MATS-30004

Quantitative Methods Dr Huw Owens www.personalpages.manchester.ac.uk/staff/huw.owens

Planning Projects

• These lectures introduce a type of business model that helps plan the way a complex project is evaluated • What is a project?

– A project is a piece of work that has a clear start and finish and which is composed of any number of component parts or

activities

.

• E.g. the construction of buildings and roads, computerising a system or the design and launch of a new product.

• We would like to schedule the project to find out when each activity should start and finish and whether there is any flexibility in the timing.

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Planning Projects

• Large projects may have many hundreds of activities and will be planned using specialist software.

• To understand the input and output you need to know a little about the underlying techniques.

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Introduction

• A

project

defines a combination of interrelated activities that must be executed in a certain order before the entire task can be completed. • An

activity

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. in a project is usually viewed as a job requiring time and • These are the

project evaluation and review technique (PERT)

and the

critical path method (CPM)

• These techniques were developed by two groups almost simultaneously Company as an application to construction projects and was later extended to a more advanced status by Mauchly Associates.

was developed by the U.S. Navy by a consulting firm for scheduling the research and development activities for the Polaris missile program

. CPM

. was developed by E. I. Du Pont de Nemours &

PERT

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• Although PERT and CPM were developed independently, they are similar in principle.

• Today, PERT and CPM actually comprise one technique and the differences, if any, are only historical. Consequently, both technique are referred to as “project 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; – constructing a network diagram with each arc representing the activity; 26/04/2020 5

• •

Scheduling

– constructing a time chart showing the start and the finish times for each activity as well as its relationship to other activities in the project; – pinpointing the critical (in view of time) activities that require special attention 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 progress reports; – updating the network.

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Introducing the Project

• • • • • • • • UoM software wants to move premises and the project manager for the move has identified the following activities and their duration: A. Draw up the requirements study – 4 weeks B. Find suitable premises – 4 weeks C. Refurbish the building – 6 weeks D. Install cabling for the computer network – 2 weeks E. Move the furniture – 1 week F. Install the computers – 2 weeks G. Print the stationery with the new address – 2 weeks 26/04/2020 7

• – However, the activities of the project cannot be done in any order.

E.g. UoM can’t refurbish the building before they have found suitable premises and they can’t install the computers until the cabling is complete.

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The precedence table

• To define the order of the activities (i.e. which activities must come before others) we must draw up a

precedence

or

dependence

table.

• This lists each activity with any activities that

immediately

precede it.

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The precedence table

Activity A Requirements study B Find suitable premises C Refurbish building D Install cabling E Move in furniture F Install computers G Print stationery 26/04/2020 Preceding Activities none A B B C D,E B 10

Drawing the network

• • • • • • The next stage is to draw the network from the precedence table.

This often involves trial and error!

Start by drawing the network on the left-handside of the page with a single circle or node called the start node, which is labelled 1.

From this start node we draw an arrow for every activity that has no preceding activities.

For the UoM example project only activity A has no preceding activities so we only need one arrow.

At the end of this arrow we draw another node (node 2) to represent the completion of activity A.

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A

1 2

• We then draw an arrow out of node 2 for each activity that lists activity A as a preceding activity. In our example it is only activity, B.

A B

1 2 3

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• The rest of the network is constructed in the same way.

• Activities C,D and G list B as a preceding activity so three arrows must leave node 3:

4

C A B D

1 2 3 5

G

6

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• •

1

We continue until we have included all the activities in the network:

4

C E A B D F

2 3 5 6

G

Notice that as both

D

and

E

precede

F their arrows must end at the same node (node 5)

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• • • The network also has a single start node and a single final node (node 6) representing the completion of the entire project.

It is also convention to ensure that all arrows go from a lower numbered node to a higher numbered node.

There are two occasion when you might have trouble drawing a network, these both require us to introduce an activity into the network called a

dummy activity

.

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Logical dummies

• To illustrate the need for a logical dummy activity suppose an extra activity (

H

) of two weeks duration which needs to be done after the cabling is added to precedence table of the UoM project Activity Preceding Activities A Requirements study none B C D E Find suitable premises Refurbish building Install cabling Move in furniture A B B C F Install computers G Print stationery H Install Telephones D,E B D 26/04/2020 16

Logical Dummies

 A logical dummy activity is required when an activity appears more than once in the preceding activity column but with different preceding activities.

 E.g. activity D appears on its own as a preceding activity of H, but with E as a preceding activity of F. The part of the network which shows D and E preceding F is

4

E F D

3 5

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Logical Dummies

• • If we try and include activity

H

we run into trouble.

D

precedes

H

but if we draw imply that both not true.

E

and

D H

going out of node 5 we must happen before

H

, which is • The problem is that both

D

and

E

precedes

H

.

precede

F

but only

D

• To get round this we introduce a new node (that we label 3* as it’s extra) at the end of

D

but before the end of

E

.

• If

H

is drawn going out of the new node it will have not

E

as a preceding activity, as required.

D

but 26/04/2020 18

• A dotted line from node

3*

to node

5

has been drawn to ensure that the logic of the network is sound. As it does not represent a real activity and takes zero time we say it represents a

dummy activity

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4 3

D

3*

E

5

F H

19

Example

• Draw the UoM network including activity

H

.

• Solution • Introduce node 3* into the network and maybe space the nodes a little differently so the arrows go from left to right.

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1

A

2

B

3 4

C D

3*

E

5

H G F

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6

21

Uniqueness dummies

• Project Planning computer software usually refers to the activities by their start and end nodes. – E.g. 2-3 refers to the activity that starts at node 2 and finishes at node 3 (activity B in our previous example) • This means that in order to identify each activity uniquely we can only allow one arrow between any two nodes.

• However, a problem arises when two or more activities have exactly the same preceding activities and themselves precede the same other activities.

– E.g. C and D have the same preceding activities (A and B) and they both precede E.

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Precedence Table

• If we try to draw this network both C and D connect nodes 2 and 3 as shown below, so activity 2-3 could mean either C or D and is not uniquely defined.

Activity C D E

A

Preceding Activities A,B A,B C,D

B 2 C D 3 E

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• To resolve this situation a new node, 2*, is introduced between nodes 2 and 3 which gives an alternative route through the network for one of the activities.

• The arrow representing one of the activities can then go from node 2 to the new node and we can draw a dotted arrow from the new node to node 3 to represent a dummy activity as shown below:

A C E B 2 3 D 2*

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Scheduling Activities

• • • In the original specification of the UoM project we gave estimated durations for each of the activities.

We want to use the network to find out when each activity must be started and/or finished and how tight the timing is for each.

To do this we use the complete UoM network (including H) and include the duration of each activity.

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1

A 4

2

B 4

4

C 6

3

D 2

3*

E 1

5

H 2 G 2 F 2

6

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Earliest and Latest Event Times

• • • To find the earliest and latest event times we will traverse the network twice.

Firstly, we start at the start node (node 1) and work forwards calculating the earliest possible times at which each node can occur.

Secondly, we start at the finish node (node 6) and work backwards calculating the latest possible time at which each node can occur.

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Earliest event times

• Node 1 is the start node so we will say that it happens at time 0 (write 0 above node 1).

• Node 2 follows on from activity A, which starts at time 0 weeks and takes 4 weeks to complete. So, the earliest time for node 2 is 4 weeks (write a 4 above node 2).

• Node 3 follows activity B which, from the 4 you’ve written above node 2, starts at time 4 weeks and takes 4 weeks to complete so the earliest event time for node 3 is 8 weeks (write 8 above node 3).

• In the same way node 3* follows activity D and so occurs 2 weeks later than node 3, at 10 weeks, and node 4 occurs 6 weeks after node 3 at 14 weeks.

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• So far, the process has been simple as only one activity arrow has preceded each node.

• Node 5 has two activities preceding it. The two activities

E

and the

dummy

must both be completed before node 5 occurs.

• The earliest time that activity E can be completed is the earliest time for node 4 (14) plus the duration of activity E (1); that is 14+1 = 15.

• In the same way the earliest time at which the dummy activity can be completed (node 3*) plus the duration of the dummy activity gives; 10+0 = 10.

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• The earliest event time for node 5 is the latest of these two times (that is 15) which we can now write above node 5.

14 0 1

A 4

4 2

B 4

8

C 6 D 2

3 4 10 3*

0 E 1

15 5

F 2

??

6

•

H 2 G 2

In general, when a node has two or more preceding activities they must all finish before the node can occur so the earliest event time is the taken by the paths into the node

maximum

of the times 26/04/2020 30

Example

• What is the earliest event time for node 6?

• Solution • Activities F,G and H immediately precede node 6. If we take them one at a time: • F can be finished on time – Earliest event time of node 5 (15) + duration of F (2) = 17 • G can be finished at – Earliest event time of node 3 (8) + duration of G (2) = 10 • H can be finished at – Earliest event time of node 3* (10) + duration of H (2) = 12 • The earliest event time of node 6 is the greatest of these, 17.

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0 1

A 4

4 2 14

B 4

8

C 6 D 2

3 4 10 3*

0 E 1

15 5

F 2 H 2 G 2

17 6

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Latest Event Times

• We find the latest event times in a similar way but we start at the end node and work backwards.

• For each node we want the latest time that it can be completed for the project to be completed on time (17).

• The latest event time for node 6 is 17 weeks, the overall duration of the project. • The latest event times are usually written beneath the nodes.

• Activity F follows node 5 and takes 2 weeks. To ensure completion by the latest event time of node 6 (17) activity F must be started by 17-2 = 15 weeks at the latest.

• In general, the latest event time for a node must allow all the immediately following activities to finish by the latest event time of their end nodes.

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Example

0 0 1

A 4

4 4 2 14

B 4

4 8

C 6

3

D 2

14 10 3*

0 E 1

15 5 8 15

H 2

15

G 2 F 2

17 6 17

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Example

• Calculate the latest event time for node 3.

• 3 activities C,D and G immediately follow node 3. Activity C must end by the latest event time of node 4 (14), activity D by the latest event time of node 3* (15) and activity G by the latest event time of node 6 (17).

• The durations of C,D and G are 6, 2 and 2 weeks, respectively. So C must start by time 14-6=8, D time 15 2 = 13 and G by time 17-2 = 15. • Node 3 must therefore occur at the earliest of these times; that is, week 8.

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Calculating Floats

• We can now us the networks to find out which activities are flexible in time and which aren’t.

– E.g. Activity D runs from node 3 to node 3*. The earliest start time for activity D is the earliest event time of node D (8 weeks) and the latest finish time for activity D is the latest event time of node 3* (15 weeks).

– This means there are 15-8=7 possible weeks in which to do activity D although it has a duration of 2 weeks.

• We say that activity D has a

float

of 5 weeks.

• By this we mean that activity D could start up to 5 weeks later than the earliest start time preceding it or, that its duration could be increased by 5 weeks without delaying the completion of the project.

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Calculating floats

• In general, the float of an activity is calculated as • Float = latest event of the following node – earliest event time of preceding node – duration of the activity • Now calculate the float of activity E.

• Solution • The latest event time of node 5 is 15, the earliest event time of node 4 is 14 and the duration of activity E is 1 so, the float = 15-14-1=0 • Activity E has a zero float which means that any delay in its start time or extension of its duration will increase the total duration of the project.

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Critical activities and Critical Paths

• Any activity with a zero float is called a

critical activity.

• Critical activities

must

start at the earliest start time of the preceding node and there is no flexibility in their duration.

• Table CP1 shows all the activities in the UoM project, their earliest start times, latest finish times and floats.

• Notice that activities A,B,C,E and F are critical activities.

• They form a continuous path through the network called the

critical path

.

• The critical path is the longest route through the network from the start node to the final node.

• There may be more than one critical path. 26/04/2020 38

Activity

A B C D E F G H 26/04/2020

Critical Paths (Table CP1)

8 14 15 8 10 4 8

Earliest Start Time Latest Finish Time

0 4 4

Duration

8 14 4 6 15 15 17 17 17 2 1 2 2 2

Float

4-0-4 = 0 8-4-4=0 14-8-6=0 15-8-2=5 15-14-1=0 17-15-2=0 17-8-2=7 17-10-2=5 39

Summary of PERT/CPM Critical Path Procedure

• Step 1 • Step 2 • Step 3 • Step 4 • Step 5 Develop a list of activities that make up the project; Determine the immediate predecessor activities for each activity listed in the project; Estimate the completion time for each activity; Draw a network depicting the activities and immediate predecessors listed in Steps 1&2; Using the network and the activity time estimates, determine the earliest start and finish times for each activity by making a forward pass through the network. The earliest finish time for the last activity in the project identifies the total time required to complete the project; 26/04/2020 40

• Step 6 • Step 7 • Step 8 • Step 9 26/04/2020 Using the project completion time identified in Step 5 as the latest finish time for the last activity, make a backward pass through the network to identify the latest start and finish times for each activity; Use the difference between the latest start time and the earliest start time for each activity to identify the slack time available for the activity; Find the activities with zero slack; these are the critical path activities; Use the information from Steps 5&6 to develop the activity schedule for the project.

41

Gantt Charts

• It’s not always obvious from the network diagram which activities should be happening at any particular time.

• A Gantt chart (named after its originator) has a row for each activity and a column for each time period and so shows the passage of time more clearly.

• By convention, the critical activities are shown at the top of the chart.

• The Gantt chart for UoM project is shown below.

• Notice that the times at which the critical activities must happen are shown as solid areas. • The times needed by non-critical activities are shown by a solid outline starting at their earliest start times whilst their floats appear as dotted outlines afterwards.

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Gantt Chart for UoM Project

A B C D E F G H 26/04/2020 43

Gantt Charts

• The Gantt Chart enables us to see at a glance how many resources are required at any particular time.

– E.g. if an activity requires only one person, the chart for the UoM project tells us that we need one person up to time 8, three people from time 8 to time 10, two between times 10 and 12 and just one from time 12 onwards.

– The chart can be used to adjust the timing of non-critical activities to spread the workload more evenly.

• Whilst a non-critical activity can be moved to the right within the bounds of the dotted outline without delaying the project, remember that this may have knock-on effects for other activities.

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Extensions

• So far we have covered the basics of project planning using networks.

• For instance, we have assumed that the duration of each activity is known. • In practice, we usually only have an estimate and we need to consider the effects of changes in the duration of each activity.

• One way of doing this is to make three estimates of each activity’s duration – A pessimistic one – A realistic one – An optimistic one 26/04/2020 45

Crashing

• This information is then used (with some probability theory) to calculate the probability of completing the project by a particular date.

• Sometimes it may be worthwhile paying to reduce the duration of an activity – E.g. by paying overtime or hiring extra machines because it will reduce the total project duration and save money.

• This is known as

crashing

.

• The time by which the duration of an activity can be reduced is known as the

crash time

and the costs of such a reduction are the crash costs.

• These can be compared to see which activities would be most worthwhile to crash.

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L3 Consideration of Time-cost Trade-offs

• • PERT/CPM can answer questions such as • the total project completion time, • the critical activities, • and the slack times of the non-critical activities. – Thus, the project manager has a clearer picture and a greater control over the project. The project schedule is based on the given cost and finish time of the individual activities.

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

These are the considerations of activity.

time-cost trade-offs

. To obtain the minimum cost or the minimum time, we need to know the possible reduction in time and extra cost for reduction per unit time for each 26/04/2020 47

Completion of projects at minimum cost

• By adding more resources, a project may be sped up. • Usually, the purpose of speeding up is to save money on project overheads, to avoid penalty clauses in contracts or, sometimes, to earn bonuses for early completion. • Complications arise when the critical activities are sped up more and more, causing other activities also become critical. • We will discuss the algorithm through the following example. 26/04/2020 48

Notation

• ES = Earliest Start Time • EF = Earliest Finish Time (EF=ES+t) • LS = Latest Start Time (LS=LF-t) • LF = Latest Finish Time • Slack = LS - ES = LF - EF ES EF Activity [0, 5] A (LS, LF

)

5 1 Expected completion time 2 26/04/2020 49

Example

• A project consisting of 8 activities is described in the following table. The cost 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: – calculate the normal completion of the project, its cost, and the critical path; – calculate and plot on a graph paper the cost/time function for the project and state: – the minimum cost and the associated time; – the shortest time and the associated cost.

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Example

Activity A (1-2) B (1-3) C (1-4) D (2-4) E (2-5) F (3-6) G (4-6) H (5-6) Normal completion time (days) 6 8 5 3 5 12 8 6 Shortest completion time (days) 4 4 3 3 3 8 6 6 Cost of reduction per day (£) 80 90 30 - 40 200 50 -

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Example

• Firstly, set-up the network according to the description of the project. • Then using the PERT/CPM scheduling technique discussed earlier, we establish the ES, EF, LS, LF times and the critical activities. • These are shown in the following network and table.

1 B [0,8] 8 (0,8) C [0,5] 5 A [0,6] (7,12) 6 (3,9) 2 3 12 F [8,20] (8,20 ) 4 [6,9] D 3 (9,12) E [6,11] 5 (9,14) G [9,17] 8 (12,20) 5 [11,17] H 6 (14,20) 6 26/04/2020 52

1 [0,8] B 8 (0,8) C [0,5] 5 A [0,6] (7,12) 6 (3,9) 2 3 12 F [8,20] (8,20 ) G [9,17] 4 [6,9] D 3 (9,12) E [6,11] 8 (12,20) 5 (9,14) 5 [11,17] H 6 (14,20) 6 26/04/2020 53

Example

The Table shows that the normal completion time is 20 days and the critical activities are B(1-3) and F(3-6). The cost of completing the project at normal speed is £5800 + 20



£160 = £9000

26/04/2020 Activity A B C D E F G H ES 0 0 0 6 6 8 9 11 EF 6 8 5 9 11 20 17 17 LS 3 0 7 9 9 8 12 14 LF 9 5 12 14 20 20 20 Slack 3 0 3 3 0 7 3 3 Critical? Yes Yes 54

Example

• • 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, the activity 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 of time to speed up is determined based on – (1) the reduction should reduce the project completion 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 completion time to 17 days. • As indicated in the following diagram, all activities except C become critical because of this. 26/04/2020 55

Example

1 B [0,5]

5

(0,5) C [0,5] 5 (4,9) A [0,6] 6 (0,6) 2 3 F 12 [5,17] (5,17 ) G [9,17] 4 [6,9] D 3 (6,9) E [6,11] 8 (9,17) 5 (6,11) 5 [11,17] H 6 (11,17) 6 • • The new cost accordingly is now: £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 is to 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.

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Example

1 B [0,4]

4

(0,4) C [0,5] 5 A [0,5] (3,8)

5

(0,5) 2 3 F [4,16] 12 (4,16 ) 4 [5,8] D 3 (5,8) E [5,10] 5 (5,10) G [8,16] 8 (8,16) 5 [10,16] H 6 (10,16) 6 • • 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 F and A which have 4 days and 1 day, respectively, to spare. • We can only reduce one day on both of these and the total completion time is now reduced to 15 days. Activity C still has 2 days slack time while all the others are critical.

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Example

1 B [0,4] 4 (0,4) C [0,5] 5 A [0,4] (2,7)

4

(0,4) 3

11

F (4,15 ) 2 4 [4,7] D 3 (4,7) E [4,9] 5 (4,9) [4,15] G [7,15] 8 (7,15) 5 [9,15] H 6 (9,15) 6 • • 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, 3 days, and 2 days available respectively). • Reducing these activities by two days decreases the total project time to 13 days.

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Example

3 F [4,13] [0,4]

9

(4,13 ) B 4 (0,4) C [0,5] G [7,13] 6 1 5 A [0,4] (2,7) 4 (0,4) [4,7] D 3 (4,7) 4 E [4,7]

6

(7,13) [7,13] H 6 (7,13) 2 5

3

(4,7) • • 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 time reduction is available on the critical path 1-2-5-6. • For the purpose of plotting the required cost/time graph, we summarise in the following table the completion times and costs of the project.

26/04/2020 59

26/04/2020

Example

9300 9200 9100 9000 8900 8800 8700 8600 8500 20 Completion time (days) 20 17 16 15 13 Cost (£) 9000 8790 8800 8920 9180

Completion time vs Cost

17 16

Days

15 13 Cost (£) 60

Example

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

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

• One way of finding the minimum time for completion of a project is to start with the normal completion network and gradually make reductions in critical activities until minimum time is reached, like the method used in the last example.

26/04/2020 62

• However, if it is the minimum time that is of interest, then there is another and more efficient way of proceeding, i.e., • 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 • Consider the activities which are not critical and allow the most expensive of 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.

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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: – calculate and state the time for completion on a normal basis; – calculate and state the critical path on this basis, and the cost; – calculate and state the cost of completion in the shortest possible time.

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Activity A(1-2) B(1-3) C(1-4) D(2-3) E(2-5) F(3-4) G(4-5) H(4-6) I(4-7) J(5-6) K(6-8) L(7-8) 26/04/2020 Completion time (days) Cost of activity (£1,000) Possible reduction time (days) Extra cost for reduction (£/day) 5 8 15 4 12 6 7 11 10 8 9 10 6 10 17 5 15 8 9 13 12 14 25 13 1 2 4 1 3 2 3 2 1 3 2 2 300 200 700 400 200 200 400 300 600 300 100 500 65

Activity A B C D E F G I H J K L 26/04/2020 Earliest Start Time 0 0 0 5 5 9 15 15 15 22 30 25 Latest Finish Time 5 9 15 9 22 15 22 30 29 30 39 39 Duration 11 10 8 9 10 5 8 15 4 12 6 7 1 0 0 5 0 0 4 4 0 0 4 Float LF ES-Dur 0 66

• According to the above table, the PERT/CPM network can be generated as follows.

2 E [5,17] 4 12 D [5,9] (10,22) J [22,30] [0,5] A 5 (0,5) 1 B 8 (5,9) 5 8 (22,30) [0,8] 3 [15,22] [15,26] (1,9) C [0,15] (9,15) 6 [9,15] G F 7 (15,22) 15 (0,15) 4 H 10 I [15,25] (19,29) 11 (19,30) 7 6 [25,35] L 10 (29,39) K [30,39] 9 (30,39) 8 26/04/2020 67

• As it is indicated, the project will require 39 days to complete under the normal situation. • • The above network shows that there are two critical paths, i.e., A-D-F-G-J-K, and C-G-J-K. The cost on the normal basis is  £(All costs) + 39  £166,500 £500 = £147,000 + £19,500 = • To find the minimum completion time, we first reconstruct the PERT/CPM network by crashing all the activities, i.e., using the shortest completion time for each activity. 26/04/2020 68

2 E 3 D 9 [4,7] [0,4] A 4 (0,4) [0,6] B 1 (1,7) 6 C (4,7) [0,11] (7,11) 11 3 4 [4,13] (8,17) [7,11] F [11,17] G 6 (11,17) (0,11) 4 5 [11,19] I 8 (13,21) [11,19] J 6 [17,23] (17,23) H 8 (15,23) 7 6 [19,27] L 8 (21,29) K [23,29] 6 (23,29) 26/04/2020 8 69

• • By crashing all the activities, the minimum completion time of the project is found to be 29 days. • Money can be saved by allowing the slowing down of those non-critical 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 following table lists all the non-critical activities and their costs to speed up.

26/04/2020 70

Non-critical activities B E H I L Cost to speed up (£/day) 200 200 300 600 500 • Obviously, the most expensive non-critical activity to speed up is I. • So, we first let activity I to slow down to its normal completion time of 10 days. • This will result in the following network diagram. 26/04/2020 71

2 E 3 D 9 [4,7] [4,13] (8,17) J [17,23] K [23,29] [0,4] 5 6 8 (0,4) (4,7) 6 (17,23) 6 (23,29) A 1 4 B 6 [0,6] 3 [11,19] (1,7) C (7,11) [0,11] 4 [7,11] G F 6 (11,17) 11 H [

11,21

] (0,11) I 4 [11,17] (

11

,21) 8 (15,23)

10

7 [

21,29

] L 8 (21,29) • The change of the completion time of activity I from 8 days to 10 days makes activities 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 following diagram.

26/04/2020 72

2 E [4,13] 3 D 9 [4,7] (8,17) J [17,23] K [23,29] [0,4] 5 6 (0,4) (4,7) 6 (17,23) 6 (23,29) A 4 [0,6] 3 [11,17] B [11,

22

] [21,29] 1 (1,7) [7,11] 6 (7,11) G 6 H

11

(

12

,23) C [0,11] 4 F (11,17) L 8 (21,29) 11 7 (0,11) 4 10 I [11,21] (11,21) • Thirdly, we allow activity E to take its full 12 days, leading to the diagram below.

26/04/2020 73 8

[0,4] A 4 (0,4) [0,6] B 1 (1,7) 6 C 2 E [4,

16

] 3 D

12

[4,7] (

5

,17) (4,7) [0,11] (7,11) 3 4 [7,11] F 11 [11,17] G 6 (11,17) (0,11) 4 5 10 I [11,21] (11,21) J 6 [11,22] [17,23] (17,23) H 11 (12,23) 7 6 [21,29] L 8 (21,29) K [23,29] 6 (23,29) 26/04/2020 8 74

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

2 E [4,16] 3 12 D [4,7] [0,4] A 4 (0,4) [0,

7

] B 1 (

0

,7)

7

C (4,7) [0,11] (7,11) 11 3 4 (5,17) [7,11] F [11,17] G 6 (11,17) (0,11) 4 5 H 10 I [11,21] (11,21) 11 J 6 [11,22] [17,23] (17,23) (12,23) 7 6 [21,29] L 8 (21,29) K [23,29] 6 (23,29) 8 26/04/2020 75

• • • • • • • • • • No further savings are possible as the three non-critical activities are all now at normal duration. Hence, the least possible cost of completing in 29 days is:  £(All costs) + 29  £500 + 1  £300 + 1  + 4  + 1  £200 £700 £400 + 2  + 1  + 2  + 3  + 2  £200 £400 £300 £100 £500 • = £147,000 + £14,500 + £6,400 = £167,900 26/04/2020 76