Construction Engineering

CVE 581- Construction Engineering Lecture Note by Olayinka Okeola University of Ilorin. Ilorin. Nigeria

CVE 581- Construction Engineering Lecture Note by Olayinka Okeola

Preface The course is designed to introduce students to processes involved in project planning, a prerequisite to project execution. The civil engineering students in particular need a basic background in construction engineering knowledge and skills to practically build and flourish upon after graduation through more practical experiences. Construction engineering is planned to expose the students to the civil engineering projects planning, management and execution. This is to be accomplished through application of theoretical basic of project scheduling using CPM and PERT approaches and project appraisal with engineering economics. Good relevant textbooks in which practical exercises and homework were taken from are listed under references and also recommended for further studies. Project scheduling using CPM/PERT and projects appraisal with engineering economics were applied to civil engineering infrastructure. Occupational Health and Safety (OHS) and quality control in construction works were discussed with case studies. Interactive sessions form part of the course delivery to ensure students participation and better understanding. At the end of the course, the students are expected to:         

Know the key members of the building team, their roles and responsibilities. Understand project planning in term of cost, time and scope of work. Know work breakdown structure (WBS) and its application. Understand project scheduling: CPM and PERT approaches. Know the techniques of economic appraisal of projects. Understand ownership and operational costs of construction equipment. Understanding basic consideration in construction equipment selection. Know the importance of safety, duty of care and quality control within the context of engineering construction works. Be well informed on commercial and open softwares for project management with emphasis on the usage of Microsoft Project.

Acknowledgements My esteemed students whose inspirations have brought colour and meaning to this course delivery effort and specifically the following students for their time and interest: Lukman Adebayo and Adedamola Afolabi (for equations editing and diagrams) and Blessing Akapa & Muktar Afodun (for effecting corrections in the equations, texts and diagrams).

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October 2012 January 2013 (Revision)


1. PROJECT MANAGEMENT An Overview of Construction Industry The importance of the construction industry in the national development cannot be overemphasized considering the fact that at least 50% of the investments in various development plans is primarily in construction. It is the next employer of labour after agriculture in underdeveloped countries, about 10% of labour force. In developed countries, activities in the industry especially the building/civil works are used as indices of economic growth and buoyancy or recession. There are two Million businesses in the US operating in the constructing industries. The value of the UK construction industry was £56.3 billion in 1998 representing 9% of the GDP. The output of the industry in Nigeria accounts for over 70% of GDP and therefore is a stimulator of national economy. The construction industry also encapsulates large numbers of specialty trade contractors such as electrical, Plumbing, Heating, and painting. The industry can be divided into 3 major types: (1) Building construction done by general contractors (2) Heavy construction also done by general contractor and specialty contractors (3) construction done by specialty trade contractors such as electrician, plumber, etc. In most cases general contractors usually assume responsibility for the entire construction projects and may subcontract aspect that required special skills or equipment. Construction types can thus be broadly classified as follows: 1. General Building: commercial buildings, offices, stores, educational facilities, government buildings, hospitals, medical facilities, hotels, apartments, housing, etc. 2. Manufacturing: auto assembly, electronic assembly, textile plants, etc. 3. Power: thermal and hydroelectric power plants, waste-to-energy plants, transmission lines, substations, cogeneration plants, etc. 4. Water Supply: dams, reservoirs, transmission pipelines, distribution mains, irrigation canals, desalination and portability treatment plants, pumping stations, etc. 5. Sewerage/Solid Waste: sanitary and storm sewers, treatment plants, pumping plants, incinerators, industrial waste facilities, etc. 6. Industrial Process: pulp and paper mills, steel mills, nonferrous metal refineries, pharmaceutical plants, chemical plants, food and other processing plants, etc. 7. Petroleum: refineries, petrochemical plants, offshore facilities, pipelines, etc

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Owner or Developer: In business term, this is known as the client and may be an individual or group of person, local authority, Government or in general anybody recognize as a legal entity. The client engages the services of a designer which is usually an architect to carry out his wishes. On upon the recommendation of tender by the architect or surveyor can sign a contract with a contractor. Architect: This is the man who in front of law has full right to act as design procedure by interpreting the client requirement into specific design and generally take over the task of seeing that they are carried to a logical conclusion. Right from take off of the project he is responsible to act and supervise on behalf of the client. He can appoint clerk of works for the owner whose main responsibility is to ensure compliance with design and material specification. Structural Engineers: The engineer carries out structural analysis and design of every member of building part depending on scope and complexity. They are usually private consultant commission by the client or Architect for the project. Mechanical and Electrical Engineer (M&E): The engineer is responsible for the design of electrical and mechanical services. The design should be completed with details and specification that are readily interpreted by the contractor. They are also private consultants.

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Building Team Building project as a whole is made up of many fragmented units with different professionals performing different functions because of the diversity and complexity involve in building operation. They share some sort of feedback relationship because the work of each professional group is depended or related to the other; and are classified and explained as follows:


Quality Surveyor (QS): The QS carry out quantities of all item of work along with cost implication. They prepare the bill of quantity (BOQ) used in tendering by interested contractors. They can be engaged by both the client and contractor depending on the scope of work. Contractor: The contractor carries out interpretation and turns the architect design into reality. They can engage the services of subcontractor and specialist to contribute to the successful execution of the project.

PROJECT PLANNING Project planning is fundamentally concern with the cost, time and scope of work. A project has a specific set of objectives and a definite schedule, budget, and set of performance requirements (Park and Tippett, 1999). Project plans tend to emerge gradually. They continuously modified and refined in terms of content, structure, and level of detail. The most useful tool for accomplishing all these tasks is Work Breakdown Structure (WBS). A WBS connects the entire project together and portrays scope graphically. It then allow resources to be assigned, permits estimate of times and costs to be developed. Subsequently provide the basis for the schedule and budget. Scope Management (SM) SM is basically identifying what need to be done. This is the first step in project planning process. The WBS identifies all major elements of work and then subdivides into smaller tasks until each task has become a comfortable size to estimate, execute and monitor. The WBS does not show the sequence in which work is to be done until after the schedule has been developed. Activity (or task) consume resources and define lengths of time and expected cost. Time Management (TM) TM is basically identifying how long it will take to complete a whole job. The next step in project planning after identifying all the activities (i.e. elements of work) is to estimate how long each activity will take. All the activities are then combined to get how long the entire project would take to complete. This take into consideration estimates of when individual activity is scheduled for completion. Some basic terminologies in time management and scheduling are:

Project Cost Estimation Refer students to the following: Olayinka Okeola (2012) Civil Engineering Practice: A monograph on Civil Engineering Practice. Published by Scribd.com. Available online at: http://www.scribd.com/doc/84984097/Civil-Engineering-Practice

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Cost Management (CM) CM aims at identifying how much it costs to get the job done. This goes beyond calculating the overall cost of a project. It also consists of a budget creation and time-scaling of the overall project expenditure. Estimating budget for project cost involves both the direct and indirect costs. A direct cost is an expenditure specifically and directly incurred by the execution of the project. These include labour, materials, training, travel and other miscellaneous costs. An indirect cost is a cost relating to supporting the facilities, general services, and organizational environment within which the project team function. Example includes fringe benefits, facilities general and administrative costs.

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Critical activity: This is activity with no latitude in both start and finish times. That is if a critical activity does not finish by the prescribed time, the project overall time line will be affected. Critical path: This is the longest path in the network and all activities therein are critical activities. Duration: This is the window of time within which an activity is expected to complete. Float (or slack): This is the flexibility or latitude with regard to when a specific activity can be worked on. Forward pass/backward pass: This is the techniques for analyzing the amount of float in the execution of individual activities. Activities with no float are known as critical activities. Parallel Activities: These are activities that can be done during the same period of time. Predecessor Activities: An Activity that must be completed before the Activity under confederation can begin. Series Relationship: Two Activities that must be done in a particular order, i.e. one must be done before the other.


Interactive Class Session 1. Name 3 basic construction types. 2. Name key members of building construction team. 3. Explain project planning within the context of work breakdown structure (WBS), scope, time, and cost management. 4. Enumerate the components of project costs. 5. The construction industry in the Nigeria’s quest for MDGs accomplishments. 6. The construction industry and the need for the recognition of professionalism.

2. PROJECT SCHEDULING Network Techniques

The project scheduling cycle starts with the WBS as shown in Fig. 2.1(Park and Tippett, 1999). Scheduling is a dynamic process as long as the project is progressing. The most common techniques are: Gantt or Bar Chart (Fig. 2.2), Milestone Chart, Line Balance and Networks. Gantt method: This method uses a bar chart to schedule and monitor progress of the project. It was the first used technique and it is still widely use. Gantt method cannot indicate dependencies and interrelationships among tasks which are very crucial to project management.

NETWORK ANALYSIS Program Evaluation and Review Techniques (PERT) and Critical Path Method (CPM) are network Analysis as the techniques are commonly called were independently developed in the late 1950s out of the need of Western Industrial and Military establishments to plan, schedule and control complex project. However the techniques were able to address the shortcomings in Gantt method. Framework for PERT and CPM

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3. Draw the “Network” connecting all the activities. Each activity should have unique event numbers. Dummy arrows are used when required to avoid giving the same numbering to 2 activities. 4. Assign time and/or cost estimates to each activity 5. Then compute the longest path through the network i.e. which is the critical path. 6. Use the Network to plan, schedule, monitor and control the project.

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Fig.2.1: Project scheduling and estimating

There are essential 6 steps which are common to both the techniques. These are: 1. Define the project and all of its significant activities or tasks. The project (made up of several tasks) should have only a single start activity and a single finish activity. 2. Develop the relationships among the activities. Decide which activities must precede and which must follow others.


Figure 2.2: Typical Gantt chart (Adapted from Park and Tippett, 1999) Preparing Activity Network

5 key questions to ask when preparing activity network are: 1. Is this a start activity? 2. Is this a finish activity? 3. What activity precedes this? 4. What activity follows this? 5. What activity is concurrent with this? Drawing the CPM/PERT Network

Each activity in a CPM/PERT Network is represented by an arrow symbol. Each activity is preceded and succeeded by an event which is represented as a circle and numbered. i.e. 2 Definition of term in a network Networks are combination of all project activities and the events. An event is the starting or ending points for one or more activities, and an activity is the work required to precede from one event or point in time to another. Activity uses up resources and consumes time. Logic Emphasis in Network Construction Network construction should be based on logical or technical dependencies among activities for e.g: Before activity “beam reinforcement” can be started, the activity “formwork” must be completed. Also network should be built on the basis of time logic.

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Types of Network Activity-on-node (AON): Nodes represent activities and arrows show precedence relationship. Activity-on-arrow (AOA): Arrows represent activities and nodes are events for points in time.

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Sequence of Activities Network construction requires two inputs i.e. are the events represent the start or the completion of an activity? Event completion is generally preferred.


Example 1:

A simple network considers a list of 4 activities in a building project:

Activity

Description

immediate Predecessors ─ Site clearing Removal of top soil A Setting out B Foundation excavation B, C

A B C D

Immediate predecessors for a particular activity are the activities that, when completed, enable the start of the activity in question. Activities A & B can start work since neither depends upon the completion of prior activities. Activity C cannot be started until B has been completed Activity D cannot be started until both activities A and C have been completed. Scheduling with Activity Time Activity Immediate predecessor A _ B _ C A D A E A F E G D, F H B, C I G, H

Completion Time (week) 5 6 4 It can be observed the total time required 3 to complete all activities is 51 weeks. We 1 can also see that several of the activities 4 can be conducted simultaneously for 14 example A and B. Hence total completion 12 time can thus reduce 2 Total 51

Earliest Start & Earliest Finish Time

The utmost interest in the network is the longest path known as the critical path. By starting at the network’s origin (i.e. nodes) and using a starting time of 0, we can compute an earliest start (ES) and earliest finish (EF) time for each activity in the network. The expression EF = ES + t can be used to find the earliest finish time for a given activity. See Figure 2.3 for numenclatures. For e.g.: For activity A, ES = 0 and t = 5, Thus EF = 0 + 5 = 5 ES Time rule: The earliest start time for an activity leaving a particular node is equal to the largest of the earliest finish times

ES

C(5, 9) 4(8,12) LF LS Fig. 2.3: Activity, Duration, ES, EF, LS, LF Nomenclatures

7

2

3

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Activity

EF


Latest Start and Latest Finish time

We need backward pass calculation to find the critical path. Starting at the completion point (node 7) and using a latest finish time (LF) of 26 for activity I, we can trace back through the network computing a latest start (LS) and latest finish time for each activity. The expression LS = LF - t can be used to calculate latest start time for each activity. For e.g., for activity I, LF = 26 and t = 2. Thus LS for activity I is LS = 26 – 2 =24. Network with LS and LF time

Latest finish time rule The LF time for an activity entering a particular node is equal to the smallest of the latest start times for all activities leaving the node. Slack (or Free or Float) Time Slack is the length of time an activity can be delayed without affecting the completion date for the entire project. For e.g., slack for C = 3 weeks, i.e. activity 3 can be delayed up to 3 weeks. LF – EF = 12 – 9 = 3 LS – ES = 8–5 = 3 LF – ES – 4 = 12 – 5 – 4 = 3 Activity Schedule Example continue Activity A B C D E F G H I

ES 0 0 5 5 5 6 10 9 24

LS 0 6 8 7 5 6 10 12 24

EF 5 6 9 8 6 10 24 21 26

LF 5 12 12 10 6 10 24 24 26

Slack Critical Path 0 Yes 6 – 3 – 2 – 0 Yes 0 Yes 0 Yes 3 – 0 Yes

7

6

I (24, 26) 2

Questions: 1. What is the total time to complete the project?  26 weeks if the individual activities are completed on schedule. 2. What are the schedule starts and completion times for each activity?  ES,EF,LS,LF are as given for each activity 3. What activities are critical and must be completed as schedule in order to keep the project on time?  Critical path activities: A, E, F, G, I. 5. How long can non-critical activities are delayed before they cause a delay in the project’s completion time.  Slack times available for all activities are as given.

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1. Slack or float shows how much allowance each activity has. That is how long it can be delayed without affecting completion date of project 2. Critical path is a sequence of activities from start to finish with zero slack. Critical activities are activities on the critical path 3. Critical path identifies the minimum time to complete project 4. If any activity on the critical path is shortened or extended, the project time will be shortened or extended accordingly. 5. Effort therefore should be put in trying to control activities along critical path so that project can meet due date. 6. Resources should not be wasted on non-critical activity because it will not shorten the project time

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Importance of Float and Critical Path


Activity on node (AON) format for CPM Network t (Critical Path)

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Figure 2.4: Activity on node (AON) format for CPM Network (Adapted from Park and Tippett, 1999)

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Figure 2.4 depicts a simple CPM network drawn in the activity on node (AON) format. The nodes or circles denote activity. The connecting arrows show precedence between activities. There are various possible routes through the network. The longest route defines the predicted total project duration. The activities on this longest route, or critical path, are called critical activities. Any delay in the execution of these critical activities translates into a delay in overall project completion. Other non-critical activities have a certain amount of leeway (or slack) associated with them which allows the project manager some flexibility in scheduling resources. Thus the project manager should pay special attention to critical activities in making tradeoff decisions, as these are the ones that can potentially extend the project duration. However, it should be noted that non-critical activities can potentially become critical if they experience delays greater than their original slack, so the project manager cannot become complacent


Interactive Class Session Que 1. An existing access culvert to COMSIT building on campus is to be demolished. A new 10m long culvert is to be constructed in its place using 800mm diameter precast ring culvert. a. Prepare a work breakdown structure (WBS) that follow sequence of implementation. b. Draw a Gantt chart of implementation. Que 2. Ken Johnston, the data processing manager for Stanley Morgan Banks is planning a project to install a new management information system. He is now ready to start the project, and wishes to finish in 20 weeks. After indentifying the 14 separates’ activities needed to carry out this project, as well as their precedence relationships and estimated duration (in weeks). Ken has constructed the following project network.

3 4

H 5

D

6

I

6 A 0

3

START

B

M

E 7 F

C

4

4

G

J

0

4

FINISH

K 3

6

N 5

L

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1. Find all the paths and path lengths through the project network. Which of these paths is a critical path? 2. Find the earliest times, latest times and slack for each activities. Will Ken be able to meet his deadline if no delays occur? 3. Use the information from part (2) to determine which of the paths a critical path is. What does this tell Ken about? Which activities he should focus most of his attention on for staying on schedule? 4. Use the information from path (2) to determine what the duration of the project would be if the only delay is that: i. Activity I take 2 extra weeks ii. What if only delay is that activity H takes 2 extra weeks? iii. What if the only delay is that activity J takes 2 extra weeks?

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5


Solution: (1) The paths and path’s length through Stanley Morgan Bank’s project network

PATH START A START A START B START C START C

LENGTH 6+4+3+6 = 19weeks 6+5+6 = 17weeks 3+7+4+6 = 20weeks 4+4+3+5 = 16weeks 4+6+5+5 = 20weeks

D H M FINISH I M FINISH E J M FINISH F K N FINISH G L N FINISH

Since the critical path is the paths with the longest path lengths thus, paths (iii) and (v) from the above table are the critical paths. i.e. (iii) START

B

E

J

M

(v) START

C

G

L

N

FINISH FINISH

Critical path with path lengths of 20 weeks each

(2) Earliest times, Latest time and slacks for each activity EARLIEST TIMES; ES and EF in weeks For activity A: ES = 0 EF = 6 Activity B; ES = 0 EF = 3 Activity C; ES = 0 EF = 4 Activity D; ES = 6 EF = 10 Activity E; ES = 3 EF = 10 Activity F; ES = 4 EF = 8 Activity G; ES = 4 EF = 10 Activity H; ES = 10 EF = 13 Activity I; ES = 6 EF = 11 Activity J; ES = 10 EF = 14 Activity K; ES = 8 EF = 11 Activity L; ES = 10 EF = 15 Activity M; ES = 14 EF = 20 Activity N; ES = 15 EF = 20

NOTE ES = Earliest start time and EF = Earliest finish time. EF = ES + duration for each activity. (Note that the earliest start time for an activity leaving a particular node is equal to the largest of the earliest finish time)

LATEST TIMES; LS and LF in weeks LF = 7 LF = 3 LF = 4 LF = 11 LF = 10 LF = 12 LF = 10 LF = 14 LF = 14 LF = 14 LF = 15 LF = 15 LF = 20 LF = 20

NOTE LS = Latest start time and LF = Latest finish time. LF = LS – duration for each activity (Note 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)

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LS = 1 LS = 0 LS = 0 LS = 7 LS = 3 LS = 8 LS = 4 LS = 11 LS = 9 LS = 10 LS = 12 LS = 10 LS = 14 LS = 15

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For Activity A: Activity B; Activity C; Activity D; Activity E; Activity F; Activity G; Activity H; Activity I; Activity J; Activity K; Activity L; Activity M; Activity N;


SLACKS AND CRITICAL PATHS This is to be noted that the activity with zero slack is on the critical path. Slacks = LF – EF [i.e. Latest finish minus Earliest finish] in weeks For Activity A; Slack= 7 – 1 = 1 Activity B; Slack = 3 – 3 = 0 Activity C; Slack = 4 – 4 = 0 Activity D; Slack = 11 – 10 = 1 Activity E; Slack = 10 – 10 = 0 Activity F; Slack = 12 – 8 = 4 Activity G; Slack = 10 – 10 = 0 Activity H; Slack = 14 – 13 = 1 Activity I; Slack = 14 – 11 = 3 Activity J; Slack = 14 – 14 = 0 Activity K; Slack = 15 – 11 = 4 Activity L; Slack = 15 – 15 = 0 Activity M; Slack = 20 – 20 = 0 Activity N; Slack = 20 – 20 = 0

ACTIVITY

A B C D E F G H I J K L M N

SLACKS

1 0 0 1 0 4 0 1 3 0 4 0 0 0

CTRITICAL PATHS

No Yes Yes No Yes No Yes No No Yes No Yes Yes Yes

Comment: Ken will be able to meet his deadline of 20 weeks if no delays occur to any of the activities with the zero slack i.e. activities that are on the critical path. (3) Critical paths from the answer in part (b) are the activities with the zero slack which are; (i) START (ii) START

B C

E G

J L

M N

FINISH FINISH

Comment: For staying on schedule; Ken must focus most on each of the critical activity in the critical paths as shown above. This means that activities B , C, E, G, J, L, M and N must be handled with no delay in order to stay on schedule. (4) Using the information in part (b) to determine the project duration with the following cases: i. Activity : If activity should takes 2 extra weeks, it would not delayed the project duration of completion since activity

has a slack of 3 weeks which means that the activity can be delayed for 3 weeks without

affecting the project completion time. ii. Activity : If activity should take 2 extra weeks, it would delay the project completion time because activity has a slack of 1 week thereby making the duration 21 weeks. iii. Activity : If activity takes 2 extra weeks, the project would be delayed by 2 weeks since j is a critical path

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(slack = 0) thereby making the duration 22 weeks.


Stanley Morgan’s Bank Project Network below. Note: Activity-on-Arrow Network (AON)

4 H (10, 13)

D (6, 10) 4 (7, 11)

START

9

5(9, 14)

M (14, 20)

J (10, 14)

6 (1, 7) B (0, 3) 3(0, 3)

0

4 (0, 4)

I (6, 11)

1

A (0, 6)

3 (11, 14)

2

4 (10, 14)

E (3, 10)

6 (14, 20)

5

7(3, 10)

10 N (15, 20)

C (0, 4) F (4, 8) 4(8, 12)

3

6

K (8, 11) 3(12, 15)

8

FINISH (20, 20)

5 (15, 20)

L (10, 15) G (4, 10) 6 (4, 10)

5 (10, 15) 7

THE PERT (PROBABILISTIC) APPROACH The PERT is used in handling uncertainty. The PERT approach is useful because it can accommodate the variation in event completion times based on an expert’s or expert committee’s estimates. For each activity, 3 time estimates are used: The most optimistic (to) The most likely (tm) The most pessimistic (tp) The duration of an activity is calculated using the following formula: PERT assumes that the form of the probability distribution is a beta distribution in order to calculate the mean (µ), 2 also known as expected time te and variance (δ ) of probability distribution.

µ

t o  4t m  t p 6

t t     p o   6 

2

2

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Although the t’s are randomly distributed, the average or expected project length Te approximately follows a Normal Distribution. Since a lot of information about Normal Distribution is known, we can make several statistically significant conclusions from the calculations.

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Probability Computation:


The probability that a project is completed with specified time is



X  

  t e  project mean time,

δ  project standard mean time, X  (proposed) specified time

Example

to

tm

tp

µ

δ

δ2

10 20 4 2 8 8 4 2 6 2

22 20 10 14 8 14 4 12 16 8

22 20 16 32 20 20 4 16 38 14

20 20 10 15 10 14 4 11 18 8

2 0 2 5 2 2 0 2.32 5.33 2

4 0 4 25 4 4 0 5.4 28.4 4

d (15, 25) a (20,4)

c

(8,4)

(10,4)

Critical path = 43

f

b (20, 0)

j

e

(14,4) g (4, 0)

(10,4)

i (18, 28.4)

h (11, 5.4) Figure 5: The complete network

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A B C D E F G H I J

Immediate Predecessor A B, C B, C B, C C G,H D, E

Page

Activity


Critical Path Analysis (PERT Method) Activity LS ES Slack Critical a 0 0 0 Yes b 1 0 1 c 4 0 4 d 20 20 0 Yes e 25 20 5 f 29 20 9 g 21 20 1 h 14 10 4 I 25 24 1 J 35 35 0 Yes Question Refer to Fig. 5, Assume a project manager promised to complete the project in 50 days. What are the chances of meeting that deadline? Answer Calculate Z Known Parameter: µ = scheduled date = 20 +15 + 8 = 43

  2 

4  25  4 

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X  50 X   50  43 Z    5.745 The probability value of Z = 1.22 is 0.888  The project manager can meet the deadline Question What deadline is 95% sure of meeting? Answer: Z value associated with 0.95% is 1.645 X  From Z 



X  Z    5.745(1.645)  43  52.45 days  There is 95% chance of finishing the project in 52.45 days

Used where times can be estimated with certainty Minimizing cost is more important

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Example: Construction project

Require 3 estimates of activity time Calculates mean & variance of time Used where time cannot be estimated with certainty Meeting time target or estimating percentage completion is more important Example: Research and Development

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3 4

PERT Same as CPM

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Table 2.1: Comparism between CPM & PERT CPM 1. Uses network, calculate slack, identify critical path & activities, guides to monitor and controlling of project 2. Use one value of activity time


Interactive Class Session 1. Give basic comparison between CPM (critical path method) and PERT (program evaluation and review technique). 2. Work Example: Bill Fredlund, president of Lincoln Log Construction, is considering placing a bid on a building project. Bill has determined that five tasks would need to be performed to carry out the project. Using the PERT three-estimate approach, Bill has obtained the estimates in the next table for how long these tasks will take. Also shown are the precedence relationships for these tasks. There is a penalty of $500,000 if the project is not completed in 11 weeks. Therefore, Bill is very interested in how likely it is that his company could finish the project in time. (a) Construct the project network for this project. (b) Find the estimate of the mean and variance of the duration of each activity. (c) Find the mean critical path. (d) Find the approximate probability of completing the project within 11 weeks. (e) Bill has concluded that the bid he would need to make to have a realistic chance of winning the contract would earn Lincoln Log Construction a profit of about $250,000 if the project is completed within 11 weeks. However, because of the penalty for missing this deadline, his company would lose about $250,000 if the project takes more than 11 weeks. Therefore, he wants to place the bid only if he has at least a 50 percent chance of meeting the deadline. How would you advise him? TIME REQUIRED (WEEKS)

A

OPTIMISTIC ESTIMATE

MOST LIKELY ESTIMATE

PESSIMISTIC ESTIMATE

3

4

5

4

1/9

1/3

B

A

2

2

2

2

0

0

C

B

3

5

6

4.83

1/4

1/2

D

A

1

3

5

3

4/9

2/3

E

B,D

2

3

5

3.17 1/4

1/2

17

(a)

The project network for the above project is as shown next page

1.056 2

16

IMMEDIATE PREDECESSOR

Page

TASK


(4,1/9)

(2,0)

(4.83,1/2)

START

FINISH (3.17, 1/4)

(3,4/9)

(b)

To obtain the estimate of the mean and variance

For activity A;



3  44  5 4 6

mean  

t o  4t m  t p

 t p  to     6 

and variance  2  

6

2

2

For activity E;

 for activity B;

2

 22 2   0  6 

2

1 52 2    4  6  (c) Obtaining the mean critical path of the project:

Paths of the project

for activity C;



3  45  6  4.83 6

 63 2  

2

1  4  6 

For activity D;



1  43  5 3 6 2

4  5 1  2    9  6 

Mean path length

START

A

B

C

FINISH

START

A

B

E

FINISH

START

A

D

E

FINISH

Activity path: Start path.

A

B

C

4 + 2 + 4.8 = 10.8

Finish is the activity with the mean critical

(d) Appropriate probability of completing the project within 11 weeks will be; x Z ; x  11 weeks 

17

2  42  2  2 6

2  43  5  3.17 6

Page

1  53 2    9  6 


For the mean critical path;

  4  2  4.83  10.83 2 

1 1 13 0  9 4 36 13  0.6 36

  2  Therefore

Z

11  10.83  0.2833 0 .6

Checking from the probability tables; the probability of completing the project = 0.61153 = 61% This means that the approximate probability of the project completion within 11 weeks will be 61%. (e)

For the duration of meeting a 50% chance of the deadline of 11 weeks for the project;

For at least a 50% chance; Probability = 1 – Pr (50%)

P50 = 1 – (0.5) P50 = 0.5

Checking from the table, The value of Z corresponding to a probability of 0.5 is 0 That is Z = 0 Substituting into the equation;

Z

x 

where   10.83 and   0.6; 0

x  10.83 0 .6

x  10.83 weeks.

Page

18

This implies that Bill has a 50% chance at least of meeting the bid deadline in 10.83 weeks which is less than the project deadline of 11 weeks. So, Bill can go ahead and make his bid and have a realistic chance of winning the contract and earn Lincoln Log Construction a profit of about $250,000.


3. OCCUPATIONAL HEALTH AND SAFETY (OHS) IN THE CONSTRUCTION INDUSTRY Occupational Health and Safety (OHS) in construction is all about protecting people from injury at work or from becoming ill through appropriate precautions. A safe, healthy and productive workforce is needed to achieve the objective of national development and economic growth since construction industry is a stimulator for national economy. However, each sector of the industry has its distinct hazards and risks determined by the peculiarities of its labour process. We are going to examine: 1. the concept of hazard and risk as cornerstone of effective OHS management. 2. the laws regulating OHS matters. 3. the state of OHS practices on the University of Ilorin construction sites and typical active site outside the campus. The aim is to investigate current safety practices on the site; identify the nature and occurrence of hazards; and the extent to which safety measures are employed.

Required material: 1. Okeola, O.G. (2009) Occupational Health and Safety Assessment in the Construction Field. Proceedings of 1st Annual Civil Engineering Conference, (Aug. 26th – 28th 2009). Pp 32 - 40. Department of Civil Engineering, University of Ilorin, Nigeria Available online at: http://www.scribd.com/doc/76525917/Occupational-Health-and-Safety-OHSAssessment-in-the-Construction-Industry-By-Okeola-O-G 2. Safety, health and welfare on construction sites: A training manual. Geneva, International Labor Office, 1995 /Trainers manual/, /Occupational safety/, /Occupational health/, /Construction industry/. 13.04.2 ISBN 92-2109182-1

Attention:

Page

19

Students are urged to participate actively in all the interactive component of this topic both in the class and on construction sites proposed for visitation


HOMEWORK Que. 1: Sharon Lowe, vice president for marketing for the Electronic Toys Company, is about to begin a project to design an advertising campaign for a new line of toys. She wants the project completed within 57 days in time to launch the advertising campaign at the beginning of the Christmas season. Sharon has identified the six activities (labeled A, B, . . . , F) needed to execute this project. Considering the order in which these activities need to occur, she also has constructed the following project network. Using the PERT three-estimate approach, Sharon has obtained the following estimates of the duration of each activity. a. find the estimate of the mean and variance of the duration of each activity. b. Find the mean critical path. c. Use the mean critical path to find the approximate probability that the advertising campaign will be ready to launch within 57 days. d. Now consider the other path through the project network. Find the approximate probability that this path will be completed within 57 days. e. Since these paths do not overlap, a better estimate of the probability that the project will finish within 57 days can be obtained as follows. The project will finish within 57 days if both paths are completed within 57 days. Therefore, the approximate probability that the project will finish within 57 days is the product of the probabilities found in parts (c) and (d). Perform this calculation. What does this answer say about the accuracy of the standard procedure used in part (c)? Optimistic Estimate 12 days 15 days 12 days 18 days 12 days 2 days

Most Likely Estimate 12 days 21 days 15 days 27 days 18 days 5 days

A

C

Pessimistic Estimate 12 days 39 days 18 days 36 days 24 days 14 days

E

F

FINISH

B

D

20

START

Page

Activity A B C D E F


Que. 2 Alfred Lowenstein is the president of the research division for Better Health, Inc., a major pharmaceutical company. His most important project coming up is the development of a new drug to combat AIDS. He has identified 10 groups in his division which will need to carry out different phases of this research and development project. Referring to the work to be done by the respective groups as activities A, B, …, J, the precedence relationships for when these groups need to do their work are shown in the following project network. To beat the competition, Better Health’s CEO has informed Alfred that he wants the drug ready within 22 months if possible. Alfred knows very well that there is considerable uncertainty about how long each group will need to do its work. Using the PERT three-estimate approach, the manager of each group has provided a most likely estimate, an optimistic estimate, and a pessimistic estimate of the duration of that group’s activity. Using PERT formulas, these estimates now have been converted into estimates of the mean and variance of the probability distribution of the duration of each group’s activity, as given in the following table (after rounding to the nearest integer). 1. Find the mean critical path for this project. 2 Use this mean critical path to find the approximate probability that the project will be completed within 22 months. 3. Now consider the other three paths through this project network. For each of these paths, find the approximate probability that the path will be completed within 22 months. 4. What should Alfred tell his CEO about the likelihood that the drug will be ready within 22 months?

Activity A B C D E F G H I J

Duration Estimated Mean Estimated Variance 4 months 5 months 6 months 10 months 4 months 8 months 3 months 6 months 8 months 12 months 4 months 6 months 3 months 5 months 7 months 14 months 5 months 8 months 5 months 7 months

A

E I C

F

D

G J

B

H

21

FINISH

Page

START


Interactive Class Session (A)

The Use of Computer Software in project management using PERT/CPM approaches i. General introduction ii. Using both MS Project and other Excel template to handle the COMSIT culvert project on page 5.

(B)

Uncertain activity duration i. Dealing with uncertainty activity duration. ii. Time-cost tradeoff for individual activities Crashing Crashing an activity Crashing the project

(C)

An overview of CPM/PERT (1)

How is the uncertainty in activity duration handled by the PERT? i. Using 3 different types of activity duration namely most likely, optimistic and pessimistic estimates. ii. PERT assumes that the form of the probability distribution is a beta distribution which is used to calculate mean and variance.

(2)

What are the major deficiencies/shortcomings for some applications? These include: i. Approximations of means and variance of activity durations ii Approximation of the probability of meeting the deadline iii Dealing with overlapping activities iv. Incorporating the allocation of resources to activities.

Required material: Chapter 22: Project Management with PERT/CPM www.mhhe.com/hillier Introduction to Operation Research. 9th Edition F. S. Hillier G. J. Lieberman McGraw-Hill International Edition

Page

22

Attention: Lecture delivery will largely be interactive, hence Students are urged to participate actively and ask questions or make contributions.


4. ENGINEERING ECONOMICS Techniques for the economics appraisal of project

Engineering project planning involves making choices among physically feasible alternatives; generally these choices are made on economic grounds by placing a monetary value on each alternative course of action (or project) The concept of interest

Because of interest money is worth more today than sometime in the future. Interest can be defined as the money that must be paid for the use of borrowed money. For public sector projects the government normally set the minimum acceptable rate. Cash flow series

Because of interest money has different values through time. Cash flow diagram provide a rationale, simple way of showing how money flows through time. A normal cash flow transaction consists of series of disbursement or receipts. The present worth of any stream of payments can be found by calculating the present value of each individual payment and summing the results. Once the present worth is found, one can make other equivalence calculations, such as calculating the future worth by using the interest factors. Cash flow transaction can be grouped into 3 categories: (1) equal cash flow series, (2) linear gradient series, and (3) geometric gradient series. Figure 4.1: Three major categories of cash flow Uniform series: This is transaction arrangement of equal cash flows at regular intervals known as an equalpayment (fig. 6a). Linear gradient series: The transaction arrangement in which the cash flow in a series increases (or decreases) by a fixed amount (for 6b). Geometric gradient series: This is another type gradient series in which the cash flow is determined, not by fixed amount like N500, but by some fixed rate, expressed as a percentage (fig. 6c). DERIVATION AND USE OF BASIC INTEREST FORMULA.

If one were to deposit NP in a bank today, and earn compound interest for n time period, how much would be available to receive at the end of period n?

Page

Present worth and future worth Compound interest is interest computed on both the principal and interest outstanding for a given time period. Define as follow: i = interest rate for a given compound interest time period, expressed as a decimal; n = number of time periods p = present amount of money; f = future amount of money at the end of period n.

23

Compound interest: Single payment


Calculation as follow: End of period

F = p (1+i)n (4.1) (1+i)n = single-payment compound amount factor = (F/p, i, n) (4.2) OR F = P(F/P, i, n)

Amount Available

0 1 2 3

P P(1+i) P(1+i) (1+i) 2 P(1+i) (1+i)

. . .

. . .

n

P(1+i)

n-1

=p = P(1+i) 2 = p(1+i) 3 = p(1+i)

(1+i)

= p(1+i)

Equation (1) can be used to solve for P in terms of F:

P = F (1+i)-n (4.3) -n (1+i) = single-payment present worth factor = (P/F, i, n) (4.4) OR P = F(P/F, i, n)

n

Example: Compute the present amount (P) that would be equivalent to the two receipts (in Naira) shown in the cash flow diagram. The compound interest rate for a period is 5%.

2000

P

1

2

3

1000

4

5

6

7

8

P  2000 P F , 5%, 3  1000 P F , 5%, 8

Solution

Ref. Eqtn. (4.3) Ref. Eqtn. (4.3)  3  2000 1  0.05  1000 1  0.058

 2000 0.863838  1000 0.676839  1727 .68  676.84  N 2404.52 Present Value of Arithmetic Gradient Series

(n-1)G

These are annual series with constant increasing values such that A1  0 2G

A3  A2  G  2G

G



1

2

2

3

P=? 1 An  An 1  G  n  1G The cash flow diagram is as shown and G is called the gradient

P

0

1

G 2

4

G 1  i n  in  1 i 2 1  i n

P  G P G , i , n 

A2

A1

... n

The present value is

A4 A3

3

3

4

24



3G

Page

A2  A1  G  G

…(4.5)


STANDARD CASH FLOW SERIES Uniform series

The uniform series present worth factor (pwf); i.e. (P/A, i, n) finds the present worth of a uniform series of n receipts or payments of amount A each period while the uniform series capital recovery factor, (crf); i.e. (A/P,i, n) permit calculation of the uniform series amount, A, that will recover an initial capital investment, P, in n periods. The uniform series compound amount factor (caf); i.e. (F/A, i, n), is used to find the future worth of a uniform series, A, while the uniform series sinking factor (sff), (A/F, i, n), permits calculation of the uniform series amount, A, that must be deposited to finance a future capital investment, F, in period n. Uniform series are common in Civil and Environmental engineering and few examples include: 1. Operation and maintenance costs 2. Estimated benefit stream from a project 3. Annualized capital costs

Uniform series Equivalence Formulas

1. Present worth factor

2. Capital Recovery Factor A=?

A n

0 P?

P

 i 1  i n  A  P  ……….(4.7)  1  i n  1

 1  i n  1 P  A  ……….(4.6)  i 1  i n  P= A( P/A, i, n)

A = P (A/P, i, n )

3. Compound Amount Factor

4. Sinking Fund Factor A=?

A 0

n

0

n

F=?

 1  i   1  F  A  i   ……….(4.8)

F=?

F = A (F/A, i, n)

A = F ( A/F, i, n)

Page

  i A  F  ……….(4.9)  1  i n  1

25

n


Capitalized Cost Major infrastructural facilities are expected to keep on delivering their service in perpetuity. Examples include highways, dam, auditorium etc. Such facilities require periodical maintenance, renovation or replacement to maintain an acceptable level of service delivery. Capitalized cost of such project is the amount of money required to set aside now to provide for service delivery in perpetuity. Let A= amount to be spent to maintain perpetual service. The initial principal must therefore be large enough such that A = Pi……………… (4.10) =P ( A/P, I, ∞) Rearranging (4.10) : P = A/i =A (P/A, I, ∞) Where: (A/P, I, ∞) is the uniform series capital recovery factor for an inﬁnite series, equal to the interest rate, i and (P/A, I, ∞) is the uniform series present worth factor for an infinite series, which is = 1/i. In the case of construction project, there is first an initial cost of construction and secondly a periodic maintenance cost that doesn’t incurred annually, rather after certain number of years. The initial cost is easily handled, however the periodic maintenance (cost) would have to be resolved into its equivalent annual series before it can be capitalized. The geometric gradient series

The geometric gradient series is useful in cases of “double compounding”. For example when benefits from a project grow at a compound rate and are themselves subject to compound interest (e.g. hydroelectric generation benefits may grow at a compound rate equal to that of a region’s economy, while the benefit itself is compounded financially. Example: A project has a net income of $50 the first year, increasing by $100 every year for the next four years. What is the net present worth of this project at an interest rate of 10%? Answer: The annual income is determined according to an arithmetic series above. The following cash flow diagram presents the model for this example.

150

50 1

0

2

250

3

350

4

100 0

50

+ 4

0

Page

300 200

26

The cash flow can be broken into two cash flows as shown


From Eqtn. 4.5

From Eqtn. 4.6

P  PV  G P G , i, n   A  , i, n  A  where : G  100, A  50, and n  4 PV  100 P G , 10%, 4   50P A , 10%, 4   100 4.378   503.170   $596 .3

Interactive Class Session QUE 1: A city wants to set aside enough money to build, operate, and renovate a sewage treatment plant in perpetuity. An engineering company estimates that the plant will cost an immediate $20 million to build and will require $5 million every 20 years to replace major equipment and $10 million every 50 years to pay for major structural renovation. It is estimated that operation and maintenance costs will be $1.5 million every year. What amount will the city need to set aside? Interest earned on the annuity is 7%. Answer: The cash flow diagram is as shown below

20

50 60

$1.5m

$10m 80

P =? $20m

$20m

$5m $5m $5m $1.5m $1.5m $1.5m

$10m

∞ P=?

∞

100

+

$10m $10m $10m ∞

27

$10m $1.5m

Page

$5m


Spreading out each $5m and $10m to the left over the preceding 20years and 50years respectively, we have an annual infinite series as shown below:

$5m

$10m

A1

n=20 A1 A1 A1

A1 A1 A1 A1

n=50

0 A2 A2 A2

A2 A2 A2 A2

Therefore;

A  A 1  F  , i, n  F  Where

A   , i, n  is the sinking fund factor F 

So that:

A  A1  $5,000,000 ,7%,20  F   5,000,000  5,000,000 

i

1  i n  1 0.07

1  0.07 20  1

 5,000,000  0.02439

A1  $ 121,964.63 A  A 2  $10,000,000  , 7%, 50  F  

1  i n  1 0.07

1  0.0750  1

 10,000,000 2.46 10 3 A 2  $24,598.50

28

 10,000,000 

i

Page

 10,000,000 


 A Present Worth    i  So that P1  $121,964  P2  $24,598.5 

1  $1,742,342.86 0.07

1  $351,407.14 0.07

For $1.5million operation and maintenance, series present worth factor is applied so that:

P  P3  A  , i%, n  A  P  Where  , i%, n  = series present worth factor A  Therefore,

P3  A 

1  i n  1 i 1  i n

 $1,500,000 

1  0.07 50  1 0.07 1  0.07 50

 $20,701,119 .44

Therefore, the city needs to set aside $42,794,869.44 for the project.

Total cost = P0  P1  P2  P3 where P0  $20,000,000

 20,000,000  1,742,342.86  351,407.14  20,701,119.44  $42,794,869.44

QUE 2: A flood control project has a construction cost during the first year of $10 million, during the second year $6 million, and during the third year $2 million. It is completed at the end of the third year and thereafter incurs an annual operating cost of $200000 per year. Benefits from the project also begin during the fourth year and are valued at $1.5 million in that year, growing at a 2% compound rate of increase out to the planning horizon of 50 years. The interest rate is taken to be 6%. Carefully draw the cash flow diagram. What is the present worth of cost? What is the present worth of benefit? Is this a viable project economically?

PWC

n = 50 $200,000

$10m

6 $200,000

$6m

5

$200,000

$2m

4

Page

3

$200,000

2

$200,000

1

$200,000

0

29

$1.5m $1.5m $1.5m


Present worth of cost (PWC)

P  P  P   P   P  PWC  $10m  , 6%, 1  $6m  , 6%, 2  $2m  , 6%, 3    $200,000  , 6%, 47   , 6%, 3  F  F  F   A   F 

P=?

n=1

i = 6%

Present worth P = A (P/A, i, n)

Eqn. 4.3

$10m

So that. PWC  10,000,000 

 200,000 

1

1  0.06 1

 6,000,000 

1

1  0.062

 2,000,000 

1

1  0.063

1  0.0647  1  1 47 0.061  0.06  1  0.063  1  i n  1 P  A  n  i 1  i  

 $9,433,962.3  $5,339,978.64  $1,679, 238.57  $2,617,769.74  $19,070,949 .21

Present worth of Benefit

P  P  PWB  $1.5m  , 2%, 47   , 2%, 3  A  F 

 1  0.0247  1  1 PWB  1,500,000    47 3  0.021  0.02  1  0.02 Eqt. 4.3

 1  i n  1 PA   eqt. 4.6 n  i 1  i   PWB  $42,809,58 3.93 PWB - PWC  $42,809,58 3.93 - $19,070,94 9.21 PWNB  $23,738,63 4.72

shown.

Page

is greater than PWC by an appreciable amount as

30

The project is viable economically because PWB


DEPRECIATION Introduction In practice, economics decisions are complicated by such considerations as taxes, possible tax deductions (one of which is depreciation of an asset) and inflation. Depreciation represents the decline in market value of properly (for example equipment) due to age, wear, deterioration and obsolescence. Depreciation can result from: 1. Physical deterioration occurring from wear and tear of the properly (i.e. machine) 2. Economic decline or obsolescence occurring over the passage of time. Depreciation is defined as the allocation of the cost of an asset over its useful or depreciable life for tax purposes minus salvage value. Thus

D total = P – S Where :

D total = total depreciation allowable over time, P = initial asset value or cost S = salvage value

There are three methods of calculating depreciation of equipment. The book value of an asset in year n is its initial cost of value minus any depreciation charges taken up to that time. Straight-line depreciation

The annual depreciation charge is assumed a constant amount chosen to produce a book value of S at the end of year N:

SLD t 

1 P  S  N

SLDt = straight line depreciation taken at the end of year t and N = useful lifetime. Example A wheelbarrow is purchased for N10.000 and is to be depreciated over five years. Salvage value at the end of year 5 is N1000. Using straight line depreciation, what is the annual depreciation charge and schedule of book value?

SLT 

Solution :

1 10,000  1000  N1800 5

Book Value (N)

Depreciation Amount (N)

1 2 3 4 5

10,000 8,200 6,400 4,600 2,800 1,000 (salvage value)

1800 1800 1800 1800 1800

6

Page

Year

31

Annual Depreciation is N 1800 Which gives a schedule for book value as follows :


Sum of the years digits depreciation

This method uses larger depreciation amounts early in time and thus allows larger tax deduction early resulting in early savings. The method uses the sum of the years digits for the planning duration period N :

Total sum of the years digits  1  2  3    ( N  1)  N 

N ( N  1) 2

And depreciation for the end of year t is calculated thus: useful life remaining at beginning of year t P  S  SOYDD t  total sum of the year ' s digits

N  t 1 P  S  ( N / 2)( N  1) Where SOYDDt = sum-of-the-years’- digits depreciation at the end of year t. 

Example Compute the SOYDDt depreciation for the same example P = N 10,000, Solution:

S = N1000,

N=5

SOYDD1  1  2  3  4  5 N 5 ( N  1)  (5  1)  15 2 2



SOYDD 2  

5 11 10,000  1000  3000 15 

SOYDD 5 





5 1  1 10,000  1000  600 15 Sum = N 9000

Total depreciation is N 9000, leaving only salvage value at the end of year N = 5. Note the higher SOYD depreciation values in year1 and 2 as compared with straight line depreciation (N 1800 every year) Declining balance depreciation

Declining balance (DB) depreciation is based on taking a constant current book value as the depreciation amount every year. Double declining balance depreciation (DDBD) uses the fraction 2/N

Worked Example A bulldozer with initial cost of $20000 is to be depreciated over 8 years using straight–line depreciation. Salvage value is estimated at $2000. What is the annual depreciation charge and schedule of book value? Also compute the depreciation and schedule of book value using (1) SOYD (2) DDB 1 P - S N Where: N = 8years, SLDt 

N = useful lifetime; P = initial lifetime P = $20,000

and

S = $2,000

32

2 2 2 (book value)  ( BVt )  ( P  total depreciation taken to date) N N N

Page

DDBDt 


SLDt 

1 20,000  2,000  $2,250 8 Year 1 2 3 4 5 6. 7 8 ≥

Book value

Depreciation Amount

$20,000 $17,750 $15,500 $13,250 $11,000 $8,750 $6,500 $4,250 $4,250 (Salvage value)

$2,250 $2,250 $2,250 $2,250 $2,250 $2,250 $2,250 $2,250 $2,250

1. Sum of Years’ Digit Method (SOYD) P = $20,000,

S = $2,000, N = 8

SOYD 

N N  1 2

8 SOYD  8  1  36 2

Where: N = useful lifetime;

P = initial lifetime

 N - t   1  P - S SOYDDt    N N  1   2  (8 - 1)  1 20,000 - 2000 8 8  1 2 8  18,000  $4,000 36

SOYDD 2  

7 18,000  $3,500 36 (8 - 3)  1 20,000 - 2000  8 8  1 2

SOYDD 3  

6  18,000  $3,000 36

SOYDD 4  

(8 - 2)  1 20,000 - 2000 8 8  1 2

(8 - 4)  1 20,000 - 2000 8 8  1 2

5 18,000  $2,500 36

Schedule of year book Year

Book value ($)

SOYDDt ($)

1 2 3 4 5 6 7 8

20,000 16,000 12,500 9500 7000 5,000 3500 2,000 (Salvage Value)

4,000 3,500 3,000 2,500 2,000 1,500 1,000 500

33



Page

SOYDD1 


(8 - 5)  1 20,000 - 2000 8 8  1 2

SOYDD 5  

4  18,000  $2,000 36

SOYDD 6  

3 18,000  $1,500 36

SOYDD 7  

SOYDD 8 

(8 - 6)  1 20,000 - 2000  8 8  1 2

(8 - 7)  1 20,000 - 2000 8 8  1 2

2 18,000  $1,000 36 (8 - 8)  1 20,000 - 2000  8 8  1 2

1  18,000  $500 36 SUM = $4,000 + $3,500 + $3,000 + $2,500 + $2,000 + $1,500 + $1,000 + $500 = $18,000 

2. Double Declining Balance Depreciation (Ddbt) 2 P - depreciation to date   2 BVt N N 2 DDBD1  20000 - 0   $5,000 8 2 DDBD 2  20000 - 5000   $3,750 8 2 DDBD 3  15000 - 3750   $2,812.5 8 2 DDBD 4  11250 - 2812.5  $2,109.38 8 2 DDBD 5  8437.5 - 2109.38  $1,582.03 8 2 DDBD 6  6328.12 - 1582.03  $1,186.52 8 2 DDBD 7  4746.09 - 1186.52  $889.89 8 2 DDBD 8  3559.57 - 889.89  $667.42 8 DDBDt 

Schedule of year book DDBDt

1 2 3 4 5 6 7 8 ≥8

$20,000 $5,000 $15,000 $3,750 $11,250 $2,812.50 $8,437.50 $2,109.38 $6,328.12 $1582.03 $4,746.09 $1,186.52 $3,559.57 $889.89 $2,669.68 $667.42 $2,000 (salvage value)

34

BVt

Page

Year


HOMEWORK Que. 1 A dam initially cost $30 million and must be renovated every 50 years at a cost of $10 million. Interest is 7%. Find the capitalized cost sufficient to construct and maintain the project in perpetuity. Que. 2 A public utility commission requires a trust fund be established by a private company wishing to build and maintain a water treatment plant for a small city. The company must deposit enough money to build the plant and then to operate and renovate the facility in perpetuity. The plant will cost N10, 000,000 to build, has an annual operating expense of 600,000, and must be renovated every 20 years at a cost of 1,000,000. The trust fund earns 6% interest. What amount the company put in the trust fund?

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35

Que. 3 A city water department has decided that two alternatives are available to meet the need for an additional 40 million gallons of water per day. One is to build a dam and reservoir, the other is to undertake a strong program of water conservation/education. The dam and related facilities can be constructed over three year at a cost of $20 million in year 1, $10 million in year 2, and $5 million in year 3. Thereafter, maintenance is $0.5 million per year. The conservations program will cost $50 million the first year, $4 million the second year, and $3 million the third year. Thereafter, $2 million per year will have to be spent to maintain low water use. The planning horizon for both alternatives is 50 year; any costs beyond this length of time are ignored and an interest rate of 7% is to be used. Which alternative should be undertaken?


5. QUALITY CONTROL IN CONSTRUCTION Introduction Quality in construction has assumed critical importance due to numerous instances of building collapses, the risks associated with construction, and the need to ensure reliability and maintainability of completed structure. There are high costs associated with poor quality of materials and equipment, including schedule delays, rework, and loss of productivity. The Standard Organization of Nigeria (SON) is an organization established and charged with responsibility of formulating national standard of quality assurance system. Much of the success of the organization is in manufacturing industry with no impact on the construction industry. Management of quality is the organizational function responsible for defect prevention [1]. There are various quality control techniques depending on complexity and scope of project. To overcome the pressures of time and completion, contractors and suppliers must have quality management plans and systems that are effective. Quality monitoring in construction is continuous. It extends through the organization and definitely beyond inspection. Quality Management The accomplished quality of a building construction project is a function of intrinsic quality of materials used and effectiveness of the quality control exercised during the construction by the team involved. The term “supplier quality” implies conformance to materials requirement in specification. All engineered, fabricated, and bulk materials supplied by vendors and subcontractors are included in this definition. The role of quality management for a construction company is not an isolated activity, but intertwined with all the operational and managerial processes of the company [2]. Total quality management (TQM) is often defined as a complete management philosophy that permeates every aspect of a company and position quality as a strategic issue. The application of ISO standards has received much attention in advance countries. However, the ISO 9000 standards series can form the basis for an efficient and advantageous quality management system in the construction industry [2]. Management commitment to quality and to continuous quality improvement is very important in each phase of the building process. ISO 9001:2000 requires the following in relation to management commitment: 1. Communicating about the importance of meeting customer as well as statutory and regulatory requirements; 2. Establishing the quality policy; 3. Ensuring that quality objectives are established; 4. Conducting management reviews; 5. Ensuring the availability of resources.

The pursuit of quality starts with the client and continues through all the process and stages of construction and the team involve. However recognizing that the Quality Control (QC) is a cost which is variable with scope of the project, the client with his architect or engineer may evaluate the level of QC supervision so desire and how it is to be achieved.

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Owner or Developer

36

However it has been argued that many of the concepts in ISO 9001 are experienced as being too abstract and difficult to comprehend in construction process. Others were of the opinion that it seems difficult for a company to improve its competitiveness and be more efficient by the use of ISO 9001 alone in view of the many stages of the construction process encompassed and the diverging interests represented. The key personnel in material quality management in building construction include Architect, Structural Engineer, Mechanical and Electrical Engineer, Quantity Surveyor, and Contractor. It is not in all building works that all these personnel are involved. The key personnel on building team and their roles in the chain of quality control management are enumerated as follow. Note that what is been discussed is based on the author (as a Chief Civil Engineer) establishment of minimum standard in quality control requirement from contractors on the University projects [3]:


Architect Right from the take off of the project, he is responsible to act and supervise on behalf of the client. He normally appoints a clerk of works for the client whose main responsibility is to ensure contractor compliance with design and material specification. Structural Engineer

The engineer is responsible for quality of the design. The design should encapsulate details and specification that are readily interpreted by the contractor. The Engineer has based the design on stated standard of materials to be used such as for example structural steel, iron rod (plain or twisted), granite among others. Depending on the scope and complexity of the project, the following quality control measures are the basic requirement for the contractor compliance [3]:         

Mixing of concrete should be by mechanical concrete mixers Functional vibrators should be provided and use on all concrete casting. At least six cube moulds (steel) should be provided on site for cube samples. Coarse materials should be granite chippings. Fine materials should be good river sand. All sandcrete blocks should be machine vibrated and of high quality All major concrete work should proceed on approval of the structural Engineer. Cube samples must be taken for all major reinforced concrete works. Cube samples must be crushed as and at when due and results promptly submitted to the structural Engineer.  Contractor must ensure correct sizes of reinforcement from source before site delivery.  From the consignment of reinforcement, tensile test may be required on specific sizes before use as directed by the Structural Engineer. Mechanical and Electrical Engineer (M&E)

The engineer is responsible for quality of the design of all electrical and mechanical services. The design should be complete with details and specification that are readily interpreted by the contractor. Depending on the scope and complexity of the project the following quality control measures are the basic requirement for the contractor compliance[3]:  Samples of electrical and mechanical conduit pipes must be certified before usage.  Conduit pipes layout for both electrical and mechanical services must be certified correct before concrete casting.  Samples of electrical cables and fittings must be certified okay before usage and there should be no cable joining within conduit pipes.  All mechanical fittings/fixtures must be certified okay before usage. The Contractor

The contractor is the sole party to the contract and has a direct control on the quality of the project. He has a duty to perform in accordance with the plans and specifications of the contract. It must be recognized that the contractor is production-oriented for a profit. Therefore, it is imperative on the client to ensure or transfer the responsibility of ensuring all quality control measures are spelt out for contractor compliance.

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The effective site control of incoming materials is pre-establishing responsibility. One individual (or commissioned consultant), probably the appointed clerk of works may be responsible for determining whether materials can be accepted, returned, or replaced. The authority to accept or reject should be vested in an organization or individual who is not influenced by compromising issues such as schedule, cost, or vendor pressure to accept non-conforming material.

37

On-Site Quality Control


Discourse: Critique the quality of commissioned projects on the University campus Critique the quality of ongoing projects in Ilorin by the government of AbdulFatah government. References 1. Kolawole, JO (1998) The role of the client in quality assurances and management in the construction industry. Nigerian Journal of Construction Technology and Management. Vol. 1, No 1, 46 50 2. Chin-Keng, T and Hamzah, A. (2011) Study of quality management in construction projects. Chinese Business Review, Vol. 10, No. 7, 542-552 3. Okeola, OG (2009). “Quality Control Management in Building Construction”. Paper delivered at Workshop organized by Kwara State Universal Basic Education Board (KWSUBEB) Ilorin. Held at NOCBUL Hotel, Ilorin, Kwara State. 21st -22nd April, 2009.

Bibliography

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38

1. Journal of Professional Issues in Engineering. Vol. 112, No 4, 1996 2. Journal of Performance of Constructed Facilities. Vol. 3, No 2, 1989


6. CONSTRUCTION EQUIPMENT INTRODUCTION Plant equipment and tools are synonymous with major construction operations and are categorized into 3. They are also estimated according to the category. Small tools and consumable:

These are tools such as blades, drill bits and consumable like electrodes etc commonly use on project. They are normally priced as a percentage of total labour prices. Equipment shared by a number of works: These kinds of equipment item are kept at the site over a period of time as long the project last and are used on the project Equipment used for specific tasks:

These are capital intensive and are used in for example excavation. These types of equipment are not kept on site; however, they are brought in, employed for a particular task, and thereafter demobilized from site. Excavation equipment, cranes, hoisting, equipment highly specialized and costly items such as concrete saw, are all fall in to this category. Construction equipment is very crucial in the operation of construction industry. Therefore estimating the cost of acquiring and operating this category construction equipment is very important in preparation of BEME for major civil engineering project. In the case of contractors in the heavy civil construction industry, the cost of acquisition and operation of equipment are key part of doing business in a profitable manner. TOTAL COST OF EQUIPMENT Total costs of construction equipment are made up of two major components: ownership costs and operating costs. Ownership cost are fixed cost (aside the one-time initial capital cost of purchasing) that are incurred each year, irrespective of whether the equipment is operated or idle, while operating costs are the cost, incurred only when the equipment is used. Both costs have different method of calculation. Ownership cost:

Ownership cost is fixed costs which are basically annual in nature. They include: initial capital cost, depreciation, investment cost, insurance cost, taxes and storage cost. Initial cost

The cost is incurred for getting the equipment into contractor’s yard or site, and having the equipment ready for operation. Normally this cost is made up of the following: factory price; Shipping cost; related charges (e.g. taxes) and assembling & erection costs. Investment cost

This cost represent the annual cost of capital invested in a machine, however if a loan is taken to purchase an equipment, the equipment cost is simply the interest charged on these funds. The average value of equipment can be determined as follow:

P

IC (n  1) 2

_______Eqt 6.1

Where:

IC = the total initial cost P =average value n =the useful life in years.

P

IC (n  1)  s (n  1) _______Eqt 6.2 2n

Where:

s = salvage value

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39

Equation 6.1 assumes that a unit of equipment will have no salvage value at the end of its useful life. However if a unit of equipment has salvage value when it is disposed of, the average value during its life can be obtained from:


Insurance tax and storage costs

Insurance cost represents the cost incurred due to fire, theft, accident, and liability insurance for the equipment. Tax cost is in real term the cost of property tax and licenses for the equipment, although this is non-operation in Nigeria. Storage cost includes the cost of rent and maintenance for equipment storage, yard, security guard wages, and employees involved in moving equipment in and out of storage as well as associated direct overhead TOTAL OWNERSHIP COST Total equipment ownership cost is calculated as the sum of depreciation investment cost insurance cost, tax and storage cost. Normally after all elements of ownership costs have been calculated they can be summed up to yield total ownership cost per hour of operation. COST OF OPERATING CONSTRUCTION EQUIPMENT Operating costs of construction equipment represents a significant cost category and therefore very important for considerations. They are basically the costs associated with the operation of a piece of equipment and are only incurred when the equipment is actually used. The fall into the following categories: - (1) maintenance and repair costs (2) tire cost (3) consumable costs. Maintenance and repairs cost constitutes the largest amount of operating expense for construction equipment. generally this cost got higher as the equipment year got older, the tire cost represent the cost of tire repair and replacement the consumables are the items required for the operating of a piece of equipment that literally gets consumed in the course of its operation. These include but not limited to fuel, lubrication, electrodes and other petroleum products. The cost of equipment mobilization and demobilization can be large and are therefore always important items in any job where substantial of equipment are used.

BASIC CONSIDERATION FOR EQUIPMENT SELECTION Equipment categories

The construction equipment is categorized in to three main categories as follow: 1. Earthmoving equipment: Bulldozers; Front-end loaders; Motor graders; Scrapers and Trucks. 2. Excavating equipment: excavators; Backhoes and Front shovels. 3. Lifting equipment:  Hoisting: Mobile, crawler, and tower cranes  Material moving: Forklifts, concrete pumps  People moving: Personnel lifts Specific consideration

First and foremost equipment needs are function of complexities and uniqueness of specific work events. For example equipment need for road construction (e.g. grader, roller, paver etc) are different for say a 20-story building (e.g. concrete mixer, crane and trucks). From this example, the basic consideration in equipment selection is in appropriating the right equipment with work activity. A typical mechanical operation is required for the three equipment categories. It is instructive to ensure that appropriate equipment is matched with work activity to prevent failure. Failure to match machine appropriately with work activity would result in operational inefficiency which may shorten the useful life. Failure prevention is also facilitated through proper and timely service, maintenance and repair.

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The desired productivity is the key major influence on the three categories of equipment selection. The physical properties of clay, gravel, sand, silt or their combination require to be moved or excavated will dictate the type and capacities of equipment selected for a particular task. The equipment productivity would be affected by the ease or difficulty of removing or excavating soil material and subsequently would determine the capacity and type of bucket, blades and accessories. The type and condition of working surface and travel distance will determine the choice of tires or tracks. Lifting production is heavily dependent on ground and on-structure craft support efficiency. Lifting capacity and vertical hoist speed are the primary equipment influences on lifting production.

40

Desire productivity


Equipment failure

All equipment can fail structurally or mechanically due to overloading or overstressed beyond the physical capabilities of its component. Stability failure also can result due to overloading or imbalance positioning on uneven surface during operation. TYPICAL CONSTRUCTION EQUIPMENT Excavator Excavator combines digging and lifting abilities. They come in a wide range of bucket sizes, boom length and operating speed. Grader The grader is mainly used to move soil to create a level surface and can be used for light surface excavation. The grader is a long tractor-driven piece of equipment with a blade mounted underneath. The blade is used to push dirt straight ahead or to the side at a desired level. Backhoe Backhoes can be used as a hoe or a loader and can accommodate many different accessories and attachments for different operations. One of the backhoe’s greatest strengths is that many attachments can be used to increase its versatility on a job site. The backhoe is ideal for light underground utility construction. The hoe can be used for trenching and lifting like the excavator. Tower Crane Tower cranes are one of the greatest construction equipment achievements. They are designed to work in congested areas. The cranes are a lifting device on top of a tower or mast. When a pick is made, the same lifting forces occur as for any other crane. Counterweight must be provided to balance the load. Compression is transferred down to the ground by the tower instead of outriggers, tires, or tracks. Lifting from up in the air can be a lot more demanding and complicated than lifting while sitting on the ground. Manufacturers classify tower cranes as top slewing, bottom slewing, self-erecting, and special application. Trucks Trucks are very crucial part of the earthmoving and material moving process. They are basically a tractor and a trailer with sides. There are available a wide range of trucks based on hauling conditions and needs. They are normally sized by their trailer volume. Trucks are typically used with excavators and loaders for excavation and soil haul off or delivery. There are two basic considerations for choosing a truck trailer: (1) method of dumping and (2) class of material hauled. Trucks can dump from the rear, from the bottom (belly dump) or from the side depending on the type of material and work activity.

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NOTE: The following pictures of relevant construction equipment within the scope of the CVE 581 course are adapted from Douglas, et. al., (2000)

41

Trucks are dependent on another piece of equipment for loading. Trucks load are rated by volume and weight. Typical cycle fixed times include loading, dumping, and required spotting times. The loading time equals number of cycles required to load the truck times the estimated cycle time. The number of loader cycles to fill a truck equals volume of the truck divided by the volume of the loader bucket per cycle. Trucks are usually loaded by front-end loaders or excavators. Spotting to load or dump and wait or delay times are influenced by job conditions


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Tracked loader

42

Belly dump truck


Motor grader

Page

43

Backhoe


Excavator

Interactive Class Session

Page

44

1. Discuss the: (1) reasons for using construction equipment on site and (2) methods of acquisition of equipment. 2. Give three types of equipment for each of the following construction operations: (1) Excavations only (2) Combined excavation and loading (3) Combined haul and deposit and (4) Combine excavation, load, haul and deposit 3. Discuss the safety procedures required in the following construction processes (1) transportation of construction loads (2) earthmoving (3) masonry works and (4) concreting.


REFERENCES

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Olayinka Okeola Dr Okeola is a licensed professional engineer under the Nigerian law and had over 10 years experience in structural engineering design and in tendering/contract administration, checking of consultant submission and project monitoring, supervision and reporting. He has written monographs, instructional guides and joint technical papers. Dr Okeola is a Peace Ambassador and has received the Pillars of Nation Building Award to 2012 Distinguished Academic ICON from the Strategic Institute for Natural Resources and Human Development of Nigeria (SINRHD). His favorite books include: Giving by Bill Clinton; Happiness by Matthieu Ricard, Social Ethics (Morality & Social Policy) by Mappes & Zembaty, and The 7 habits of highly effective people by Stephen R. Covey. Contact Info: [email protected]; [email protected] +234-703-230-7770

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1. Ardalan, A. (2000) Economic and Financial Analysis for Engineering & Project Management. Technomic Publishing Company, Inc USA 2. Douglas, DG, Calin, MP and Richard, CR (2000) Constructions Equipment Management for Engineers, Estimators and Owners. Taylor & Francis Group, LLC. USA. Copyright 2000 3. Heerkens, GR (2000) Project management. McGraw-Hill, USA. 4. Kerzner, H. (2003) Project Management: A System Approach to Planning, Scheduling & Controlling. John Wiley and Son, Inc. New Jersey, USA 5. Bernett, LF (2003) The Management of Construction: A project Life Cycle Approach. Butterwort- Heinemann Inc. UK. 6. Okeola, OG and Salami, AW. (2012). A pragmatic approach to the Nigeria’s engineering infrastructure dilemma. Epistemics in Science, Engineering and Technology.(2):1 pp 55-61 7. Okeola, O.G. (2009) Occupational Health and Safety Assessment in the Construction Field. Proceedings of 1st Annual Civil Engineering Conference, (Aug. 26th – 28th 2009). Pp 32-40. Department of Civil Engineering, University of Ilorin, Nigeria. 8. Okeola, O.G. (2009). “Quality Control Management in Building Construction”. Paper delivered at Workshop organized by Kwara State Universal Basic Education Board (KWSUBEB) Ilorin. Held at NOCBUL Hotel, Ilorin, and Kwara State. 21st -22nd April, 2009. 9. Park, CS and Tippett, DD (1999)“Engineering Economics and Project Management” Mechanical Engineering Handbook. Ed. Frank Kreith. Boca Raton: CRC Press LLC, 1999 10 Revelle, CS, Whitlatch, EE and Wright, JR (1997)Civil and Environmental Systems Engineering. Prentice Hall, Upper Saddle River, New Jersey.

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