Gate 2004

GATE 2004 PI : Production And Industrial Engineering Duration: Three hours Maximum Marks : 150 Read the following instructions carefully.

1.

This question paper contains 90 objective questions. Q.1–30 carry one mark each and Q.31–90 carry two marks each.

2.

Answer all the questions.

3.

Questions must be answered on special machine gradable Objective Response Sheet (ORS) by darkening the appropriate bubble (marked A,B,C,D) using HB pencil against the question number on the left hand side of the ORS. Each question has only one correct answer. In case you wish to change an answer, erase the old answer completely using a good soft eraser.

4.

There will be NEGATIVE marking. For each wrong answer, 0.25 marks from Q.1–30 and 0.5 marks from Q.31–90 will be deducted. More than one answer marked against a question will be deemed as an incorrect response and will be negatively marked.

5.

Write your registration number, name and name of the Centre at the specified locations on the right half of the ORS.

6.

Using HB pencil, darken the appropriate bubble under each digit of your registration number.

7.

Using HB pencil, darken the appropriate bubble under the letters corresponding to your paper code.

8.

No charts or tables are provided in the examination hall.

9.

Use the blank pages given at the end of the question paper for rough work.

10.

Choose the closest numerical answer among the choices given.

11.

This question paper contains 28 printed pages. Please report, if there is any discrepancy. Q. 1 − 30 carry one mark each.

Q.1

The given system of equation is: 4x + 6y + 3z = 20 x + 2y + 3z = 26 6x + 10y + 9z = 72 The system of equations:

Q.2

(A)

Has no solution

(B)

Defines a line in (x, y, z) space

(C)

Has a unique solution

(D)

Defines a plane in (x, y, z) space

The standard deviation of a Poisson distribution with mean λ is λ

(A)

(B)

(C) Q.3

0

λ2 (D)

1

Among the following material properties at room temperature, P − Ductility Q − Modulus of Elasticity R − Hardness S − Thermal conductivity microstructure sensitive properties are: (A)

P and Q (B)

(C)

Q and R R and S

(D) Q.4

P and R

A wooden block weighing 40N is resting on a horizontal table. A horizontal force of 10N is applied to the block. The coefficient of static friction between the table and the block is 0.3. The magnitude of the frictional force in Newton is (A)

10 (C)

(B) 15

12 (D)

20

Q.5

Which of the following motions of cam roller−follower leads to the lowest jerk? (A)

Uniform (B)

(C)

Cycloidal (D)

Q.6

Simple harmonic Parabolic

Consider the frictionless leak proof vertical piston cylinder mechanism indicated below. The weight of the piston is w. Initially, the valve connecting the mechanism and the tank is closed and the piston is stationary in the position indicated. The valve is opened slowly and the piston is found to gradually move up. Which of the following statement is correct?

Q.7

(A)

Positive work is done by the piston

(B)

Positive work is done by the atmospheric pressure

(C)

Tank pressure is always equal to atmospheric pressure

(D)

Tank pressure is always greater than atmospheric pressure

A well machined steel plated of thickness L is kept such that the wall temperatures are Th and Tc as seen in the figure below. A smooth copper plated of the same thickness L is now attached to the steel plate without any gap as indicated in the figure below. The temperature at the interface is Ti . The temperature of the outer walls are still the same at Th and Tc . The heat transfer

rates are q1 and q 2 per unit area in the two cases respectively in the direction shown. Which of the following statements is correct?

(C)

Th > Ti > Tc and q 1 < q 2 Th < Ti < Tc and q 1 = q 2 Ti = ( Th + Tc ) / 2 and q 1 > q 2

(D)

Ti < ( Th + Tc ) / 2 and q 1 > q 2

(A) (B)

Q.8

A machine shaft of diameter 30 mm which is subjected a torsional load is replaced by a hollow shaft of outer diameter 40 mm and inner diameter of 25 mm. Both the shafts are made of the same material. Based on the maximum shear stress theory, the factor of safety increases approximately by: (A)

1.5

(B)

(C) Q.9

3.0

2.0 (D)

Electrolyte used for fine hole drilling by electrochemical machining is: H 2 SO 4

(A) (B) (C)

NaBr NaCl

(D)

KI

4.0

Q.10

Flash and gutter are provided in drop forging dies at the following stage: (A)

Blocking (B)

(C)

Preforming (Edging) Finishing

(D) Q.11

Fullering

The current in Amperes used in resistance spot welding of plain carbon steel sheets (1 to 3 mm thick) lies within the range: 10 − 50

(A)

50 − 500

(B)

500 − 5000

(C)

(D)

5000 − 50000 Q.12

The metal powder used in Thermit welding of steel is: (A)

Al

(B)

(C) Q.13

Pb

Cu (D)

W

Wood flour is added to core sand to improve: (A)

Collapsibility of core

(B)

Dry strength of core (C)

Shear strength of core

(D)

Tolerance on casting Q.14

Cold shut (lap) may occur in products obtained by: (A)

Casting (B)

(C)

Machining (D)

Q.15

Forging Welding

In deep drawing of sheet metal, spring loaded stripper clamps the work until: (A)

The punch penetrates twice the stock thickness

Q.16

Q.17

(B)

It removes wrinkles on the product edges

(C)

Shedder removes the work from the tools

(D)

Punch is completely withdrawn from the work material

Chaplets are placed between mould and core surfaces in order to : (A)

Reduce directional solidification

(B)

Help local alloying of molten metal

(C)

Help easy removal of core from casting

(D)

Prevent core movement due to buoyancy

The flanks of teeth of rack−type gear cutter used for cutting involute gear profiles are: (A)

Cycloidal (B)

(C)

Circular Involute

(D) Q.18

Q.19

Straight

Reamer is designed to have even number of flutes to: (A)

Balance the cutting forces

(B)

Conform to shop floor standard

(C)

Enable measurement of the reamer diameter

(D)

Help in regrinding of reamer

In PERT analysis, which of the following distributions is assumed for the duration of the activities? (A)

Normal

(B)

Beta (C)

Poisson Exponential

(D)

Q.20

Which of the following are among the principles of Theory of Constraints? P − Balance capacity and do not balance flow Q − Transfer batch and production batch can vary over time R − Output from a bottleneck determines the output of the system (A)

Only P (B)

(C)

Only Q Only P and Q

(D)

Only Q and R Q.21

In an Assignment Problem for assigning n jobs to n machines, the number of decision variables and constraints are respectively: n 2 and 2n

(A) (B)

2n and 2n n 2 and n

(C) (D)

Q.22

2n and n

Which among the following are parts of Total Quality Management (TQM) of an organization ? P − Quality Policy Q − Quality Circles R − Continuous Improvement S − Rapid Prototyping (A)

Only P, Q and R

(B)

Only P, Q and S (C)

Only P, R and S P, Q, R and S

(D)

Q.23

The number of Basic Solutions in a linear programming problem with n decision variables and m inequality constrains (of ≤ type) with non negative RHS values is nC

(A)

m mC

(B)

( m+n ) C

(C)

m

( m+n ) C

(D)

Q.24

n

n

Which of the following are constructive heuristics for facilities layout? P − CRAFT Q − ALDEP R − CORELAP (A)

Only P (B)

(C)

Only P and Q Only Q and R

(D)

Only P and R Q.25

Q.26

Copyright protection in India is for a period of (A)

Sixty years

(B)

Life of the author

(C)

Life of the author plus fifty years

(D)

Life of the author plus sixty years

Find the odd one in the following set of distributions: (A)

Exponential Gamma

(C)

Geometric (D)

Chi − Square

(B)

Q.27

For a Time Series coming from the model X(t) = a + ε (t) where a is an unknown constant and ε (t) terms are pure noise, having zero mean and same variance for all t, the forecast for X(t + 1) based on data {X(k); 1 ≤ k ≤ t}, for minimizing the expected squared error in forecasting, we should use

Q.28

(A)

Exponential Smoothing

(B)

Two period moving average

(C)

Three period moving average

(D)

Simple Average

Which of the following is NOT part of Inventory Carrying Cost (A)

Cost of storage space

(B)

Cost of Obsolescence

(C)

Cost of insurance

(D)

Cost of inwards goods inspection

Q. 29 − 30 are “Matching” exercises. Choose the correct one out of the following alternatives A, B, C, D Q.29

Group 1 Group 2

P Q R S

Release Inspection Put Delay

1 2 3 4

MTM −2 MTM −1 Process Chart Outline Process Chart

P − 1, Q −3, R − 2, S − 4

(A)

(B)

P − 1, Q − 4, R −2, S − 3 P − 2, Q − 4, R − 1, S − 3

(C)

P − 2, Q − 3, R − 4, S − 1 Q.30

Group 1 Group 2

P Taguchi Q Wald

1 Continuous Sampling Plan 2 Orthogonal Arrays

(D)

R Dodge S Stewart

3 Xbar and R charts 4 Sequential Sampling Plan P − 2, Q − 4, R − 1, S − 3

(A)

(B)

P − 4, Q − 3, R − 1, S − 2 P − 1, Q −3, R − 2, S − 4

(C)

(D)

P − 1, Q −4, R − 2, S − 3 Q. 31 − 90 carry two marks each Q. 31

A set of four numbers has a distinct MODE, which is less than the

MEDIAN. The largest number is twice the MEDIAN. The RANGE is 6 and the MEAN is 6. The MEDIAN is (A)

4

(B)

(C)

Q.32

6

5 (D)

10

If X1 and X 2 are independent random variables, each having uniform

distribution on the interval (0,1), then ( X1 + X 2 ) has the following type of distribution: (A)

Normal (B)

(C)

Uniform Beta

(D)

Q.33

Triangular

X1 is Normal with a mean of 2 and a standard deviation of 3, X 2 is

Normal with a mean of 3 and a standard deviation of 4. X1 and X 2 are independent. Find the expectation of ( 2 X1 + 3X 2 ) 2 . (A)

169 (B)

196

(C)

349 (D)

Q.34

376

The ends of a solid Cast Iron rod and solid Steel rod, each of 20 mm diameter, are bonded together. Their lengths are 30mm and 40 mm respectively. The bonded rod is subjected to an axial compressive force of 100N. The approximate ratio of the change in length of the Cast Iron rod to the change in length of the Steel rod is: (A)

Q.35

1:1

(B)

2:3

3:2

(D)

3:4

(C)

The average speed of a flywheel is 1200 rpm and its polar moment of inertia is I (kg. m 2 ). The fluctuation of speed is 4%. The average speed of the flywheel is to be halved and the fluctuation of speed is to be brought down to 2%. Assuming that the fluctuation of energy in the flywheel remains constant, the polar moment of inertia of the flywheel has to be changed to (A)

2I (C)

Q.36

(B) 4 2I

4I (D)

8I

A riveted joint (with four rivets numbered 1 to 4) as shown in the

figure is subjected to an eccentric loading. The least and the most stressed rivets respectively are:

(A)

2 and 4 (B)

(C)

4 and 2 1 and 3

(D)

Q.37

3 and 1

The loading on a simply supported beam of cross section 100 mm wide × 200 mm deep is shown in the figure. The bending stress at C is

(A)

19 Mpa (B)

(C)

25 Mpa 32 Mpa

(D) Q.38

39 Mpa

A pinion and gear are in mesh with each other. The gear ratio is 2 and the moment of inertias of the pinion and the gear are 3 kg. m 2 and 5 kg. m 2 respectively. For the pinion to have an angular acceleration of 4 rad/ s 2 , the torque to be applied to the pinion shaft is (A)

5 N.m (B)

(C)

14 N.m (D)

Q.39

10 N.m 17 N.m

In the linkage shown in figure, O1A = 50 mm, AB = 60 mm, O 2 B = 50 mm, BC = 50 mm, CD = 60 mm, O 3 D = 100 mm. In the position shown in the figure, if O1A has momentarily and angular velocity of 2 rad/s without any angular acceleration, then the velocity and acceleration of D will be:

(A) (C)

100 mm/s, 100 mm/ s 2 200 mm/s, 100 mm/ s 2 200 mm/s, 300 mm/ s 2

(B) (D)

200 mm/s, 400 mm/ s 2 Q.40

An ideal gas enters a steady flow device at points 1 and 2 shown in

figure. It leaves the device at point 3. The velocities and elevation differences are negligible. Mass flow rates, temperatures and the heat input rate are indicated in the figure. Specific heats at constant pressure and volume for the gas can be assumed to be constant at 1000 J/kg. K and 714 J/kg.K respectively. The temperature of the gas leaving at section 3 ( T3 ) is approximately equal to:

(A)

453 K (B)

(C)

467 K 480 K

(D)

550 K

Q.41

The pressure of water in a pipe is measured by a manometer as shown

in the figure. The ambient pressure is Pa (Pascal). The densities of mercury and water are ρ m (kg/ m 3 ) and ρ w (kg/ m 3 ) respectively. The acceleration due to gravity is g (m/ s 2 ). The correct expression for the pressure of water at the point P in the pipe is:

Q.42

(A)

Pa + (ρ m h 3 g )

(B)

Pa + (ρ m h 3 g ) + (ρ w h 2 g )

(C)

Pa − (ρ m h 3 g ) − (ρ w h 2 g )

(D)

Pa − (ρ m h 3 g ) − (ρ w h 2 g ) + 2 (ρ w h 1 g )

A lever having 90  bend is to be produced by drop forging using mild steel bar as raw material. The various operations to be performed on it during forging are: P − Cutting Q − Bending R − Fullering S − Blocking cum finishing T − Edging The correct sequence for performing the operations is : (A)

P−Q−R−T−S R−T−Q−S−P

(B)

(C)

T−R−S−Q−P

(D)

R−Q−T−S−P Q.43

Determine the load required to punch a 20 mm diameter hole in 2 mm thick sheet. The properties of the material of the sheet are: Tensile strength Yield strength in tension

580 MPa 410 MPa

Shear strength

350 MPa

Yield strength in shear

250 MPa

The load in kN is: (A)

Q.44

31.40

(B)

43. 98

64.40

(D)

91.10

(C)

A two−pin gauge is to be designed for inspecting the relative positions of two holes shown in the figure. The positional tolerance on the holes is specified in the figure. The nominal diameter of each of the gauge pins in mm is:

(A)

14.87

(B)

14.91

14.97

(D)

15.01

(C)

Q.45

The drawing shows a machined shaft. The purpose of providing rounded grooves at M and N is to

Q.46

(A)

Supply lubricating oil for bearing mounting

(B)

Facilitate facing operation

(C)

Facilitate grinding of bearing mounting surface

(D)

Reduce stress concentration

A cylinder of 25 mm diameter and 100 mm length is turned with a tool, for which the relation VT 0.25 = 55 is applicable. The cutting velocity is 22 m/min. For a tool feed of 0.046 mm/rev, the number of tool regrinds required to produce 425 cylinder is (A)

12

(B) (C)

43

22 (D)

85

Q.47

The hole size f a bush is 20 +−00..03 00 mm. What is the shaft size (in mm) if the required minimum and maximum interferences are 0.03 and 0.08 mm respectively? 20 −−00..03 05

(A)

20 +−00..06 03

(B) 20 ++00..08 00

(C) (D)

20 +−00..08 06

Q.48

A casting of size 100 mm × 100 mm × 50 mm is required. Assume volume shrinkage of casting as 2.6%. If the height of the riser is 80 mm and riser volume desired is 4 times the shrinkage is casting, what is the appropriate riser diameter in mm? (A)

Q.49

14.38

(B)

20.34

28.76

(D)

57.52

(C)

A hole of 40 mm diameter is pierced in a steel sheet of 4 mm thickness without shear on the tool. Shear strength of steel is 400 N/mm 2 and penetration is 25%. What is the expected percentage load reduction if a shear of 1 mm is provided on the punch? (A)

Q.50

25.00

(B)

33.33

50.00

(D)

66.67

(C)

Orthogonal machining of a steel work−piece is done with a HSS tool of zero rake angle. The ratio of the cutting force and the thrust force on the tool is 1 : 0.372. The length of cut chip is 4.71 mm while the uncut chip length is 10mm. What are the shear plane angel φ and friction angle β in degrees? Use Merchant’s theory. (A)

32.49, 10.22

(B)

25.22, 20.41 (C)

64.78, 20.41

(D)

25.22, 23.21 Q.51

A cylindrical cup of 48.5 mm diameter and 52 mm height has a corner radius of 1.5 mm. Cold rolled steel sheet of thickness 1.5 mm is used to produce the cup. Assume trim allowance as 2 mm per 25 mm of cup diameter. What is the blank size in mm? (A)

102.0

(B)

115.4

120.5

(D)

128.5

(C)

Q.52

To measure the angle of a 90  V−block, two standard rollers of diameter 10 mm and 20 mm are used. For each roller, the height of the top of the roller, when placed on the V−block, is measured and the height difference in the cases of the two rollers is found to be 12.235 mm. What is the error in the angle of the V− block in degrees? (A)

Q.53

0.128

(B)

1.028

1.431

(D)

2.570

(C)

A steel billet of 180 mm thickness is rolled to 95 mm thickness in a three high reversible rolling mill. The roll diameter is 650 mm. The coefficient of friction between the rolls and the hot material is assumed as 0.20. How many rolling passes are required? (A)

2 (C)

Q.54

(B)

4

5

(D)

7

Two fillet welds of length 100 mm each are used to support flat (b) on the bracket (a). Flat (b) carries only a tensile load. The leg length of both the fillet welds is 5 mm. The properties of weld metal are: Tensile strength

480 MPa

Yield strength in tension

410 MPa

Yield strength in shear

230 MPa

Reduction in area

40%

Find the load carrying capacity P of the welded joint.

(A)

230.0 kN (B)

(C)

162.6 kN 480.0 kN

(D) Q.55

80.0 kN

In a single sampling plan for attributes, with sample size n >1, and acceptance number c = 0, the O. C. curve (with p = defective fraction)

Q.56

(A)

Has maximum steepness at p = 0

(B)

Has maximum steepness at p = 1

(C)

Has exactly one point of inflection

(D)

Has negative derivative for all p in the range 0 ≤ p ≤ 1

The average weekly demand for an item is 10 units/week. The fixed cost of ordering is Rs. 98 per order. The inventory holding cost is Rs. 0.58 per unit per week. Suppose the permitted values of reorder interval are only: 1 or 2 or 4 or 8 weeks. Among these values of reorder interval (in weeks), which one yields the least total relevant cost per week? (A)

1 (C)

Q.57

(B) 4

2 (D)

8

Determine the maximum value of the objective function in the problem Maximize X1 X 2 X 3 X1 + 2 X 2 + 3X 3 = 18 X1 , X 2 , X 3 ≥ 0 (A)

24 (C)

Q.58

(B) 54

36 (D)

60

With compound interest at the rate of (100)p% per annum, it takes n(p) years for an amount to get doubled, where

(1 + p ) n ( p ) = 2 As p → 0+, the product of p and n(p) goes to (A)

0.693 (B)

(C)

0.707 0.732

(D) Q.59

0.750

Consider a static n job single machine sequencing problem. Job j (j = 1, 2, …., n) has processing time P j and material value R j (Rupees). Let R(t) = total material value of jobs remaining in the system at time t. We want to minimize the time− average of R(t). Then the required priority dispatching rule must pick the job with minimum value of : (A)

Pj

(B)

(− R j )

R j / Pj

(D)

Pj / R j

(C)

Q.60 −71 are “Matching” exercises. Choose the correct one out of the following alternatives A, B, C, D Q.60

Requirement in the design of jigs and fixtures

Recommended

device P − Heavy clamping force on the work piece

1 − Clamp with a

floating pad Q − Clamping on rough surfaces of work piece

2 − Slip renewable

bus R − Four work pieces in a line to be bush machined in one loading 4 − Equalizing clamp

3 − Indexing

S − Drilling and reaming of work piece in

5 − Strap clamp

one loading P −5, Q − 1, R − 4 , S −2

(A)

(B)

P − 1, Q − 4, R − 5, S − 3 P − 5, Q − 1, R − 2, S − 3

(C)

(D)

P − 4, Q − 5, R − 3, S − 2 Q.61

Mechanism Machine (P) Pantograph

(1)

Automobile (Q) Quick return

(2)

Lathe (R) Ackerman

(3)

Engraver (S)

Toggle (4)

Shaper (5)

(A)

Press

P − 3, Q − 1, R − 2, S − 4

(B)

P − 3, Q − 4, R − 1, S − 5 (C)

P − 2, Q − 4, R − 5, S − 3

(D)

P − 2, Q − 3, R − 4, S −5 Q.62

Products Process (P)

Electrical switches Pultrusion

(1)

(Q)

Plastic Buckets

(2)

Blow Moulding (R)

FRP Channels

(3)

Transfer Moulding (S)

Plastic Bottles

(4)

Injection Moulding (5)

Vacuum Forming

P − 3, Q − 4, R − 2, S −1

(A)

(B)

P − 2, Q − 5, R − 4, S −3 P − 3, Q − 4, R − 1, S − 2

(C)

(D)

P − 2, Q − 5, R − 3, S − 1 Q.63

Product Process (P)

Gears (1)

(Q)

Isostatic Compaction

HSS Tools

(2)

Roll Compaction (R)

Filters (3)

(S)

High Energy Rate Compaction

Bimetal bearing

(4)

Close Die Compaction (5) (A)

Vibratory Compaction

P − 2, Q − 4, R − 1, S −3

(B)

P − 1, Q − 3, R − 4, S − 5 (C)

P −4, Q − 1, R − 5, S − 2 P − 3, Q − 2, R − 1, S − 4

(D)

Q.64

Group 1 Group 2 P − Impulse turbine

1 − Pulsating

flow Q − Axial flow turbine

2

−

Volute

casing R − Centrifugal pump

3 − Adjustable

vanes S − Reciprocating pump

4

−

Large

mass flow rates 5 − Nozzle (A)

P − 4, Q − 3, R − 5, S −2

(B)

P − 5, Q − 4, R − 2, S − 1 (C)

P −5, Q − 4, R − 3, S − 2

(D)

P − 3, Q − 2, R − 1, S − 4 Q.65

Group 1 Group 2 P − Mechanized warehousing 1 − AGV Q − To inspect more than one dimension in one setting 2 − AS/RS R − Automatic transportation of parts in a manufacturing system CMM S − Make range of products in small batches 4 − PLC

3

−

5 − FMS P − 1, Q − 3, R − 4, S − 5

(A)

(B)

P − 2, Q − 3, R − 1, S − 5 P − 2, Q − 3, R − 4, S − 1

(C)

(D)

P − 3, Q − 1, R − 2, S − 5 Q.66

Alloying element Primary influence on alloy P − Aluminium

1

−

3

−

4

−

Free cutting action Q − Lead 2 − Work hardenability R − Manganese Fine grained structure S − Tungsten Corrosion resistance 5 − Deoxidizer P −5, Q − 1, R − 2, S − 3

(A)

(B)

P − 4, Q − 2, R − 1, S − 3 P − 3, Q − 2, R − 1, S − 4

(C)

(D)

P − 5, Q − 1, R − 4, S − 3

Q.67

NC code

Machine

action P − G03 stop

1 − Spindle

Q − G41

2 − Constant

surface speed R − M05

3 − Tool nose

radius compensation left S − M08

4 − Circular

interpolation counter clockwise 5

−

Coolant on (A)

P − 4, Q − 2, R − 3, S − 5

(B)

P − 4, Q − 3, R − 1, S − 5 (C)

P − 5, Q − 4, R − 1, S − 2

(D)

P − 1, Q − 3, R − 2, S − 4 Q.68

Tolerance Symbol

Explanation 1 − Perpendicularity of a line

(axis) with reference to a datum surface 2 − Circular run out radial 3 − Symmetry of a median plane 4 − Cylindricity 5 − Parallelism of a line (axis) with reference to a datum line (A)

P − 2, Q − 3, R − 1, S − 4

(B)

P − 3, Q − 1, R − 5, S − 2 (C)

P − 4, Q − 3, R − 1, S − 2 P − 4, Q − 5, R − 2, S − 1

(D)

Q.69

Group 1 Group 2 P

Linked Lists

1

Random Access Q

Arrays 2

R

LIFO

Large Files

3

Index based search S

Stack 4

(A)

Dynamic Allocation P − 1, Q − 3, R − 2, S − 4

(B)

P − 4, Q − 1, R − 3, S − 2 (C)

P − 2, Q − 1, R − 4, S − 3

(D)

P − 4, Q − 3, R − 1, S − 2 Q.70

Group 1 Group 2 P

Maximum Flow Problem

1

Forward Recursion Q

Probabilistic Dynamic Programming

R

Shortest Path problem

2

Degeneracy 3

Backward Recursion S

Assignment Problem

4

Labelling Algorithm (A)

P − 4, Q − 1, R − 3, S − 2

(B)

P − 3, Q − 2, R − 1, S − 4 (C)

P − 2, Q − 3, R − 4, S − 1 P − 4, Q − 3, R − 1, S − 2

(D)

Q.71

Group 1 Group 2 P

Worker −Task Assignment

1

Dynamic

Programming Q

Knapsack Problem

2

Nonlinear constrained optimization R

Machining parameter Optimization

3

Quadratic

Objective

4

Linear

Objective

Assignment Problem S

Allocating n locations to n facilities

Assignment Problem P − 4, Q − 1, R − 2, S − 3

(A)

(B)

P − 4, Q − 2, R − 1, S − 3 P − 3, Q − 4, R − 2, S − 1

(C)

(D)

P − 2, Q − 1, R − 3, S − 4 Data for Q.72−73 is given below. Solve the problems and choose correct answers A weld is made using MIG welding process with the following welding parameters: Current

200 A

Voltage

25 V

Welding speed

18 cm/min.

Wire diameter

1.2 mm

Wire feed rate

4 m/min

Thermal efficiency of the process

65%

Q.72

The heat input per unit length of the weld in kJ/cm is: (A)

0.18 (B)

(C)

0.28 10.83

(D) Q.73

16.66

The area of cross section of weld bead in mm 2 is: (A)

16.3 (B)

(C)

25.1 30.3

(D)

38.6

Data for Q.74−75 is given below. Solve the problems and choose correct answers There are two identical machines, each with a memory less failure rate, λ , = 0.2 failure/hour. These two machines are repaired by a common repairman, with a memory less repair (service−completion) rate, µ = 0.4 repairs/hour. Assume that the repairman follows FCFS order for service. The system is operating in steady state. Q.74

What is the repairman’s utilization (i.e. fraction of time he is busy)? (A)

0.50 (B)

(C)

0.60 0.67

(D) Q.75

0.75

What is the utilization of each one of the machines? (A)

0.5 (C)

(B) 0.7

0.6 (D)

0.8

Data for Q.76−77 is given below. Solve the problems and choose correct answers

In an M/M/1 queue, let λ = Arrival rate and µ = Service rate. Q.76

Suppose the Expected number of customers waiting in the queue in steady state equals 4.00, then 4µ ( µ − λ ) = λ

(A)

(B)

4 λ (µ − λ ) = µ 2 4 (µ − λ ) = λ

(C)

(D)

4 λ2 = µ 2 Q.77

Let N denote the number of customers in the system in steady state (a random variable). What is the mean (Expectation) of the conditional distribution of N, given that N is positive? (A)

λ2 /(µ − λ ) 2

(C)

1 /(µ − λ ) 2 λ /( µ − λ )

(B)

µ /(µ − λ )

(D)

Data for Q.78 − 79 is given below. Solve the problem and choose correct answers In a 2−machine flowshop, the processing times of jobs j on machines A and B (in that order) p(A,j), p(B,j) are given as follows and we are told that the job sequence 1−2−3−4−5 does minimize the makespan. job j p(A,j) p(B,j) Q.78

1 3 5

3 6 10

4 8 θ = p(B,4)

Under the given conditions it is necessary and sufficient that: (A) (B) (C) (D)

Q.79

2 4 4

8≥θ≥ 3 10 ≥ θ ≥ 3 θ≥ 8 θ≥ 3

For θ = 10, the minimal makespan is:

5 4 3

(A)

25

(B)

(C)

36

28 (D)

40

Data for Q.80−81 is given below. Solve the problems and choose correct answers A newspaper stall owner at the beginning of every week buys a certain number of copies, S of a certain weekly for the new week, and disposes off the left over copies (if any) of the last week’s issue. Suppose the demand for the weekly during the week has the probability mass function (0.02)(0.98) n ; n ≥ 0 Q.80

If S = 50, then what is the Expected value of the demand in a weak that is NOT met? (A)

17.84

(B) (D)

Q.81

21.39

(C)

24.92

48.38

Suppose that on every copy sold during the week, the owner makes a profit of Rs. 1.20, and on every copy not sold during the week, he loses Rs 5.00. Then the optimal value of S is (A)

8 (C)

(B) 12

10 (D)

14

Data for Q.82−84 is given below. Solve the problems and choose correct answers Consider the Liner Programming problem: Maximize Z = 3X1 + 4 X 2 Subject to X1 + X 2 ≤ 12 2 X1 + 3X 2 ≤ 30 X1 + 4 X 2 ≤ 36

X1 , X 2 ≥ 0 Slack variables X 3 , X 4 and X 5 are associated with the 1st, 2nd and 3rd inequalities respectively. The optimal solution is X1 = 6, X 2 = 6 and Z = 42. Q.82

The variables in the optimal basis are X1 , X 2 , X 3

(A)

(B)

X1 , X 2 , X 4 X1 , X 2 , X 5

(C)

(D)

X3 , X 4 , X5 Q.83

The marginal value of the third resource at the optimum is (A)

0

(B)

(C) Q.84

36

12 (D)

42

If the Right Hand Side (RHS) value in the third constraint changes to 30, then (A)

The problem has an alternate optimum solution

(B)

The optimal solution is degenerate

(C)

The dual has unbounded solution

(D)

The dual is infeasible

Data for Q.85 − 86 is given below. Solve the problem and choose correct answers You are given a transportation problem with two supply points (i = 1,2) and three demand points (j = 1 to 3). The unit costs of transportation are indicated in top left hand corner of each position in the matrix. A feasible solution ( X11 = 20, X12 = 10, X 21 = 10, X 23 = 30 ) is also given in the table 12

8

20

10

6

Q

10

11 2

30

Q.85

The set of all the values of Q for which the given solution is the only optimum solution is (A)

Q>0

(B)

Q≥ 1 (C)

Q>2

(D)

Q>3 Q.86

For Q = 2, which of the following is NOT true? (A)

The problem has unique optimal solution

(B)

The given feasible solution is optimal

(C)

The problem has a degenerated optimal solution

(D)

The problem has non basic solutions which are optimal

Data for Q.87−88 given below. Solve the problems and choose correct answers Your are given a project network with six activities (A to F), their durations, and the cost to reduce the duration of each activity by 1 day. An activity − on − are network is shown Activity Duration (days) Cost to reduce duration by 1 day (Rs)

Q.87

The number of critical paths for the project is :

D 9 40

E 9 20

(A)

1 (C)

Q.88

(B) 3

2 (D)

4

The minimum amount (Rs.) to be spent to reduce the project duration by one day is: (A)

20 (C)

(B) 120

60 (D)

130

Data for Q.89−90 given below. Solve the problems and choose correct answers You are given a maximum flow problem with five nodes (Node 1 is the source and Node 5 is the sink). The data (a,b) represents the actual flow on the arc and the arc capacity respectively.

Q.89

Relative to the given feasible flow, the number of distinct flow augmenting path is: (A)

0 (C)

Q.90

(B) 2

1 (D)

3

The value of the minimum cut for the problem is: (A)

0 (C)

(B) 27

24 (D)

42

************************

SOLUTIONS 1.

(A) The determinant of coefficients of equation has zero value. 4 6 3 1 2 3 = 4(18−30) − 6 (9 −18) + 3(10 − 12) = 0 6 10 9

2.

(A) Poisson’s probability distribution function P(x = k) =

λk e − λ k!

where, k = 0, 1, 2, 3, 4 …… e = 2.7183 base of natural logarithm. λ = mean of the poisson distribution.

Properties of the poisson distribution are :

mean = λ variance = λ standard deviation =

λ

It is a discrete probability distribution and is widely used in statistical work. The distribution finds application in a wide variety of situations in which some kind of event occurs repeatedly but haphazardly. The distribution is used where the chance of any individual event being successful is small. The distribution is used to describe behaviour of rare events such as : (i)

Number of telephone calls arriving on exchange per unit time (ii)

Number of customers arrive at a shop.

(iii)

Number of welding defects occurring in the long length of weld.

etc. 3.

(D)

The internal structure of a material, simply called the structure is studied at various levels of observation. The magnification and the resolution of the physical aid [lens, microscope] used are a measure of the level of observation. Depending on level, material structure is classified as : macrostructure, microstructure, substructure, crystal structure, electronic structure and nuclear structure. Macro structure of a material is studied with naked eye. Microstructure generally refers to the structure as observed under the optical microscope with magnification upto 1500 times linear. Human eye can distinguish upto 0.1 mm i.e. distinguishes only when two lines are separated by distance more than 0.1 mm. The optical microscope can resolve details upto a limit of about 0.1 µ m. Substructure refers to the structure obtained by using a microscope with a much higher magnification and resolution than the optical microscope. In an electron microscope a magnification of 10 6 times linear is possible. 4.

(B)

Frictional force = µ W = 0.3 × 40 = 12 N. The applied force 10 N is insufficient to move the block forward as it is less than the frictional force 10 N.

5.

(C)

6.

(B)

7.

(D)  T + Tc   and q 1 > q 2 Ti <  h 2   k s = thermal conductivity of steel = 54 W/m −  C . k c = thermal conductivity of copper = 386 W/m −  C . q 1 = heat transfer rate per unit area in first case q 2 = heat transfer rate per unit area in second case. Applying heat conduction formula, ∴

k ( T − Tc ) Th − Tc q 1 = s h = L (L / k s ) (1) k ( T − Ti ) k c ( Ti − Tc ) q 2 = s h = L L Th − Tc =  l L    + k  k  s c  (2)

From equations (1) and (2) , q 1 > q 2 . Solving equation (2) for Ti

k s ( Th − Ti ) = k c ( Ti − Tc ) k s Th + k c Tc = Ti ( k c + k s )  k T + k c Tc Ti =  s h  kc + ks

∴

   

Ti = intermediate temperature is arithmetic mean of two extreme temperature when both plate materials are same. For example, when both plates are made of steel, k T + k s Tc Th + Tc Ti = s h = ks + ks 2 When two dissimilar metals are there, it will be less than arithmetic mean. Assuming Th = 100  C and Tc = 30  ; 54 × 100 + 386 × 30 ≅ 38.5  C 54 + 386 100 + 30 = 65  C which is less than 2 Ti =

8.

(B)

Zp = polar modulus =

polar moment of inertia max imum radius

T = f s × Zp f s is the maximum shear stress occurring at the greatest radius on application of twisting torque (T). Zp1 = polar modulus for a solid shaft of a diameter d = 3 cm. π × d4 π π = 32 = d 3 = × ( 3) 3 d/2 16 16 d = 3 cm. 27 = π 16 Zp 2 = polar modulus of hollow shaft

[

π d 2 4 − d1 4 = 32 d2 / 2

]

d2 =

OD = 4 cm, d1 = ID = 2.5 cm π × ( 4 4 − 2.5 4 ) π 16 = = × 54 4 16 Torque carrying capacity increase with Z p Zp 2 Zp1 9.

=

π / 16 × 54 =2 π / 16 × 27

(C) The properties essential for an electrolyte for electrochemical machining are : 1.

High electrical conductivity

2.

Low viscosity and high specific heat.

3.

Chemical stability

4.

Resistance to formation of passivating film on the workpiece surface.

5.

Non corrosive and non toxic

6.

Inexpensive and easy availability

Salt solution with large proportion of water satisfy many of the above conditions and therefore are generally used. Electrolytes generally used are : (i)

NaCl or KCl upto 0.25 kg/lit. This is widely used because of its low cost

and stable conductivity over a broad range of pH values. However it is corrosive and produces a large sludge. It cannot be used on tungsten carbide or Molybdenum. (ii)

sodium nitrate upto 0.5 kg/lit. It is less corrosive but forms a passive film

on the workpiece. Hence it is not used as a general purpose electrolyte. It is used for machining aluminium and copper.

10.

(C) Final and exact shape and size is given in finishing impression. To ensure complete filling of cavities, an extra material is added which is squeezed out in the form of flash and gutter. The sequence of operations in a multimpression drop die forging is − fullering, edging, bending, blocking and finishing.

11.

(D) In resistance welding, a low voltage (typically 1V) and very high current (typically 15000A) is passed through the joint spot for a very short time (typically 0.25 s). This high amperage heats the joint due to the contact resistance at the joint and melt it. The pressure at the joint is continuously maintained and the metal fuses together under this pressure. The heat generated in resistance welding: H = K I2 R t K = a constant to account for the heat losses from the weld joint. The critical variable in a resistance welding process is the contact resistance between the work piece plates and their resistance themselves. The contact resistance is affected by the surface finish on the plates as the rougher surface have higher contact resistance. The contact resistance increases with oxide or other contaminants present on the surface −The lower resistance of the joint requires a very high current to provide enough heat to melt it. The average resistance is of the order of 100 micro ohm, as a result the currents required would be of the order of tens of thousands of ampere. With a 10,000 A current passing for 0.1s, the heat generated is −

H = I 2 R t = (10,000) 2 × (100 × 10 −6 ) × 0.1 = 1000 J. This is typical for the welding of 1 mm thick sheet. The actual heat required for welding /melting − weld area of 5 mm dia and 1.5 mm thick is 440 J. The rest of the heat (560 J) is utilized in heating the surrounding areas and lost at other points. 12.

(A)

13. 14.

15.

(D)

18.

(A)

19.

(B)

(B) 16.

17.

(A)

(D)

(D)

The β distribution is not necessarily symmetric, the degree of skewness depends on the location of t m to t o and t p . Thus the range specified by the optimistic time ( t o ) and pessimistic time ( t p ) estimates is assumed to enclose every possible estimates of the duration of the activity. The most likely time ( t m ) may  to + tp   and may occur to its left or to its right, not coincide with the midpoint   2   as shown below:

Because of these properties, it is justified to assume that duration of each activity may follow Beta ( β ) distribution with its unimodal point occurring at t m and its end points at t o and t p .  to + tp   is given half weight of the most In Beta distribution , the mid point   2   likely point t m . Thus the expected or mean value of activity duration ( t e ) (t o + t p ) + 2 t m t + 4t + t o m p 2 = = 3 6 Based on analogy of normal distribution,

∴

6σ ≈ t p − t o  tp − to σ=  6 

    1

Variance of a activity time σ 2 =  ( t p − t o )   6 

2

20. (D) 21. (A) Assignment model :

Facilities

1

jobs 2

n

1

C11

C12

C1n

2

C 21

 

  C n1

n

 

 

Cn2

….

supply 1

C 2n

1

  C nn

  1

Demand

1

1

…

1

x ij = [0 if the ith facilities not assigned to jth job = 1 if the ith facility is assigned to jth job. ] n

n

Minimize ∑ ∑ C ij x ij j=1 i =1

subject to constraints (i)

Constraints on facilities. let n = 4

x 11 + x 12 + x 13 + x 14 = 1 x 21 + x 22 + x 23 + x 24 = 1 x 31 + x 32 + x 33 + x 34 = 1 x 41 + x 42 + x 43 + x 44 = 1 (ii) Constraints on jobs

x 11 + x 21 + x 31 + x 41 = 1 x 12 + x 22 + x 32 + x 42 = 1 x 13 + x 23 + x 33 + x 43 = 1 x 14 + x 24 + x 34 + x 44 = 1 Thus for n jobs and n machines problems, number of constraints are 2 n (8) and variables equal to n 2 .(16) 22.

(A)

23.

(C) If there are m equality constraints and (m + n) is the number of variables (including slack and surplus variables), a start for the optimal solution is made by putting n unknown [ out of (m + n) unknowns] equal to zero and then solving for the m equations in the remaining m unknowns, provided that solution exists and is unique. The n zero variables are called non basic variables and remaining ‘m’ variables are called basic variables which form a basic solution. If the solution yields all non negative values of variables, it is called basic

feasible solution; otherwise, it is infeasible. The step reduces the number of alternatives for the optimal solution from infinite to a finite number whose maximum limit is m+n C

25.

28.

31.

34.

m

=

( m + n )! m! n!

(D)

(D)

(C)

(C) PL AE P = Applied force = 100 N δ = change in length =

L = length A = cross section. E = Elasticity modulus. δ s = change in length in steel =

P × 40

2 × 10 5 × A E s = Elasticity modulus for steel = 2 × 10 5 N/mm 2

 P × 30  δc =    1 × 10 5 × A  δ s 40 1 × 10 5 2 = × = δ c 30 2 × 10 5 3 ∴

δc δs

=

3 2

26.

(D)

27.

(A)

29.

(A)

30.

(A)

32.

(A)

33.

(A)

35.

(D) 2 π N 2 π ×1200 = = 40 π 60 60 I = moment of inertia of the flywheel ω=

ω1 = maximum speed ω 2 = minimum speed e = maximum fluctuation of energy. k = coefficient of fluctuation of speed. ∴

e=

1 2 1 2 1 Iω − Iω = I ( ω12 − ω 22 ) 2 1 2 2 2

 ω1 + ω 2   ( ω1 − ω 2 ) = I 2   ω + ω2 where ω = 1 2

= Iω( ω1 − ω 2 )  ω − ω2  = Iω 2  1  ω   e = I ω2 k ω 0 = mean speed in first case ω1 = mean speed in second case =

ω0

(given) 2 k 0 = coefficient of fluctuation in first case = 4% = 0.04 k 1 = coefficient of fluctuation in second case = 2% = 0.02

As fluctuation of energy (e) remains same, e = I 0 ω 02 k 0 = I1ω12 k 1 ∴

ω =  0 I 2  ω1 I1

2

 k  × 0  k   1

 0.04  = ( 2) 2 × =8  0.02 

∴ The polar moment of inertia of the flywheel has to be changed to 8I. 36.

(D)

37.

(C)

Moment at C = 0.125 m × 80 kN = 10 kNm. Taking moment at A and equating it to zero, we have VB × 0.8 − 80 × 0.4 − 10 = 0 ∴

VB = 52.5 kN VA + VB = 80 VA = 27.5

∴ Bending moment at C: 27.5 × 0.4+10 = 21 kNm = 21 × 10 6 Nmm M = f. M z f = bending stress. M z = section modulus.

∴ ∴

bd 3 I Mz = = 12 y max d / 2 bd 2 = 6 (100 )( 200 ) 2 21 × 10 6 = ×f 6 2 21 = × f 3

f = 31.5 N/mm 2 ≈ 32 MPa 38.

(D) Torque required to accelerate a geared system:

B rotates G times the speed of pinion A. Therefore, gear ratio G =

NB NA

=

1 (given) 2

where N B and N A are speeds of pinion A and gear B. ∴ Since the gear B turns G times the pinion A, the rate of change of angular speed of gear B with respect to time (i.e. angular acceleration of gear B, α B ) must be equal to G times the rate of change of angular speed i.e. α A . αB = G αA .

∴

I A and I B are mass moment of inertia of pinion A and gear B. I A = 3 kg.m 2 I B = 5 kg.m 2 ∴

Torque required for pinion to accelerate itself TA = I A .α A

and torque required to accelerate gear B itself TB = α B .I B = G.α A .I B In order to provide torque TB on the gear B, the torque applied on the pinion must be G TB . Therefore, torque applied to pinion A in order to accelerate gear B TAB ~ _ G TB = G 2 α A I B ∴ Total torque which must be applied to pinion A in order to accelerate the

geared system T = TA + TB = I A .α A + G 2 I B α A = (I A + G 2 I B ) α A α A = angular acceleration of pinion = 4 rad/ s 2   1 2 T = 3 +   × 5  × 4 = 17 N.m   2   39.

43.

(D)

41.

(C)

42.

(B)

(B) F = p × t × τs F = cutting force p = cutting perimeter = π d = 20 π τ s = shear strength = 350 MPa = 350 N/m m 2 . Yield shear strength is of no use as it only initiates shearing but does not complete it. Shear strength = 350 N/m m 2 is a ultimate shear strength. Tensile strength and yield strength in tension are irrelevant for shearing operation. t = thickness = 2 mm F = 20 π × 2 × 350 =43982 N = 43.98 kN.

44.

(D) The condition specified here is maximum material conditions. i.e. shaft is of maximum size and hole of minimum size. Variation on centre to centre distance, given is 0.08. So pin of 15.01 mm size can be used.

45.

(C)

46.

(D) V = 22 m/mim. Given: VT 0.25 = C 1

T =  C  0.25 =  C  V V

∴

4

4

55 T = tool life =   = 39 min.  22  V=

πd N m/min 1000

22 =

π × 25 × N 22 25 = × ×N 1000 7 1000

N = 280 rpm. L 100 = = 7.8 min. f × N 0.046 × 280

t = machining time per piece =

Number of cylinder machined per grind =

T 39 = =5 t 7.8

∴ to produce 425 cylinders, 85 grinds are required. 47.

(D) (i) max. interference = large shaft size − smallest hole size 0.08 = large shaft size − 20.00 ∴ maximum shaft size = 20.08 (ii)

minimum interference = smallest shaft size − largest hole size 0.03 = smallest shaft size −20.03

∴

smallest shaft size = 20.06 +0.08

shaft size 20 + 0.06

48.

(C) VC = volume of casting = 10 × 10 × 5 cm 3 = 500 cm 3 Volume of shrinkage of casting =

2.6 × 500 = 13cm 3 100

As the required riser volume is four times the volumetric shrinkage, Vr = riser volume = 4 × 13 = 52 cm 3 π Vr = d 2 × h 4 cm.

∴

52 =

d = dia. of riser, h = height of riser = 8

π × d2 × 8 4

∴ d = 2.876 cm

= 28.76 mm 49.

(C)

Fmax = Max. punch force required without shear. F = punch force with shear. k = coefficient of penetration = 0.25 (given) s = amount shear provided = 1 mm The basic assumption underlying the computation of force is energy required for shearing with and without shear is same.

Fmax .(k.t) = F (kt +s) ∴ ∴

kt  0.25 × 4 1 =  = = Fmax  kt + s  ( 0.25 × 4 + 1) 2 F

Expected load with shear is just half of without shear. Hence expected percentage of load reduction is 50%.

50.

(B)

r = cutting ratio = tan φ =

length of cut chip 4.71 = = 0.471 uncut chip length 10

r − cos γ 1 − r sin γ

where, φ = shear angle γ = tool rake angle = 0  (given) 0.471 cos 0

∴

tan φ =

∴

φ = 25.22  .

1 − 0.471 sin 0 

From diagram, tan( λ − γ ) = ∴

= 0.471

FN FC

= 0.372

λ − γ = tan −1 ( 0.372 ) = 20.4

Since γ = 0 ; λ = 20.4 51.

(B) d = 48.5 mm r = 1.5 mm d = 32.33 r D = blank diameter h = cup height = 52 mm (given) D = d 2 + 4dh D = d 2 + 4dh − 0.5 r

when when 15 ≤

d ≥ 20 r

d ≤ 20 r

d ≤ 15 r

when 10 ≤

D = d 2 + 4dh − r

D = ( d − 2 r ) 2 + 4 d ( h − r ) + 2 πr ( d − 0.7 r )

when

D = d 2 + 4dh D = ( 48.5) 2 + 4 ( 48.5) × 52 = 111.5 mm. Trim allowance is 2 mm for 25 mm cup diameter ∴

Trim allowance =

48.5 × 2 = 3.9 mm 25

∴ Blank diameter = D + trim allowance = 111.5 + 3.9 = 115.4 mm. 52.

(D) Let α = half included angle of the Vee block. C1 = centre of small cylinder C 2 = centre of large cylinder  r  h 1 = OC1 + C1T1 =  1 + r1   sin α  r h 2 = OC 2 + C 2 T2 = 2 + r2 sin α ( r2 − r1 ) sin α 5 12.235 = (10 − 5) + sin α 5 7.235 = sin α 5 sin α = 7.235 h 2 − h 1 = ( r2 − r1 ) +

∴

α = 43.71

∴ The included angle of given V block = 2 α = 87.42  C ∴

Error in the angle = 90  − 87.42  = 2.58 

d ≤ 10 r

53.

(B) ∑ ∆h = total required reduction = h 0 − h f =180 − 95 = 85 mm.   1  maximum reduction per pass = ∆h max = D 1 −  1 + µ 2  D = roll diameter = 650 mm (given) µ = coefficient of friction between roll and the hot billet = 0.2 (given)   1 ∆h max = 650 1 −  = 12.62 mm  1 + 0.2 2  85 ∑ ∆h = = 6.73 Number of passes required = ∆h max 12.62 For two high reversible mill, 7 passes are required. But two reductions are possible in one pass in three high reversible mill. Rolling passes −4.

54.

(B) The fillet welds are of two types: (i) transverse fillet weld which are assumed to fail in tension (ii)

parallel fillet welds which are assumed to fail in shear.

The plane of maximum shear stress in the conventional 45  fillet weld is the 45  throat when subjected to parallel load and the 67

1 throat when subjected to a 2

transverse load. Throat distance = h sin 45  = h / 2 (i)

Transverse weld subjected to load P, Tensile stress =

(ii)

P 2 L × ( h sin 45  )

=

P 2hL

Parallel weld subjected to load P P Shear stress intensity = ( 2 L ) × h × sin 45  P Shear stress = 2hL given problem is of parallel weld 230 N/mm 2 =

∴

55.

(D)

P =162.6 kN

P 2 × 5 × 100

56.

(D) R = weekly demand rate = 10 C 3 = ordering cost

= 98

C1 = inventory holding cost per unit = Rs.0.58 per week q0 =

2C 3 R C1

=

2 × 98 × 10 = 58.13 0.58

optimum cost will be when q 0 = 58.13 units. (i)

q = 40 (four week)

Inventory carrying cost =

1 × 40 × 0.58 2

Total cost = 109.6 109.6 cost per week = = Rs. 27.4 4 (ii)

q = 80

inventory carrying cost =

1 × 80 × 0.58 = 27.6 2

Total inventory cost for meeting 8 week demand = 98 + 27.6 = 125.4 ∴

inventory cost per week =

125.4 = Rs. 15.67 8

Reorder interval should be of 8 weeks to minimize the inventory cost. 57.

(B) X1 + 2 X 2 + 3X 3 = 18

∴

58.

X1 = 6 ;

2X 2 = 6 ;

3X 3 = 6 ;

X1 = 6 ;

X 2 = 3;

X3 = 2;

(C) This formula is used to find investment doubling period for a given interest

n × p = 72 p = interest rate in percent n = number of years to double the investment At rate of interest 6% it will take about 12 years to double the amount by compounding. A n = A 0 (1 + p ) n

A n = Amount after n years A 0 = Initial investment 59.

(D)

60. 61.

62.

(C)

65.

68.

(B)

70. (A)

72.

(D) arc power input = arc voltage × arc current (watts) Heat input rate =

current ( I ) × arc voltage travel speed 200 × 25 Watts × 60 = 18 cm / min = 16667 J/cm = 16.67 kJ/cm

67.

(B)

69.

(C)

(A)

(D)

(B) 66.

(C)

71.

(B) 63.

64.

(A)

(A)

73.

(B) Weld metal deposited per minute = wire cross section × wire feed rate d = diameter of wire = 1.2 mm = 0.12 cm. feed rate = 4 m/min = 400 cm/min. ∴ weld metal deposited =

π ( 0.12) 2 × 400 = 4.52cm 3 / min 4

As the volume of material deposited is equal to volume of weld bead, Cross section of weld bead × welding speed = weld metal deposition rate A × 18 4.25 = 4.52 A = 0.2511 cm 2 = 25.11 mm 2

∴

74.

(A)

Resource utilisation

76.

λ 0 .2 = = 0 .5 µ 0 .4

(A) Wq = expected number of customers in queue =

λ µ (µ − λ )

as Wq = 4 (given) λ 4= µ (µ − λ ) 4µ ( µ − λ ) = λ 77.

(A)

78.

(D) Even when θ is greater than 10, optimal scheduling is not affected

79.

(C)

θ = 10 M/C A job sequence 1 2 3 4 5 82.

In 0 3 7 13 21

M/C B Out 3 7 13 21 25

In 3 8 13 23 33

Our 8 12 23 33 36

(C) One need not to solve the problem. Simply substituting the values of X1 = 6, X 2 = 6 in three constraints: X1 + X 2 + X 3 = 12 (1) 2 X1 + 3X 2 + X 4 = 30 (2) X1 + 4 X 2 + X 5 = 36 (3) X 3 = 0 , X 4 = 0, X 5 = 6 X 3 , X 4 are non basic variables. Since X 5 assumes a positive value, it must be in the optimal basis

83.

(A) Third resource is surplus and hence it marginal value is zero.

84.

(B) The optimal solution becomes degenerate by changing RHS value of third constraint to 30, but it will optimum. The X 5 turn out to be zero i.e. X 5 = 0

When basic variable (in this case X 5 ) assumes a zero value, solution becomes degenerate. 85.

(C)

86.

(A) when Q = 2, the given problem will have alternate optimal solution, given below. 12

8

11

10

2

30 6

10

30

It is an optimal solution, but degenerate optimal solution as number of occupied cells are less than (m + n − 1) ⇒ 3 + 2 − 1 = 4 87.

(B) Two paths : 1 − 2 − 3 − 4 − 5 1−2−4−5 with the duration of 23 days.

88.

(B) Cost of reduction of duration is least for D and E activities. If the duration of D and E activities is reduced by one day each, projected duration time can reduced from 23 days to 22 days. **********************