Gate Mechanical
Gate Mechanical
Gate Mechanical
Engineering
Forecasting
Forecasting is the first major activity in the planning. lt involves careful study of past data and present
scenario. The main purpose of forecasting is to estimate the occurrence, timing or magnitude of future
events.
Types of Forecasting in following areas like as capital and cash planning, inventory
planning, enrolment of student in college, product planmng,
Short, medium and long terms are the usual categories in behalf sales planning and sales force decisions etc.
of which the forecasting can be defined but the 'actual meaning Medium range forecasting needs judgement as well as time
of each will vary according to the situation that is being studied. series analysis. Combination of collective opinion, regression
analysis, correlation of different index and inflation etc. may
Long Range Forecasting be useful in forecasting.
Long range forecasting have more than 5 yr time period. It is
difficult to model and foresee event for more than 5 yr. It is
Short Range Forecasting
mainly due to economic uncertainty and variation in the These forecastings are defined, as the range of forecasting is
behaviour of the interrelated process. These are useful in the less from t h to 1 yr. In most cases, it is for one season, a few
various-areas like capital planning, plant location, plant layout months or few weeks. Short range forecasting is based on past
or expansion, new product planning, research and development data.
planning and technology management etc. The methods of
forecasting used in these cases are qualitative techniques. Measuring Forecast Error
Medium Range Forecasting 1. Mean Absolute Deviation (MAD)
The time period of medium range forecasting is generally 1 to In this method, we calculate as the average of absolute value
5 yr. As the range offorecast shortens 5 to 1 yr, the accuracy of difference between actual and forecasted values. The
of forecast increases. Estimate of reliability of forecast may negative sign in this difference is ignored as overestimate as
be useful in mediumrange forecast. This forecast is very useful well as underestimate are both off-target and thus undesirable
776 B
# AT # Ttsf,rp rz Mechanical Engi neeri ng
\{o, - r)' (t r -i
. MSE= t=l h :l-l>
./+r \n)i=uur_n
D,
Example 1. The forecast for April was 100 units with a - aro
smoothing constant 0.20 and using first order exponential Forecast of the load on the centre in the 8th month
smoothing. What is the July forecast? What do you = 829
think about a 0.20 smoothing forecast? (II) On the basis of three month moving weighted
average Forecast for 8th month
P 4 6 0+4=4 0
When number of jobs are waiting in queue before an o 5 1 4+5=9 2
operational facility (such as boring machine), there is need to R 3 8 9+3=12 4
decide the sequence of processing all the waiting jobs. This is s '7
10 12+7 =19 9
defined as 'the decision regarding order in which jobs in waiting T 2 -) 19+2=21 18
are processed on an operational facility or work-centre is called Total flowtime = 4 +9 + 12 + 19 + 21
sequencing.'
= 65 days
Total flow time 65
_=_
Mean flow time =
If the priority ruIe ofjob sequencing is known, we can sequence Number of 5 jobs
thejobs in a particular order. Sequencing does not tell us day = 13 days
and time at which a particular job is to be done. This aspect is Total lateness ofjobs = 0 +2 + 4 + 9+ 18 = 33 days
covered in detailed scheduling. Total lateness ofjobs
This is defined as 'the formation of starting and finish time of Average lal.eness oljobs = Nurb., ol Job,
all jobs at each operational facility.' l
33
= t =6.6days
Scheduling Rules
T 0
When many jobs are waiting before an operational facility,
R 0
we have some rule to decide the priority while sequencing. .P 5
It will be clear by following rules: o t1
1. First Come First Serve (FCFS). s 18
2. Shortest Processing Time (SPT)
or Shortest Operation Time (SOT) rule Totalflowtime = 2+5+9+14+21 =51 days
3. Earliest Due Date (EDD) rule
4. Last Come First Serve (LCFS) Mean flow time 10.2 days
5. Random Schedule Rule (RSR) Total lateness of iobs =3+7+11=21 davs
6. SlackTime Remaining (STR) rule
lndustrial Engineering & 779
r*if,ffi
,rlri;i(Pi$x,if.t-r,li:il ;r i, r :,,: :i,
i:t::;i iEl::i:i,iii,
! $1sl*;i ;,f;i#ri..
iili:lifrtrj:Ed;iri;i
iil
,:!:]:;;ii]:;,ii;:irii ii:.j ;ii#flilii;
T 2 3 0+2=2 0
P 4 6 2+4=6 0
a 5 7 6+5=11 4
R 3 8- 1l+3=14 6
s '7
10 14+7 =21 1t
On the basis of Last Come First Serve (LCFS) Slack Time Remaining (srR)
In this rule or method, the job has to be given priority which
In this method, highest priority gives to that job which has has minimum (di- pi).
arrived most recently.
#"ffififfiHlir iiilfl*uu#t*;r
r:::F$*i t
:ii,sun*i! :i:: ;111&.$tt i,::: ,.1$ry,F."ry l
'flt1uEy"s"r;""
.:: iri:r:::l;? i: l:ri i:r! ii i: i:iri:i, r:r
:$*-s.i&#i i+!{lf. .#$iiii:: ,I'liti:lPfiirtiFilr.,i ':, :'hofie€l ,l
::r:i:d{i,,,i::
"F.)F..,',+.u^
ri::.#r..}. g-,)r::ir: t::i : :'r, : ::i:.:i t; :; :. i:s:!,i lri$;l*r(rt,:4j, I
:::: ii:t::::
lit +L
::-:i!a:i:!:
::i+:idii::iirr:i F.rltF.
l; l:1
. +z.l
.I :ii:tlriffi "S;*d;rlrr;
I :airi-I
,iitlti+Y A T 2 3 0+2=2 0
T 2 J 0+2=2 0 P 4 6*,, 2+4=6 0
s '7 10 2+7 =9 0
o 5 7 6+5=11 4
n J 8 9+3=72 4
,s l 10 11+7=18 8
O 5 7 12+5=l'7 10
P 11 + 4 --21
R 2 8 18 +2=20 t2
4 6 15
Example 6. When n jobs are performed on two machines five Minimum time to complete all jobs by processed sequence =
jobs are performed first on machine M, and then on 30 min. Total idletime for machineMn= 30 min.
machine Mr. Time in hour taken by each job on each If process is recycled then, idle time to machine M = 3O 2g
machine is given below. ; -
=2 min
Inventory
Inventory is defined as the lisr of movable gbods which helps
directiy or indirectly in the production of goods for
Step 1 The smallest processing time between two sale.
machines is 1 which corresponds to job Q and then the We can also defined inventory as a comprehensive list of
job Q is processed firsr. movable items which are required for manufacturing the
products and to maintain the plant facilities in working
Step 2 Now, write remaining table as conditions.
Inventory is divided into two parts
-
1. Direct Inventories
The inventories, which play adirect role in manufacturing of
a product'and become an integral part of the finished proJuct
Now, minimum processing time is 2 corresponds to job are called direct inventories.
P on machine M,. Raw materiais, work in progress, purchased parts and finished
Then, it is processed in the last. goods are direct inventories.
2. Indirect Inventories
Mafhiiirt Jsb This is defined as the materiai which helps the raw material to
T get converted into the finished products. But they are not
Mr 9 3 l0 integral part of the finished product. Tools and supplies
M2 7 8 +^ (material used in running the plant but do not go into the
The srnallest processing time is 3 on machine M, product) are indirect inventories.
corresponds tojob S.
Then, it is assigned at number 2. Inventory Control
a s P Inventory control means making the desired item of required
Now, reduced table is written as quality and in required quantity available to various
departrnents when needed.
I
Determining Inventory Level
2 The main object of inventory control is to establish level of
In this table, the smallest processing time is 4 on machine inventory which will serve to minimize the product,s cost and
M, corresponds to job L So, it is written at 2nd position maxirnize company's reverlue. *. +
from last. The amount of inventory a company should carry, is detemined
by four basic variables.
o s R T P
1. Order quantity
Sequencing ofjobs are given in order as 2. Reorder point
Q-->S-+R-->T-+P 3. Lead time
sqqrreiira *i:i';ili,!:i':
'Mi
4. Safety stock
,ofg.g6,;, l*me. :ii.l.ltl€:it ?ftii$,.i 'Prosest$g IlfE :IdIe,
,l iri,.:, ,:riiAt,::. ,:m:r':, i,:,,:titlis,i,::, iilat: tiiire 1. Order quantity
O 0 1 I I 6 '7 1
Order quantity is the volume of the stock at which order is
s 1 -) 4 1 8 t5 placed. Order quantity means total quantity of buy or sell order.
R 4 9 13 l5 7 22 Sometimes traders don't want to teil others how much quantity
T 13 t0 23 23 4 27 1 he is buying or selling, hence he may choose to disclose only a
P 23 5 28 28 2 30 I part of his total order quantity i.e., minimum )AVo of total
order quantity.
lndustrial Engineering 1 751
Safety stock = K
3. Carrying cost
where, K = a factor based on acceptable:frequency of Carrying or holding costs are the costs incurred maintaining
stock-out in a given number of years
the stores in the firrn Annual carrying cost is calculated by
TC=C,rS*9rS+!xC,xi
EOQ = 'q'-
where, C, = ao.t of placing an order =cxs*QtS+IxC,
-u"" q 2 '
= annual consumption of,the product
,S where, notations are described previously.
C, = ur,it cost of an item
i = interest rate charged per unit per year Exam ple 7. A company requires 16000 units of raw material
costing 7 2per unit. Cost ofplacing an order is {45 and
Crx i = Co the,Carrying costs are 20Vo per year per unit of average
C, is known as the holding cost of an item for a year. inventory. Determine-
Graphical presentation of various costs with order
quantity. i. The economic order quantity.
1. Ordering cost 2. Cycle time.
3. Total variable cost of managing the inventory.
Sol. First we should write all giu"n auta with their usual
notations
S = 16000, C,=7 2/unit
Co=745, i=20%o
1. Economic Order euanrity (EOe)
Ec-s 2x45x16000
- Vc,ri = 2x0.2
= 1891.36
But EOQ is a quantity. So, we should write EOe in
2. terms of whole number = 1898 units
Number of orders per year
tl,
o Annual consumption
6f
C
c Econornic order quantity
16000
= 1898-=8'42 = 9
a
o
o Il4
: vr - -Xl2=-
= e'
6
E
9 3
c = 1.33 months
3. Total variable cost of managing the inventory
C
_- -:gxS xi+!xC.
q 2 -u
PointA represents economic order quantity andhere, total cost
= Jlrl6ggg*]!98 r. 2x0.2
is minimum.
= 7158.94
lndustrial Engineering l, 783
Example 8. A company needs 2000 units per month. Cost of Example 9. An item is produced at the rate of 50 items
placing an order is { 40. In addition to { 0.50 the carrpng per day. The demand occurs at the rate of 25 items
costs are 20Vo per unit of average inventory per year, the per 'day. If the setup cost is < 200 per setup and
purchase price of < 10.00 per unit. Find the economic holding cost is < 0.01 per unit of item per day. Find
lot size to be ordered and the total minimum cost. the economic lot size for one run, assuming that
Sol. In this problem the demand is given in month. So, we shortages are not permitted. Also, find the time of
calculate the annual demand S = 12 x 2000 = 24000 units cycle and minimum total cost for bne run.
Inventory carrying cost is 2070 per unit of average Sol. First we should write all the given data with their usual
inventory in addition to { 50. notations.
So, the C,.i is given by Given, p= 50 items per day
C,.i=0.50+207ox70 S= d = 25 items per day
= { 2.50 / unit of average inventory co={2oo
c'=740 Cu'i=Cn={0'01
1. So, economic order quantity is calculated by
So, EOQ =
zCuS 2C,S
=
[,-4'1.,,, .,, ['-' j
\ P.) \ P)
= 876.35 = 877 units
Total minimum cost = Cost of material + Total variable 2x200x25
cost of managing the inventorY / l<\
o.o1xl r-11
x10+ ,,
24ooo ,40+817
, xz.so
\ s0/
=24000 z
2x200x25x2
= 7 242190.89
Economic Order Quantity when replenishment is not
instantaneous (Gradual Replenishment or Finite l*
= V4U0x50xl00
Replenishment)
Let p = dally production rate or rate of = 20 x 10"60
replenishment = l4l4.Zl = l4l5
d = daily compution rate: q^ l4l5
2. Cvcle time = a= --::-
/ ,) " .\ '25 =56.6=)/ days
Intro Exercise I
1. Forecasting provides guidelines on 11. ltem Erequires four numbers of item C. product prequires
(a) how many of the products produced are likely to be two numbers of items B and five number of item C. lf five
demanded by the customers numbers of product P are to be manufactured then. the
(b) material requirements schedules number of item C required will be
(c) amount of business, the firm can expect during the (a) 65 (b) 50
planning period (c) 75 (d) 30
(d) All of the above 12. The lead time consumption is 600 units. The annual
2. The order in which different jobs are being taken up in a consumption is 8000 units. The company has a policy of
machine or process is called EOQ ordering and maintenance of 200 units as saiety
(a) routing (b) sequencing stock. The Reorder Point (ROp) is
(c) scheduling (d) aggregate planning (a) 500 units (b) 700 units
(c) 800 units (d) None of these
3. lf forecast and actual demand for march are 60 and 70
respeclively. lf cl = 0.1, then forecasted value of April is 13. The rnonthly demand is t g000 of sales. Annual carrying
(a) 70 (b) 61 cost is ( 8400. The ordering cost per order is { 600. The
(c) 6e (d) 64 EOQ. is
4. The time at which the processing of job in the machine is
(a) one month of sales (b) two months of sales
(c) 26 days of sales (d) 28 days of sales
completed is
(a) processing time (b) due date 14. The fixed per cost of land is { 50000. The variable cost is
(c) completion time (d) flow time (
t0 per unit production. lf selling price is 20 per unit {
5. The flxed cost of {
25000 and break even quality of and production is 6000 units then, margin of safety is
(a) 1000 units
35000 units are estimated for a production. The profit at (b) 2000 units
a sales volume of 50000 units is (c) 3000 units (d) None of these
(a) 1 0000 (b) 1 0s00 15. The demand and forecast for a month are 12000 and
(c) 10714 (d) None of these 10000 respectively. By using single exponential smoothing
6. Earliest Due Date EDD sequencing of jobs in method forecast for next month is (given smoothing
a single
facility, would coefficient = 0.25)
(a) minimise the mean tardiness (a) 10500 (b) 10510
(b) minimise the mean lateness (c) 1 0230 (d) 10600
(c) minimise the maximum tardiness 16. Four jobs are to be processed on a machine as per data
(d) None of the above : listed in the table.
7. Short Processing Time (SpT) sequencing of jobs in a iro$: :ottp #;
single facility would minimise the
7
(a) mean lateness
10
(b) maximum tardiness 19
(c) Both (a) and (b) 15
(d) mean tardiness
(l) lf the Earliest Due D*eLe (EDD) rute is used to
8. Annual carrying cost, for a given annual demand sequencing the jobs then, the number of jobs delayed
(a) will increase with the increase in the number of orders is
placed per annum (a) 1 (b) 2
(b) will decrease with the increase in the number of orders (c) 3 @)4
placed per annum (ll)Using the Short Processing Time (SpT) rute totat
(c) is independent of number of orders piaced per annum tardiness is
(d) will increase with the decrease in lead time (a) 8 (b) 7
(c) 11 (d) 13
9. When the ordering cost is increased to .l 6 times, the EOe
will be increased to 17. In a manufacturing shop each job must pass through two
(a) 2 times (b) 4 times machines (M, and M* in lhal order). The processing time
(c) 8 times (d) remains same (in hours) for these jobs is
18. An oil engine manufacturer purchases lubricant at rate of (a) { 1200 (b) < 11s8
{ +z per piece from a vendor. Requirement of these (c) t 1305 (d) < 14e1
lubricant is 1800 per year. lf cost per placement of an 20. Consider the
order is { 16 and inventory carrying charge per rupee per
year is only 20 paise. The order quantity per order is
(a) 80 (b) 83 0 <01<650 20
(c) 82 (d) 85 650<02<800 18.50
19. A manufacturing ccmpany purchases 9000 parts of a 800<q 17.50
machine for its annual requirement, ordering one month
usage at a time. Each part cost 20. The ordering cost{ The monthly demand for the product is 300 units, the
per order is { 15 and carrying chargeis are 15o/o of average cost of carrying inventory is 2o/" of the unit price of the
inventory per year. item and cost of ordering is { 400. The optimum order
(l) The economical ogder quantity is quantity is
(a) 300 units (b) 350 units (a) 700 (b) e00
(c) 250 units (d) 400 units (c) 800 (d) 828
2x16C".5
- 5000
= 1000 units
EOQ'= = 4 (EOQ)
I-5. (a) Given, D712000, F, = 10000
10. (a)
u= 0.25
11. (a)
We know that
L6. (I) (c) Arrange in ascending order ofdue dare. (II) For M' job J, For M, Job J,
tr)Ue
date
ffi Tirie ','1 " ''r',,'D,€lh)'pet:dfly-,,.,r, .', ,
ffi
M2 2x9000x15
Select the job in which least time is taken bymachine. EoQ= =
20x0.15
For job A, Mrtakes least time, so write it on right
side C, = 300 units
(II) (c) In case of inventory each month
A
After this F
M1 M2
q= 9000
D =75ounits
F A
; . ,,.,
Total variable cosr = C. r.
After this C but by machine M, +
U F A = 15"9ooo*zoxo.tsx75o
Complete it
750 2
=180+1125=<1305
c E D B F A
20. (d) Quantity 44 Unitprice
;Ioh ,,Mt: ',,',,tMl, 0<0r <650 20
llfliir, 650 < 02 < 800
l:::-l In '. 18.50
C 0 0+9=9 9 9+14=23 800 < 03 fi.so
E 9 9 + 12=21 23 23+15=38
D 2t 21+28=49 49 49 +21 =10 C'=7400
B 49 49+33=82
82+20=102 82 S = 300 units
F 82 + I7 =99 82 102+8=ll0 102 i=2Vo=0.2
A 9999+16=115 115 ll5 +7 =122 At C, = yL50
Optimal time is 122 h.
(I) For M, job J, For M* Job J,
= 828 units
Then, for M,lob "/, (Time),. 828 > 800
= (Time)o,, of Mrfor job J, .'. Optimum order quantity = 828 units
See the 3rd row 49 > 38
lndustrial Engineering V 787
Linear Programming
Linear programming is defined as 'it is the optimization (either maximization or minimization) of a linear
function of variables subject to constraint of linear inequalities.'
Features of Linear there is no case then, it can be negative. For the products
Q, R and S the non-negative conditions can be
,f
Example l. A firm manufactures two products X and I, both Putting the value ofy in Eq. (i),
have to be processed on two machines P, and pr. Product 4x+6x2=24
X requires 4 h each on both machines, while product I , 4x=12
requires 6 h on machine P, and 2 h on machine pr. The x=3
available hours on machines P, and P, are 24 and 16 Then, the coordinates of O are (3,2).
respectively. The profit per unit is estimated at { 100 for The points m, O and r gives the optimum solution.
productXand( 125for product Y. For point ru,
What quantityof each product to be produced to maximize Z= I00x+125y
profit? = 100x0+125x4
Sol. Step 1. First of all, we formulate the given problem in = 500
form of mathematical equation. For point O,
Let we assume, Z=100x+125y
x = number of units of product X = 100x3+125x2
. y = number of units of product I = 550
Z = incremental profit of the firm For point /,
The objective function.is maximize = 100x+I25y
Z= l00x+lZ5y = 100x 4+125x0
Inequalities are written as ..':, = 400
4x + 6y a24 Our object is maximized Z.
4x+ 2yS16 Then, it is given by point O. Three units of product and two
In graphical method, inequalities are converted into units of product I should be produced by firm.
equations.
Then, Example 2. Maximize
4x+6y =24 ...(i) Z=3x+2y
4x+
2y = 16 ...(ii) Subject to
Then, the constraint Eqs. (i) and (ii) are represented on 8x+6y)48
the graph. 3x+2y<6
The product X is taken on X-axis and the product I on .r.) > 0
Y-axis. Sol. First we should convert inequalities into equalities and
Step 2. Draw straight lines on rhe graph using the plot the graph.
constraints equations and to mark the feasible solution 8x+6Y={3
on the graph.
-xv
Y-axis
A*8=1 ...(i)
s (0, 8) Coordinates are given by (6, 0) and (0, 8).
3x+2y', = 6
f
E
o
*
o_ 4x+2y=i$ "'(ii)
Coordinates are given by (2, 0) and (0, 3).
(0,4)
m '*1=l
Region of feasible region Y8
7
4x+6y=24 6
5
X-axis
(4,0) h(6,0) 4
Product X
2
The coordinates O are calculated by solving Eqs. (i) and
1
(ii).
4x+6Y = 24 .. (iii)
4x +2Y = l$ ... (iv) There is no common feasible region by
Subtracting Eq. (iv) from Eq. (iii), 8x+6.v>48and3x+2y36
4Y=8 Therefore, there cannot be a feasible solution. Thus, this
.._a linear programming problem gives infeasible solution.
-t-z
lndustrial Engineering E 789
Simplex Method 5. Ifthe consraints of primal are of less than equal to ( < )
type as in maximization problerrl all the constraints of
When the decision variables are more than two, graphical the dual become greater than equal to (2) type.
method becomes inadequate and linear programming problem 6. The column coefficients of primal constraints are changed
is solved by simplex method. into row coefficients of the dual constraints.
Simplex method is defined as an algebraic procedure that 7. The value of slack variables in the index row of the dual
through a series of repetitive operations. progressively corresspond to the values of decision'variables in the
approaches an optimal solution. prirnal.
Some terms related to describe simplex method- 8. The optimal solutions are same in both cases.
O_ in terms of unit
,s -y
/ v\
Profit=Sll-=l-F
Jl
\
cosl
Angle of Incidence (0)
It is defined as the angle between the lines of total cost and
total revenue. An direct measure is profit volume ratio. It
"eJ e
is given by
,,r"01
Sales-Variable cost
"o.t Profit volume ratio =
J Sales
Qr.t Quantity
Profit-Volume Graph
Margin Of Safety (MOS) (P-V Graph)
It is the difference between operating sales and break even
Profit volume graph is a plot similar to break even point
sales.
analysis. It is useful for comparing different processes or
Margin of safety = Present sales - Break even sales
systems. The chart represents quantity on X-axis and profit on
Margin of safety I-axis. If there is no production then, profit is negative and
Margin of safety ratio =Present sales equal to the fixed cost. The slope of line starting with (zero
Higher is this ratio, more sound in the economics of the firm. production of fixed cost) point is dependent on profit_volume
ratio.
lndustrial Engineering 1 791
Mathematical Formation of
Profit
(or loss
Transportation Problem
.t
if negative Let
j
= index for origin
I
= index for destination
x..U = quantity shipped'from origin I to
destinationT
s, = total capacity of origin i
d = total demand of destinationT
J
c.. = unit cost of shipping one unit from
U
By the graph
origin i to destinationT
Linear programming formulation of a transportation problem
Fixed,cost
Profit-volume ratio =
Break even quantity
Fixed cost+Profit
Minimize z=iicu*ii
- i=1 j=l
Quantity at profit
Subjected tt iy,j = d,foralti (where,7 =1,2,...n)
Example 4.
A company is producing certain tlpe of circuit l=l
breakers. The fixed cost of land, building is n
I'' = I"
minimum production level? mn
If the firm is operating at present so that production is
20000 units, what is the firm's profit and margin of safety?
Sol. Fixed cost (F) = ( 80000 X,i}0 for all l, where i=l,2 ... m
Variable cost (V) = 10 ( for all j, whereT = l, 2, ... n
Sales price (S) = t 2O
Let break even quantity is Q then, Decision Rule
F+QV=QS (I) IfIsupply=LDemand
F 80000 Then, go to next step.
Q= S-V- = 20-10 =8000units (ID IfLsupply>LDemand
The minimum production level of firm should be Then, add a dummy destination with zero transportation
8000 units. cost.
Margin of safety in terms of unit (IID If X supply < I Demand
= 20000 - 8000 Then, add a dummy source with zero transportation cost.
= 12000 units
In terms of rupees = 12000 x Sales price Feasible Solution
= 12000 x20 A basic feasible solution for tlre general transportation problem
= ( 240000 must consist of(m+ n 1) occupiedcells.
Then, profit product of 20000 units -
where, m = numbet of destinations
= Total revenue - Total cost rz = number of resources
= 20000 x 20- (80000+ 20000 x l0)
= 400000 - 280000 Non-Feasible Solution
= ( 120000
When the number of occupied cells is less ttran (m + n - l).
Transportation Model then solution is non-feasible. It is also known as degeneracy.
Example 5. kt see it by an example. Highest of all calculated penaltycosts is S, (or S). Therefore,
allocation is to made in row of source Sri So, tie route will
be S3D5 becausetransportation cost is riinimum.
Supply
Now, matrix is written as
250
200
(150) 16- 15 = 1
7#tso
In this method, we calculate the difference between
the two least cost routes for each row and column. 200
The difference is called as penalty cost for not using
the least cost route.
Now, select maximum penalty cost. It may be in row 150
13-0=13 17-16=1
15-0=15 16--15=1
15-0=15
17-16=1
'15-13 t7-16 17-16 16_15 0-0
penaltv 17 - 16 17 - 16 't
=2 =1 =1 =1 =0 6 - 15
cost =1 =1 = 1
lndustrial Engineering * 791
There is tie between all routes, we break the tie by arbitrarily After allocations are done, the matrix is written as
selecting any routed then SrD, is least cost route.
Supply
250
s2
100 ,
Intro Exercise 2
1. ln order the linear programming techniques provide valid 3. ln simplex method of linear programming, the objective
results, then the row of the matrix consists ol
(a) relation between factors must be linear (positive) (a) slack variables
(b) relation between factors must be linear (negative) (b) coefficient of the objective function, which is the profit
(c) results do not depend upon relation between factors contribution per unit of each of the pioducts
(d) Both (a) and (b) (c) names of the variables of the problem
(d) None of the above
2- ln the graphical method of linear programming problem,
every corner of the feasible polygon indicates 4. Which of the following is not correct about LPP?
(a) optimum solution (a) All constraints must be linear relationships
(b) a basic feasible solution (b) Objective functions must be linear
(c) Both (a) and (b) (c) All the constraints and decision variables must be of
(d) None of the above either<or>type
(d) All decision variables must be non-negative
794 & #e'{# {xs{*ffi Mechanicat Engineering
(c) maximize Z=3wt+5wz A company has one surplus truck in each of the cities A,
w, a9, wrl 4 B, C and one deficit truck in each of cities R Q, R and S.
The distance between cities in kilometre is shown in matrix
w., + 3wra 6 beloiiv. By assigning the trucks from cities is surplus to
wf wr2 0 cities in deficits, lf minimum distance is covered by
(d) None of the above vehicles.
19. There are 4 machines and 4 operators. Operator charges PORS
on different machines as A 7 5 10 17
(Charge on machine in {) 20 10
B 5 13
I ll lll lv C 6 5 3 8
1 6 7 7 8 (l) Correct assigning orders is as
(Operator) 2 7 I I 7 (a) A*P (b) A-o
B-Q B_P
3 8 6 7 6
I
C_R c-s
4 8 7 6 (c) A-O (d) A-s
lf one operator use pne machine then, minimum cost to B_P B-Q
operate all machines is c-H c-P
(a) { 21 (b) <25 (ll) Minimum distance covered IS
15. (c) First of all, convert all inequalities into equation 2xr=g
-xl+3x2= l0 xt= 4
* -* :') and xz= 2
x1+ xz= 6 So, coordinates ofpoint B is (4,2).
Draw the lines Corner points are (0, 10/3), (4,2), (2, 4) and (2,0).
For minimize,
x2
Z=-xr*2x,
(0,6)
Ar (0, 10/3),
Z = O+2x9=20
JJ
(0,10/3) At(4,2), **Z=-4+4=0
At(2,4), Z=-2+8=6
At(2,0), Z=-2+2x0=-2
16. (d)
(2,0) (6,0)
Now, find the values of A and B.
For A,
-xl+3x2= 70 ...(i)
-xl+xz= 6 ...(ii)
Adding Eq. (i) and Eq. (iii), (20,10)
4xr= 16
xz= 4
796 h #P"'{ff '{ cs{qpr: Mechanical Engineering
Corner points are (12,12), (18, lZ), (20,10) and Step 2 Least element in each column subtract from
(20,0). other elements.
Maximize Z=2x, + 3x, 'IIIIIIIV
At(12,12), Z=24+36=60
At (20,10), Z = 40 + 30 = 70 I I 1
x1 O R
(1,0)
There is no common region between given constraints.
A l 5 10 t7
B 5 t3 20 10
So, solution is infeasible.
C 6 5 J
f8. (a) Constraints 8
D1 0 0 0 0
xt+3xz> 3
xr+2xr> 5 Step 1 It is not a square matrix because one row
xyx2'x3>0 is less than the column so, add a dummy row.
Minimize Z = 4xt+ 6r, + 1 gx,
Step 2 Select least element in each row and
Dual is given by
subtract it from other elements in that row. The
Maximize Z=3wr+ 5w,
number of lines is four, hence it is optimal
Constraints wt < 4,wrS9 condition.
3wr+wr36
wPwr) 0 A 2 E 5 t2
le. (b) IV B E *o3 15 5
1 6 7 7 8 C 3 2 E 5
(Operator) 2 7 8 9 7 Dl a * g E
-l 8 6 7 6 P R
4 8 7 6 I Order is
Step 1 Select least number in each row and subtract A-->Q
that number from all other elements.
B-+P
IIIIIIIV C+R
I 0 I 1 2
D, -->S
2 0 I 2 0 (II) (d) Minimum distance= 5+5+3
3 2 0 1 0
4
=13km
2 1 0 .,
lndustrial Engineering X 797
Assignment Models
Assignment problem is defined as a particular case of transpofiation problem in which the objective is to
assign a number of resources to an equal number of activities at minimum cost or maximum profit.
Assigning p jobs to p different machines one job to each machine or g task to q facilities one task to each
facility.
1 2 -') I 5 0 3 2 2 t0t I J
2 2 0 5 4 I 4 4 >s< tr
x
1
J 2 I 0 J 2 5 u 4 1 -1
4 5 0 2 -') I
Queueing Theory Waiting time in the system = Waiting time + Service time
The size of a queueing system is defined as the total number Lq = averagenumber of customers in the queue system
of customers that might require service from time to time. The n = mean number of units in the queueing system
size of input source is generally assumed to be infinite. including the one being served
p-).
7. _ I _ I _,
Average waiting time in the queue is given by p-). 3-2
rr.q_ __
f,_ = 60 min
p(p - }.t
Example 3. Patrons arrive at a reception counter, an
8. Average time in the queue system (1.e., waiting + service average interval rate 5 min and receptionist takes an
time) is given by average of 3 min for one patron. Determine average
I queue length.
w' = p-1,
Sol. i" = I5 = 12 patrons per hour
9. Probability density function of waiting time. (Probability
that a customer shall have to wait between time /, and tr.) b = ry = 2o patrons per hour
3
is given by
We know that average queue length
p= l', 11,
_7"yr-,r-^,,41 k?o 12x12 144
Jrt
p(p-I) 20x(20-t2) 20x8
Example 2. Customers arrive at one person barber shop _9
according to a Poisson process with a mean time arrival 10
30 min. Customers spends on an average of 20 min in PERT (Programme Evaluation and Review Technique) and
the barber chairs. CPM (Critical PathMethod) are scheduling techniques which
Determine- are used to plan, schedule, control a project consisting of
1. Probability that a new arrival need not to wait for number of interrelated activities.
the barber to free.
These techniques give a picture which defines the job to be
2. Number of customers in the shop.
done, integrates them in a logical sequence and provides a
3. Waiting time for a customer.
system of control over the progress of the plan.
4. Total time spend by customer in the shop.
800 * A;*FS Fx*f*r: Mechanical Engineering
Activity
All projects consist of number ofjob operations or tasks which
are called activities. An activity is any tirne or resource
consuming part of the project which has defineable start and This network have five events and its activities to be performed
finish. It may be process such as moulding, flrushing, cutting, are designated as at e and m is dummy activity. Event a is
forging and material handling or material procurement cycle. start ofthe project and event 5 is end ofthe project.
An activity is shown by an arrow in a network diagram. It
begins in start event which is also known as tail event and end Earliest5tart Time (EST)
in completion event or head event. An activity is normally given It is earliest
possible time at which an activity can start and
a name like A, B, C etc. which are marked below the arrow calculated by moving from first to last event in the network
and the estirrated time to accomplish the activity is marked diagram.
above the affow.
Earliest Finish Time
H
Tail event Head event
(EFT)
It is the earliest possible time at which an activity can finish.
It can be written as
Start End
EFT = EST + Duration of that activity
Progranilne Evaluation
and Review Technique
(PERT) critical path on the network and also calculate the total
project duration.
PERT finds applications in planning and control of complex Sol. First of all, write all the possible paths in network
set of activities, functions and relationships. The main object diagram.
of the PERT is controlling the critical activities of the project (D l--t2 -s 6
that can be completed within the scheduled time limit. PERT '(ID 1 -+2-+3-+6
gives an idea to indentifying critical activities right at the (ilI)1+2->3->5-+6
planning stage so that management is aware of these potentially (IV)1 -+ 3 -+ 6
troublesome areas and exercise proper control over them, so (V) 1-+3-+5-+6
as to complete the project within the schedule time. (VI)1 -+ 4 --> 5 -+ 6
PERT uses three time estimates for each activity. Now, calculate total duration on each path
(I) l-+2-+6=3+5=8days
1. Optimistic Time (rJ (II) 1 -+ 2 -+3 -+6 = 3 + 10 + 6= 19 days
(III)1 -+ 2 -+ 3 -+5 -+ 6 = 3 + 10 + 4 + l= 18 days
It is the shortest time in which
an activity can be completed (IV)1 -+ 3 -+ 6 = 4 + 6 = 10 days
assuming that everything goes exceptionally well. It has low (V) 1 -->3 + 5 -+ 6 = 4 + 4 + I =9 days
probability of occurrence. (VI)1 -+ 4 -+ 5 -+ 6 = 14+ 1 + 1 = 16 days
Now, the path have long duration that is critical path.
2. Most Likely Time (r_) Critical path is I --> 2 -+ 3 --> 6.
This is the most likely time required to complete the activity Critical path should be shown by double arrow in the
taking into consideration, all favourable and unfavourable network diagram.
elements. This estimate of time lies between:the optimistic
and pessimistic time.
',Srt$r#,:
the probability of occurrence of t * is 4 times that of t o and t A 1 5 J
,. A
Siandard deviation is given by B 2 + 3
.2 C J 5 4
t, *to \ -L D A 2 10 9
6) ='o 6 E C 4 6 5
F B,D,E 5 13 6
^ ( ,, -,.,\'
VarianceY=(a,)'=[=U G A 2 6 4
J H GF 0 6 3
802 * #eY* "f
xat*r: Mechanical Engineering
From the three estimates, obtain the expected times of Now, write all the possible path in the network diagram
all the activities and slack of all the events. Also, find (I) A-+ G-+H
the critical path. (II) A.-+ D -+ F -s H
Sol. First of all, make the table and calculate activity expected (III)B -+ F -+ H
time. (IV)C -+ E -s F -; H
to+4tm+tp Now, calculate the total duration on each path
We know that activity expected time t"= (I) A -+ G -+ H =3 + 4+ 3 = 10days"
(II) A -+ D -+ F ) H =3 + 8 + 7 + 3 =21 days
Sri (III)B -+ F --> H = 3 +7 + 3 = 13 days
l+12+5
A 1 5 3
6
(IV)C+ E-+F-)H= 4+5+J +3= 19days
So, critical path is
2 +72+ 4
B 2 4
6" A-+D-+F-+H
3+16+5 This bracket indicates as
C 3 5 4
6 EST = Earliest Start Time
2+36+.10 LFT = Latest Finish Time
D A 2 10 9
4+20+6
llTtl
E C 4 6 5
6
5 +24+13
F B,D,E 5 t3 6 -'
6
2+16+ 6
G A 2 6 4
6
H G,F 0 6 3 _O+12+6 _1
t4T6l
Event 1 2 J 4 5 6
Slack 0 0 0 2 0 0
Intro Exercise 3
1. CPM is 4. A dummy activity
(a) time oriented technique (a) does not require any time
(b) activity oriented technique (b) is represented by a dotted line
(c) event oriented technique
(d) Both (b) and (c) (c) is artificially introduced
(d) All of the above
2- PERT is
(a) event oriented 5. The critical path of a network represents
(b) activity oriented
(c) cost oriented (d) Both (a) and (b) (a) minimum time required for completion of project
(b) maximum time required for completion of project
3. Critical activities are those for which
(a) float=+1 (b) float = 0 (c) minimum cost required for completion of project
(c) float < 1 (d) float > 1 (d) maximum cost required or completion of project
lndustrial Engineering $ 803
6. One time estimate is (V) Number of hours for which the cashier remains busy
(a) PERT (b) cPM in 8 days
(c) Both (a) and (b) (d) None of these (.a)6h (b) 7h
(c) 6.4 h (d) 5h
7. Which of the following is not correct?
(Vl) Average length of non-empty queue is
(a) The critical path of a project network represents the
(a) 4 (b) 3
minimum time needed to complete the project
(c) 6 (d) 5.
(b) Critical path is the longest path in a project network
(c) A delay in the completion of critical activities need not
(Vll) Average cost due to waiting on a part of cashier, iI
cashier is value at <1 00/h
cause a delay in the completion of the whole project
(d) The sum of the variances of the critical activity times (a) t 15 (b) < 20
gives the variance of the overall project completion (c) ( so (d) < 25
time 14. Consider the following network:
8. lf an activity has zero slack, it implies that
(a) it is a dummy activity
(b) it lies on the critical path
(c) there are more than one critical paths
(d) All of the above '
9. An activity in the network
(a) represents a task which has a definite beginning and
a definite end Distance between different stations is shown on each link
(b) cannot staft unless all its immediate procedures are in km.
completed (l) The shortest route from 1 to 8 is
(c) have float zero, if it is critical, otherwise total float as (a) 1-+3-+5+8
positive in case of non-critical activity (b)1-+4-+7-+8
(d) All of the above (c) 1+2-,5-i.7)B
10. The probability of obtaining a total of 6 in a single throw (d) None of the above
to two dice (ll) Shortest length is
(a) 5/36 (b) e/36 (a) 9 km (b) 11 km
(c) 1/13 (d) 7t13 (c) 10 km (d) 12 km
11. ln a queue, the mean arrival rate is equal to 10 per hour 15. A network is shown in figure.
and the mean service rate is 15 per hour. The expected
queue length is
(a) 1.57 (b) 1.33
(c) 3.2 (d) 2-75'
12. ln a single channel queue, if mean waiting time in system
is 60 min, the mean waiting time in queue is 40 min, then The critical path is
mean rate of service will be (a) 1-+2+5+6-+7-+ 8-+9
(a) 2 per hour (b) 3 per hour (b) 1 -+2-+3-+4-+8-+9
(c) 1 per hour (d) 4 per hour (c) 1 + 2 -+ 3 -+5 -+ 6 -+ 7 -+ 8 -+ I
13. At a cashier counter,8 customers arrive on an average (d) 1 -+ 2 -+ 3 -+ 4 8 -, 9
17*--r
every 5 min. While cashier can serve 10 customers in 16. Variance for critical path is as
5 min. Select correct option. a ----) b= 4
1i1 Expected number of units in the system b -------) c= 16
(a) 4 (b) 5 c -----) d=4
(c) 6 (d) 8 d -------> e= 1
(ll) Average number of units in the queue Then, standard deviation of critical path
(a) 3 (b) 3.2 a -----J eis
(c) 2 (d) 4 (a) 6 (b) 5
(lll) Expected time per unit in the system is (c) 4 (d) 3
(a) 2.5 min (b) 3 min 17. Time estimate of a PERT activity are given below.
(c) 3.2 min (d) 4 min Pessimistic time = 14 min, most likely time = 10 min,
(lV) Expected time, a customer spend in the queue optimistic time = 8 min. Then, the expected time of activity
IS
(a) 4 min (b) 3.2 min
(a) 10 min (b) 11 min
(c) 3 min (d) 2 min (c) 10.33 min (d) 12 min
4 {;e #"# E'xetc}y: Mechanical Engineering
.'. Probabilitv = :
5 )e 3.2
96x96
'36 *120x24 =
pGr-I)
(III) (a) Expected time per customer in system
ll.(b) p=l-lo-2
' u 15 3 1
2/3 - -l
il20-961- 24 "
u
and L" = --P- = U-).
' 1-p t-213 =z 60
Lqsu33
=L-
I = )-Z 4 =
-24 =2.5min
= = 1.33
(rv) (d) Expected time, a customer spend in queue
I 7"96
12. (b) -Isq=w -w I
=60-40=20
p(u-i.) 120x24- 30 "
i = 2min
l, = 20 per minute (V) (c) Cashier busy in 1 day shift
60
= 3perhour = ;(,-1)=1
i= \ u/ u
lndustrial Engineering $ 805
14. A shortest route is a route in network which length is less 17. rc) r
D =
14 min. l=m l0 min, /o=8min
than other route of the network. We know that
tp+4tm+to
_
te6-6 _ 14+4x10+g
62
: = 10.33min
-6
18. (a)
a
b
C
I -+4 -+7 -+ 8= 5 +'7 +3 = 15 km The paths and {uration are given below.
(D (c) Route of minimum length is
a-+c-+e-->g- 3+5+7+2 = 17days
I->2-->5-r7+8. a-->c-+f-+h =3+ 5+5+10 =23days
(ID (b) Minimum length of route is 1l km. b -+ d ) e -) g=4 +5 + 7 + 2 = 18 days
I.5. (c) b -s d + f ) h = 4 + 5 + 5 + 10= 24 days
Hence, critical path is
b -+ d -sf--s h
le. (d) First to all, draw the network diagram
\
lndustrial Engineering ft tOZ
unit Exercise 1
(1 MarkQuestions) 6. Given that Q = procurement cost per order,
D = number of units demanded per year,
1. Match List I with List II and select the correct answer
/1 = holding cost per unit year, i - rate of interest,
using the codes given below the lists.
P = purchase price per unit. The proeurement quantity
per order (Q) is given by
P. Decision making under 2.0.D
complete certainity (a) Q= (b) Q=
H +iP
Q. Decision making under 2. Maximum criterion
risk 2.0
R. Decision making under Transportation model (c) Q= (d) Q= D(H +iP)
complete uncertainity
S. Decision makingbasedon Decision tree 7. Consider the following statements:
expert opinion The assignment problem is seen to be the special case of
Codes the transportation problem in which
Pa R S I:tm = n
(a)3 4 1 2 2. all a.= |
(b)4 3 2 1 3. x..= I
(c)3 4 2 1
u
(The symbols have the usual meaning.)
(d)4 3 1 2
Which of these statements are correct?
, In exponential smoothing method of forecasting, the (a) 1,2 and3 (b) 1 and 2
forecast for higher values of the smoothing constant
(c) 2 and3 (d) 1 and 3
(a) will be more sensitive to forecast of the previous period
(b) will be more sensitive to changing patterns in demand 8. Ifthe fixed cost ofthe assets for a given period doubles,
(c) will not be affected by the forecast of the previous then how much will the break even quantity becomes?
period (a) Half the original value
(d) will notbe affectedbythechanging patterns in demand (b) Same as the original value
(c) Twice the original value
3. Which one of the following statements is not correct for (d) Four time the original value
the exponential smoothing method of demand
forecasting? 9. While solving a linear programming problem by simplex
'
method, if all ratios of the right hand side (b,) to the
(a) Demand for the most recent data is given more
coefficient, in the key rcw (a) becomes negative, then
weightage
the problem has which of the following qpes of solution?
(b) This method requires only the current demand and (a) An unbound solution
forecast demand (b) Multipte solutions
(c) This method assigns weight to all the previous data (c) Aunique solution
(d) This method gives equal weightage to all the periods (d) No solution r ,.
4. In a transportation problem, the matbrials are transported 10. Consider the following statements:
from 3 plants to 5 warehouses. The basic feasible solution 1. For the application of optimality test in case of
must contain exactly, which one of the following allocated transportation model, thenumber of allocations should
cells? be equal tom+ n, where misthe number of rows and
(a) 3 (b) s n is the number of columns of the matrix.
(c) 7 (d) 8 2. Transportation problem is a special-case of a linear
programming problem.
5. A production line is said to be balanced when
(a) there are equal number of machines at each work 3. In case of assignment problem, the first step is to make
a square matrix by adding a dummy row or a dummy
station
(b) there are equal number of operators of each work column.
station Which of these statements is/are correct?
(c) the waiting time for service at each station is the same (a) I,2 and 3 (b) 1 and 2
(d) the operation time at each station is the same (c) 2 and 3 (d) Only 2
808 * {;,4{# {rs{r*rzMechanical Engineering
11. Match List I (OR-technique) with List II (Model) and 1,6. Match List I (OR-technique) with List II (Application)
select the correct answe.r using the codes given below the and select the correct answer using the codes given below
lists. the lists.
The break even point increases 17. For.,a M/}y'/I: -/FCFS queue, the mean arrival rate
f. ifthe fixed cost per unit increases. is equal to 10 per hour and the mean service rate is
2. if the variable cost per unit decreases. 15 per hour. The expected queue length is
3. ifthe selling price per unit decreases. (a) 1.33 (b) l.s3
Which of these statements is/are correct? (c) 2.75 (d) 3.20
(a) Only 1 (b) 1 and 2 18. Match List I (Methods) with List II (Applications) and
(c) 2 and3 (d) I and 3 select the correct answer using the codes given below the
13. A single bay car wash with a Poisson arrival rate and lists.
exponential service time has cars arriving at an average ,EffiI
rate of
10 min apart and an average service time of
Di#,It
P. Break even analysis l. To provide dillerent
4 min. What is the system utilization?
(a) 1.00 facilities at different
(b) 0.67
(c) 0.40 locations
(d) 0.24
Q. Transportation problem 2. To take action from
14. Which one of the following statement is not correct? among the paths with
(a) Assignment model is a special casq of a linear uncertaintiy
programming problem R. Assignment problem 3. To choose between
(b) In queueing models, Poisson arrivals and exponential
different methods of
services are assumed
(c) In transportation manufacture
problems, the non-square matrix is
made square by adding a dummy row or a dummy
S. Decision tree 4. Todeterminethe location
column ofthe additional plant
(d) In linear programming problems, dual of a dual is a
Codes
primal
P AR S
15. Consider the following statements: (a)4 3 1 2
1. A linear programming problem with three variables (b)3 41 2
and two constraints can be solved by graphical (c)3 4 2 1
method. (d)4 3 2 I
2. For solutions of a linear programming problem with
19. In the solution of a linear programming problem
mixed constraints, Big-M method can be employed. by
simplex method, if during an iteration, all ratios of right
3. In the solution process of a linear programming
hand side b,to the coeffrcient of entering variable a are
problem using Big-M method, when an artificial
found to be negative, it implies that the problem has
variable leaves the basis, the column of the artificial
(a) infinite number of solutions
variable can be removed from all subsequent tables.
(b) infeasible solution
Which of these statements are correct?
(a) 1,2 and3 (c) degeneracy
(b) I and2
(c) i and 3 (d) 2 and 3 (d) unbound solution
lndustrial Engineering i 809
20. Consider the following statements regarding the 27. Which one of the following is an inventory system
characteristics of the standard form of a linear that keeps a running record of the amount in storage
programming problem. and replenishes the stock when it drops to a certain
1. All the constraints are expressed in the form of level by ordering a fixed quantity?
equations. (a) EOQ (b) Periodic
2. The right hand side of each constraint equation is (c) Peripheral (d) ABC
non-negative.
28. The critical path of a network is the path that
3. A1l the decision variables are non-negative.
(a) takes the shortest time
Which of these statements are correct? (b) takes the longest time
(a) 1, 2 and 3 (b) I and2 (c) has the minimum variance
(c) 2 and 3 (d) 1 and 3 (d) has the maximum variance
21. Routing in production planning and control refers to the 29. Customers arrive at a counter randomly at the rate of
(a) balancing of, toad on machines 6 customers per hour. The service is provided at the
(b) authorization of work to be performed counter by a server. The mean time of the service is 4
(c) progress of work pbrformed min per customer. The services are exponentially
(d) sequence of operations to be performed distributed. What is theprobabilitythat anewly anived
'tustomer has to wait?
)', Dummy activities are used in a network to
(a) 0.4 (b) 0.6
(a) facilitate computation of slacks
(c) 0.66 (d) 0.8
(b) satisfy precedance requirements
(c) determine project completion time 30. Which one of the following statements is correct?
(d) avoid use ofresources Queueing theory is applied best in situation where
(a) arrival rate ofcustomers equal to service rate
?1 Which one of the following is true in respect of production (b) average service time is greater than average
control for continuous or assembly line production?
arrival time
(a) Control is achieved by PERI network (c) there is only one channel of arrival at random
(b) Johnson algorithm is used for sequencing and the service time is constant
(c) Control is on one work centre only (d) the arrival and service rate cannot be analyzed
(d) Control is on flow of identical components through through any standard statistical distribution
several operations
31. PERT considers the following time estimates:
24. Which one of the following is assumed for timing the 1. Optimistic time
activities in PERT network?
2. Pessimistic time
(a) cr distribution
3. Most likely time
(b) B distribution
Which of these statements are correct?
1c1 Binomial distribution (.a) 1,2 and3 (b) 1 and 2
(d) Erlangian distribution (d)
(c) Only 3 1 and 3
,,8
Thethreetime estimates of aPERI activityare; optimistic
32. Which one of the following statements is not correct?
time = 8 min, most likely time = 10 min and pessimistic
(a) PERT is probabilistic and CPM is deterministic
time : 14 min. The expected time of the activity would be
(b) In PERT, events are used and in CPM, activities
(a) 10.00 min (b) 10.33 min
are used
(c) 10.66 min (d) 11.00 min
(c) In CPM, the probability to complete the project
26. In a network, what is total float equal to? in a given time duration is calculated
(u) LFI - EST, + r,_,
33. What is the additional time available for the
(b) ESl -LET,+ t" performance of an activity in PERT and CPM
(c) ES, -LIit,- t,_, calculated on the basis that all activities will start at
(d) Lry-EST,-r,, their earliest start time, called?
where, LFT = latest finish time of an activity; (a) Slack (b) Total float
EST = earlieststart timeof an activity; 4_j=time ofactivity (c) Free float (d) Independentfloat
i - j.
-f'w|r*r;
810 E {;4 lti Mechanical Engineering
34. In case of solution of a two variables, linear programming 39. Which one of the following is correct?
problem by graphical method, one constraint line comes
In the basic EOQ model, if lead time increases from 5 to
parallel to the objective function line. Which one of the 10 days, the EOQ will
following is correct?
(a) double
The problem vvill have
(a) infeasible solurion (b) decrease by a factor oftwo
(b) unbounded solution (c) remain the same
(c) degenerate solution (d) Data is insufficienr
(d) infinite number of optimal solution 40. The solution in a transportation model (of dimension
35. Match List I (Techniques/methods) with List II (Models) m x n) is said to be degenerate, if it has
and select the correct answer using the codes given below (a) exactly (m + n - 1) allocations
the iists. (b) fewer than (m + n - l) allocations
Li*tII (c) more than (m + n - 7) allocations
P. Vogel's approximation 1. Assignment model (d) (.m x n) allocation
method 41. Economic order quantity is the quantity at which the cost
Q. Floods technique 2. Transportation model ofcarrying is
R. Two phase method 3. PERT and CPM (a).".minimum
S. Crashing 4. Linearprogramming (b) equal to the cost ofordering
(c) less than the cost ofordering
Codes
P (d) cost of overstocking
AR S
(a)3 4 I 2 42. A shop owner with an annual constant demand of A unit
(b)2 t 4 3 has ordering costs of { P per order and carrying costs
(c)3 I 4 2 { l per unit per year. The economic order quantity for a
(d)2 4 1 J purchasing model having no shortage may be determined
36. In CPM, the project iluration can be reduced by from
crashing
24P D4AP
(a) one or more non-critical activities ft)
(b) one or more critical activities V1
(c) one or more dummy activities
(d) activities having independent float (c)\i,
EAP
(dr ^t_ m
!P
37 . ln the inventory control, if the yearly demand for a certain a_
material is fixed, the economic order quantity gives +J' If the annual demand of an item becomes half, ordering
minimum cost doub :, holding cost one-fourth and the unit cost
(a) inventory carrying cosl. per year twice, then what is the ratio of the new EOe and the
(b) acquisition cost per year earlier EOQ?
(c) total cost per year I
(d) number oforders per year (at (b) I-
2 {r tl2
38. If D = annual demand for a material (units per year) (c) (d)
xD 2
Q = quantity ofmaterial ordered at each order point
(unit per order) 44. If critical path of a project is 20 months with a standard
C = cost of carrying one unit in inventory for one deviation 4 months, what is the probability that the project
year (rupees per unit per year) will be complet I in 24 months?
S = average cost of completing an order for a material (a) 15.85% (b) 68.37o
(rupees per order) (c) 84.27o (d) 95.50E"
TSC = total annual stocking costs for a material (rupees
per year) 45. If the earliest starting time for an activity is g weeks, the
Then, the Economic Order euantity (EOe) is latest finish time is 37 weeks and the duration time of the
activity is 11 weeks, then the total float is equal to
(a)
EDS
t/ c (b)2DC
s
(a) 18 weeks (b) 14 weeks
(c) 56 weeks (d) 40 weeks
EDC
(c)
Vs (d) 2DS
c
lndustrial Engineering X 811
46. The variance of the completion time for a project is the 54. In the EOQ mode, if the unit ordering cost is doubled,
sum of variance of the EOQ
(a) all activity time (a) is halved
(b) non-critical activity time (b) is doubled
(c) critical activity time (c) increases 1.414 times
(d) decreases 1.414 times
(d) activity times of first and last activities of the project
The inter-arrival time at a tool crib are dxponential with an
47. The indirect cost of a plant is T 400000 per year. The
average time of 10 min and the length of the service time
direct cost is t 20 per product. Ifthe average revenue per
6 min assumed to be exponential. The probability that a
product is ( 60, the break even point is
person arriving at the booth will have to wait, is equal to
(a) 10000 product (b) 20000 product
(a) 0.15 (b) 0.40
(c) 40000 product (d) 60000 product
(d)
(c) 0.42 0.6
48. In order, for a transportation matrix which has six rows
and four columns not to degenerate, what is the number 56. The symbol used for transport in work study is
of occupied cells in the matrix? IGATE 20031
62. Consider a single server queueing model with poission A schedule that minimizes the total inventory cost is
arrivals ()u = 4lh) and exponential service (p = 4/h). Ttre
number in the system is restricted to a maximum of 10.
(a) T-+S -+Q -sR -+p (b) p+R __>S _+e__+T
(c) T.-+R +,S + e -+p (d) p -+ e -+R-+S_+T
The probability that a person who comes in leaves without
joining the queue is 67, The expected time (/,) of a pERI activity in terms of
IGATE 200s1
(a) tltl (b) t/10 optimistic time (r,), pessimistic time (rr) and most likely
(c) 1/9 (d) time (/,) is given by , TGATE 2009I
1/2
63. The number of customers arriving at a railwayreservation to+4tt +t, to+4t,,+tt
'e6 (bl t"=
counter is Poisson distributed with an arrival rate of ---
8 customers per hour. The reservation clerk at this counter to+4tt+tD to+4tD+tt
takes 6 min per customer on an average with an tL, I
ea- 'el
J J
exponentially distributed service time. The average
number of the customers in the queue will be 68. In a CNC program block, N002 G02 G91 X4OZ4O ...,
IGATE 20061
G02 and G91 refer to TGATE 20101
(a) 3 (b) (a) circular interpolation in counterclockwise direction
3.2
(c) 4
and incremental dimension
@) 4.2
(t) circular interpolation in counterclockwise direction
64, In an MRP system, component demand is '
hnd absolute dimension
IGATE 20061 (c) circular interpolation in clockwise direction and
(a) forecasted incremental dimension
(b) established by the master production schedule (d) circular interpolation in clockwise direction and
(c) calculated by the MRp system from the master absolute dimension
production schedule
69- The demand and forecast for February are 12000 and
(d) ignored
10275 respectively. Using single exponential smoothening
65. In an M/M/l queueing system, the number of arrivals in method (smoothening coefftcient = 0.25), forecast for the
an interval oflength Zis a poisson random variable (i.e., month of March is IGATE 20101
the probability of there being n arrivals in an interval of (a) 431 (b) 9s87
-),7,a,
e-"' ()tT)n (c) 10706 (d) 11000
"nl ris
length r.The probabilitydensity funcrion
70. Littile's law is a relarionship between IGATE 20101
/(r) of the inter-arrival time is given by TGATE 200g1 (a) stock level and lead time in an inventory system
(b) waiting time and length of the queue in a queueing
(a) t"2 (r-^'\ (b) t -),2r
---- system
)t'
(c) number of machines and jobs due dates in a
e scheduling problem
(c) ?,' e-)"r (d) (d) uncertainty in the activity time and project completion
)"
time
66. A set of jobs is to be processed on a single machine. The
5
processing time (in days) is given in the table below. The 71. Vehicle manufacturing assembly line is an example of
holding cost for each job is { K per day. TGATE 20031 (a) product layout (b) process layout
(c) manual layout (d) fixed layour
P IGATE 20101
72. Simplex method of solving linear programming problem uses
O
(a) all the points in the feasible region TGATE 20101
R (b) only the corner points ofthe feasible region
s (c) intermediate points within the infeasible region
T (d) only the interior points in the feasible region
lndustrial Engineering n 813
Unit Exercise 2
(2 Marks Quesfions) 7. Consider the following sets of tasks to complete the
assembly of an engineering component:
1. For a small scale industry, the fixed cost per month is
< 5000. The variable cost per product is {
20 and sales price
is { 30 per piece. The break even production per month will
be
(a) 300 (b) 460
(c) 500 (d) 10000
2. A new facilityhas to be designed to do all the welding for
3 products A, B and C. Per unit wglding time for each
product is 20 s, 40 s and 50 s respectivelyr. Daily demand
forecast for product A is 450, for B is 360 and for C is
240. A welding line can operate efficiently for 220 min
in a day. Number of welding lines required are The expected production rate is 3000 units per shift of
(a) 5 (b) 4 8 h duration. The minimum number of workstations that
(c) 3 (d) 2 are needed to achieve this production level is
(a) 4 (b) 8
3. At a self-service store, a cashier can serve 10 customers
(c) 10 (d) 11
in 5 min. On an average, 15 customers arrive every
10 min. If the arrivals are as per Poisson distribution and 8. In the network shown below, the critical path is along
services as per exponential distribution, the probability
that the cashier would be idle is
(a) 0.5 (b) 0.7s I\
t\
(c) 0.25 (d) zero t\
t,
D,
4. Which one of the following information combinations I
I
11. A company intends to use exponentiai smoothing (a) 26 days (b) 27 days
technique for making a forecast for one of its products. (c) 30 days (d) indeterminable
The previous year's forecast has been 7g units and the
actual demand for the corresponding period turned out 17. Match List I (OR-technique) with Lisr II (Model) and
to be 73 units. If the value of the smoothing constant cr, is select the correct answer using the codes given below the
0.2, the tbrecast fbr the next period will be lists.
(a) 73 units (b) 75 units
(c) 77 units (d) 78 units Branch and bound technique PERI and CPM
Expected yalue approach Integer
t2. Time estimates of an activity in a pERT network
are; optimistic time to = 9 days; pessimistic time programming
Smoothing and Levelling
tp= 2l days and rnost likely time t* = 15 days. The Queueing theory
Exponential distribution Decision theory
approximate probability of completion of this activity
in 13 days is Codes
(a) 16To lb) 34Io PaRS
(c') 50c'/o (d) 84Vo
(a)2 t 4 3
(b)2 413
13. A dealer sells a radio set at { 900 and makes 807o profit (c)3 4 1 2
on his investment. If he can sell it at { 200 more. his (d).3 1 4 2
profit as percentage of investment will be
(a) 160 (b) 18. Process I requires 20 units of fixed cost and 3 units of
180
variable costs per piece, while process II required 50 units
(c) 100 (d) 120 offixed costs and I unit ofvariable cost per piece. For a
t4. Estimated time T" and variance of the activities Von the company producing 10 pieces per day
critical path in a PERTnetwork are given in the following (a) process I should be chosen
table: (b) process II should be chosen
(c) either ofthe two processes should be chosen
(d) a combination ofthe process I and process II should
be chosen
't) Consider the below network. Activity times are given in 28. Based on the given graph, the economic range of
number of day. The earliest expected occurrence time (/r) batch sizes to be preferred for General Purpose (GP)
for event 50 is machine, NC machine and Special Purpose (SP)
machine will be
NC
---sP
o)
E
'iA
ll
Er
f,o
(6
Batch size +
Codes
(a) 22 (b) 23
(c) 24 (d) 2s
GP NC SP
(a)2s 4
23. Using the exponential smoothing method of forecasting, (b)t4 J
what will be the forecast for the fourth week, if the actual (d3 2 4
and forecasted demand for the third week is 480 and 500, (d)14 2
respectively and a= 0.2?
(a) 400 (b) 4e6 29. There are two machines M, and Mrwhich process jobs
(c) 500 (d) s04 A, B, C, D, E and F. The proceSsing sequence for these
jobs is M, followed by Mz. Consider the following data in
24. Consider the 3 activities of a CPM network as shown this regarde:
below Earliest and latest occuffences times of these events Process time required in minutes-
are given on the nodes. The total float on activity 20-30 :.p
Job B: C D
is
11
ES=5 ES=10 ES=18 gg=28
l1 10
32. The earliest time of the completion of the last event in c -+ d = 4 timeunits, d -) e = I time unit. The standard
the below network in weeks is deviation ofthe criticalpath a -+ e is
(a) 3. (b) 4
(c) 5 (d) 6
38. Consider the data given in the following table:
i €g
* fi,.:rllj
I 500 500
2 650 650
33. The demand for ia product in the month of March turned Give the fact that production in regular and overtime is
out to be 20 units against an earlier made forecast of limited to 650 and 150 respectively, the balance demand
20 units. The actual demand for April and May turned to of 100 units in the 4th period can be met by
be 25 and 26 units respectively. What will be the forecast (a) using overtime in period 2
sm,..'hin, (b) using regular production in period 1
:"t,::: T:1': :::11 :::i::::::ilTi (c) subcontracting
(a) 20 units (b) 22 units (d) using any of the steps indicated in A, B and, C
(c) 26 units (d) 28 units
39. Aproduct is manufactured byprocessing on the fourWork
34. Manufacturing a product requires processing on four Station (WS). The capacityof each machine on these work
machines A, B, C:, D in the ofi,er A-+B-+C-+D. The stations is given in the diagram as shown below. In the
capacities of four naachines are A = 1 00, B = 110, C 120
= diagram Ml, M2A, M2B, Mj, MoA and M.,B are the
and D=130 units prer shift. If the expected output isgO%
machines and 500, 2i5,275,560,200 and 200 are their
of the system capar:ity, then what is the expected output? capacities in number of products made per shift. If the
(a) 90 units (b) 99 units products made in this system are 5Vo defective, then what
(c) 108 units (d) 117 units wiil be the output from this system?
t
35. Process X has a fixed cost of 40000 per.month and a
WS,
variable cost of( 9 per unit. Process Ihas a fixed cost of
< 16000 per month and a variable cost of T 24 per unit.
At which value, total cost of processes X and y will be
equal?
(a) 800 (b) 1200
(c) 1600 (d) 2000
36. A furniture company is maintaining a constant work force
which can produce 3000 tables per quarter. The annual
demand is 12000 units and is distributed seasonally in
accordance with the quarterly indexes O, = 0.80,
Qz= I.40, O: = 1.00 and Qo = 0.80. Inveniories are (a) 380 (b) 47s
accumulated when demand is less than the capacity and
(c) 522 (d) s32
are used up during periods of strong demand to supply
the total demand. To take into account any seasonal 40. In a forecasting situation, exponential smoothing with a
demand the inventories on hand at the beginning of the smoothing constant a = 0.2 is to be used. If the demand
first quarter should be at least for nth period is 500 and the actual demand for the
(.a) zero (b) 600 corresponding period turned out to be 450, what is the
(c) 1200 (d) 2400 forecast for the (n + l)th period?
41. A capstan lathe is used to mass produce, in batches of 48. An operations consultantfor an automatic car wash wishes
200, a particular component. The direct material cost is to plan for enough capacity tohandle 60 cars perhour. Each
( 4 per piece, the direct labour cost is { 3 per piece and car will have a wash time of 4 rrrinutes, but there is to be a
the overhead costs are 400Vo of the labour costs. What is 25Vo allowance for setup time, delays and payment
the production cost per piece? transactions. How manycar wash stalls shouldbe installed?
(a) 19t (b) < 23 (a) 3 (b) 4
(c) { 16 (d) < 1s (c) 5 (d) 6
42. Atime standard for a data entry clerk is to be set. A job is 49. Process 1 requires 20 units of fixed cost and 3 units of
rated at 120Vo , it takes 30 s to enter each record and the variable costs per piece, while process II required 50 units
allowances are 15Vo. What is the normal time? offixed costs and 1 unit ofvariable cost per piece. For a
(a) 25 s (b) 30 s company producing 10 pieces per day
(c) 36 s (d) 40 s (a) process I should be chosen
43. The standard time of an operation has been calculated as (b) process II should be chosen
10 min. The worker was rated at SOVo.If the relaxation (c) either ofthe two processes should be chosen
and other allowances were257o, then the observed time (d) a combination ofthe process I and process II should
would be be chosen
(a) 12.5 min (b) 10 min 50. Annual demand for a product costing ( 100 per piece is
(c) 8 min (d) 6.5 min t 900. Ordering cost per order is { 1 00 and inventoryholding
44. Two machines of the same production rate are available cost is { 2 per unit per year. The,economic lot size is
for use. On machine 1, the fixed cost is { 100 and the (a) 200 (b) 300
variable cost is { 2 per piece produced. The corresponding (c) 400 (d) s00
numbers for the machine are { 200 and ( 1 respectively.
51. At a self service store, a cashier can serve 10 customers in
For certain strategic reasons both the machines are to be
5 min. On an average 15 custormers arrive every 10 min. If
used concurrently. The sale price ofthe first 800 units is
the arrivalsare as per Poisson ,distribution and services as
< 3.50 per unit and subsequently it is only t 3.00. The
per exponential distribution, threprobability that the cashier
break even production rate for each machine is
would be idle is
IGATE 20031
(a) 0.50 (b) 0.7s
(a) 75 (b) 100 (c) 0.25 (d) None of these
(c) 150 (d) 600
\,, The sale of cycles in a shop in f,our consecutive months are
45. A residential school stipulates the study hours as
given as 70,68, 82,95. Exponrontiallysmoothing average
8.00 pm to 10.30 pm. Warden makes random checks on
method with a smoothing factor of 0.4 is usedin forecasting.
a certain student 1 1 occasions a day during the study hours
The expected number of sales in the next month is
over a period of 10 days and observes that he is studying
on 71 occasions. Using 95%o confrdence interval, the IGATE 20031
(a) -
1
1
6
day (b) ; day
J
The project can be completed IGAT'E 20031
(a) between 18, 19 days (b) between 20,22days (c) I day (d) 3 days
(c) between 24,26 days (d) between 60,70 days
60. A company has an annual demand of 1000 units, ordering
56. A standard machine, tool and an automatic machine tool cost of < 100/ order and carrying cost of{ 100/unit. If
are being compared for the production of a component. the stock out costs are estimated to be nearly { 400 each
Following data refersi to the two machines: time the company runs out-of-stock, the safety stock
justified by the carrying cost will be
$tr[qdaid-, Artomh{ic,
,,meffiinertool maehinetool IGATE 20041
Setup time 30 min 2h {a) 4 (b) 20
Machining ti 22 mtn 5 min
(c) 40 (d) 100
per piece 61. A companyproduces two types of toys p and
e. production
Machine rate { 2t00 per hour t 800 hour time of Q is twice that of p and the company has a maximum
The break even prod,uction batch size above which the of 2000 time units per day. The supply of raw material is
automatic machine t,ool will be economical to use. will just sufficient to procluce I 500 toys (of any type) per
day.
be Toy tlpe Q requires an electric switch which is available
IGATE 20041
(a) 4 (b) s @_600 pieces per day only. The company makes a profit
(c) 24 (d) 22s of t 3 and { 5 on gpe P and
e respectively. For maximi zation
ofprofits, the dailyproduction quantities ofp and toys
5/- A soldering operation was Lvork sampled over two rlays e
should respectively be IGATE 20041
16 h during which an r:ruployee soidered 10g joints. Actual
(a) 1000,500 (b) s00,1000
working time was 90% of the total time and the
performance rating
(c) 800,600 (d) 1000,1000
estin-rated to be 120 Vo" If the
r7yn5
contract provides allov,zance of 2AVo of the iotai time 62. A manufacturing shop processs sheet metal jobs, wherein
available, the standard time fbr the operaticn would be each job mlrst pass through two machines (M, and M,in
IGATE 20041 that order). The processing time (in hour) for these jobs
(a) E min (b) 8.9 min is
(c) 10 min (d) 12 min
Madriae, .Iobs r,'..:
58. An electronic
equipiment manufacturer has decided to add
a component sub-a.ssembly operation that can produce -F ) ft s'. T U
80 units during a regular 8 h shift. This operation consists M1 15 32 8 27 11 t6
ofthree activities a.s below M2 6 19 13 20 t4 1
Activity Standard time (in min) The optirnal make span (in hour) for these jobs is
M. Mechanical assernbly t2
IGATE 20061
E. Electric wiring t6 (a) 120 (b) 11s
L Test 3 (c) 109 (d) 79
lndustrial Engineering I 819
The optimum order quantity is IGATE 20061 68. A stockist wishes to optimize the number of perishable
(a') 447 (b) 47r items he needs to stock in any month in his store. The
(c) 500 (d) > 600 demand distribution for this perishable item is
64. A component can be produced by any of the four processes, Demand (in unit)
I, [, ru and IV. Procpss I has fixed cost of { 20 and Prohahilitv 0.10 0.35 0.35 0.20
variable cost of{ 3 per piece. Process II has a fixed cost
The stockist pays { 70 for each item and he sells each
of{ 50 and variable cost of{ 1 per piece. Process III has af( 90. lf the stock is left unsold in any month, he can
a fixed cost of <40.00 and variable cost of{ 2 per piece.
sell the item at ( 50 each. There is no penalty for
Process IV has fixed cost of t 10 and variable cost { 4
unfulfilled demand. To maximize the expected profit, the
per piece. If company wishes to produce 100 pieces of
optimal stock level is IGATE 20061
the component, from econoinic point of view it should (a) 5 units (b) 4 units
choose IGATE 200sI (c) 3 units (d) 2 units
(a) process I (b) process II
(c) process III (d) process IV 69. The table gives details of an assembly line.
.Iry:sf,&,ltilitiifi ,::i:i,: ri:; f -::: :i:IIl ,rtrIE:,. ;,,..Y1, ,rtlV,I
65. A welding operation is time studied during which an t
operator was pace rated as l207o.The operator took, on Total task time at
an average, 8 min for producing the weld-joint. If a total the work station 1 9 10 9 6
of l07a allowances are allowed for this operation. The (in minute)
expected standard production rate of the weld-joint (in What is the line efficiency of the assembly line?
unit per 8 days) is TGATE 200s1 IGATE 2006I
(a) 45 (b) s0 (a) 70Vo (b) 75Vo
(c) 55 (d) 60 (c) 80Vo (d) 857c
66, A company has two factories S,, S, and two warehouses 70. The maximum level of inventory of an item is 100 and it
D,., Dr.. The supplies from S, and S, are 50 and 40 units is achieved with infinite replenishment rate. The inventory
respectively. Warehouse D, requires a minimum of becomes zero over one and half month due to consumption
20 units and a maximum of 40 units. Warehouse D, at a uniform rate. This cycle continues throughout the
requires a minimum of 20 units and, over and above, it year. Ordering cost is t 100 per order and inventory
can take as much as can be supplied. A balanced carrying cost is ( 10 per*iem per month. Annual cost (in
transportation problem is to be fornlulated for the above { ) of the plan, neglecting material cost, is
situation. The number of supply points, the number of IGATE 20071
demand points, and the total suppiy (or total demand) in (a) 800 (b) 2800
the balanced transportation problem respectively are (c) 4800 (d) 6800
IGATE 200s1 7t- In a machine shop, pins of 15 mm diameter are produced
(a) 2, 4,90 (b) 2,4, trc at a rate of 1000 per month and the same is consumed at
(c) 3,4,90 (d) 3, 4, tto a rate of 500 per month. The production and consumption
continue simultaneously till the maximum inventory is
67. A firm is required to procure three items (P, Q and R).
reached. Then inventory is allowed to reduce to zero due
The prices quoted for these items (in {) by suppliers S,,
to consumption. The lot size of production is 1000. If
S, and S. are given in table. The management policy
backlog is not allowed, the maximum inventory level is
requires that each item has to be supplied by only one
gupplier and one supplier supply only one item. The IGATE 20071
(a) 400 (b) s00
minimum total cost (in t) of procurement to the firm is
(c) 600 (d) 700
820 il r;,*,'! #. {s.sg{*Ft Mechanical Engineering
of the shortest path from node p to node G. Let d,,be the xr) O,xr)0 IGATE 20091
length ofdirected arc from node i to node l.
(a) The LPP has a unique oprimal solution
(b) The LPP is infeasible
(c) The LPPis unbounded
(d) The LPP has multiple oprimal solution
79. Six jobs arrived in a sequence as given below.
I 4
II 9
Let ,1 be the length of the shortest path from p to node7. m 5
Which of the following equations can be used to find S"? IV 10
IGATE 20081 V 6
(a) So = Min {S, So} VI 8
(b) Sc = Min {S, _ d4s, Sn_ d*o}
(c) Sc = Min {Sn * deo, So + doo} Average flow time (in days) for the above jobs using
Shortest Processing Time lSpT) rule is
(d) Sc =Min {dq6, dpc}
IGATE 20091
76. The product structure of an assembly p is shown in the (a) 20.83 (b) 23.16
figure. (c) 125.00 (d) r39.00
lndustrial Engineering * tZt
80. A box contains 2 washers, 3 nuts and 4 bolts. Items are 82. The project activities, precedence relationships and
drawn from the box at random one at a time without durations are described in the table. The critical path of
replacement. The probability of drawing 2 washers first the project is IGATE 20101
followed by 3 nuts and subsequently the 5 bolts is
IGATE 201.01
(a) 21315 (b) 1/630
(c) tlt260 (d) 1t2s20
8L. Annual demand for window frames is 10000. Each frame
costs { 200 and ordering cost is { 300 per order. Inventory
holding cost is (
40 per frame per year. The supplier is
willing to offer 27o discotnt if the order quantity is 1000
or more and 4Vo if order quantity is 2000 or more. If the
total cost is to be minimized, the retailer should
IGATE 20101
(a) order 200 frames every time
(b) accept 2Vo discourt (a) P -+R -+T -+V (b) 0 -+S -+T ->V
(c) accept 47o discount (c) P -+R -+U -+W (d) Q -+S -+ U -+W
(d) order Economic Order Quantity (EOQ)
Jpn 2010 80
Expected sales per year - 2500, lead time : 3 days,
Feb 2010 65
number of working days = 250,
March 2010 90
Ordering cost = { 12.50/order, holding cost = 20Vo of April2010 10
average inventory
May 2010 80
Lot size (units) price per unit (t) June 2010 100
I to 259 4.00 July 2010 85
August 2010
260 to999 3.00 60
ber 2010 75
1000 and above 2.00
15. Forecasted value of a car in September month, if using
9. Optimum order size is
3 months moving average method is
. (a) 1000 (b) 300 (a) 81.66 (b) 83.33
(c) 250 (d) 280 (c) 83.34 (d) 7s
10. Reorder level is 16. Forecasted value for October month, if o = 0.2 is
(a) 50 units (b) 40 units ' (a) 80.33 or 81 (b) 75
(c) 30 unirs (d) 20 units (r) 78 (d) 7e
Statements for Linked Answer euestions 11 and 12 Common Data for Questions 17 to 19
.A news paper boy buys paper for { 1.40 and sell them In a machine production shop breakdown at an average
for { 2.45 each. He cannot return the unsold papers. 2 per hour. The non-productive time of any machine cosls
Daily demand has the following distribution: { 30 per hour. If cost of repair man is { 5b per hour and
repair rate is { 3 per hour.
Probabiiitr, 0.0 1.05 0.050 0. l0 0.15 0. i5 a.t2 0.10 0.10 o.o7 0.0( 0.0t
Demand 25 26 27 28 29 30 31 32 33 34 35 3(
17. Number of machine not working at any polnt of tirne is
(number of (a) 3 (b) 2
news paper)
(c) 4 (d) s
11. How many papers he should order per day 18. Average time that a machine is waiting to the repairman
(a) 30 is
(b) 2e (a) 7t3
(c) 28 (d) 27 (b) 2
(c) 2/3 (d) 4t3
12. How much he saved per day (if order optimum numbers
19. Expected cost ofthe system per hour is
of paper)
(a) { 110 (b) { 120
(a) { 29.50 (b) t 31.50 i (c) t 115 (d) < 10s
(c) t 30.50 (d) < 32.s0
Statements for Linked Answer euestions 2O and2l
or Answer Questions 13 a bre_ak even analysi fbr one type ofproduct is given below
In a machine shop, certain type of machines breakdown at {
Fixedcost=t10000C0
s
an average rate of 6 per hour. The breakdowns are Variable cost : { l0 per unit
accordance with Poission process. The estimated cost of idle Revenue = { 15 per unit
machine is { 5per hour. Two repairmen A and B with
different skills are being hired as repairmen. Repairmen A 20. The break even quanrity is
takes 6 min on an average to repair a machine and his (a) 300000 '"(b) 150000
wages are t
8 per hour whereas the repairmhn B takes fire (c) 225000 (d) 200000
minutes to repair and the are { 10 per hour. 21. Profitability for 300000 units is
13. What is cost/per day when service taken from repairman (a) t 500000 (b) { 600000
A (c) t a00000 (d) < 300000
(a) 1244 (b) < 248 Statements for Linked Answer euestions 22 and 23
(c) T 236 (d) < 230 A company wants to launch a new product. The fixed cost
14. Which repairman's service should be used? of the new product is t 40000 and variable cost per unit is
< 1000. The revenue function for the sales of x uniis is given
(a) repairman A (b) repairman B by 10000 r- 500 -r2
(c) Cannot predict (d) data is insuffrcient
)) The profit function is
Common Data for Questions 15 and 16 (a) -50012 + 9000x - 40000
A dealership for honda city car sells a particular model of (b) -50x2 + 900x * 4000
the car in various months of the year. (c) -50012 +1200x - 40000
(d) -50x2 + 9000r- 36000
lndustrial Engineering \ 823
23. For break even point, the value ofx is/are ommon Data for Question 30 and 31
(a) 8 (b) 10 Present the following activities in the form of a PERT
(c) Both (a) and (b) (d) Neither (a) nor (b) networkl
4 l--1 1 1 4
Job number i 1 2 J 5
2-4 ti 16 t2
MachineA 5 1 6 9 5
3-5 J 1 5
Machine B 2 1 4 5 J
4-5 0 0 0
Machine C 3 7 5 6 7 4-6 J 9 6
5-1 -l 9 6
24. The sequence that minimized the [otal elapsed time is
(b) 2-+5-+3-s4-+1 5-8 4 8 6
(a) 5-+4+3-+2-+l t2
(c) 2-->5-+4-+3-+1 (d) l-->3->4'+2-+5 7-9 4 8
8-9 2 8 5
25. Total idle time lor machine C is 9.10 4 16 10
(a) 18 h (b) 13 h 6-10 4 8 6
(c) l2 h (d) 11 h
30. Project completion time is
Common Data for Questions 26 and27 (a) 44 days (b) 48 days
Four technicians are required to do four differentjobs. (c) 46 days (d) 42 days
Tbchnieian Hour, to, corrrpl*ts iob 31. The standard deviation of the project is
Job l Joli,2 Joh 3 Jotr 4 (a) 3.12 (b) 2.8
A 20 36 3l 21 (c) 3.8 (d) 4.2
at 45 22 Statements for Linked Answer Questions 32 and 33
B L+ 34
C 45 38 t8 Consider a linear programming problem with two variables
^!-z
and two constraints. The objective lunction is maximize
D 31 40 35 28
xtx xz. The corner points ofthe feasible region are (0, 0),
Job 4 is given to technicran (0,2) (2,0) and (4/3,4/3).
(a) B (b) A
(c) c (d) D 32. If an additional constraint X, + Xr< 5 is added the optimal
solution is
27. Total work time is (a) (5/3, s/3) (b) (4t3,4t3)
(1) 103 h (2) 96 h'
(.c') (512,512'S (d) (s, 0)
(3) 103 h (4) r07 h
Let I and Y, be the decision variable ofthe dual u, and
Statements for Linked Answer Questions 28 and 29 y. be the slack variables of the dual of the given linear
A small project has six activities.The time duration and piogramming problem. The optimum dual variables are
immediate predecessor is given as below. IGATE 200s1
(a) Y, and Y, (b) Yr and V,
(c) YrandV, 4i (d) Yr and V,
5
Statements for Linked Answer Questions 34 and 35
6 Consider a PERI network for a project involving six tasks
5 (a to/):
4 ?ask Fredectsscr Expected ta*ktime Yaria*ceof the tr
-) ' {i* da.ys} l
,'timb {in daysi
4 a 30 25
b a 40 64
The critical path is C a 60 81
29. Project duration is 34. The expected completion time of the project is
(a) 20 days (b) 22 days (a) 238 days (b) 224 days
(c) 24 days (d) 18 days (c) 171 days (d) 155 days
X {",4Y# Tas€z+ra Mechanical Engineering
-4 -6 0 0 0
J 3 2 1 0 6
t 2 J 0 i 6
x v .t t RHS
1. (c) 2. (b) 3. (d) 18. (b) le. (d) 20. (a) 21. (d)
,,) (b)
4. (c) Large value of cr is selected for fluctuating demand 23. (b) 24. (b)
while lower value of cx, is selected for smooth demand
pattern tn+4t-+t, 8+4x10+14
2s. (b)
3 plants ------+ 5 warehouses
"66
m=3 .n= 5 = 10.33 min
Degeneracy occurs when number of allocations are 26. (d) 27. (a) 28. (b)
lessthanm+n-l 29. (a) Arrival rate = ?t = 6
=3+5-l=7 60
(d) 6. (c) 7.(a) Servicerate=!t=T =15
8. (c) Selling costlitem x n = Fixed cost
+ Variable cost/item x n Probabilityto wait = = 0.4
*
Cr. n = Cr+ Cr- n
30. (c) 31. (a) 32. (c) 33. (c)
Here n break even quantity
34. (d) 3s. (b) 36. (b) 37. (b)
,= 'o 38. (a)
C"_C,
If Co becomes twice then n will be twice. 39. (c) Lead time has no effect on EOQ. It affects the reorder
point.
9. (a) 10. (c) 11. (a)
40. (b) 41. (b) 42. (c)
12. (d) We know that
CF EDC.
BEP = 43. (d) EoQ = .l---:--!
'!c,
cs -c,
This relation provides that 1 and 3 are correct. C,
D' = D/2, Co' =ZCo, C r= 4
13. (b) Arrivalrate l.= l0 =6/h
Eoe'=
Servicerate P=T=15/h
60
W=2Eoe
7"6 !:a = 4?
= ts =o'+ 44. (c) Weknow, , = =,
P=
[
System utilization
and 0(1) = fr.842 = 84.2 Vo
880+870+890
3
where,D=annualdemand
C, = ordering cost/order ny
"'pon"r,,",
r*;,i::,ll'li",n'0,
C, = holding or carrying cost/unit per year i's = c[,'.y4+(1 -o")Fo
50. (a) Normal time = Observed timex performancerating factor 880 = cr . 890 + (1 -.q) - 876
Standard time = Normal time + Allowances
a = 2l'7
51. (d) cr = 0.3 62. (a) Given, ?," -
Actual demand = !,= 59
lL=4
= Fr=64
Forecast
Maximum number = 10
Weknowi{+i = 0. }, + (1 - a) F,
)"
= 0.3 x 59 + (1 *O.3) x 64 Trafllcintensity=p= = I
= 62.5
[
10
s2. (b) We know, L P-. =|
n=l)
60 II
(d) Arrival rate= =6person/h Probability that a person who comes in but leaves
-I0
without joining the queue is
60
Service rate = ;6 = I0person/h 1
p=?"tp=*=0.0 'll ll
'10 63. (b) Given, i"=8,F=60/6=10
56. (a) s7. (b)
i"8
58. (c) Forecast F,=25 i
s P = u = lo =0.8
Actual sale !, = 20 Average number of customers in queue is
a=0.2
oz (0.8)2
F,, *t=6y.)r+(l_ 0-)Fn N=l-P=ffi=3''
=0.2x20+(l-0.2)x25 64. (c) 6s. (c)
=4+20=24 66. (a) Job order should be in asCending order ofprocessing
time
I
=-2 I -----+ 2 ----s 3 ------+ 5
7-+,S)Q+R-+p
67. (a) 68. (c)
60. (d)
61. (c)
69.. (c) )p"b = 12000, Fr.o: 1027 5 , a" = 0.25
)1 = 860' 1l2 = 880
FM*.h = O(.)Feb + (1 - o) Fo"o
)3 = 870,.)a=890
= 0.25 x 12000 + (1 - 0.25) x t0275
Fq = 876
Yr+Y,+lr, = 10706
k_-
rs
--)
- . 70. (b) 71. (a) 72. (b)
lndustrial Engineering B tZZ
Un it Exerc ise 2
1. (c) Fixed cost/month 5. (d) As product has to be operated on all these machines
cr'= 5ooo 306
Variable cost of one piece .'. System efficiency = 360 = 0.85 ct 857o
2. (c) A s, B ------+ 40
20 s, C ------+ 50 s
BEp =x =
1f$ = ro
->demand fdrecast for
Daily 10. (a) Capacity = 60 cars per hour = I car per minute
A -+ 450 B -+ 360 C -+ 240 Time required to wash one car is
Time required to weld product A ,..5
20
=2+2>< I00 =2.4min
= lx+so
60
= 150 min
Number of cars washed in I h one stall is
4tJ 60
Similarlv. for B =
"60 _ x360 =240 mjn
2.4
60
and for C 9x240 So, number of stalls = N =2.4 = 3
= 60 = 200 min
11. (c) F, = a . r-n_t * (l - d") F,
Total time required 1
=0.2x73+(l-0.2)x18
= 150 + 240 + 200 = 590 min = 77 units
Number of welding lines required
= 15 days
= 2.68 =3 -'o
Standard deviation = 'n -9
3.
15 66 =21 =2
(c) Arrival rate = -Io =1.5/min=)"
*-i 13*15
z=_=_=_l
Service rate
10
= ;-) :2/min =B o2
andP(-1)=0.1586 (from chart)
-
p: l/[= 1.5
:0.75 = 0.16 or 16%o
^
Probability that cashierlryould be idle is 100x900
1-p=l-0.15:0.25 13. (d) CP: ,oO**. =500
c. 30000 SP ir he *"jt#o'*'33r
4. @)
' C, -C, 40- lo ll1;,
SP-CPxloo
40000 Profit =
BEP^= CP
' 40*15 =1600 11oo-5oo
tsEP-=
5oooo - xtoo =t2Ovo
' 40-20 =2500 14. (c) Standard deviation
500
60000
BEP4=
40_30 =6000
Clearly break even point is lowest for case (1) .
828 g #eX"# E'aa{*yt Mechanical Engineering
J4,r4a1 =3
where, 02 = variance
21. (b) Hour utilized per day= 16 - rc x # = t2h
Hour utilized per week = 12 x 5 = 60 h
7- _
x- x 43-40
z---:.=l |
60x60
6J Number of items produced on machine =
P (1) = 9.314 or 81.4Vo
a
10
15. (c) Ei=S Li=26 7.=8 Tr= 32 = 360
EDC
\/E: :
2x900x100
le. (b) EOQ =
2
29. (a) Find smallest value, min (Mt, M) if Mris smallest .'. Output from system
then allocate on extreme left side.
= 400 x 0.95 = 380
If Mris smallest, then allocate on extremeright side
M1
40. (c) Demand for nft period = 500
Actual demand for nft period = 450
c A B F E D
Forecast for (n + 1)ft period is
30. (b) 31. (d) Fn+t.- Jn +(1-cr).F-n
cx,.v
32. (d) ), = 60/10 = 6, 1t= $913 = 29 = 0.2 x 450 + 1l - 0.2) x 500
Q=),lp=6/20=0.3 = 490
4=8 4= 33 41. (a) Total cost = Material cost + Labour cost
+ Overhead cost
200x400
= 4x200+3 x 200+ 3, ,*
10 T,= 40 Te= 46
^
o = { 3800
3800
Production cost per piece = ZOO
= ttg
f.= 13 f,= 23
42. (c) Normal time = Observed time x Rating factor
Earliest time of completion = 46 weeks
33' (b) ron'
frltut..ti;:H'*,,
=30x 1.2=36s
43. (b) Normal time= Standard time x Rating factor
=
=20
= 10x0.8=8min
Observed time = Normal time
Fru, = 0 )onn, + (1 - o) Foon,
+ Percentage allowance of normal time
=0.2x25+0.8x20 / "5\l=lomin
=21 =81 l+"loo,
FJu," = Cr( . )uay + (1 - Cx,) Fru,
\
= 0.2 x26 + 0.8 x2l 44. (a) Machine (1) Machine (2)
=22 Fixedcost = { 100 Fixedcost={200
34. (a) ForA= 100x0.9=90, for B=110x0.9=99 Variable cost ={ 2 /piece Variable cost = { l/piece
For C= 120 x0.9= 108, for D= 130 x 0.9 =lll Let machine (1) produces n quantities as production
Minimum output is 90 units for machine A. rate of both machine is same.
35. (c) 40000+9x=16000+29x .'. Machine (2) will also produce n quantities. Hence,
BEP,r = 1600 total cost in production
36. (b) Demand in first quarter = 0.8 x 3000 = 2400
= (100 +2xn) + (200 + 1 xn)
Demand in second quarter = 1.40 x 3000 =4200
=300+3n
Demand in
third quarter = 1.00 x 3000 = 3000 Given, sale price = ( 3.50/ unit, if n < 800
Demand in
fourth quarter = 0.80 x 3000 =2400 and { 3.00/ unit, if r > 800
As demand in second quarter :4200 Consider Ist case then at break even point,
So. surplus required for consumption 300+3n=3.50x2n
=
4200 - 3000 = 1200 n = i5
Production in every quarter is 3000 and total demand $.goo)
45. (c) Number of total observation in 10 days
+ surplus required in first quarter
= 2400 + 1200 = 3600 = 11 x 10= 110
So, inventory required at beginning oflst quarter Studying occasions = 71
= - 3600 3000 = 600 71
37. (c) Standard deviation ofcritical path a -+ e is
Probability of studying = 1 10
= 0.6455
Study hours in one day = 2.5 h
= ,[116-.4..,. 1 =5 (8.00 pm to 10:30 pm)
38. (b) For 10 days = 25 h
39. (a) Defective products = 57o Minimum number of hours of studying in 10 days
Percentage of items without any defects = 25 x 0.6455 = 6.1 37 h 1
= 100-5=95Vo 46. (b) Time taken to wash one car = 3 x 1.2 = 3.6 min
Station WSo has minimum number of material Number of cars which can be washed in a stall/hour
= 200 + 200 = 4OO 60
= 36 = 16.67
830 1 {;A"{# Ys*t#{,: Mechanical Engineering
UO
Number of car wash stalls = 16.61 = O
Z =3P+50
mrx Hence, optimum orde
c r quantity = 441
64. (b) Number ofpieces produced = 100
Total cost = Fixed cost + Variable cost
x Number of pieces
TCr=29+3x100=320
TCr= 5g + 1x 100= 150
TCr= 49 +2xlO0=240
TCo:19 +4x 100=410
Process II is corresponding to minimum cost.
63. (a) Annual demand = 2500 units/yr Step 1 Subtract minimum value in each column from
Ordering cost = { lOOiorder all values on that column.
Holding cost = 257o of unit price
"{w{:*rz
832 A #A {* Mechanical Engineering
s1 s2 s3
..=!1{xlox12
2
P
< 6000
O Total annual cost = 800 + 6000 = { 6800
R
71. (b) Maximum inventorylevel = G -a1.9
Step 2 Apply some procedures for rows.
s1 s2 s3 where, P = 1000 per month
Q = 1000, d = 500 per month
P 1000
o = (1000 * 500), ,O* = rOO
72.(d)Week 1 2 3 4
R
Cost
P st t20 76. (c)
O s2 t40 77. (c) Lets see a deterministic model-
R s3 125
Minimum cost
=
= 120+ 140 + 125
={385
68. (b) o
69. (c) Line efficiency
Total time used
- Number of workstation x Cycle time
78. (d) *t14 . .. (i) Total cost : Purchase cost + Holding cost
xz< 6 ... (ii) + Ordering cost
3xr+2xr< 18 .. . (iii) T (C) = Annual demand x Cost of one item
+ Average inventory x Holding cost / item
x2
i + Number of orders x Ordering cost per
I
order
loooot3oo
= loooo x zoo +S1{x+o+
2 387
= 7 2015492
For Q = 1000
I(C) =
98 1000 10000
10000x200x - x4O+ --._ x300
Objective function
{
-+ 2
100 1000
Z=3xr+2x, = 1983000
t-D_
21321 Lowest total cost occurs at Q:)ggg
-X-X_X-X-X
9 816 5
I
Hence he should accept 47o discount.
| / t260 82. (d) For f -------) At junction take greater value
81. (c) Annual demand = 10000 For f <- At junction take lower value
Ordering cost = (
300/ order Critical path = Q -+ S -+ U -+ W
Cost of each frame = { 200
TL= 4
Holding cost = { 40 frame per year rE= 16, TL- 22
Tr= 3
If O> 1000, discowt=27o
If Q>2000, discount = 47o -\
\.
Economic order
. quantitr:
EDt" Te= o (z
lE=24
il-ar- Tt= o .-7 TL= 24
2x10000x300 u,5
{wrc
T-= 4 Tr= 3
= 387 unit Tt= o Tt- 14
J 13 t3 8ooj<tz
4
1
2 8 16
1
4
7. (a) Given. 5= 4
c,=tlzo
Xr=0 Ly =43 X xy: -3 Lx?=10 c'={1oo
We know that
i=20%
Zry -
na+bLx S = 2400 units
43=4a+bxO zC,S 2x120x2400
EOQ = =
a = 70.75 c.i 100x0.2
Lxy=qYx+bLxz
--3 = a x 0 + bx x 10 m = 240 =170
b = -0.3 , = Z+Ox
l^ 11
Demand function y = 10.15 - 0.3 x 8. (b) .When company purchases 800 units
4. (c) From now 2 months later x = 4 Then, total managing variable cost
Then, l=10.15-0.3x4 C,,xS . q
^ I
= 10.75 - l.2O = q1
= 9.55
-+-L,
l2ox24oo 1x
(a) For maximum profit- - + 8oox tooxo.2
Number of ash tray = a .8002
= 360+8000=<8360
Number of tea tray = y
When company purchases EOQ
Constraints are
l1x+2Oy<30000 r = Q+)oc, i
15x+5y < 30000
Maximize Z=20x+30y l2ox24oo
Now, inequalities are converted into linear equation - 170
+ 1 x 17ox
2
loox 0.2
10x+20y = 30000 = 73394
x+2Y = 3000 ..(i) if company purchase
Annual saving
3;r+Y = 6999 . .. (ii) EOQ= 8360-3394=4965
9. (a) Given,
(0, 6000) S = 2500 units
Leadtime=3days
Number of working days = 250 days
Ordering cost =t 12.50 / order
Holding cost = 2O7o of fverage inventory
Lot size (unit) Price per unit ( t)
I to2594.00
(2000,0) (s000,0)
260 to 9993.00
The feasible region is bounded by four corner points
1000 and above 2.00
1. (0,1s00), (0,0) (2000,0) (1800,600)
So, maximizeZ=20x + 3Oy Now,
Ar (0, 1500), 2x12.50x2500
Z:O+30x1500=45000 Qn
2x0.25
= .,[25ooo
At (0,0), : 354 approx
Z=0
At (2000, 0), Qr, < b, (1000), so we calculate Qn,
Z=20x2000+30x0=40000 2x12.50x2500
Ar (1800, 600), Qoz= = 289 approx
Z = l80O x 20 + 30 x 600 = 54000
lndustrial Engineering i 835
M" (d)
I+4x2+.3 /
-*-**._-
6
-- 3
f:-r)'
i-"7- I
\O J
5+6x4+7 *6
-*-- 6
[7*-ttl 4
6
L6) X
3 + 4x5-r7
-*__"-___ =5 t') 'i f7--3\2 16
r,
\6J 36
5+4x? +9 _
ill i
6
2+ 4x4+ 6
'----_*n r0
6 {:i
4*4x5+6 _ \
*---._**.- ( 6* 4','
tc
6 0
l6 j
4+4x6+8
"._-_**=o \2 t6
t8
i'8-a
l*-.-,----_
6 6
\ti.) 36
2t"4x3+4 .-1
-r 0
(+*z\'
l--*r* I =*-
4
--*--- 6
\6 ) 36
lndustrial Engineering * 839
5,TL=15
1=z:N 1 0+4=4 +2
2 4+7=ll .,
3 11+8=19 1
19 +2=21
/,)
4