ncp 26 for pgpcm of nicmar

PGPM 22: PROJECT PLANNING & CONTROL NCP 26/ CBCM 4/IDM 21: CONSTRUCTION PROJECT MANAGEMENT TECHNIQUES

Prof. Sanjay BhoyarFaculty,

NICMAR,Pune

SANJAY BHOYAR NICMAR, PUNE

Project: Meaning

A non-routine non-repetitive work undertaken to create an unique product/ service/ structure, within stipulated cost and time.

A project is temporary, yet unique Definite Objective Definite Timeframe

Has time, cost & other resources constraints


Project Life-Cycle:


Need for Project Management: Complexity in projects Timely completion of projects Resources limitations Cost of capital Price escalation Increased outlay Longer spans Risks have increased Financers require workable plan Setting targets against which performance can be

measured


Project Management:

Application of knowledge, skills, tools & techniques of management to project activities, to achieve the project objectives with the available resources

Major functions: Planning Organizing Controlling

Quality & Performance

Cost Time

Project scope


Project Planning:

Defining the scope of the project Identifying major project activities

Design Procurement Construction

Integration of project activities Project team Project schedules Estimating project costs Financial planning Site planning

Project Management Systems:

The systems around which the managerial functions like planning, organizing & control are structured

a WBS an OBS, & Responsibility matrix Project Schedules Cost accounts – for cost aggregation & control Budgets PMIS



Work Breakdown Structure (WBS): A systematic division of a project into

work packages, sub-work packages & activities

Project: Overall scope of the project

Work packages: Project deliverables

Sub-work packages: Sub-division of deliverables

Activities: Discrete work steps

Project

Work packages

Sub Work packages

Activities


Construction & development of a Residential complex

Preliminary works

Amenities Service facilities

Drainage Bldg B Club house Swimming poolBldg ‘A’

Sub-structure Super-structure

Buildings

-Setting out

-Excavation

-PCC

-RCC footing

-RCC for PB

-Plinth filling

-RCC column-RCC slab & beams-Brickwork-Plastering-Plumbing-Flooring-Doors & windows-Electrification-Painting


Road Project

Subways/Underpasses

Minor Bridges/c/d works

Major BridgesRoad

Earthwork Pavement Drainage Lighting

Wearing surfaceSub-baseBase Sub-grade

MetallingCompacting

Hydropower Project

1. Dam

2. HRT (Transfer Tunnel)

3. U/S Surge Shaft

4. Penstock Shaft

5. Power House

6. TRT

7. Sub-station Workshop

8. Operations Building

9. Transmission Line


1. Dam1.1 Diversions

1.2 Dam wall construction

1.3 Spillway section

1.4 Gates

1.5 D/s works

1.6 Delivery tunnel

2. HRT 2.1 ADIT

2.2 Tunnel construction

2.2.1 Survey & profile marking

2.2.2 Drilling & blasting

2.2.3 Defuming & mucking

2.2.4 Bottom cleaning

2.2.5 Rock bolting

2.2.6 Rib erection, legging & backfill concrete

2.2.7 Shotcrete

3. U/S Surge Shaft

4. Penstock ShaftSANJAY BHOYAR NICMAR, PUNE

5. Power house

5.1 Civil Works

5.1.1ADIT

5.1.2 Underground cavern

5.1.3 Main access tunnel

5.1.4 Transformer chamber

5.1.5 Penstock valve chamber

5.1.6 Access tunnel

5.1.7 D/S surge chamber

5.2 Turbines & Valves

5.2.1 Turbines

5.2.2 Penstock guard valve

5.2.3 Turbine inlet valve

5.2.4 Bypass valve

5.3 Electrical Works

5.3.1 Generators

5.3.2 Switch gear

5.3.3 Power transformer

6. TRT

7. Sub-station Workshop

8. Operations Building

9. Transmission Line

SANJAY BHOYARNICMAR, PUNE


Project Scheduling:

Part of Planning

Determining the sequence & timings of activities of a project

Time-frame for the project

Effective utilization of resources

Time-cost relationship


Steps in Project Scheduling: Identifying the activities

Estimating the work content

Estimating durations of activities

Establishing the logic

Drawing the schedule

Analysis of schedule

Updating the schedule


Scheduling Techniques:

Non-Network techniques Bar chart/ Gantt chart Linear scheduling Time- chainage diagram Line of balance (LOB) technique

Network scheduling techniques CPM PERT


Network scheduling:

Series of nodes & arrows AOA / AD / ADM AON / ND / PDM

Activity: A discrete work step required for project completion Consumes time & resources

Event: An instantaneous occurrence during the progress of work Signifies specific occurrence/ substantial completion of

work Does not require resources or time


Activity-on-Arrow network (AOA): Suitable for both, event oriented as well as

activity oriented network scheduling techniques (PERT & CPM)

jiActivity

duration

1

53

2 4

6

Project startProject finish

A

B

EG

C

D

F


Activity-on-Node network (AON): Suitable for activity oriented scheduling

technique (CPM)

B

A C

D

E G

FSTART FINISH


Activity Preceding activity Succeeding activity

Duration

A - E 3

B - D 5

C - H,F,G 6

D B H 4

E A K,L 7

F C N 6

G C N 9

H C,D L 4

K E M 7

L E,H P 2

M K - 5

N F,G P 3

P L,N - 8

CPM Problem


A-O-A Network

4

7

10

2 5

1 3 6 9

8

11 12

A

B

C

D

E

F

G

H

K

L

M

N

P

3

5

6

4

7

6

9

4

7

2

5

3

8

d1

d2

d3


A-O-N Network

A

B D

C

G

F

H

E

N

L

K

P

M

START FINISH


Critical Path Method (CPM): Adequate knowledge & information about the work Certainty about

The scope & nature of work, Work content Productivity of resources

Similar works carried out in the past Time & cost estimates are fairly accurate

Deterministic & activity oriented scheduling


Scheduling by CPM:

Both, A-O-A & A-O-N can be used

Determination of activity times & floats

Identification of critical activities & critical path

Time- cost optimization


CPM analysis (using A-O-N): Activity times: EST = [EFTprec] max

EFT = EST + t

LFT = [LSTsucc] min

LST = LFT - t

Act succprec


CPM analysis… Float:

Excess time available for an activity Indicates the criticality of an activity Float = Time available – time required

Total float: TF = LST – EST or LFT - EFT

Free float: FF = ESTsucc - EFT

Act succprec


CPM analysis… Critical activity:

Lesser the float, more critical the activity Normally, activities with zero total float

Critical path: The longest path in the network Determines the Project Duration Comprises of critical activities only


730 A

44 3

0

850 B

33 5

0

660 C

00 6

0

1295 D

38 4

0

15156 G

06 9

0

15126 F

39 6

3

16139 H

312 4

0

14103 E

47 7

0

181815 N

015 3

0

181513 L

316 2

3

211710 K

414 7

0

262618 P

018 8

0

262217 M

421 5

4

STARTFINISH

0

0

26

26

LF

EFES Act

TF

LS t

FF

Description

CP: C-G-N-PPD = 26

A-O-N … PDM Precedence

Relationships:

Finish to start (FS)

Start to start (SS)

Start to finish (SF)

Finish to finish (FF)

A B

A B

A B

A B

+5

-3

+3

+4


PDM problem:

Determine the critical path & project duration

Activity Duration Predecessor Relationship with lead/ lag time

K 10 - -

L 6 K SS + 2

M 9 K FS

N 5 K FS

P 8 L FS - 2

Q 7 M SS + 5

R 4 N FS

S 12 N FS

T SF + 5

T 10 P FS

U 9 Q FS

R FS

V 7 S FS

T FS

U FF +2


LF

EFES Act

TF

LS t

Description

343427 V

027 7

323122 U

123 9

272414 T

317 10

272715 S

015 12

231915 R

419 4

232215 Q

116 7

17146 P

39 8

151510 N

010 5

201910 M

111 9

1182 L

35 6

10100 K

00 10

START0

0

2-2

5

5

FINISH

34

342

CP: K-N-S-VPD = 34


Program Evaluation & Review Technique ( PERT): Lack of knowledge & information about the work Uncertainty about

The scope & nature of work Work content Working conditions Productivity of resources

No accurate estimates of time & cost Suitable for non-repetitive type of projects

Probabilistic & event oriented scheduling technique


Scheduling by PERT:

Only A-O-A network can be used

Determination of event times & slack

Identification of critical events & critical path

Determination of probability of project completion (an event) by given date


PERT analysis: Activity time estimates Three time estimates

Optimistic time (to) Pessimistic time (tp) Most likely time (tm) Beta distribution

Expected/ mean duration of an activity

Variance of an activity

6

.4 pmoe

tttt

2

2

6

op tt


PERT analysis…Project completion (event) time: Governed by Longest (Critical) path Normal probability distribution

Average Expected project duration Te = EOT of the last event

Variance & standard deviation of project duration Var proj = ∑ var along Longest (Critical) Path

projproj var


PERT analysis…Probability calculations:

Normal value z = (Ts – Te)/ σproj

Area under cumulative normal distribution curve gives the probability of completing the project on or before the target time ‘Ts’


Activity Preceding activity

to(weeks)

tm(weeks)

tp(weeks)

A - 2 3 5

B - 3 5 7

C - 6 7 9

D A 5 7 9

E A 3 4 7

F B 1 2 3

G C 7 10 13

H D 3 5 8

K B,G 5 8 9

L B,G 2 4 6

M H 4 6 8

N E,F,K 1 3 4

P M,N 3 6 8

Q L 8 9 10

PERT Problem


Determine the critical path & the expected project duration

What is the probability of completing the project within 30 weeks?

What is the probability of completing the project within 35 weeks?

What is the project duration for 90 % certainty of on-time project completion?



to tm tp te σ²

A - 2 3 5 3.17 0.25

B - 3 5 7 5 0.44

C - 6 7 9 7.16 0.25

D A 5 7 9 7 0.44

E A 3 4 7 4.33 0.44

F B 1 2 3 2 0.11

G C 7 10 13 10 1

H D 3 5 8 5.16 0.69

K B,G 5 8 9 7.67 0.44

L B,G 2 4 6 4 0.44

M H 4 6 8 6 0.44

N E,F,K 1 3 4 2.83 0.25

P M,N 3 6 8 5.83 0.69

Q L 8 9 10 9 0.11


10

20

30

50

80

40

70

60

100

90

110

0

0

3.17

A

3.17

B5

C7.16

G

10

D7

H5.16

E4.33

F

2

L

4

Q

9

P

5.83

M6

N

2.83

10.17 15.33

5

7.16 17.16

24.83

27.67

33.50

21.16

33.50

24.50

27.67

24.83

17.167.16

21.6716.50

9.50

17.16 K7.67

i jAct

te

EOT EOT

LOT LOT

Average expected project duration, Te = 33.50

CP: 10-40-60-80-100-110 (C-G-K-N-P)

Var proj = 2.63 ; σ proj = 1.62

PERT Network

Probability of completing project in 30 weeks Te = 33.50 weeks; σproj = 1.62; Ts=30 weeks z = (Ts – Te)/ σproj

Z = (30-33.50)/1.62 = -2.16 Corresponding, p = 0.0154 = 1.54%

Probability of completing project in 35 weeks Te = 33.50 weeks; σproj = 1.62; Ts=35 weeks z = (Ts – Te)/ σproj

Z = (35-33.50)/1.62 = 0.93 Corresponding, p = 0.8238 = 82.38%

Duration for 90% certainty For, p = 0.90 => Z = 1.28 z = (Ts – Te)/ σproj

1.28 = (Ts – 33.50)/ 1.62 Ts = 35.57 weeks



Project costs: Direct costs

Attributed to a particular activity Indirect costs

Cannot be attributed to an activity Project overheads

Penalties / incentives

Time-cost trade-off Determining the minimum possible total cost for a particular

project duration.


Time- cost trade-off: Determining the minimum possible total cost for a

particular project duration.

Compression: Reducing project duration

Compressing critical activities Increase in Direct cost Reduction in Indirect cost

Crashing: Ultimate stage of compression


Time - Cost trade-off:

Project duration

Indirect Cost

Total Cost

MinimumTotal Cost

Optimum PD

Direct Cost

Cost

Normal PDMinimum PD


Steps in compression:

Determine activity compression potential & cost slope

Draw network for normal durations List all paths, with their durations Identify CP Compress activities along CP (s) Continue till at least one CP is fully

compressed (crashed)



NT(weeks)

CT(weeks)

NC(x Rs 1000)

CC(x Rs 1000)

A - 6 4 64 76

B - 4 3 30 33

C A 9 5 54 74

D A 3 2 28 34

E B 7 4 70 91

F B 5 4 82 92

G C 4 4 47 47

H D,E 6 5 39 45

K F 3 2 27 30

L G 9 6 86 98

M F,H 5 3 55 65

N K 2 1 27 29

Indirect cost is Rs. 10000 per week

Numerical problem:


Activity NT CT NC CC Compression potential

Cost slope

A 6 4 64 76 2 6

B 4 3 30 33 1 3

C 9 5 54 74 4 5

D 3 2 28 34 1 6

E 7 4 70 91 3 7

F 5 4 82 92 1 10

G 4 4 47 47 0 0

H 6 5 39 45 1 6

K 3 2 27 30 1 3

L 9 6 86 98 3 4

M 5 3 55 65 2 5

N 2 1 27 29 1 2


A(6)

START FINISH

E(7)

D(3)

C(9)

K(3)

H(6)

G(4)

N(2)

M(5)

L(9)

B(4)

F(5)

Paths Duration along path @ each stage

0 1 2 3 4

A-C-G-L 28*

A-D-H-M 20

B-E-H-M 22

B-F-M 14

B-F-K-N 14

A-O-N Network(normal durations)


Paths Duration along path @ each stage

0 1 2 3 4 5

A-C-G-L 28* 25* 22* 21* 20* 19

A-D-H-M 20 20 20 20 18 16

B-E-H-M 22 22 22* 21* 20* 19

B-F-M 14 14 14 13 12 11

B-F-K-N 14 14 14 13 13 13

A(6)

START FINISH

E(7)

D(3)

C(9)

K(3)

H(6)

G(4)

N(2)

M(5)

L(9)

B(4)

F(5)A-O-N Network(normal durations)


Compression table:Stage Act CS Dur

of comp

PD Increase

in DC

DC IC Penalty

TC

0 - - - 28 - 609 280 889

1 L 4 3 25 12 621 250 871

2 C 5 3 22 15 636 220 856

3 C 51 21 08 644 210 854

B 3

4 A 61 20 11 655 200 855

M 5

5 A 61 19 11 666 190 856

M 5

NPD

Min PD

OPD


A(6)

START FINISH

E(7)

D(3)

C(5)

K(3)

H(6)

G(4)

N(2)

M(5)

L(6)

B(3)

F(5)

A(4)

START FINISH

E(7)

D(3)

C(5)

K(3)

H(6)

G(4)

N(2)

M(3)

L(6)

B(3)

F(5)

A-O-N Network(Optimum PD)

A-O-N Network(Minimum PD)


Resources scheduling:

Scheduling the work for effective utilization of resources

Resources allocation Assigning the resources for each activity

Resources aggregation Daily resource requirement

Resources smoothing Time constrained scheduling

Resources leveling Resource constrained scheduling


Resource utilization factor (RUF)/ Effective force ratio (EFR):

RUF or EFR =

Idle Force ratio (IFR) = 1- EFR

Higher the RUF, more effective the resource utilization

Total resource-days required

Peak resource requirement x project duration


Activities, their durations & resource requirements are given below. Schedule the work for the most effective resource utilization.Activity Preceding activity Duration (days) Resource rate

A - 6 5

B - 4 4

C A 9 6

D A 3 5

E B 7 4

F B 5 2

G C 4 6

H D,E 6 3

K F 3 2

L G 9 5

M F,H 5 7

N K 2 6


Days 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 ∑

DRR (ESS)9 9 9 9 11 11 17 17 17 12 12 11 15 15 9 9 9 13 13 12 12 12 5 5 5 5 5 5 293

A/5

B/4

C/6

D/5

E/4

F/2

G/6

H/3

K/2 N/6

M/7

L/5


Days 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 ∑

DRR (ESS)9 9 9 9 11 11 17 17 17 12 12 11 15 15 9 9 9 13 13 12 12 12 5 5 5 5 5 5 293

DRR (LSS) 5 5 5 5 5 5 10 10 10 10 10 10 10 10 15 15 15 9 11 10 10 10 10 14 14 14 18 18 293

A/5

B/4

C/6

D/5

E/4

F/2

G/6

H/3

K/2 N/6

M/7

L/5

ESS

A/5

B/4

C/6

D/5

E/4

F/2

G/6

H/3

K/2 N/6

M/7

L/5

LSS


Days 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 ∑

DRR (ESS)9 9 9 9 11 11 17 17 17 12 12 11 15 15 9 9 9 13 13 12 12 12 5 5 5 5 5 5 293

DRR (Trial 1) 9 9 9 9 11 11 12 12 12 12 12 13 17 17 9 9 9 9 9 8 12 12 12 12 12 5 5 5 293

A/5

B/4

C/6

D/5

E/4

F/2

G/6

H/3

K/2 N/6

M/7

L/5

ESS

A/5

B/4

C/6

D/5

E/4

F/2

G/6

H/3

K/2 N/6

M/7

L/5

TRIAL1


Days 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 ∑

DRR (ESS)9 9 9 9 11 11 17 17 17 12 12 11 15 15 9 9 9 13 13 12 12 12 5 5 5 5 5 5 293

DRR (Trial 2) 9 9 9 9 9 9 12 12 12 12 12 11 11 11 9 9 9 11 11 10 12 12 12 12 12 11 11 5 293

A/5

B/4

C/6

D/5

E/4

F/2

G/6

H/3

K/2 N/6

M/7

L/5

ESS

A/5

B/4

C/6

D/5

E/4

F/2

G/6

H/3

K/2 N/6

M/7

L/5

TRIAL 2


Resource Histograms

0

2

4

6

8

10

12

14

16

18

20

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

Time


Time chainage diagram:

Suitable for repetitive type of projects Same activities are sequentially repeated at

regular interval/ stage/ location

Resource work continuity


Sample problem: On a 12 km long road project, following four activities are

repeated, sequentially. Prepare time-chainage schedule.

Activity Progress rate (km/day)

Duration for 12 km

Buffer Start buffer

End buffer

Start time

Finish time

Subgrade 1.0 12 - - - 0 12

Sub-base 0.5 24 2 2 - 2 26

Base 0.5 24 2 2 - 4 28

Wearing course

1.5 8 2 - 2 22 30


12

11

10

9

8

7

6

5

4

3

2

1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1617 18 19 20 21 22 23 24 25 26 27 28 29 30

Sub-grade

Sub-base

Base

Wearing

Time-chainage Diagram

Time (days)

Cha

inag

e (k

m)


Project Monitoring & Control: Monitoring:

Establishing progress of the project Periodic Review of current status

Control: Comparing the planned versus actual progress Identifying deviations & reasons for such deviations Analyzing the effect on time, cost & performance

parameters of the project Corrective actions Updating plans


Base Plan-schedule-budget -scope

Deviations -schedule-budget -scope

Actual status-schedule-budget -scope

Impact on Schedule, Budget & Scope

Corrective Actions

Revised Plan-schedule-budget -scope

Raplace Base plan

Project Control


Project time-cost control:

S- curve Project budget plan Actual cost of work performed Budgeted cost of work performed

Earned value Planned cost allocated to the completed work EV = BCWP EV = ∑ [Budgeted cost of activity x % activity completion]


Project performance status:

Schedule variance SV = BCWP - BCWS

Cost variance CV = BCWP - ACWP

Schedule performance index SPI = BCWP/ BCWS

Cost performance index CPI = BCWP/ ACWP

Project performance index = SPI x CPI Estimated cost to project completion

ECPC = BCPC/ CPI Estimated time to project completion

ETPC = STPC/ SPI

Sample problem:

Task Duration(weeks)

Preceding task

Budgeted cost (Rs in lacs)

Status by end of 5th week

Actual expenditure (Rs in lacs)

% complete

A 3 - 6 100 8

B 5 A 20 40 6

C 3 A 30 33.33 12

D 4 A 24 50 9

E 2 A 16 50 7

F 4 C 16 0 0

G 3 D,E 9 0 0

H 2 B,F,G 18 0 0


Task Duration(weeks)

Weightages

Budgeted cost (Rs in lacs)

Status by end of 5th week

Actual expenditure (Rs in lacs)

% complete

% project Progress

A 3 11.54 6 100 11.54 8

B 5 19.24 20 40 7.70 6

C 3 11.54 30 33.33 3.85 12

D 4 15.38 24 50 7.70 9

E 2 7.69 16 50 3.85 7

F 4 15.38 16 0 0 0

G 3 11.54 9 0 0 0

H 2 7.69 18 0 0 0

100 34.64SANJAY BHOYARNICMAR, PUNE

Weeks 1 2 3 4 5 6 7 8 9 10 11 12

Weekly progress (%)3.85 3.85 3.85 15.38 15.38 11.54 11.54 11.54 7.70 7.70 3.85 3.85

% progress (ESS)3.85 7.70 11.54 26.92 42.3 53.85 65.38 76.92 84.62 92.31 96.15 100

Weekly cost (ESS)2 2 2 28 28 20 14 11 7 7 9 9

Cumulative cost (ESS)2 4 6 34 62 82 96 107 114 121 130 139

ESSA/2

B/4

C/10

D/6

E/8

F/4

G/3

H/9


Status of progress: At end of 5th week

Task Duration

Budgeted cost(BC)

Scheduled Start

Scheduled Finish

% completion Cost Earned value(BCWP)Sched

uled(WS)

Actual(WP)

Scheduled(BCWS)

Actual(ACWP)

A 3 6 0 3 100 100 6 8 6

B 5 20 3 8 40 40 8 6 8

C 3 30 3 6 66.67 33.33 20 12 10

D 4 24 3 7 50 50 12 9 12

E 2 16 3 5 100 50 16 7 8

F 4 16 6 10 0 0 0 0 0

G 3 9 7 10 0 0 0 0 0

H 2 18 10 12 0 0 0 0 0

62 42 44


Review date: 5 weeks from start WS = 42.3 % WP = 34.64 % BCWS = 62 ACWP = 42 BCWP = 44 = EARNED VALUE CV = BCWP – ACWP = 44 – 42 = +2 (cost savings) SV = BCWP – BCWS = 44 – 62 = -18 (behind schedule) CPI = BCWP/ACWP = 44/42 = 1.05 SPI = BCWP/BCWS = 44/62 = 0.71 Project performance index = SPI x CPI = 0.7455 ECPC = BCPC / CPI = 139/1.05 = 132.68 lacs ETPC = STPC / SPI = 12/0.71 = 16.91 weeks



Table below gives the budgeted monthly cashflow requirement for a construction project. At end of Oct’ 06, the actual project expenditure is Rs 114 lakhs & the progress is 50% Draw ‘s’ curve, & calculate project performance indices. Estimate the delay in project completion.

Month May 06

Jun 06

Jul 06 Aug 06

Sep 06 Oct 06

Nov 06 Dec 06

Jan 07

Feb 07

Mar 07

Cash-flow

8.5 11.5 13.5 15.5 30.5 27.0 22.5 20.0 20.0 12.5 2.5

Cumm CF

8.5 20 33.5 49 79.5 106.5 129 149 169 181.5 184

Numerical Problem


S curve

0

20

40

60

80

100

120

140

160

180

200

time

cos

t


Review date: Oct’ 06 ACWP = 114 WS = (6/11)x 100 = 54.54 % WP = 50 % BCWP = (79.5+106.5) / 2 = 93 = EARNED VALUE BCWS = 106.5 CV = BCWP – ACWP = -21 SV = BCWP – BCWS = -13.5 CPI = BCWP/ACWP = 0.8158 SPI = BCWP/BCWS = 0.8732 ECPC = BCPC / CPI = 225.55 ETPC = STPC / SPI = 12.6 months Delay in project completion = ETPC – STPC = 1.6


Contact NICMAR:

www.nicmar.ac.in 020-27291342

Prof. Sanjay Bhoyar [email protected] 9552546301; 020-66859116

http://www.nicmar.ac.in/

mailto:[email protected]

ncp 26 for pgpcm of nicmar

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