Project Scheduling by PERT/CPM

Project Scheduling by PERT/CPM

Reference Books:

Anderson, Sweeney, and Williams, AN INTRODUCTION TO MANAGEMENT SCIENCE, QUANTITATIVE APPROACHES TO DECISION MAKING, 7

edition, West Publishing Company,1994

Hamdy A. Taha, OPERATIONS RESEARCH, AN INTRODUCTION, 5

edition, Maxwell Macmillan International, 1992

Daellenbach, George, McNickle, INTRODUCTION TO OPERATIONS RESEARCH TECNIQUES, 2

edition, Allyn and Bacon. Inc, 1983

Lawrence Lapin, QUANTITATIVE METHODS for Business Decisions with Cases, 4

edition Harcourt Brace Jovanovich, Inc., 1988

T. A. Burley and G O’sullivan, OPERATIONAL RESEARCH, MacMillan Education Ltd., 1990

Lecture 1

1. 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 in a project is usually 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 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;

1

• constructing a network diagram with each arc representing the activity;

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.

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 event represents 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

4 3

2

Tail Head 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 dummy activity is introduced in such situations);

D B

B A

A

In this case, D is the dummy activity.

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

• What activities must be completed immediately before this activity can 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 procedure involves 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 Table below. Construct the network diagram.

Activity Immediate Predecessor A - B - C B

D A, C

A 4

1

2

A 3 D

C

D

B C B

3

Example 2:

Construct the network based on the Table of information .

Activity

Immediate Predecessor

A - B - C B

D A, C

E C F C

G D, E, F

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

A D

1 C

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 C and node 3 in order to add E correctly. This is shown below.

A D

1 E

C B

3

5

4

2

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

A D

E

G

F 1

C B

3

5

4

2

6

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

A D

E

G

F 1

C B

3

5

4

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 the completion 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.

Example

The owner of a shopping centre is considering modernising and expanding the current 32-business shopping complex. He hopes to add 8 to 10 new business or tenants to the shopping complex. The specific activities that make up the

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

Activity

Activity Description

Immediate Predecessor

Completion Time (weeks)

A Prepare architectural drawings - 5

B Identify potential new tenants - 6

C Develop prospectus for tenants A 4

D Select contractor A 3

E Prepare building permits A 1

F Obtain approval for building permits E 4

G Perform construction D, F 14

H Finalise contracts with tenants B, C 12

I 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 problem specification.

A

H

I G

F E

D

C

B

14

12 3

1 4 4

6

2 5

7 6

5

4

3 2

1

Completion time

for activity H

Starting at the network’s source node (node 1) we will have to compute the earliest start time and the earliest finish time for each activity in the network. Let’s assume that

ES = earliest start time for a particular activity EF = earliest finish time for a particular activity t = expected completion time for the activity

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

EF = ES + t

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

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

A

[0, 5]

5

2

1 Expected completion time

Activity

EF ES

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

Earliest Start Time Rule

The earliest start time for an activity leaving a particular node is equal to 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 the network, which now looks as follows:

7

[5,8]

[6,10]

[10,24]

[5,6]

[0,5]

[24,26]

[5,9]

[9,21]

[0,6]

A

H

I G

E F

D

C

B

14

12 3

4 1

4

6

2 5

7 6

5

4

3 2

1

As has demonstrated, proceeding in a forward pass through the network, we can establish 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 earliest finish time for the last activity. In the case of the shopping centre, the total time required for project completion is 26 weeks.

We now continue the algorithm for finding the critical path by making a backward pass calculation. 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 time for each activity. Let

LS = latest start time for a particular activity LF = 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 will put 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.

[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]

A A

H

I G

E F

D

C

B

14

12 3

1 4 4

6

2 5

7 6

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. The amount 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 of information (the project schedule) for the shopping centre project.

Activity ES LS EF LF Slack Critical Path?

A 0 0 5 5 0 Yes

B 0 6 6 12 6

C 5 8 9 12 3

D 5 7 8 10 2

E 5 5 6 6 0 Yes

F ^{6 6 10 10 0} Yes

G 10 10 24 24 0 Yes

H 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 are completed 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 in order 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 a delay 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 permissible range 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 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;

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 the network to identify the latest start and finish times for each activity;

Step 7 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;

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 answer

questions such as the total project completion time, the critical activities, and the

slack times of the non-critical activities. This, obviously, will give the project

manager a clear picture for his 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 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 activity.

4.1 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. The complication which arises is that as the critical activities are sped up more and more, other activities also become critical. We will discuss the algorithm through the following example.

Example:

A project consisting of 8 activities are 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:

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

2) 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.

Activity

Normal completion time (days)

Shortest completion time (days)

Cost of reduction per day (£)

A (1-2) 6 4 80

B (1-3) 8 4 90

C (1-4) 5 3 30

D (2-4) 3 3 -

E (2-5) 5 3 40

F (3-6) 12 8 200

G (4-6) 8 6 50

H (5-6) 6 6 -

We first 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.

11

A

F

B

C G

H D

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

6 3

6 5

(3,9)

3

6 4 8

1

5

2

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 3

F 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 activities are 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, 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.

The new cost accordingly is now:

A

F

B

C G

H D

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

6 3

6 5

(0,6)

3

6 4 8

1

5 2

£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.

A

F

B

C G

H D

[4,16]

(4,16 ) [0,4]

(0,4)

[8,16]

[0,5]

(3,8) (8,16)

[10,16]

[5,8]

[0,5]

(5,8) 12 4

5 5

3

6 4 8

1

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.

A

F

B

C G

H D

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 4 8

1

5 2

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). Reduction of two days on these activities makes the total projection time to 13 days.

A

F

B

C G

H D

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

4 3

6 3

(0,4)

3

6 4 6

1

5 2

13

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.

Completion time (days)

Cost (£)

20 9000 17 8790 16 8800 15 8920 13 9180

Completion time 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, and that 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

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.

However, if it is the minimum time that is of interest, then there is another and more efficient 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 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.

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)

Possible reduction time

(days)

Extra cost for reduction

(£/day)

A(1-2) 5 6 1 300

B(1-3) 8 10 2 200

C(1-4) 15 17 4 700

D(2-3) 4 5 1 400

E(2-5) 12 15 3 200

F(3-4) 6 8 2 200

G(4-5) 7 9 1 400

H(4-6) 11 13 3 300

I(4-7) 10 12 2 600

J(5-6) 8 14 2 300

K(6-8) 9 25 3 100

L(7-8) 10 13 2 500

Answer:

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

E

D J K

A B

H L

F C

I [5,17]

(10,22) 12

[30,39]

[22,30]

[5,9]

[0,5] 4

(0,5) (5,9) 8 (22,30) 9 (30,39)

[25,35]

[15,26]

[15,22]

[0,8]

(1,9)

(9,15) 7 (19,30)

8 11 (29,39)

(15,22) 6

[0,15]

15 10

(0,15) [15,25]

(19,29) 10

5

[9,15]

G 2

8 6

5

3 1

7 4

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

(2) 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

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

(3) 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.

E

D J K

A B

H L

F C

I [4,13]

(8,17) 9

[23,29]

[17,23]

[4,7]

[0,4] 3

(0,4) (4,7) 6 (17,23) 6 (23,29)

[19,27]

[11,19]

[11,17]

[0,6]

(1,7)

(7,11) 6 (15,23)

6 8 (21,29)

(11,17) 4

[0,11]

11 8

(0,11) [11,19]

(13,21) 8 4

[7,11]

G 2

8 6

5

3 1

7 4

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 follow table lists all the non-critical activities and their costs to speed up.

Non-critical activities

Cost to speed up (£/day)

B 200 E 200 H 300 I 600 L 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.

E

D J K

A B

H L

F C

I [4,13]

(8,17) 9

[23,29]

[17,23]

[4,7]

[0,4] 3

(0,4) (4,7) 6 (17,23) 6 (23,29)

[21,29]

[11,19]

[11,17]

[0,6]

(1,7)

(7,11) 6 (15,23)

6 8 (21,29)

(11,17) 4

[0,11]

11 8

(0,11) [11,21]

(11,21) 10

4

[7,11]

G 2

8 6

5

3 1

7 4

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.

E

D J K

A B

H L

F C

I [4,13]

(8,17) 9

[23,29]

[17,23]

[4,7]

[0,4] 3

(0,4) (4,7) 6 (17,23) 6 (23,29)

[21,29]

[11,22]

[11,17]

[0,6]

(1,7)

(7,11) 6 (12,23)

6 11 (21,29)

(11,17) 4

[0,11]

11 8

(0,11) [11,21]

(11,21) 10

4

[7,11]

G 2

8 6

5

3 1

7 4

17

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

E

D J K

A B

H L

F C

I [4,16]

(5,17) 12

[23,29]

[17,23]

[4,7]

[0,4] 3

(0,4) (4,7) 6 (17,23) 6 (23,29)

[21,29]

[11,22]

[11,17]

[0,6]

(1,7)

(7,11) 6 (12,23)

6 11 (21,29)

(11,17) 4

[0,11]

11 8

(0,11) [11,21]

(11,21) 10

4

[7,11]

G 2

8 6

5

3 1

7 4

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

E

D J K

A B

H L

F C

I [4,16]

(5,17) 12

[23,29]

[17,23]

[4,7]

[0,4] 3

(0,4) (4,7) 6 (17,23) 6 (23,29)

[21,29]

[11,22]

[11,17]

[0,7]

(0,7)

(7,11) 6 (12,23)

7 11 (21,29)

(11,17) 4

[0,11]

11 8

(0,11) [11,21]

(11,21) 10

4

[7,11]

G 2

8 6

5

3 1

7 4

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:

Σ£(All costs) + 29 × £500 + 1 × £300

+ 1 × £200

+ 4 × £700

+ 1 × £400

+ 2 × £200

+ 1 × £400

+ 2 × £300

+ 3 × £100

+ 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 have the same starting and ending nodes.

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

Activity

Immediate predecessor

H B, C

I B, C

G D, E

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

Activities Immediate predecessor A - B - C A D A

E C, B

F C, B

G 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:

Activities Time (months) A 4 B 6 C 2 D 6 E 3 F 3 G 5 a. Find the critical path.

b. The project must be completed in 1½ 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

3 7

6

7

3

10

8 A

B

D

C

E

F

H

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 eight activities. The immediate predecessor and activity times are shown below.

Activity Immediate predecessor Time (weeks)

A - 3

B - 6

C A 2

D B, C 5

E D 4

F E 3

G B, C 9

H 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 on campus. The complex would provide a new gymnasium for intercollegiate basketball games, expanded office space, classrooms, and intramural facilities. The activities that would have to be undertaken before beginning constructing are shown below.

Activity Description

Immediate predecessor

Time (weeks)

A Survey building site - 6

B Develop initial design - 8 C Obtain board approval A, B 12

D Select architect C 4

E Establish budget C 6

F Finalise design D, E 15

G Obtain financing E 12

H 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 initial design plans? What is the expected completion time for the project?

7. Hamilton Country Parks is planning to develop a new park and recreational area on a recently purchased 100-acre tract. Project development activities include cleaning playground and picnic areas, constructing road, constructing a shelter house, 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 7

9

3

6

6

0

3

2 6

3 I H

G F

E D A

C B

Activity time in weeks

a. What is the critical path for this network?

b. Show the activity schedule for this project.

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

3 5

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 1400

B 2 1 1200 1900

C 5 3 2000 2800

D 5 3 1500 2300

E 6 4 1800 2800

F 2 1 600 1000

G 2 1 500 1000

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

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 interoffice communications for a particular company. Contained in the proposal is a list of activities that must be accomplished to complete the new office system project.

Information about the activities is shown below.

Activity Description

Immediate predecessor

Time Normal

(weeks) Crash

Cost Normal

($000’s) Crash

A Plan needs - 10 8 30 70

B Order equipment A 8 6 120 150

C Install equipment B 10 7 100 160

D Setup training lab A 7 6 40 50

E Conduct training D 10 8 50 75

F 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 desired completion time at the least possible cost? Work through the network, and attempt 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?