Project Scheduling by PERT

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    Project Scheduling by PERT/CPM

    Reference Books:

    Anderson, Sweeney, and Williams, AN INTRODUCTION TO MANAGEMENTSCIENCE, QUANTITATIVE APPROACHES TO DECISION MAKING, 7thedition,West Publishing Company,1994

    Hamdy A. Taha, OPERATIONS RESEARCH, AN INTRODUCTION, 5 th edition,Maxwell Macmillan International, 1992

    Daellenbach, George, McNickle, INTRODUCTION TO OPERATIONS RESEARCHTECNIQUES, 2ndedition, Allyn and Bacon. Inc, 1983

    Lawrence Lapin, QUANTITATIVE METHODS for Business Decisions with Cases,4thedition Harcourt Brace Jovanovich, Inc., 1988

    T. A. Burley and G Osullivan, OPERATIONAL RESEARCH, MacMillan EducationLtd., 1990

    Lecture 1

    1. Introduction

    A projectdefines a combination of interrelated activities that must be executed in acertain order before the entire task can be completed. An activity in a project isusually viewed as a job requiring time and resources for its completion.

    Project management has evolved as a field with the development of two analytical

    techniques for planning, scheduling, and controlling of projects. These are the project

    evaluation and review technique (PERT) and the critical path method(CPM).

    These techniques were developed by two groups almost simultaneously. CPM was

    developed by E. I. Du Pont de Nemours & Company as an application to construction

    projects and was later extended to a more advanced status by Mauchly Associates.

    PERT was developed by the U.S. Navy by a consulting firm for scheduling the

    research and development activities for the Polaris missile program.

    Although PERT and CPM were developed independently, they are similar inprinciple. Today, PERT and CPM actually comprise one technique and thedifferences, if any, are only historical. Consequently, both technique are referred to asproject scheduling techniques.

    Project scheduling by PERT-CPM consists of three basic phases:

    Planning

    breaking down the project into distinct activities; determining the time estimates for these activities;

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    constructing a network diagram with each arc representing the activity;

    Scheduling

    constructing a time chart showing the start and the finish times for eachactivity as well as its relationship to other activities in the project;

    pinpointing the critical (in view of time) activities that require specialattention if the project is to be completed on time.

    Showing the amount of slack (or float) times for the non-critical activities;

    Controlling

    Using the network diagram and the time chart for making periodic progressreports;

    updating the network.

    2. Network Diagram Representations and Network Construction

    The network diagram represents the interdependencies and precedence relationships

    among the activities of the project. An arrow is commonly used to represent an

    activity, with its head indicating the direction of progress in the project. An eventrepresents a point in time that signifies the completion of some activities and the

    beginning of new ones. The following diagram shows an example, where activities (1,3) and (2, 3) must be completed before activity (3, 4) can start.

    1

    43

    2

    TailHead event

    Rules for constructing a network diagram:

    1. Each activity is represented by one and only one arrow in the network;2.No two activities can be identified by the same head and tail events (a

    dummyactivity is introduced in such situations);

    D

    B

    B

    A

    A

    In this case, D is the dummy activity.

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    3. To ensure the correct precedence relationship in the network diagram, thefollowing questions must be answered as every activity is added to thenetwork:

    What activities must be completed immediately before this activitycan start?

    What activities must follow this activity? What activities must occur concurrently with this activity?

    Example 1:The Galaxy plc is to buy a small business, Tiny Ltd. The whole procedureinvolves four activities:

    A. Develop a list of sources for financing;B. Analyse the financial records of Tiny Ltd;C. Develop a business plan (sales projections, cash flow projections,

    etc.);

    D. Submit a proposal to a lending institution.

    The precedence relationship of these four activities is described as in the Tablebelow. Construct the network diagram.

    Activity Immediate Predecessor

    A -B -C BD A, C

    A41

    2

    3A D

    C

    D

    CBB

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    Example 2:Construct the network based on the Table of information .

    ActivityImmediate

    Predecessor

    A -B -C BD A, CE CF CG D, E, F

    For activities A, B, C, and D, the network portion is as follows:

    AD

    1C

    B

    3

    4

    2

    When activity E, which has C as its immediate predecessor, is to be added, we come

    cross a problem because activities A and C both end at node 3. If activity E is to

    happen after C, it has to be after A as well in this arrangement, which is not true

    according to the specification. The solution is to add a dummy activity between Cand node 3 in order to add E correctly. This is shown below.

    AD

    E1

    CB

    3

    54

    2

    The first completed 7-activity network is shown as follows:

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    AD

    E

    G

    F

    1

    CB

    3

    5 4

    2

    6

    It is seen that activities E and F share the same head and tail events, which is inconflict with Rule 2. In such situations, dummy activities should be introduced.

    AD

    E

    G

    F

    1

    CB

    3

    54

    2

    6

    7

    The above network describes correctly the relationships among the 7 activities.

    Lecture 2

    3. Determination of the Critical Path

    An activity is said to be critical if a delay in its start will cause a delay in thecompletion date of the entire project. A non-critical activity is an activity that has

    time to spare (known as slack or float time) within the entire project. A critical path

    is a sequence of connected critical activities that leads from the source node to

    the sink node.

    We will discuss the determination of the critical path through the following example.

    ExampleThe owner of a shopping centre is considering modernising and expanding the

    current 32-business shopping complex. He hopes to add 8 to 10 new businessor tenants to the shopping complex. The specific activities that make up the

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    expansion project, together with information on immediate predecessor andcompletion time, are listed in the following table.

    ActivityActivity

    DescriptionImmediate

    PredecessorCompletion Time

    (weeks)

    A Prepare architectural drawings - 5B Identify potential new tenants - 6C Develop prospectus for tenants A 4D Select contractor A 3E Prepare building permits A 1F Obtain approval for building permits E 4G Perform construction D, F 14H Finalise contracts with tenants B, C 12I Tenants move in G, H 2

    Total 51

    We are now asked to answer the following questions:

    1) What is the total completion time of the project?2) What are the scheduled start and completion time for each activity?3) Which activities are critical and must be completed exactly as scheduled in

    order to keep the project on schedule?4) How long can the non-critical activities to be delayed before they cause a

    delay in the completion time for the project?

    To solve the problem, we need first construct the network according to the problemspecification.

    A

    H

    I

    GFE

    D

    C

    B

    14

    12

    3

    41

    4

    6

    2

    5

    76

    5

    4

    3

    2

    1

    Completion timefor activity H

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    Starting at the networks source node (node 1) we will have to compute the earliest

    start timeand the earliest finish timefor each activity in the network. Lets assumethat

    ES = earliest start time for a particular activity

    EF = earliest finish time for a particular activityt = expected completion time for the activity

    The earliest finish time can be calculated by the following expression for a givenactivity:

    EF = ES + t

    For example, for activity A ES = 0 and t = 5; thus the earliest finish time for activityA is EF = 0+5 = 5.

    We will write ES and EF directly on the network in brackets. Using activity A as anexample, we have

    A[0, 5]

    5

    2

    1Expected completiontime

    Activity

    EFES

    Since activities leaving a node cannot be started until all immediate precedingactivities have been completed, the following rule determines the earliest start timefor activities.

    Earliest Start Time Rule

    The earliest start time for an activity leaving a particular node is equalto the largest of the earliest finish time for all activities entering the

    node.

    Using this rule, the earliest start and finish times for each activity are written onto thenetwork, which now looks as follows:

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    [5,8]

    [6,10][10,24][5,6][0,5]

    [24,26]

    [5,9][9,21]

    [0,6]

    A

    H

    I

    G

    FE

    D

    C

    B

    14

    12

    3

    41

    4

    6

    2

    5

    76

    5

    4

    3

    2

    1

    As has demonstrated, proceeding in a forward pass through the network, we canestablish the earliest start time and then the earliest finish time for each activity. This

    process gives the earliest completion time of the entire project, which is the earliestfinish time for the last activity. In the case of the shopping centre, the total timerequired for project completion is 26 weeks.

    We now continue the algorithm for finding the critical path by making a backwardpasscalculation. Starting at the sink node (node 7)and using a latest finish time of 26

    weeks for activity I, we trace back through the network, computing a latest start time

    and latest finish timefor each activity. Let

    LS = latest start time for a particular activityLF = latest finish time for a particular activity

    The latest start time is given by the following expression:

    LS = LF - t

    The latest start and finish times are also to be displayed on the network, but we willput them within a pair of round brackets. The following rule determines the latest

    finish time for any activity in the network.

    Latest Finish Time Rule

    The latest finish time for an activity entering a particular node is equal to the

    smallest of the latest start times for all activities leaving the node.

    The PERT/CPM network with both [ES, EF] and (LS, LF) for the example is shown

    below.

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    [5,8]

    [6,10]

    [10,24][5,6][0,5]

    (7,10)

    (6,10)(10,24)

    (0,5) (5,6)

    (8,12)(24,26)

    (12,24)

    (6,12)

    [24,26]

    [5,9][9,21]

    [0,6]

    AA

    H

    I

    G

    FE

    D

    C

    B

    14

    12

    3

    41

    4

    6

    2

    576

    5

    4

    3

    2

    1

    From the above diagram, we find the amount of slack or free time associated with

    each of the activities. Slack is defined as the length of time an activity can be

    delayed without affecting the total time required to complete the project. Theamount of slack is computed as follows:

    Slack = LS - ES = LF - EF

    Activities with zero slack are the critical path activities.

    According to the finished PERT/CPM network, we arrive at the following table ofinformation (the project schedule) for the shopping centre project.

    Activity ES LS EF LF Slack Critical Path?

    A 0 0 5 5 0 YesB 0 6 6 12 6

    C 5 8 9 12 3

    D 5 7 8 10 2

    E 5 5 6 6 0 YesF 6 6 10 10 0 YesG 10 10 24 24 0 YesH 9 12 21 24 3

    I 24 14 26 26 0 Yes

    We can now answer the questions we were asked before:

    1) What is the total completion time of the project?

    The project can be completed in 26 weeks if the individual activities arecompleted on schedule.

    2) What are the scheduled start and completion time for each activity?

    See the above Table.

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    3) Which activities are critical and must be completed exactly as scheduled inorder to keep the project on schedule?

    A, E, F, G, and I are the critical path activities.

    4) How long can the non-critical activities to be delayed before they cause adelay in the completion time for the project?

    Table above shows the slack time associated with each activity. It is evident that

    it is the critical paths that determine the project completion time;

    changing time of the non-critical activities within the permissiblerange will not affect the project completion time; but changing time of

    the critical activities may cause the project completion time to change.

    Now, let us summarise the PERT/CPM critical path procedure.

    Step 1 Develop a list of activities that make up the project;Step 2 Determine the immediate predecessor activities for each activity listed

    in the project;Step 3 Estimate the completion time for each activity;Step 4 Draw a network depicting the activities and immediate predecessors

    listed in Steps 1&2;Step 5 Using the network and the activity time estimates, determine the

    earliest start and finish times for each activity by making a forwardpass through the network. The earliest finish time for the last activity in

    the project identifies the total time required to complete the project;Step 6 Using the project completion time identified in Step 5 as the latest

    finish time for the last activity, make a backward pass through thenetwork to identify the latest start and finish times for each activity;

    Step 7 Use the difference between the latest start time and the earliest starttime for each activity to identify the slack time available for theactivity;

    Step 8 Find the activities with zero slack; these are the critical path activities;Step 9 Use the information from Steps 5&6 to develop the activity schedule

    for the project.

    Lecture 3

    4. Consideration of Time-cost Trade-offs

    From the shopping centre example, it is seen that the PERT/CPM can answerquestions such as the total project completion time, the critical activities, and theslack times of the non-critical activities. This, obviously, will give the projectmanager a clear picture for his control over the project. The project schedule is basedon the given cost and finish time of the individual activities.

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    In practice, we sometimes demand more than this. We may be interested incompleting a project at minimum cost, or completing a project in minimum time.

    These are the considerations of time-cost trade-offs. To obtain the minimum cost orthe minimum time, we need to know the possible reduction in time and extra cost forreduction per unit time for each activity.

    4.1 Completion of projects at minimum cost

    By adding more resources, a project may be sped up. Usually, the purpose of speedingup is to save money on project overheads, to avoid penalty clauses in contracts or,sometimes, to earn bonuses for early completion. The complication which arises isthat as the critical activities are sped up more and more, other activities also becomecritical. We will discuss the algorithm through the following example.

    Example:

    A project consisting of 8 activities are described in the following table. Thecost for completion of these 8 activities is 5800 excluding the site overhead.The overhead cost of general site activities is 160/day. We are asked to:

    1)calculate the normal completion of the project, its cost, and the criticalpath;

    2)calculate and plot on a graph paper the cost/time function for the projectand state:

    the minimum cost and the associated time;

    the shortest time and the associated cost.

    ActivityNormal completion

    time (days)Shortest completion

    time (days)Cost of reduction per

    day ()

    A (1-2) 6 4 80B (1-3) 8 4 90C (1-4) 5 3 30D (2-4) 3 3 -E (2-5) 5 3 40F (3-6) 12 8 200G (4-6) 8 6 50H (5-6) 6 6 -

    We first set-up the network according to the description of the project. Then using thePERT/CPM scheduling technique discussed earlier, we establish the ES, EF, LS, LFtimes and the critical activities. These are shown in the following network and table.

    11

    A

    F

    B

    C G

    HD

    E

    [8,20]

    (8,20 )[0,8]

    (0,8)

    [9,17][0,5]

    (7,12) (12,20)[11,17][6,9]

    [0,6](9,12)

    (14,20)[6,11]

    (9,14)

    12

    8

    5

    63

    6

    5

    (3,9)

    3

    6

    841

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    Activity ES EF LS LF Slack Critical?

    A 0 6 3 9 3

    B 0 8 0 8 0 Yes

    C 0 5 7 5 7

    D 6 9 9 12 3

    E 6 11 9 14 3F 8 20 8 20 0 Yes

    G 9 17 12 20 3

    H 11 17 14 20 3

    The above shows that the normal completion time is 20 days and the critical activitiesare B(1-3) and F(3-6). The cost of completing the project at normal speed is

    5800 + 20 160 = 9000

    Now, we wish to speed up the project so that the project will cost the least. The rule

    is to speed up firstly the critical activity that cost the least to do so. Obviously, theactivity to speed up is B, which costs 90 for speeding up one day. According to the

    project description, the activity B can be shortened by 8-4=4 days. The amount oftime to speed up is determined based on (1) the reduction should reduce the projectcompletion time the most; and (2) the reduction should cause as many activities to

    become critical as possible. Let us speed up 3 days for B. This reduces the completiontime to 17 days. As indicated in the following diagram, all activities except C becomecritical because of this.

    The new cost accordingly is now:

    A

    F

    B

    C G

    HD

    E

    [5,17]

    (5,17 )[0,5]

    (0,5)

    [9,17][0,5]

    (4,9) (9,17)[11,17][6,9]

    [0,6](6,9)

    (11,17)[6,11]

    (6,11)

    12

    5

    5

    63

    6

    5

    (0,6)

    3

    6

    841

    52

    9000 - 3 160 + 3 90 = 8790

    In order to achieve any further saving, it is necessary to reduce time along all the

    critical paths simultaneously. The cheapest way this can be done in this example isto save one day on activities A and B the same time. This action further reduces the

    project completion time into 16 days. The critical paths remain the same.

    12

    A

    F

    B

    C G

    HD

    E

    [4,16]

    (4,16 )[0,4]

    (0,4)

    [8,16][0,5]

    (3,8) (8,16)[10,16][5,8]

    [0,5](5,8)

    (10,16)[5,10]

    (5,10)

    12

    4

    5

    53

    6

    5

    (0,5)

    3

    6

    841

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    The total cost under this circumstance is

    8790 - 1 160 + 1 90 + 1 80 = 8800

    Now, to reduce time further on all the critical paths, we need to consider activities Fand A which have 4 days and 1 day, respectively, to spare. We can only reduce oneday on both of these and the total completion time is now reduced to 15 days. ActivityC still has 2 days slack time while all the others are critical.

    A

    F

    B

    C G

    HD

    E

    [4,15]

    (4,15 )[0,4]

    (0,4)

    [7,15][0,5]

    (2,7) (7,15)[9,15][4,7]

    [0,4](4,7)

    (9,15)[4,9]

    (4,9)

    11

    4

    5

    4

    3 6

    5

    (0,4)

    3

    6

    841

    52

    The total cost is now

    8800 - 1 160 + 1 80 + 1 200 = 8920

    The project can still be sped up by reducing time on activities E, F, and G (2 days, 3days, and 2 days available respectively). Reduction of two days on these activitiesmakes the total projection time to 13 days.

    A

    F

    B

    C G

    HD

    E

    [4,13]

    (4,13 )[0,4]

    (0,4)

    [7,13][0,5]

    (2,7) (7,13)[7,13][4,7]

    [0,4](4,7)

    (7,13)

    [4,7](4,7)

    9

    4

    5

    43

    63

    (0,4)

    3

    6

    641

    52

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    The total cost in this case is

    8920 - 2 160 + 2 (200 + 50 + 40) = 9180

    13 days is the minimum completion time for the project because no further timereduction is available on the critical path 1-2-5-6.

    For the purpose of plotting the required cost/time graph, we summarise in thefollowing table the completion times and costs of the project.

    Completion time(days)

    Cost()

    20 9000

    17 879016 880015 892013 9180

    Completion t ime VS Cost

    8500

    8600

    8700

    8800

    8900

    9000

    9100

    9200

    20 17 16 15 13

    Days

    () Cost ()

    It is evident that the minimum cost for completing the project is 8800 in 17 days, andthat the minimum possible completion time is 13 days costing 9180.

    This concludes the example.

    Lecture 4

    4.2 Completion of projects in minimum time

    In some circumstances the primary interest when completing a project is to use the

    least possible time even if this does not mean the least possible cost. One example for

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    this is the situation when the equipment being used for the project is urgently neededfor more profitable work else where.

    One way of finding the minimum time for completion of a project is to start with thenormal completion network and gradually make reductions in critical activities until

    minimum time is reached, like the method used in the last example.

    However, if it is the minimum time that is of interest, then there is another and moreefficient way of proceeding, i.e.,

    1) Crash every activity and look at the resulting network

    this will certainly give us minimum completion time

    but it may be a highly wasteful way of achieving the minimum time

    2) Consider the activities which are not critical and allow the most expensiveof these to slow down as much as possible without the duration of the

    project being increased above the desired minimum.

    Note: To crash an activity is to use the shortest possible time available for the

    activity.

    Example:The data shown in the following table relates to a contract being undertaken.There are also site costs of 500 per day.

    You are required to:

    (1) calculate and state the time for completion on a normal basis;(2) calculate and state the critical path on this basis, and the cost;(3) calculate and state the cost of completion in the shortest possible time.

    Activity Completion time(days)

    Cost of activity(1,000)

    Possiblereduction time

    (days)

    Extra cost forreduction(/day)

    A(1-2) 5 6 1 300B(1-3) 8 10 2 200C(1-4) 15 17 4 700D(2-3) 4 5 1 400E(2-5) 12 15 3 200F(3-4) 6 8 2 200G(4-5) 7 9 1 400H(4-6) 11 13 3 300I(4-7) 10 12 2 600J(5-6) 8 14 2 300K(6-8) 9 25 3 100L(7-8) 10 13 2 500

    Answer:

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    (1) According to the above table, the PERT/CPM network can be generated asfollows.

    E

    DKJ

    A

    B

    H LFC

    I

    [5,17]

    (10,22)12

    [30,39][22,30][5,9][0,5]

    4

    (0,5) (5,9) (22,30) (30,39)98

    [25,35][15,26][15,22][0,8]

    (1,9)(9,15) (19,30)78

    (29,39)11(15,22)6[0,15]

    1510

    (0,15)[15,25]

    (19,29)10

    5

    [9,15]G

    2

    865

    3

    1

    7

    4

    As it is indicated, the project will require 39 days to complete under the normalsituation.

    (2) The above network shows that there are two critical paths, i.e., A-D-F-G-J-K, andC-G-J-K. The cost on the normal basis is

    (All costs) + 39 500 = 147,000 + 19,500 = 166,500

    (3) To find the minimum completion time, we first reconstruct the PERT/CPMnetwork by crashing all the activities, i.e., using the shortest completion time foreach activity.

    E

    DKJ

    A

    B

    H LFC

    I

    [4,13]

    (8,17)9

    [23,29][17,23][4,7][0,4]

    3

    (0,4) (4,7) (17,23) (23,29)66

    [19,27][11,19][11,17][0,6]

    (1,7)(7,11) (15,23)66

    (21,29)8(11,17)4[0,11]

    118

    (0,11)[11,19]

    (13,21)8

    4

    [7,11]G

    2

    865

    3

    1

    7

    4

    By crashing all the activities, the minimum completion time of the project is found tobe 29 days. Money can be saved by allowing the slowing down of those non-critical

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    activities which are most expensive to speed up. It should be mentioned that the

    slowing down of the non-critical activities should not increase the minimum

    completion time of the project. The follow table lists all the non-critical activitiesand their costs to speed up.

    Non-criticalactivities

    Cost to speed up(/day)

    B 200E 200H 300I 600L 500

    Obviously, the most expensive non-critical activity to speed up is I. So, we first letactivity I to slow down to its normal completion time of 10 days. This will result in

    the following network diagram.

    E

    DKJ

    A

    B

    H LFC

    I

    [4,13]

    (8,17)9

    [23,29][17,23][4,7][0,4]

    3

    (0,4) (4,7) (17,23) (23,29)66

    [21,29][11,19][11,17][0,6]

    (1,7)(7,11) (15,23)66

    (21,29)8(11,17)4[0,11]

    118

    (0,11)[11,21]

    (11,21)10

    4

    [7,11]G

    2

    865

    3

    1

    7

    4

    The change of the completion time of activity I from 8 days to 10 days makesactivities I and L critical without increasing the minimum project completion time.

    Secondly, we let activity H take its full 11 days, which leads to the followingdiagram.

    E

    DKJ

    A

    B

    H LFC

    I

    [4,13]

    (8,17)9

    [23,29][17,23][4,7][0,4]

    3

    (0,4) (4,7) (17,23) (23,29)66

    [21,29][11,22][11,17][0,6]

    (1,7)(7,11) (12,23)66

    (21,29)11(11,17)4[0,11]

    11

    8

    (0,11)[11,21]

    (11,21)10

    4

    [7,11]G

    2

    865

    3

    1

    7

    4

    17

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    Thirdly, we allow activity E to take its full 12 days, leading to the diagram below.

    E

    DKJ

    A

    B

    H LFC

    I

    [4,16]

    (5,17)12

    [23,29][17,23][4,7][0,4]

    3

    (0,4) (4,7) (17,23) (23,29)66

    [21,29][11,22][11,17][0,6]

    (1,7)(7,11) (12,23)66

    (21,29)11(11,17)4[0,11]

    11

    8

    (0,11)[11,21]

    (11,21)10

    4

    [7,11]G

    2

    865

    3

    1

    7

    4

    Fourthly, we allow activity B to take 7 days, which makes B critical as indicated inthe following diagram.

    E

    DKJ

    A

    B

    H LFC

    I

    [4,16]

    (5,17)12

    [23,29][17,23][4,7][0,4]

    3

    (0,4) (4,7) (17,23) (23,29)66

    [21,29][11,22][11,17][0,7]

    (0,7)(7,11) (12,23)67

    (21,29)11(11,17)4[0,11]

    118

    (0,11)[11,21]

    (11,21)10

    4

    [7,11]G

    2

    865

    3

    1

    7

    4

    No further savings are possible as the three non-critical activities are all now atnormal duration. Hence, the least possible cost of completing in 29 days is:

    (All costs) + 29 500 + 1 300

    + 1 200

    + 4 700

    + 1 400

    + 2 200

    + 1 400

    + 2 300+ 3 100

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    + 2 500= 147,000 + 14,500 + 6,400 = 167,900

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    Exercise 2

    Scheduling with PERT/CPM

    1. Consider the PERT/CPM network shown below.

    1

    2 3

    4

    5

    6 7

    A

    B

    C

    D

    E

    F

    G

    H

    I

    J

    a. Add the dummy activities that will eliminate the problem that the activities havethe same starting and ending nodes.

    b. Add dummy activities that will satisfy the following immediate predecessorrequirements:

    ActivityImmediate

    predecessor

    H B, CI B, CG D, E

    2. Construct a PERT/CPM network for a project having the following activities:

    Activities Immediate predecessor

    A -B -C AD AE C, BF C, BG D, E

    The project is completed when activities F and G are both complete.

    3. Assume that the project in problem 2 has the following activity times:

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    Activities Time (months)

    A 4B 6C 2

    D 6E 3F 3G 5

    a. Find the critical path.b. The project must be completed in 1 years. Do you anticipate difficulty in

    meeting the deadline? Explain.

    4. Consider the following project network (the times shown are in weeks):

    1

    2

    3

    4

    5

    6

    5

    37

    6

    7

    3

    10

    8A

    B

    D

    C

    E

    FH

    G

    a. Identify the critical path.b. How long it will take to complete the project?c. Can activity D be delayed without delaying the entire project? If so, how many

    weeks?d. Can activity C be delayed without delaying the entire project? If so, how many

    weeks?e. What is the schedule for activity E?

    5. A project involving the installation of a computer system consists of eightactivities. The immediate predecessor and activity times are shown below.

    Activity Immediate predecessor Time (weeks)

    A - 3B - 6C A 2D B, C 5E D 4F E 3G B, C 9H F, G 3

    a. Draw the PERT/CPM network for this project.

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    b. What are the critical path activities?c. What is the expected project completion time?

    6. Piccadily College is considering building a new multipurpose athletic complex oncampus. The complex would provide a new gymnasium for intercollegiate

    basketball games, expanded office space, classrooms, and intramural facilities. Theactivities that would have to be undertaken before beginning constructing areshown below.

    Activity DescriptionImmediate

    predecessorTime(weeks)

    A Survey building site - 6B Develop initial design - 8C Obtain board approval A, B 12D Select architect C 4

    E Establish budget C 6F Finalise design D, E 15G Obtain financing E 12H Hire contractor F, G 8

    a. Develop a PERT/CPM network for this project.b. Identify the critical path.c. Develop the activity schedule for the project.d. Does it appear reasonable that construction of the athletic complex could begin

    1 year after the decision to begin the project with the site survey and initialdesign plans? What is the expected completion time for the project?

    7. Hamilton Country Parks is planning to develop a new park and recreational area ona recently purchased 100-acre tract. Project development activities includecleaning playground and picnic areas, constructing road, constructing a shelterhouse, purchasing picnic equipment, and so on. The PERT/CPM network shown

    below is being used in the planning, scheduling, and controlling of this project.

    1

    2

    3

    4

    5

    6 79

    3

    6

    6

    0

    3

    2

    6

    3

    I

    H

    G

    F

    E

    D

    A

    C

    B

    Activity timein weeks

    a. What is the critical path for this network?b. Show the activity schedule for this project.

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    c. The park commissioner would like to open the park to the public within 6months from the time the work on the project is started. Does this opening dateappear feasible? Explain.

    8. Consider the project network with activity times shown in days:

    1

    2

    3

    4 6

    5

    A

    B

    C

    D

    E

    F G

    35

    2

    5

    2

    6

    2

    The crash data for this project are as follows:

    Time (days) Total Cost ($)

    Activity Normal Crash Normal Crash

    A 3 2 800 1400B 2 1 1200 1900

    C 5 3 2000 2800D 5 3 1500 2300E 6 4 1800 2800F 2 1 600 1000G 2 1 500 1000

    a. Find the critical path and the expected project completion time on the normalbasis.

    b. What is the total project cost using the normal times?c. Find out the minimum project completion time using the crashing method.d. What is the minimum cost associated with the crashed project completion time?

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    9. Office Automation, Inc., has developed a proposal for introducing a new

    computerised office system that will improve word processing and interofficecommunications for a particular company. Contained in the proposal is a list ofactivities that must be accomplished to complete the new office system project.

    Information about the activities is shown below.

    Activity DescriptionImmediate

    predecessorTime

    Normal(weeks)

    CrashCost

    Normal($000s)

    Crash

    A Plan needs - 10 8 30 70B Order equipment A 8 6 120 150C Install equipment B 10 7 100 160D Setup training lab A 7 6 40 50E Conduct training D 10 8 50 75F Test system C, E 3 3 60 60

    a. Show the network for the project.b. Develop an activity schedule for the project.c. What are the critical path activities, and what is the expected project completion

    time?d. Assume that the company wishes to complete the project in 6 months or 26

    weeks. What crashing decisions would be recommended to meet the desiredcompletion time at the least possible cost? Work through the network, andattempt to make the crashing decisions by inspection.

    e. Develop an activity schedule for the crashed project.f. What is the added project cost to meet the 6-month completion time?