CONSTRUCTION MANAGEMENT AND ADMINISTRATION UNIT III Unit III

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 Probability Z = tp x Time X -3 s -2 s -1 s 0 1 s 2 s 3 s Area, % 0. 135 2. 28 15. 87 50 84. 13 97. 73 99. 865

Activity Immediate predecessor Opt. Time Most likely Pessimistic Time a - 10 22 22 20 b - 20 20 c - 4 10 16 10 d a 2 14 32 15 e b, c 8 8 20 10 f b, c 8 14 20 14 g b, c 4 4 h c 2 12 16 11 I g, h 6 16 38 18 j d, e 2 8 14 8 Expected Time

Expected Time Std. Dev Variance • What is the probability of completing Project in 40 days? • Expected project completion time =43 days 20 2 4 20 0 0 • SD of activities in critical path = sqrt of (4+25+4) = 5. 745 10 2 4 • X=(scheduled date-expected date)/SD 15 5 25 10 2 4 • X= (40 -43)/5. 745= -0. 522 14 2 4 • The Probability for x=-0. 500 is 0. 3085 4 0 0 • Probability of completing project in 40 days is 31% approx 11 2 5 18 5 28 8 2 4 20 20 a, 20 0 20 21 j 8 3 10 14 43 43 e 10 f 14 2 g 4 c 10 Critical path: a-d-j 4 d 15 1 b 20 35 35 h 11 6 i 18 5 24 25

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 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…. .

Direct Cost DIRECT COST MANPOWER SUBCONTRACTOR DEPARTMENTAL MACHINERY MATERIAL BASIC MATERIAL CONSUMABLES OWN HIRED

Indirect Cost STAFF EXPENSES LABOUR EXPENSES PF &OTHER ITEMS WELFARE EXPENSES OTHER RATES & TAXES VARIABLE COST WATER & POWER CONVEYANCE EXPENSES INDIRECT COST OPE HIRE & RENTAL CHARGES POSTAGE & STATIONERY FIXED COST INFRASTRUCTURE

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

Project Acceleration üWhat 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 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 ) Crash cost Crashing activity Activity cost Slope = crash cost per unit time Normal cost Crash time Normal Activity Normal time Activity 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 Direct cost Indirect cost Optimal Project time

Cost optimization depends on… • Rate of crashing or Cost slope • Critical path • Indirect cost Project Time-Reduction ü variety of terms used to the process of shortening project time durations – ‘‘Least-cost expediting, ’’ – ‘‘project compression, ’’ and – ‘‘time-cost trade-off’’

ü Which activity to be expedited to reduce total project time? ü 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 ü When the date of project completion is to be advanced, it is the network critical path that must be shortened ü When a longest path is shortened, the floats of other activity paths leading to the same event are reduced commensurately ü continued shortening of the original critical path will lead, sooner or later, to the formation of new critical paths and new critical activities ü 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 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

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 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 Indirect Cost = Rs. 250/day

Activity Normal Time, Tn Crash Time, Tc Normal Cost, Cn Crash Cost, Cc Cost Slope, Rs/Day 200 1 -2 4 3 400 600 2 -3 5 2 300 750 150 2 -4 7 5 360 540 90 3 -4 4 2 500 1000 250 4 13 Indirect Cost = Rs. 250/day Step 1: determine project cost with normal time 9 9 3 0 4 0 1 4 4 5 2 Total Cost = Normal cost + Crash cost + Indirect Cost =(400+360+500)+0+(250*13) =(1560)+(3250) =4810 P 1(13, 4810) 7 4 Critical Path: 1 -2 -3 -4 13

Activity Normal Time, Tn Crash Time, Tc Normal Cost, Cn Crash Cost, Cc Cost Slope, Rs/Day 200 1 -2 4 3 400 600 2 -3 5 2 300 750 150 2 -4 7 5 360 540 90 3 -4 4 2 500 1000 250 Indirect Cost = Rs. 250/day Step 2: Crash least cost slope activity in Critical path by 1 day 8 8 3 0 0 1 4 4 4 2 Total Cost = Normal cost+ Crash cost + Indirect Cost =(1560)+(150*1)+250*12 =4710 P 2(12, 4710) 5 -1 4 7 12 12 4 Critical Path: 1 -2 -3 -4

Activity Normal Time, Tn Crash Time, Tc Normal Cost, Cn Crash Cost, Cc Cost Slope, Rs/Day 200 1 -2 4 3 400 600 2 -3 5 2 300 750 150 2 -4 7 5 360 540 90 3 -4 4 2 500 1000 250 Indirect Cost = Rs. 250/day Step 3: Crash least cost slope activity in Critical path by 2 days 3 0 0 4 1 4 5 -2 4 2 Total Cost = Normal cost+ Crash cost + Indirect Cost =(1560)+(150*2)+250*11 =4610 P 3(11, 4610) 7 7 4 7 11 11 4 Critical Path: 1 -2 -3 -4 &1 -2 -4

Activity Normal Time, Tn Crash Time, Tc Normal Cost, Cn Crash Cost, Cc 1 -2 4 3 400 600 2 -3 5 2 300 750 150 2 -4 7 5 360 540 90 3 -4 4 2 500 1000 250 Indirect Cost = Rs. 250/day 6 Step 4: Crash activity 1 -2 by 1 day 0 0 1 6 3 3 4 -1 Cost Slope, Rs/Day 200 3 4 5 -2 2 Total Cost = Normal cost+ Crash cost + Indirect Cost =(1560)+(200*1+150*2)+250*10 =4560 P 4(10, 4560) 7 10 10 4 Critical Path: 1 -2 -3 -4 &1 -2 -4

Activity Normal Time, Tn Crash Time, Tc Normal Cost, Cn Crash Cost, Cc 1 -2 4 3 400 600 2 -3 5 2 300 750 150 2 -4 7 5 360 540 90 3 -4 4 2 500 1000 250 Indirect Cost = Rs. 250/day 0 0 1 3 4 -1 3 5 5 Step 5: Crash activities 2 -3&2 -4 by 1 day further Cost Slope, Rs/Day 200 3 4 5 -3 2 Total Cost = Normal cost+ Crash cost + Indirect Cost =(1560)+(200*1+150*3+90*1)+250*9 =4550 P 5(9, 4550) 7 -1 9 9 4 Critical Path: 1 -2 -3 -4 &1 -2 -4

Activity Normal Time, Tn Crash Time, Tc Normal Cost, Cn Crash Cost, Cc Cost Slope, Rs/Day 200 1 -2 4 3 400 600 2 -3 5 2 300 750 150 2 -4 7 5 360 540 90 3 -4 4 2 500 1000 250 Indirect Cost = Rs. 250/day 5 Step 6: determine project cost with max. crash time 5 3 0 1 3 3 4 -1 2 2 Total Cost = Direct cost + Indirect Cost =(Normal cost+ Crash Cost) + Indirect Cost ={(1560)+(200*1+150*3+90*2+250*1)}+250*8 =4640 P 6(8, 4640) 5 8 4 Critical Path: 1 -2 -3 -4 &1 -2 -4 8

Total Cost Time P 1 4810 13 P 2 4710 12 Step 8: Plot the graph of Total cost and Time P 3 4610 11 P 4 4560 10 Step 9: Determine Optimum cost & Optimum Time 4900 4800 4700 4600 4500 4400 4300 4200 4100 4000 5 6 P 5 4550 9 P 6 4640 8 Step 7: Tabulate the calculated total cost and time values Optimum cost is Rs. 4550 and Optimum time is 9 days 7 8 9 10 11 12 13 14

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

Activity Normal Time, week Crash Time, week Normal Cost, Rs. Crash Cost, Rs. 1 -2 3 2 12000 16000 Cost Slope, Rs/week 4000 1 -3 6 3 18000 24000 2 -4 2 1 20000 24000 3 -4 4 2 16000 21000 2500 4 -5 5 4 30000 35000 Step 2: determine project cost with least cost slope critical activity 3 5 Indirect Cost = Rs. 3000/week 2 2 0 0 1 3 6 -3 2 3 3 3 7 12 7 4 5 4 Total Cost = Normal cost+ Crash cost+ Indirect Cost =(12000+18000+20000+16000+30000)+(3*2000)+(3000*12) =96000+36000 =138000 P 2 (12, 138000) 12 5 Critical Path: 1 -3 -4 -5 What next? . . . & Why?

Activity Normal Time, week Crash Time, week Normal Cost, Rs. Crash Cost, Rs. 1 -2 3 2 12000 16000 Cost Slope, Rs/week 4000 1 -3 6 3 18000 24000 2 -4 2 1 20000 24000 3 -4 4 2 16000 21000 2500 4 -5 5 4 30000 35000 Step 3: determine project cost with critical activity 3 -4 crashing 2 weeks 3 3 2 0 0 1 3 6 -3 2 3 3 3 4 -2 5 Indirect Cost = Rs. 3000/week 5 4 10 10 5 5 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 P 3 (10, 137000) What next? . . . & Why?

Activity Normal Time, week Crash Time, week Normal Cost, Rs. Crash Cost, Rs. 1 -2 3 2 12000 16000 Cost Slope, Rs/week 4000 1 -3 6 3 18000 24000 2 -4 2 1 20000 24000 3 -4 4 2 16000 21000 2500 4 -5 5 4 30000 35000 Step 4: determine project cost with critical activity 4 -5 crashing 1 week 3 3 0 0 1 2 3 6 -3 2 3 3 3 4 -2 5 Indirect Cost = Rs. 3000/week 5 4 9 9 5 -4 5 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 P 4 (9, 139000) What next? . . . & Why?

Step 5: Tabulate the calculated total cost and time values Step 6: Plot the graph of Total cost and Time Total Cost Time P 1 141000 15 P 2 138000 12 P 3 137000 10 P 4 139000 09 Step 7: Determine Optimum cost & Optimum Time 144000 142000 140000 138000 136000 134000 132000 130000 Optimum cost is Rs. 137000 and Optimum time is 10 weeks 6 7 8 9 10 11 12 13 14 15 16 17 18

Determine the optimum time duration and optimum cost. 4 A, 10 1 2 B, 2 C, 8 F, 15 D, 5 5 6 G, 6 E, 7 3 Activity Normal Time, month Crash Time, month Normal Cost, 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

Activity Normal Time, month Crash Time, month Normal Cost, Rs. Crash Cost, Rs. Cost Slope, Rs/month 1 -2 2 1 30000 32000 1 -3 8 6 40000 46000 3000 1 -4 10 5 50000 75000 2 -5 5 3 10000 15000 2500 3 -5 7 6 25000 26000 1000 4 -6 15 10 70000 100000 6000 5 -6 6 4 15000 0 4 10 0 1 8 5 2 2 12 8 3 10 10 14 2 12 23000 4000 Indirect Cost = Rs. 10, 000/month 15 15 19 5 4 25 6 7 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 P 1 (25, 490000) What next? . . . & Why?

Activity Normal Time, month Crash Time, month Normal Cost, Rs. Crash Cost, Rs. Cost Slope, Rs/month 1 -2 2 1 30000 32000 1 -3 8 6 40000 46000 3000 1 -4 10 5 50000 75000 2 -5 5 3 10000 15000 2500 3 -5 7 6 25000 26000 1000 4 -6 15 10 70000 100000 6000 5 -6 6 4 15000 10 -4 0 0 1 2 2 8 6 4 2 10 6 15 5 15 8 8 3 8 23000 4000 Indirect Cost = Rs. 10, 000/month 5 15 6 21 7 Total Cost = Normal cost+ Crash cost+ Indirect Cost =(2, 40, 000)+(4*5, 000)+(21*10, 000) =2, 40, 000+2, 10, 000 =4, 70, 000 P 2 (21, 470000) What next? . . . & Why?

Activity Normal Time, month Crash Time, month Normal Cost, Rs. Crash Cost, Rs. Cost Slope, Rs/month 1 -2 2 1 30000 32000 1 -3 8 6 40000 46000 3000 1 -4 10 5 50000 75000 2 -5 5 3 10000 15000 2500 3 -5 7 6 25000 26000 1000 4 -6 15 10 70000 100000 6000 5 -6 6 4 15000 10 -5 0 0 1 2 2 8 5 4 2 9 5 15 5 14 7 8 3 8 23000 4000 Indirect Cost = Rs. 10, 000/month 5 14 6 20 7 -1 Total Cost = Normal cost+ Crash cost+ Indirect Cost =(240000)+(5*5000+1*1000)+(20*10000) =2, 40, 000+26, 000+2, 000 =4, 66, 000 P 3 (20, 466000) What next? . . . & Why?

Activity Normal Time, month Crash Time, month Normal Cost, Rs. Crash Cost, Rs. Cost Slope, Rs/month 1 -2 2 1 30000 32000 1 -3 8 6 40000 46000 3000 1 -4 10 5 50000 75000 2 -5 5 3 10000 15000 2500 3 -5 7 6 25000 26000 1000 4 -6 15 10 70000 100000 6000 5 -6 6 4 15000 10 -5 0 0 1 2 2 8 -2 2 7 3 6 5 15 -2 5 5 6 5 4 23000 4000 Indirect Cost = Rs. 10, 000/month 12 5 12 6 18 7 -1 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 P 4 (18, 464000) What next? . . . & Why?

Activity Normal Time, month Crash Time, month Normal Cost, Rs. Crash Cost, Rs. Cost Slope, Rs/month 1 -2 2 1 30000 32000 1 -3 8 6 40000 46000 3000 1 -4 10 5 50000 75000 2 -5 5 3 10000 15000 2500 3 -5 7 6 25000 26000 1000 4 -6 15 10 70000 100000 6000 5 -6 6 4 15000 10 -5 0 0 1 2 B, 2 8 -2 2 7 3 6 5 15 -3 5 5 6 5 4 23000 4000 Indirect Cost = Rs. 10, 000/month 12 5 12 6 -1 17 6 17 7 -1 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 P 5 (17, 464000) What next? . . . & Why?

Activity Normal Time, month Crash Time, month Normal Cost, Rs. Crash Cost, Rs. Cost Slope, Rs/month 1 -2 2 1 30000 32000 1 -3 8 6 40000 46000 3000 1 -4 10 5 50000 75000 2 -5 5 3 10000 15000 2500 3 -5 7 6 25000 26000 1000 4 -6 15 10 70000 100000 6000 5 -6 6 4 15000 10 -5 0 0 1 2 2 8 -2 2 7 3 6 5 15 -4 5 5 6 5 4 23000 4000 Indirect Cost = Rs. 10, 000/month 12 5 12 6 -2 16 6 16 7 -1 Total Cost = Normal cost+ Crash cost+ Indirect Cost =(240000)+(5*5000+1*1000+2*3000+4*6000+2*4000)+(16*10000) =464000 P 6 (16, 464000) What next? . . . & Why?

Tabulate the calculated total cost and time values Plot the graph of Total cost and Time Determine Optimum cost & Optimum Time Total Cost Time P 1 490000 25 P 2 470000 21 P 3 466000 20 P 4 464000 18 P 5 464000 17 P 6 464000 16 500000 490000 480000 470000 460000 450000 Optimum cost is Rs. 464000 and Optimum time is 16 months 15 17 19 21 23 25 27 29

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 stages 1. Measuring the progress of current activities 2. Updating sub-project plans 3. 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? 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

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 timerecovery 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 method D 4 Completed activity D 4 2 Completed event G 2 Partially Completed activity 4 3 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

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

Network Updating method • Progress of work of a project at the end of 6 th week Activity S. No. Duration Work done value (in 1000$) Code Description original Balance of work Total % 1 0 -1 D 5 2 100 60 2 1 -6 E 2 2 10 Nil 3 6 -8 F 1 1 5 Nil 4 5 -8 G 3 3 15 Nil 5 0 -2 A 4 0 20 100 6 2 -4 B 3 2 30 33 7 0 -3 C 2 0 20 100 8 3 -7 H 3 3 45 Nil 9 7 -8 J 2 2 15 Nil 260 42. 3% Total

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

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 -3 1. Define terms 1. 2. 3. 4. Project Normal time, Normal cost Crash Cost, Crash time Cost slope 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