1. M. Sundaram Tenn. Tech1 MANAGING PROJECTS USING NETWORK TECHNIQUES

2. M. Sundaram Tenn. Tech2 Project Management  Project Management is one of the world’s most in- demand skill sets and is one of the fastest growing professional disciplines in North America.  Project Management is used by large corporations, governments, and smaller organizations to standardize and reduce the tasks necessary to complete a project in the most effective and efficient manner.  Engineers that master in project management skills may lead improvement initiatives that result in measurable growth in return on investment, economic value added, sales growth, customer satisfaction and retention, market share, time to market, employee satisfaction, and employee motivation.  PMI provides certification in project management

3. M. Sundaram Tenn. Tech3 Network based Techniques-Outline  Project Planning – An Introduction  Development of Project Network  Identifying Critical paths  Probabilistic Analysis in PERT Networks  How to use the Normal table  Project Cost Control  Resource Allocation  EVA for monitoring progress of Projects

4. M. Sundaram Tenn. Tech4 PERT and CPM PERT (Program Evaluation Review Technique) was developed by a joint team set up by the U.S. Navy Special Projects Office that included representatives from Lockheed Aircraft Corporation (Prime contractor of the POLARIS program) and from the consulting company Booze, Allen, and Hamilton. The objective of this team was to develop an integrated planning and control system for the Polaris missile submarine program which would help avoid the time and cost overruns that had plagued other such development programs. An important feature of the PERT approach is its statistical treatment of the uncertainty in activity time estimate which involved the collection of three separate time estimates and the calculation of probability estimates of meeting specified schedule dates. The three estimates used are: Optimistic time, Pessimistic time, and Most likely time. PERT networks are usually to manage projects that have several uncertain activities

5. M. Sundaram Tenn. Tech5 PERT and CPM- Contd. CPM (Critical Path Method) evolved from a parallel joint effort initiated originally at DuPont and later expanded to include Remington Rand Univac and Mauchly Associates. The two key differences of this approach from PERT: (1) the use of only one time estimate for each activity (and thus no statistical treatment of uncertainty) and (2) the inclusion, as an integral part of the overall scheme, of a procedure for time/cost tradeoff to minimize the sum of direct and indirect project costs. An important common feature of both PERT and CPM is the use of a network diagram for project representation in which arrows represent activities ("activity-on­arrow"). A modification of this approach involves the representation of activities by circles, with arrows indicating precedence ("activity-on-node").

6. M. Sundaram Tenn. Tech6 Applications of PERT and CPM Construction projects (e.g.) buildings, highways, houses, and bridges.) Preparation of bids and proposals for large projects. Maintenance planning of oil refineries, ships, chemical plants and other large scale operations. Planning for relocating a facility Manufacture and assembly of large and complex products such as airplanes, ships, and mainframe computers. Simple projects such as home remodeling, moving to a new house, and home cleaning and painting. Design and development of new products Facilities planning and implementation of new layouts in manufacturing. Development of computer Software packages

7. M. Sundaram Tenn. Tech7 What is a design project? Design Project – Unique sequence of activities (work tasks) required to be performed in developing a product. cost Design project performance time Changing the length of any side of the project triangle affects the other sides!

8. M. Sundaram Tenn. Tech8 Project Planning A Simple Project Network DesignFabricate 132

9. M. Sundaram Tenn. Tech9 Managing a design project? Design problem –FUNCTION (customer & company requirements) Solution - FORM (manufacturing specifications) Activities (decision making processes) Develop a project plan then execute the plan

10. M. Sundaram Tenn. Tech10 Why should we plan a design project? WHAT ? ……...scope of work tasks WHEN ? ……...schedule HOW MUCH?..budget WHO?………...organization chart, responsibilities table Without a clear roadmap, how will you get where you need to go? In planning a design project we make decisions which answer the following questions

11. M. Sundaram Tenn. Tech11 Work breakdown structure

12. M. Sundaram Tenn. Tech12 Table 3.4 Example scope of work for section 1. 1.0 Design Problem Formulation 1.1 Visit Site, Meet with customers to determine desired attributes and parameters 1.2 Complete QFD/HOQ Determine requirements, engineering characteristics 1.3 Satisfaction Curves, Determine the satisfaction curves for each engineering characteristic. 1.4 Create EDS List in-use purposes for the product List product performance requirements 1.5 Conduct Benchmarking Research existing products that are currently available Contact manufacturers and request brochures Analyze the competition for functionality and performance 1.6 Contact Customers Make phone calls to determine pros and cons of current unit Set an appointment time to witness existing product operation 1.7 Determine parameters State problem definition parameters State design variables State solution evaluation parameters for satisfaction curves 1.8 Determine Schedule Refine work breakdown structure Assign a time value to each task Prepare Gantt chart 1.9 Calculate Budget Determine total number of engineering hours Determine total number of expert faculty hours Sum all hours and material cost 1.10 Outline Work Scope 1.11 Prepare for and conduct design review meeting. 1.0 Design Problem Formulation 1.1 Visit Site, Meet with customers, determine desired attributes and parameters 1.2 Complete QFD/HOQ Determine requirements, engineering characteristics 1.3 Satisfaction Curves, Determine the satisfaction curves for each engineering characteristic. 1.4 Create EDS List in-use purposes for the product List product performance requirements 1.5 Conduct Benchmarking Research existing products that are currently available Contact manufacturers and request brochures Analyze the competition for functionality and performance Scope of work (partial)

13. M. Sundaram Tenn. Tech13 Responsibilities table Who does what assist responsible

14. M. Sundaram Tenn. Tech14 Project schedule

15. M. Sundaram Tenn. Tech15 Project Budget

16. M. Sundaram Tenn. Tech16 Elements of Project Management Project team Individuals from different departments within company Matrix organization Team structure with members from different functional areas depending on skills needed Project manager Leader of project team

17. M. Sundaram Tenn. Tech17 Project Planning Statement of work Written description of goals, work & time frame of project Activities require labor, resources & time Precedence relationship shows sequential relationship of project activities

18. M. Sundaram Tenn. Tech18 Project Control All activities identified and included Completed in proper sequence Resource needs identified Schedule adjusted Maintain schedule and budget Complete on time

19. M. Sundaram Tenn. Tech19 A Gantt Chart Popular tool for project scheduling Popular tool for project scheduling Graph with bar for representing the time for each task Graph with bar for representing the time for each task Provides visual display of project schedule Provides visual display of project schedule Also shows slack for activities Also shows slack for activities Amount of time activity can be delayed without delaying project Amount of time activity can be delayed without delaying project

20. M. Sundaram Tenn. Tech20 A Gantt Chart |||||||||| Activity Design house and obtain financing Lay foundation Order and receive materials Build house Select paint Select carpet Finish work 0246810 MonthMonth 13579135791357913579 Figure 6.2

21. M. Sundaram Tenn. Tech21 CPM/PERT- A Little History Critical Path Method (CPM) DuPont & Remington-Rand (1956) Deterministic task times Activity-on-node network construction Project Eval. & Review Technique (PERT) US Navy, Booz, Allen & Hamilton Multiple task time estimates Activity-on-arrow network construction

22. M. Sundaram Tenn. Tech22 The Project Network Network consists of arcs & nodes 132 ArcNode Figure 6.3

23. M. Sundaram Tenn. Tech23 Network Construction In AON, nodes represent activities & arrows show precedence relationships In AOA, arrows represent activities & nodes are events for points in time An event is the completion or beginning of an activity A dummy shows precedence for two activities with same start & end nodes

24. M. Sundaram Tenn. Tech24 Project Network for a House 3 20 1 3 11 1 12467 3 5 Lay foundation Design house and obtain financing Order and receive materials Dummy Finish work Select carpet Select paint Build house Figure 6.4

25. M. Sundaram Tenn. Tech25 Concurrent Activities 23 Lay foundation Order material (a)Incorrect precedence relationship (b)Correct precedence relationship 3 42 Dummy Layfoundation Order material 1 20

26. M. Sundaram Tenn. Tech26 Example Problem -1 Develop an activity-on-arrow (AOA) type network for the precedence relationship shown below. ActivityPredecessor A- BA CA DB,C EC FD,E

27. M. Sundaram Tenn. Tech27 Solution

28. M. Sundaram Tenn. Tech28 Example Problem -2 Develop an A-O-A type network from the description below. Activity Predecessor A- B- CA DB EA FC,D GC,D,E HF

29. M. Sundaram Tenn. Tech29 Solution

30. M. Sundaram Tenn. Tech30 Critical Path A path is a sequence of connected activities running from start to end node in network A path is a sequence of connected activities running from start to end node in network The critical path is the path with the longest duration in the network The critical path is the path with the longest duration in the network Project cannot be completed in less than the time of the critical path Project cannot be completed in less than the time of the critical path

31. M. Sundaram Tenn. Tech31 The Critical Path A:1-2-3-4-6-7 3 + 2 + 0 + 3 + 1 = 9 months B:1-2-3-4-5-6-7 3 + 2 + 0 + 1 + 1 + 1 = 8 months C:1-2-4-6-7 3 + 1 + 3 + 1 = 8 months D:1-2-4-5-6-7 3 + 1 + 1 + 1 + 1 = 7 months 3 20 1 3 11 1 12467 3 5 Lay foundation Design house and obtain financing Order and receive materials Dummy Finish work Select carpet Select paint Build house

32. M. Sundaram Tenn. Tech32 The Critical Path A:1-2-3-4-6-7 3 + 2 + 0 + 3 + 1 = 9 months B:1-2-3-4-5-6-7 3 + 2 + 0 + 1 + 1 + 1 = 8 months C:1-2-4-6-7 3 + 1 + 3 + 1 = 8 months D:1-2-4-5-6-7 3 + 1 + 1 + 1 + 1 = 7 months 3 20 1 3 11 1 12467 3 5 Lay foundation Design house and obtain financing Order and receive materials Dummy Finish work Select carpet Select paint Build house The Critical Path

33. M. Sundaram Tenn. Tech33 The Critical Path 3 20 1 3 11 1 12467 3 5 Lay foundation Design house and obtain financing Order and receive materials Dummy Finish work Select carpet Select paint Build house 12467 3 5 3 20 1 3 11 Start at 3 months Start at 5 months 1 Finish at 9 months Start at 8 months Activity Start Times

34. M. Sundaram Tenn. Tech34 Early Times ES - earliest time activity can start Forward pass starts at beginning of CPM/PERT network to determine ES times EF = ES + activity time ES ij = maximum (EF i ) EF ij = ES ij - t ij ES 12 = 0 EF 12 = ES 12 + t 12 = 0 + 3 = 3 months 3 20 1 3 11 1 12467 3 5 Lay foundation Design house and obtain financing Order and receive materials Dummy Finish work Select carpet Select paint Build house

35. M. Sundaram Tenn. Tech35 Computing Early Times ES 23 = max EF 2 = 3 months ES 46 = max EF 4 = max 5,4 = 5 months EF 46 = ES 46 + t 46 = 5 + 3 = 8 months EF 67 = 9 months, the project duration 3 20 1 3 11 1 12467 3 5 Lay foundation Design house and obtain financing Order and receive materials Dummy Finish work Select carpet Select paint Build house

36. M. Sundaram Tenn. Tech36 Computing Early Times 3 20 1 3 11 1 12467 3 5 Lay foundation Design house and obtain financing Order and receive materials Dummy Finish work Select carpet Select paint Build house 12467 3 5 3 20 1 3 11 1 (ES = 0, EF = 3) (ES = 5, EF = 8) (ES = 3, EF = 5) (ES = 3, EF = 4) (ES = 5, EF = 6)(ES = 6, EF = 7) (ES = 8, EF = 9) (ES = 5, EF = 5) Early Start and Finish Times

37. M. Sundaram Tenn. Tech37 Late Times LS - latest time activity can start & not delay project Backward pass starts at end of CPM/PERT network to determine LS times LF = LS + activity time LS ij = LF ij - t ij LF ij = minimum (LS j )

38. M. Sundaram Tenn. Tech38 Computing Late Times LF 67 = 9 months LS 67 = LF 67 - t 67 = 9 - 1 = 8 months LF 56 = minimum (LS 6 ) = 8 months LS 56 = LF 56 - t 56 = 8 - 1 = 7 months LF 24 = minimum (LS 4 ) = min(5, 6) = 5 months LS 24 = LF 24 - t 24 = 5 - 1 = 4 months 3 20 1 3 11 1 12467 3 5 Lay foundation Design house and obtain financing Order and receive materials Dummy Finish work Select carpet Select paint Build house

39. M. Sundaram Tenn. Tech39 Computing Late Times LF 67 = 9 months LS 67 = LF 67 - t 67 = 9 - 1 = 8 months LF 56 = minimum (LS 6 ) = 8 months LS 56 = LF 56 - t 56 = 8 - 1 = 7 months LF 24 = minimum (LS 4 ) = min(5, 6) = 5 months LS 24 = LF 24 - t 24 = 5 - 1 = 4 months 3 20 1 3 11 1 12467 3 5 Lay foundation Design house and obtain financing Order and receive materials Dummy Finish work Select carpet Select paint Build house

40. M. Sundaram Tenn. Tech40 12467 3 5 3 20 1 3 11 1 ES = 3, EF = 5 LS = 3, LF = 5 ( ) ES = 5, EF = 8 LS = 5, LF = 8 ( ) ES = 3, EF = 4 LS = 4, LF = 5 ( ) ES = 0, EF = 3 LS = 0, LF = 3 ( ) ES = 5, EF = 5 LS = 5, LF = 5 ( ) ES = 5, EF = 6 LS = 6, LF = 7 ( ) ES = 8, EF = 9 LS = 8, LF = 9 ( ) ES = 6, EF = 7 LS =7, LF = 8 ( ) Early and Late Start and Finish Times

41. M. Sundaram Tenn. Tech41 Activity Slack Activities on critical path have ES = LS & EF = LF Activities not on critical path have slack S ij = LS ij - ES ij S ij = LF ij - EF ij S 24 = LS 24 - ES 24 = 4 - 3 = 1 month

42. M. Sundaram Tenn. Tech42 Activity Slack Data 3 20 1 3 11 1 12467 3 5 Lay foundation Design house and obtain financing Order and receive materials Dummy Finish work Select carpet Select paint Build house Activity LSESLFEFSlack *1-200330 *2-333550 2-443541 *3-455550 4-565761 *4-655880 5-676871 *6-788990 * Critical path

43. M. Sundaram Tenn. Tech43 Activity Slack Data 3 20 1 3 11 1 12467 3 5 Lay foundation Design house and obtain financing Order and receive materials Dummy Finish work Select carpet Select paint Build house ActivityLSESLFEFSlacks *1-200330 *2-333550 2-443541 *3-455550 4-565761 *4-655880 5-676871 *6-788990 * Critical path 12467 3 5 3 20 1 3 11 1 Activity Slack S = 0 S = 1 S = 0

44. M. Sundaram Tenn. Tech44 Other Methods of Determining the Critical Path Simple Method Tabular Method Enumeration of all paths

45. M. Sundaram Tenn. Tech45 Simple Method

46. M. Sundaram Tenn. Tech46 Enumeration of all paths

47. M. Sundaram Tenn. Tech47 Probabilistic Time Estimates Reflect uncertainty of activity times Beta distribution is used in PERT

48. M. Sundaram Tenn. Tech48 Probabilistic Time Estimates Reflect uncertainty of activity times Beta distribution is used in PERT a = optimistic estimate m = most likely time estimate b = pessimistic time estimate where Mean (expected time): t = a + 4 m + b 6 Variance:  2 = b - a 6 2

49. M. Sundaram Tenn. Tech49 Example Beta Distributions P(time) Time amtbamtb m = t Time Time ba

50. M. Sundaram Tenn. Tech50 Kat Tech Company System changeover 2 4 6 17359 8 Manual Testing Dummy System Training Dummy System Testing Orientation Position recruiting System development Equipment installation Equipment testing and modification Final debugging Job training a b cde f g h i jk l m

51. M. Sundaram Tenn. Tech51 Activity Estimates 1 - 2681080.44 1 - 336961.00 1 - 413530.44 2 - 500000.00 2 - 6 241252.78 3 - 5 23430.11 4 - 534540.11 4 - 822220.00 5 - 7371171.78 5 - 824640.44 7 - 800000.00 6 - 914741.00 7 - 91101394.00 TIME ESTIMATES (WKS)MEAN TIMEVARIANCE ACTIVITY ambt σ 2 2 4 6 17359 8

52. M. Sundaram Tenn. Tech52 2 4 6 17359 8 Early and Late Times For Activity 1-2 a = 6, m = 8, b = 10 t = = = 8 weeks a + 4 m + b 6 6 + 4(8) + 10 6  2 = = = week b - a 6 2 10 - 6 6249

53. M. Sundaram Tenn. Tech53 2 4 6 17359 8 Early and Late Times ACTIVITY t σ 2 ESEFLSLFS 1 - 280.4408191 1 - 361.0006060 1 - 430.4403252 2 - 500.0088991 2 - 6 52.7881316218 3 - 5 30.1169690 4 - 540.1137592 4 - 820.0035141611 5 - 771.789169160 5 - 840.4491312163 7 - 800.00131316163 6 - 941.00131721258 7 - 994.00162516250

54. M. Sundaram Tenn. Tech54 Kal Tech Company 2 4 6 17359 8 ES = 9, EF = 16 LS = 9, LF = 16 ES = 0, EF = 8 LS = 1, LF = 9 ES = 0, EF = 6 LS = 0, LF = 6 ES = 6, EF = 9 LS = 6, LF = 9 ES = 0, EF = 3 LS = 2, LF = 5 ES = 3, EF = 7 LS = 5, LF = 9 ES = 9, EF = 13 LS = 12, LF = 16 ES = 8, EF = 8 LS = 9, LF = 9 ES = 13, EF = 13 LS = 16, LF = 16 ES = 3, EF = 5 LS = 14, LF = 16 ES = 16, EF = 25 LS = 21, LF = 25 ES = 13, EF = 17 LS = 21, LF = 25 ES = 8, EF = 13 LS = 16, LF = 21 8 5 4 637 9 3 2 40 0

55. M. Sundaram Tenn. Tech55 Kal Tech Company 2 4 6 17359 8 ES = 9, EF = 16 LS = 9, LF = 16 ES = 0, EF = 8 LS = 1, LF = 9 ES = 0, EF = 6 LS = 0, LF = 6 ES = 6, EF = 9 LS = 6, LF = 9 ES = 0, EF = 3 LS = 2, LF = 5 ES = 3, EF = 7 LS = 5, LF = 9 ES = 9, EF = 13 LS = 12, LF = 16 ES = 8, EF = 8 LS = 9, LF = 9 ES = 13, EF = 13 LS = 16, LF = 16 ES = 3, EF = 5 LS = 14, LF = 16 ES = 16, EF = 25 LS = 21, LF = 25 ES = 13, EF = 17 LS = 21, LF = 25 ES = 8, EF = 13 LS = 16, LF = 21 8 5 4 637 9 3 2 40 0  2 =  2 +  2 +  2 +  2 = 1.00 + 0.11 + 1.78 + 4.00 = 6.89 weeks 13355779 Total project variance

56. M. Sundaram Tenn. Tech56 Probabilistic Network Analysis Determine probability that project is completed within specified time where  = t p = project mean time  =project standard deviation x =proposed project time Z =number of standard deviations x is from mean Z =Z =Z =Z = x -  

57. M. Sundaram Tenn. Tech57 Normal Distribution Of Project Time  = t p Timex ZZProbability

58. M. Sundaram Tenn. Tech58 Kal Tech Company What is the probability that the project is completed within 30 weeks?  = 25 Time (weeks) x = 30 P( x  30 weeks)

59. M. Sundaram Tenn. Tech59 Kal Tech Company What is the probability that the project is completed within 30 weeks? Example 6.2  = 25 Time (weeks) x = 30 P( x  30 weeks)  2 = 6.89 weeks  = 6.89  = 2.62 weeks Z = = = 1.91 x -   30 - 25 2.62

60. M. Sundaram Tenn. Tech60 Kal Tech Company What is the probability that the project is completed within 30 weeks?  = 25 Time (weeks) x = 30 P( x  30 weeks)  2 = 6.89 weeks  = 6.89  = 2.62 weeks Z = = = 1.91 x -   30 - 25 2.62 From Table A.1, a Z score of 1.91 corresponds to a probability of 0.4719. Thus P(30) = 0.4719 + 0.5000 = 0.9719

61. M. Sundaram Tenn. Tech61 Kal Tech Company What is the probability that the project is completed within 22 weeks?  = 25 Time (weeks) x = 22 P( x  22 weeks) 0.3729

62. M. Sundaram Tenn. Tech62 Kal Tech Company What is the probability that the project is completed within 22 weeks?  2 = 6.89 weeks  = 6.89  = 2.62 weeks Z = = = -1.14 x -   22 - 25 2.62  = 25 Time (weeks) x = 22 P( x  22 weeks) 0.3729

63. M. Sundaram Tenn. Tech63 What is the probability that the project is completed within 22 weeks?  2 = 6.89  = 6.89  = 2.62 Z = = = -1.14 x -   22 - 25 2.62 From Table A.1, a Z score of -1.14 corresponds to a probability of 0.3729. Thus P(22) = 0.5000 - 0.3729 = 0.1271  = 25 Time (weeks) x = 22 P( x  22 weeks) 0.3729 Kal Tech Company

64. M. Sundaram Tenn. Tech64 Another Example problem For the PERT network given below, Determine the critical path and the expected length of the critical path. i. Compute the probability of completing the project in 20 days. ii. What is the likely project duration that the project manager can be confident with 95% certainty? ActivityNodeamb A 1  2 117 B 1  3 147 C 1  4 228 D 2  5 111 E 3  5 2514 F 4  6 258 G 5656 3615

65. M. Sundaram Tenn. Tech65 Solution

66. M. Sundaram Tenn. Tech66 Project Crashing Crashing is reducing project time by expending additional resources Crashing is reducing project time by expending additional resources Crash time is an amount of time an activity is reduced Crash time is an amount of time an activity is reduced Crash cost is the cost of reducing the activity time Crash cost is the cost of reducing the activity time Goal is to reduce project duration at minimum cost Goal is to reduce project duration at minimum cost

67. M. Sundaram Tenn. Tech67 Housebuilding Network 12 80 412 44 4 12467 3 5

68. M. Sundaram Tenn. Tech68 12 80 412 44 4 12467 3 5 Housebuilding Network $7,000 – $6,000 – $5,000 – $4,000 – $3,000 – $2,000 – $1,000 – – ||||||| 02468101214Weeks Normal activity Normal time Normal cost

69. M. Sundaram Tenn. Tech69 12 80 412 44 4 12467 3 5 Housebuilding Network $7,000 – $6,000 – $5,000 – $4,000 – $3,000 – $2,000 – $1,000 – – ||||||| 02468101214Weeks Crash cost Crashed activity Normal activity Normal timeCrash time Normal cost

70. M. Sundaram Tenn. Tech70 Housebuilding Network $7,000 – $6,000 – $5,000 – $4,000 – $3,000 – $2,000 – $1,000 – – ||||||| 02468101214Weeks Crash cost Crashed activity Normal activity Normal timeCrash time Normal cost Slope = crash cost per week Total crash cost$2,000 Total crash time5 = = $400 per week

71. M. Sundaram Tenn. Tech71 Normal Activity and Crash Data 12467 3 5 TOTAL NORMALCRASHALLOWABLECRASH TIMETIMENORMALCRASHCRASH TIMECOST PER ACTIVITY(WEEKS)(WEEKS)COSTCOST(WEEKS)WEEK 1-2127$3,000$5,0005$400 2-3852,0003,5003500 2-4434,0007,00013,000 3-4000000 4-5415001,1003200 4-612950,00071,00037,000 5-6415001,1003200 6-74315,00022,00017,000 $75,000$110,700

72. M. Sundaram Tenn. Tech72 Normal Activity and Crash Data 12467 3 5 TOTAL NORMALCRASHALLOWABLECRASH TIMETIMENORMALCRASHCRASH TIMECOST PER ACTIVITY(WEEKS)(WEEKS)COSTCOST(WEEKS)WEEK 1-2127$3,000$5,0005$400 2-3852,0003,5003500 2-4434,0007,00013,000 3-4000000 4-5415001,1003200 4-612950,00071,00037,000 5-6415001,1003200 6-74315,00022,00017,000 $75,000$110,700 12 80 4 44 4 12467 3 5 $400 $500 $3,000$7,000 $200 $7,000

73. M. Sundaram Tenn. Tech73 Normal Activity and Crash Data 12467 3 5 TOTAL NORMALCRASHALLOWABLECRASH TIMETIMENORMALCRASHCRASH TIMECOST PER ACTIVITY(WEEKS)(WEEKS)COSTCOST(WEEKS)WEEK 1-2127$3,000$5,0005$400 2-3852,0003,5003500 2-4434,0007,00013,000 3-4000000 4-5415001,1003200 4-612950,00071,00037,000 5-6415001,1003200 6-74315,00022,00017,000 $75,000$110,700 12 80 4 44 4 12467 3 5 $400 $500 $3,000$7,000 $200 $7,000

74. M. Sundaram Tenn. Tech74 Normal Activity and Crash Data 12467 3 5 TOTAL NORMALCRASHALLOWABLECRASH TIMETIMENORMALCRASHCRASH TIMECOST PER ACTIVITY(WEEKS)(WEEKS)COSTCOST(WEEKS)WEEK 1-2127$3,000$5,0005$400 2-3852,0003,5003500 2-4434,0007,00013,000 3-4000000 4-5415001,1003200 4-612950,00071,00037,000 5-6415001,1003200 6-74315,00022,00017,000 $75,000$110,700 7 80 412 44 4 12467 3 5 $500 $3,000$7,000 $200 $7,000 Crash cost = $2,000

75. M. Sundaram Tenn. Tech75 Normal Activity and Crash Data 12467 3 5 TOTAL NORMALCRASHALLOWABLECRASH TIMETIMENORMALCRASHCRASH TIMECOST PER ACTIVITY(WEEKS)(WEEKS)COSTCOST(WEEKS)WEEK 1-2127$3,000$5,0005$400 2-3852,0003,5003500 2-4434,0007,00013,000 3-4000000 4-5415001,1003200 4-612950,00071,00037,000 5-6415001,1003200 6-74315,00022,00017,000 $75,000$110,700 7 80 412 44 4 12467 3 5 $500 $3,000$7,000 $200 $7,000 Crash cost = $2,000

76. M. Sundaram Tenn. Tech76 Normal Activity and Crash Data 12467 3 5 TOTAL NORMALCRASHALLOWABLECRASH TIMETIMENORMALCRASHCRASH TIMECOST PER ACTIVITY(WEEKS)(WEEKS)COSTCOST(WEEKS)WEEK 1-2127$3,000$5,0005$400 2-3852,0003,5003500 2-4434,0007,00013,000 3-4000000 4-5415001,1003200 4-612950,00071,00037,000 5-6415001,1003200 6-74315,00022,00017,000 $75,000$110,700 7 70 412 44 4 12467 3 5 $500 $3,000$7,000 $200 $7,000 Crash cost = $2,000 + $500 = $2,500

77. M. Sundaram Tenn. Tech77 Crashing costs increase as project duration decreases Crashing costs increase as project duration decreases Indirect costs increase as project duration increases Indirect costs increase as project duration increases Reduce project length as long as crashing costs are less than indirect costs Reduce project length as long as crashing costs are less than indirect costs Time-Cost Relationship

78. M. Sundaram Tenn. Tech78 Time-Cost Tradeoff Cost ($) Project duration CrashingTime M inimu m cost = optimal project time Total project cost Indirect cost Direct cost

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83. M. Sundaram Tenn. Tech83 Earned -Value Analysis (EVA) EVA is a widely used project control method that compares actual Vs budgeted expenses on a period-by-period basis. Project control is comparing project progress to the plan so that corrective action can be taken when deviation from planned performance occurs. The first step in performing an earned -value analysis is to calculate the budget cost of work scheduled(BCWS) for each time period. - For example if weekly periods are chosen, the total budgeted amount for a task will be distributed evenly during the scheduled period. The Table next page shows a project with three tasks. Task “A” is budgeted for $3,000. If it is scheduled for three weeks, the budget cost/week is $1,000.

84. M. Sundaram Tenn. Tech84 Earned -Value Analysis (EVA)- Contd. The next step is to determine the actual cost of work performed (ACWP). This may be entered weekly into the weekly column labeled ACWP. The next step is to calculate the budgeted cost of work performed (BCWP) by estimating the percent completion for each task and multiplying the total budget amount. Next the schedule variance and the cost variance for each week will have to be calculated as below. - Schedule variance = BCWP – BCWS - Cost Variance = BCWP – ACWP

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88. M. Sundaram Tenn. Tech88 Example Problem Project team Alpha has accumulated $5,000 of expenses as of the end of week#10. The budgeted cost of work scheduled for Week #10 is $6,500. The project has one task, which is about 45% complete at the end of week #10. Calculate the schedule and cost variances. Is the project ahead (or behind), and under (over) budget?

89. M. Sundaram Tenn. Tech89 Solution ACWP=$5000 BCWS=$6500 BCWP (for end of wk #10) = 45%(6500) = $2925 Schedule Variance = BCWP-BCWS = 2925-6500 = $-3575 Cost Variance = BCWP-ACWP = 2925-5000 = $-2075 The project is behind by $3575 of work. Also, for the work performed, it is over budget by $2075.

90. M. Sundaram Tenn. Tech90 Another Problem Team Delta is working on a project that has two work tasks, each worth $5,000, which are 45% and 35% complete as of Week #5. The actual expenses are $2,500 as of the end of Week #5. The budgeted cost of work scheduled for Week #5 is $6,500. Calculate the schedule and cost variances. Is the project ahead (or behind), and under (over) budget?

91. M. Sundaram Tenn. Tech91 Solution ACWP=$2500 BCWS=$6500 BCWP (end of wk #5)=45%(5000)+35%(5000) =2250+1750 = $4000 Schedule Variance = BCWP-BCWS = 4000-6500 = $-2500 Cost Variance = BCWP-ACWP = 4000-2500 = $ 1500

92. M. Sundaram Tenn. Tech92 Resources  Manpower  Money  Machines  Materials  Method

93. M. Sundaram Tenn. Tech93 Resource Allocation  What to do if resources are limited? Allocate the limited resources so that the project may be completed on time using the resources optimally. Two or more activities may require the same resource simultaneously resulting in increased project length.  What to do if resources are unlimited? Very rarely a project may have the luxury of having unlimited resources. If this pleasant situation were to arise, the responsibility of the project team lies in leveling the resources so that the same amount of the resources are used every period.