1. CONSTRUCTION MANAGEMENT AND ADMINISTRATION UNIT - III

2. 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

3. 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

4. Normal Distribution of Project Time  = t p Time x ZZ Probability X-3 s -2 s -1 s 01s1s 2s2s 3s3s Area, %0.1352.2815.875084.1397.7399.865

5. Activity Immediate predecessor Opt.Time Most likely Time Pessimistic Time a-1022 b-20 c-41016 da21432 eb,c8820 fb,c81420 gb,c444 hc21216 Ig,h61638 jd,e2814 Expected Time 20 10 15 10 14 4 11 18 8

6. Std.Dev 2 0 2 5 2 2 0 2 5 2 Expected Time 20 10 15 10 14 4 11 18 8 3 g 4 0 b 20 c 10 a,20 2 1 d 15 6 4 j 8 f 14 e10 5 i 18 2035 h 11 20 21 10 24 43 25 14 35 20 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

7. 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

8. 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.

9. Project cost Project cost broadly divided into 1.Direct Cost and 2. Indirect cost 1.Direct Cost It consists of expenditures which are chargeable to and can be identified specially with the activities of the project e.g. material cost, labor cost.. 2.Indirect cost It consists of expenditures which cannot be clearly allocated to individual activities of the project e.g. establishment charges, insurance charges, administration charges…..

10. Direct Cost DIRECT COSTMANPOWER SUB- CONTRACTOR DEPARTMENTALMATERIALBASIC MATERIALCONSUMABLESMACHINERYOWNHIRED

11. 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 COSTINFRASTRUCTURE

12. Project cost cost time Direct cost Indirect cost Total cost Total cost = Direct cost+ Indirect cost

13. Project Acceleration What 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

14. Cost time analysis in network planning Normal time (T n ) – 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

15. Activity crashing Activity cost 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 )

16. 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 cost time Direct cost Indirect cost Total cost Optimal Project time Min Total cost

17. 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’’

18. 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….

19. Procedure for Cost Optimization 1.Determine the total cost from network of normal durations 2.Calculate the cost slope of all activities 3.Starting with the network of normal durations, crash the critical activity having the least cost slope 4.Redraw the network considering the above crashing and determine the project duration and total cost 5.Successively crash critical activities and determine respective project time durations and total costs 6.Determine the total cost for the all/max. crash network 7.Tabulate project time durations and corresponding total costs and draw the total cost vs. time curve 8.Identify least total cost which is the optimum cost. The time corresponding to this cost is the optimum time

20. 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 ActivityNormal Time, Tn Crash Time, Tc Normal Cost, Cn Crash Cost, Cc 1-243400600 2-352300750 2-475360540 3-4425001000 12 3 4 Indirect Cost = Rs.250/day

21. 12 3 4 0 4 4 5 7 4 9 13 9 4 0 Total Cost = Normal cost + Crash cost + Indirect Cost =(400+300+360+500)+0+(250*13) =(1560)+(3250) =4810P1(13, 4810) ActivityNormal Time, Tn Crash Time, Tc Normal Cost, Cn Crash Cost, Cc Cost Slope, Rs/Day 1-243400600 2-352300750 2-475360540 3-4425001000 Indirect Cost = Rs.250/day 250 90 150 200 Critical Path:1-2-3-4 Step1: determine project cost with normal time

22. 12 3 4 0 4 4 5-1 7 4 8 12 8 4 0 Total Cost = Normal cost+ Crash cost + Indirect Cost =(1560)+(150*1)+250*12 =4710P2(12, 4710) ActivityNormal Time, Tn Crash Time, Tc Normal Cost, Cn Crash Cost, Cc Cost Slope, Rs/Day 1-243400600 2-352300750 2-475360540 3-4425001000 Indirect Cost = Rs.250/day 250 90 150 200 Step2: Crash least cost slope activity in Critical path by 1day Critical Path:1-2-3-4

23. 12 3 4 04 4 5-2 7 4 7 11 7 4 0 Total Cost = Normal cost+ Crash cost + Indirect Cost =(1560)+(150*2)+250*11 =4610P3(11, 4610) ActivityNormal Time, Tn Crash Time, Tc Normal Cost, Cn Crash Cost, Cc Cost Slope, Rs/Day 1-243400600 2-352300750 2-475360540 3-4425001000 Indirect Cost = Rs.250/day 250 90 150 200 Step3: Crash least cost slope activity in Critical path by 2days Critical Path:1-2-3-4 &1-2-4

24. 12 3 4 0 3 4-1 5-2 7 4 6 10 6 3 0 Total Cost = Normal cost+ Crash cost + Indirect Cost =(1560)+(200*1+150*2)+250*10 =4560P4(10, 4560) ActivityNormal Time, Tn Crash Time, Tc Normal Cost, Cn Crash Cost, Cc Cost Slope, Rs/Day 1-243400600 2-352300750 2-475360540 3-4425001000 Indirect Cost = Rs.250/day 250 90 150 200 Step4: Crash activity 1-2 by 1day Critical Path:1-2-3-4 &1-2-4

25. 12 3 4 03 4-1 5-3 7-1 4 5 9 9 5 3 0 Total Cost = Normal cost+ Crash cost + Indirect Cost =(1560)+(200*1+150*3+90*1)+250*9 =4550P5(9, 4550) ActivityNormal Time, Tn Crash Time, Tc Normal Cost, Cn Crash Cost, Cc Cost Slope, Rs/Day 1-243400600 2-352300750 2-475360540 3-4425001000 Indirect Cost = Rs.250/day 250 90 150 200 Step5: Crash activities 2-3&2-4 by 1day further Critical Path:1-2-3-4 &1-2-4

26. 12 3 4 0 3 3 2 5 4-1 5 8 8 5 3 0 Total Cost = Direct cost + Indirect Cost =(Normal cost+ Crash Cost) + Indirect Cost ={(1560)+(200*1+150*3+90*2+250*1)}+250*8 =4640P6(8, 4640) ActivityNormal Time, Tn Crash Time, Tc Normal Cost, Cn Crash Cost, Cc Cost Slope, Rs/Day 1-243400600 2-352300750 2-475360540 3-4425001000 Indirect Cost = Rs.250/day 250 90 150 200 Critical Path:1-2-3-4 &1-2-4 Step6: determine project cost with max. crash time

27. Total CostTime P1481013 P2471012 P3461011 P4456010 P545509 P646408 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

28. 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 ActivityNormal Time, week Crash Time, week Normal Cost, Rs. Crash Cost, Rs. 1-2321200016000 1-3631800024000 2-4212000024000 3-4421600021000 4-5543000035000 14 2 5 Indirect Cost = Rs.3000/week 3

29. ActivityNormal Time, week Crash Time, week Normal Cost, Rs. Crash Cost, Rs. 1-2321200016000 1-3631800024000 2-4212000024000 3-4421600021000 4-5543000035000 14 2 5 Indirect Cost = Rs.3000/week 3 15 Total Cost = Normal cost+ Indirect Cost =(12000+18000+20000+16000+30000)+(3000*15) =96000 + 45000 =141000P1 (15, 141000) Step1: determine project cost with normal durations Critical Path:1-3-4-5 10 66 38 00 3 6 2 4 5 What next?...& Why? 5

30. ActivityNormal Time, week Crash Time, week Normal Cost, Rs. Crash Cost, Rs. Cost Slope, Rs/week 1-2321200016000 1-3631800024000 2-4212000024000 3-4421600021000 4-5543000035000 2500 4000 2000 4000 14 2 5 Indirect Cost = Rs.3000/week 3 5000 12 Total Cost = Normal cost+ Crash cost+ Indirect Cost =(12000+18000+20000+16000+30000)+(3*2000)+(3000*12) =96000+6000+36000 =138000P2 (12, 138000) Step2: determine project cost with least cost slope critical activity Critical Path:1-3-4-5 77 33 35 00 3 6-3 2 4 5 What next?...& Why? 2

31. ActivityNormal Time, week Crash Time, week Normal Cost, Rs. Crash Cost, Rs. Cost Slope, Rs/week 1-2321200016000 1-3631800024000 2-4212000024000 3-4421600021000 4-5543000035000 2500 4000 2000 4000 14 2 5 Indirect Cost = Rs.3000/week 3 5000 10 Total Cost = Normal cost+ Crash cost+ Indirect Cost =(12000+18000+20000+16000+30000)+(3*2000+2*2500)+(3000*10) =96000+11000+30000 =137000P3 (10, 137000) Step3: determine project cost with critical activity 3-4 crashing 2weeks Critical Path:1-3-4-5 &1-2-4-5 55 33 33 00 3 6-3 2 4-2 5 What next?...& Why?

32. ActivityNormal Time, week Crash Time, week Normal Cost, Rs. Crash Cost, Rs. Cost Slope, Rs/week 1-2321200016000 1-3631800024000 2-4212000024000 3-4421600021000 4-5543000035000 2500 4000 2000 4000 14 2 5 Indirect Cost = Rs.3000/week 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 =139000P4 (9, 139000) 55 33 3 3 00 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?

33. Total CostTime P114100015 P213800012 P313700010 P413900009 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

34. Determine the optimum time duration and optimum cost. ActivityNormal Time, month Crash Time, month Normal Cost, Rs. Crash Cost, Rs. 1-2(B)213000032000 1-3 (C)864000046000 1-4 (A)1055000075000 2-5 (D)531000015000 3-5 (E)762500026000 4-6 (F)151070000100000 5-6 (G)641500023000 Indirect Cost = Rs.10,000/month 3 D, 5 1 B, 2 C, 8 A,10 2 6 4 F, 15 G, 6 5 E, 7

35. ActivityNormal Time, month Crash Time, month Normal Cost, Rs. Crash Cost, Rs.Cost Slope, Rs/month 1-22130000320002000 1-38640000460003000 1-410550000750005000 2-55310000150002500 3-57625000260001000 4-61510700001000006000 5-66415000230004000 3 5 1 2 8 10 2 6 4 15 6 5 7 00 812 Indirect Cost = Rs.10,000/month 10 2 14 1519 25 Total Cost = Normal cost+ Crash cost+ Indirect Cost =(2,40,000)+0+(25*10000) =2,40,000+0+2,50,000 =4,90,000P1 (25, 490000) 12 4 4 What next?...& Why?

36. ActivityNormal Time, month Crash Time, month Normal Cost, Rs. Crash Cost, Rs.Cost Slope, Rs/month 1-22130000320002000 1-38640000460003000 1-410550000750005000 2-55310000150002500 3-57625000260001000 4-61510700001000006000 5-66415000230004000 3 5 1 2 8 10-4 2 6 4 15 6 5 7 00 88 Indirect Cost = Rs.10,000/month 66 2 10 15 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 =4,70,000P2 (21, 470000) 8 What next?...& Why?

37. ActivityNormal Time, month Crash Time, month Normal Cost, Rs. Crash Cost, Rs.Cost Slope, Rs/month 1-22130000320002000 1-38640000460003000 1-410550000750005000 2-55310000150002500 3-57625000260001000 4-61510700001000006000 5-66415000230004000 3 5 1 2 8 10-5 2 6 4 15 6 5 7-1 00 88 Indirect Cost = Rs.10,000/month 55 2 9 14 20 Total Cost = Normal cost+ Crash cost+ Indirect Cost =(240000)+(5*5000+1*1000)+(20*10000) =2,40,000+26,000+2,00,000 =4,66,000P3 (20, 466000) 7 What next?...& Why?

38. ActivityNormal Time, month Crash Time, month Normal Cost, Rs. Crash Cost, Rs.Cost Slope, Rs/month 1-22130000320002000 1-38640000460003000 1-410550000750005000 2-55310000150002500 3-57625000260001000 4-61510700001000006000 5-66415000230004000 3 5 1 2 8-2 10-5 2 6 4 15-2 6 5 7-1 00 66 Indirect Cost = Rs.10,000/month 55 2 7 12 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 =4,64,000P4 (18, 464000) 5 What next?...& Why?

39. ActivityNormal Time, month Crash Time, month Normal Cost, Rs. Crash Cost, Rs.Cost Slope, Rs/month 1-22130000320002000 1-38640000460003000 1-410550000750005000 2-55310000150002500 3-57625000260001000 4-61510700001000006000 5-66415000230004000 3 5 1 B, 2 8-2 10-5 2 6 4 15-3 6-1 5 7-1 00 66 Indirect Cost = Rs.10,000/month 55 2 7 12 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+ =464000P5 (17, 464000) 5 What next?...& Why?

40. ActivityNormal Time, month Crash Time, month Normal Cost, Rs. Crash Cost, Rs.Cost Slope, Rs/month 1-22130000320002000 1-38640000460003000 1-410550000750005000 2-55310000150002500 3-57625000260001000 4-61510700001000006000 5-66415000230004000 3 5 1 2 8-2 10-5 2 6 4 15-4 6-2 5 7-1 00 66 Indirect Cost = Rs.10,000/month 55 2 7 12 16 Total Cost = Normal cost+ Crash cost+ Indirect Cost =(240000)+(5*5000+1*1000+2*3000+4*6000+2*4000)+(16*10000) =464000P6 (16, 464000) 5 What next?...& Why?

41. Total CostTime P149000025 P247000021 P346600020 P446400018 P546400017 P646400016 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

42. 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

43. Project Time Monitoring Time monitoring of complex projects can broadly be divided into the following three stages 1.Measuring the progress of current activities 2.Updating sub-project plans 3.Updating the project master schedule/original network

44. 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

45. 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

46. 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….…

47. Network Updating method 2 Completed activityCompleted 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

48. Network Updating method 0 A 00 4 2 3 C 2 B 3 4 1 D 5 5 7 H 3 G 3 8 6 E 2 F 1 J 2 57 44 4578 77 77 10 79 Original Time scaled Network

49. Progress of work of a project at the end of 6 th week Network Updating method S.No. ActivityDurationWork done value (in 1000$) CodeDescriptionoriginalBalance of work Total% 10-1D5210060 21-6E2210Nil 36-8F115Nil 45-8G3315Nil 50-2A4020100 62-4B323033 70-3C2020100 83-7H3345Nil 97-8J2215Nil Total26042.3%

50. Network Updating method 0 A 00 4 2 3 C 2 B 3 4 1 D 5 5 7 H 3 G 3 8 6 E 2 F 1 J 2 27 00 0033 22 22 55 49 End of 6 th week, updated Time scaled Network 2 2

51. To make a network update, information as of the cutoff date is required concerning the work that has yet to be done: New activities that must be added to the network Existing activities that are to be deleted Changes in job logic Changes in original material delivery or other resource availability dates Estimated times to complete all activities presently under way but not yet completed Changes in estimated activity durations Changes in the scope of the work Network Updating….…

52. 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

54. Assignment -3 1.Define terms 1.Project Normal time, Normal cost 2.Crash Cost, Crash time 3.Cost slope 4.Optimized cost and duration 2.Differentiate direct cost and indirect cost 3.Explain the procedure for determining optimum cost and duration for a CPM network 4.Explain network updating and rescheduling