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PROJECT MANAGEMENT WITH CPM 1

PROJECT MANAGEMENT WITH CPM 2. name activity description duration precedentA old boiler and pipes disconnection 4 - B electrical plant disassembly 1 - C new boiler's order 40 - D ele

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PROJECT MANAGEMENT WITH CPM 1

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    1. PROJECT MANAGEMENT WITH CPM 1 Let us consider the following project: In the heating plant of a palace we have to substitute an old oil boiler with a new gas boiler, equipped with electrical valves and an automatic regulator. The project may be subdivided into activities; an activity is an autonomous piece of project, executed under the responsibility of one person, one team or one firm. Activities’ durations are deterministic. Among activities there may be some finish-to-start precedences, which impose the preceding activity (or activities) shall finish before the preceded one(s) can begin. We shall treat the problem by the Critical Path Method (CPM)

    2. PROJECT MANAGEMENT WITH CPM 2 name activity description duration precedent A old boiler and pipes disconnection 4 - B electrical plant disassembly 1 - C new boiler’s order 40 - D electrical valves’ order 10 - E automatic regulator order 30 - F old boiler removal 2 A G building works (basement, windows,...) 8 B, F H electrical plant assembly 2 G I new boiler assembly 3 C,G L pipes and electrical valves assembly 3 D, I M automatic regulator assembly and conn. 1 E, L N gas plant connection 11 G O final inspection 1 H,M,N

    3. PROJECT MANAGEMENT WITH CPM 3 AMERICAN GRAPHICAL REPRESENTATION 1 – Layering: activities without predecessors are put in the first layer, then activities preceded only by activities in the first layer are put in the second layer, then activities preceded only by activities in the first or second layer are put in the third layer, and so on till all activities have been layered. In our example: A B C D E; F; G; H I N; L; M; O.

    4. PROJECT MANAGEMENT WITH CPM 4 AMERICAN GRAPHICAL REPRESENTATION 2 – Activities on arcs: activities are represented by directed arcs; if activities A1, A2 precede activities B1, B2, then precedence is represented as follows: A1 B1 A2 B2 Circles are called events and separate predecessors and successors

    5. PROJECT MANAGEMENT WITH CPM 5 AMERICAN GRAPHICAL REPRESENTATION 3 – All activities of the first layer (initial activities) take origin from the project beginning event which is generally numbered 0; in our example: A B C D E

    6. PROJECT MANAGEMENT WITH CPM 6 AMERICAN GRAPHICAL REPRESENTATION 4 – All activities of the last layer (final activities) take end on the project ending event; in our example: O

    7. PROJECT MANAGEMENT WITH CPM 7 AMERICAN GRAPHICAL REPRESENTATION 5 – In our example activities H, I, N are preceded by activity G, but only activity I is preceded by activity C: in order to correctly represent precedences, a dummy (fictitious) activity of null duration is added as follows: H G Y N C I

    8. PROJECT MANAGEMENT WITH CPM 8 AMERICAN GRAPHICAL REPRESENTATION 6 – In our example activities H, N are both preceded by activity G, and both precede activity O: in order to avoid two arcs with the same extremal events, a dummy (fictitious) activity of null duration is added as follows: G H O N X

    9. PROJECT MANAGEMENT WITH CPM 9 AMERICAN GRAPHICAL REPRESENTATION 7 – All events are numbered in such a way that every activity has its starting event with a number smaller than the one of its ending event: i < j below: H

    10. PROJECT MANAGEMENT WITH CPM 10 EXAMPLE AMERICAN REPRESENTATION F G N X O A B Y H C I M D L E

    11. PROJECT MANAGEMENT WITH CPM 11 MINIMUM PROJECT COMPLETION TIME 8 – The problem consists in finding the minimum project completion time. The model is a linear program: in our example: MIN (t9 – t0) s.t.: t0=0; t1-t0=4; t2-t0=1; t2-t1=2; t3-t2=8; t4-t0=40; t4-t3=0; t5-t0=10; t5-t4=3; t6-t0=30; t6-t5=3; t7-t3=11; t8-t3=2; t8-t6=1; t8-t7=0; t9-t8=1. As we have a maximum path problem a specific algorithm is better than simplex!

    12. PROJECT MANAGEMENT WITH CPM 12 SPECIFIC ALGORITHM The specific algorithm consists of two phases. In the first forward phase it finds, for every event i, the minimum time ti, i.e., the time before which the event cannot take place because of preceding activities. In the second backward phase it finds, for every event i, the maximum time Ti, i.e., the time after which the event causes a delay in project completion because of following activities.

    13. PROJECT MANAGEMENT WITH CPM 13 SPECIFIC ALGORITHM FORWARD PHASE S1: i:=0, ti:=0; S2: i:=i+1, ti:=MAX{th+dur(hi) | h€Prec(i)} S3: if i is the project ending event ti=project completion time and STOP, else go to S2 SPECIFIC ALGORITHM BACKWARD PHASE S1: i:=n (ending event), Ti:= Tf = tn (final time) S2: i:=i-1, ti:=Min{tj-dur(ij) | j€Succ(i)} S3: if i is the project beginning event STOP, else go to S2

    14. PROJECT MANAGEMENT WITH CPM 14 EVENTS’ TOTAL FLOAT The difference Ti-ti is defined as the total float fi of event i, and is the time that event i may delay with respect to minimum time without delaying the whole project completion. Events with null float are critical; all paths from beginning to ending event which cross only critical events are called critical paths; all activities on a critical paths are critical activities, which shall not increase their duration, otherwise the whole project delays.

    15. PROJECT MANAGEMENT WITH CPM 15 EVENTS’ MIN TIME, MAX TIME AND FLOAT i ti Ti fi 0 0 0 0 1 4 26 22 2 6 28 22 3 14 36 22 4 40 40 0 5 43 43 0 6 46 46 0 7 25 47 22 8 47 47 0 9 48 48 0

    16. PROJECT MANAGEMENT WITH CPM 16 EUROPEAN GRAPHICAL REPRESENTATION 1 – Layering: activities without predecessors are put in the first layer, then activities preceded only by activities in the first layer are put in the second layer, then activities preceded only by activities in the first or second layer are put in the third layer, and so on till all activities have been layered. In our example: A B C D E; F; G; H I N; L; M; O.

    17. PROJECT MANAGEMENT WITH CPM 17 EUROPEAN GRAPHICAL REPRESENTATION 2 – Activities on nodes: activities are represented by nodes; if activity A precedes activity B, then precedence is represented by an arrow, as follows:

    18. PROJECT MANAGEMENT WITH CPM 18 EUROPEAN GRAPHICAL REPRESENTATION 3 – If there is only one activity in the first layer (initial activity) then the graph takes origin from it; if there are more activities in the first layer then a fictitious initial activity of null duration, preceding all activities of the first layer, is added. In our example we have the second situation:

    19. PROJECT MANAGEMENT WITH CPM 19 EUROPEAN GRAPHICAL REPRESENTATION 4 – If there is only one activity in the last layer (ending activity) then the graph ends with it; if there are more activities in the last layer then a fictitious ending activity of null duration, successor of all activities of the last layer, is added. In our example we have the first situation:

    20. PROJECT MANAGEMENT WITH CPM 20 EXAMPLE EUROPEAN REPRESENTATION

    21. PROJECT MANAGEMENT WITH CPM 21 EUROPEAN ACTIVITIES TIME VARIABLES 5 –Four variables are defined for each activity: ES=early start: time before which the activity cannot start because of preceding activities EF=early finish: time before which the activity cannot end because of preceding activities LS=late start: time after which the activity shall not start otherwise it delays the project LF=late finish: time after which the activity shall not finish otherwise it delays the project

    22. PROJECT MANAGEMENT WITH CPM 22 EUROPEAN SPECIFIC ALGORITHM The specific algorithm consists of two phases. In the first forward phase it finds, for every activity i, the early start ES(i) and the early finish EF(i). In the second backward phase it finds, for every activity i, the late finish LF(i) and the late start LS(i).

    23. PROJECT MANAGEMENT WITH CPM 23 EUROPEAN ALGORITHM FORWARD PHASE Starting from the first layer and going on layer by layer till the last one; For every activity i of the current layer we have: If the activity is the initial one: ES = 0, else: ES(i) = MAX {EF(h) | h € Prec(i)} EF(i) = ES(i) + dur(i)

    24. PROJECT MANAGEMENT WITH CPM 24 EUROPEAN ALGORITHM BACKWARD PHASE Starting from the last layer and going on layer by layer till the first one; For every activity i of the current layer we have: If the activity is the ending one: LF(i) = LS(i) else: LF(i) = MIN {LS(j) | j € Succ(i)} LS(i) = LF(i) - dur(i)

    25. PROJECT MANAGEMENT WITH CPM 25 EUROPEAN ACTIVITIES FLOATS 6 –Two floats are defined for each activity: TF(i)=total float=EF(i)-ES(i)=LF(i)-LS(i): increase of the activity duration, or delay in the activity start, which does not cause any delay in the whole project completion time FF(i)=free float= MIN {ES(j)| j € Succ(i)} - EF(i): increase of the activity duration, or delay in the activity start, which does not cause any delay in the successor activities’ start

    26. PROJECT MANAGEMENT WITH CPM 26 EUROPEAN CRITICAL ACTIVITIES 7 – Activities with null total float are critical; paths from the initial activity to the ending activity which cross only critical activities are critical paths EF(ending activity) = project completion time

    27. PROJECT MANAGEMENT WITH CPM 26 EUROPEAN GRAPHICAL REPRESENTATION 8 – The four characteristic times are usually written near every activity in the following format: ES EF LS LF

    28. PROJECT MANAGEMENT WITH CPM 28 EXAMPLE EUROPEAN RESULTS 1 Activity ES EF LS LF tf ff A 0 4 22 26 22 0 B 0 1 27 28 27 5 C 0 40 0 40 0 0 D 0 10 33 43 33 33 E 0 30 16 46 16 16 F 4 6 26 28 22 0 G 6 14 28 36 22 0 H 14 16 45 47 31 31

    29. PROJECT MANAGEMENT WITH CPM 29 EXAMPLE EUROPEAN RESULTS 2 Activity ES EF LS LF tf ff I 40 43 40 43 0 0 L 43 46 43 46 0 0 M 46 47 46 47 0 0 N 14 25 36 47 22 22 O 47 48 47 48 0 0

    30. PROJECT MANAGEMENT WITH CPM 30 EXAMPLE EUROPEAN RESULTS 3 A graphical presentation of results is the Gantt diagram, a cartesian diagram where every activity i is represented by a horizontal bar, where the extreme points have abscissas respectively equal to ES(i) and EF(i). If useful bars may be protract with a segment equal either to tf(i) or to ff(i). See our example Gantt diagram where no floats are evidenced.

    31. PROJECT MANAGEMENT WITH CPM 31 EXAMPLE GANTT DIAGRAM A B C D E F G H I L M N

    32. PROJECT MANAGEMENT WITH CPM 32 LIMITED RESOURCE PROJECT PLANNING Consider a new example, where activities are characterized by name, duration, precedent activities and amount of used resource. This resource is a hired resource, like for instance operators, trucks, machines, equipments, etc. We want to plan the project in the case this resource is limited. It is obvious that the elaboration shall be different from the one we employed in the previous example, where we considered unlimited resources.

    33. PROJECT MANAGEMENT WITH CPM 33 LIMITED RESOURCE EXAMPLE DATA activity duration predecessors resource A 7 - 3 B 5 A 2 C 2 B 4 D 3 - 6 E 2 D 5 F 1 E, H 1 G 4 - 3 H 6 G 5 I 3 H 2

    34. PROJECT MANAGEMENT WITH CPM 34 LIMITED RESOURCE EXAMPLE GRAPH

    35. PROJECT MANAGEMENT WITH CPM 35 If the amount of resource is infinite, activity times are the following: activity ES EF LS LF A 0 7 0 7 B 7 12 7 12 C 12 14 12 14 D 0 3 8 11 E 3 5 11 13 F 10 11 13 14 G 0 4 1 5 H 4 10 5 11 I 10 13 11 14

    36. PROJECT MANAGEMENT WITH CPM 36 Use of unlimited resource (resource limit=7) 14 12 10 G H 8 6 D E H I 4 F I 2 C A 0 B

    37. PROJECT MANAGEMENT WITH CPM 37 LIMITED RESOURCE PARALLEL TECHNIQUE Activities are scheduled according to a decision set (DS), which is a dynamical list built up for every instant of time. Activities put in DS(t) are ordered according to increasing parallel total float, based on unlimited resource activity times, given by: TF(k):=LS(k)-t Activities at the top of the list have higher urgency to be scheduled.

    38. PROJECT MANAGEMENT WITH CPM 38 PARALLEL TECHNIQUE ALGORITHM S1: t:=0; put in DS(0) all activities which can start at time 0; S2: order all activities in DS(t); S3: extract from DS(t), in the order, all activi- ties which can be scheduled at t because of limited resource, schedule them and cancel them from DS(t); put the remai- ning activities into DS(t+1) S4: t:=t+1; add to DS(t) all activities which can start at time t because of preceding ones; if DS(t)=F STOP else go to S2;

    39. PROJECT MANAGEMENT WITH CPM 39 PARALLEL TECHNIQUE ALGORITHM Take the example: DS(0):={A,D,G}, ordered {A,G,D} schedule A and G at time 0, put D into DS(1) t:=1; no activity added to DS(1); DS(1)={D} is already ordered; D cannot be scheduled because of missing resource; t:=2, 3 without scheduled activities; t:=4; H is added to DS(4), ordered {H,D}; t:=5, 6 without scheduled activities; t:=7; B is added to DS(7), ordered {H;B;D}

    40. PROJECT MANAGEMENT WITH CPM 40 PARALLEL TECHNIQUE ALGORITHM H and B are scheduled at time 7; DS(8)={D} t:=8, 9, 10, 11, 12 no scheduled activities; t:=13 C and I are added to DS(13), ordered {D,I,C}; D is scheduled at time 13; DS(14)={I,C}; t:=14, 15 no scheduled activities; T:=16 E is added to DS(16), ordered {E,I,C}; E and I are scheduled at time 16; DS(17)={C}; t:=17 no scheduled activities; t:=18 F is added to DS(18), ordered {C,F}; C and F are scheduled at time 18; DS(19)=F; STOP.

    41. PROJECT MANAGEMENT WITH CPM 41 Use of limited resource=7 (parallel technique): 14 12 10 8 6 B F 4 G E D C 2 H A I 0

    42. PROJECT MANAGEMENT WITH PERT 41 PROJECT EVALUATION AND REVIEW TECHNIQUE (PERT) It cosiders finish-to-start precedences and random activity durations. Conventionally the method considers for every activity k three estimates: ak minimum duration of activity k bk maximum duration of activity k mk most probable duration of activity k From the three estimates we assume the mean activity duration: tk = (ak + bk + 4mk)/6 and the standard deviation: ?k = (bk - ak)/6

    43. PROJECT MANAGEMENT WITH PERT 43 PERT TECHNIQUE Here we solve a problem of unlimited resources CPM, taking for every activity the mean duration. The resulting completion time Tf is considered as the expected project duration. Then we consider the sum of variances ?k2 of activities on the critical path as the variance of project duration ?f2 . We assume the project duration has a gaussian distribution of mean Tf and variance ?f2 . It is obvious that results are easy to obtain but strongly approximated.

    44. PROJECT MANAGEMENT WITH PERT 44 PERT SIMULATION TECHNIQUE For every activity a duration probability distribution is given. From every distribution a random value is extracted, then CPM is solved and we obtain a realization of project completion time. This is iterated a sufficient number of times (50-100 is generally enough), i.e., a sample is taken, then the sample distribution of project completion time is obtained. The result is more precise than the one of the conventional method.

    45. PROJECT MANAG. OTHER TECHNIQUES 45 CPM-cost: activity duration may be modified with change of activity cost. The objective is to minimize total project cost. MPM (Metra Potential Method): precedence may be of the finish-to-start, start-to-start, finish-to- finish and start-to-finish type

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