Quantitative Methods for Business chapter 14 pot

To avoid activities F and H having the same event node marking their start and the same event node marking their completion we can introduce a dummy activity, an activ-ity that takes no

Getting from A to B – project planning using

networks

14

Chapter objectives

This chapter will help you to:

■ construct network diagrams

■ apply critical path analysis (CPA)

■ identify slack time available for project activities

■ use the Program Evaluation and Review Technique (PERT)

■ conduct cost analysis of projects using network diagrams

■ become acquainted with business uses of CPA/PERTMany business operations involve planning and coordinating a project –

a series of interlinked tasks or activities all of which have to be performed

in the correct sequence and within the least amount of time in order toachieve a successful conclusion to the venture This is not only typical oflarge-scale projects such as you would find in industries like constructionand shipbuilding, but also occurs on a more modest basis in administra-tive processes such as organizing events like conferences and concerts

To help plan and execute projects successfully managers can turn

to network analysis, a system of representing and linking the activities

involved in a project Once they have designed a network they can use

critical path analysis (CPA) to establish the minimum duration of the

project by identifying those tasks whose completion on time is essential,

the critical activities Beyond this they can bring into consideration the

probability distributions that reflect the chances of the activities being

completed by specific times using the program evaluation and review technique (PERT) In this chapter we will look at these techniques.

14.1 Network analysis

We can apply network analysis to any project that consists of series

of distinct activities provided that we know which activities must be pleted before other activities can commence This is called the prece-dence of the activities because it involves identifying the activities that

com-must precede each activity These are crucial because the point of a network

diagram is to show how the activities in a project are linked

The diagrams used for network analysis are built up using a singlearrow for each activity The arrow begins at a circle representing the point

in time at which the activity begins and finishes at another circle that

represents the completion of the activity These circles are known as event nodes We would represent a single activity as shown in Figure 14.1.

In Figure 14.1 the direction of the arrow tells us that the event node on the right marks the completion of the activity The activity islabelled using a letter and the events nodes are numbered Networkdiagrams consist of many arrows and many event nodes, so logical layout and labelling is important

Networks should be set out so that the beginning of the project is resented by the event node on the extreme left of the diagram and thecompletion by the event node on the extreme right In compiling themyou should try to ensure that the event nodes are labelled sequentiallyfrom left to right from event number 1, the start of the project

rep-All the arrows should point from left to right either directly or atangle and certainly not from right to left This is usually straightfor-ward because the purpose of an arrow in a network is to represent the

position of the activity in relation to other activities and not its duration.

A network diagram is not intended to be to scale

Figure 14.1

A single activity

Example 14.1

Avia Petitza is an independent airline flying short-haul routes in South America Thenew general manager is keen to improve efficiency and believes that one aspect of theiroperations that can be improved is the time it takes to service planes between flights.She has identified the activities involved and their preceding activities and compiledthe following table:

Produce a network diagram to portray this project

Activity A has no preceding activities so we can start the network with this as shown

in Figure 14.2(a) Four subsequent activities, B, C, D and I depend on activity A being completed so we can extend the network as shown in Figure 14.2(b) Activity Gmust follow activity C and activity J follows activity I as depicted in Figure 14.2(c) Activity E follows activity B and activity F follows activity E as shown in Figure14.2(d)

Before activity K can take place activity H must also have finished This presents aproblem because we cannot have two or more activities starting at the same event nodeand finishing at the same event node If we did we would not be able to use the diagramfor the sort of scheduling analysis you will meet in the next section of this chapter

To avoid activities F and H having the same event node marking their start and the

same event node marking their completion we can introduce a dummy activity, an

activ-ity that takes no time nor uses any resources Its role is merely to help us distinguish two activities that might otherwise be confused in later analysis To distinguish between

a dummy activity and real activities it is portrayed as a dotted line You can see a dummyactivity in Figure 14.2(e) used to ensure that the event node that marks the conclusion

of activity H is not the same as that marking the end of activity F

A Drive service vehicles to plane None

The dummy activity in Figure 14.2(e) helps us to identify two

sep-arate activities A dummy activity used in this way is an identity dummy.

Dummy activities are also used to resolve logical difficulties that arise

in some networks A dummy activity used to do this is a logical dummy.

2 1

9

H E

(e)

H E

The final diagram, Figure 14.2(f), incorporates activities K, L and M Activities K and

L share the same finishing event as activities D and J This event represents the ning of activity M, whose closing event marks the conclusion of the project We can alsonumber the events now the diagram is complete

begin-Figure 14.2

Stages in the compilation of the network diagram for aircraft servicing in Example 14.1

Both types of dummy activity serve only to clarify the network andavoid ambiguity The important distinction between real and dummyactivities is that the latter do not take time or use resources.

Compiling network diagrams often brings an understanding of the way

in which activities that make up a project fit together It enables you toclarify uncertainties about the planning of the project This is some-thing usually achieved by constructing drafts of the network beforeproducing the final version

Whilst an understanding of the sequence is important, there is muchmore to successful project planning Ascertaining the minimum time inwhich the project can be completed and scheduling the activities over

D B

D B

Figure 14.3

Incorrect (a) and correct (b) networks for Example 14.2

C Pour hot water in Chai’s mug A

D Pour hot water in Bru’s mug A, B

time are typically the central issues The technique that enables us to dothis is critical path analysis.

14.2 Critical path analysis

If we know the duration of each activity we can use a network diagram

to find out the least amount of time it will take to finish the project

A network diagram shows the way that activities are linked; in effect itportrays the project as a series of paths of activities Since every activitymust be finished for the project to be completed, the minimum dur-ation of the project is the length of time required to carry out the most

time-consuming path of activities This path is known as the critical path

since any delay in the completion of activities along it will prolong theentire project Finishing those activities on the critical path on time is

therefore critical for the project; they are known as critical activities.

Critical path analysis involves increasing the amount of information

in the network by enhancing the role of the event nodes In drawing anetwork diagram the event nodes are in effect the punctuation marks

in the diagram; they bring order to the sequence of arrows In criticalpath analysis they become distinct points in time For each event node

we assign an earliest event time (EET) and a latest event time (LET) These

are written in the circle that represents the event node beneath theevent number, as illustrated in Figure 14.4

The circle on the left in Figure 14.4 is the event node that marks thepoint in time when activity X begins The number in the upper part ofthe circle is the event number, 20 The number below it to the left is theearliest event time and the number below it on the right is the latestevent time These numbers tell us that the earliest time that activity Xcan start is time period 9 and the latest time it can start is time period

11 From the equivalent figures in the event node to the right of activity

X you can tell that the earliest time the activity can be completed is timeperiod 13 and the latest time it can be completed is time period 15

To work out the earliest and latest times for the events in a network weuse the activity durations Starting with the event node at the beginning

Figure 14.4

Earliest and latest

event times for a

single activity

13 15

20 11 9

of the network and working through the network from left to right wewrite in the earliest time that each event node in the network can occur.

This is referred to as making a forward pass through the network

When doing this you need to remember that where there are two or

48 5

2 2

Figure 14.5

Network for ground servicing with earliest event times

A Drive service vehicles to plane 2

more activities leading to the same event node it is the duration of thelongest of the activities that determines the earliest time for the eventsince all the activities must be completed before the event is reached Wewrite the earliest event times in the lower left-hand side of the event nodes.

If you look carefully at Figure 14.5 you will see that the earliest eventtime for event 1 is 0 reflecting the fact that 0 minutes of project timehave elapsed at the beginning of the project The earliest event timefor event 2 is 2, reflecting the 2 minutes needed for the completion ofthe only activity between event node 1 and event node 2, activity B,driving the service vehicles to the plane

The earliest event time for event 8 is perhaps less obvious The figureentered, 33, has been worked out based on the longest route to it Theactivities leading to event node 8 are activity F, with its associateddummy activity between events 7 and 8, and activity H The earliest timethat activity F can start is the earliest event time on the event marking itsbeginning; the 13 in event 4 If we add the 20 minutes that activity F,cleaning the cabin, takes to the earliest event time for event 4 we get 33minutes as the earliest time activity F can be completed Event 4 alsomarks the beginning of activity H, loading the food and beverages If weadd the time this activity takes, 10 minutes, to the earliest event time ofactivity 4 we get 23 minutes This would be the earliest event time forevent 8 if we did not need to complete activity F, but since both activity

F and activity H have to be completed before event 8 can occur, the est time we can get there must allow for the longer activity, activity F, to

earli-be concluded, hence the earliest event time for event 8 is 33 and not 23.The event node indicating the completion of the project on theextreme right of the network has an earliest event time of 59 minutes.This is the minimum duration of the entire project; given the activitydurations it cannot be completed in a lesser amount of time We nowneed to turn our attention to the latest event times, as comparing thesewith the earliest event times will enable us to identify the critical pathfor the project

Once we have established the minimum project duration, in the case

of Example 14.3, 59 minutes, we assume that the project manager willwant to complete it in that time We now undertake the same sort of task

to find the latest event times as we used to find the earliest event times,but this time we start on the right-hand side and work back throughthe network ascertaining the latest time each event can occur if theproject is to be finished in the minimum time This is referred to as

making a backward pass through the network.

The latest event times for Example 14.3 are included in Figure 14.6.Each event now has a number entered in the lower right-hand side ofits node

Looking at Figure 14.6 you can see that the latest event time for thelast event, event 11, is 59 minutes If the project is to be completed in

59 minutes then the latest time we must reach this point is 59 minutes.The latest event time of event 10 is 57 minutes, 59 minutes less the

2 minutes we must allow for activity M to be completed

Several activities conclude at event 10, including activities G and L.Activity G begins at event 5 The latest time event time of event 5 is 27 min-utes, sufficient to allow the 30 minutes necessary for the completion ofactivity G in time for event 10 to be reached in 57 minutes, which in turnallows for the 2 minutes to complete activity M and thus conclude theproject in 59 minutes Activity L begins at event 9 Since activity L, detach-ing the stairway, takes 3 minutes, the latest event time for event 9 is 54minutes, sufficient to allow the 3 minutes for the completion of activity L

so that event 10 can be reached in 57 minutes and thus leave 2 minutesfor activity M to finish in time for completing the project in 59 minutes

If you study Figure 14.6 carefully you will see that some events, such

as event 5, have the same earliest and latest event times while otherevents, such as event 6, have different ones In the case of event 6 theearliest event time is 7 minutes whereas the latest event time is 52 min-utes This implies that activity I can be completed by 7 minutes but itdoesn’t have to be finished until 52 minutes have elapsed In otherwords, there is time to spare for the completion of activities I and J;

they have what is called slack or float.

48 5

33

33 39 39

59 59 11

In contrast, the latest event time for event 5 is the same as its earliestevent time, 27 minutes In this case the earliest time event 5 can bereached is the same as the latest time it has to be reached if the project

is to be completed in 59 minutes This implies there is no slack or floatfor activities C and G; they must be undertaken at the earliest feasible

opportunity as their completion on time is critical for the conclusion

of the project, they are critical activities If you look at the event nodes

that have the same earliest and latest event times you will see that theyconnect activities A, C, G and M These activities are driving servicevehicles to the plane, unloading and loading baggage, and driving

service vehicles from the plane They form the longest or critical path

through the network; a delay in the execution of any of them will result

in the project being prolonged beyond 59 minutes On the other hand,any reduction in the duration of any of them will reduce the duration

of the project, but only up to a point; at some point another path willbecome critical

By definition the critical activities in a network have no slack or float,but other activities will A project manager is likely to be very interested

in the float available for these activities, as it indicates the degree of latitude available in scheduling them

To work out the total float for an activity we need to know the latest

event time for the event that marks its completion, its finishing event,

and the earliest event time for the event that represents its beginning,

its starting event If we subtract the duration of the activity from the

dif-ference between these times what is left over is the total float for theactivity We can summarize this procedure as:

Total float  Latest event time (finishing event)

 earliest event time (starting event)

 activity durationUsing abbreviations:

TF LET (F)  EET (S)  ADLook carefully at Table 14.1 in Example 14.4 and you will see thatsome activities have a zero float These are the activities on the criticalpath – A, C, G and M In contrast there is a float of 6 minutes in the pathB–E–F–K–L, which is the uppermost path in Figure 14.6 This is floatassociated with the path rather than an individual activity; once used up

by, say, taking 9 minutes rather than 3 minutes for activity B, attachingthe stairway, it would not be available for the subsequent activities E, F,

K and L, which as a consequence would become as critical as the ties on the critical path

activi-Activity H, loading the food and beverages, has a total float of 16minutes, 6 of which are shared with the other activities along its path;

B, E , K and L The remaining 10 minutes are specific to activity H; itmust be completed by the same time as activity F, cleaning the cabin, ifthe project is to be finished in the minimum time, yet it takes 10 min-utes less to perform

Activity D, refuelling, has a float of 40 minutes, which cannot beshared with the other activities along its path, A and M, as they are both critical The relatively large amount of float for this activity allowsthe project manager considerable flexibility in scheduling this activity.Similar flexibility exists for activities I and J, the water operations,which share a total float of 45 minutes

At this point you may find it useful to try Review Questions 14.1 to

14.11at the end of the chapter

14.3 The Program Evaluation and

Review Technique (PERT)

So far we have assumed that the activities making up a project eachhave fixed durations In reality this is unlikely to be the case In the

Example 14.4

Find the total floats for the ground servicing activities in Example 14.1

Table 14.1

Total floats for activities in Example 14.1

Activity LET (F) (1) EET (S) (2) AD(3) TF(1)–(2)–(3)

ground servicing project introduced in Example 14.1 factors such asthe state of equipment, the time of the day or night, and weather con-ditions will influence the time it takes to complete the activities Ratherthan being fixed, activity durations are more likely to be variable PERTallows us to take this into account.

In using PERT we assume that the duration of an activity is a ous variable that follows a continuous probability distribution Welooked at continuous probability distributions like the normal distri-bution in Chapter 13 The distribution used in PERT as the model for

continu-activity durations is the beta distribution Like the normal distribution it

is continuous, but unlike the normal distribution it can be skewed orsymmetrical and it has a minimum and a maximum value The keycharacteristics or parameters of the beta distribution are the minimum

or optimistic duration, a, the maximum or pessimistic duration, b, and the most likely duration, m.

You can see two types of beta distribution in Figure 14.7 The upperdiagram portrays a symmetrical distribution with a minimum of 0 and a

maximum of 10 The most likely value of this distribution is 5 minutes,midway between the extremes In the asymmetrical beta distribution inthe lower diagram the minimum and maximum durations are the same

as in the upper diagram but the most likely value is rather lower

If you know the optimistic (a), most likely (m) and pessimistic (b)

durations of an activity you can work out its mean duration using theexpression:

The standard deviation of the activity duration is:

The variance of the activity duration is the square of its standard deviation:

The mean duration of an entire project is the sum of the mean ations of the activities that are on the critical path, the critical activities.The variance of the project duration is the sum of the variances of thecritical activities and the standard deviation of the project duration isthe square root of its variance Note that you cannot get the standarddeviation of the project duration by adding together the standard devi-ations of the durations of the critical activities; it is the variances thatsum, not the standard deviations

The mean project duration is the sum of the mean durations of the critical activities

In this project the critical activities are A, C, G and M The means of these activities are:

␮C 22 (4 * 25) 31  25.5

6

␮A 1 (4 * 2) 3  2

6

The mean duration of the critical path is the sum of these means:

␮CP  ␮A ␮C ␮G ␮M 2  25.5  29  2  58.5 minutes

The standard deviation of the critical path is the square root of the variance of thecritical path, which is the sum of the variances of the critical activities, A, C, G and M.The variances of these activities are:

␴A2 ((3  1)/6)2 4/36  1/9

␴C2 ((31  22)/6)2 81/36  9/4

␴G2 ((32  25)/6)2 49/36

␴M2 ((3  1)/6)2 4/36The variance of the duration of the critical path is:

The standard deviation of the duration of the critical path is: