Project Management and Planning

Transcription

1 Project Management and Planning Project: Is defined as a set of interrelated activities that must be executed in a particular order to complete an entire task. Concerned With: Time of completion Costs Available resources Gantt Chart: First scientific technique for planning a project Two Modern Techniques fixed. CPM - Critical Path Method - Developed by Kelly & Walker (1959) - Activity Times are PERT - Program Evaluation & Review Technique - Booze, Allen, Hamilton & US Navy ( 1959). Activity times are probabilistic. - Both used widely in large complex projects. - Construction industry - widely used (80%) - One time large projects - Spaceship construction Requirements for Projects 1. Well defined activities Install Roof weeks Install Ducts for A/C & Heating weeks. Some Activities may be started & finished independently 3. Some activities may require completion of other activities Roof can be installed Truss is built. 4. If an activity is started, it must be completed without interruption

2 Network Representation of a Project Examples (i) Project and activities represented by a network Network is a graph showing each activity to be performed, its predecessor & successor Activity: Represented by an arrow - generally needs some resource for its performance - Use letters Nodes: Activities end or start at points called nodes Represented by circles/ Use numbers Event: A point in time Dummy Activity: Uses no resources & is used for showing precedence requirements only 1 A B 3 (ii) When A is started it must be finished. Needs two nodes. B cannot be started before A is finished. 1 A B

3 Two activities cannot start and end at same nodes. Use dummy activity. We must use one of the following: 1 A B 3 1 B A 3 Precedence; Both B and A need to No precedence; B starts after and A Start after 1 Starts after 1 ****************************************** Establishment of Critical Path Critical Path: Minimum time needed for completion of project. Method: Sequence of activities with no slack time - Add times of all these activities Slack: Is determined by finding earliest/latest start and finish times. Earliest Starting Time: Earliest time an activity can be started when all preceding activities are completed as quickly as possible. Earliest Finish Time: EF = ES + t, (t = time for an activity) Latest Starting Time: Latest time when an activity can be started (for finishing project in appropriate time) Latest Finish Time: LF = LS + t or LS = LF - t We first develop a network diagram. Then we find ES, EF moving forward in the network and LS, LF by moving backward in the network.

4 Then Slack = LS - ES or Slack = LF - EF Example Activity Predecessor Time A * 0 B A C B 4 D C 6 E C 3 F C 5 G D,E,F 5 H G 7 First draw a network flow diagram. A,0 1 B, C,4 D,6 3 G,8 4 7 H,7 8 9 E,3 6 F,5 5 At each node, one or more activities can start or finish. In Forward Pass, we start at a node and find ES and EF time for each activity (a) ES for activity starting at a node is EF. This is maximum of all ES that End at this node.

5 (b) EF for activity is (ES + time) Forward Pass Calculations Node 1: Only A starts and none ends A: ES = 0 EF = ES+ t = = 0 Node : B starts and A ends B: ES = EF for A = 0 EF = ES + t = 0 + = Node 3: C starts and B ends C: ES = EF for B = EF = + 4 = 6 Node 4: D, E, F start and C ends. EF for all is 6 D: EF = = 1 E: EF = = 9 F: EF = = 11 Nodes 5 and 6 are Dummy nodes Node 7 : G starts and D, E, and F end. As more than one activities end, we must Use maximum EF as ES for G ES for G = Maximum of {EF for D, E, F} = 1 EF for G = ES + time = = 0 Node 8: H starts and G ends EF for H = = 7 Node 9: None starts and H ends. End of the project. Backward Pass

6 We move back from the last node to the first node for finding LS and LF Recall LF = latest finish time for an activity And LS = LF - time of activity Node 9 H ends /None starts H; LF = Project completion time = 7 LS = 7-7 = 0 Node 8 G ends/ H starts G; LF = LS for H = 0 LS = 0-8 = 1 Node 7 D, E, F end / G starts. D; LF = LS for G = 1 LS = 1-3 = 9 E; LF = LS for G = 1 LS = 1-6 = 6 F; LF = LS for G = 1 LS = 1-5 = 7 Node 6 & 5 These are Dummy Nodes Node 4 D, E, F end/ E starts C; LF = Minimum LS {D, E, F} = Minimum {9, 6,7} = 6 LS = 6-4 = Node 3 B ends /E starts B; LF= LS for C = LS = - = 0 Node A ends/b starts A; LF = LS for B = 0 LS = 0-0 = 0 Node 1 A ends/none starts Done

7 Alternate Way of writing ES, EF, LS, and LF: We write the times on the network flow diagram for each activity in a box as shown. ************************************************* Slack for each activity: Difference between LS and ES or LF and EF Slack = LS - ES or Slack = LF - EF Using LS and ES, slack for each activity is shown below: Activity ES LS Slack Critical Path A * B * C 0 * D * E F 6 7 1

8 G * H * Activities on the Critical Path have no Slack Critical Path: A, B, C, D, G, H Critical Path Time: = 7 All other paths take less time Example: A, B, C, E, G, H = 4 days Estimating activity time We have assumed, so far, that the time for completing an activity is known for sure. This may not be true in real life. Time for completion of an activity is uncertain and follows some probability function. We are interested in finding probabilities for completion of a project. One commonly used probability function for activity time is Beta distribution. The parameters for Beta distribution can be estimated relatively easily. Estimates of most likely, optimistic and pessimistic times are needed. Example: Most likely time (m) = 6 days Optimistic time (a) Pessimistic time (b) = 3 days = 15 days We can use normal distribution for finding various probabilities. Mean for Normal = {a + 4m + b)}/{6} = {3 + 4(6) + (15)}/{6} = 7days Standard Deviation = {b - a}/{6} = {15-3}/{6} = days

9 Project Completion Time A project consists of many activities, each of which approximately follow a normal distribution with known mean and standard deviation. For project completion time, we need to consider only activities on the Critical Path. Let t i = mean completion time of the ith activity s i = variance of completion time of ith activity Then Mean completion time of project = T m = t i Variance of completion time = s = s i Determination of Probability of Completion Time for a Project Let us suppose that the mean completion time of a project is 7 days and variance is 36 days. The probability that the project can be completed in 9 days or more can be calculated using the normal distribution. Let x be completion time z = (x - Mean)/(standard Deviation) = (9-7)/ (6) = Using Normal tables P(x>9) = P(z>0.333) = This means that there is 37.07% chance that project will finish in more than 9 days. Generally, there is penalty for not completing a project in time. If it is so, one may consider

10 assigning extra resources to some of the activities to make sure that project will finish in time. But this extra cost must be balanced against the possibility of incurring An Example 3 J,9 B,1 A,8 1 D,3 4 E,1 I,7 6 8 C,7 5 H, K,6 7 L,10 Find ES, LS, EF, and LF Find Slack Critical Path: A, E, K, and L Critical Path time is = 36 days (i) Find probability that project will take 34 days or less. T m = t i = = 36 days

11 s = s i = = Activity Predecessor a m b t i s i A * B * C * D A E A F A G C,F H C,F I B,D J B K E,G,I L H,K ( 34 36) / ( ) z = (x - mean)/(s) = = P { x < 34 } = P { z < } = % Only % chance that project can be completed in 34 days or less.

12 (ii) Find probability that project will be completed in 39 days or more. ( 39 36) / ( ) z = = 1.14 P { x > 39} = P { z > 1.14} = 11% Only 11% chance that project will take 39 days or more. TIME COST TRADEOFF It may be possible to reduce time needed to complete an activity by utilizing more resources (workers, machinery etc.) ---- This will result in extra costs (labor, overtime, etc.). These are increases in direct costs. Indirect costs such as supervisory, management, other overhead may be reduced because of reduced time for completing the project. In addition, there may be savings due to avoidance of penalty costs. MINIMUM COST schedule is obtained by balancing the two costs.

13 Procedure Let NT NC CT CC = normal time = completion time of an activity with normal allocation of resources = normal cost = cost associated with NT = crash time = shortest possible completion time of an activity with extra resources. = crash cost = cost/day associated with CT (assumed linear) 1. Prepare network diagram. Find ES, LS, EF, LF, slack, critical path and normal time (normal cost)schedule 3. Determine direct cost of decreasing activity time by one unit for all the activities on the critical path ---- choose activity with minimum direct cost. 4. Weigh it against the decrease in indirect cost by one unit. 5. If Indirect cost reduction > direct cost; reduce the activity by time one unit.

14 CAUTION: Make sure that critical path does not change. 6. Go step # 3 and repeat till minimum cost schedule is obtained. An Example A,3 C, D,4 4 F,3 1 G,4 H, B,4 E,4 3 Activity Normal Time Crash Time Crash Cost/Day A 3 $00 B 4 $150 C * D 4 3 $300 E 4 3 $450 F 3 1 $50 G 4 $0 H 4 3 $300 Critical Path is A, D, G, H Normal Time for critical path = = 15 days

15 I. Minimum direct cost activity on the critical path is A (00/day) Savings/day = 500 We use extra resources and reduce activity time of A by one day (3 to days) Critical path time = 14 days Savings = = 300 Note: Critical path is still the same II Minimum direct cost activity is G (Note: A cannot be reduced) We can take days out of G Savings = (500) - (0) = 560 Critical path time is 1 days All other paths also take 1 days. This means if one day is reduced on one path it must be reduced on all the other paths. III Use minimum direct cost activity Take 1 day out of B; cost is 150 Take 1 day out of H; cost is 300 Savings = ( ) = 50 Critical path is 11 days IV Take 1 day out of B; cost is 150 Reduce D by 1 day; cost is 300 Reduce F by 1 day; cost is 50 Total cost is 700; more than savings of 500/day V Optimal (Minimum cost) critical path time is 11 days with total savings of or $910

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