1. 1
OPERATION
MANAGEMENT
PRACTICAL PROBLEMS
Tips and tools for solving.T.VENKATARAMANAN.FCA(COST)

2. Old syllabus
2

Contains only important problems

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. 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
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. 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. 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. 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. 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. 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
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. 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. 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
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. 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. 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. 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. 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. 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. 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. 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

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
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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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
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
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. 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. 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. 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. 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
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. 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. 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. 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. 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. 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. 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. (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. (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
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. 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. 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. (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. (b) Write short notes on :
(i) Line Balancing
(ii) Work Simplification
(iii) Progressing
(iv) Six Sigma quality programme
59

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

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. 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. (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. 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. 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. 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. 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. 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. 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
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. 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. 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. 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. 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

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
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. (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. 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. (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

83. 83

84. 84

85. 85

86. Widescreen Test Pattern (16:9)
Aspect Ratio
Test
(Should appear
circular)
4x3
86
16x9