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CONSTRUCTION MANAGEMENT AND ADMINISTRATION

CONSTRUCTION MANAGEMENT AND ADMINISTRATION . UNIT - III . Unit III: list of topics. Normal distribution curve and network problems Project cost Project Time Acceleration Cost time analysis in network planning Time compression of critical path Updating, Rescheduling

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CONSTRUCTION MANAGEMENT AND ADMINISTRATION

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  1. CONSTRUCTION MANAGEMENT AND ADMINISTRATION UNIT - III

  2. Unit III: list of topics • Normal distribution curve and network problems • Project cost • Project Time Acceleration • Cost time analysis in network planning • Time compression of critical path • Updating, Rescheduling • 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. Probability Z  = tp x Time Normal Distribution of Project Time

  5. 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 a,20 20 d 15 35 1 20 35 4 j 8 43 e10 43 b 20 0 21 20 f 14 6 2 c 10 g 4 i 18 Critical path: a-d-j 5 3 h 11 25 24 14 10

  6. 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

  7. What deadline are you 95% sure of meeting Xvalue 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.

  8. Project cost Project cost broadly divided into • Direct Cost and 2. Indirect cost • 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.. • 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…..

  9. Direct Cost

  10. Indirect Cost

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

  12. 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

  13. 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

  14. Crashing activity Crash cost Activity cost Normal Activity Normal cost Crash time Normal time Activity time Activity crashing • 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 ) Slope = crash cost per unit time

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

  16. 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’’

  17. 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….

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

  19. 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 3 1 2 4 Indirect Cost = Rs.250/day

  20. 200 150 90 250 Indirect Cost = Rs.250/day Step1: determine project cost with normal time 9 9 3 5 4 4 13 13 4 0 0 1 2 4 4 7 Critical Path:1-2-3-4 Total Cost = Normal cost + Crash cost + Indirect Cost =(400+300+360+500)+0+(250*13) =(1560)+(3250) =4810 P1(13, 4810)

  21. 200 150 90 250 Indirect Cost = Rs.250/day Step2: Crash least cost slope activity in Critical path by 1day 8 8 3 12 5-1 4 12 4 4 0 0 1 2 4 4 7 Critical Path:1-2-3-4 Total Cost = Normal cost+ Crash cost + Indirect Cost =(1560)+(150*1)+250*12 =4710 P2(12, 4710)

  22. 200 150 90 250 Indirect Cost = Rs.250/day Step3: Crash least cost slope activity in Critical path by 2days 7 7 3 5-2 4 11 11 0 0 4 4 1 2 4 4 7 Critical Path:1-2-3-4 &1-2-4 Total Cost = Normal cost+ Crash cost + Indirect Cost =(1560)+(150*2)+250*11 =4610 P3(11, 4610)

  23. 200 150 90 250 Indirect Cost = Rs.250/day 6 6 Step4: Crash activity 1-2 by 1day 3 4 5-2 10 3 0 10 3 0 1 2 4 4-1 7 Critical Path:1-2-3-4 &1-2-4 Total Cost = Normal cost+ Crash cost + Indirect Cost =(1560)+(200*1+150*2)+250*10 =4560 P4(10, 4560)

  24. 200 150 90 250 Indirect Cost = Rs.250/day 5 5 Step5: Crash activities 2-3&2-4 by 1day further 3 4 5-3 9 3 0 9 0 3 1 2 4 4-1 7-1 Critical Path:1-2-3-4 &1-2-4 Total Cost = Normal cost+ Crash cost + Indirect Cost =(1560)+(200*1+150*3+90*1)+250*9 =4550 P5(9, 4550)

  25. 200 150 90 250 Indirect Cost = Rs.250/day 5 5 Step6: determine project cost with max. crash time 3 4-1 2 8 3 8 3 0 0 1 2 4 3 5 Critical Path:1-2-3-4 &1-2-4 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)

  26. Step7: Tabulate the calculated total cost and time values Step8: Plot the graph of Total cost and Time Step9: Determine Optimum cost & Optimum Time Optimum cost is Rs.4550 and Optimum time is 9days

  27. 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 2 1 4 5 3 Indirect Cost = Rs.3000/week

  28. Step1: determine project cost with normal durations Indirect Cost = Rs.3000/week 3 8 5 2 15 0 0 10 10 15 3 2 1 4 5 5 6 6 6 4 3 Critical Path:1-3-4-5 Total Cost = Normal cost+ Indirect Cost =(12000+18000+20000+16000+30000)+(3000*15) =96000 + 45000 =141000 P1 (15, 141000) What next?...& Why?

  29. 4000 2000 4000 2500 5000 Step2: determine project cost with least cost slope critical activity Indirect Cost = Rs.3000/week 2 3 5 2 12 12 0 0 3 2 7 7 1 4 5 5 3 3 6-3 4 3 Critical Path:1-3-4-5 Total Cost = Normal cost+ Crash cost+ Indirect Cost =(12000+18000+20000+16000+30000)+(3*2000)+(3000*12) =96000+6000+36000 =138000 P2 (12, 138000) What next?...& Why?

  30. 4000 2000 4000 2500 5000 Step3: determine project cost with critical activity 3-4 crashing 2weeks Indirect Cost = Rs.3000/week 3 3 2 10 0 0 5 5 10 3 2 1 4 5 5 3 3 6-3 4-2 3 Critical Path:1-3-4-5 &1-2-4-5 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) What next?...& Why?

  31. 4000 2000 4000 2500 5000 Step4: determine project cost with critical activity 4-5 crashing 1week Indirect Cost = Rs.3000/week 3 3 2 9 0 0 5 5 9 3 2 1 4 5 5-4 3 3 6-3 4-2 3 Critical Path:1-3-4-5 &1-2-4-5 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) What next?...& Why?

  32. Step5: Tabulate the calculated total cost and time values Step6: Plot the graph of Total cost and Time Step7: Determine Optimum cost & Optimum Time Optimum cost is Rs.137000 and Optimum time is 10weeks

  33. 4 Determine the optimum time duration and optimum cost. F, 15 D, 5 2 5 6 G, 6 E, 7 3 A,10 1 B, 2 C, 8 Indirect Cost = Rs.10,000/month

  34. 4 15 5 2 5 6 6 7 3 Indirect Cost = Rs.10,000/month 10 10 10 0 0 14 2 15 19 25 25 12 4 1 2 8 12 8 4 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) What next?...& Why?

  35. Indirect Cost = Rs.10,000/month 6 6 4 15 10-4 0 0 10 2 15 15 21 21 1 5 2 2 5 6 6 8 8 7 8 8 3 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) What next?...& Why?

  36. Indirect Cost = Rs.10,000/month 5 5 4 15 10-5 0 0 9 2 14 14 20 20 1 5 2 2 5 6 6 7 8 7-1 8 8 3 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) What next?...& Why?

  37. Indirect Cost = Rs.10,000/month 5 5 4 15-2 10-5 0 0 7 2 12 12 18 18 1 5 2 2 5 6 6 5 8-2 7-1 6 6 3 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) What next?...& Why?

  38. Indirect Cost = Rs.10,000/month 5 5 4 15-3 10-5 0 0 7 2 12 12 17 17 1 5 2 B, 2 5 6 6-1 5 8-2 7-1 6 6 3 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) What next?...& Why?

  39. Indirect Cost = Rs.10,000/month 5 5 4 15-4 10-5 0 0 7 2 12 12 16 16 1 5 2 2 5 6 6-2 5 8-2 7-1 6 6 3 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) What next?...& Why?

  40. Tabulate the calculated total cost and time values Plot the graph of Total cost and Time Determine Optimum cost & Optimum Time Optimum cost is Rs.464000 and Optimum time is 16months

  41. 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

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

  43. 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

  44. What is Network update? The method used for displaying progress of activities on the planning charts, corresponding to a given time is called Network updating Why Network update? • To continue to provide realistic management guidance, it becomes necessary to incorporate the changes and deviations into the working operational program

  45. Network Updating….… • 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

  46. Network Updating method D 2 4 Completed activity Completed event D G 2 3 4 4 Partially Completed activity Still to start activity 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

  47. Network Updating method Original Time scaled Network 5 7 7 9 D E F 6 1 5 2 1 10 10 G 5 8 7 7 3 0 0 4 4 A B 0 2 4 4 3 7 7 C H J 3 7 2 3 4 5 7 8 2

  48. Network Updating method • Progress of work of a project at the end of 6th week

  49. Network Updating method End of 6th week, updated Time scaled Network 2 7 4 9 D E F 6 1 2 1 5 2 5 5 G 5 8 2 2 3 0 0 0 0 A B 0 2 4 4 2 3 2 2 C H J 3 7 3 2 2 0 0 3 3

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