A
27 B 57 C
28 D 58 D
29 C 59 C
30 C 60 A