Network analysis: P.E.R.T,C.P.M & Resource Allocation Some important definition: 1. Activity : It is a particular work of a project which consumes some resources (in ) & time. It is shown as & represented by a capital letter A,B,C etc. 2. Event : it denotes the start (tail) & end (head) of an activity & represented by a circle both at the start & end of the activity. Events are no as 1, 2, 3 etc. so that it moves from lower to higher no. while presenting an activity. It does not consume any time or resource as it it is an indicator only. This is known as Fulkerson s rule. e.g. Activity A B C D Event No. 1-2 1-3 3-5 4-7 etc 3. Every project or network diagram starts with a single event & ends at a single event. All starting activities will starts from that single event. Similarly all end activities will merge to that single event. 4. Sequence : activities of a project are linked in a sequence known as immediate predecessor i.e. an activity(s) must be completed first to start the following Activity(s). The activity so starts from predecessors is known as successor. 5. Between two events there must be only one activity. 6. Path or Route : a series (one by one) of activities taken together to link the first & end event of an project. Time of a path = time of the activities in that path. 7. Critical Path: The longest time duration path is known as critical path & represented by thick line or double lines. All activities lying in this critical path are called critical activities as any delay in their execution will lead to a delay in the completion of the entire project. 8. Total cost of a project = cost of all the activities in the project + Indirect cost or Overheads during the project duration 9. Forward Pass of event : adding activity time from left to right. Start time of a project is = 0. Earliest finish time (EFT) = Earliest finish time of previous or tail event + time of the activity. In case of convergent event, select the highest EST out of different paths. 10. Backward Pass of event : Subtract time from EFT of last event Latest start time (LST) of an event = LFT of head event Activity time. Activity Event Activity Activity Activity In case divergent event, select the lowest time out of different LST in backward pass. 11. In the critical path all forward pass time = backward pass time Examples : Draw the network diagram, Identify the critical path by applying all the computation method i.e. Activity wise, Forward pass & path method. Find project durations & total costs from the following examples 1 to 24.
Example 1: Indirect costs are 300 per day Activity A B C D E Predecessor Immediate -- -- A B C, D Duration (Days) 4 7 3 5 2 Costs ( ) 500 700 800 200 400 Example 2: Indirect costs are 2,500 per day Activity A B C D Immediate Predecessor - A A B Duration (days) 7 3 9 4 Cost ( ) 5,000 3,000 8,000 4,000 Example 3: Indirect costs are 75,000 per month. Example 4: Example 5: Activity A B C D E F Predecessor - A A A A,B A,B,E, C Duration (months) 5 2 1 3 4 2 Costs (Rs. In Lakhs) 5 7 9 6 4 5 Activity: 1 2 1 3 1 4 2 3 2 5 3 5 4 5 Duration: 4 2 5 7 7 2 5 Activity 1-2 1-3 2-4 3-4 3-5 4-9 5-6 5-7 6-8 7-8 8-9 8-10 9-10 Time in weeks 4 1 1 1 6 5 4 8 1 2 1 8 7 Example 6: Crossing of activities is avoidable Example 7: Example 8: Example 9: Activity A B C D E F Predecessor - - - B A EC Duration (days) 3 4 5 2 3 4 Activity A B C D E F G H I Predecessor - - - A B, D B, D A G, F E, C Duration (months) 7 8 5 4 2 5 7 6 3 Activity A B C D E F G H I J Predecessor - - - A BD A A E GH CF Time 2 4 7 3 5 6 5 4 7 4 Act A B C D E F G H I J K L M Predecessor - A A A C C BF BF E DG E HI JKL Time: weeks 2 4 3 5 7 5 8 6 7 3 5 4 6
Example 10: when crossing is Un-avoidable Activity A B C D E F G H Predecessor - - A A B B C, F D, E Duration (months) 7 5 4 5 3 6 5 4 Definition 12. Dummy Activity is required in a network diagram when two or more activities are there between two same events or to maintain the logical sequence. It is denoted as. It does not consume any time or resource. When a Dummy Act is required: Between 2 same events there can be only one Activity Example 11: Example 12: Activity A B C D Predecessor - - - A,B,C Duration(days) 3 5 6 4 Activity A B C D Predecessor - - A A Duration (Weeks) 2 3 4 5 1. Example 13: Example 14: Activity A B C D Predecessor - A A B,C Duration (Weeks) 3 5 7 4 Activity A B C D Predecessor - A A A Duration (Weeks) 5 3 4 6 When a Dummy Act is required: To Maintain logical sequence Example 15: Example 16: Activity Description of Activity Predecessor Duration(month) A Land Development -- 6 B Factory Layout - 3 C Preparation for Building A 4 D Order for Plant & Machinery A & B 2 Activity Description of Activity Predecessor Duration(month) A Bring Raw Materials -- 2 B Labour Training - 4 C Machine installation B 5 D Product Designing A & C 4 E Store the Materials A 3
Example 17: Activity A B C D E F G Predecessor - - A A BCD C BCD Duration(weeks) 3 4 7 3 5 2 5 Example 18: Activity A B C D E F G Predecessor - - B B C A A, D Duration months 2 5 4 5 2 3 5 Example 19: Activity A B C D E Predecessor - - A B A, B Duration 3 5 4 2 7 Example 20: Activity A B C D E F Predecessor - A A A B, C B, D Duration (Month) 3 4 5 2 7 5 Example 21: Example 22: Activity A B C D E F G H I Predecessor - A A A C B, C D,C E, F E, G Duration 7 4 3 5 7 6 4 3 5 Activity A B C D E F G H I Predecessor - - - A C B, C D, E E, F E, F Duration 7 4 3 5 7 6 4 3 5 Example 23: A Project which is about to start comprises the following activities : Activity Immediately Duration Activity Immediately Duration Preceding in Preceding in Activities weeks activities weeks A -- 4 J H 17 B A 13 K H 2 C A 5 L J, K 3 D C 11 M F, L 3 E C 3 N B, M 3 F D, E 4 O I, M 2 G -- 3 P O 3 H A, G 5 Q N, P 4 I G 4 Ignoring the holiday periods, the project must be completed by the end of week 38. If the project is delayed beyond this date it is estimated that it will cost the firm Rs. 4 lakhs a week. Required : (a) Draw a critical path network to represent the project and determine the critical path. What is the earliest time, at which the project can be completed and what penalty cost (if any) will be incurred? (b) Activity K is a two week course to train new salesmen. The hotel which will be used for the course has been booked for weeks 12 &13. In the light of your analysis should this booking be changed?
Example 24: The following information is known for a project. Draw the network and find critical path. Capital letters denote activities and numbers in bracket denote activity times. This must be completed: A (30) B (7) B B C (10) C D (14) E (10) Before this can start C D G K D G E F This must be completed F (7) F F G (21) G H (7) I (12) K (30) Before this can start H I L I L J(15) J L (15) Definition 13. Forward Pass & Backward Pass of activity & Computation of idle time for a PERT : Tail Event E/L Activity ( t ) Head Event E/L Slack is the idle time of an event and float is the idle time of an activity. Floats are 3 types : Total, Free & Independent, as described below A. Total float for an activity = L.S.T. E.S.T. Or, LFT - EFT, of activity Where, L.S.T. of an activity = L.F.T. Activity time E.F.T. of an activity = E.S.T. + Activity time It defined as total idle time available to non-critical activities in a sequence without any interference of critical path. Prepare a separate table to calculate E.F.T & L.S.T as the these results for the non-critical activity are different from the time shows in the diagram. Float for Critical activity = 0. B. Head or Tail event slack = Latest time of the event Earliest of that event Consider this data from the network diagram. C. Free float = Total float Head event slack. It is part of total float available to last non-critical activity in a sequence, provided other non-critical activities are completed in right time. D. Independent float = Free float Tail event slack. It is part of total float for an non-critical activity, not on shearing basis. F. Interfering Float = Latest event time of the head event earliest event time of that. = H.E.S. PERT Problems : compute the float of activity 1. 2. Activity A B C D E F Predecessor Immediate -- -- A B C, D C,D Duration (Days) 4 7 3 5 2 1 Activity A B C D Immediate Predecessor - A A B Duration (days) 7 3 9 4 3. Activity: 1 2 1 3 1 4 2 3 2 5 3 5 4 5 Duration: 4 2 5 7 7 2 5 4. Draw network diagram from following activities and find critical path and all slack of activities : Job A B C D E F G H I J K Job(days) time 13 8 10 9 11 10 8 6 7 14 18 Immediate Predecessor -- A B C B E D, F E H G,I J
Resource Allocation & Smoothing with Time graph: 5. A Project has the following time schedule : Activity Months lab/ day Activity Months lab/ day 1 2 2 4 3 7 5 8 1 3 2 5 4 6 3 4 1 4 1 8 5 8 1 5 2 5 4 7 6 9 4 6 3 6 8 6 8 9 3 8 (a) All float of activity. (b) Show the Time Graph & Resource Smoothing. 6. The following information is available: Activity No. of days No. of men required per day A 1-2 4 2 B 1-3 2 3 C 1-4 8 5 D 2-6 6 3 E 3-5 4 2 F 5-6 1 3 G 4-6 1 8 i) Draw the network and find the critical path. ii) What is the peak requirement of Manpower? On which day(s) will this occur? iii) If the maximum labour available on any day is only 10, when can the project be completed? Expected time analysis: Three Time Estimate for PERT. An Act time can be divided into: A. Optimistic time, (t o or a). This is the minimum time to perform the activity, assuming that everything goes well. Given weight =1 B. Pessimistic time,(t p or b ). This is the maximum time that is required to perform the activity, under extremely bad conditions. However, such condition do not include acts of nature like earthquakes, flood, etc. Given weight =1 C. Most likely time,( t m or m). This is the most occurring duration of the activity. Statistically, it is the modal value of duration of the activity. Given weight =4 T e or Average time of Act = (t o + 4t m + t p ) 6 S.D. of an activity= (t p -t o ) 6. Variance of an activity = { (t p -t o ) 6} 2 Steps: (1) Compute T e or Average time of each Act. (2) Compute Variance of each activity. (3) Draw the network.. (4) Compute the project duration & critical path on the basis of T e of the activity. This project duration is known as µ (mu). (5) Compute Variance of the critical path & project = Variances of critical activities in the critical paths. So S.D. of the project = δ = variance of critical activities in the project. In case of more than one critical paths, the Principle critical path is the path with highest variance. (6) Draw the normal distribution diagram. Shade the required area by applying the concept of normal (Z) distribution. Remember probability or area is measured from mean. Value of Z = { Observe value (X) Expected value (µ) } Standard deviation of project(δ) (7) Find the probability on the basis of Z table & normal distribution diagram.
7. A project consists of the following activities and trend time estimates : Activity Least time Great time Most likely time (days) (t o ) (days) (t p ) (days) (t m ) 1 2 3 15 6 1 3 2 14 5 1 4 6 30 12 2 5 2 8 5 2 6 5 17 11 3 6 3 15 6 4 7 3 27 9 5 7 1 7 4 6 7 2 8 5 (a) What is the probability that the project will be completed by 27 days. Given : Pb (- X 27) =0.6368 OR, Pb (µ X 27) =0.1368 OR, Pb (- Z 0.35) =0.6368 OR, Pb (0 Z 0.35) =0.1368. Area can be represented as area bounded in integration. (b) Find the probability of completing within 22 to 29 days. (c) What is the duration if the probability of completion is (i) 96.5% (ii) 43.64% confidence? (d) If the average duration for activity 6-7 increases to 7 days, what will be its effect on the expected project completion time which will have 95% confidence interval. Find the principle critical path. (e) Calculate node variances? Find the probability of completing the event 6 by 22 days. (f) Find the probability of completing the event 5 by 16 days by applying both forward & backward pass. 8. A construction company is preparing a PERT network for laying the foundation of a new art museum. Given the following set of activities, their predecessor requirements and three time estimates of completion time : Time Estimates (Weeks) Activity Predecessors Optimistic Pessimistic Most Likely A none 2 4 3 B none 8 8 8 C A 7 11 9 D B 6 6 6 E C 9 11 10 F C 10 18 14 G C,D 11 11 11 H F,G 6 14 10 I E 4 6 5 J I 3 5 4 K H 1 1 1 The contract specifies a 5,000 per week penalty for each week the completion of the project extends beyond 37 weeks. Find the probability that the penalty is max 20,000? 9. YZ Ltd. Planning a project to introduce a new product, as listed the following activities : Activity A B C D E F G H I J Preceding activity -- -- A A A C D B, D, E H I, G, F Expected time (week)6 3 5 4 3 3 5 5 2 3 (a) Draw the critical path network for the project and determine the critical path and its duration. (b) If the start of activity B is delayed by 3 weeks. Activity E by 2 weeks and activity G by 2 weeks, how is the total time for the project affected? (c) Assume that time given in the above table are the expected times of the activities, the durations of which are normally distributed with the following standard deviation. Activity A B C D E F G H I J Std. deviation 1 0.5 1 1 0.5 0.5 1 1 0.5 1 Ignoring the delay referred to in (b) and the possible effect of uncertainty in non-critical activities, determine a 95% confidence interval for the expected time on the critical path.
Critical Path Analysis: CPM or Crashing : Objective: Reduce the project duration & total cost. Total cost of a project = direct costs of all the activities + overhead per day X no. of days. To reduce project duration, only critical path activities should be reduced. By reducing the non critical activities, project duration cannot be reduced. If the duration of critical activity(s) is reduced then additional resources will be required at additional cost e.g. overtime of labour. This is known as crash cost or marginal cost which is a part of direct costs of activities. So, if a critical activity is reduced, marginal cost (MC) will increase but overhead will be saved (i.e. marginal revenue or MR). Therefore, when MR MC, management earns higher profit by crashing. This process is known as crashing of project duration or timecost trade off. Steps to follow: 1. Find the cost slope or crash cost per day-crashed or expediting cost for each activity. Cost slope = (crash cost normal cost) ( normal duration crash duration) 2. Draw the network, find CP. 3. Prepare Time Schedule of paths and its durations. Find the minimum float between CP & nearest non CP. 4. Prepare decision schedule. Decisions are taken on the basis of (i) write the critical activities can be crashed. (ii) select minimum cost slope among critical activity(s) (i) find the max crash duration possible for the critical activity. (ii) Compare it with minimum float as in step 3. (iii) Take a decision about number of days to crash until a new critical path arises. (iv) In case more than one critical paths, find combination of activities in such a way that all the critical paths are crashed, otherwise the project will not reduced. To find the combination, apply theory of flow of water or electricity. 5. Repeat step 2,3,4 at end of each decision. When MC>MR or a critical path is exhausted, optimum duration & minimum cost of the project is achieved. 6. Prepare cost schedule. 7. If the customer will pay the additional cost (i.e. MC- MR), reduce the project duration to its minimum project durations, until a CP is exhausted. 8. When a minimum cost slope critical activity is common among critical & non critical paths, crash that critical activity to its maximum available crash duration ( without considering minimum float in time schedule), as no new critical path will arise. 9. For a tie in minimum cost slope activities among critical activity(s), crash the common critical activity(s) to its maximum crash duration available. 10(a). Find the minimum & optimum project duration. Activity Work nature Normal Crash Immediate Indirect Days Cost( ) Days Cost ( ) Predecessor cost per day A Lab recruit 8 1,200 5 1,800 none B lab Training 6 2,900 2 3,600 A C mat purchase 10 3,800 4 5,000 none 450
10(b). C 12 3,800 6 5,000 none 380 10(c). C 15 3,800 2 6,400 none 380 Effect of common activity 11. Find the minimum and optimum project duration if indirect cost per day is 4,80,000. Activity Normal Crash Immediate Days Cost ( ) Days Cost ( ) Predecessor A 8 12,00,000 4 18,00,000 none B 6 29,00,000 3 37,00,000 A C 10 38,00,000 2 54,00,000 A D 5 20,00,000 1 32,00,000 B 12. A small project is having seven activities. The relevant data about these activities is given below Activity Dependence Normal Crash Normal Crash Duration duration cost cost (Days) (days) ( 000) ( 000) A -- 7 5 500 900 B A 4 2 400 600 C A 5 5 500 500 D A 6 4 800 1,000 E B, C 7 4 700 1,000 F C, D 5 2 800 1,400 G E, F 6 4 800 1,600 (i) Find Optimum and Minimum duration, If indirect cost is 3,50,000 per day. (ii) What is the percentage decrease in cost to complete the project in 21 days? 13. The following table relates to a network: Activity Normal Time Crash Time Normal Cost Crash (Days) (Days) ( ) ( ) 1-2 5 4 30,000 40,000 2-3 6 4 48,000 70,000 2-4 8 7 1,25,000 1,50,000 2-5 9 6 75,000 1,20,000 3-4 5 4 82,000 1,00,000 4-5 7 5 50,000 84,000 The overhead cost per day is 5,000 and the contract includes a penalty clause of 15,000 per day if the project is not completed in 20 days. (i) (ii) Draw the network and calculate the normal duration and its cost. Find out: (1) The lowest cost and the associated time. (2) The lowest time and the associated cost
14. ABC Ltd. is in the process of undertaking a contract. An article clark has produced the following wrong diagram of the project : D 6 4 F E 10 S 6 4 B 4 The activities are mentioned as A,B,C,D,E & F. The starting event is S. The numbers given along the activity are the normal expected completion time in days. Dotted lines are representing Dummy activities. Each separate activity costs 20,000 per day and there is further charge of 50,000 for everyday the project is in progress. You are required to (i) Draw the Network diagram & compute the EST & EFT for each activity. (ii) Indicate the critical path and the total cost. (iii) Calculate the optimum plan for the project if the time of activities (b), (c) and (d) can be progressively reduced to one day at an extra cost as under : Activity B C D Extra cost per day-saved ( ) 12,500 15,000 30,000 15. A small project consists of jobs as given in the table below. The crash cost (in per month) of each job is also given: Job (i-j) Normal duration (in month) 1-2 9 1-3 8 1-4 15 2-4 5 3-4 10 4-5 2 C Minimum (crash) Duration (in month) 6 5 10 3 6 1 Cost of Crashing ( in lacs per month) 20 25 30 10 15 40 (i) What is the normal project length and the minimum project length? (ii) Determine the minimum crashing cost schedules ranging from normal length down to and including the minimum length duration, i.e. if L = Length of the schedule, find the costs of schedules which are L, L-1, L-2, and so on. (iii) Overhead costs total 61,20,000 p.m. What is the optimum length schedule in terms of both crashing and overhead cost? List the schedule duration of each job for your solution. Typical Examples: 16. From a network given below, find 3 different ways of reducing the project duration by four weeks. 5 weeks 3 weeks 5 weeks 2 5 9 8 2 weeks A 1 6 Weeks 3 weeks 1 week 3 6 2 weeks 2 weeks 2 weeks 4 3 weeks 7
17. A project Manager has to manage various projects. For each project given below, you are required to advise him whether to use PERT or CPM and briefly state the reason: (i) Project K is yet to begin. The manager has recently successfully handled similar projects. He has able to break down the project into smaller modules and knows when he may comfortably finish each module. (ii) Project L has been sanctioned some fixed amount. Though the manager is familiar about what time it will take, he expects pressure towards the end to finish the project slightly earlier, by deploying additional resources of the company. (iii) Project M is new to the manager. He has never handled such a project. He can break up the project into smaller modules, but even then, he is not sure of their exact times. (iv) Project N has a limitation on the skilled workforce available. But the manager knows from earlier experience, the slack on each event in the project. He is confident of handling the bottleneck of labour. (v) Project 0 is a research project, bound to produce immense benefit to the company in future. 18. A company had planned its operations as follows: Activity 1-2 2-4 1-3 3-4 1-4 2-5 4-7 3-6 5-7 6-8 7-8 Duration(days) 7 8 8 6 6 16 19 24 9 7 8 (i) Draw the network and find the critical paths. (ii) After 15 days of working, the following progress is noted: (a) Activities 1-2, 1-3 and 1-4 completed as per original schedule. (b) Activity 2-4 is in progress and will be completed in 4 more days. (c) Activity 3-6 is in progress and will need 17 more days to complete. (d) (e) (f) The staff at activity 3-6 are specialised. They are directed to complete 3-6 and undertake an activity 6-7, which will require 7 days. This rearrangement arose due to a modification in a specification. Activity 6-8 will be completed in 4 days instead of the originally planned 7 days. There is no change in the other activities. Update the network diagram after 15 days of start of work based on the assumption given above. Indicate the revised critical paths along with their duration. 19. The NRB Company is planning to design, develop and market a new racing cycle. The project is composed of the following activities: (a) Activity Description Predecessors Time(weeks) A Design frame ` none 4 B Design wheels none 3 C Design gears none 3 D Design handlebars C 2 E Test steering A,B,D 1 F Test gears A,B,D 2 G Performance test E, F 3 H Manufacturing layout A,B,D 3 I Manufacture demonstrators H 5 J Prepare advertising G 2 K Prepare users manuals G 4 L Distribute to dealers I,J.K 2 Construct the network, determine the critical path and the duration of the above Project. (b) NRB Company s management would like to get the new bicycle to their dealers in 15 weeks by assign more designers to design the gears from other design activity. Would it help if they transfer workers from frame designed activity & allow overtime to get the frame designed in only 3 weeks? Also state from what other designing activity should the designers be taken from?
20. At the end of activity 6 7, a product is to be launched and the date has been announced for the inaugural function, based on the normal duration of project. Activities have been subcontracted by project manager to contractors A, B, C, D, E, F, G & H as indicated in the table below. Each subcontractor offers a discount on his contract price for each day given to him in addition to the normal days indicated in the network. What will be the maximum discount that the project manager may earn for the company without delaying the launch of the product? 1 6 2 7 5 2 3 9 5 14 4 6 8 6 7 Activity 1-2 1-3 1-4 2-5 3-5 4-6 5-6 6-7 Contractor A B C D E F G H Discount ( 000 Per day) 300 200 1,200 500 400 1,000 600 500 PERT & Simulation 21. A project consists of 7 activities. The time for performance of each of the activities is a random variable with the respective probability distribution as given below : Activity Immediate predecessor Time (in days) and its probability A - 3 4 5 0.20 0.60 0.20 B - 4 5 6 7 8 0.10 0.30 0.30 0.20 0.10 C A 1 3 5 0.15 0.75 0.10 D B, C 4 5 0.80 0.20 E D 3 4 5 6 0.10 0.30 0.30 0.30 F D 5 7 0.20 0.80 G E, F 2 3 0.50 0.50 Simulate the project using random numbers to determine the activity times. Find the critical paths and the project duration for each run. A B C D E F G Run-1 97 95 12 11 90 49 57 Run-2 80 06 14 99 16 89 96