OPERATION MANAGEMENT PRACTICAL PROBLEMSTips and tools for solving.T.VENKATARAMANAN.FCA(COST)

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Old syllabus 2

Contains only important problems

Terms :1)EBQ 2)ELS 3)EMQ4)ERL = 2AS/CI(1- D/P) A=ANNUAL DEMAND S=SETUP/ORDERING COST

D= DEMAND RATE P= PRODUCTION RATE C= COST/UNIT I=INVENTORY CARRYIG COST

Problems from economic batch economic batch production

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CALCULATE :EOQ & ROP/ROL

DATA 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

2)2* 40 + 5* 40=280

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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 RUNS3)MINIMUM INVENTORY HOLDING COST

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1)ERL = 2 * 24,000 * 3240/1* 12 = 3600 2) interval between 2 orders= 3600/2000* 30=

54 days3)minimum inventory holding cost = 3600/2

*12= 21,600

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) Cost of unit Rs 500. 2)average monthly demand =2,0002) Ordering cost Rs 100 4)inv.carrying cost 20% pa.3) Normal usage 100 units /week 6)maxi usage 200/week4) Mini usage 50/week8) lead time 6 -8 weeks

Ans: 1) EOQ = 102 2)DIFFERENTIAL COST /BENEFIT=130,000- 61450 = 68,550 –ACCEPTABLE

3)ROL=1600,MAX.LEVEL =1402,MINI LEVEL= 900

Problem No. 6

CALCULATE 1)EOQ 2)EXTRA COST IF ORDERING QUANTITY IS 4,000.

3)MINIMUM CARRYING COST

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 )

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Compute EOQ & RELATED TOTAL COST MONTHLY

CONSUMPTION=250 UNITS CARRYING COST 10% OF

PRICE WHICH IS RS 10. ORDERING COST RS 15/=

Ans: EOQ = 300 UNITS ORDERING COST = 150 CARRYING COST = 150 COST OF MATERIAL

=30,000 TOTAL COST = 30,300

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PROBLEM 3/1619

The demand for a component is random .It has been estimated that the monthly demand Has a normal distribution with a mean of 680 & a std. deviation of 130 units.The price/unit 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 cost of controlling the inventory at 97.5 % service level

Solution : annual demand = 680 * 12 = 8160. EOQ = (/ 2 *8160 * 20)/2.5 = 361

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Failures-calculate the number of failures expected in a year & the mean time between failures:

Testing time 100hrsSamples tested 50 unitsFailures 2 unitsAverage usage - 2 hrs /dayTotal sales in a year=500 units

Total testing time 50*100 =5000unit hrsLost during testingassuming 50%= 2 * 100/ 2

= 100Net unit hrs =5000-100= 4900Expected failures = 2/4900*500 *365 *2=149 unitsMean time between failures = (4900/2) * 2* 365= 3.36 unit year/failure.

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

H.R.Planning11

No. oper

Hrs/day

No,of days /pm

Std.prodn /mon

Std lab hrs /unit

Absenteeism /loss

Unit produced

Idle time man hrs

15 8 25 300

8 30 240

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Solution 12

1) No of days /month 25 (2)hrs /day -82) No. of operators 153) Man Hrs /month 15*25* 8 = 3,0004) Hrs lost on a/c of absenteeism 30 * 8 = 2405) Absenteeism % 240/3000 * 100= 8%6) Idle time 2767) Total 5168) Actual hours utilized = 3000 – 516 =24849) Actual production = 240 units10)Labor efficiency =240 * 8 /2484 =77.3%11)Labor utilization =1920/3000 *100=64%

Overall productivity

Units produced -240No.operators 15Units/operator/month -16

Selection of incentive scheme-advise 13 Production /

day200 units

Selling price/unit

Rs8

DM Rs 2

DL 1

OH( INCLUDING SELLING)

Rs 800/day

decrease in SP Rs1 Sales increases by50%

The workers are willing to produce

More if wages are

Increased proportionately

Suitable incentive scheme

Costing Rs 100/day

To administer

Solution 14

RS proposed

Sales 200*8 1600 300*7 2100

DM 200*2 400 300*2 600

DL 200*1 200 NOT VARIABLE

present 200

OH200*4 800 OH+EXTRA 100

900

ADDL-50% L 100

TOTAL 1400 1800

PROFIT 200 300

Labor remuneration 15

TermWhat is a scanlon plan?Definitiontype of gain sharing program in which employees receive a bonus if the ratio of labor costs to the sales value of production is below a set standard.

Compute the amount available to be paid as bonus under Scanlon plan for 2007

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Information relating to the 3 previous years

Year Sales revenue Rs

Total salaries &wages Rs

2004 120,000 36,000

2005 125,000 35,000

2006 135,000 35,100

2007 150,000 36,000

Ratio s&w/sales

36/120= 0.30

35/125=0.28

35.1/135=0.26

Total =0.84

Average=.84/3=0.28

solution17

Expected labor cost = 0.28*150,000 = 42,000Salaries & wages saved =42,000 – 36,000= 6,000Assuming 30% is set aside for equalization 6000-1800= 4200 is available for scanlon bonus for 2007

From the following data for a m/c in a factory

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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% efficiencyWhat 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?

solution19

1) 8*25*75/100=150*60/30=300

2)Std m/c time =22m/c productivity increased by 20%Revised std. time =22*100/120=110/6=18.33 min.Actual time /unit= 18.33= 8 = 26.33Production = 150 * 60/26.33= 3423)Operator efficiency reduced by 20% i.e 10 minProduction = 150 * 60/32= 281 units

20-GUN METAL BUSH REQUIRES OPERATIONS ON LATHE ,MILLING & DRILLING MACHINE.THE OPERATORS EFFICIENCIES ,STD. TIMES ,M/C TOOLS AVAILABLE ARE AS FOLLOWS

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/weekWhat will be the % of spare capacity available in each type of m/c tool

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solutionm/c lathe Milling Drilling Remarks

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 2.5% of 9 0.79% of 14

0.79% of 721

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DETAILS

A ltd. Manufactures a component which

A LTD. Manufactures a component which passes through 3 m/c s P Q & R.THE STANDARD TIME OPERATOR EFFICIENCY etc ARE AS FOLLOWS:

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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 week.Compute the no. of machines & available time if any for other jobs.

Following table gives details regarding the number of cakes packed per day.if std. time per packing is 4 min& shift duration is 480 min..

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.

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Worker

No. of soap cakes packed / day

A 800 CALCULATE

B 600 (1)wages payable /worker

C 100 (2)average cost of packing per soap /day

D 700

Min. ,compute:Worker productivityGroup& comment on the same.

Solution 25

Worker

A B C D Total/group

WAGES 200 200 200 200

Incentive

80 60 10 70

Total 280 260 210 270 1020

Average cost

1020/22000.46

Labour productivity

80/12667%

60/12500%

10/1283.33%

70/12583.33%

55/12458.3%

1)Productivity of CIs poor.2) Average or groupIs ok

Productivity of material26

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/output/A

200 400

Input/output/B

300 200

RM usage

1000 kg 1000kg

Labor 300Mhrs 250Mhrs @ Rs 5/=

Per Hr

Ec energy

1000kw 1500kw @ Rs 1.5/

Kwhrs

RM price/kg

20 30

Compute productivity ofRM, L, & E.ENERGYCOMMENT ON RELATIVE ADVANTAGE of P & Qp/195

Productivity=v of output/v of input

solution27

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

Problem No.2A work shop has 25 numbers of identical m/c. The failure pattern of the m/cs are given below Lapsed time

in months after maintenance

Probability of failure

1 .10 It costs Rs 160 to attend a failed m/c

2 .15 Compute yearly cost of servicing

3 .15 m/c s failed

4 .15

5 .20

6 .25

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Solution 2

Expected time before failure :.1 *1+.15*2 +.15*3 +.15*4 + .2*5 + .25*6 = 3.95 monthsNo. of repair /m/c / annum12/3.95 = 3.038Yearly cost of servicing = 3.038 *25 *160 = 12,152

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Problem 3 Alternative set ups A & B are available for

producing a component on a m/c whose operating expenses /Hr is Rs 25/=Set up A Set up B

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

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Solution 3

Setup A B IN CASE OF

PRODN/HR 20 40 3000 HRS

COMP /SETUP

20,000 30,000 A B

HRS/SET UP

1,000 750

OPTG COST Rs25000 Rs18750 75,000 75,000

SET UP COST

500 600 1500 2400

TOTAL Rs25,500 Rs 19350 76,500 77,400

COST / UNIT

1.275 0.645 1.275 0.645

B IS BENEFICIAL31

PROBLEM NO 4. An article is processed on 3 machines as shown below:

Machines

Time in minutes

Processing

total Prepn. Time min/day

Cleaning time min/day

A 2 2.5 4.5 15 10

B 3 10 13 30 10

C 2 5 7 25 10IF JIGS FOR M/C B& C ARE REDESIGNED LOADING & UNLOADING TIME COULD BE REDUCED TO 2 & 1 MIN 1)rsply.calculate no of pieces /day of 8 hrs. 2) unless production 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.

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SOLUTION NO4.

(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= 40Output 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.

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PROB NO 5.

A COMPANY MANUFACTURES ITEMS A & B THE DETAILS ARE AS FOLLOWS:

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 hrFormulate 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

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Prob no.6 Sales forecasting

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

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Solution prob No.6year Devia

tion(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 +Bxa = 90/5= 18B= 50/10=5Y=18+ 5x= 18+ 5*3= 33= 33,000 sets

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Sum No.7 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

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Problem-8 –location of plant.

Two locations A & B are considered for location of a medical testing equipmentFixed costs Rs25 30 laks.Variable 300 250 .average selling price of equipmentRs 550/ unitCompute Range of annual production & sales volume for which location is most suitable

Ans: locations A BTOTAL COST 25L +300X 30L+ 250XCONTRIBUTION 250 300BE POINT 10,000 10,000LOSS WILL BE LESS FOR A <10,000FOR ABOVE 10,000 UNITS B IS BETTER.

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Deatails/units

8000 12000

ATotal costRs Ls

49 61

B total cost

50 60

PROBLEM -9-OPERATION TIME

A SHAFT OF 3000 MM LENGTH requires machining on lathe .If the spindle executes 1500 rpm & the feed is .20 mm per revolution ,how long does the cutter take to pass down the entire length of the shaft.

Ans: number of revolutions to pass 3000 mm length = 3000/.20 = 15,000Time required = 15,000/1500= 10 .min

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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 .

Compute EOQ from the following

Ans : EOQ 300 UNITS TOTAL COST Rs 30, 300/=

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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/

The data collected from a transport corporation about the number of breakdowns for months over the past 2 years are as follows:

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No of break downs

0 1 2 3 4

No of months occurred

3 7 11 2 1

solution42

No break dns

Freq in months

Freq in %

Expected value(1 * 3)

Break dn cost /mon

Preventive maintenance

0 3 0.125 0 1.625*3000

3000 +

1 7 0.292 0.292 =4875 1375

2 11 0.458 0.916 4375/=

3 2 0.083 0.249

4 1 0.042 0.168

total 1.625

Preventive maintenance is suitable

Problem 10

A manufacturing line consists of 5 work stations in series.The individual work station capacities are given. The actual output of the unit / shift is 540.

Work stn. No:

1 2 3 4 5

Capacity/shift

700 650 700 650 600

Calculate 1) system capacity 2)EFFICIENCY OF THE PRODN.LINE

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Problem 11

Machine A & B are capable of manufacturing a particular product . the details are as follows:

Machine

A B REMARKS

INVESTMENTRs

100,000

150,000

INTEREST ON LOAN 9% FOR BOTHOperating cost/hrA-Rs12 B- Rs 10Production/hr 5 pieces 9 piecesHrs worked /year 4000 4000

1) Find out which m/c is2) suitable for regular3) production4) Which m/c will give5) lower cost of prodn.for 6) 5000 pieces.7) At what level 8) cost of production will9) Be the same for both

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Problem no 12

A factory has capacity to provide 3999 hrs /week.It has capacity to produce A BANNUAL COSTS Rs 15,000Maximum sales 5,000 4,000Variable cost Rs 9 12Selling price 15 18Hrs reqd./ unit 5 6Calculate 1) product mix for max profit 2)

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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 0.3km away. The lift trucks can move three loaded pallets per trip and travel at an average speed of8 km. per hour (allowing for loading, unloading, delays and travel). If 640 pallet loads must bemoved during 8 hours shift, how many lift trucks are required? Assume single shift working and300 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.

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Answer 3. (a)The choice of material handling equipment is essentially based on technical suitability and Economic considerations. In the first stage we check up whether the equipment offers meets the technical criteria/parameters mentioned in the specification i.e the load to be lifted/carried, the speed of movement,maneuverability, turning radius etc. Once we are satisfied that the equipment meets the technical parameters , we check up the cost aspects and select the equipment having the lowest life time cost. We thus take into account the cost of initial acquisition receiving costs incurred during the life cycle of the equipment as annual operating cost and 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 discounted so as to have a common basis for comparison. While the above approach is suitable forequipments 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.

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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/shift\ 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

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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 thefollowing charges shown below . The firm has an opportunity to accept an order for either (1) 50units at Rs. 20/unit or (2) 150 units at Rs. 12/unit.Machine Fixed Cost (Rs) Variable Cost(Rs)A 50 4/unitB 200 2/unitC 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 resultantprofit?(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 andRs. 100 thereafter. Will the answers to (i) and (ii) above vary? If so, what will be the revisedanswer?

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Answer 4. (a)The general factors to be considered for any Industry location are the following :1. Proximity to raw material sources2. Availability of critical input required for the process3. Proximity to the Market4. Availability of skilled labour5. Special tax and other financial benefits available in a location6. 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 :

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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 transportation 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 other factor. 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.

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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= 750Machine B 200 + 50 × 2 = 300, 1000 – 300= 700Machine C 400 + 50 × 1 = 450, 1000 – 450= 550Since 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 machinesMachine Rs. Profit Rs.Machine A 50 + 150 × 4 = 650, 1800 – 650 =1,150Machine B 200 + 150 × 2 = 500, 1800 – 500= 1,330Machine C 400 + 150 × 1 = 550 ,1800 – 550= 1,250Hince 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 + 2xTotal revenue at x units = 12x.At Breakeven point.200 + 2x = 12x i.e 10x = 200x = 20 Hence 20 units is the Breakeven Volume.

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(iv) The fixed cost for machine A being a step function, the total cost of manufacturing of 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.

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(b) A Company adopts a counterseasonal product strategy to smooth production requirements. Itmanufactures its spring product line during the first four months of the year and would like toemploy a strategy that minimises production costs while meeting the demand during these fourmonth. The Company presently has on its rolls, 30 employees with an average wage of Rs. 1,000per months. The Company presently has on its rolls, 30 employees with an average wage ofRs. 1,000 per month. Each unit of the product requires 8 man-hours. The Company works onsingle shift basis (8 hrs. shift/day). Hiring an employee costs Rs. 400 per employee per occasionand discharging an employee costs Rs. 500 per person per occasion. Inventory carrying costs areRs. 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 22February 600 19March 800 21April 400 21The 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.

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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 Total1. Workers required 22 500= 23 19 600= 32 21 800= 38 21 400= 192. Labour cost 23,000 , 32,000; 38,000 ;19,000 ;=1,12,0003. Hiring costs 9 × 400 6 × 400= 3,600 = 2,400 6,0004. Lay off costs 7 × 500 19 × 500= 3,500 = 9,500 13,000 Total 1,31,000Plan II : Maintain a steady work force and use of inventory plus stock onJanuary February March April Total1. Workers used 30 30 30 302. Labour cost 30,000 30,000 30,000 30,000 1,20,0003. Units produced 660 570 630 6304. Inventory costs 5 × 160 5 × 130 5 × 190 = 800 = 650 = 950 24005. Shortage costs 100 × 40 = 4,000 ; 4,000Total 1,26,400 On the basis of costs, plan II would be the choice. Moreover this strategy would result in higher workermorale, smoother production, and generally, a higher quality product.

Problem no.13

5 jobs are required to be processed through 2 m/cs. SequentiallyThe 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 jobs2) For what period m/c 2 will remain idle3)When does job –B GETS COMPLETED.

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Q. 6. (a) The following tasks are to be performed on an assembly line in the sequence and times specified :Task Time (Seconds) Tasks that must precedeA 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 forecasted demand of400 units per an 8-hour day?(iii) Use the longest task time rule and balance the line in the minimum number of stations toproduce 400 units per day.(iv) Evaluate the Line Efficiency and the Smoothness Index

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(i) Schematic Diagram : D FA C H

E GB

A-20 B-40

D-45 E= 20F-25 G-10H-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.

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(b) Write short notes on :(i) Line Balancing(ii) Work Simplification(iii) Progressing(iv) Six Sigma quality programme

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(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 None2 D 45 27 E, F 70 70F 25 2 None3 B 40 32 E E 20 12 G 70 70 G 10 2 None4 H 35 37 None 35 70Total 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.

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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 operationmay vary. Therefore if the line is not balanced , a few operators will be over loaded whose operationtimes 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. Thisalso considers the sequence in which the operations will have to be carried out. Then Line Balancingtechnique groups operations in such a manner that each group will have generally equal total tasktimes. Each group is called a work station and assigned to an operator. Thus all the operators ineach work station will have balanced load.(ii) Work Simplification involves subdivision of an operation into its constituent elements in order tosimplify 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, shieldingthem from adverse factors inseparable from actual operations. The work of the progress departmentstarts where the planner’s ends. The Progress Controller/Progress Chaser , as he is called, is notconcerned with methods of carrying out operations(which are under the purview of the processplanning department) but the ‘doing’ of them at proper time in the correct order and at the lowestanticipated cost. The duties of the Progress Chaser would vary widely from unit to unit. In a well

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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 ofwarning to all concerned.(iv) Six Sigma Quality program is company- wide approach for continuous improvement in quality ofproducts and services. It measures the degree to which the process deviates from the standardsand 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 sourcesof variation and improve quality as well as productivity.It is a statistical measurement which tells us how good our products, services and process are andenables us to benchmark our operations with the best in the field. It thus helps us to establish ourcourse in the race for total customer satisfaction.A process at 6- Sigma level normally produces 3-4 non conformances in a million operations. Thisis supposed to be the best-in-class quality. Thus 6-Sigma is essentially a philosophy of workingsmarter. This means making fewer mistakes in everything we do. As we discover and eliminate thesources of variation , the non conformances are eliminated and the process capability improves.

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(c) The following cost have been recorded :Particulars Rs.Incoming materials inspection 10000Training of personnel 30000Warranty 45000Process planning 15000Scrap 9000Quality laboratory 30000Rework 25000Allowances 10000Complaints 14000What are the costs of prevention, appraisal, external failure and internal failure?

Problem on transportation65

Factory –ware house –cost ,demand & supply schedule

F W-1 W-2 W-3 W-4 SUPPLY

F-1 6 8 8 5 30

F-2 5 11 9 7 40

F-3 8 9 7 13 50

DEMAND

35 28 32 25 120

w

VOGL’S METHOD66

Factory –ware house –cost ,demand & supply schedule

F W-1 W-2 W-3 W-4 SUPPLY

COL /PENALTY

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

DEMAND

35 28 32 25 120

w

ROW PLTY 1 1 1 2/0

SELECT HIGHIEST PENALTYALLOT for cell F2W4--25

VOGL’S METHOD67

Factory –ware house –cost ,demand & supply schedule

F W-1 W-2 W-3 W-4 SUPPLY

COL /PENALTY

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

DEMAND

35-35 28 32 25 120

w

ROW PLTY 1 1 1 2/0

SELECT HIGHIEST PENALTYALLOT for cell F2W1--35

VOGL’S METHOD68

Factory –ware house –cost ,demand & supply schedule

F W-2 W-3 W-4 SUPPLY

COL /PENALTY

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

DEMAND

28 32 25-25 120

w

ROW PLTY 1 1 1 2/0

SELECT HIGHIEST PENALTYALLOT for cell F2W1--35

2)3 being the highiestAllot 25 units to F1-W4

VOGL’S METHOD69

Factory –ware house –cost ,demand & supply schedule

F W-2 W-3 SUPPLY

COL /PENALTY

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

DEMAND

28 32-32 120

w

ROW PLTY 1 1 1 2/0

SELECT HIGHIEST PENALTYALLOT for cell F2W1--35

2)3 being the highiestAllot 25 units to F1-W4

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IBFS F/W W1 W2 W3 W4 Supply

F1 6/ 8/5 8 5/25 30

F2 5/35 11/5 9 7 40

F3 8 9/18 7/32 13 50

Demand

35 28 32 25 120

Cost/units /unit Total cost

5 35 175

8 5 40

11 5 55

9 18 162

7 32 224

5 25 125 =781

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 .

Optimality test71

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 path4)The loop starts from unallocated cells moves horizontaly or vertically & finally forms a loop ending with the unallocated cell5)See the next slide

Evaluation process & final solution

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2)For F1W3=8-8+11-9+ 23)F2W3=9-11+9-7=04)F2W4=7-5+8-11=-45)F3W1=8-5+11-9=56)F3W4=13-5+8-11+9-7=7F2W4 GIVES – VE VALUE

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IBFS

F/W W1 W2 W3 W4 Supply

F1 6/ 8/10 8 5/20 30

F2 5/35 11/0 9 7 5 40

F3 8 9/18 7/32 13 50

Demand

35 28 32 25 120

W-1 is the least Costed un allocated cellBy moving one unit From W-2 to W-1 The cost /unit will come down by Re 1But this will out Shoot supply Therefore we move I unit from F2W1 TO F2W2.The change in cost =-8+6-5+11= 4

Cost= 5* 35=175 9*18= 162 7*32= 224 8*10= 80 5*20= 100

7*5= 35 255 + 297 +224 = 776

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IBFS

F/W W1 W2 W3 W4 Supply

F1 6/ 8/10 8 5/20 30

F2 5/35 11/0 9 7/5 40

F3 8 9/18 7/32 13 50

Demand

35 28 32 25 120

W-1 is the least Costed un allocated cellBy moving one unit From W-2 to W-1 The cost /unit will come down by Re 1But this will out Shoot supply Therefore we move I unit from F2W1 TO F2W2.The change in cost =-8+6-5+11= 4

Thus by moving 5 units from F2W2 to F2w4 the cost can be brought down by5* 1 =Rs 5This method can be repeated till all the values are +.It means no further reductionis possible.

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MODI METHOD.77

F\W W1 W2 W3 W4 supply

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

DEMAND

35 28 32 25 150

V V1 V2 V3 V4

Occupied cellsF1W2=8=U1+V2F1W4=5=U1+V4F2W1=5=U2+V1F2W2=11=U2+V2F3W2=9=U 2+V2F3W3=7=U2+V3

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Answer 7. (c)Particulars Rs.Training of personnel 30000Process planning

15000Total cost of prevention

45000Incoming materials inspection

10000Quality laboratory

30000Total cost of appraisal 40000Scrap 9000Rework

25000Total cost of internal failure

34000Warranty

45000Allowances 10000Complaints 14000Total cost of external failure

69000

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(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 whichshould not be included in the analysis?(ii) Compute the basic time for the job and the standard time if a relaxation allowance of 12%, a

contingency allowance of 3% and an incentive allowance of 20% are applicable for the job.

80

Basic time = 7.722 minsAllowance = (12+3) = 15% (Incentive allowance not considered)Standard time = 7.722 × 1.15 = 8.88 min (approx)

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(c) A work sampling study is to be made of a data entry operator pool. It is felt that operators are idle 30% of the time. How many observations should be made in order to have 95.5% confidence that the accuracy is within +/- 4%.

Answer 9. (c)n = {4p(1-p)}/s2p = 0.31– p = 0.7s = 0.04n = 4 × 0.3 × 0.7/(.04)2= 0.84/.0016= 525.

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83

84

85

Widescreen Test Pattern (16:9)

Aspect Ratio Test

(Should appear circular)

16x9

4x3

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