Operation management problems

8,346 views
8,047 views

Published on

Published in: Business, Technology
0 Comments
1 Like
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
8,346
On SlideShare
0
From Embeds
0
Number of Embeds
14
Actions
Shares
0
Downloads
146
Comments
0
Likes
1
Embeds 0
No embeds

No notes for slide
  • Problem no 2 in the book
  • Operation management problems

    1. 1. 1 OPERATION MANAGEMENT PRACTICAL PROBLEMS Tips and tools for solving.T.VENKATARAMANAN.FCA(COST)
    2. 2. Old syllabus 2  Contains only important problems
    3. 3. 3 Problems from economic batch economic batch production Terms :1)EBQ 2)ELS 4)ERL = 2AS/CI(1- D/P) D= DEMAND RATE CARRYIG COST 3)EMQ A=ANNUAL DEMAND P= PRODUCTION RATE S=SETUP/ORDERING COST C= COST/UNIT I=INVENTORY
    4. 4. CALCULATE :EOQ & ROP/ROL 4   ANNUAL demand 10,000 units  CARRYING COST P.A of component x = Rs 40/=  Ordering cost Rs 320/order.  250 WORKING DAYS IN THE YEAR  LEAD TIME =5 days  Safety stock = 2 days  Ans:  1)400 DATA  2)2* 40 + 5* 40=280
    5. 5. 5 HMT bearings is committed to supply 24,000. bearings p.a to M.ltd.on a daily basis .It costs as Re1/= as inventory holding per month.set up cost per run=3240/= compute 1)ERL SIZE.2)INTERVAL BETWEEN 2 RUNS 3)MINIMUM INVENTORY HOLDING COST 1)ERL = 2 * 24,000 * 3240/1* 12 = 3600 2) interval between 2 orders= 3600/2000* 30= 54 days 3)minimum inventory holding cost = 3600/2 *12= 21,600
    6. 6. From the following data calculate 1) EOQ 2)If suppliers is willing to supply 1500units on a quarterly basis a discount of 5% 3)ROL,maxi level, & minimum level 1) 2) 3) 4) Cost of unit Rs 500. 2)average monthly demand =2,000 Ordering cost Rs 100 4)inv.carrying cost 20% pa. Normal usage 100 units /week 6)maxi usage 200/week Mini usage 50/week8) lead time 6 -8 weeks Ans: 1) EOQ = 102 2)DIFFERENTIAL COST /BENEFIT=130,000 = 68,550 –ACCEPTABLE 3)ROL=1600,MAX.LEVEL =1402,MINI LEVEL= 900 Problem No. 6
    7. 7. CALCULATE 1)EOQ 2)EXTRA COST IF ORDERING QUANTITY IS 4,000. 3)MINIMUM CARRYING COST 7  DATA   monthly demand 4000 units   Price of component x = Rs 20/=   Ordering cost Rs 120/order.  Holding cost 10% Ans: 1)2400 2)5440-4800=640 3)2400(oc= cc )
    8. 8. Compute EOQ & RELATED TOTAL COST 8    MONTHLY CONSUMPTION=250 UNITS  Ans:  EOQ = 300 UNITS CARRYING COST 10% OF PRICE WHICH IS RS 10.  ORDERING COST = 150 ORDERING COST RS 15/=  CARRYING COST = 150  COST OF MATERIAL =30,000  TOTAL COST = 30,300
    9. 9. PROBLEM 3/161 9 The demand for a component is random .It has been estimated that the monthly de Has a normal distribution with a mean of 680 & a std. deviation of 130 units.The pric is Rs 10/= & ordering cost Rs 20/= carrying cost is 25% p.a. The procurement lead time is constant & is one week.find the EOQ,expected total c of controlling the inventory at 97.5 % service level Solution : annual demand = 680 * 12 = 8160. EOQ = (/ 2 *8160 * 20)/2.5 = 361
    10. 10. Failures-calculate the number of failures expected in a year & the mean time between failures: 10 Testing time 100hrs Samples tested 50 units Failures 2 units Average usage - 2 hrs /day Total sales in a year=500 units Total testing time 50*100 =5000unit hrs Lost during testingassuming 50%= 2 * 100/ 2 = 100 Net unit hrs =5000-100= 4900 Expected failures = 2/4900*500 *365 *2= 149 units Mean time between failures = (4900/2) * 2* 36 = 3.36 unit year/failure.
    11. 11. 11 H.R.Planning From the following data available calculate: 1)%absenteeism 2)efficiency of utilization of labor 3)productive Efficiency of labor.4) over all productivity of labour in terms units/man/month No. Hrs ope /da r y No, of day s /pm Std .pr odn /mo n Std lab hrs /uni t Ab sen teei sm /los s Uni t pro duc ed Idle tim e ma n hrs
    12. 12. Solution 12 1) No of days /month 25 (2)hrs /day -8 Overall productivity 2) No. of operators 15 3) Man Hrs /month 15*25* 8 = 3,000 Units produced -240 4) Hrs lost on a/c of absenteeism 30 * 8 = 240 No.operators 15 5) Absenteeism % 240/3000 * 100=Units/operator/month -16 8% 6) Idle time 276 7) Total 516 8) Actual hours utilized = 3000 – 516 =2484 9) Actual production = 240 units 10)Labor efficiency =240 * 8 /2484 =77.3% 11)Labor utilization =1920/3000 *100=64%
    13. 13. Selection of incentive scheme-advise 13 Production / day 200 units Selling price/unit Rs8 DM Rs 2 DL 1 OH( INCLUDING SELLING) Rs 800/day decrease in SP Rs1 The workers are willing to produce More if wages are Increased proportionately Suitable incentive scheme Costing Rs 100/day Sales increases by50% To administer
    14. 14. 14 Solution RS DM 200*2 DL 200*1 OH200*4 TOTAL 1600 300*7 2100 400 300*2 600 present 200 OH+EXTRA 100 900 ADDL-50% L Sales 200*8 proposed 100 200 NOT VARIABLE 800 1400 1800
    15. 15. Labor remuneration 15 Term What is a scanlon plan? Definition type of gain sharing program in which employees receive a bonus if the rati labor costs to the sales value of production is below a set standard.
    16. 16. Compute the amount available to be paid as bonus under Scanlon plan for 2007 16 Information relating to the 3 previous years Year Sales revenue Rs Total salaries &wages Rs Ratio s&w/sales 2004 120,000 36,000 36/120= 0.30 2005 125,000 35,000 35/125=0.28 2006 135,000 35,100 35.1/135=0.26 2007 150,000 36,000 Total =0.84 Average=.84/3= 0.28
    17. 17. solution 17 Expected labor cost = 0.28*150,000 = 42,000 Salaries & wages saved =42,000 – 36,000= 6,000 Assuming 30% is set aside for equalization 6000-1800 = 4200 is available for scanlon bonus for 2007
    18. 18. From the following data for a m/c in a factory 18 details Hrs worked /day 8 Working days in a month 25 No . of operator 1 Std minutes /unit.m/c time 22 Operator time /efficiency 8 min/efficiency 100% Total time /unit 30 min If plant is operated @75% efficiency What is the output Per month (2)if m/c Productivity is increased by 20% What is the output ? 3)If operator efficiency decreased by 20% What is the output?
    19. 19. solution 19 1) 8*25*75/100=150*60/30=300 2)Std m/c time =22 m/c productivity increased by 20% Revised std. time =22*100/120=110/6=18.33 min. Actual time /unit= 18.33= 8 = 26.33 Production = 150 * 60/26.33= 342 3)Operator efficiency reduced by 20% i.e 10 min Production = 150 * 60/32= 281 units
    20. 20. LATHE ,MILLING & DRILLING MACHINE.THE OPERATORS EFFICIENCIES ,STD. TIMES ,M/C TOOLS AVAILABLE ARE AS FOLLOWS 20 Type of m/c Operator efficiency Std. man hours m/c availability Lathe 75% 0.15 95% Milling m/c 80% 0.20 75% Drilling 80% 0.10 75% The factory operates 8 hours for 6 days in a week.If u want to produce 2000 bushes/wee What will be the % of spare capacity available in each type of m/c tool
    21. 21. solution m/c Milling Drilling Hrs in a week8*6 48 48 48 Available Hrs 95% 75% 75% stdHrs/bus 0.15 0.20 0.10 Operator eff 75% 80% 80% Opt time/b 0.20 hrs 0.25 0.125 Prodn./week 228 144 288 1*2/5 No of m/cs 8.77 13.88 6.94 2000/228 etc % spare 21 lathe Remarks 2.5% of 9 0.79% of 14 0.79% of 7
    22. 22. 22
    23. 23. 23 A LTD. Manufactures a component which passes through 3 m/c s P Q & R. THE STANDARD TIME OPERATOR EFFICIENCY etc ARE AS FOLLOWS: DETAILS MACHINES Std. hrs /component Operator efficiency P 0.16 80% Q 0.23 100% R 0.09 90% The factory operates 2 shifts of 8 hrs ,6 days in a week to produce 4000units, per w Compute the no. of machines & available time if any for other jobs.
    24. 24. 24 A soap factory is following a piece rate system for its packing section, the rate being Re0.10/=there is a guaranteed wage of Rs200/per day. Following table gives details regarding the number of cakes packed per day.if std. time per packing is 4 min& shift Min. ,compute: Worker No. ofduration is 480 min.. day soap cakes packed / A 800 CALCULATE B 600 (1)wages payable /worker C 100 /day (2)average cost of packing per soap D 700 Worker productivity Group & comment on the same
    25. 25. Solution 25 Worker A B C D WAGE S 200 200 200 200 Incentiv e 80 60 10 70 Total 280 260 210 270 1)Productivity of C Is poor. 2) Average or group Is ok 1020 1020/2200 0.46 Averag e cost Labour 80/12 producti 667% Total/gr oup 60/12 500% 10/12 83.33% 70/12 583.33 55/12 458.3%
    26. 26. Productivity of material 26  A factory manufactures 2 products A & B using 2 materials P & Q . Selling price of A& B are Rs 70 & 30/unit resply. Material .P Material Q INPUT/ou 200 tput/A 400 Input/outp 300 ut/B 200 RM usage 1000 kg 1000kg Labor 300Mhrs 250Mhrs Compute productivity of RM, L, & E.ENERGY COMMENT ON RELATI ADVANTAGE of P & Q p/195 @ Rs 5/= Per Hr Productivity= v of output/v of input
    27. 27. solution 27 Details Product A Product B output 200 300 Value 14,000 9,000 23,000 Material 20,000 Productivity 23/20= 1.15 Labour 1500 23/1.5= 15.33 Elec 1500 15.33
    28. 28. Problem No.2 A work shop has 25 numbers of identical m/c. The failure pattern of the m/cs are given below Lapsed time in Probability of months after failure maintenance 1 It costs Rs 160 to attend a failed m/c 2 .15 Compute yearly cost of servicing 3 .15 m/c s failed 4 28 .10 .15 5 .20 6 .25
    29. 29. Solution 2 29 Expected time before failure : .1 *1+.15*2 +.15*3 +.15*4 + .2*5 + .25*6 = 3.95 months No. of repair /m/c / annum12/3.95 = 3.038 Yearly cost of servicing = 3.038 *25 *160 = 12,152
    30. 30. Problem 3 30  Alternative set ups A & B are available for producing a component on a m/c whose operating expenses /Hr is Set up A Set up B Rs 25/= No,of components / set up 20,000 30,000 Calculate Mfg rate /piece Set up cost p.a Rs 500 Rs 600 Assuming 3000hrs in a year Production rate /hr 20 pieces 40 pieces Which one will be economical
    31. 31. Solution 3 Setup B IN CASE OF PRODN/HR 20 40 3000 HRS COMP /SETUP 20,000 30,000 A HRS/SET UP 1,000 750 OPTG COST Rs25000 Rs18750 75,000 75,000 SET UP COST 31 A 500 600 1500 2400 Rs 19350 76,500 77,400 0.645 1.275 0.645 TOTAL Rs25,500 B IS BENEFICIAL COST / 1.275 B
    32. 32. PROBLEM NO 4. An article is processed on 3 machines as shown below: Machin Time in Proces es minute sing s total Prepn. Time min/da y Cleanin g time min/da y A 2 2.5 4.5 15 10 B 3 10 13 30 10 C 2 5 7 25 10 IF JIGS FOR M/C B& C ARE REDESIGNED LOADING & UNLOADING TIME COUL REDUCED TO 2 & 1 MIN 1)rsply.calculate no of pieces /day of 8 hrs. 2) unless prod is increased by 20% new jigs would not be worth while.3)incase of large vol suggest changes to present arrangement & estimate new production rate. 32
    33. 33. SOLUTION NO4. 33 (a)THAT WHICH TAKES THE LONGEST TIME IS THE CRITICAL JOB. i.e. job no B,which takes 13 minutes. Production time = 8* 60 =480 min. (-) prep & cleaning for job b =30 +10= 40 Output from M/C B =440/13 = 33.84 (b)If jigs are redesigned time for b will become 12 from 13. Output of B = 440/12 = 36.66.increase % = 3/33* 100= 9% So not worthwhile ( c) if production to be increased one more m/c B TO BE ADDED.
    34. 34. PROB NO 5. 34 A COMPANY MANUFACTURES ITEMS A & B THE DETAILS ARE AS FO DETAILS PRO X Y Remarks Profit 35 25 Material 3kg 2kg Max 350 kg Labor 4 hrs 3 hrs Max 600 hrs m/c hr 2 2 550 m/c hr Formulate under simplex method 1) the objective function & linear constraints 2)The equations after introducing slack variables (b)State the various methods of solving linear programming problem
    35. 35. Prob no.6 Sales forecasting 35 The annual sales of TV SETS in Chennai are as under: year Sales in (000) 2004 3 2005 14 2006 36 2007 4 2008 33 Find the linear trend equation to sales figure & Estimate the sales for the year 2009
    36. 36. Solution prob No.6 36 year Deviati on(d) sales D*D D* S 2004 -2 3 4 -6 2005 -1 14 1 -14 2006 0 36 0 0 2007 1 4 1 4 2008 2 33 4 66 N=5 Sum=0 Sum of y=90 Sum ofD2= 10 Sumof DS = 50 Y=a +Bx a = 90/5= 18 B= 50/10=5 Y=18+ 5x = 18+ 5*3= 33 = 33,000 sets
    37. 37. Sum No.7 37 As on 1st aug the following jobs are to be processed .the processing time & due dates are As follows : Jobs A B C D PROCESS TIME IN DAYS QD 2 6 7 12 DUE DATE AUG 12 AUG 7 AUG 4 AUG8 Sequence jobs on minimum ratio TIME REMAINING 11 6 3 7 Time needed to complete the jobsA-2 ,B-6, C-7, D-12 CRITICAL RATIO A-11/2=5.5,B1.,C-3/7=0.43, D=0.58 SEQUENCE C D B A
    38. 38. Problem-8 –location of plant. 38 Two locations A & B are considered for location of a medical testing equipment Fixed costs Rs25 30 laks. Variable 300 250 .average selling price of equipmentRs 550/ unit Compute Range of annual production & sales volume for which location is most suitab Ans: locations A TOTAL COST 25L +300X CONTRIBUTION 250 BE POINT 10,000 LOSS WILL BE LESS FOR FOR ABOVE 10,000 UNITS B Deatails/un 30L+ 250X its 300 ATotal 10,000 costRs Ls A <10,000 B B IS BETTER.total cost 8000 12000 49 61 50 60
    39. 39. PROBLEM -9-OPERATION TIME 39 A SHAFT OF 3000 MM LENGTH requires machining on lathe .If the spindle execu 1500 rpm & the feed is .20 mm per revolution ,how long does the cutter take to pas down the entire length of the shaft. Ans: number of revolutions to pass 3000 mm length = 3000/.20 = 15,000 Time required = 15,000/1500= 10 .min
    40. 40. 40 Compute EOQ from the following 1) Consumption per month-250 units. 2) Carrying cost 10% of purchase price of Rs 10/=ordering cost Rs 15 per order. Compute EOQ & related total cost . Ans : EOQ 300 UNITS TOTAL COST Rs 30, 300/=
    41. 41. 41 The data collected from a transport corporation about the number of breakdowns for months over the past 2 years are as follows: No of break downs 0 1 2 3 4 No of months occurre d 3 7 11 2 1 Each break down costs Rs 3000/= On an average.Preventive maintenance of Rs 1375/= break downs can be reduced to an average of one per month.Which policy is suitable/
    42. 42. solution 42 No break dns Freq in Freq in Expect Break dn month % ed cost /mon s value(1 * 3) Prevent ive mainten ance 0 3 0.125 0 1.625*3000 3000 + 1 7 0.292 0.292 =4875 1375 2 11 0.458 0.916 3 2 0.083 0.249 4 1 0.042 0.168 Preventive maintenance is suitable total 1.625 4375/=
    43. 43. Problem 10 43 A manufacturing line consists of 5 work stations in series.The individual work statio capacities are given. The actual output of the unit / shift is 540. Work stn. No: 1 2 Capacit 700 650 y/shift Calculate 1) system capacity 3 4 5 700 650 600 2)EFFICIENCY OF THE PRODN.LINE
    44. 44. Problem 11 44 Machine A & B are capable of manufacturing a particular product . the details are as follows: 1) Find out which m/c is Machin A B REMA 2) suitable for regular e RKS 3) production INVEST 100,000 150,000 4) Which m/c will give MENTR 5) lower cost of prodn.for s INTEREST ON LOAN 9% FOR BOTH 6) 5000 pieces. Operating cost/hrA-Rs12 B- Rs 10 7) At what level Production/hr 5 pieces 9 pieces 8) cost of production will Hrs worked /year 4000 4000 9) Be the same for both
    45. 45. Problem no 12 45 A factory has capacity to provide 3999 hrs /week. It has capacity to produce A ANNUAL COSTS Rs 15,000 Maximum sales 5,000 Variable cost Rs 9 Selling price 15 Hrs reqd./ unit 5 Calculate 1) product mix for max profit 2) B 4,000 12 18 6
    46. 46. 46 Q. 3. (a) Explain how you would choose a material handling equipment from amongst alternative offers. (b) Prasad Timber Works uses forklift trucks to transport lumber from factory to a storage area 0.3 km away. The lift trucks can move three loaded pallets per trip and travel at an average speed of 8 km. per hour (allowing for loading, unloading, delays and travel). If 640 pallet loads must be moved during 8 hours shift, how many lift trucks are required? Assume single shift working and 300 working days in a year. (c) State the machine tool to be used for following operations : (i) Melting steel for making castings. (ii) Picking up bits of iron and steel in a scrap yard. (iii) Squeezing a piece of hot metal in a die. (iv) Making a small hole in a block of metal. (v) Making keyways on inside surface of the bore of a pulley.
    47. 47. Answer 3. (a) The choice of material handling equipment is essentially based on technical suitability an Economic considerations. In the first stage we check up whether the equipment offers m the technical criteria/parameters mentioned in the specification i.e the load to be lifted/ca the speed of movement,maneuverability, turning radius etc. Once we are satisfied that t equipment meets the technical parameters , we check up the cost aspects and select th equipment having the lowest life time cost. We thus take into account the cost of initial a receiving costs incurred during the life cycle of the equipment as annual operating cost repair/maintenance costs, and the salvage value of the equipment at the end of its life. The cash out flows and inflows occurring during the various periods are suitably discoun as to have a common basis for comparison. While the above approach is suitable for equipments offering identical performance/output, we decide the issue on cost per unit handled, in case of equipment having differing output parameters, subject to of course their meeting the technical criteria specified. 47
    48. 48. Answer 3. (b) Total distance travelled by fork lift truck per trip = (0.3+0.3) km = 0.6 km(up and down No. of trips that can be made by the truck per shift = 8km/0.6km × 8hrs = 106.66 trips No. of pallet loads carried per shift by each truck = 106.66 × 3 = 319.98 = 320 Total no. of fork lift trucks required for 640 pallet loads = 640/320 = 2 fork lift trucks. Answer 3. (c) (i) Electric Arc Furnace. (ii) Electromagnet (iii) Forging machine (iv) Drilling machine (v) Slotting machine 48
    49. 49. Q. 4. (a) What factors will have to be considered in choosing the location for the following industries? (i) Aluminium industry. (ii) Thermal power plant. (iii) Large furniture(domestic and office)manufacturing unit. (b) Empire Glass Company can produce a certain insulator on any three machines which have the following charges shown below . The firm has an opportunity to accept an order for either (1) 50 units at Rs. 20/unit or (2) 150 units at Rs. 12/unit. Machine Fixed Cost (Rs) Variable Cost(Rs) A 50 4/unit B 200 2/unit C 400 1/unit (i) Which machine should be used if 50 units order is accepted and how much profit will result? (ii) Which machine should be used if the 150 units order is accepted and what will be the resultant profit? (iii) What is the break-even volume for machine B when the price is Rs. 12/unit? (iv) Suppose the fixed cost for machine A is a stepped function with Rs. 50 up to 40 units and Rs. 100 thereafter. Will the answers to (i) and (ii) above vary? If so, what will be the revised answer? 49
    50. 50. Answer 4. (a) The general factors to be considered for any Industry location are the following : 1. Proximity to raw material sources 2. Availability of critical input required for the process 3. Proximity to the Market 4. Availability of skilled labour 5. Special tax and other financial benefits available in a location 6. Central/state/municipal regulations. While all the above factors are important for all Industries, some factors will be dominant for some industries as explained below : 50
    51. 51. 1. Aluminium Industry is a power intensive industry. Hence the region/location 2. where availability of power is a very critical consideration for the choice of location Likewise, proximity to raw material source namely bauxite is also a vital consideration. 2. For thermal plant proximity to coal mines is very important since transportatio of huge quantity of coal every day is very costly and difficult. Equally important is the availability of abundant quality of water for the boiler. 3. For furniture industry proximity to the Market is a crucial factor apart from othe While transporting finished furniture, damages may take place and also it will be bulky and occupy more space and hence costly. Therefore furniture units are located nearer towns and cities nearer to offices and houses. 51
    52. 52. Answer 4. (b) (i) For 50 unit order at Rs. 20/unit.Costs for various machines : Machine Rs. Profit Rs. Machine A 50 + 50 × 4 = 250, 1000 – 250= 750 Machine B 200 + 50 × 2 = 300, 1000 – 300= 700 Machine C 400 + 50 × 1 = 450, 1000 – 450= 550 Since Machine A gives the highest profit of Rs. 750 it is to be preferred. (ii) For 150 unit order at Rs. 12/unit.Costs for various machines Machine Rs. Profit Rs. Machine A 50 + 150 × 4 = 650, 1800 – 650 =1,150 Machine B 200 + 150 × 2 = 500, 1800 – 500= 1,330 Machine C 400 + 150 × 1 = 550 ,1800 – 550= 1,250 Hince Machine B to be preferred. (iii) Breakeven Volume for Machine B at Rs. 12/unit. Let X be the No. of units to be produced.Total costs at ‘X’ units = 200 + 2x Total revenue at x units = 12x. At Breakeven point. 200 + 2x = 12x i.e 10x = 200 x = 20 Hence 20 units is the Breakeven Volume. 52
    53. 53. (iv) The fixed cost for machine A being a step function, the total cost of manufacturing o 50 units with machine. A = 100 + 50 × 4 = Rs. 300, which is also the cost of production with machine B. Thus either of the two machines A or B could be chosen to produce 50 units. A will be Rs. 700, which is higher than the production cost on machine B. Hence the answer in this case will not vary. 53
    54. 54. (b) A Company adopts a counterseasonal product strategy to smooth production requirements. It manufactures its spring product line during the first four months of the year and would like to employ a strategy that minimises production costs while meeting the demand during these four month. The Company presently has on its rolls, 30 employees with an average wage of Rs. 1,000 per months. The Company presently has on its rolls, 30 employees with an average wage of Rs. 1,000 per month. Each unit of the product requires 8 man-hours. The Company works on single shift basis (8 hrs. shift/day). Hiring an employee costs Rs. 400 per employee per occasion and discharging an employee costs Rs. 500 per person per occasion. Inventory carrying costs are Rs. 5/unit/month and shortage costs are Rs. 100/unit/month and shortage costs are Rs. 100/ unit/month. The Company forecasts the demand for the next four months as below : Month (Demand units) No. of working daysin the month jan 500 22 February 600 19 March 800 21 April 400 21 The Company is thinking of adopting one of the following pure strategies : Plan I : Vary work force levels to meet the demand. Plan II : Maintain 30 employees and use inventory and stockouts to absorb demand fluctuations. Which strategy would you recommend? You may assume nil inventory at the start. 54
    55. 55. 55 The overall costs of both the strategies are computed in the following tables. Plan I, varying the work force levels to suit the production needs : January February March April Total 1. Workers required 22 500 = 23 19 600 = 32 21 800 = 38 21 400 = 19 2. Labour cost 23,000 , 32,000; 38,000 ;19,000 ;=1,12,000 3. Hiring costs 9 × 400 6 × 400 = 3,600 = 2,400 6,000 4. Lay off costs 7 × 500 19 × 500 = 3,500 = 9,500 13,000 Total 1,31,000 Plan II : Maintain a steady work force and use of inventory plus stock on January February March April Total 1. Workers used 30 30 30 30 2. Labour cost 30,000 30,000 30,000 30,000 1,20,000 3. Units produced 660 570 630 630 4. Inventory costs 5 × 160 5 × 130 5 × 190 = 800 = 650 = 950 2400 5. Shortage costs 100 × 40 = 4,000 ; 4,000 Total 1,26,400 On the basis of costs, plan II would be the choice. Moreover this strategy would result in higher worker morale, smoother production, and generally, a higher quality product.
    56. 56. Problem no.13 56 5 jobs are required to be processed through 2 m/cs. Sequentially The following table gives processing times in hours Job A B C D E M/C 1 2 7 5 6 5 M/C 2 4 8 8 7 3 1)Calculate minimum completion time of all jobs 2) For what period m/c 2 will remain idle 3)When does job –B GETS COMPLETED.
    57. 57. Q. 6. (a) The following tasks are to be performed on an assembly line in the se and times specified : Task Time (Seconds) Tasks that must precede A 50 –;B 40 –;C 20 A ;D 45 C;E 20 C;F 25 D;G 10 E;H 35 B, F, G (i) Draw the schematic diagram. (ii) What is the theoritical minimum number of stations required to meet a fore demand of 400 units per an 8-hour day? (iii) Use the longest task time rule and balance the line in the minimum numbe produce 400 units per day. (iv) Evaluate the Line Efficiency and the Smoothness Index 57
    58. 58. (i) Schematic Diagram : D A C E B A-20 B-40 F H G D-45 E= 20 F-25 G-10 H-35 (Activity on Node) 20 10 (ii) Theoretical minimum number of stations to meet D = 400 is : N = T/C = 245/{(60 seconds 480minutes) 400units} = 245 ÷ 72 = 3.4 day 4. 58
    59. 59. (b) Write short notes on : (i) Line Balancing (ii) Work Simplification (iii) Progressing (iv) Six Sigma quality programme 59
    60. 60. 60 (iii) Line Balancing : Station Task Task Time Unassigned Time Feasible Station Cycle time (Seconds) (Seconds) Remaining Time (Seconds) Task (Seconds) 1 A 50 22 C 70 70C 20 2 None 2 D 45 27 E, F 70 70F 25 2 None 3 B 40 32 E E 20 12 G 70 70 G 10 2 None 4 H 35 37 None 35 70 Total 245 280 (iv) Line Efficiency = (245/280) × 100 = 87.5% Smoothness Index = (70 − 70)2 + (70 −70)2 +(70 −70)2 +(70 −35)2 = (35)2 = 35%. Answer 6.
    61. 61. 61
    62. 62. Answer 6. (b) (i) This technique is employed to ensure balanced flow of production specially in Assembly lines. When products require a number of operations to be performed , the time taken for each operation may vary. Therefore if the line is not balanced , a few operators will be over loaded whose operation times are long and others will be idling away their time. This technique decides the cycle time for the products based on the output required per shift. This also considers the sequence in which the operations will have to be carried out. Then Line Balancing technique groups operations in such a manner that each group will have generally equal total task times. Each group is called a work station and assigned to an operator. Thus all the operators in each work station will have balanced load. (ii) Work Simplification involves subdivision of an operation into its constituent elements in order to simplify operations and eliminate wasteful motions. It reduces fatigue and improves productivity. It covers all aspects of work, i.e equipment , layout , procedures , methods etc. (iii) Progressing is the gentle but firm direction of ‘activities planned’ to proper channels, shielding them from adverse factors inseparable from actual operations. The work of the progress department starts where the planner’s ends. The Progress Controller/Progress Chaser , as he is called, is not concerned with methods of carrying out operations(which are under the purview of the process planning department) but the ‘doing’ of them at proper time in the correct order and at the lowest anticipated cost. The duties of the Progress Chaser would vary widely from unit to unit. In a well 62
    63. 63. organized manufacturing activity, his duties consists of : (A) recording of actual production or output against planned production; (B) assessment of causes leading to activities falling behind schedule; (C) reporting to appropriate authority; (D) where possible , to foresee all that may lead to failure on schedules and to sound a note of warning to all concerned. (iv) Six Sigma Quality program is company- wide approach for continuous improvement in quality of products and services. It measures the degree to which the process deviates from the standards and takes efforts to improve the process to achieve customer satisfaction. The objective of Six Sigma Quality programme are two –fold : (i) to improve the customer satisfaction and reducing and eliminating gaps/defects and (ii) to continuously improve processes throughout the organization with a view to reduce sources of variation and improve quality as well as productivity. It is a statistical measurement which tells us how good our products, services and process are and enables us to benchmark our operations with the best in the field. It thus helps us to establish our course in the race for total customer satisfaction. A process at 6- Sigma level normally produces 3-4 non conformances in a million operations. This is supposed to be the best-in-class quality. Thus 6-Sigma is essentially a philosophy of working smarter. This means making fewer mistakes in everything we do. As we discover and eliminate the sources of variation , the non conformances are eliminated and the process capability improves. 63
    64. 64. (c) The following cost have been recorded : Particulars Rs. Incoming materials inspection 10000 Training of personnel 30000 Warranty 45000 Process planning 15000 Scrap 9000 Quality laboratory 30000 Rework 25000 Allowances 10000 Complaints 14000 What are the costs of prevention, appraisal, external failure and internal failur 64
    65. 65. Problem on transportation 65 Factory –ware house –cost ,demand & supply schedule F w W-1 W-2 W-3 W-4 SUPPL Y F-1 6 8 8 5 30 F-2 5 11 9 7 40 F-3 8 9 7 13 50 28 32 25 120 DEMAN 35 D
    66. 66. VOGL’S METHOD 66 Factory –ware house –cost ,demand & supply schedule F w W-1 W-2 W-3 W-4 SUPP LY COL /PENA LTY F-1 6 8 8 5 30 ½ F-2 5/35 11 9 7 40/5 2/4 F-3 8 9 7 13 50 1/1 28 32 25 120 DEMA 35 ND ROW PLTY 1 1 1 2/0 SELECT HIGHIEST PENAL ALLOT for cell F2W4--25
    67. 67. VOGL’S METHOD 67 Factory –ware house –cost ,demand & supply schedule F w W-1 W-2 W-3 W-4 SUPP LY COL /PENA LTY F-1 6 8 8 5 30 ½ F-2 5/35 11 9 7 40/5 2/2 F-3 8 9 7 13 50 1/1 32 25 120 DEMA 35-35 28 ND ROW PLTY 1 1 1 2/0 SELECT HIGHIEST PENAL ALLOT for cell F2W1--35
    68. 68. VOGL’S METHOD 68 Factory –ware house –cost ,demand & supply schedule F w W-2 W-3 W-4 SUPP LY COL /PENA LTY F-1 8 8 5/25 30/5 1/3 F-2 11 9 7 40/5 2/2 F-3 9 7 13 50 1/2 32 25-25 120 DEMA 28 ND ROW PLTY 1 1 1 2/0 SELECT HIGHIEST PENAL ALLOT for cell F2W1--35 2)3 being the highiest Allot 25 units to F1-W4
    69. 69. VOGL’S METHOD 69 Factory –ware house –cost ,demand & supply schedule F w W-2 W-3 SUPP LY COL /PENA LTY F-1 8/5 8 30/5 1/3/0 F-2 11/5 9 40/5 2/2/2 F-3 9/18 7/32 50 1/2/2 32-32 120 DEMA 28 ND ROW PLTY 1 1 1 2/0 SELECT HIGHIEST PENAL ALLOT for cell F2W1--35 2)3 being the highiest Allot 25 units to F1-W4
    70. 70. IBFS F/W F1 W1 W2 W3 W4 Supply 6/ 8/5 8 5/25 30 F2 5/35 11/5 9 7 40 F3 8 9/18 7/32 13 50 Deman d 35 28 32 25 120 Cost/units Total cost 5 35 175 8 5 40 11 5 55 9 18 162 7 70 /unit 32 224 5 25 125 =781
    71. 71. 71 Optimality test The initial basic solution obtained by any one of the three methods may still have some better solution . To find out whether there exists a better solution we may carry out the following tests : 1)stepping stone method. 2)MODI method .
    72. 72. Stepping stone method: 72 1)Verify whether the IBFS has (4+3-1 )= 6 occupied cell. 2)Then evaluate unoccupied /unallocated or Non-Basic cells. This done by finding out the net increase or decrease in cost by moving one unit from occupied cell to unallocated cell. 3)This is done by forming a loop or closed path 4)The loop starts from unallocated cells moves horizontaly or vertically & finally forms a loop ending with the unallocated cell 5)See the next slide
    73. 73. Evaluation process & final solution 73 2)For F1W3=8-8+11-9+ 2 3)F2W3=9-11+9-7=0 4)F2W4=7-5+8-11=-4 5)F3W1=8-5+11-9=5 6)F3W4=13-5+8-11+9-7=7 F2W4 GIVES – VE VALUE
    74. 74. IBFS F/W W2 W3 W4 Supply F1 6/ 8/10 8 5/20 30 F2 5/35 11/0 9 7 5 40 F3 74 W1 8 9/18 7/32 13 50 Deman 35 28 32 25 120 d Cost= 5* 35=175 9*18= 162 7*32= 224 8*10= 80 5*20= 100 7*5= 35 255 + 297 +224 = 776 W-1 is the least Costed un allocated cell By moving one unit From W-2 to W-1 The cost /unit will come down by Re 1 But this will out Shoot supply Therefore we move I unit from F2W1 TO F2W2. The change in cost -8+6-5+11= 4
    75. 75. IBFS F/W W1 W2 W3 W4 F1 6/ 8/10 8 5/20 F2 5/35 11/0 9 7/5 F3 8 9/18 7/32 13 Deman d 35 28 32 25 W-1 is the least Costed un allocated cell By moving one unit From W-2 to W-1 The cost /unit will Supply come down by Re 30 1 But this will out 40 Shoot supply Therefore we 50 move I unit from F2W1 120 TO F2W2. The change in cost cost can be brought=down by5* -8+6-5+11= 4 Thus by moving 5 units from F2W2 to F2w4 the 1= This method can be repeated till all the values are +.It means no further reductionis 75
    76. 76. 76
    77. 77. MODI METHOD. 77 FW W1 W2 W3 W4 suppl y U F1 6 8/5 8 5/25 30 U1 F2 5/35 11/5 9 7 40 U2 F3 8 9/18 7/32 13 50 U3 DEMA 35 ND 28 32 25 150 V V2 V3 V4 V1 Occupied cells F1W2=8=U1+V2 F1W4=5=U1+V4 F2W1=5=U2+V1 F2W2=11=U2+V2 F3W2=9=U 2+V2 F3W3=7=U2+V3
    78. 78. 78 Answer 7. (c) Particulars Rs. Training of personnel Process planning Total cost of prevention Incoming materials inspection 10000 Quality laboratory 30000 Total cost of appraisal Scrap Rework Total cost of internal failure 34000 Warranty 45000 Allowances Complaints 30000 15000 45000 40000 9000 25000 10000 14000
    79. 79. (c) An industrial engineer, deputed to conduct a time study for a job, has, after observation, divided (i) Are there any outliers in the data, i.e., probable errors in reading or recording data which 79 should not be included in the analysis? (ii) Compute the basic time for the job and the standard time if a relaxation allowance of 12 contingency allowance of 3% and an incentive allowance of 20% are applicable for the job.
    80. 80. Basic time = 7.722 mins Allowance = (12+3) = 15% (Incentive allowance not considered) Standard time = 7.722 × 1.15 = 8.88 min (approx) 80
    81. 81. (c) A work sampling study is to be made of a data entry operator pool. It is felt t operators are idle 30% of the time. How many observations should be made in to have 95.5% confidence that the accuracy is within +/- 4%. Answer 9. (c) n = {4p(1-p)}/s2 p = 0.3 1– p = 0.7 s = 0.04 n = 4 × 0.3 × 0.7/(.04)2 = 0.84/.0016 = 525. 81
    82. 82. 82
    83. 83. 83
    84. 84. 84
    85. 85. 85
    86. 86. Widescreen Test Pattern (16:9) Aspect Ratio Test (Should appear circular) 4x3 86 16x9

    ×