SK MONDAL INDUSTRIAL ENGINEERING PDF

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S k mondal industrial engineering free download GATE Exam, engineering job, mechanical job, online test, free pdf. Industrial Engineering By S K Mondal, MiB .. thanks for uploading material. . can u please upload full pdf of production engineering. S.K. MONDAL's INDUSTRIAL ENGINEERING IES GATE PSU's TANCET & GOVT CONCEPTS) FOR MECHANICAL ENGINEERING – PDF FREE DOWNLOAD.


Sk Mondal Industrial Engineering Pdf

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Work Measurement (time study) Predetermined Motion Time System. This book is written by. S. K. Mondal 1. S K Mondal's. Industrial Engineering. IES. Industrial Engineering By S K Mondal. Pages Industrial Engineering and Management E-Book Auth: Rhona C. Free Sage Publications, Inc Industrial. S K Mondal's Industrial Engineering Contents Chapter 1: Forecasting + PDF allowance = Observed time × Performance rating + PDF allowance = 16 × +.

Which one of the following statements is correct? Which one of the following methods can be used for forecasting the sales potential of a new product?

Decision making under 1. Delphi approach complete certainty B. Decision making under 2. Maximax criterion risk C. Decision making under 3 Transportation mode complete uncertainly D. Decision making based on 4. Decision tree expert opinion Codes: Moving average method of forecasting demand gives an account of the trends in fluctuations and suppresses day-to-day insignificant fluctuations.

Working out moving averages of the demand data smoothens the random day-to-day fluctuations and represents only significant variations. Which one of the following is a qualitative technique of demand forecasting?

Moving average 1. Assembly B. Line balancing 2. download C. Economic batch size 3. Forecasting D.

Johnson algorithm 4. Sequencing Codes: The demand for a product in the month of March turned out to be 20 units against an earlier made forecast of 20 units. The actual demand for April and May turned to be 25 and 26 units respectively. A company intends to use exponential smoothing technique for making a forecast for one of its products. The previous year's forecast has been 78 units and the actual demand for the corresponding period turned out to be 73 units.

It is given that the actual demand is 59 units, a previous forecast 64 units and smoothening factor 0. What will be the forecast for next period, using exponential smoothing? Consider the following statements: Is a modification of moving average method 2. Assigns the highest weight age to the most recent observation Which of the statements given above are correct?

Consider the following statement relating to forecasting: The time horizon to forecast depends upon where the product currently lies its life cycle. Opinion and judgmental forecasting methods sometimes incorporate statistical analysis. In exponential smoothing, low values of smoothing constant, alpha result in more smoothing than higher values of alpha. Which of the statements given above are correct? Which one of the following statements is not correct for the exponential smoothing method of demand forecasting?

Line Balancing 1. Value analysis B. Product Development 2. Exponential smoothing C. Forecasting 3. Control chart D. Quality Control 4. Selective control 5. Rank position matrix Codes: For a product, the forecast for the month of January was units.

The actual demand turned out to be units. Which of the following is the measure of forecast error? To meet short range changes in demand of a product, which of the following strategies can be considered? Overtime 2. Subcontracting 3.

Building up inventory 4. New investments Select the correct answer from the codes given below: Regression is used for short and medium range. Delphi is used for long range forecasting. Exponential smoothing is a modification of weightage moving average method. What are moving average and exponential smoothing models for forecasting? A dealership for Honda city cars sells a particular model of the car in various months of the year.

Using the moving average method, find the exponential smoothing forecast for the month of October Take exponential smoothing constant as 0.

Refer theory part of this book. How are forecasting methods classified? The past data about the load onPage a machine centre is as given below: Most organisations are not in a position to wait unit orders are received before they begin to determine what production facilities, process, equipment, manpower, or materials are required and in what quantities.

Most successful organizsation nticipate the future and for their products and translate that information into factor inputs required to satisfy expected demand. Forecasting provides a blue print for managerial planning. Forecasting is the estimation of the future on the basis of the past.

In many organizations, sales forecasts are used to establish production levels, facilitate scheduling, set inventory levels, determine man power loading, make downloading decisions, establish sales conditions pricing and advertising and aid financial planning cash budgeting and capital budgeting. A good forecast should have the following attributes. It should be accurate, simple, easy, economical, quick and upto date. Following are the basic steps involved in a systematic demand forecast.

Explain simple exponential smoothing method of forecasting demand. What are its limitations? Comment on the forecast model. The recent data is given more weightage and the weightages for the earlier periods are successfully being reduced.

Let x1 is the actual historical data of demand during the period t. The forecast for April was units with a smoothing constant of 0. What do you think about a 0. Using exponential smoothing average: The sum of absolute deviation for the five data is? It also includes the determination of path that the work shall follow and the necessary sequence of operations which must be done on the material to make the product.

Routing procedure consist of the following steps: The finished product is analysed thoroughly from the manufacturing stand point, including the determination of components if it is an assembly product. Such an analysis must include: The following activities are to be performed in a particular sequence for routing a product 1. Analysis of the product and breaking it down into components.

Taking makes or downloads decisions. Determination of operations and processing time requirement. Determination of the lot size. Scheduling Introduction Scheduling is used to allocate resources over time to accomplish specific tasks. It should take account of technical requirement of task, available capacity and forecasted demand.

Forecasted demand determines plan for the output, which tells us when products are needed. The output-plan should be translated into operations, timing and schedule on the shop-floor. S K Mon ndal Chapter 2 The Planning and Sched duling Fun nction Loa ading The customer c order for eachh job has ceertain job contents, whhich need too be performmed on variou us work cennters or facillities. Durin ng each plan nning periodd, jobs orders are assig gned on facilitties. This ulltimately deetermines tthe work-loaad or jobs tto be perforrmed in a planned p period d.

The e assignme ent of spec cific jobs to o each operational fa acility duriing a plann ning period iss known ass loading. Seq quencin ng When n number off jobs are wa aiting in queue before an a operation nal facility such as, a milling m machiine , there is a need to decide the sequen nce of proccessing all the waiting jobs. Sequeencing is baasically an order o in whhich the jobs, waiting b before an op perational facility, f are prrocessed.

Foor this, priorrity rule, prrocessing tim me, etc. Th he decision n regardin ng order in n which job bs-in-waiting are pro ocessed on an oper rational fa e is called as sequen acility or work-centre w ncing. Dettailed Schedul S ling Once the priorityy rule of job sequencingg is known, we can sequ uence the joobs in a parrticular order..

This orderr would deteermine whiich job is do one first, then which thhe next one is and so on.. However, sequencing s does not telll us the day y and time a at which a particular p joob is to be doone.

This asspect is covvered in detailed scheduling. In this, estima ates are prrepared regardding setup and processsing time a at which a job is due to start an nd finish. S K Mondal Chapter 2 Detailed scheduling encompasses the formation of starting and finishing time of all jobs at each operational facility.

Expediting Once the detailed schedule is operationalized, we need to keep a watch over the progress in the shop-floor. This is necessary to avoid a deviation from the schedule.

In case of deviation from the schedule, the causes of deviation are immediately attended to. For example, machine breakdown, non-availability of a tool, etc. Therefore, continuous follow up or expediting is needed to overcome the deviations from schedule. The objective of expediting is to complete the jobs as per the detailed schedule and overcome any special case causing delay, failure, break-down, non-availability of material and disruption of detailed schedule.

Short-term Capacity Input-output Control Schedules are made so that jobs are completed at a specific time on every facility. For this, each facility has certain capacity to perform. In real situation, the utilization of the capacity of each facility may be different from the planned one. This difference should be monitored carefully because under-utilization of capacity means waste resource and over- utilization may cause disruption, failure, delays, or even breakdown.

Therefore, in case of discrepancy in input and output of the capacities, some adjustments in schedule are needed. Short-term capacity control involves monitoring of deviation between actual and planned utilization of the capacity of an operational facility. There are two types of schedules used: Master Schedules and Shop or Production Schedule. Master schedule: The first step in scheduling is to prepare the Master Schedule.

A master schedule specifies the product to be manufactured, the quality to be produced and the delivery date to the customer. It also indicates the relative importance or manufacturing orders. The scheduling periods used in the master schedule are usually months. Whenever a new order is received, it is scheduled on the master schedule taking into account the production capacity of the plant. Based on the master schedule, individual components and sub-assemblies that make up each product are planned: S K Mondal Chapter 2 The objectives of master schedule are: It helps in keeping a running total of the production requirements.

With its help, the production manager can plan in advance for any necessity of shifting from one product to another or for a possible overall increase or decrease in production requirements.

It provides the necessary data for calculating the back log of work or load ahead of each major machine. After an order is placed in the master schedule, the customer can be supplied with probable or definite date of delivery.

Shop or production schedule: After preparing the master schedule, the next step is to prepare shop or production schedule. This includes the department machine and labour- load schedules, and the start dates and finish dates for the various components to be manufactured within the plant.

A scheduling clerk does this job so that all processing and shipping requirements are relatively met. For this, the following are the major considerations to be taken case of: Objectives of Production Schedule: It meets the output goals of the master schedule and fulfils delivery promises.

It keeps a constant supply of work ahead of each machine. It puts manufacturing orders in the shortest possible time consistent with economy. The Scheduling Problem List Scheduling Algorithms This class of algorithms arranges jobs on a list according to some rule.

The next job on the list is then assigned to the first available machine. Random List This list is made according to a random permutation. Longest Processing Time LPT The longest processing time rule orders the jobs in the order of decreasing processing times. Whenever a machine is free, the largest job ready at the time will begin processing.

This algorithm is a heuristic used for finding the minimum make span of a schedule. It schedules the longest jobs first so that no one large job will "stick out" at the end of the schedule and dramatically lengthen the completion time of the last job.

Shortest Processing Time SPT The shortest processing time rule orders the jobs in the order of increasing processing times. Whenever a machine is free, the shortest job ready at the time will begin processing. This algorithm is optimal for finding the minimum total completion time and weighted completion time. S K Mondal Chapter 2 number of jobs in the system, minimizing the mean waiting time of the jobs from the time of arrival to the start of processing, minimizing the maximum waiting time and the mean lateness.

Let t[i] and w[i] denote the processing time and the weight associated with the job to be done in the sequence ordered by the WSPT rule. Earliest Due Date EDD In the single machine environment with ready time set at 0 for all jobs, the earliest due date rule orders the sequence of jobs to be done from the job with the earliest due date to the job with the latest due date.

Let d[i] denote the due date of the ith job in the ordered sequence. Let d[i] and t[i] denote the due date and the processing time associated with the ith job to be done in the ordered sequence. Other Algorithms Hodgson's Algorithm Hodgson's Algorithm minimizes the number of tardy jobs in the single machine environment with ready time equal to zero.

Let E denote the set of early jobs and L denote the set of late jobs. Initially, all jobs are in set E and set L is empty. Step 1: Order all jobs in the set E using EDD rule. Step 2: If no jobs in E are late, stop; E must be optimal. Otherwise, find the first late job in E. Let this first late job be the kth job in set E, job [k].

Step 3: Out of the first k jobs, find the longest job. Remove this job from E and put it in L. Return to step 2. Their sequence of arrival, processing time and due-date are given in the table below. Compare the results. S K Mon ndal Chapter 2 Soluttion: Th hen the neext arrived job is sched duled, and soo on. This approoach gives fo ollowing seq quence of job bs for the giiven problem m: S K Mo ondal Cha apter 2 Th his rule givees highest prriority to th he job having earliest du ue-date: Most M recent job b is the last arrived joob.

The schheduling of jobs on thiis rule is ex xplained thhrough the earlier examp ple. The rule giv Ta ves priority y of jobs in a random order. Let th he random b be: S K Mondal Chapter 2 It is observed that SPT sequencing rule for single machine and many jobs performs better than other rules in minimizing total flow time, average flow time, and average lateness of jobs.

Identify the job with the smallest processing time on either machine. If the smallest processing time involves: Machine 1, schedule the job at the beginning of the schedule. S K Mon ndal Chapter 2 Machine e 2, schedulle the job tow ward the ennd of the sch hedule. Iff there is soome unsched duled job, goo to 1 otherw wise stop.

Solvinng an equivaalent two-m machine prob blem with processing p tiimes: Ana alysis of o Resu ult The present p sequ uence is ana alyzed for tim me on mach hines as folloows: Therefore,, only after 1 min, next job J1 will start on M1 and J6 will go on M2.

Fo or this, let tiji be the pro ocessing tim me of job i on machine jj. Att lease one of o the follow wing conditioons must bee satisfied beefore we can n use this allgorithm: The processing time on R a and S is calcculated as folllows: Ex xample: Check k for necessaary conditioons: Now, let us frame two hypotthetical macchines R and d S on whicch the processing times are: Using g Johnson's algorithm the optimum m sequence e for two ma achines R and a S and ssix jobs is: The tiime calculattions are ass follows: Calcu ulation of Machine M Id dle Time: J1 and J2.

Ea ach job is to be processeed on m macchines: M1, M Th here are tw wo different sequences, one each for each job b. It is nott permissiblle to have altternative seequences. Only one job can be perfformed at a time on th he two mach hines. The proocessing timme is know wn and is deterministi d ic.

The probblem is to find the seequence of proocessing so as to minimmize the tota al elapsed tiime in the system. Graphical G method m is ussed to solvee this proble em. It can b be illustrateed with an example. Tw wo jobs J11 and J2 are to be processed d on five machines m M1, M2, The pr rocessing time t and jo ob sequenc ces are as ffollows: Jo ob 1 Jo ob 2 Fin nd the totall minimum elapsed tim me using gra aphical apprroach.

So olution: S K Mon ndal Chapter 2 Step 3: Shade the commoon area for eeach machin nes Fig. Step 4: Start from f origin. The on nly conditioon to be av voided is to cross a sh haded area by the diagonnal line.

The linne moving horizontally h ns that J1 iss processed and J2 y i. A diagonnal line meaans that both J1 and J2 2 are processed. The shhaded portiion is avoid diagonal line, because at any ded to be crrossed by d time both b J1 and J2 cannot b be processed me machine. Note the t idle timee for each joob from grap ph. For Job J 2: Use Hodgson's H algorithmm to schedu ule five job bs for whic ch the pro ocessing tim me ti and due-date d d di are as follows: S K Mo ondal Cha apter 2 Steps: Hig gden Algoritthm step ex xplained bellow.

Forrm a string of jobs into first late joob. Rem move this jo ob from string of jobs a and put in th he new latee job in the string s and rep peat Steps 1 to 4. Step 5: Sin nce at this sttage there is i no late job b, we will sttop.

He ence, solutio on is: S K Mon ndal Chapter 2 Exam mple: Use Johnson's J algorithm m to schedu ule six jobss and two machines: Using g Johnson's rule: Select S task with least processing g time in thhe string oof the given n jobs. If itt is on m machine A, place p at the left-end oth herwise on right-end. Remove R thatt task from string and a apply rule again.

Repeat R stepss 1 — 2 till all jobs are over. Exam mple: Use Jackson'ss Extension n of Johnsson's Rule e to schedu ule five jo obs on three e machines s. S K Mo ondal Cha apter 2 Soolution: Sin nce machine B is domiinated by machine m A: Hence, above problemm is converted to fit into o 2 machinee n job as follows: Ussing Johnson's rule the optimal seq quence is: Machi M ine Lo oadin ng Alllocating thee job to woork centres is referred tto as "Mach hine Loadingg", while alllocation of job bs to the enttire shop is called "Shop Loading"..

S K Mondal Chapter 2 and equipment. The load capacity of a machine may be expressed in terms of pieces for a given length of time or in time for a given number of pieces. In either case, the capacity may be determined very readily from the standard time values of the operations performed by the machine. A machine load chart is a chart for showing the work ahead for various machines and processes. A typical machine load chart is shown in figure.

Success case

Here, the load is expressed in terms of the number of hours for a given number of pieces. A bar represents a task. It is shown along the horizontal axis which indicates time scale. A Typical Machine Load Chart Despatching After the schedule has been completed, the production planning and control department makes a master manufacturing order with complete information including routing, the desired completion dates within each department or on each machine and the engineering drawings.

From this master manufacturing order, departmental manufacturing orders can be made up giving only the information necessary for each individual foreman. These include inspection tickets and authorization to move the work from one department to the next when each department's work is completed. When a foreman of a particular department receives the manufacturing order, he is authorized to begin production in his department. This function is purely a clerical function and requires voluminous paper work.

S K Mondal Chapter 2 1. Initiate the work by issuing the current work order instructions and drawings to the different production departments, work stations, machine operators or foremen. The various documents despatched include: It is not raw material it is material from store] 2.

Release materials from stores. Release production tooling, that is, all tools, jigs, fixtures and gauges for each operation before operation is started. Keep a record of the starting and completion date of each operation. Getting reports back from the men when they finish the jobs. Works order documents. The usual formats of various works order documents used by the despatcher are the route sheet card , operation sheet and machine loading chart. Work order 2. Machine load chart 3. Material requisition form [but not raw material] 4.

Move ticket 5. Inspection ticket The term "despatching" is not much heard. In this system, one shop order copy, known as "traveller", circulates through the shop with the parts.

Product Development What is Product Development? Product development is the process of designing, creating, and marketing an idea or product. The product can either be one that is new to the market place or one that is new to your particular company, or, an existing product that has been improved. In many instances a product will be labelled new and improved when substantial changes have been made. The Product Development Process All product development goes through a similar planning process.

Although the process is a continuous one, it is crucial that companies stand back after each step and evaluate whether the new product is worth the investment to continue.

That evaluation should be based on a specific set of objective criteria, not someone's gut feeling. Even if the product is wonderful, if no one downloads it the company will not make a profit. Brainstorming and developing a concept is the first step in product development. Once an idea is generated, it is important to determine whether there is a market for the product, what the target market is, and whether the idea will be profitable, as well as whether it is feasible from an engineering and financial standpoint.

Once the product is determined to be feasible, the idea or concept is tested on a small sample of customers within the target market to see what their reactions are. A manufacturing shop processes sheet metal jobs, wherein each job must pass through two machines M1 and M2, in that order. The processing time in hours for these jobs is: The optimal make-span in hours of the shop is: Routing in production planning and control refers to the a Balancing of load on machines [IES] b Authorization of work to be performed c Progress of work performed d Sequence of operations to be performed IES The routing function in a production system design is concerned with.

Is a general timetable of manufacturing 2. Is the time phase of loading 3. Is loading all the work in process on 4. Machines according to their capacity Which of the statements given above are correct? Preparation of master production schedule is an iterative process 2. Schedule charts are made with respect to jobs while load charts are made with respect to machines 3. MRP is done before master production scheduling Which of the statements given above are correct?

Which of the following factors are to be considered for production scheduling? Sales forecast 2. Component design 3. Route sheet 4. Time standards Select the correct answer using the codes given below: Planning and scheduling of job order manufacturing differ from planning and scheduling of mass production manufacturing.

In mass production manufacturing, a large variety of products are manufactured in large quantity. Production scheduling is simpler, and high volume of output and high labour efficiency are achieved in the case of: A manufacturer's master product schedule of a product is given below: Week-l Week-2 Week-3 Planned Production: Before the start of week-1, there are components of type A in stock.

The lead time to procure this component is 2 weeks and the order quantity is Number of components A per product is only one. The manufacturer should place the order for a components in week-l b components in week-3 c components in week-l and components in week-3 d components in week-5 Machine Loading IES Which one of the following charts gives simultaneously, information about the progress of work and machine loading?

Which one of the following is required for the preparation of the load chart machine? Despatching function of production planning and control refers to: Which one of the following statements is not correct? Which one of the following statements is correct in relation to production, planning and control? Consider the following statement [IES] Dispatching 1.

Is the action of operations planning and control 2. Releases work to the operating divisions. Conveys instructions to the shop floor.

Of these statements: The value engineering technique in which experts of the same rank assemble for product development is called [IES] a Delphi b Brain storming c Morphological analysis d Direct expert comparison IES Which one of the following is the preferred logical sequence in the development of a new product?

Page 41 Consider the following of aspects: Functional 2. Operational 3. To simulate sales function 2. To utilize the existing equipment and power 3. The following activities are to be performed in a particular sequence for routing a product [IAS] 1. Analysis of the product and breaking it down into components 2. Determination of the lot size 3.

Determination of operations and processing time requirement 4. Taking makes or downloads decisions The correct sequence of these activities is a 1, 2, 3, 4 b 3, 1, 2, 4 c 3, 1, 4, 2 d 1, 4, 3, 2 Scheduling IAS Which of the following are the objectives of scheduling? Reducing average waiting time of a batch 2. To meet the deadline of order fulfillment 3.

To improve quality of products 4. To increase facility utilization Select the correct answer using the code given below: Which one of the following is the correct definition of critical ratio in scheduling?

Consider the following advantages: Very flexible 2. Simple to understand 3. Detailed operation can be visualized Which of these are the advantages of master scheduling? Activities involved in production planning and control system are listed below: Planning 2.

Loading 3. Scheduling 4. Despatching 5. Routing 6. Follow up The correct sequence of these activities in production planning and control system is: Conventional production planning techniques cannot be used for managing service operations. Service operations cannot be inventoried. Which of the following pairs are correctly matched? Project scheduling — Critical path analysis 2. Batch production — Line of balance scheduling 3. Despatching — Job order 4. Routing — Gantchart Select the correct answer using the codes given below: Consider the following advantages [IAS] 1.

Lower in-process inventory 2. Higher flexibility in rescheduling in case of machine breakdown 3. Lower cost in material handling equipment When compared to process layout, the advantages of product layout would include a 1 and 2 b 1 and 3 c 2 and 3 d 1, 2, and 3 IAS Operation process chart 1.

Scheduling project operations B. Flow process chart 2. To study backtracking and traffic congestion C. Flow diagram 3. To analyze indirect costs such as material handling cost D. PERT chart 4. To study relations between operations Codes: In which one of the following types of industrial activities, the problem of loading and scheduling becomes more difficult?

Which of the following factors necessitate a change in schedule? Change in Board of Directors 2. Capacity modification 3. Lack of capital 4. Change in priority 5. Unexpected rush orders [IAS] Select the correct answer using the codes given below: Johnson's rule is applicable for planning a job shop for [IAS] a n machines and 2 jobs b 2 machines and n jobs c n machines and n jobs d 1 machine and n jobs Machine Loading IAS Dispatching function of production planning and control refers to: In a low volume production, the dispatching function is not concerned with issuing of which one of the following?

IES-7 7. S K Mondal Chapter 2 to the operating facility through the release of orders and instruction in accordance with a previously developed plan of activity time and sequence establish by scheduling section of the production planning and control department. At firsst we converrt it equivallent two-ma achine probllem. Proc cessing tim me in min nute of siix jobs on n two machines are given below w.

Use Johnson's rule e to schedu ule these jobs. Minim mum processing time off 1 min is foor J3 on M2 and J6 on M1. Place J6 at the fiirst and J3 at a the end oof the sequen nce. Now, remove J3 and a J6 from m the consid deration. Wee have the foollowing job bs: S K Mo ondal Cha apter 2 Ouut of all thee remaining g processingg times, J4 on M2 is leeast and equ ual to 3 miinutes. So, pla ace it at thee last of thee sequence.

It is in the last becausse of being least processsing time on n M2 and nott on M1. Affter elimina ating J4 fro om the abovve list, we have J1, J2 and J5. O Out of all remaining r pro mes, J1 on M1 is least and is equa ocessing tim al to 4 min.. Therefore,, place this job at the beginning of the list. Affter placing J4 at the end and J1 1 in the beginning we e have the folllowing sequuence: Noow, the rem maining jobss are J2 and d J5.

Looking at their processing times, it iss observed tha at the leasst time is 6 min. Thereffore, place J2 at the beginning of left-most l slo ot of sequen nce and J5 at the righ ht-most slot of the sequ uence. The woorkpiece visiit stations successively s y as they aree moved aloong the line usually by some kind of transportattion system,, e. Objectiv O ve in Lin ne Bala ancing Problem m In an assemb bly line, th he problem is to desiggn the work k station. Each E work station is designed to co omplete feww processing g and assemmbly tasks.

The objective in the deesign is to assign processes and tassks to indiv vidual statioons so that the total tiime requireed at each woork station is approxim mately samme and nearrer to the desired d cycle time or production p ratte. In case, all the work elem ments which h can be grouped at any y station have same sta ation time, the en this is a case of perrfect line ba ow would bee smooth in this case.

Prooduction flo Hoowever, it isi difficult to achieve this in reeality. When n perfect line balanciing is not ach hieved, the station timme of slowesst station woould determmine the prooduction ratte or cycle tim me.

Let L us conssider a fiv ve-station assembly system in which the station tim mes are 12, 16, 13, 11 and 15 miinutes resppectively. The T slowesst station is i station 2, which takkes 16 min n. Woork carrier r enters att station: Now a w work carrie er Fig. A Att station 1 cannot leav ve station 1 after 12 minutes ass station 2 is not freee after 12 miinutes of woork on a prreviously arrrived work carrier.

On nly after 16 6 minutes itt is free to pu ull work carrrier from station 1. Simmilarly, in each e cycle, station s 3,4 and a 5 would d be idle for 3, 2 and 4 mmin. Sin nce, idle tim me at any sttation is thee un-utilized d resource, the objectiv ve of line ba alancing is to minimise thhis.

The workpieces jobs are consecutively launched down the line and are moved from station to station. At each station, certain operations are repeatedly performed regarding the cycle time maximum or average time available for each work cycle. The total work load necessary for assembling a workpiece is measured by the sum of task times tsum.

Due to technological and organizational conditions precedence constraints between the tasks have to be observed. These elements can be summarized and visualized by a precedence graph.

It contains a node for each task, node weights for the task times, arcs for the direct and paths for the indirect precedence constraints. Any type of ALBP consists in finding a feasible line balance, i.

Because of the long-term effect of balancing decisions, the used objectives have to be carefully chosen considering the strategic goals of the enterprise. From an economic point of view cost and profit related objectives should be considered. However, measuring and predicting the cost of operating a line over months or years and the profits achieved by selling the products assembled is rather complicated and error-prone.

In the most simple case, the line efficiency is defined as follows: Sum of the task times for all the assembly tasks for the product. For example, operation of super-finishing cannot start unless earlier operations of turning, etc. While designing the line balancing problem, one has to satisfy the precedence constraint. This is also referred as technological constraint, which is due to sequencing requirement in the entire job.

Another constraint in the balancing problem is zoning constraint. If may be either positive zoning constraint or negative zoning constraint. Positive zoning constraint compels the designer to accommodate specified work-elements to be grouped together at one station.

For example, in an automobile assembly line, workers are doing work at both sides of automobile. Therefore, at any station, few operations have to be combined. Many times, operation and inspection are grouped together due to positive zoning constraint. In a negative zoning constraint few operations are separated from each other. For example, any work station, which performs spray painting, may be separate from a station, which performs welding, due to safety considerations.

Therefore, following constraints must be following in a line balancing problem: Precedence relationship. Zoning constraints if any. Restriction on number of work stations n , which should lie between one and total number of work elements N. Work Element i The job is divided into its component tasks so that the work may be spread along the line.

Work element is a part of the total job content in the line. Let N be the maximum number of work element, which is obtained by dividing the total work element into minimum rational work element.

Minimum rational work element is the smallest practical divisible task into which a work can be divided. Also, all TiN are additive in nature.

Where N is total number of work elements? Work Stations w It is a location on the assembly line where a combination of few work elements is performed. Total Work Content Twc This is the algebric sum of time of all the work elements on the line. Station Time Tsi It is the sum of all the work elements i on work station s. Cycle Time Tc Cycle time is the rate of production. This is the time between two successive assemblies coming out of a line. Cycle time can be greater than or equal to the maximum of all times, taken at any station.

Necessary clarification is already given in the previous example. Delay or Idle Time at Station Tds This is the difference between the cycle time of the line and station time. Precedence Diagram This is a diagram in which the work elements are shown as per their sequence relations.

Any job cannot be performed unless its predecessor is completed. A graphical representation, containing arrows from predecessor to successor work element, is shown in the precedence diagram.

Every node in the diagram represents a work element. Therefore, the effort is done to minimise the balance delay. Due to imperfect allocation of work elements along various stations, there is idle time at station.

Therefore, balance delay: Line Efficiency LE It is expressed as the ratio of the total station time to the cycle time, multiplied by the number of work stations n: Smoothness Index SI Smoothness index is a measure of relative smoothness of a line: Methods of Line Balancing It is not possible to date to have an approach, which may guarantee an optimal solution for a line balancing problem. Many heuristics exist in literature for this problem. The heuristic provides satisfactory solution but does not guarantee the optimal one or the best solution.

We would discuss some of the heuristics on a sample problem of line balancing, as given below: Let us consider the precedence diagram of 13 work elements shown below. The time for each work element is at the top of each node: Larges st Cand didate Rule R Step 1: List alll work elem ments i in descending d order of theeir work elements TiN value.

Decide e cycle time Tc. Assignn work elem ment to the station. Sta art from thee top of the list of unasssigned elemen nts. Select only feasiible elemen nts as per the preced dence and zoning constraints. Selecct till the sta ation does not n exceed cy ycle time. Continnue step 3 foor next station. Till alll work elem Problem 1: Refer the probblem shown in figure beelow. Decidee cycle time e. Th me Tc musst satisfy: LLet us selecct 5 stations design.

List w work elemeents in desccending 5 order of their worrk element. Connstruct preecedence diaagram. Mak ke a colummn I, in whhich includee all work elem ments, whiich do not have h a preccedence work elementt.

Make colu umn II in which list all elements, e n column I. Continue till all work which follow elements in elem ments are exhausted. A feasible cy ycle time iss selected. Number N of sta ations would d be: Asssign the worrk elementss in the work k station soo that total sstation timee is equal to or o slightly leess than the e cycle timee.

Reppeat step 4 for unassignned work ellements. Seven co olumn inittial assignm ment No ow, selecting g cycle timee as equal too 18 secondss we follow these t steps:: Th he cycle timee must lie between 68 fo or one statioon to 10 min n. Now, regroup elements in columns I and II till we get 17 min. Thus, elements 1, 2, 3, 4 are selected at station I. We proceed in the same way for remaining elements: This is apparent when we consider the following grouping: Draaw the precedence diag gram.

Forr each workk element, determine d th he positionaal weight. It I is the totaal time on thee longest path from thee beginning of the operation to thee last operattion of the nettwork. Ran nk the work elementts in desceending orde er of rankeed position nal weight R.

Callculation of RPW would d be explain ned in the ex xample to foollow. An electronic equipment manufacturer has decided to add a component subassembly operation that can produce 80 units during a regular 8-hour shift.

What is the line efficiency of the assembly line? Balance delay 4. Mechanical assembly 12 E. Line inefficiency which results from the idle time due to imperfect allocation of work among stations C. Which one of the following is true in respect of production control for continuous or [IES] assembly line production? C and D. Week 1 2 3 4 5 6 Demand Ignore lead times for assembly and sub-assembly. The actual output turns to be units per day. Every product needs all the four operations to be carried out.

Test 3 For line balancing the number of work stations required for the activities M. Per unit welding time for each product is 20 s.

Number of welding lines required is a 5 b 4 c 3 d 2 Daily demand forecast for product A is E and T would respectively be a 2. B and C. A company has four work centres A. The machines are laid down in order A.

A welding line can operate [IES] efficiently for minutes a day. The designed capacity of each of the four machines is Production capacity per week for component R is the bottleneck operation. What is [IES] the system efficiency? A production system has a product type of layout in which there are four machines laid in series. The product structure of an assembly P is shown in the figure.

The system capacity would be a products per day b products per day c products per day d products per day Each machine does a separate operation. A new facility has to be designed to do all the welding for 3 products: Starting with zero inventory.

The minimal number of workstations that are needed to achieve this production level is: How many car wash stalls should be installed? Each car will have a wash time of 3 minutes. Other machines will be on empty position. If products are to go out after assembly at the rate of 60 per shift.

A work shift is for 8 hours duration.

An operations consultant for an automatic car wash wishes to plan for enough capacity of stalls to handle 60 cars per hour. Consider the following sets of tasks to complete the assembly of an engineering component: Total time for one assembly is sec.

Also mention some of the steps you would recommended to improve the line balance. Elemental Time Min Immediate Predecessors 1 0. The following elemental data pertains to the assembly of a new model toy: Element No.

Although the process is a continuous one. Once the product is determined to be feasible. Once an idea is generated. Which one of the following is the preferred logical sequence in the development of a [IES] new product? Even if the product is wonderful. That evaluation should be based on a specific set of objective criteria. Question with Answer IES. The product can either be one that is new to the marketplace or one that is new to your particular company.

Brainstorming and developing a concept is the first step in product development. In many instances a product will be labelled new and improved when substantial changes have been made. Product development is the process of designing. The Product Development Process All product development goes through a similar planning process.

The value engineering technique in which experts of the same rank assemble for [IES] product development is called a Delphi b brain storming c morphological analysis d direct expert comparison 1. Break-even point analysis is also used to make a choice between two machine tools to produce a given component. It usually refers to the number of pieces for which a business neither makes a profit nor incurs a loss.

In other words. Whose Variable cost is less that one is economical. Whose Fixed cost is less that one is economical. Contribution is the measure of economic value that tells how much the sale of one unit of the product will contribute to cover fixed cost. But if Q comes out negative then. At BEP. Whose both Fixed and Variable cost is less that one is economical. Margin of Safety: It is the difference the operating sale break-even sales.

Case A When fixed cost are a large portion of total cost. Case B When variable cost are high a reduction is variable cost may be more effective in generating profits than changes in the total volume or per-unit prices. Profit volume ratio: High VC Case A requires a large volume of output to reach break even.

Higher is the angle of incidence faster will to the attainment of considerable profit for given increase in production over BEP. Process I requires 20 units of fixed cost and 3 units of variable cost per piece. If each unit is sold at Rs. Type I: Brazed type of original cost Rs. Type II: Throwaway tip square of original cost Rs.

The break even quantity of product would be a b c d 2. All variable costs are directly variable with production. Fixed investments for manufacturing a product in a particular year is Rs. The variable cost per unit for this product is [IAS ] Rs. The break even quantity between the two alternatives is a 25 b 50 c 75 d 5.

The break-even production [IES] per month will be a b c d 6. The estimated sales for this period is 2. On a lathe. Two alternative methods can produce a product first method has a fixed cost of Rs.

The variable cost per product is Rs. For a small scale industry. For a [IES] company producing 10 piece per day a process I should be chosen b process II should be chosen c either of the two processes could be chosen d a combination of process I and process II should be chosen Two [IES] types of carbide tools are available.

At that volume. Variable cost Sale Higher is the profit volume ratio. The second method has a fixed cost of Rs.

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Last year. If orders are placed once a month to meet an annual demand of 6. It is possible to have more than one break-even point in break even charts. The indirect cost of a plant is Rs 4. The break even analysis shows that the break even quantities using automatic plant vs traditional plant are in the ratio of 1: The fully automatic plant can turn out a shoe at a unit variable cost of Rs. If the fixed cost of the assets for a given period doubles.

At what [IES] production quantity. If the average revenue per product is Rs To determine the location of the additional plant Code: The direct cost is Rs 20 per product.

Process X has fixed cost of Rs. If the annual estimated sales is Rs. The per unit revenue used in [IES] the break even calculation is a Rs. Assignment problem 3. To choose between different methods of manufacture D. Transportation problem 2. There is yet another choice of traditional manufacture that needs in investment of Rs.

To provide different facility at different locations B. They can use a fully automatic robot-controlled plant with an investment of Rs. The fixed costs for a year is Rs. Which one of the following information combinations has lowest break-even point? Break even analysis 1.

To take action from among the paths with uncertainty C. Decision tree 4. A larger margin of safety in break-even analysis is helpful for management decision. If the total investment is Rs. Fixed cost Codes: Cost for new design 5. Break-Even Point C.

Based on the given graph. If Break. If the margin of safety is large. Profit B. The company makes a profit of Rs. Process II has a fixed cost Rs. Two jigs are under consideration for a drilling operation to make a particular part.

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Toy type Q requires an electric switch which is available pieces per day only. A component can be produced by any of the four processes I. Jig A costs Rs. Process I has a fixed cost of Rs. For maximization of profits. Process IV has a fixed cost of Rs. It has a variable cost of Rs. Two machines of the same production rate are available for use. The quantity of parts to be manufactured at which either jig will prove equally costly is [IAS] a b c d A standard machine tool and an automatic machine tool are being compared for the [GATE] production of a component.

On machine 1. The supply of raw material is just sufficient to produce toys of any type per day. The marginal cost is the maximum value at which the product selling price must be fixed to recover all the costs. A company produces two types of toys: P and Q. A company sells The Production time of Q is twice that of P and the company has a maximum of time units per day. Following data refers to the two machines. If the company wishes to produce pieces of the component.

III and IV. Jig B costs Rs. Marginal cost in linear break-even analysis provides the management [IAS] with useful information for price fixing.

Process III has a fixed cost of Rs. The breakeven production batch size above which the automatic machine tool will be economical to use. The [GATE] breakeven production rate for each machine is a 75 b c d Semi-variable 3.

List I Element of cost A. Water and electricity Codes: So it will give us minimum BFQ. And Fixed cost and variable cost both are minimum in case of a. The variable cost per unit associated with automated assembly line VA. Direct labour C. Interest on capital B. For certain strategic reasons both the machines are to be used concurrently.

The sale price of the first units is Rs. Variable 2. Shop floor control 2. The term "Capacity" of a plant is used to denote the maximum rate of production that the plant can achieve under given set of assumed operating conditions. Rescheduling orders based on production priorities C. Master production 3. Listing products to be assembled and when to be delivered B. Production planning and control Capacity planning 1. Material requirement 4.

Short term capacity planning involves decisions on the following factors: Capacity planning is concerned with determining labour and equipment capacity requirements to meet the current master production schedule and long term future needs of the plant. Capacity planning forms the second principal step in the production system. Closure tolerances schedule D. Monitor progress of orders and report their status planning 5. In PERT all activity time is probabilistic.

The activity time is CPM applications were relatively less uncertain and were. In CPM we actually know the Activity time. Employs Beta-distribution for the time.

PERT and CPM applications started overlapping and now they are used almost as a single techniques and difference between the two is only of the historical or academic interest. While the CPM is Activity. Program Evaluation and Review Technique. Critical Path Method. With the passage of time.

The principal feature of PERT is that its activity time estimates are probabilistic. In CPM activity time is deterministic. The other difference: PERT is Event.

A group of students is given the project of designing. No expenditure of manpower or resources may be associated with an event. It is the beginning.

A project is a combination of interrelated activities which must be performed in a certain order for its completion. An event is a point in time within the project which has significance to the management. An event triggers on activity of the project.

The activity or a unit of work. The project can be broken down into the following sub-parts. Let us consider a very simple situation to illustrate the W.

Design Fabrication Testing Terminology Activity: It is a time consuming effort that is required to perform part of a work. Critical Path: Critical path is the on the net work of project activities which takes longest time from start to finish.

An activity beginning and ends with events. The completion time for the project has a normal distribution about the expected completion time. An activity with zero slack is known as critical activity. The process of dividing the project into these activities is called the Work-Break-Down structure W.

This is called the scheduled completion time TS and for the backward pass computation. In such cases the critical path is the path of least float. But in a PERT network. Negative Float and Negative slack.

Cost associated to crash is crash cost. Now there may be three cases: Network Construction i What activities must be completed before a particular activity starts? The process of reducing on activity time by adding fresh resource and hence usually increasing cost. The latest allowable occurrence time TL for the end event in a CPM network is usually assumed to be equal to the earliest expected time TE for that event. Float and Slack: That the float of an activity has the same significance as the slack of the events.

It is useful for proper representation in the network. An activity that consumes no time but shows precedence among activities. TL for the end event is taken equal to Ts. Crashing is needed for finishing the task before estimated time.

The assumption is that successive activities take their expected time.

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Start from first event and go upto last end. The assumption is that all predecessor activities are started at their earliest start time. Start from last event and come upto first. LS 4 ES previous i ]. It is defined as the amount of time on activity can be delayed without affecting the duration of the project. It is used to denote the amount of time an activity can be delayed without delaying the earliest start of any succeeding activity. Total Float: It is the maximum time.

It is important when the network of the project runs on earliest time. Expected time i. I or J may or may not end in same time. If an activity reaches next stage at the latest time. K or L will not start until both I and J finished. The most likely time need not be the midpoint of to and tp and hence the frequency distribution curve may be skewed to the left.

Activity P depends only on N not on activity M. Probability tensity function What is the probability that the activity will be completed in this expected time? Variance is the measure of this uncertainty. Greater the value of variance. Probability of Meeting the Scheduled Dates The standard normal distribution curve.

Two activities can be identified by the same beginning and end events. Optimistic time 2. Total slack along the critical path is not zero 4. The earliest occurrence time for event '2' is 32 weeks and the latest occurrence time for event '2' is 37 weeks. Most likely time Which of the statements given above are correct? Dummy activities are used in a network to a facilitate computation of slacks c determine project completion time 4.

Each activity is represented by one and only one arrow in the network. Which of the following are the guidelines for the construction of a network diagram? It consists of activities with uncertain time phases 2.

If the activity time is 11 weeks. Consider an activity having a duration time of Tij. If the earliest starting time for an activity is 8 weeks. It is similar to electrical network Which of the statements given above are correct? Pessimistic time 3.

E is the earliest occurrence time and L the latest occurrence time see figure given. This is evolved from Gantt chart 3. Dummy activity consumes no time or resource. PERT considers the following time estimates 1. Dangling must be avoided in a network diagram. There can be more than one critical path in PERT network 5. Which one of the following networks is correctly drawn? Ei Of these statements a 1. A project consists of three parallel paths with durations and variances of According to the standard PERT assumptions.

Tij 1. The essential condition for the decompression of an activity is a The project time should change due to decompression b After decompression the time of an activity invariably exceeds its normal time. Tij 3. Stage Coach C. The variance is four days. Time estimates of an activity in a PERT network are: In PERT. Dynamic Problem 3. In a PERT network. Which one of the following statements is not correct? A PERT network has three activities on critical path with mean time 3. Critical Path B.

PERT is based on the approach of multiple time estimates for each activity. The probability that the project will be completed in 20 days is a 0. Vogel's Approximate Method Levelling and smoothing are the techniques related to resource scheduling in CPM. What is the expected time of the activity? In a small engineering project. Assignment Problem 2. Critical path method 4. Estimated time Te and variance of the activities 'V' on the critical path in a PERT new work are given in the following table: Has zero total slack Code: Crashing 4.

Vogel's approximation method 1. Critical path 2. Transportation model C. Floods technique 2. For the PERT network shown in the given figure. Constructed only to establish sequence D. Dummy activity 1. Two phase method 3. PERT activity 3. It is built on activity oriented diagram C. Assignment model B. Consider a PERT network for a project involving six tasks a to f For the network below.

Which of the following equations can be used to find sG? The expected completion time of the project is a days b days c days Let dij be the length of directed arc from node i to node j. The expected time of the activity would be [IES] a Exponential distribution B C D 4 3 1 4 1 2 Smoothing and Leveling 3. Poisson distribution 5. Control chart for fraction rejected D. Control charts for variables B. Expected value approach 2. Branch and Bound technique 1. Control chart for number of non-conformities C.

Decision theory Codes: Binomial distribution 2. The change in critical path required rescheduling in a PERT network. Queuing theory D. A Project consists of activities A to M shown in the net in the following figure with the duration of the activities marked in days [GATE] The project can be completed a between The three time estimates of a PERT activity are: Normal distribution 4.

Exponential distribution 4. Beta distribution 3. Integer programming d. Some of the activities cannot be completed in time due to unexpected breakdown of equipments or non-availability of raw materials. In CPM network critical path denotes the a path where maximum resources are used b path where minimum resources are used [IAS] c path where delay of one activity prolongs the duration of completion of project d path that gets monitored automatically For the network shown in the given figure.

In a network. The critical path of a network is the path that: Latest start time of an activity in CPM is the [IES] a latest occurrence time of the successor event minus the duration of the activity b earliest occurrence time for the predecessor event plus the duration of the activity c latest occurrence time of the successor event d earliest occurrence time for the predecessor event [IES] In CPM. The earliest time of the completion of the last event in the above network in weeks is a 41 b 42 c 43 d 46 [IES] In the network shown below.

For the network shown in the figure. The probability of completion of the project in 24 days is a The critical path is along a b c d The variance of the completion time for a project is the sum of variances of a All activity times b Non-critical activity times [IES] c Critical activity times d Activity times of first and last activities of the project For short duration project. The outbreak of natural calamity necessitates updating Which of the above statements are correct? The number of activities that need to be crashed to reduce the project duration by 1 day is [GATE] a 1 b 2 c 3 d 6 A project has six activities A to F with respective activity durations 7.

In d both activity is dummy it is also not correct. All the activities can be crashed with the same crash cost per day. C-D and E-F. Consider the following statements regarding updating of the network: The network has three paths A-B. For large duration project. Updating is caused by overestimated or underestimated times of activities 4.

What is the standard deviation of the project completion time along the critical path? If the standard deviation of the corresponding activity are s1, s2 and s3, Solution: Conventional Question [ESE]: The network diagram of the project is given in Fig below. The activity times in days are shown on the arcs.

Activity Time A building project consists of 10 activities; their estimated duration is given below. Activity Duration 5 2 6 4 4 2 3 7 8 2 Draw the network and compute i Event times ii Activity time iii Total float and determine iv Critical path Solution:.

A small plant layout job consists of 10 steps their precedence relationship and activity times are identified as follows. F 35 Draw the network complete the forward and backward passes what activities make up the critical path?

Which activity has the most slack? Table 1 gives the different activities associated with a project consisting of 12 tasks A. I Slack of A Also determine the critical path time. Critical path A. ABC Analysis is done for items on stock and the basis of analysis is the annual consumption in terms of money value. Non-moving items ABC Analysis: The common and important of the selective inventory control of ABC analysis.

Control of A -item: Operation Control Inventory control Theory A fundamental objective of a good system of operation control of Inventories is to be able to place an order at the right time. Finished goods inventories 5. Control of B. Production inventories. Miscellaneous inventory Another way of classifying industrial inventories are i Transition inventory ii Speculative inventory iii Precautionary inventory Selective inventory control Different type of inventory analysis?

In-Process inventories semi-finish products at various production stages 4. Control of C-item: MRO Maintenance. Repair and operating supplies eg. Unit cost of inventory i Costs paid to the supplies for procuring one unit. For batch production it is Set-up costs. For discount model cost of inventory is considered.

Ordering costs: Total cost to procure 1 time. Annual Demand. Industrial Engineering Carrying costs or holding costs. Shortage or stock-out costs. Infinite production Rate. If Buffer stock is zero then. That so why EOQ model is very useful. Given Annual usage. We therefore conclude that unless the demand is highly uncertain the EOQ model gives fairly satisfactory decision values.

The unit cost is Rs. If the cost of one procurement is Rs. He picked on one item BV Delivery always occurs the day after ordering and the average carrying cost Rs From the current record he found that the average annual demand was valves.

The accounting information system revealed that the carrying cost Rs. What is the EOQ? Find the number of cases per order.

Assuming zero safety stock calculate EOQ. Storage cost is negligible. The cost per item is Rs. The current policy adopted in the company Is this an optimal policy? What would be the annual savings if the EOQ concept was applied? What do you infer?

The cost of each item is Rs. Conventional Question [CSE] Explain the salient feature of the following inventory models i Deterministic models ii Probabilistic models iii Model under uncertainty In a deterministic model the ordering cost is Rs If the annual demand is units determine EOQ.

It cost Rs 40 to place an order and estimated inventory carrying cost. It orders raw material for the brackets in lots of units from a supplies. Calculate the variation in percentage in their order quantity from optimal.

Estimated demand for the year is units. How many units in a batch should they produce? In each batch once the production starts they can make 80 units per day.

Set-up cost of the manufacturing process is per setup. The demand during the production period is 60 units per day. Carrying cost is Rs. Inventory control for deterministic demand lead time zero. This demand is fixed and known. The customer uses its item in assembly operation and has no storage space. The company can produce this item units per months. The holding a is Rs 0.

Because it will save Rs. Is it profitable to avail the discount? The annual demand are and respectively a Find the EOQ. Is it advisable to accept the discount? Comment on the result. Step — II: The unit prices of the items are Rs. The order cost for each of the items is Rs. The ordering cost is Rs. Economy in downloading.

Select the correct answer using the code given below: The inventory carrying cost includes a expenditure incurred for payment of bills c receiving and inspecting 86 [IAS] b placing an order d obsolescence and depreciation.

This iteration process continues. Which of the following are the benefits of inventory control? Improvement in customers relationship. Which one of the following correctly represents the average inventory turnover ratio for raw materials?

Elimination of the possibility of duplicate ordering. Group A constitutes costly items. ABC analysis is based on Pareto's principle 2. Classifying items in A.

A items require: In the ABC method of inventory control. Simulation can be l1sed for inventory control. ABC analysis in materials management is a method of classifying the inventories [IES] based on a The value of annual usage of the items b Economic order quantity c Volume of material consumption d Quantity of materials used [IES] 8.

What is the [IES] usual percentage of such items of the total items? Annual Usage of items Items No. Out of the following item listed below. Annual usage X Unit cost Rs. ABC analysis is based on Pareto distribution.

Of these statements a 1 alone is correct b 1 and 3 are correct c 2. Band C categories for selective control in inventory [IES] management is done by arranging items in the decreasing order of a total inventory costs b item value c annual usage value d item demand.

Selective control manages time more effectively. In ABC analysis. Cost of holding materials B. Setup costs do not include a labour cost of setting up machines c maintenance cost of the machines [GATE] b ordering cost of raw material d cost of processing the work piece In the EOQ model.

Carrying cost 2. If the annual demand of an item becomes half. Cost of receiving order C. Economic order quantity 3. Procurement lead time D. Procurement cost 1. Reorder point 4. Economic Order Quantity is the quantity at which the cost of carrying is a minimum b equal to the cost of ordering c less than the cost or ordering d cost of over-stocking [IES] Break-even analysis Code: Which one of the following is an inventory system that keeps a running record of the amount in storage and replenishes the stock when it drops to a certain level by ordering a fixed quantity?

A shop owner with an annual constant demand of 'A' units has ordering costs of Rs. In inventory control theory. The economic order [GATE] quantity is a b 2. An item can be downloadd for Rs Market demand for springs is 8. Cost of negotiation with suppliers Which of these costs are related to inventory carrying cost? The cost of making a download order is Rs. Cost of obsolescence 3. Cost of scrap 4.

If the annual demand is units. If the demand for an item is doubled and the ordering cost halved. Cost of insurance 5. Consider the following costs: A company downloads these springs in lots and sells them. Cost of inspection and return of goods 2. The cost of storage of springs is Rs. Annual demand for a product costing Rs. Details of cost for make or download decision are shown in the given graph.

Procurement cost 3. Given that. Ordering cost per order is Rs. Inventory carrying cost 2. Which one of the following ranges would lead to the economic decision? download A. A discount is offered for volume of download above 'V'. The economic lot size is [IES] a b c d Set up cost Select the correct answer using the codes given below: The given figure shows the details of stock-level in the periodic review inventory control system. DE B Ordered quantity 2.

FH C Safety stock 3.Decision tree expert opinion Codes: Let d[i] denote the due date of the ith job in the ordered sequence.

Work order 2. It consists of activities with uncertain time phases 2. Apportionment of sequential work activities into work stations 2. Employs Beta-distribution for the time. MPS is designed to meet the market demand both the firm orders and forecasted demand in future in the taken planning horizon. S K Mo ondal Cha apter 2 Steps: For a [IES] company producing 10 piece per day a process I should be chosen b process II should be chosen c either of the two processes could be chosen d a combination of process I and process II should be chosen