GATE Mechanical Engineering notes on industrial engineering

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1 Industrial Engineering Introduction As per American Institute of Industrial Engineers (AIIE) industrial engineering is defined as follows lsquoIt is a branch of engineering concerned with the design improvement and installation of integrated systems of people materials equipment and energy lsquo 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

x More efficient use of resources x Less waste per unit of input applied x 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 vs Industrial Engineering

frac34 Production management familiarizes a person with concepts and techniques specific to the analysis and management of a production activity

frac34 Industrial engineering deals with the analysis design and control of productive systems ie the system produces either a product or a service

Eg The training of an aircraft pilot is analogous to management education whereas the designing of the aircraft is analogous to industrial engineering education

frac34 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 ndash the decision is considered good if it increases revenue as compared to costs

2 The principle of time perspective ndash the decision should take into account the long term and short term effects on costs and revenue

3 The Opportunity cost principle ndash decision making should carefully measure the sacrifices required by the various alternatives

4 The discounting principle ndash 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 10200

= 120

20300

= 115

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 x Nearness to Raw material (reduces transportation

cost) x Transport facilities (like road rail etc) x Nearness to markets (reduces transportation cost

and the danger of damage to the finished product before reaching the customer)

x Availability of Labour (supply of stable and trained work force)

x Availability of water resources (for industries such as paper and chemical industries)

x Climatic conditions x Financial and other aids x 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

frac34 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

frac34 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 amp Process Planning Line Balancing Line balancing means balancing the line ie 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 6666 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 frac34 The main task of line balancing is to ensure that the

tasks are evenly distributed among men machinery and thereby ensuring minimum idle time

x 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

x 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) frac34 For intermittent flow pattern Heuristic method is

used as they are simple and involve less time and money

frac34 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 3810 = 4 The maximum number of stations may be equal to as many as the number of tasks or steps ie 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 2810=3 and the maximum number of work stations may be 6 ie 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

frac34 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 850 each The firm uses 100000 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 50000 Labour costs will increase by Rs 125000 Factory overheads currently running Rs 500000 pa may increase by 12 Raw materials used to make the part will cost Rs 600000

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 50000 Additional labour costs Rs 125000 Raw materials costs Rs 600000 Additional overheads costs

Rs 60000 ( 012 X 500000)

Total cost to manufacture Rs 835000 Cost to manufacture one part

Rs 835000 100000 = Rs 835

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

x This type of planning is called man-variant process planning and is the commonest type of planning used for production today

x 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

x Production cost or time are the dominant criteria in process selection but machine utilization and routing often affects the plans chosen

Automated Process Planning

x Man variant planning is often a boring and tedious job with errors in human judgement

x To eliminate this automated process planning has come to aid of industrial process planning

x Figure (a) below shows a completely automated process planning system

x 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- x Planning phase- Forecasting order writing

product product design material control tool control loading etc

x Action phase- Dispatching x 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 x In a job shop the material in-process follows

different processing patterns in batches through batch facilities

x The material does not flow in a serial fashion x 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 eg 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 eg sales etc Forecasting plays an important role in planning for the future The purpose of forecasting is-

x To determine the production volume and rate x To prepare production budget for production

processes etc Basic elements of forecasting are- x Trends x Cycles x Seasonal variations x Irregular variations Forecasting techniques-

x Historic estimate x Sales force estimate x Market survey x Delphi method x Judgemental techniques x Prior knowledge x Forecasting by past average x Forecasting from last period sales x Moving average method x Weighted moving average method x Exponential smoothening method x Correlation analysis x 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

119890119904119905119894119898119886119905119890119889 119904119886119897119890119904 119891119900119903 119905ℎ119890 119899119890119909119905 119901119890119903119894119900119889= 119886119907119890119903119886119892119890 119904119886119897119890119904 119891119900119903 119901119903119890119907119894119900119906119904 119901119890119903119894119900119889

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 periodrsquos 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 4875 4 30 4625 1989 1 35 4375 2 45 40 3 35 375 4 25 3625 1990 1 35 35 2 45 35 3 35 35 4 30 3625

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

119865119904 = 040(95) + 030(105) + 020(90) + 010(100)= 975

Let the actual sales for month 5 came out to be 110 then the month 6 forecast will be

119865119904 = 040(110) + 030(95) + 020(105) + 010(90)= 1025

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

119899119890119908 119891119900119903119890119888119886119904119905 = 120572(119897119886119905119890119904119905 119904119886119897119890119904 119891119894119892119906119903119890)+ (1 minus 120572)(119900119897119889 119891119900119903119890119888119886119904119905)

The term lsquoαrsquo 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

x The smoothing constant indicates the amount by which the new forecast responds to the latest sales figure and its value lies between 01 to 03

To find out the smoothing constant that gives the equivalent of an N-period moving average use the relation

120572 =2

119873 + 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

x Quantity of work to be done x Quality of work to be completed x Availability of tools equipments and personnel x Sequence of operations to be performed x 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 amp 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-

x The criteria which judges the best of the several alternatives

x Characteristics of the various alternatives being judged

x 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

x RHS of each constraint should be non negative x Each decision variable of the problem should be

non negative x Inequalities in constraints should be converted

to equalities x Set m = no of equality constraints and n= no of

variables x Put (n-m) variable equal to zero x (n-m) = non basic variable x m= basic variable

Special case

x Infinite solution When a non basic variable in an optimal solution has a zero value for Δj row then the solution is not unique

x Unbounded solution When all replacement ratios are either infinite (or) negative then the solution terminates This indicates the problem has unbounded solution

x 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 d 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

119905119900119905119886119897 119905119903119886119899119904119901119900119903119905119886119905119894119900119899 119888119900119904119905 = sum sum 119862119894119895119883119894119895

119898

119895=1

119899

119894=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

x Total number of allocations is exactly equal to (m+n-1)

x These allocations are in independent positions Unbalanced TP

x If total supply from all sources equals total demand in all destinations then the TP is balanced otherwise unbalanced

x 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

x Matrix must be a square matrix x The optimal solution to the problem would

always be such that there would only be one assignment in the given row or column

x Hungarian method is used to solve an AP

Steps for applying Hungarian method x Subtract the smallest element of each row to

the every element of corresponding row

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x Subtracting the smallest element of each column to every element in corresponding columns

x Now all zeroes are to be covered with minimum number of lines

x 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 x Size of queue x Pattern of arrivals x Customerrsquos behaviour

2 Queue discipline 3 Service mechanism

x Single queue one server x Single queue several server x Several queue one server x Several queue several server

4 Capacity of the system Operating characteristics of a Queuing system

x 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

x 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 ie excluding the customer being serviced or Lq = L-1

x 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

x 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

x The server utilization factor 119875 = 120582120583 It is the proportion of time that a service actually stands with a customer Here 120582 = average number of customers arriving per unit time and 120583 = 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 gt 120583 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 120582 le 120583 there shall be no queue and hence no waiting time The proportion of time the service facility would be idle is 1 minus 120582

120583

Probabilistic Queuing system It is assumed that customers joining the queuing system arrive in a random manner (variable) and follow a Poissonrsquos 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 120582119899= 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 MM1 (infin FIFO) Single service channel Poissonrsquos input exponential service no limit on the system capacity First In First Out

119875119899 = 120588119899(1 minus 120588) 119908ℎ119890119903119890 120588 =120582120583 lt 1 119886119899119889 119899 ge 0

x inter arrival time = 1λ x traffic intensity facto 120588 = 120582120583 x average number of customers in system 119871 =

120588(1 minus 120588) x average number of customer of queue or

average queue length 119871119902 = 119871 minus 120588 x Waiting time in system 119908 = 1

120583minus120582

x Average or expected waiting time in queue 119908119902 = 119871119902

120582

x Probability that the service facility is idle 1198750(119905) = (1 minus 120588)

x Probability that the service facility has n customers at time t 119875119899(119905) = 1205881198991198750(119905)

x Average length of non empty queue 119871119899 = 11minus120588

x The fluctuation of queue length 119907(119899) = 120588(1minus120588)2

x Probability of n arrivals in time t 119875(119899119905) =119890minus120582(120582119905)119899

119899

x Probability that the waiting time in the queue is greater than or equal to t 119875(119908119902 ge 119905) =

120582 119890minus(120583minus120582)119905

120583

x Probability that waiting time in system is greater than or equal to t 119875 = 119890minus(120583minus120582)119905

x Probability that waiting time in system is less than or equal to t 119875 = 1 minus 119890minus(120583minus120582)119905

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 eg 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 eg 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 u 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 = Q2

Safety stock = k

Average Consumption during lead 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 u Average lead time x When no stock outs are desired x Buffer stock = Maximum demand during lead time

(DDLT) Average Demand During Lead Time (DDLT) x When demand rate varies about the average

demand during a constant lead time (LT) period Reorder Level (ROL) = Average (DDLT) u LT + BS

Inventory Cost The costs that are affected by firmrsquos 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 u 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

12 | wwwmindvisin

Annual ordering cost= oCQ

u 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 u i u a2

where Cu = Unit purchase cost i = Interest rate Shortage Cost When an item cannot be supplied on consumerrsquos 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 u 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 x The ordering cost is constant x The rate of demand is known x Lead time is fixed x Purchase price of the item is constant x Replenishment is made x 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 QDQ 2

u u Cu u i

EOQ (Qq) = 0

u

2C DC iu

where Co = Cost of placing an order Cu = Unit purchase cost i = interest rate D= Annual consumption of the product

x Optimum number of order placed per year

no= o

DQ

= u

o

DC i2C

u

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) 0198 (B) 73 (C) 90 (D) 99

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

13 | wwwmindvisin

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) 03 (B) 05 (C) 07 (D) 09

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 119901 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) (016 024) (B) (012 028) (C) (008 032) (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 times rating factor + allowances (D) normal time times 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 119860 to 119872 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 1199091 and 1199092 respectively The profit is given is 21199091 + 51199092 The production constraints are

1199091 + 31199092 le 40 31199091 + 1199092 le 24 1199091 + 1199092 le 10

1199091 gt 0 1199092 gt 0 The maximum profit which can meet the constraints is (A) 29 (B) 38 (C) 44 (D) 75

13 Market demand for springs is 800000 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) 2828 (C) 4000 (D) 8000

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 04 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

800 pm to 1030 119901119898 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) 85 hours (B) 139 hours (C) 161 hours (D) 184 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 119877119890 1 respectively For certain strategic reasons both the machines are to be used concurrently The sales price of the first 800 units is Rs 350 per unit and subsequently it is only Rs 300 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)rArr (B) 119879 (C) (D) 120571

18 A company produces two types of toys 119875 and 119876

Production time of 119876 is twice that of 119875 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 119876 requires an electric switch which is available 600 pieces per day only The company makes a profit of

14 | wwwmindvisin

Rs 3 and Rs 5 on type 119875 and 119876 respectively For maximization ofprofits the daily production quantities of 119875 and 119876 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 unityear If the stock‐out cost are estimated to be nearly Rs 400 each 119905 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 lsquofirst come first servedrsquo 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 119872 119864 and 119879 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) 89 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 119875 and 119876 with the following

characteristics Product Deman

d (Units)

Order cost ( 119877119904 order)

Holding Cost (Rs unityear)

119875 100 50 4 119876 400 50 1

The economic order quantity (EOQ) of products 119875 and 119876 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 (120572) is taken as 02 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 11987627 and 11987628 Consider a linear programming problem with two variables and two constraints The objective function is Maximize1198831 + 1198832 The corner points of the feasible region are (00) (02) (20) and (43 43) 27 If an additional constraint 1198831 + 1198832 le 5 is added the

optimal solution is

(A) (553rsquo3

) (119861) (43rsquo

43)

(C) (552rsquo2

) (D) (5 0)

28 Let 1198841 and 1198842 be the decision variables of the dual

and 1199071 and 1199072 be the slack variables of the dual of the given linear programming problem The optimum dual variables are

(A) 1198841 and 1198842 (B) 1198841 and 1199071 (C) 1198841 and 1199072 (D) 1199071 and 1199072 29 A company has two factories 1 1198782 and two

warehouses 1198631 1198632 The supplies from 1198781 and 1198782 are 50 and 40 units respectively Warehouse 1198631 requires a minimum of 20 units and a maximum of 40 units Warehouse 1198632 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)2490 (119861)24110

15 | wwwmindvisin

(C)3490 (119863)34110 30 A project has six activities (A to 119865) 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

119875 robability

80 020

100 025 120 030

140 025 The reorder level is 125 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 (120582 = 4hour) and exponential service (120583 =4hour) 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) 111

(B) 110

(C) 19 (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 120572 is

(A) 17 (B) 1

5

(C) 27 (D) 2

5

36 An assembly activity is represented on an Operation

Process Chart by the symbol (A) (B) 119860 (C) 119863 (D) 119874 37 The standard deviation of the critical path of the

project is (A)radic151 days (119861)radic155 days

(C)radic200 days (119863)radic238 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 119875 robability 010 035 035 020

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 119904 (D) 2 unit 119904 41 Consider the following data for an item

Annual demand 2500 units per year Ordering cost Rs 100 119901 er order Inventory holding rate 25 of unit price Price quoted by a supplier

Order quantity (units) Unit price (Rs)

lt 500 10 ge 500 9

The optimum order quantity (in units) is (A) 447 (B) 471 (C) 500 (D) ge 600 42 An manufacturing shop processes sheet metal jobs

wherein each job must pass through two machines (119872119897 119886119899119889 1198722 119894119899 that order) The processing time (in hours) for these jobs is

16 | wwwmindvisin

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

(119875 119876 119886119899119889 119877) The prices quoted for these items (in Rs) by suppliers 1198781 1198782 and 1198783 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) 32 (C) 4 (D) 42 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 ldquoLeast Unit Cost Techniquerdquo 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 119901 er it em and Rs 24 119901er 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 11987650 and 11987651 Consider the Linear Programme (119871119875) Max 4119909 + 6119910 Subject to 3119909 + 2119910 le 6

2119909 + 3119910 le 6 119909 119910 ge 0

50 After introducing slack variables 119904 and 119905 the initial

basic feasible solution is represented by the table below (basic variables are 119904 = 6 and 119905 = 6 and the objective function value is 0)

minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6

119909 119910 119904 119905 RHS After some simplex iterations the following table is obtained

0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2

119909 119910 119904 119905 RHS From this one can conclude that (A) the 119871119875 has a unique optimal solution

(B) the 119871119875 has an optimal solution that is not unique

(C) the 119871119875 is infeasible (D) the 119871119875 is unbounded 51 The dual for the 119871119875 in 119876 50 is (A) Min 6119906 + 6119907 (B) Max 6119906 + 6119907 subject to

3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0

(C) Max 4119906 + 6119907 subject to 3119906 + 2119907 ge 6 2119906 + 3119907 ge 6

119906 119907 ge 0 subject to 3119906 + 2119907 le 4 2119906 + 3119907 le 6 119906 119907 ge 0 (D) Min 4119906 + 6119907 subject to 3119906 + 2119907 le 6 2119906 +3119907 le 6 119906 119907 ge 0 52 The product structure of an assembly 119875 is shown in

the figure

Estimated demand for end product 119875 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 119877 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 119875 to node 119866 Let 119889119894119895 be the length of directed arc from node 119894 to node 119895 Let 119878119895 be the length of the shortest path ffom 119875 to node 119895 Which of the following equations can be used to find 119878119866

(A) 119878119866 = Min 119878119876 119878119877 (B) (119861) 119878119866 = Min 119878119876 minus 119889119876119866 119878119877 minus 119889119877119866 (C) 119878119866 = Min 119878119876 + 119889119876119866 119878119877 + 119889119877119866 (D) (119863) 119878119866 = Min 119889119876119866 119889119877119866

54 A moving average system is used for forecasting

weekly demand 1198651(119905) and 1198652(119905) are sequences of forecasts with parameters 1198981 and 1198982 respectively where 1198981 and 1198982(1198981 gt 1198982) denote the numbers of weeks over which the moving averages are taken The actual demand shows a step increase from 1198891 to 1198892 at a certain time Subsequently (A) neither 1198651(119905) nor 1198652(119905) will catch up with the value 1198892 (B) both sequences 1198651(119905) and 1198652(119905) will reach 1198892 in the same period

(C) 1198651(119905) will attain the value 1198892 before 1198652(119905) (D) 1198652(119905) will attain the value 1198892 before 1198651(119905)

55 For the standard transportation linear programme

with 119898 source and 119899 destinations and total supply equaling total demand an optimal solution (lowest cost) with the smallest number of non‐zero 119909119894119895 values (amounts ffom source 119894 to destination j) is desired The best upper bound for this number is

(A) 119898119899 (B) 2 (119898 + 119899) (C) 119898 + 119899 (D) 119898 + 119899 minus 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 119870 per day

Job Processing time 119875 5 119876 2 119877 3 119878 2 119879 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 1198721198721 queuing system the number ofarrivals

in an interval oflength 119879 is a Poisson random variable (ie the probability of there being arrivals in an

interval of length 119879 is 119890minus120582119879(120582119879)119899

119899) The probability

density function 119891(119905) of the inter‐arrival time is

(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582

Common Data For 11987658 and 11987659 Consider the following PERT network

The optimistic time most likely time and pessimistic time ofall the activities are given in the table below

Activity Optimistic time (days)

Most likely time (days)

Pessimistic time (days)

1 minus 2 1 2 3 1 minus 3 5 6 7 1 minus 4 3 5 7 2 minus 5 5 7 9 3 minus 5 2 4 6 5 minus 6 4 5 6 4 minus 7 4 6 8 6 minus 7 2 3 4

18 | wwwmindvisin

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) 033 (B) 055 (C) 077 (D) 166 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) 2083 (B) 2316 (C) 12500 (D) 13900 61 Consider the following Linear Programming Problem

(LPP) Maximize 119885 = 31199091 + 21199092 Subject to 1199091 le 4

1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0

(A) The LPP has a unique optimal solution (B) The 119871119875119875 is infeasible (C) The 119871119875119875 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 (119905119890) of a PERT activity in terms of

optimistic time 1199050 pessimistic time (119905119901) and most likely time (119905119897) is given by

(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3

Common Data For 11987665 and 11987666 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 (119878119875119879) 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) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 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 Littlersquos 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

19 | wwwmindvisin

(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 =025) forecast for the month of March is

(A) 431 (B) 9587 (C) 10706 (D) 11000 Common Data For 11987673 and 11987674

One unit of product 1198751 requires 3 kg ofresources 1198771 and 1 kg ofresources 1198772 One unit of product 1198752 requires 2 kg of resources 1198771 and 2 kg of resources1198772 The profits per unit by selling product 1198751 and 1198752 are Rs 2000 and Rs 3000 respectively The manufacturer has 90 kg of resources 1198771 and 100 kg of resources 1198772

73 The unit worth of resources 1198772 ie dual price of

resources 1198772 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 lsquokanbanrsquo 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 Poissonrsquos

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 11987677 and 11987678 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

Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2

77 The critical path for the project is (A) 119886 minus 119887 minus 119890 minus 119892 minus ℎ (B) 119886 minus 119888 minus 119892 minus ℎ (C) 119886 minus 119889 minus 119891 minus ℎ (D) 119886 minus 119887 minus 119888 minus 119891 minus ℎ 78 If the duration of activity 119891 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

2 | wwwmindvisin

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 x Nearness to Raw material (reduces transportation

cost) x Transport facilities (like road rail etc) x Nearness to markets (reduces transportation cost

and the danger of damage to the finished product before reaching the customer)

x Availability of Labour (supply of stable and trained work force)

x Availability of water resources (for industries such as paper and chemical industries)

x Climatic conditions x Financial and other aids x 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

3 | wwwmindvisin




















Line flow


L type flow


Circular flow


U type flow




Processing upwards


4 | wwwmindvisin











Steps or

elemental tasks



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



duration







5 | wwwmindvisin











Rs 60000 ( 012 X 500000)


Rs 835000 100000 = Rs 835


6 | wwwmindvisin








Event Symbol

Operation

Storage

Temporary Storage

Transport

Inspection












7 | wwwmindvisin












119890119904119905119894119898119886119905119890119889 119904119886119897119890119904 119891119900119903 119905ℎ119890 119899119890119909119905 119901119890119903119894119900119889= 119886119907119890119903119886119892119890 119904119886119897119890119904 119891119900119903 119901119903119890119907119894119900119906119904 119901119890119903119894119900119889

Example

Period No Sales 1 7 2 5 3 9 4 8 5 5 6 8


6 = 7


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


8 | wwwmindvisin


Year Period Sales


1987 1 50 2 60 3 50 4 40 1988 1 50 50 2 55 50 3 40 4875 4 30 4625 1989 1 35 4375 2 45 40 3 35 375 4 25 3625 1990 1 35 35 2 45 35 3 35 35 4 30 3625


Month 1

Month 2

Month 3

Month 4

Month 5


119865119904 = 040(95) + 030(105) + 020(90) + 010(100)= 975


119865119904 = 040(110) + 030(95) + 020(105) + 010(90)= 1025


119899119890119908 119891119900119903119890119888119886119904119905 = 120572(119897119886119905119890119904119905 119904119886119897119890119904 119891119894119892119906119903119890)+ (1 minus 120572)(119900119897119889 119891119900119903119890119888119886119904119905)





120572 =2

119873 + 1











9 | wwwmindvisin












Special case








119905119900119905119886119897 119905119903119886119899119904119901119900119903119905119886119905119894119900119899 119888119900119904119905 = sum sum 119862119894119895119883119894119895

119898

119895=1

119899

119894=1












10 | wwwmindvisin


















120583


119875119899 = 120588119899(1 minus 120588) 119908ℎ119890119903119890 120588 =120582120583 lt 1 119886119899119889 119899 ge 0




120583minus120582


120582






119899


120582 119890minus(120583minus120582)119905

120583




11 | wwwmindvisin



Safety stock = k







12 | wwwmindvisin


u D




TVC = oC QDQ 2

u u Cu u i

EOQ (Qq) = 0

u

2C DC iu



no= o

DQ

= u

o

DC i2C

u















13 | wwwmindvisin
















1199091 + 31199092 le 40 31199091 + 1199092 le 24 1199091 + 1199092 le 10











(A)rArr (B) 119879 (C) (D) 120571



14 | wwwmindvisin






6 day (B) 1

3 day
















d (Units)



119875 100 50 4 119876 400 50 1




optimal solution is

(A) (553rsquo3

) (119861) (43rsquo

43)

(C) (552rsquo2

) (D) (5 0)





(A)2490 (119861)24110

15 | wwwmindvisin





119875 robability

80 020

100 025 120 030







(A) 111

(B) 110

(C) 19 (D) 1

2



(A) 17 (B) 1

5

(C) 27 (D) 2

5






7 9 7 10 9 6








lt 500 10 ge 500 9



16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





3 | wwwmindvisin




















Line flow


L type flow


Circular flow


U type flow




Processing upwards


4 | wwwmindvisin











Steps or

elemental tasks



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



duration







5 | wwwmindvisin











Rs 60000 ( 012 X 500000)


Rs 835000 100000 = Rs 835


6 | wwwmindvisin








Event Symbol

Operation

Storage

Temporary Storage

Transport

Inspection












7 | wwwmindvisin












119890119904119905119894119898119886119905119890119889 119904119886119897119890119904 119891119900119903 119905ℎ119890 119899119890119909119905 119901119890119903119894119900119889= 119886119907119890119903119886119892119890 119904119886119897119890119904 119891119900119903 119901119903119890119907119894119900119906119904 119901119890119903119894119900119889

Example

Period No Sales 1 7 2 5 3 9 4 8 5 5 6 8


6 = 7


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


8 | wwwmindvisin


Year Period Sales


1987 1 50 2 60 3 50 4 40 1988 1 50 50 2 55 50 3 40 4875 4 30 4625 1989 1 35 4375 2 45 40 3 35 375 4 25 3625 1990 1 35 35 2 45 35 3 35 35 4 30 3625


Month 1

Month 2

Month 3

Month 4

Month 5


119865119904 = 040(95) + 030(105) + 020(90) + 010(100)= 975


119865119904 = 040(110) + 030(95) + 020(105) + 010(90)= 1025


119899119890119908 119891119900119903119890119888119886119904119905 = 120572(119897119886119905119890119904119905 119904119886119897119890119904 119891119894119892119906119903119890)+ (1 minus 120572)(119900119897119889 119891119900119903119890119888119886119904119905)





120572 =2

119873 + 1











9 | wwwmindvisin












Special case








119905119900119905119886119897 119905119903119886119899119904119901119900119903119905119886119905119894119900119899 119888119900119904119905 = sum sum 119862119894119895119883119894119895

119898

119895=1

119899

119894=1












10 | wwwmindvisin


















120583


119875119899 = 120588119899(1 minus 120588) 119908ℎ119890119903119890 120588 =120582120583 lt 1 119886119899119889 119899 ge 0




120583minus120582


120582






119899


120582 119890minus(120583minus120582)119905

120583




11 | wwwmindvisin



Safety stock = k







12 | wwwmindvisin


u D




TVC = oC QDQ 2

u u Cu u i

EOQ (Qq) = 0

u

2C DC iu



no= o

DQ

= u

o

DC i2C

u















13 | wwwmindvisin
















1199091 + 31199092 le 40 31199091 + 1199092 le 24 1199091 + 1199092 le 10











(A)rArr (B) 119879 (C) (D) 120571



14 | wwwmindvisin






6 day (B) 1

3 day
















d (Units)



119875 100 50 4 119876 400 50 1




optimal solution is

(A) (553rsquo3

) (119861) (43rsquo

43)

(C) (552rsquo2

) (D) (5 0)





(A)2490 (119861)24110

15 | wwwmindvisin





119875 robability

80 020

100 025 120 030







(A) 111

(B) 110

(C) 19 (D) 1

2



(A) 17 (B) 1

5

(C) 27 (D) 2

5






7 9 7 10 9 6








lt 500 10 ge 500 9



16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





4 | wwwmindvisin











Steps or

elemental tasks



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



duration







5 | wwwmindvisin











Rs 60000 ( 012 X 500000)


Rs 835000 100000 = Rs 835


6 | wwwmindvisin








Event Symbol

Operation

Storage

Temporary Storage

Transport

Inspection












7 | wwwmindvisin












119890119904119905119894119898119886119905119890119889 119904119886119897119890119904 119891119900119903 119905ℎ119890 119899119890119909119905 119901119890119903119894119900119889= 119886119907119890119903119886119892119890 119904119886119897119890119904 119891119900119903 119901119903119890119907119894119900119906119904 119901119890119903119894119900119889

Example

Period No Sales 1 7 2 5 3 9 4 8 5 5 6 8


6 = 7


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


8 | wwwmindvisin


Year Period Sales


1987 1 50 2 60 3 50 4 40 1988 1 50 50 2 55 50 3 40 4875 4 30 4625 1989 1 35 4375 2 45 40 3 35 375 4 25 3625 1990 1 35 35 2 45 35 3 35 35 4 30 3625


Month 1

Month 2

Month 3

Month 4

Month 5


119865119904 = 040(95) + 030(105) + 020(90) + 010(100)= 975


119865119904 = 040(110) + 030(95) + 020(105) + 010(90)= 1025


119899119890119908 119891119900119903119890119888119886119904119905 = 120572(119897119886119905119890119904119905 119904119886119897119890119904 119891119894119892119906119903119890)+ (1 minus 120572)(119900119897119889 119891119900119903119890119888119886119904119905)





120572 =2

119873 + 1











9 | wwwmindvisin












Special case








119905119900119905119886119897 119905119903119886119899119904119901119900119903119905119886119905119894119900119899 119888119900119904119905 = sum sum 119862119894119895119883119894119895

119898

119895=1

119899

119894=1












10 | wwwmindvisin


















120583


119875119899 = 120588119899(1 minus 120588) 119908ℎ119890119903119890 120588 =120582120583 lt 1 119886119899119889 119899 ge 0




120583minus120582


120582






119899


120582 119890minus(120583minus120582)119905

120583




11 | wwwmindvisin



Safety stock = k







12 | wwwmindvisin


u D




TVC = oC QDQ 2

u u Cu u i

EOQ (Qq) = 0

u

2C DC iu



no= o

DQ

= u

o

DC i2C

u















13 | wwwmindvisin
















1199091 + 31199092 le 40 31199091 + 1199092 le 24 1199091 + 1199092 le 10











(A)rArr (B) 119879 (C) (D) 120571



14 | wwwmindvisin






6 day (B) 1

3 day
















d (Units)



119875 100 50 4 119876 400 50 1




optimal solution is

(A) (553rsquo3

) (119861) (43rsquo

43)

(C) (552rsquo2

) (D) (5 0)





(A)2490 (119861)24110

15 | wwwmindvisin





119875 robability

80 020

100 025 120 030







(A) 111

(B) 110

(C) 19 (D) 1

2



(A) 17 (B) 1

5

(C) 27 (D) 2

5






7 9 7 10 9 6








lt 500 10 ge 500 9



16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





5 | wwwmindvisin











Rs 60000 ( 012 X 500000)


Rs 835000 100000 = Rs 835


6 | wwwmindvisin








Event Symbol

Operation

Storage

Temporary Storage

Transport

Inspection












7 | wwwmindvisin












119890119904119905119894119898119886119905119890119889 119904119886119897119890119904 119891119900119903 119905ℎ119890 119899119890119909119905 119901119890119903119894119900119889= 119886119907119890119903119886119892119890 119904119886119897119890119904 119891119900119903 119901119903119890119907119894119900119906119904 119901119890119903119894119900119889

Example

Period No Sales 1 7 2 5 3 9 4 8 5 5 6 8


6 = 7


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


8 | wwwmindvisin


Year Period Sales


1987 1 50 2 60 3 50 4 40 1988 1 50 50 2 55 50 3 40 4875 4 30 4625 1989 1 35 4375 2 45 40 3 35 375 4 25 3625 1990 1 35 35 2 45 35 3 35 35 4 30 3625


Month 1

Month 2

Month 3

Month 4

Month 5


119865119904 = 040(95) + 030(105) + 020(90) + 010(100)= 975


119865119904 = 040(110) + 030(95) + 020(105) + 010(90)= 1025


119899119890119908 119891119900119903119890119888119886119904119905 = 120572(119897119886119905119890119904119905 119904119886119897119890119904 119891119894119892119906119903119890)+ (1 minus 120572)(119900119897119889 119891119900119903119890119888119886119904119905)





120572 =2

119873 + 1











9 | wwwmindvisin












Special case








119905119900119905119886119897 119905119903119886119899119904119901119900119903119905119886119905119894119900119899 119888119900119904119905 = sum sum 119862119894119895119883119894119895

119898

119895=1

119899

119894=1












10 | wwwmindvisin


















120583


119875119899 = 120588119899(1 minus 120588) 119908ℎ119890119903119890 120588 =120582120583 lt 1 119886119899119889 119899 ge 0




120583minus120582


120582






119899


120582 119890minus(120583minus120582)119905

120583




11 | wwwmindvisin



Safety stock = k







12 | wwwmindvisin


u D




TVC = oC QDQ 2

u u Cu u i

EOQ (Qq) = 0

u

2C DC iu



no= o

DQ

= u

o

DC i2C

u















13 | wwwmindvisin
















1199091 + 31199092 le 40 31199091 + 1199092 le 24 1199091 + 1199092 le 10











(A)rArr (B) 119879 (C) (D) 120571



14 | wwwmindvisin






6 day (B) 1

3 day
















d (Units)



119875 100 50 4 119876 400 50 1




optimal solution is

(A) (553rsquo3

) (119861) (43rsquo

43)

(C) (552rsquo2

) (D) (5 0)





(A)2490 (119861)24110

15 | wwwmindvisin





119875 robability

80 020

100 025 120 030







(A) 111

(B) 110

(C) 19 (D) 1

2



(A) 17 (B) 1

5

(C) 27 (D) 2

5






7 9 7 10 9 6








lt 500 10 ge 500 9



16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





6 | wwwmindvisin








Event Symbol

Operation

Storage

Temporary Storage

Transport

Inspection












7 | wwwmindvisin












119890119904119905119894119898119886119905119890119889 119904119886119897119890119904 119891119900119903 119905ℎ119890 119899119890119909119905 119901119890119903119894119900119889= 119886119907119890119903119886119892119890 119904119886119897119890119904 119891119900119903 119901119903119890119907119894119900119906119904 119901119890119903119894119900119889

Example

Period No Sales 1 7 2 5 3 9 4 8 5 5 6 8


6 = 7


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


8 | wwwmindvisin


Year Period Sales


1987 1 50 2 60 3 50 4 40 1988 1 50 50 2 55 50 3 40 4875 4 30 4625 1989 1 35 4375 2 45 40 3 35 375 4 25 3625 1990 1 35 35 2 45 35 3 35 35 4 30 3625


Month 1

Month 2

Month 3

Month 4

Month 5


119865119904 = 040(95) + 030(105) + 020(90) + 010(100)= 975


119865119904 = 040(110) + 030(95) + 020(105) + 010(90)= 1025


119899119890119908 119891119900119903119890119888119886119904119905 = 120572(119897119886119905119890119904119905 119904119886119897119890119904 119891119894119892119906119903119890)+ (1 minus 120572)(119900119897119889 119891119900119903119890119888119886119904119905)





120572 =2

119873 + 1











9 | wwwmindvisin












Special case








119905119900119905119886119897 119905119903119886119899119904119901119900119903119905119886119905119894119900119899 119888119900119904119905 = sum sum 119862119894119895119883119894119895

119898

119895=1

119899

119894=1












10 | wwwmindvisin


















120583


119875119899 = 120588119899(1 minus 120588) 119908ℎ119890119903119890 120588 =120582120583 lt 1 119886119899119889 119899 ge 0




120583minus120582


120582






119899


120582 119890minus(120583minus120582)119905

120583




11 | wwwmindvisin



Safety stock = k







12 | wwwmindvisin


u D




TVC = oC QDQ 2

u u Cu u i

EOQ (Qq) = 0

u

2C DC iu



no= o

DQ

= u

o

DC i2C

u















13 | wwwmindvisin
















1199091 + 31199092 le 40 31199091 + 1199092 le 24 1199091 + 1199092 le 10











(A)rArr (B) 119879 (C) (D) 120571



14 | wwwmindvisin






6 day (B) 1

3 day
















d (Units)



119875 100 50 4 119876 400 50 1




optimal solution is

(A) (553rsquo3

) (119861) (43rsquo

43)

(C) (552rsquo2

) (D) (5 0)





(A)2490 (119861)24110

15 | wwwmindvisin





119875 robability

80 020

100 025 120 030







(A) 111

(B) 110

(C) 19 (D) 1

2



(A) 17 (B) 1

5

(C) 27 (D) 2

5






7 9 7 10 9 6








lt 500 10 ge 500 9



16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





7 | wwwmindvisin












119890119904119905119894119898119886119905119890119889 119904119886119897119890119904 119891119900119903 119905ℎ119890 119899119890119909119905 119901119890119903119894119900119889= 119886119907119890119903119886119892119890 119904119886119897119890119904 119891119900119903 119901119903119890119907119894119900119906119904 119901119890119903119894119900119889

Example

Period No Sales 1 7 2 5 3 9 4 8 5 5 6 8


6 = 7


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


8 | wwwmindvisin


Year Period Sales


1987 1 50 2 60 3 50 4 40 1988 1 50 50 2 55 50 3 40 4875 4 30 4625 1989 1 35 4375 2 45 40 3 35 375 4 25 3625 1990 1 35 35 2 45 35 3 35 35 4 30 3625


Month 1

Month 2

Month 3

Month 4

Month 5


119865119904 = 040(95) + 030(105) + 020(90) + 010(100)= 975


119865119904 = 040(110) + 030(95) + 020(105) + 010(90)= 1025


119899119890119908 119891119900119903119890119888119886119904119905 = 120572(119897119886119905119890119904119905 119904119886119897119890119904 119891119894119892119906119903119890)+ (1 minus 120572)(119900119897119889 119891119900119903119890119888119886119904119905)





120572 =2

119873 + 1











9 | wwwmindvisin












Special case








119905119900119905119886119897 119905119903119886119899119904119901119900119903119905119886119905119894119900119899 119888119900119904119905 = sum sum 119862119894119895119883119894119895

119898

119895=1

119899

119894=1












10 | wwwmindvisin


















120583


119875119899 = 120588119899(1 minus 120588) 119908ℎ119890119903119890 120588 =120582120583 lt 1 119886119899119889 119899 ge 0




120583minus120582


120582






119899


120582 119890minus(120583minus120582)119905

120583




11 | wwwmindvisin



Safety stock = k







12 | wwwmindvisin


u D




TVC = oC QDQ 2

u u Cu u i

EOQ (Qq) = 0

u

2C DC iu



no= o

DQ

= u

o

DC i2C

u















13 | wwwmindvisin
















1199091 + 31199092 le 40 31199091 + 1199092 le 24 1199091 + 1199092 le 10











(A)rArr (B) 119879 (C) (D) 120571



14 | wwwmindvisin






6 day (B) 1

3 day
















d (Units)



119875 100 50 4 119876 400 50 1




optimal solution is

(A) (553rsquo3

) (119861) (43rsquo

43)

(C) (552rsquo2

) (D) (5 0)





(A)2490 (119861)24110

15 | wwwmindvisin





119875 robability

80 020

100 025 120 030







(A) 111

(B) 110

(C) 19 (D) 1

2



(A) 17 (B) 1

5

(C) 27 (D) 2

5






7 9 7 10 9 6








lt 500 10 ge 500 9



16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





8 | wwwmindvisin


Year Period Sales


1987 1 50 2 60 3 50 4 40 1988 1 50 50 2 55 50 3 40 4875 4 30 4625 1989 1 35 4375 2 45 40 3 35 375 4 25 3625 1990 1 35 35 2 45 35 3 35 35 4 30 3625


Month 1

Month 2

Month 3

Month 4

Month 5


119865119904 = 040(95) + 030(105) + 020(90) + 010(100)= 975


119865119904 = 040(110) + 030(95) + 020(105) + 010(90)= 1025


119899119890119908 119891119900119903119890119888119886119904119905 = 120572(119897119886119905119890119904119905 119904119886119897119890119904 119891119894119892119906119903119890)+ (1 minus 120572)(119900119897119889 119891119900119903119890119888119886119904119905)





120572 =2

119873 + 1











9 | wwwmindvisin












Special case








119905119900119905119886119897 119905119903119886119899119904119901119900119903119905119886119905119894119900119899 119888119900119904119905 = sum sum 119862119894119895119883119894119895

119898

119895=1

119899

119894=1












10 | wwwmindvisin


















120583


119875119899 = 120588119899(1 minus 120588) 119908ℎ119890119903119890 120588 =120582120583 lt 1 119886119899119889 119899 ge 0




120583minus120582


120582






119899


120582 119890minus(120583minus120582)119905

120583




11 | wwwmindvisin



Safety stock = k







12 | wwwmindvisin


u D




TVC = oC QDQ 2

u u Cu u i

EOQ (Qq) = 0

u

2C DC iu



no= o

DQ

= u

o

DC i2C

u















13 | wwwmindvisin
















1199091 + 31199092 le 40 31199091 + 1199092 le 24 1199091 + 1199092 le 10











(A)rArr (B) 119879 (C) (D) 120571



14 | wwwmindvisin






6 day (B) 1

3 day
















d (Units)



119875 100 50 4 119876 400 50 1




optimal solution is

(A) (553rsquo3

) (119861) (43rsquo

43)

(C) (552rsquo2

) (D) (5 0)





(A)2490 (119861)24110

15 | wwwmindvisin





119875 robability

80 020

100 025 120 030







(A) 111

(B) 110

(C) 19 (D) 1

2



(A) 17 (B) 1

5

(C) 27 (D) 2

5






7 9 7 10 9 6








lt 500 10 ge 500 9



16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





9 | wwwmindvisin












Special case








119905119900119905119886119897 119905119903119886119899119904119901119900119903119905119886119905119894119900119899 119888119900119904119905 = sum sum 119862119894119895119883119894119895

119898

119895=1

119899

119894=1












10 | wwwmindvisin


















120583


119875119899 = 120588119899(1 minus 120588) 119908ℎ119890119903119890 120588 =120582120583 lt 1 119886119899119889 119899 ge 0




120583minus120582


120582






119899


120582 119890minus(120583minus120582)119905

120583




11 | wwwmindvisin



Safety stock = k







12 | wwwmindvisin


u D




TVC = oC QDQ 2

u u Cu u i

EOQ (Qq) = 0

u

2C DC iu



no= o

DQ

= u

o

DC i2C

u















13 | wwwmindvisin
















1199091 + 31199092 le 40 31199091 + 1199092 le 24 1199091 + 1199092 le 10











(A)rArr (B) 119879 (C) (D) 120571



14 | wwwmindvisin






6 day (B) 1

3 day
















d (Units)



119875 100 50 4 119876 400 50 1




optimal solution is

(A) (553rsquo3

) (119861) (43rsquo

43)

(C) (552rsquo2

) (D) (5 0)





(A)2490 (119861)24110

15 | wwwmindvisin





119875 robability

80 020

100 025 120 030







(A) 111

(B) 110

(C) 19 (D) 1

2



(A) 17 (B) 1

5

(C) 27 (D) 2

5






7 9 7 10 9 6








lt 500 10 ge 500 9



16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





10 | wwwmindvisin


















120583


119875119899 = 120588119899(1 minus 120588) 119908ℎ119890119903119890 120588 =120582120583 lt 1 119886119899119889 119899 ge 0




120583minus120582


120582






119899


120582 119890minus(120583minus120582)119905

120583




11 | wwwmindvisin



Safety stock = k







12 | wwwmindvisin


u D




TVC = oC QDQ 2

u u Cu u i

EOQ (Qq) = 0

u

2C DC iu



no= o

DQ

= u

o

DC i2C

u















13 | wwwmindvisin
















1199091 + 31199092 le 40 31199091 + 1199092 le 24 1199091 + 1199092 le 10











(A)rArr (B) 119879 (C) (D) 120571



14 | wwwmindvisin






6 day (B) 1

3 day
















d (Units)



119875 100 50 4 119876 400 50 1




optimal solution is

(A) (553rsquo3

) (119861) (43rsquo

43)

(C) (552rsquo2

) (D) (5 0)





(A)2490 (119861)24110

15 | wwwmindvisin





119875 robability

80 020

100 025 120 030







(A) 111

(B) 110

(C) 19 (D) 1

2



(A) 17 (B) 1

5

(C) 27 (D) 2

5






7 9 7 10 9 6








lt 500 10 ge 500 9



16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





11 | wwwmindvisin



Safety stock = k







12 | wwwmindvisin


u D




TVC = oC QDQ 2

u u Cu u i

EOQ (Qq) = 0

u

2C DC iu



no= o

DQ

= u

o

DC i2C

u















13 | wwwmindvisin
















1199091 + 31199092 le 40 31199091 + 1199092 le 24 1199091 + 1199092 le 10











(A)rArr (B) 119879 (C) (D) 120571



14 | wwwmindvisin






6 day (B) 1

3 day
















d (Units)



119875 100 50 4 119876 400 50 1




optimal solution is

(A) (553rsquo3

) (119861) (43rsquo

43)

(C) (552rsquo2

) (D) (5 0)





(A)2490 (119861)24110

15 | wwwmindvisin





119875 robability

80 020

100 025 120 030







(A) 111

(B) 110

(C) 19 (D) 1

2



(A) 17 (B) 1

5

(C) 27 (D) 2

5






7 9 7 10 9 6








lt 500 10 ge 500 9



16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





12 | wwwmindvisin


u D




TVC = oC QDQ 2

u u Cu u i

EOQ (Qq) = 0

u

2C DC iu



no= o

DQ

= u

o

DC i2C

u















13 | wwwmindvisin
















1199091 + 31199092 le 40 31199091 + 1199092 le 24 1199091 + 1199092 le 10











(A)rArr (B) 119879 (C) (D) 120571



14 | wwwmindvisin






6 day (B) 1

3 day
















d (Units)



119875 100 50 4 119876 400 50 1




optimal solution is

(A) (553rsquo3

) (119861) (43rsquo

43)

(C) (552rsquo2

) (D) (5 0)





(A)2490 (119861)24110

15 | wwwmindvisin





119875 robability

80 020

100 025 120 030







(A) 111

(B) 110

(C) 19 (D) 1

2



(A) 17 (B) 1

5

(C) 27 (D) 2

5






7 9 7 10 9 6








lt 500 10 ge 500 9



16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





13 | wwwmindvisin
















1199091 + 31199092 le 40 31199091 + 1199092 le 24 1199091 + 1199092 le 10











(A)rArr (B) 119879 (C) (D) 120571



14 | wwwmindvisin






6 day (B) 1

3 day
















d (Units)



119875 100 50 4 119876 400 50 1




optimal solution is

(A) (553rsquo3

) (119861) (43rsquo

43)

(C) (552rsquo2

) (D) (5 0)





(A)2490 (119861)24110

15 | wwwmindvisin





119875 robability

80 020

100 025 120 030







(A) 111

(B) 110

(C) 19 (D) 1

2



(A) 17 (B) 1

5

(C) 27 (D) 2

5






7 9 7 10 9 6








lt 500 10 ge 500 9



16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





14 | wwwmindvisin






6 day (B) 1

3 day
















d (Units)



119875 100 50 4 119876 400 50 1




optimal solution is

(A) (553rsquo3

) (119861) (43rsquo

43)

(C) (552rsquo2

) (D) (5 0)





(A)2490 (119861)24110

15 | wwwmindvisin





119875 robability

80 020

100 025 120 030







(A) 111

(B) 110

(C) 19 (D) 1

2



(A) 17 (B) 1

5

(C) 27 (D) 2

5






7 9 7 10 9 6








lt 500 10 ge 500 9



16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





15 | wwwmindvisin





119875 robability

80 020

100 025 120 030







(A) 111

(B) 110

(C) 19 (D) 1

2



(A) 17 (B) 1

5

(C) 27 (D) 2

5






7 9 7 10 9 6








lt 500 10 ge 500 9



16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





16 | wwwmindvisin

Machine

Jobs P Q R S T U

M1 15 32 8 27 11 16 M2 6 19 13 20 14 7



















2119909 + 3119910 le 6 119909 119910 ge 0



minus4 minus6 0 0 0 119904 3 2 1 0 6 119905 2 3 0 1 6


0 0 0 2 12 119904 53 0 1 minus13 2 119910 23 1 0 13 2




3119906 + 2119907 ge 4

17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





17 | wwwmindvisin

2119906 + 3119907 ge 6 119906 119907 ge 0



the figure


Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200


















(A) 1205822(119890minus1205822119905) (B) 119890minus1205822119905

1205822

(C) 120582119890minus120582119905 (119863) 119890minus120582119905

120582







18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





18 | wwwmindvisin






1199092 le 6 31199091 + 21199092 le 18 1199091 ge 0 1199092 ge 0







(A) 119905119890 = 119905119900+4119905119879+119905119901

6

(C) 119905119890 = 119905119900+4119905119879+119905119901

3

(B) 119905119890 = 119905119900+4119905119901+119905119879

6

(D) 119905119890 = 119905119900+4119905119901+119905119879

3



Due date

1 4 6 2 7 9 3 2 19 4 8 17






days) 119875 - 3 119876 - 4 119877 119875 5 119878 119876 5 119879 119877 119878 7 119880 119877 119878 5 119881 119879 2 119882 119880 10








19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





19 | wwwmindvisin












Activity

Predecessors

Durations (days)

119886 3 119887 119886 4 119888 119886 5 119889 119886 4 119890 119887 2 119891 119889 9 119892 119888 119890 6 ℎ 119891 119892 2





Education

GATE Mechanical Engineering notes on industrial engineering