construction management and administration

CONSTRUCTION MANAGEMENT AND ADMINISTRATION

UNIT - III

Unit III: list of topics

1. Normal distribution curve and network problems

2. Project cost

3. Project Time Acceleration

4. Cost time analysis in network planning

5. Time compression of critical path

6. Updating, Rescheduling

7. Simple problems of civil engineering works

Normal distribution curve and network problems

• Estimating the Probability of Completion Dates

– Determine the expected durations of all activities– Compute critical path– Assess the expected project completion time (along critical path) and SD– Determine the abscissa of normal curve in SD units as x= (scheduled date – expected date)/SD– The probability of completing the project in x units of time = Area under

the normal curve from - ∞ to x

• Estimating the …..% sure of meeting a deadline

Normal Distribution of Project Time

= tp Timex

Z

Probability

X -3s -2s -1s 0 1s 2s 3sArea, % 0.135 2.28 15.87 50 84.13 97.73 99.865

Activity Immediate predecessor Opt.Time Most likely

TimePessimistic

Time

a - 10 22 22

b - 20 20 20

c - 4 10 16

d a 2 14 32

e b,c 8 8 20

f b,c 8 14 20

g b,c 4 4 4

h c 2 12 16

I g,h 6 16 38

j d,e 2 8 14

Expected Time

20

20

10

15

10

14

4

11

18

8

Std.Dev

20

2

52

2

0

2

5

2

Expected Time

20

20

10

15

10

14

4

11

18

8

3

g 4

0 b 20

c 10

a,20

2

1 d 15

6

4j 8

f 14

e10

5

i 18

20 35

h 11

20 21

10 24

43 43

2514

3520

Variance

4

0

4

25

4

4

0

5

28

4

Critical path: a-d-j

• What is the probability of completing Project in 40days?

• Expected project completion time =43days

• SD of activities in critical path = sqrt of (4+25+4) = 5.745

• X=(scheduled date-expected date)/SD

• X= (40-43)/5.745= -0.522

• The Probability for x=-0.500 is 0.3085

• Probability of completing project in 40days is 31% approx

Assume, PM promised to complete the project in 50 days. What are the chances of meeting that deadline? Calculate X, where X = (s-m) / s

s = 50; m= 43; s =5.745;

X = (50 – 43) / 5.745 = 1.22 •The probability value of X = 1.22, is 0.888

1.22

What deadline are you 95% sure of meeting X value associated with 0.95 is 1.645 S = Exp.D + 5.745 (1.645) = 43 + 9.45 = 52.45 days Thus, there is a 95 percent chance of finishing the project by 52.45 days.

Project costProject cost broadly divided into 1. Direct Cost and 2. Indirect cost

2. Direct CostIt consists of expenditures which are chargeable to and can be identified specially with the activities of the project e.g. material cost, labor cost..

3. Indirect costIt consists of expenditures which cannot be clearly allocated to individual activities of the project e.g. establishment charges, insurance charges, administration charges…..

Direct Cost

DIRECT COST

MANPOWER

SUB-CONTRACTOR DEPARTMENTAL

MATERIAL

BASIC MATERIAL CONSUMABLES

MACHINERY

OWN HIRED

Indirect Cost

INDIRECT COST

VARIABLE COST

STAFF EXPENSES

LABOUR EXPENSES

PF &OTHER ITEMS

WELFARE EXPENSES

OTHER RATES & TAXES

WATER & POWER

CONVEYANCE EXPENSES

OPE

HIRE & RENTAL CHARGES

POSTAGE & STATIONERY

FIXED COST INFRASTRUCTURE

Project costco

s t

time

Direct cost

Indirect cost

Total cost

Total cost = Direct cost+ Indirect cost

Project AccelerationWhat are the reasons for project time acceleration?

– The contractor may wish to achieve job completion by a certain date to avoid adverse weather, to beat the annual spring runoff, to free workers and equipment for other work

– Financial arrangements may be such that it is necessary to finish certain work within a prescribed fiscal period

– The prime contractor may wish to consummate the project ahead of time to receive an early completion bonus from the owner

– Project work schedules must be adjusted to accommodate adverse job circumstances (local political/social issues)

– Revisions are often essential to meet contract time requirements– On a job in progress, the owner may desire an earlier completion date

than originally called for by the contract and may request that the contractor quote a price for expediting the work

Cost time analysis in network planning• Normal time (Tn)

– It is time for performing an activity with the normally available resources• Normal Cost (Cn)

– It is the minimum direct cost when the activity is performed in normal time duration

• Crashing– Reducing project time by expending additional resources

• Crash time (Tc)– It is the minimum time in which an activity can be performed

• Crash cost (Cc)– It is the direct cost corresponding to the crash time

Activity crashingAc

tivi ty

cos

t

Activity time

Crashing activity

Crash time

Crash cost

Normal Activity

Normal time

Normal cost

Slope = crash cost per unit time

Rate of crashing or Cost slope: It is the ratio of difference between crash and normal cost to the difference between normal and crash time

=(Cc - Cn) /(Tn –Tc )

Cost Optimization• Optimized cost

– The curve for total cost has a point where the tangent is horizontal. At this point, the total cost is minimum and is called the “optimum cost”. The time duration corresponding to the optimum cost is called optimum time

cos t

time

Direct cost

Indirect cost

Total cost

Optimal Project time

Min Total cost

Cost optimization depends on…• Rate of crashing or Cost slope• Critical path• Indirect cost

Project Time-Reduction variety of terms used to the process of shortening project time

durations– ‘‘Least-cost expediting,’’– ‘‘project compression,’’ and – ‘‘time-cost trade-off’’

Which activity to be expedited to reduce total project time?

The time required to reach any future network event, terminal or otherwise, is determined by the longest time path/critical path from the current stage of project advancement to that event

When the date of project completion is to be advanced, it is the network critical path that must be shortened

When a longest path is shortened, the floats of other activity paths leading to the same event are reduced commensurately

continued shortening of the original critical path will lead, sooner or later, to the formation of new critical paths and new critical activities

When multiple critical paths are involved, all such paths must be shortened simultaneously if the desired time advancement of the event is to be achieved

What sequence to be followed in activity crashing?

Least Cost slope activities first….

Procedure for Cost Optimization1. Determine the total cost from network of normal durations2. Calculate the cost slope of all activities3. Starting with the network of normal durations, crash the critical activity

having the least cost slope4. Redraw the network considering the above crashing and determine the

project duration and total cost5. Successively crash critical activities and determine respective project time

durations and total costs6. Determine the total cost for the all/max. crash network7. Tabulate project time durations and corresponding total costs and draw

the total cost vs. time curve8. Identify least total cost which is the optimum cost. The time

corresponding to this cost is the optimum time

Determine the optimum time duration and optimum cost. Also plot a curve of total cost vs. time and indicate on it the optimum time and optimum cost

Activity Normal Time, Tn

Crash Time, Tc

Normal Cost, Cn

Crash Cost, Cc

1-2 4 3 400 600

2-3 5 2 300 750

2-4 7 5 360 540

3-4 4 2 500 1000

1 2

3

4

Indirect Cost = Rs.250/day

1 2

3

40 4

4

5

7

4

9

13 13

9

40

Total Cost = Normal cost + Crash cost + Indirect Cost=(400+300+360+500)+0+(250*13)=(1560)+(3250)=4810 P1(13, 4810)


Crash Time, Tc

Normal Cost, Cn

Crash Cost, Cc

Cost Slope, Rs/Day

1-2 4 3 400 600

2-3 5 2 300 750

2-4 7 5 360 540

3-4 4 2 500 1000


250

90

150

200

Critical Path:1-2-3-4

Step1: determine project cost with normal time

1 2

3

4

0 4

4

5-1

7

4

8

12 12

8

40

Total Cost = Normal cost+ Crash cost + Indirect Cost=(1560)+(150*1)+250*12=4710 P2(12, 4710)


Crash Time, Tc

Normal Cost, Cn

Crash Cost, Cc

Cost Slope, Rs/Day

1-2 4 3 400 600

2-3 5 2 300 750

2-4 7 5 360 540

3-4 4 2 500 1000


250

90

150

200

Step2: Crash least cost slope activity in Critical path by 1day


1 2

3

4

0 4

4

5-2

7

4

7

11 11

7

40

Total Cost = Normal cost+ Crash cost + Indirect Cost=(1560)+(150*2)+250*11=4610 P3(11, 4610)


Crash Time, Tc

Normal Cost, Cn

Crash Cost, Cc

Cost Slope, Rs/Day

1-2 4 3 400 600

2-3 5 2 300 750

2-4 7 5 360 540

3-4 4 2 500 1000


250

90

150

200

Step3: Crash least cost slope activity in Critical path by 2days

Critical Path:1-2-3-4 &1-2-4

1 2

3

40 3

4-1

5-2

7

4

6

10 10

6

30

Total Cost = Normal cost+ Crash cost + Indirect Cost=(1560)+(200*1+150*2)+250*10=4560 P4(10, 4560)


Crash Time, Tc

Normal Cost, Cn

Crash Cost, Cc

Cost Slope, Rs/Day

1-2 4 3 400 600

2-3 5 2 300 750

2-4 7 5 360 540

3-4 4 2 500 1000


250

90

150

200

Step4: Crash activity 1-2 by 1day


1 2

3

4

0 3

4-1

5-3

7-1

4

5

9 9

5

30

Total Cost = Normal cost+ Crash cost + Indirect Cost=(1560)+(200*1+150*3+90*1)+250*9=4550 P5(9, 4550)


Crash Time, Tc

Normal Cost, Cn

Crash Cost, Cc

Cost Slope, Rs/Day

1-2 4 3 400 600

2-3 5 2 300 750

2-4 7 5 360 540

3-4 4 2 500 1000


250

90

150

200

Step5: Crash activities 2-3&2-4 by 1day further


1 2

3

40 3

3

2

5

4-1

5

8 8

5

30

Total Cost = Direct cost + Indirect Cost=(Normal cost+ Crash Cost) + Indirect Cost={(1560)+(200*1+150*3+90*2+250*1)}+250*8=4640 P6(8, 4640)


Crash Time, Tc

Normal Cost, Cn

Crash Cost, Cc

Cost Slope, Rs/Day

1-2 4 3 400 600

2-3 5 2 300 750

2-4 7 5 360 540

3-4 4 2 500 1000


250

90

150

200

Critical Path:1-2-3-4&1-2-4

Step6: determine project cost with max. crash time

Total Cost TimeP1 4810 13

P2 4710 12

P3 4610 11

P4 4560 10

P5 4550 9

P6 4640 8

5 6 7 8 9 10 11 12 13 144000410042004300440045004600470048004900

Optimum cost is Rs.4550 and Optimum time is 9days

Step7:Tabulate the calculated total cost and time values

Step8:Plot the graph of Total cost and Time

Step9:Determine Optimum cost & Optimum Time

Determine the optimum time duration and optimum cost. Also plot a curve of total cost vs. time and indicate on it the optimum time and optimum cost

Activity Normal Time, week

Crash Time, week

Normal Cost, Rs.

Crash Cost, Rs.

1-2 3 2 12000 16000

1-3 6 3 18000 24000

2-4 2 1 20000 24000

3-4 4 2 16000 21000

4-5 5 4 30000 35000

1 4

2

5

Indirect Cost = Rs.3000/week

3


Crash Time, week

Normal Cost, Rs.

Crash Cost, Rs.

1-2 3 2 12000 160001-3 6 3 18000 240002-4 2 1 20000 240003-4 4 2 16000 210004-5 5 4 30000 35000

1 4

2

5


3

15 15

Total Cost = Normal cost+ Indirect Cost=(12000+18000+20000+16000+30000)+(3000*15)=96000 + 45000=141000 P1 (15, 141000)

Step1: determine project cost with normal durations


10 10

6 6

3 8

0 0 3

6

2

4

5

What next?...& Why?

5


Crash Time, week

Normal Cost, Rs.

Crash Cost, Rs.

Cost Slope, Rs/week

1-2 3 2 12000 16000

1-3 6 3 18000 24000

2-4 2 1 20000 24000

3-4 4 2 16000 21000

4-5 5 4 30000 350002500

4000

2000

4000

1 4

2

5


3

5000

12 12

Total Cost = Normal cost+ Crash cost+ Indirect Cost=(12000+18000+20000+16000+30000)+(3*2000)+(3000*12)=96000+6000+36000=138000 P2 (12, 138000)

Step2: determine project cost with least cost slope critical activity


7 7

3 3

3 5

0 0 3

6-3

2

4

5

What next?...& Why?

2


Crash Time, week

Normal Cost, Rs.

Crash Cost, Rs.

Cost Slope, Rs/week

1-2 3 2 12000 16000

1-3 6 3 18000 24000

2-4 2 1 20000 24000

3-4 4 2 16000 21000

4-5 5 4 30000 350002500

4000

2000

4000

1 4

2

5


3

5000

10 10

Total Cost = Normal cost+ Crash cost+ Indirect Cost=(12000+18000+20000+16000+30000)+(3*2000+2*2500)+(3000*10)=96000+11000+30000=137000 P3 (10, 137000)

Step3: determine project cost with critical activity 3-4 crashing 2weeks

Critical Path:1-3-4-5 &1-2-4-5

5 5

3 3

3 3

0 0 3

6-3

2

4-2

5

What next?...& Why?


Crash Time, week

Normal Cost, Rs.

Crash Cost, Rs.

Cost Slope, Rs/week

1-2 3 2 12000 16000

1-3 6 3 18000 24000

2-4 2 1 20000 24000

3-4 4 2 16000 21000

4-5 5 4 30000 350002500

4000

2000

4000

1 4

2

5


3

5000

9 9

Total Cost = Normal cost+ Crash cost+ Indirect Cost=(12000+18000+20000+16000+30000)+(3*2000+2*2500+1*5000)+(3000*9)=96000+16000+27000=139000 P4 (9, 139000)

5 5

3 3

3 3

0 0 3

6-3

2

4-2

5-4

Step4: determine project cost with critical activity 4-5 crashing 1week

Critical Path:1-3-4-5 &1-2-4-5

What next?...& Why?


P2 138000 12

P3 137000 10

P4 139000 09

6 7 8 9 10 11 12 13 14 15 16 17 18130000

132000

134000

136000

138000

140000

142000

144000

Optimum cost is Rs.137000 and Optimum time is 10weeks

Step5:Tabulate the calculated total cost and time values

Step6:Plot the graph of Total cost and Time

Step7:Determine Optimum cost & Optimum Time

Determine the optimum time duration and optimum cost.

Activity Normal Time, month

Crash Time, month

Normal Cost, Rs.

Crash Cost, Rs.

1-2(B) 2 1 30000 32000

1-3 (C) 8 6 40000 46000

1-4 (A) 10 5 50000 75000

2-5 (D) 5 3 10000 15000

3-5 (E) 7 6 25000 26000

4-6 (F) 15 10 70000 100000

5-6 (G) 6 4 15000 23000

Indirect Cost = Rs.10,000/month

3

D, 51 B, 2

C, 8

A,10

2 6

4F, 15

G, 65

E, 7


Crash Time, month

Normal Cost, Rs.

Crash Cost, Rs. Cost Slope, Rs/month

1-2 2 1 30000 32000 20001-3 8 6 40000 46000 3000

1-4 10 5 50000 75000 50002-5 5 3 10000 15000 25003-5 7 6 25000 26000 1000

4-6 15 10 70000 100000 60005-6 6 4 15000 23000 4000

3

51 2

8

10

2 6

415

65

7

0 0

8 12

Indirect Cost = Rs.10,000/month10 10

2 14 15 19 25 25

Total Cost = Normal cost+ Crash cost+ Indirect Cost=(2,40,000)+0+(25*10000)=2,40,000+0+2,50,000=4,90,000 P1 (25, 490000)

12 4

4

What next?...& Why?


Crash Time, month

Normal Cost, Rs.


1-2 2 1 30000 32000 20001-3 8 6 40000 46000 3000

1-4 10 5 50000 75000 50002-5 5 3 10000 15000 25003-5 7 6 25000 26000 1000

4-6 15 10 70000 100000 60005-6 6 4 15000 23000 4000

3

51 2

8

10-4

2 6

415

65

7

0 0

8 8


2 10 15 15 21 21

Total Cost = Normal cost+ Crash cost+ Indirect Cost=(2,40,000)+(4*5,000)+(21*10,000)=2,40,000+20,000+2,10,000=4,70,000 P2 (21, 470000)

8

What next?...& Why?


Crash Time, month

Normal Cost, Rs.


1-2 2 1 30000 32000 20001-3 8 6 40000 46000 3000

1-4 10 5 50000 75000 50002-5 5 3 10000 15000 25003-5 7 6 25000 26000 1000

4-6 15 10 70000 100000 60005-6 6 4 15000 23000 4000

3

51 2

8

10-5

2 6

415

65

7-1

0 0

8 8


2 9 14 14 20 20

Total Cost = Normal cost+ Crash cost+ Indirect Cost=(240000)+(5*5000+1*1000)+(20*10000)=2,40,000+26,000+2,00,000=4,66,000 P3 (20, 466000)

7

What next?...& Why?


Crash Time, month

Normal Cost, Rs.


1-2 2 1 30000 32000 20001-3 8 6 40000 46000 3000

1-4 10 5 50000 75000 50002-5 5 3 10000 15000 25003-5 7 6 25000 26000 1000

4-6 15 10 70000 100000 60005-6 6 4 15000 23000 4000

3

51 2

8-2

10-5

2 6

415-2

65

7-1

0 0

6 6


2 7 12 12 18 18

Total Cost = Normal cost+ Crash cost+ Indirect Cost=(240000)+(5*5000+1*1000+2*3000+2*6000)+(18*10000)=2,40,000+44,000+1,80,000=4,64,000 P4 (18, 464000)

5

What next?...& Why?


Crash Time, month

Normal Cost, Rs.


1-2 2 1 30000 32000 20001-3 8 6 40000 46000 3000

1-4 10 5 50000 75000 50002-5 5 3 10000 15000 25003-5 7 6 25000 26000 1000

4-6 15 10 70000 100000 60005-6 6 4 15000 23000 4000

3

51 B, 2

8-2

10-5

2 6

415-3

6-15

7-1

0 0

6 6


2 7 12 12 17 17

Total Cost = Normal cost+ Crash cost+ Indirect Cost=(240000)+(5*5000+1*1000+2*3000+3*6000+1*4000)+(17*10000)=2,40,000+=464000 P5 (17, 464000)

5

What next?...& Why?


Crash Time, month

Normal Cost, Rs.


1-2 2 1 30000 32000 20001-3 8 6 40000 46000 3000

1-4 10 5 50000 75000 50002-5 5 3 10000 15000 25003-5 7 6 25000 26000 1000

4-6 15 10 70000 100000 60005-6 6 4 15000 23000 4000

3

51 2

8-2

10-5

2 6

415-4

6-25

7-1

0 0

6 6


2 7 12 12 16 16

Total Cost = Normal cost+ Crash cost+ Indirect Cost=(240000)+(5*5000+1*1000+2*3000+4*6000+2*4000)+(16*10000)=464000 P6 (16, 464000)

5

What next?...& Why?


P2 470000 21

P3 466000 20

P4 464000 18

P5 464000 17

P6 464000 16

15 17 19 21 23 25 27 29450000455000460000465000470000475000480000485000490000495000500000

Optimum cost is Rs.464000 and Optimum time is 16months

Tabulate the calculated total cost and time values

Plot the graph of Total cost and Time

Determine Optimum cost & Optimum Time

Project Time Monitoring• As construction proceeds, diversions from the established plan and schedule

inevitably occur, some of the reasons are– Inaccurate estimation of the duration of completed activities– Unforeseen climatic conditions– Non supply of materials on time– Changes in the scope of work– Inadequate resources– Labor strikes, Bandh etc.

• Unforeseen job circumstances result in changes in – Activity durations,– Activity delays, and– Changes in project logic

• As such deviations occur and accumulate, the true job status diverges further and further from that indicated by the programmed plan and schedule

Project Time Monitoring• Time monitoring of complex projects can broadly be divided into

the following three stages1. Measuring the progress of current activities2. Updating sub-project plans3. Updating the project master schedule/original network

Measuring the progress of current activities• The state of activities is measured by comparing their actual progress against

the programmed schedule• At any point of time, activities can be classified in to

– Completed activities– In-progress activities and – Still to start activities

• The completed activities and the non-starter activities can be easily identified• Measurement of the in-progress activities is considered from two angles, i.e.

time performance and physical performance (work done quantity performance) – Time performance in terms of time units– Physical performance in terms of % of work done quantity

• The progress reports gives management only a general idea of the time status of the job

• To know the overall time status of the project as of the date of the last progress report , it needs a complete updating calculation

What is Network update?

Why Network update?

The method used for displaying progress of activities on the planning charts, corresponding to a given time is called Network updating

•To continue to provide realistic management guidance, it becomes necessary to incorporate the changes and deviations into the working operational program

• The basic objective of an update is to reschedule the work yet to be done using the current project status as a starting point

• Updating reveals the current time posture of the job, indicates whether expediting actions are in order, and provides guidance concerning how best to keep the job on schedule

• An update is also very valuable in testing the effectiveness of proposed time-recovery measures.

• Updating involves making necessary network corrections and recalculating activity and float times

• It is concerned entirely with determining the effect of schedule deviations and plan changes on the portion of the project yet to be completed

• It would keep job management continuously up-to-date on the time status of the work and would assure prompt and informed remedial action when needed

Network Updating….…

Network Updating method

2

Completed activity Completed event

Partially Completed activity

4

D

4

D

2

Still to start activity

4

G

3

New activities as a result of changes in the scope of work, should be incorporated logically into the network, and their durations written

Compute the EFT of the network to determine the minimum time required for the completion of the remaining work

Set the LFT equal to project time objective in the network. Time analyze the updated network


0A0 0

42

3C

2

B

34

1D

5

5

7H

3

G

38

6E

2

F

1

J

2

5 7

4 4

4 5 7 8

7 7

7 7

10 10

7 9

Original Time scaled Network

• Progress of work of a project at the end of 6th week


S.No.

Activity Duration Work done value (in 1000$)

Code Description original Balance of work

Total %

1 0-1 D 5 2 100 602 1-6 E 2 2 10 Nil3 6-8 F 1 1 5 Nil4 5-8 G 3 3 15 Nil5 0-2 A 4 0 20 1006 2-4 B 3 2 30 337 0-3 C 2 0 20 1008 3-7 H 3 3 45 Nil9 7-8 J 2 2 15 Nil

Total 260 42.3%


0A0 0

42

3C

2

B

34

1D

5

5

7H3

G3 8

6E

2

F

1

J2

2 7

0 0

0 0 3 3

2 2

2 2

5 5

4 9

End of 6th week, updated Time scaled Network

2

2

Rescheduling• Resource smoothing is accomplished through rescheduling• Equipment conflict can be removed by rescheduling one of the

activities involved by using its float or by adding a precedence restraint to the project network. – Adding a precedence restraint means using a dependency line to show

that one of the activities involved must follow the other rather than parallel it

• The conflict must be removed by rescheduling activities with the least possible increase in project duration and cost

‘‘Manpower leveling’’ is the process of smoothing out daily labor demands. Perfection in this regard can never be attained, but often the worst of the inequities can be removed through a process of selective rescheduling of noncritical activities

Assignment -31. Define terms

1. Project Normal time, Normal cost2. Crash Cost, Crash time3. Cost slope4. Optimized cost and duration

2. Differentiate direct cost and indirect cost3. Explain the procedure for determining optimum cost and

duration for a CPM network4. Explain network updating and rescheduling

construction management and administration

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