Middle Technical University
Electrical Engineering Technical College
Project Management
Prepared by:
Rawaa Dawood Salim
Electrical Engineering
Techniques college
2021/2022
Project Management
Introduction, Objectives, Principles and
Phases
For
Students of Fourth Stage
Computer Department
Department Of computer Engineering
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Electrical Engineering
Techniques college
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1. Overview
a. Target Population: For students of stage of second and fourth stage in technical
Colleges and institutes in foundation of technical education.
b. Rationale: Management is an art of directing workers efforts to obtain on high
production and excellent quality. This leads to avoidance of loss and profits are
increased and it provides security and safety element. This mode unit consists of main
elements of management, objectives, and steps/phases of project management.
c. Central Ideas: Project management is a set of principles, methods and steps for
effective planning of objective-oriented work, thereby establishing a sound basis for
effective scheduling, controlling and re-planning in the management of programs and
projects.
d. Objectives: The student will be able after finishing lecture on:
-
Define main elements of project management.
-
Study steps/phases of project management.
2. Pre-Test:
1. Define the term „project‟.
2. Organizing is done after planning in elements of management. State True or false
3. Project control is one of the step of project management- State True or False
4. The term “scope definition” in a steps of project management comes under
a) Project initiation
b) project planning
c) project scheduling d) project costing
5. The term “client feedback” in a steps of project management comes under
a) Project termination b) project planning c) project scheduling d) project costing
Note: Check your answers in “Answer Keys” in end of mode unit. If you obtain 75% of
solution, you cannot need to this mode unit. If your answer is poor, you will transfer to
next page.
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3. Theory:
Introduction
Project management emerged because of the growing demand for complex,
sophisticated, customized goods and services and the exponential expansion of human
knowledge. The former depends on the integration of product design with production /
distribution and the latter allows a number of academic disciplines to contribute to the
development of goods and services.
Elements of management
1. Planning
2. Organizing involves:
a. Establishing a structure to be filled by people, aimed at reaching the defined
goals and objectives.
b. Defining job content, interfaces, responsibilities, authority, and resource
allocation.
3. Staffing involves:
a. Filling the positions in the organizational structure with suitable people.
b. Keeping the positions filled, in order to execute the plan.
4. Directing (or Leading) involves:
a. Creating an environment in which individuals, working together in groups, can
accomplish well-selected aims.
b. Influencing people to contribute to reaching the goals and objectives.
c. Using leadership styles, communication, conflict resolution, delegation, etc. in order
to overcome the problems arising from people issues (attitudes, desires, motivations,
behavior in groups, etc.) on a project.
5.
Controlling (and co-ordination) involves:
a. Measuring actual performance.
b. Comparing actual- with desired results and implementing corrective actions – e.g. by
controlling the actions of the people doing the work.
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Fig. (1) Block diagram of elements of management.
Project is an assignment/task/job that has to be undertaken and completed within a set time,
budget, resources and performance specifications designed to meet the needs of stakeholder
and beneficiaries.
Project Management is the use of knowledge, skills, tools, and techniques to plan and
implement activities to meet or exceed stakeholder needs and expectations from a project.
Project Management is a set of principles, methods and techniques for effective planning of
objective-oriented work, thereby establishing a sound basis for effective scheduling,
controlling and re-planning in the management of programs and projects. In other words, it
provides an organization with powerful tools that improve the organization's ability to plan,
organize, implement and control its activities and the ways it uses its people and resources.
4. Self- Test
1. Define project management.
2. List the main elements of management.
Objectives of the Project management
The basic purpose for initiating a project is to accomplish some goals. The reason for
organizing the task as a project is to focus the responsibility and authority for the attainment
of the goals on an individual (project manager) or a small group (project team).
Project Management is a means by which to fit the many complex pieces of the project
puzzle together, both human and technical, by use of:
-
Schedules
-
Budgets, including resource allocation
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Scope (product) definition
Project Management fulfills two purposes:
1. Technical: Documentation techniques to communicate
-
The 'plan'
-
Status which compares 'planned' versus 'actual' performance
2. Human: Managerial skills to be a better 'manager' of people as well as the project
Implementation of project management technique can have significant results such as:
1. Cost reduction
2. Time reduction
3. Recourses allocation
4. Increased quality
3. Project management fulfills two purposes --------- and -------------.
4. Implementation of project management technique leading to --------, ---------, --------, and -------.
Steps/Phases of Project Management
The steps followed for project management are essentially the steps for successful project
initiation, development and completion.
1. PROJECT INITIATION
-
Concept definition, which includes identification and selection of opportunities and
identification of objectives
-
Feasibility study and justification
2. PROJECT PLANNING
-
Scope definition
-
Goal definition, includes time, money, resources and product targets
-
Project requirements - definition of deliverables
-
Project objectives - definition of major work efforts, quantifiable
-
Work break down structure
-
Analysis & break down of project into smaller pieces of work
-
Development of checklist of everything that needs to be done
-
Team building
-
Selection of project manager
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Selection of team members,
-
Use resource matrix to match skills task requirements
3. PROJECT SCHEDULING
-
Determining sequence of work
-
Building network / interdependence
-
Analysis of interdependence, estimation of total duration (CPM, PERT) and
determination of Critical Path
-
Establish milestones
-
Graph on time chart (Gantt chart)
-
Determining human resource loading
-
Establishing milestones / reporting periods
4. PROJECT COSTING
-
Estimate costs, capital / operating
-
Develop cost spreadsheets
5. PROJECT CONTROL
-
Done periodically (at milestones)
-
Time control, status, deviations from plan, replanning, new estimates
-
Cost control, Expenditure, deviations from plan, new estimates
-
Quality control, performance versus performance criteria / project requirements
6. PROJECT TERMINATION / EVALUATION
-
Post project activity
-
Statistics from monitoring progress
-
Client feedback
-
Profitability or not of the project
-
Post implementation report
5. The term “team building” in a steps of project management comes under
a) Project termination b) project planning c) project scheduling d) project imitation
Note: Check your answers in “Answer Keys” in end of mode unit.
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5. Post- Test
1. List the steps/phases of project management.
2. Controlling is last element of elements of management. State True or False
3. Implementation of project management leads to increase the cost- State True or False
4. Project Management is a means by which to fit the many complex pieces of the project
puzzle together, both human and technical, by use of --------,----------, and ----------.
5. The term “establish milestones” in a steps of project management comes under
a) Project termination b) project planning c) project scheduling d) project costing
6. References
1. Y. Bakouros and V. Kelessidis “Project management” INNOREGIO: dissemination of
innovation and knowledge management techniques, January 2000.
Answer Keys
2. J.R. Meredith and S.J. Mantel “Project Management”, J. Wiley & Sons, 1995.
3. http://www.projectmanagement.com/main.htm.
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Pre- Test
1. Project is an assignment/task/job that has to be undertaken and completed within a set time,
budget, resources and performance specifications designed to meet the needs of stakeholder
and beneficiaries.
2. True.
3. True.
4. b) project planning
5. a) project termination
Self-Test
1. Project Management is the use of knowledge, skills, tools, and techniques to plan and
implement activities to meet or exceed stakeholder needs and expectations from a project.
2. Planning, organizing, staffing, directing, and controlling.
3. Technical and human.
4. Cost reduction, time reduction, recourses allocation, and increased quality.
5. b) project planning
Post- Test
1. Project initiation, project planning, project scheduling, project costing, project control, and
project termination.
2. True.
3. False.
4. Schedules, budgets, including resource allocation, and scope (product) definition
5. c) project scheduling
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Planning and Control in Projects:
Planning
Scheduling
Controlling
For
Students of Fourth Stage
Computer Department
Department Of Computer Engineering
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Electrical Engineering
Techniques college
2021/2022
1. Overview
a. Target Population: For students of stage of second and fourth stage in technical
Colleges and Institutes in foundation of technical education.
b. Rationale: Project planning is a critical element of every successful investment. It
provides a foundation on which to base anticipated efforts. Additionally, it helps
identify investment components and illustrates these components in a project plan.
a. Central Ideas: The basic purpose of a planning and control in projects is to help
managers schedule, monitor and control large and complex projects.
b. Objectives: The student will be able after finishing lecture on:
-
Define planning, scheduling and control processes.
-
Study the main steps of planning, scheduling and control processes in projects.
2. Pre-Test:
1. Define the term “scheduling process”.
2. Contingency needs to be allowed both on the estimated effort and elapsed time because
of the likelihood of unforeseen work arising. State True or False
3. The role of the project manager falls into three areas ------, -------- and ------.
4. ------------- is the process of comparing actual performance with planned performance,
analyzing the differences, and taking the appropriate corrective action.
5. Planning process is closest to mean for:
a)
Project Scope Management
b)
Project Time Management
c)
Project Cost Management
d)
Project Quality Management
Note: Check your answers in “Answer Keys” in end of mode unit. If you obtain 75% of
solution, you cannot need to this mode unit. If your answer is poor, you will transfer to
next page.
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3. Theory:
Introduction
Planning
This process performed to define and mature the project scope, develop the project
management plan, and identify and schedule the project activities that occur within the
project.
Scheduling
Scheduling is the process of determining when project activities will take place depending on
defined durations and precedent activities. Schedule constraints specify when an activity
should start or end based on duration, predecessors, external predecessor relationships,
resource availability, or target dates.
Control
Control is the process of comparing actual performance with planned performance, analyzing
the differences, and taking the appropriate corrective action.
4. Pre-Test:
1. Define the term “planning process”.
2. Scheduling is the process of determining when project activities will take place
depending on defined ---------- and ------------.
Project Planning and Scheduling
Project Planning begins as soon as Definition allows. The process involves planning
subprojects first and hence Definition must at least have identified the sub-projects and the
major tasks involved in them. From this point, Planning and Definition tend to continue in
parallel as a series of iterations, gradually refining and hardening both Definition and Plans.
The purpose of the Project Plan at this stage is to provide detailed realistic estimates of
time, duration, resource and cost, and planning should be carried out only in sufficient detail
to allow this to be achieved. Detailed planning for allocation of tasks to individuals is carried
out progressively as the work proceeds.
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Where there are sub-projects these should be planned first and then combined to produce
the overall project plan. Produce a plan for each sub-project, or for the total project if there are
no sub-projects as follows:
1. Identify Major Activities
Break the work down into activities of the order of 20-50 days of effort, ensuring that
milestones correspond to completion of one or more of these. In practice the achievement of a
milestone is usually a good basis for identifying an activity e.g. “prepare and perform user
training”.
2. Identify and Chart Dependencies
Produce a network chart for the sub-project showing dependencies between the major
activities and dependencies on other sub-projects or external events.
3. Estimate Effort and Duration
Estimate effort and duration of each major activity.
4. Provide Contingency
At this stage estimates are likely to be 'soft' and probably expressed in ranges, because precise
details of the work are not settled. Contingency needs to be allowed both on the estimated
effort and elapsed time because of:
• The likelihood of unforeseen work arising,
• The likelihood that tasks will take longer than expected,
• The likelihood of changes to requirements or plans before publication. (Subsequent changes
should be processed through Change Control).
Contingency provision should remain evident in plans (probably as one or more contingency
'tasks'). This provision should then progressively be removed from plans during Tracking and
Control as a result of either:
• being used up by e.g. tasks taking longer than planned,
• or reaching a point where uncertainty is reduced such that a part of contingency provision
can safely be deleted. This usually means the deletion of contingency allowed, but not used,
on tasks now completed.
5. Schedule Major Activities
Determine start and end dates for each major activity and produce a bar chart or other
diagram, showing relationships between activities.
6. Calculate Resource Requirements
Calculate requirements for each time period. Identify needs for each resource type (e.g.
systems analyst, user staff) and identify needs for special skills or scarce resources.
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7. Calculate Costs
Calculate costs for the sub-project. This should include 'hardening up' items such as cabling,
training etc., for which an order of costs had been produced previously.
8. Determine Overall Costs and Benefits of the Project
The cost/benefit justification should have already been stated in the feasibility study.
This stage provides the opportunity to review the case in the light of more detailed
information.
9. Document the Project Plan
Once a viable plan has emerged (i.e. conflicts have been resolved, resource availability has
been confirmed etc.) the Project Manager should produce the Project Plan covering:
• Project Schedule. This should show major activities by sub-project on a bar chart or other
diagram. The chart should also show project milestones and target dates.
Show contingency as a single provision at the end. Include an overall project network
showing the critical path. Narrative explanation may be included for clarification.
• Major Check points and Reviews. List the dates of Checkpoint Reports, Checkpoint
Meetings, Steering Group Meeting and the Post-Implementation Review.
• Deliverables. List the major products of the project with delivery dates and acceptance
procedures.
• Resources. Summarise the resource needs from the sub-project plans.
• Costs and Benefits figures. Revise and refine as a result of completion of Definition and
Planning.
• Potential Problems. List any risks, problems or assumptions which may jeopardize the Plan,
together with actions needed to correct the situation.
10. Ensure Management Systems are in place.
3. Contingency needs to be allowed both on the estimated effort and elapsed time because of the
likelihood that tasks will take longer than expected. State True or False
4. Systems analyst and user staff come under “Calculate Resource Requirements” State True or False
Project Implementation and Control
The role of the project manager falls into three areas:
i)
Management of stakeholders
ii)
Management of the project life cycle
iii)
Management of performance
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An approach needs to be developed for each of these. Control and monitoring procedures
need to be put in place and appropriate information systems developed. The procedures which
are put into place can only be successful if:
a) there is satisfactory information to enable the team to manage the project effectively;
b) they are simple and easy to operate and understand;
c) they have the full support of the project team.
How should this relate to the three categories referred to above?
i) Management of stakeholders:
Stakeholders' interest must be monitored to ensure that:
1. their interest and support is maintained;
2. their views and ideas are being adequately reflected in the project development;
3. their personal success criteria are being pursued and achieved;
4. Environmental change is fully taken into account.
ii) Management of the project life cycle:
This is probably the most conventional view of project control. Feedback systems need to be
set up to monitor key areas. The key areas would be as follows:
1. The project timetable, with particular reference to critical event times and potential
bottlenecks. There should be feedback on activity times achieved and their effect on
the whole project. If network analysis is used, then it is vital that the network is
reworked and updated to take into account the actual performance achieved.
2. The project budget; budgetary control procedures can be used as in respect of any
other form of budget.
3. Quality and performance standards; these need to be monitored against the original
project specification subject to changes agreed with stakeholders in the course of
project development. Where possible this should all be done through positive
reporting which will require action to be taken.
iii) Management of performance:
This is the least tangible but possibly the most important of the three categories. How it is
tackled will depend upon what kind of project is being carried out.
It is unlikely that the project team will spend all of their working time together in close
proximity and under the direct supervision of the project manager. It is much more likely that
they will work apart most of the time, only meeting up occasionally and only meeting with
the project manager from time to time. Control issues that need to be considered therefore
would be:
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1. How to get the best out of the team when they are together. If you are holding
meetings then they should be purposeful and effective. They should not simply be part
of the routine. Having said that, they may be an important element in binding the team
together and in developing a team approach to planning and monitoring of
performance.
2. Ensuring people work when the team is apart. You need to set people realistic
deadlines and ensure that they see the importance of their contribution and that their
contribution is fully valued.
3. Communications are important in terms of disseminating information and keeping
everyone informed. There are views that team members should be given information
on a need to know basis but this approach can cause problems.
4. Ensuring continuing commitment by the team and adherence to the values and beliefs
being pursued by the team.
5. Change, in particular, needs to be communicated to team members quickly and
effectively. It is important to stress once again the need to look at the team and also for
the project leader to look inwards at his or her own performance.
5. The term “scope definition” in a steps of project management comes under
a) Project initiation
b) project planning
c) project scheduling d) project costing
5. Post- Test
1. Define the term ”control process”
2. List briefly the points that the manager should produce to cover “Document the Project
Plan”.
3. The role of the project manager falls into three areas ------, -------- and ------.
4. Planning process performed to define and mature the ---------.
5. Identify and Chart Dependencies les under:
a) Planning process
b) Control process
c) Planning and scheduling process
d) Staffing process
Note: Check your answers in “Answer Keys” in end of mode unit.
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6. References
1. Principles of Project Management, NPC publication
2. S. Choudhury “Project Management”, Tata McGraw Hill – 2003
3. Y. Bakouros and V. Kelessidis “Project management” INNOREGIO: dissemination of
innovation and knowledge management techniques, January 2000.
4. http://www.projectmanagement.com/main.htm
Answer Keys
Pre- Test
1. Scheduling is the process of determining when project activities will take place depending
on defined durations and precedent activities. Schedule constraints specify when an activity
should start or end based on duration, predecessors, external predecessor relationships,
resource availability, or target dates.
2. True.
3. management of stakeholders, management of the project life cycle and management of
performance
4. Control process
5. a) Project Scope Management
Self-Test
1. Planning: This process performed to define and mature the project scope, develop the
project management plan, and identify and schedule the project activities that occur within
the project.
2. Durations and precedent activities.
3. True
4. True
5. b) project planning
Post- Test
1. Control is the process of comparing actual performance with planned performance,
analyzing the differences, and taking the appropriate corrective action.
2.
• Project Schedule.
• Major Check points and Reviews.
• Deliverables.
• Resources.
• Costs and Benefits figures.
• Potential Problems.
• Appendices.
3. management of stakeholders, management of the project life cycle and management of
performance
4. project scope
5. a) Planning and scheduling process
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Project Planning Techniques
Network analysis
Critical Path Method (CPM)
For
Students of Fourth Stage
Computer Department
Department Of computer Engineering
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1. Overview
a. Target Population: For students of stage of second and fourth stage in technical
Colleges and institutes in foundation of technical education.
b. Rationale: The most common and widely used project management technique that
can be classified under the title of Network Analysis is Critical Path Method (CPM). It
was developed in the 1950's to help managers schedule, monitor and control large and
complex projects. CPM was first used in 1957 to assist in the development and
building of chemical plants within the DuPont corporation.
a. Central Ideas: The basic purpose of a network analysis is to help managers schedule,
monitor and control large and complex projects.
b. Objectives: The student will be able after finishing lecture on:
-
Define Critical Path Method (CPM).
-
Study steps of CPM.
2. Pre-Test:
1. Define the term ‗CPM.
2. CPM predicts the time required to complete the project— State True or False
3. The time between its earliest and latest start time, or between its earliest and latest finish
time of an activity is
a) delay time
b) slack time c) critical path
d) start time
4. The path through the project network in which none of the activities have slack is called
a) start time b) slack time c) critical path
d) delay time
5. Activity is an ------------- needed for the completion of a project.
Note: Check your answers in ―Answer Keys‖ in end of mode unit. If you obtain 75% of
solution, you cannot need to this mode unit. If your answer is poor, you will transfer to
next page.
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3. Theory:
Introduction
Critical Path Method (CPM) or (Calculate Schedule) is a modeling process that defines all
the project's critical activities which must be completed on time. CPM models the activities
and events of a project as a network.
Steps in CPM Project Planning
1. Specify the individual activities.
2. Determine the sequence of those activities.
3. Draw a network diagram.
4. Estimate the completion time for each activity.
5. Identify the critical path (longest path through the network)
6. Update the CPM diagram as the project progresses.
CPM Benefits
• Provides a graphical view of the project.
• Predicts the time required to complete the project.
• Shows which activities are critical to maintaining the schedule and which are not.
Critical path is the longest-duration path through the network. The significance of the critical
path is that the activities that lie on it cannot be delayed without delaying the project. Because
of its impact on the entire project, critical path analysis is an important aspect of project
planning. The critical path can be identified by determining the following four parameters for
each activity:
1. Earliest Start time (ES): the earliest time at which the activity can start given that its
precedent activities must be completed first.
2. Earliest Finish time (EF), equal to the earliest start time for the activity plus the time
required completing the activity.
3. Latest Finish time (LF): the latest time at which the activity can be completed without
delaying the project.
4.
Latest Start time (LS), equal to the latest finish time minus the time required to
complete the activity.
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4. Self-Test:
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
1. Define the term ―critical path‖.
2. List the benefits of CPM.
The slack time or Total float for an activity is the time between its earliest and latest start
time, or between its earliest and latest finish time. Slack is the amount of time that an activity
can be delayed past its earliest start or earliest finish without delaying the project.
The critical path is the path through the project network in which none of the activities have
slack, that is, the path for which ES=LS and EF=LF for all activities in the path. A delay in
the critical path delays the project. Similarly, to accelerate the project it is necessary to reduce
the total time required for the activities in the critical path.
Activity is an individual task needed for the completion of a project.
Duration is the length of time (hours, days, weeks, months) needed to complete an activity.
Float is the amount of time that an activity can slip past its duration without delaying the rest
of the project.
Free float is the excess time available before the start of the following activity.
3. The time between its earliest and latest start time, or between its earliest and latest finish
time of an activity is
a) delay time b) slack time c) critical path
B bActivity on arrow (A.)
d) start time
4. ------------ is the longest-duration path through the network.
Activity on arrow (A.O.A)
Examples
Ex1: Determine the critical path by using CPM of the following Table (project),
Activities
A
B
C
D
E
Path
1─2
2─3
2─4
3─5
4─5
Duration (day) Description
3
3
‫وصف‬
‫مختصر لكل‬
1
‫فعالية‬
2.5
2
Ans:
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Critical path =3+ 3+ 2.5= 8.5 days
Ex2: Determine the critical path by using CPM of the following Table (project),
Activities
A
B
C
D
E
F
G
Path
1─2
1─3
3─5
3─4
2─4
5─6
4─6
Duration (week) Description
1
4
3
‫وصف مختصر‬
5
‫لكل فعالية‬
2
3
6
Ans:
Critical path= 4+ 5+ 6= 15 weeks
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Dummy Activity: An imaginary activity that requires no time and is used to correctly
maintain the appropriate precedence relationships.
5. Determine the critical path by using CPM of the following Table (project),
Activities
A
B
C
D
E
Path
1─2
2─3
3─4
3─5
5─6
Duration (day)
2
3
8
5
4
Description
‫تحضير الموقع‬
‫تخطيط الموقع‬
‫انشاء االعمدة‬
‫بناء الجدران الخارجية‬
‫بناء السقف‬
5. Post- Test
1. What is slack time in critical path in CPM?
2. Which of the following statements about critical path analysis (CPA) is true?
a) The critical path is the longest path through the network
b) The critical path is the shortest path through the network
c) Tasks with float will never become critical
d) The network should remain constant throughout the project
3. In Critical Path of CPM used in project planning techniques indicates-------------.
a) time require for the completion of the project
b) delays in the project
c) early start and late end of the project
noneyour
of theanswers
above in ―Answer Keys‖ in end of mode unit.
Note:d)Check
4. Dummy Activity is an imaginary activity that requires no time and is used to correctly
maintain the appropriate precedence relationships. State True or False
5. Determine the critical path by using CPM of the following Table (project),
Activities
A
B
C
D
E
Path
1─2
2─3
2─4
3─5
4─5
Duration (day)
3
3
1
2.5
2
Description
‫وصف‬
‫مختصر لكل‬
‫فعالية‬
Note: Check your answers in ―Answer Keys‖ in end of mode unit.
6. References
1. Principles of Project Management, NPC publication
2. S. Choudhury ―Project Management‖, Tata McGraw Hill – 2003
3. W. Durfee and T. Chase, ―Project Management - Gantt Chart Tutorial‖ University of
Minnesota, 2003
4. http://www.projectmanagement.com/main.htm
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Answer Keys
Pre- Test
1. Critical Path Method (CPM) or (Calculate Schedule) is a modeling process that defines all
the project's critical activities which must be completed on time. CPM models the activities
and events of a project as a network.
2. True.
3. b) slack time
4. c) critical path
5. individual task
Self-Test
1. Critical path is the longest-duration path through the network. The main objective of a Gantt
chart is to assess how long a project should take and to establish the order in which tasks
need to be carried out by the ending of the project.
2.
• Provides a graphical view of the project.
• Predicts the time required to complete the project.
• Shows which activities are critical to maintaining the schedule and which are not.
3. b) slack time
4. Critical path
5.
Post- Test
1. The slack time or Total float for an activity is the time between its earliest and latest start
time, or between its earliest and latest finish time
2. a) The critical path is the longest path through the network
3. a) time require for the completion of the project
4. True
5.
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Activity on anode (A.O.N)
ACTIVITY
E.s
E.f
L.s
L.f
DURATION
Example 3:- by using activity on Node(A.O.N)
Determine the critical path by using CPM of the following Table (project),
Activities
A
B
C
D
E
F
G
Preceded
by
None
A
A
B, C
B
D
E,F
Duration (week) Description
1
4
3
DESCIPTION
5
OF ACTIVITY
2
3
6
Solution:E
B
A
4
C
D
F
e
PATH1:-A,B,E,G=1,4,2,6=13
WEEKS
PATH 2:-A,B,D,F,G=1,4,5,3,6=19 WEEKS
PATH3:- A,C,D,F,G=1,3,5,3,6=18 WEEKS CRTICAL PATH=19 WEEKS
30
G
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EX:-for the following table of the network below draw the network
And calculate the cp(critical path) and T.f (total float).
Activity
A
B
C
D
E
F
G
H
Duration
5
6
8
10
7
5
4
10
Preceded by
None
A
A
A
B,C,D
B,E
F,D
B,F,G
T.F
0
4
2
0
0
0
0
0
C.P
41
41
41
41
41
41
Solution:-Activity on anode (A.O.N)
5 B 11
0A5
055
9 6 15
1212
5 C 13
7 8 15
9 5 12
5 D 15
31 H 41
2 2 F 27
15 E 22
2 2 5 27
15 7 22
5 10 15
Cp=A,D,E,F,G,H=5+10+7+5+4+10=41
36
27 G 31
27 4 31
31 10 41
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EX:- find the c.p from figure below.
Solution:Path1=A, D, H, J=2+5+7+4=18
Path2=B, E, H, J=3+6+7+4=20 =cp
Path3=B, F, J=3+4+5=12
Path4=C+G+I+J=4+7+3+4=18
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Project Planning Techniques
Program Evaluation and Review Technique
(PERT)
For
Students of Fourth Stage
Computer Department
Department of computer Engineering
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1. Overview
a. Target Population: For students of stage of second and fourth stage in technical
Colleges and institutes in foundation of technical education.
b. Rationale: PERT is a technique for estimating and planning a large project. One of its
most powerful concepts is that project management is the management of
probabilities. PERT makes use of simple statistical mathematics in order to come up
with a probability distribution for the completion dates of the project milestones.
c. Central Ideas: The basic purpose of a PERT is to implement huge projects involving
thousands of contractors and reduce both the time and cost required to complete a
project.
d. Objectives: The student will be able after finishing lecture on:
-
Define Program Evaluation and Review Technique (PERT).
-
Study steps of PERT.
2. Pre-Test:
1. There are three techniques to plan the project --------, --------- and ---------.
2. What is meaning with optimistic time ( )
3. Define the term „PERT‟.
4. A strategic analysis of a PERT network concentrates on the allocation of resources to
reduce the time on the critical path. State True or False
5. Project management technique which uses three time estimates-optimistic, pessimistic
and most likely, which help in establishing the probability of completing a project within
a specified time and take calculated risk before commencing a project is---------.
a) PERT
(b) Gantt chart
(c) CPM
(d) none of the above
Note: Check your answers in “Answer Keys” in end of mode unit. If you obtain 75% of
solution, you cannot need to this mode unit. If your answer is poor, you will transfer to
next page.
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3. Theory:
Introduction
The Program Evaluation and Review Technique (PERT) is a network model that
allows for randomness in activity completion times. PERT uses three time estimatesoptimistic, pessimistic and most likely, which help in establishing the probability of
completing a project within a specified time and take calculated risk before commencing a
project. It has the potential to reduce both the time and cost required to complete a project.
What are the different steps involved in PERT planning?
PERT planning involves the following steps:
1. Identify the specific activities and milestones.
2. Determine the interdependencies and proper sequence of the activities.
3. Construct a network diagram.
4. Estimate the time (three time estimates, if probabilities are to be computed) required for
each activity.
5. Determine the critical path.
6. Update the PERT chart as the project progresses.
What are the benefits of PERT?
PERT is useful because it provides the following information:
• Expected project completion time.
• Probability of completion before a specified date.
• The critical path activities that directly impact the completion time.
• The activities that have slack time and that can lend resources to critical path activities.
• Activity starts and end dates.
4. Self-Test:
1. It has the potential to reduce both the --------- and ---------- required to complete a project.
2. Of all paths through the network, the critical path
a) has the maximum expected time.
b) has the minimum expected time.
c) has the maximum actual time
d) has the minimum actual time.
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Estimate activity times
A distinguishing feature of PERT is its ability to deal with uncertainty in activity completion
times. For each activity, the model usually includes three time estimates:
• Optimistic time (
) - generally the shortest time in which the activity can be
completed.
• Most likely time (
) - the completion time having the highest probability. This is
different from expected time. Seasoned managers have an amazing way of estimating
very close to actual data from prior estimation errors.
• Pessimistic time (
) - the longest time that an activity might require.
The expected time for each activity can be approximated using the following weighted
average:
Expected time (
=
This expected time might be displayed on the network diagram.
The standard deviation and variance for each activity are given by:
3. Define the term “Most likely time ( )”.
4. The standard deviation of an activity time is estimated as (b – a)/6, where b is the pessimistic
and is the optimistic time. State true or false
5. The probability of completing the project by time T is equal to the probability of completing the
critical path by time T. State true or false
Ex:
By
Activities
A
B
C
D
using
path
1─2
1─3
2─4
3─4
PERT
To
2
3
3
4
draw
Tm
4
16
8
6
Tp
6
17
7
8
the
network
diagram
and
determine
for the following Table. Assume D=23 weeks.
Te
4
14
7
6
Ans:
33
S
0.6
2.3
0.6
0.6
V
0.36
5.29
0.36
0.36
ES
0
0
4
14
EF
4
14
11
20
LS
9
0
13
14
LF
13
14
20
20
C.P
9
0
9
0
20=
20=
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√
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√
5. Post- Test
1. What is meaning with pessimistic time ( ).
2. The calculation of the probability that the critical path will be completed by time T
a) assumes that activity times are statistically independent
b) assumes that total time of the critical path has approximately a beta distribution
c) requires knowledge of the standard deviation for all activities in the network
d) all of the above
3. The standard deviation of an activity time is estimated as (a – b)/6, where b is the
pessimistic and a is the optimistic time. State True or false
4. Optimistic time ( ) - generally the shortest time in which the activity can be completed.
5.
By using PERT draw the network diagram and determine
,slackfor the following Table. Assume D=38 weeks.
Activities
A
B
Path
1─2
1─6
To
2
2
Tm
5
5
Tp
14
8
C
2─3
5
11
29
D
2─4
1
4
7
E
3─5
5
11
17
F
4─5
2
5
14
G
6─7
3
9
27
H
5─8
2
2
8
I
7─8
7
13
31
Te
S
V
ES
EF
LS
LF
Vc
C.P
Note: Check your answers in “Answer Keys” in end of mode unit.
6. References
1. J.R. Meredith and S.J. Mantel “Project Management”, J. Wiley & Sons, 1995.
2. http://www.projectmanagement.com/main.htm
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Answer Keys
Pre- Test
1. Gantt chart, CPM and PERT.
2. Optimistic time ( ) - generally the shortest time in which the activity can be completed.
3. PERT is a network model that allows for randomness in activity completion times. PERT
uses three time estimates-optimistic, pessimistic and most likely, which help in establishing
the probability of completing a project within a specified time and take calculated risk
before commencing a project.
4. False.
5. a) PERT
Self-Test
1. time and cost
2. a) has the maximum expected time.
3. Most likely time ( ) - the completion time having the highest probability. This is different
from expected time. Seasoned managers have an amazing way of estimating very close to
actual data from prior estimation errors.
4. True.
5. False.
Post- Test
1.
2.
3.
4.
5.
Pessimistic time ( ) - the longest time that an activity might require.
a) assumes that activity times are statistically independent
False
Shortest time
Activities
A
To
2
2
Tm
5
5
Tp
14
8
Te
6
S
2
B
Path
1─2
1─6
V
4
ES
0
EF
6
LS
0
5
C
2─3
5
11
29
13
D
2─4
1
4
7
E
3─5
5
11
17
F
4─5
2
5
14
G
6─7
3
9
H
5─8
2
2
I
7─8
7
13
LF
6
slack
0
1
1
4
16
0
5
6
19
2
7
2
6
19
0
4
1
11
2
1
6
4
19
10
20
24
14
30
19
30
6
2
4
0
10
16
24
30
14
27
8
11
4
3
1
16
5
16
7
18
2
1
30
33
30
33
31
0
15
4
16
16
31
18
33
2
√
35
√
C.P
=
=
=
=
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Where D = 4 given in question
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Crashing the project
For
Students of Fourth Stage
Computer Department
Department of computer Engineering
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For example
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EX: - the following data about the project tasks, network, and crash times/costs .calculate the cost
of the project at all tie durations until you can no longer crash the project any further.
Activity
Preceded
by
Normal
time(weeks)
A
B
C
D
E
F
G
None
A
A
B
C
E
D,F
5
10
13
3
5
10
5
Normal
cost
($)
500
1200
3600
300
1000
2400
700
Crash
time
(weeks
4
6
11
1
4
8
5
Crash
cost
($)
600
2000
4800
600
1400
5400
700
Solution:Activity
Preceded
by
Normal
time(week
s)
Normal
cost
($)
A*
B
C*
D
E*
F*
G*
None
A
A
B
C
E
D,F
5
10
13
3
5
10
5
500
1200
3600
300
1000
2400
700
∑ 9700 $
Crash
time
(weeks
4
6
11
1
4
8
5
Crash
cost
($)
Allowable
crash time
=tn-tc
600
2000
4800
600
1400
5400
700
Slope(cost/time)=
cc-cn/tn-tc
1
4
2
2
1
2
0
Slope for activity (A)= cc-cn/tn-tc
( 600 - 500 )/(5 - 4 )= 100
Slope for activity (B)= cc-cn/tn-tc
( 2000
- 1200
Slope for activity (C)= cc-cn/tn-tc
(4800
- 3600
Slope for activity (D)= cc-cn/tn-tc
( 600
Slope for activity (E)= cc-cn/tn-tc
(1400
Slope for activity (F)= cc-cn/tn-tc
( 5400 - 2400 )/(10 -8 )=1500
Slope for activity (G)= cc-cn/tn-tc
(700 - 700 )/( 5-5 )=0
30
-
300
-
1000
)/( 10 - 6 )= 200
)/( 13 - 11 )=400
)/(3 -1 )=150
)/(5 400
100
200
600
150
400
1500
0
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Draw the project by using( A.O.N)
Find C.p
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Crash activity choose the small value in slope in cp path
First crashing activity (
A
) ( 1 ) week
cost=9700+100=9800$
Duration = 4 weeks
c.p=37 weeks
First
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Crash activity choose the small value in slope in cp path
Crashing activity (
E
) (1) week
cost=9800+400=10200$
Duration = 4 weeks
c.p=36 weeks
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crashing activity (
2021/2022
c
) (1 ) week
cost=10200+ 600=10800$
Duration = 12 weeks
c.p=35 weeks
Crashing activity (
C
) (1 ) week
Duration = 11 weeks
cost=10800+600=11400$
c.p=34 weeks
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Crash activity choose the small value in slope in cp path
Crashing activity (
f
) (1 ) week
Duration = 9 weeks
cost=11400+1500=12900
c.p=33 week
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Crash activity choose the small value in slope in cp path
crashing activity ( F
) ( 1 ) week
cost=12900+ 1500
=14400$
Duration = 8weeks
c.p=32 weeks
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EX:- using table below of activities assume the following normal duration
/cost and crash duration/cost
Draw the network and find the cp and crash the project (4 weeks) and
determine the total cost.
Activity
path
1-2
1-3
2-4
2-5
3-4
4-5
Solution
Normal
Normal
time(weeks) cost
($)
8
100
4
150
2
50
10
100
5
100
3
80
Crash
time
(weeks)
6
2
1
5
1
1
Crash
cost
($)
200
350
90
400
200
100
activity
Normal
Normal
time(weeks) cost
($)
Crash
time
(weeks)
Crash
cost
($)
Allowable Slope(cost/time)=
crash time cc-cn/tn-tc
=tn-tc
1-2*
1-3
2-4
2-5*
3-4
4-5
8
4
2
10
5
3
6
2
1
5
1
1
200
350
90
400
200
100
2
2
1
5
4
2
100
150
50
100
100
80
50
100
40
60
25
10
∑580$
Slope for activity (1-2)= cc-cn/tn-tc
(200-100)/(8-6)=50
Slope for activity (1-3)= cc-cn/tn-tc
(350-150)/(4-2)=100
Slope for activity (2-4)= cc-cn/tn-tc
(90-50)/(2-1)=40
Slope for activity (2-5)= cc-cn/tn-tc
(400-100)/(10-5)=60
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Slope for activity (3-4)= cc-cn/tn-tc
(200-100)/(5-1)=25
Slope for activity (4-5)= cc-cn/tn-tc
(100-80)/(3-1)=10
8
Paths
1-2--5=18 c.p (critical path
1-2-4-5=13
1-3-4-5=12
Crash activity choose the small value in slope in cp path
First crashing activity (1-2) (2) week
cost= 580+2*50
Duration =6 weeks
Cost=680$
The cp path after first crashing (1-2-5=16)
Other path 1-2-4-5=11
1-3-4-5=12
Draw after crash (1-2) two week
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Second crashing activity (2-5) (2) week
cost= 680+2*60
Cost=800$
The cp path after first crashing (1-2-5=14)
Other path 1-2-4-5=11
1-3-4-5=12
Draw after crash (2-5) two week
8
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HW
EX: - using table below of activities assume the following normal duration
/cost and crash duration/cost
Draw the network and find the cp and crash the project (2 days) and
determine the total cost.
Activity
Preceded by
Crash
Normal
A
B
C
D
E
F
G
A,B
C
C
C
D,E,F
Time(day)
10
12
11
5
8
9
8
41
Cost($)
5000
1200
3600
300
1000
2400
700
Time(day)
10
11
9
4
6
7
7
Cost($)
5000
1300
4800
600
2000
5400
1000
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Cost Account Methods
(Break Even Analysis Method)
For
Students of Fourth Stage
Computer Department
Department of computer Engineering
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1. Overview
a. Target Population: For students of stage of second and fourth stage in technical
Colleges and institutes in foundation of technical education.
b. Rationale: Costing means looking at the amount to be spent on selling a product,
running a production process or delivering a service. This mode unit introduces of
classification of costs according variability in order to select the suitable method to
determine the selling price.
c. Central Ideas: The basic purpose of classification of costs to determine the most
suitable method of accumulating and allocating costs.
d. Objectives: The student will be able after finishing lecture on:
-
Classify the cost according the variability.
-
Determine the BEP and its effect on the costs.
2. Pre-Test:
1. The examples of semi-variable costs include; ---------, -------- and ---------.
2. The ------------ method is used to calculate the cost according the variability.
3. Material cost is considered from variable costs. State True or False
4. Define the term „Fixed cost‟.
5. For the following data:
Fixed cost = 400,000 ID
Variable cost per unit = 20 ID
Selling price per unit=100 ID
Estimating profit = 300,000 ID
a- Calculate sales value.
b- Calculate the contribution if the total sales 950,000 ID.
Note: Check your answers in “Answer Keys” in end of mode unit. If you obtain 75% of
solution, you cannot need to this mode unit. If your answer is poor, you will transfer to
next page.
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3. Theory:
Introduction
The cost is classified by variability to:
1. Variable Costs
-
Variable Costs: are costs which do vary directly with the level of output or
production. Sometime are called direct costs.
-
Costs that change in total, directly in proportion to changes in the level of activities
(volume).
-
The unit cost remains the same over a wide range of volume (referred to as the
relevant range).
-
Relevant Range is the range of activity (production volume) within which variable
unit costs are constant and fixed costs are constant and fixed costs are constant in total.
In this range, the incremental cost of one additional unit of production is the same.
-
Examples include direct materials, direct labor, and part of manufacturing overhead.
2. Fixed Costs
-
Fixed Costs: are costs which do not vary directly with the level of output or
production. Sometime are called indirect costs.
-
Costs that do not change in total regardless of changes in activity.
-
The unit cost decreases as volume increases.
-
Examples include rent, taxes, and insurance on manufacturing plant.
3. Semi-variable (Mixed) Costs
-
Costs that contain both variable and fixed costs.
-
Examples include: light, heat, and power.
Breakeven Point: It is the relationship between a variable cost, fixed cost and volume of
production (sales). No profit or loss at the breakeven point.
Benefits of BEP
1. Forecasting profit at any volume of sales.
2. Giving clear image about relationship between costs and sales per unit.
3. Help management to make decisions.
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4- Self-Test:
1. The technique used a variable and a fixed cost is called -----------.
2. Define the term „variable cost‟.
Breakeven Point (BEP) Methods
1. Equation method
- Operating profit = Sales - Total Fixed Costs - Total Variable Costs
(Quantity x Unit Selling Price) (Quantity x Unit Variable Cost)
Because at the breakeven point operating profit is zero, the equation can be as follows:
- Sales = Total Fixed Costs + Total Variable Costs
Where
is a fixed cost,
is a variable cost and
is total sales.
Ex1:For the following data:
Fixed cost = 100,000 ID
Variable cost per unit = 300 ID
Selling price per unit=500 ID
1. Find number of unit (Units of Production) to verify BEP.‫جد عدد الىحداث الىاجب بيعها لتحقيق وفطت التعادل‬
2. Find BEP. ‫جد وقطت التعادل‬
3. Find number of unit to obtain on operating profit of 1,000,000 ID. ‫جد عدد الىحداث الىاجب بيعها لتحقيق‬
‫مليىن ديىار ربح‬
4. Find the operating profit at production and selling 2,000 units.‫جد الربح المتحقق عىد اوتاج وبيع الفيه وحدة‬
Ans:
1. Operating profit = Sales - Total Fixed Costs - Total Variable Costs
Operating profit = 0 at BEP
Sales = Total Fixed Costs + Total Variable Costs
2.
3. Operating profit = Sales - Total Fixed Costs - Total Variable Costs
4. Operating profit = Sales - Total Fixed Costs - Total Variable Costs
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(
)
3. Manufacturing company sells coats at 100 $ per unit .The variable cost is 30 $ per unit and total
fixed are 21,000 $ require.
a. Find Units of Production.
b. What is the breakeven point?
To calcu late sales value )‫ (قيمت المبيعاث‬at giving estimating profit, we used the following equation:
Where
is estimating profit.
To calculate contribution )‫(المساهمت‬, we used the following equation:
(
Where
is contribution,
is total sales and
)
is selling price per unit.
Ex2: For the following data:
Fixed cost = 400,000 ID
Variable cost per unit = 20 ID
Selling price per unit=100 ID
Estimating profit = 300,000 ID
a- Calculate sales value.
b- Calculate the contribution if the total sales 950,000 ID.
Ans:
1.
2.
(
)
(
)
2. Chart method:
Ex3: For the following data:
Fixed cost = 40,000 ID
Variable cost = 20,000 ID
Sales =100,000 ID
Calculate BEP by using chart method.
Ans:
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4. To calculate contribution, the following equation used
. State True or False
5. Draw the BEP chart with labeling.
5. Post- Test
1.
2.
3.
4.
5.
a.
b.
c.
d.
Define the term “BEP”.
Examples of variable cost include rent, taxes, and insurance on manufacturing plant.
Costs that contain both variable and fixed costs are called ----------.
List benefits of BEP method.
For the following data:
Fixed cost = 60,000 ID
Variable cost = 100,000 ID
Sales =200,000 ID
Calculate BEP
Determine BEP at increasing in selling price 20% of sales value.
Determine BEP at increasing 10% of fixed cost.
Determine
BEP atyour
decreasing
20%
variableKeys”
cost. in end of mode unit.
Note: Check
answers
in of
“Answer
Note: Check your answers in “Answer Keys” in end of mode unit.
6. References
1. J.R. Meredith and S.J. Mantel “Project Management”, J. Wiley & Sons, 1995.
2. http://www.projectmanagement.com/main.htm
49
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3. Mike Holt, “Applying Overhead and Determining Break-Even Cost” Mike Holt
Enterprises, Inc,2001, www.ecmweb.com
50
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Answer Keys
Pre- Test
1.
2.
3.
4.
light, heat, and power.
BEP method.
True.
Fixed Costs: are costs which do not vary directly with the level of output or production.
Some time are called indirect costs.
5.
(
)
(
)
Self-Test
1. breakeven point
2. Variable Costs: are costs which do vary directly with the level of output or production.
Sometime are called direct costs.
3.
Ans:
1.
This means that to cover 21,000 $ of fixed costs, 300 units must be sold to break even.
2.
OR
4. False.
5.
51
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Post- Test
1. Breakeven Point: It is the relationship between a variable cost, fixed cost and volume of
production (sales). No profit or loss at the breakeven point.
2. False.
3. mixed costs
4.
- Forecasting profit at any volume of sales.
- Giving clear image about relationship between costs and sales per unit.
- Help management to make decisions.
5.
(
)
(
(
)
)
52
Limitation of Graphical Method
Graphical solution is limited to linear
programming models containing only two
decision variables.
Procedure




Step I: Convert each inequality as equation
Step II: Plot each equation on the graph
Step III: Shade the ‘Feasible Region’. Highlight the
common Feasible region.
 Feasible Region: Set of all possible solutions.
Step IV: Compute the coordinates of the corner points
(of the feasible region). These corner points will
represent the ‘Feasible Solution’.
 Feasible Solution: If it satisfies all the constraints
and non negativity restrictions.
Procedure (Cont…)

Step V: Substitute the coordinates of the corner points into the
objective function to see which gives the Optimal Value. That
will be the ‘Optimal Solution’.
 Optimal Solution: If it optimizes (maximizes or minimizes)
the objective function.
 Unbounded Solution: If the value of the objective function
can be increased or decreased indefinitely, Such solutions
are called Unbounded solution.
 Infeasible (Inconsistent) Solution: It means the solution of
problem does not exist. This is possible when there is no
common feasible region.
Example
Max
s.t.
x2
8
7
z = 5x1 + 7x2
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 , x2 > 0
6
5
4
Every point is in this nonnegative quadrant
3
2
1
1
2
3
4
5
6
7
8
9
10
x1
Example (Cont…)
Max
s.t.
x2
8
7
6
x1 < 6
5
4
3
2
(6, 0)
1
1
2
3
4
5
6
7
8
9
10
x1
z = 5x1 + 7x2
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 , x2 > 0
Example (Cont…)
Max
s.t.
x2
z = 5x1 + 7x2
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 , x2 > 0
(0, 6.33)
8
7
2x1 + 3x2 < 19
6
5
4
3
2
(9.5 , 0)
1
1
2
3
4
5
6
7
8
9
10
x1
Example (Cont…)
x2
Max
s.t.
(0, 8)
8
7
x1 + x2 < 8
6
5
4
3
2
(8, 0)
1
1
2
3
4
5
6
7
8
9
10
x1
z = 5x1 + 7x2
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 , x2 > 0
Example (Cont…)
Max
s.t.
x2
x1 + x2 < 8
8
z = 5x1 + 7x2
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 , x2 > 0
7
x1 < 6
6
5
4
3
2x1 + 3x2 < 19
2
1
1
2
3
4
5
6
7
8
9
10
x1
Example (Cont…)
x2
8
7
6
(0,6.33)
E
5
4
3
(5,3)D
2
(6,2)C
1
A (0,0)
1
(6,0)B
2
3
4
5
6
7
8
9
10
x1
Example (Cont…)
Objective Function : Max Z= 5x1+7x2
Corner Points
A – (0,0)
B – (6,0)
C – (6,2)
D – (5,3)
E – (0,6.33)
Optimal Point : (5,3)
Optimal Value : 46
Value of Z
0
30
44
46
44.33
Example
Max Z=3 P1 + 5 P2
s.t.
P1
< 4
P2 < 6
3 P1 + 2 P2 < 18
P1, P2
>0
Example
Max Z=3 P1 + 5 P2
s.t.
P1
< 4
P2 < 6
3 P1 + 2 P2 < 18
P1, P2
>0
P2
Every point is in this nonnegative quadrant
P1
0
Example (Cont…)
Max Z= 3 P1 + 5 P2
s.t.
P1
< 4
P2 < 6
3 P1 + 2 P2 < 18
P1, P2
>0
P2
(0,0)
(4,0)
P1
Example (Cont…)
Max Z= 3 P1 + 5 P2
s.t.
P1
< 4
P2 < 6
3 P1 + 2 P2 < 18
P1, P2
>0
P2
(0,6)
(0,0)
(4,0)
P1
Example (Cont…)
Max Z= 3 P1 + 5 P2
s.t.
P1
< 4
P2 < 6
3 P1 + 2 P2 < 18
P1, P2
>0
P2
(0,6)
(2,6)
(4,3)
(6,0)
(0,0)
(4,0)
P1
Example (Cont…)
P2
(0,6)
E
(2,6)
D
C
A
(0,0)
(4,3)
(6,0)
B
(4,0)
P1
Example (Cont…)
Objective Function : Max Z= 3 P1 + 5 P2
Corner Points
A – (0,0)
B – (4,0)
C – (4,3)
D – (2,6)
E – (0,6)
Optimal Point : (2,6)
Optimal Value : 36
Value of Z
0
12
27
36
30
Example
Min z = 5x1 + 2x2
s.t.
2x1 + 5x2 > 10
4x1 - x2 > 12
x1 + x2 > 4
x 1, x 2 > 0
Example
Min z = 5x1 + 2x2
s.t.
2x1 + 5x2 > 10
4x1 - x2 > 12
x1 + x2 > 4
x1, x2 > 0
x2
5
4x1 - x2 > 12
x1 + x2 > 4
4
3
2x1 + 5x2 > 10
2
1
1
2
3
4
5
6
x1
Example (Cont…)
Min z = 5x1 + 2x2
s.t.
2x1 + 5x2 > 10
4x1 - x2 > 12
x1 + x2 > 4
x1, x2 > 0
x2
5
Feasible Region
4
This is the case of
‘Unbounded Feasible Region’.
3
2
A (35/11 , 8/11)
1
B (5,0)
1
2
3
4
5
6
x1
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Example: - by using the (Graphical method) find the optimum solution of
X1, X2.
Max Z=6X1+4X2 Sub in to
X1+3X2≤2
2.5
4X1+2X2≤5
X1,X2≥0
Solution: 1-
In Eq .
at X1=0
at X2=0
B(0,0.667) 0.667
X1+3X2=2
X2=0.667 (0,0.667)
X1=2 (2,0)
feasible regin
2- In Eq.
at X1=0
at X2=0
4X1+X2=5
X2=2.5
(0,2.5)
X1= 1.25 ( 1.25,0)
A(0,0 )
D
1.25 2
c(1.25,0)
to find point D
( X1+3X2=2)*4
4 X1+2X2=5
(1)
(2)
+4 X1+12X2= 8
-4X1 - 2 X2 = -5
X2=0.3
point
A
B
C
D
sub in any eq 1or 2 X1=1.1 D=(1.1,0.3)
X1
X2
Z=6X1+4X2
0
0
Z=0
0
0.667
Z=2.668
1.25
0
Z=7.5
1.1
0.3
Z=7.8
optimum solution in point D
LINEAR PROGRAMMING
PROBLEM
Linear : form meant a mathematical expression of the type
a1x1 + a2x2 + ……….+ anxn
where a1, a2, ….., an are constants,
and x1, x2, ………, xn are variables.
Programming : refers to the process of determining
a particular program or plan of action.
Linear Programming Problem(LPP): Technique for
optimizing(maximizing/minimizing) a linear function
of variables called the ‘OBJECTIVE FUNCTION’
subject to a set of linear equations and/or inequalities
called the ‘CONSTRAINTS’ or ‘RESTRICTIONS’.
FORMULATION OF
LP PROBLEMS
LP Model Formulation
Objective function
a linear relationship reflecting the objective of
an operation
most frequent objective is to maximize profit or
to minimize cost.
Decision variables
an unknown quantity representing a decision
that needs to be made. It is the quantity the
model needs to determine
Constraint
a linear relationship representing a restriction on
decision making





Steps in Formulating the
LP Problems
1.
2.
3.
4.
5.
Define the objective. (min or max)
Define the decision variables.
Write the mathematical function for the objective.
Write the constraints.
Constraints can be in <, =, or > form.
Example
Two products: Chairs and Tables
Decision:
How many of each to make this month?
Objective: Maximize profit
Data
Profit
Contribution
Tables
Chairs
(per table) (per chair)
$7
$5
carpentry
3 hrs
4 hrs
Hours
Available
2400
Painting
2 hrs
1 hr
1000
Other Limitations:
• Make no more than 450 chairs
• Make at least 100 tables
Solution
Decision Variables:
T = Num. of tables to make
C = Num. of chairs to make
Objective Function: Maximize Profit
Maximize $7 T + $5 C
Constraints

Have 2400 hours of carpentry time available
3 T + 4 C < 2400

(hours)
Have 1000 hours of painting time available
2 T + 1 C < 1000
(hours)
More Constraints:
Make no more than 450 chairs
C < 450
Make at least 100 tables
T > 100
Non negativity:
Cannot make a negative number of chairs or tables
T>0
C>0
Model
Max 7T + 5C
Subject to the constraints:
3T + 4C < 2400
2T + 1C < 1000
C < 450
T > 100
T, C > 0
General Formulation of LPP
Max/min
z = c1x1 + c2x2 + ... + cnxn
subject to:
a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1
a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2
:
am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm
x1 ≥ 0, x2 ≥ 0,…….xj ≥ 0,……., xn ≥ 0.
xj = decision variables
bi = constraint levels
cj = objective function coefficients
aij = constraint coefficients
Example
Cycle Trends is introducing two new lightweight bicycle
frames, the Deluxe and the Professional, to be made from
aluminum and steel alloys. The anticipated unit profits
are $10 for the Deluxe and $15 for the Professional.
The number of pounds of each alloy needed per
frame is summarized on the next slide. A supplier
delivers 100 pounds of the aluminum alloy and 80 pounds
of the steel alloy weekly. How many Deluxe and
Professional frames should Cycle Trends produce each
week?
Example
Pounds of each alloy needed per frame
Deluxe
Professional
Aluminum Alloy
2
4
Steel Alloy
3
2
Solution
Define the objective
Maximize total weekly profit
Define the decision variables
x1 = number of Deluxe frames produced weekly
x2 = number of Professional frames produced
weekly
Solution
Max Z = 10x1 + 15x2
Subject To
2x1 + 4x2 < 100
3x1 + 2x2 < 80
x1 , x2 > 0
Example
A firm manufactures 3 products A, B and C. The profits are Rs.3, Rs.2,
and Rs.4 respectively. The firm has 2 machines and below is the
required processing time in minutes for each machine on each product.
Product
A
Machines
B
C
G
4
3
5
H
2
2
4
Machine G and H have 2000 and 2500 machine-minutes
respectively. The firm must manufacture 100 A’s, 200 B’s and 50
C’s, but not more than 150 A’s. Set up an LP problem to maximize
profit.
Solution
Define the objective
Maximize profit
Define the decision variables
x1 = number of products of type A
x2 = number of products of type B
x3 = number of products of type C
Solution
Max Z = 3x1 + 2x2 + 4x3
Subject To
4x1 + 3x2 + 5x3 ≤ 2000
2x1 + 2x2 + 4x3 ≤ 2500
100 ≤ x1 ≤ 150
x2 ≥ 200
x3 ≥ 50
x1, x2, x3 ≥ 0
Example
The Sureset Concrete Company produces concrete.
Two ingredients in concrete are sand (costs $6 per
ton) and gravel (costs $8 per ton). Sand and gravel
together must make up exactly 75% of the weight of
the concrete. Also, no more than 40% of the
concrete can be sand and at least 30% of the
concrete be gravel. Each day 2000 tons of concrete
are produced. To minimize costs, how many tons of
gravel and sand should be purchased each day?
Solution
Define the objective
Minimize daily costs
Define the decision variables
x1 = tons of sand purchased
x2 = tons of gravel purchased
Cont…
Min Z = 6x1 + 8x2
Subject To
x1 + x2 = 1500
x1 < 800
x2 > 600
x1 , x2 > 0
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Solve problem of LP by using (formulation, graphical method,
and simplex method)
EX: - 1- you wish to produce two products
i)
ii)
Walkman Am/Fm cassette.
Watch Tv.
2- W.m takes 4 hours of electronic works and 2 hours
assembly.
3- W.T takes 3 hours of electronic works and 1 hours
assembly.
4- There are 240 hours of electronic works time and 100
hours of assembly time available.
5-profit on a W.m 7$ and on a T.v 5$? Find the max profit
Sol: -
Formulation of Lp
Max Z=7X1+5X2 Sub in to
4X1+3X2≤240
2X1+X2≤100
X1,X2≥0
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Graphical method
Max Z=7X1+5X2 nSub to
4X1+3X2≤240
100
2X1+X2≤100
X1,X2≥0
1-
B(0,80)
In Eq .
at X1=0
at X2=0
80
4 X1+3X2=240
X2= 80 (0,80)
X1= 60 (60 ,0)
feasible regin
2- In Eq.
at X1=0
at X2=0
2X1+1X2=100
X2=100 (0,100)
X1=50 (50,0)
D
50 60
c( 50,0)
to find point D
4 X1+3X2=240
( 2X1+ X2=100)*2
(1)
(2)
+4 X1+3X2= 240
-4X1 - 2 X2 = -200
X2=40
sub in any eq 1or 2 X1=30 D=(30,40)
point
X1
X2
Z=7X1+5X2
A
0
0
Z=0
B
0
80
Z=400
C
50
0
Z=350
D
30
40
Z= 410
optimum solution in point D
Z=410,x1=30,x2=40
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Simplex method
Max Z=7X1+5X2 Sub in to
4X1+3X2≤240
2X1+X2≤100
X1,X2≥0
Sol:Z-7X1-5X2=0
4X1+3X2+S1=240
2X1+X2+S2=100
X1,X2,s1,s2≥0
At X1=0,X2=0
Variable Z
Z
1
S1
0
S2
0
S1=240 ,S2= 100 ,Z=0
X1
X2
S1
S2
-7
-5
0
0
4
3
1
0
2
1
0
1
Sol.
0
240
100
Ratio
‫تهمل‬
60
50
variable Z
X1
X2
S1
S2
Sol.
Ratio
Z
1
0
-1.5
0
3.5
350
‫تهمل‬
S1
0
0
1
1
-2
40
40
X1
0
1
1/2
0
1/2
50
100
1-(-7*0)/2=1,0-(4*0)/2=0,-5-(-7*1)/2=-5+3=-1.5,3-(4*1)/2=3-2=1
0-(-7*0)/2=0,1-(4*0)/2=1,0-(-7*1)/2=3.5,0-(4*1)/2=-2,0-(7*100)/2=350,240-(4*100)/2=40
1-(-1.5*0)/1=1,0-(0*0.5)/1=0,0-(-1.5*1)/1=1.5,3.5-(-1.5*-2)/1=0.5
350-(-1.5*40)/1=410,1-(0*1/2)/1=1,0-(0.5*1)
/1=-0.5,0.5-(-2*0.5)/1=0.5+1=1.5,50-(40*0.5)/1=30
variable Z
X1
Z
1
0
X2
0
0
X1
0
1
Z=410,x1=30,x2=40
X2
0
1
0
S1
1.5
1
-0.5
S2
1/2
-2
1.5
Sol.
410
40
30
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Engineering economics
For
Students of Fourth Stage
Computer Department
Department of computer engineering
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Engineering economics
Engineering economics: - is branch of economics used by
engineering to optimize their designs and construction
projects.
The following symbol and definitions
P: - principle, a sum of money invested in initial year, or
The present sum of money.
i :- interest rate per of unit of time expressed as a
decimal ex:- 5% = 0.05
n:- time , the number of units of times over which inters
Accumulates.
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I: - simple interest
F: - compound amount, a sum of money at end of n
units at interest i .
Simple interest
Simple interest (I) = p i n
Example:- an amount of 2500$ is deposit in a bank
offering 5% simple interest per annum what is the
interest at the end of the first year?
Solution:P= 2500$
i=0.05
n= 1
Simple interest (I ) = p i n
= 2500 * 0.05 * 1
= 125 $
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Compound interest
F= p(1+i)n
P= f/ (1+i)n
F=P(1+ )nm if interest i compound m times per period
n
Where
m= 2 if compound semiannually
m=4 if compound quarterly
m= 12 if compound monthly
m= 365 if compound daily
Example1:- if you deposit 4000$ into an account paying
6% annual interest compound quarterly how much will
be in the account after 5 years?
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Solution:P= 4000$
i=0.06
n= 5
m=4
F=p (1+ )nm
= 4000( 1+0.06/4)(5)(4)
F= 5387.42$ in the account
Example2:- if you deposit 1000$ into an account paying
12% annual interest compound annually how much will
be in the account after 4 years?
Solution:P=1000$
F=p(1+i)n
F=1000(1+0.012)4
F=1573.5$
i=12%
n=4 years
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EX3:- A business man need to have 100,000$ in 5 years
.How much must be put into his account in the bank
,10% annual interest assuming compound interest.
Solution:F=100,000$
i=10%
n=5 year
F=p (1+i)n
100,000=p (1+0.1)5
p=62092.13$
EX4:-what the interest rate a compound semiannually
will have to be earned in order for 500$ to amount to
1104 $ in ten years?
Solution:P= 500$
i=?
F=p (1+ )nm
( )1/nm = (1+ )nm/nm
(
)1/nm= (1+ )
n= 10
m=2 F=1104$
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)1/nm -1= i/m
(
(
)1/20 -1= i/2
i=0.0808, i=8.08%
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HW
HW 1:- if you deposit 8000$ into an account paying 9%
annual interest compound semiannually how much will
be in the account after 7 years?
HW 2:- an amount of 3000$ is deposit in a bank offering
8% simple interest per annum what is the interest for
the 5year?
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Layouts
For
Students of Fourth Stage
Computer Department
Department of computer Engineering
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Inventory
For
Students of Fourth Stage
Computer Department
Department of computer Engineering
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By the end of this topic, you should be able to:
1. Explain the importance of inventory control;
2. Compute the EOQ to determine how much to order
3. Compute the ROP in determining
► INTRODUCTION
Inventory means any stored resource that is used to satisfy a current
or a future need; for example, the raw materials and finished goods.
Therefore the inventory control is crucial for every company.
Reducing on-hand inventory level means reducing costs and thus
increases the cash flow. On the other hand frequent stock outs may
dissatisfy the customers. Thus a company must make a balance
between low and high inventory levels. The major factor to be
considered in achieving this balance is the cost minimization.
INVENTORY DECISIONS
There are two fundamental decisions that we have to make
when controlling inventory:
1. How much to order or produce?
2. When to order?
The major objective in controlling inventory is to minimize the
total inventory costs which include:
1. Cost of the items (purchase cost or material cost) (Cp):- : it
is usually the purchase price of the item under consideration. If
unit cost is related with the purchase quantity, it is called as
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discount price.
2. Cost of ordering (Co):- This includes the cost of order
preparation, tender placement, cost of postages, telephone
costs, receiving costs, set up cost etc.
3. Cost of carrying or holding inventory (Ch):- This represents
the cost of maintaining inventories in the plant. It includes the
cost of insurance, security, warehouse rent, taxes, interest on
capital engaged, spoilage, breakage etc.
4. Cost of stock outs. Or Shortage costs (Cs): This represents the cost
of loss of demand due to shortage in supplies. This includes cost of loss
of profit, loss of customer, loss of goodwill, penalty etc.
Economic order quantity (EOQ) with no shortage
This is the basic model in achieving an optimal ordering
quantity which Minimizes the total inventory costs. However
there are some assumptions to be upheld:1- Demand is deterministic and occurs at constant rate.
2- The unit purchase cost does not depend on the size of the
order. In other words the model does not permit quantity
discount.
3 ■ the lead time for each order is zero. Each order arrives as
soon as it is placed.
No shortages are allowed. That is all demands must be met on
time; a negative inventory is not allowed either.
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Fig
ure
1:
Inv
ent
ory usage over time in a simple model
The following mathematical equations
√
Where:Annual Demand = D
Ordering cost per order = Co
Holding or carrying cost per unit per year = Ch
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√
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Example (1): suppose the demand for a product is 6000 unit per year
and the items are withdrawn uniformly. The order cost is S1000. The
inventory holding cost is $2 per item per year. Assuming shortages are
not allowed, find:
(1)
(2)
(3)
(4)
The economic order quantity.
The number of ordering during the year (order frequency).
The interval between two orders (time between orders).
The annual cost.
Answer:√
1-
√
=2449.48≈2449 unit
2-
orders
3-
year
4-
√
√
=4898.97$
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Example (2): Suppose that the demand for a product is 2000 per year,
and the items are withdrawn uniformly. The order cost is $150 and the
inventory holding cost is $0.5 per item per month.
Assuming shortages are not allowed, find:
(1)
(2)
(3)
(4)
Tne economic order quantity.
The number of ordering during the year ( order frequency).
The interval between two orders (time between orders ),
The annual cost.
Answer:√
1-
√
=316.227≈316 unit
2-
orders
3-
4-
year
√
√
=3600000$
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Economic order quantity (EOQ) with shortage;
The following mathematical equations
(
)2
(
)
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Example (3): Suppose that the demand for a product is 4000 per year,
and the items are withdrawn uniformly. The order cost is $250 and the
inventory holding cost is $5 per item per year. If shortages cost $10 per
item per year, find:
(1) The economic order quantity.
(2) Maximum inventory.
(3) Shortage allowable quantity.
(2) The number of ordering during the year ( order frequency).
(3) The interval between two orders (time between orders ).
(4) The annual cost.
Answer:-
√
1-
√
=774.597 775 unit
2-
=
=516.6≈517 unit
3-
=258.3 ≈ 258 unit
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4=
=5.16≈ 5 order
5=
=0.193 year
6-
= 2581$
(
)2
(
)2
(
)
(
)
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Re-order Level or Ordering Point or Ordering Level
This is that level of materials at which a new order for supply of materials
is to be placed. In other words, at this level a purchase requisition is made
out. This level is fixed somewhere between maximum and minimum levels.
Order points are based on usage during time necessary to requisition
order, and receive materials, plus an allowance for protection against stock
out.
The order point is reached when inventory on hand and quantities due in
are equal to the lead time usage quantity plus the safety stock quantity.
According to Fangruo Chen, the ROP quantity reflects the level of
inventory that triggers the placement of an order for additional units.
Whereas, the quantity associated with safety stock protects the company
from stock outs or backorders. Safety stock is also known as a "buffer”
In determining the reorder point the following three
factors need to be at hand:
1. Demand - Quantity of inventory used or sold each day
2. Lead Time - Time (in days) it takes for an order to
arrive when an order is placed.
3. Safety Stock - The quantity of inventory kept on hand
in case there is an unpredictable event like delays in
lead time or unexpected demand.
Formula of Re-order Level or Ordering Point:
The following two formulas are used for the calculation of
reorder level or point.
Ordering point or re-order level = Maximum daily or weekly
or monthly usage x Lead time
The above formula is used when usage and lead time are known with
certainty; therefore, no safety stock is provided
-
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When safety stock is provided then the following formula
will be applicable:
Ordering point or re-order level = Maximum daily or
weekly
or monthly usage x Lead time + Safety stock
Examples:
Example 1:
Minimum daily requirement Time
required to receive emergency
supplies
Average daily requirement
Minimum daily requirement Time
required for refresh supplies
800 units
4 days
700 units
600 units
One month (30
days)
Calculate ordering point or re-order level Calcul
ation:
Ordering point = Ordering point or re-order level =
Maximum daily or weekly or monthly usage x Lead time
= 800 x 30
= 24,000 units
-
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Example 2:
Two types of materials
are used as
Minimum usage
20 units per
week each 40
units per
Maximum usage
week each 60
units per week
each
Normal usage
Re-order period
time Material A:
Material B
or
Lead
3 to 5 weeks 2
to 4 weeks
Calculate re order point for two types of materials.
Calculation:
Ordering point or re-order level = Maximum daily or weekly
or monthly usage x Maximum re-order period
A: 60 x 5 = 300 units
B:60x 4=240 units
-
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-
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( Productivity )
Productivity is commonly defined as a ratio between the output
volume and the volume of inputs. In other words, it measures how
efficiently production inputs, such as labor and capital, are being
used in an economy to produce a given level of output.
Productivity=
SAMPLE PROBLEMS FOR PRODUCTIVITY
Example 1
A company that processes fruits and vegetables is able to produce
400 cases of canned peaches in one half hour with four workers.
What is the labor productivity?
Solution:
Labor productivity = Quality Produced / Labors Hours
= 400 cases/ (4 workers x 1/2 hours / workers)
= 200 cases per labor hour
Example 2
A wrapping paper company produced 2,000 rolls of paper one day.
Standard price is $ 1/roll. Labor cost was $ 160, material cost was $
50, and overhead was $ 320. Determine the multifactor productivity.
Solution:
Multifactor productivity =
Quality produced at standard price/ (Labor cost + Material cost +
Overhead)
= 2,000 rolls x $ 1/ ($160+ $ 50 + $320)
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= 3.77 rolls output per dollars
Example 3
a) Find the productivity if four workers installed 720 square yards of
carpeting in Eight hours.
b) Compute for the productivity of a machine which produced 68
usable pieces in two Hours.
Solution:
a) Productivity = yards of carpeting install / Labors Hours worked
= 720 square yard / (4 workers x8 hours / worker)
= 720 yards / 32 Hours
= 22.5 yards/ hours
b) Productivity = Usable Pieces / Production Time
= 68 usable pieces / 2 hrs.
= 34 pieces/ hours
Example # 4
Determine the multifactor productivity for the combined input of the
labor and the machine time using the following:
Input:
Labor: $ 1,000
Materials: $ 520
Overheads: $ 2,000
Keep in mind the Production is 1760 unit
Solution:
Multifactor Productivity = Output / (Labor + Materials +
Overheads)
= 1,760 Units / ($ 1,000 + $ 520 + $ 2,000)
= 0.50 units/$
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Solve the Following Problems
Problem
Company has a staff of 4, each working 8 hours per day (for a
payroll cost of $ 640 / day) and overhead expenses of $ 400 / day.
Company Processes and closes on 8 outputs each day.
The company recently purchased a new system that will allow the
processing of 14 outputs per day. Although the staff, their works
hours, and pay will be same, the overheads expenses are now $ 800
per day.
Solution:
Labor productivity with the old system:
= 8 output per day/ 32 labor hours = 0.25 output per labor hour
Labor productivity with the new system:
=14 output per day/ 32 labor hours = 0.44 output per labor hours
Multifactor productivity with the old system:
=8 output per day / (640 + 400) = 0.0077 output per dollars
Multifactor productivity with the new system:
=14 output per day / (640 + 800) = 0.0097 output per dollars
Problem
A company producer of an apple boxes sold to growers has been able,
with his current equipment, to produces 240 boxes per 100 logs. He
currently purchases 100 logs per day, and each logs required 3 labor
hours to process. He believes that he can hire a professional buyer
who can buy a better quality log at the same cost. If this is the case,
he increases his production to 260 boxes per 100 logs. His labor hours
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will increase by 8 hours per day. What will be the impact on
productivity (measured in crates per labor –hour) if the buyers is
hired? What is the Growth in productivity in this case?
Solution:
a) Current labor productivity = 240 boxes / 100 logs (3 hours pert
log)
= 240/ 300
= 0.8 boxes per labor hour
b) Labor productivity with buyer = 260 boxes / (100 logs (3 hours per
logs) + 8 hours )
= 260 / 308
= 0.844 boxes per labor hours
c) Growth = (0.844 – 0.8)/0.8 x100 =
Problem
Calculate the productivity for the following operations:
a) Three employees processed 600 insurance policies last week.
They 8 hours per day, 5 days per week.
b) A team of workers made 400 units of product, which is valued by
its standard cost of $10 each (before markups for other expenses and
profit). That accounting department reported that for this job the
actual cost were $ 400 per labor, $1000 for materials and
$300 for overhead:
Solution:
a) Labor productivity = Policies processed/
Employee hours
= 600 policies /3 (40)
= 5 policies per hours
b) Multifactor productivity = Quality at standard cost
Labor + Materials + Overheads
= 400units ($10/units)/( $400 + $1000 + $ 300)
= $4000/$1700
=2.35
Problem
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Student tuition at Boering University is $ 100 per semester credit
hours. The states supplement school revenue by matching student
tuition, 100dollars. Average class size for typical three credit course
is 50 students. Labor costs are $4000 per class, material costs are $20
per student, and overhead cost are $25,000 per class.
Find:
a) What is the multifactor productivity ratio?
b) If instructors work an average, what is the labor productivity
ratio? (Keep in mind that professor delivering the lecture work
14 hours per week the semester last for 16 weeks)
Solution:
a) Value of Output = (50 student) x (3 credit hours) x ($ 100 tuition +
$ 100 state support)
Class student credit hours = $ 30,000 per class
Value of input = Labor + Materials + Overheads
= $ 4000 + ($20 per student x 50 students) + $25,000 Class
= $ 30,000 per class
Multifactor productivity = Output/ Input
= $ 30,000 per class /$ 30,000 per class
= 1.00
b) Labor productivity is the ratio of the value of output to the labor
hours. The value of output is the same as in part (a), or $ 30,000 per
class, so Labor hours of input = 14 hours x 16 week = 224hours per
class
Labor productivity = Output/ Input
= $ 30,000 per class/ 224 hours per class
= $ 133.93 per hours
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Quality control is a process that measures output relative to
standard, and acts when output doesn't meet standards.
The purpose of quality control is to assure that processes are
performing in an acceptable manner.
Companies accomplish quality control by monitoring process output
using statistical techniques
This approach places an emphasis on three aspects
1
Elements such as controls, job management, defined and well
managed processes, performance and integrity criteria, and
identification of records
Competence,
qualifications
2
3
Soft
such
elements,
as
such
knowledge,
as
skills,
personnel,
experience,
integrity,
and
confidence,
organizational culture, motivation, team spirit, and quality
relationships.
Types of control charts
There are four types of control charts; two for variables, and two for
attributes
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Attribute: counted data (e.g., number of defective items in a sample,
the number of calls per day)
Variable: measured data, usually on a continuous scale (e g., amount
of time needed to complete a task, length, width, weight, diameter of
a part)
Variables Control Charts
Mean control charts
*Used to monitor the central tendency of a process.
*X-bar charts Range control charts
*Used to monitor the process dispersion *R charts
Control Chart for Attributes
Control charts for attributes are used when the process characteristic
is counted rather than measured. Two types are available:
•
P-Chart - Control chart used to monitor the proportion of
defectives in a process
•
C-Chart - Control chart used to monitor the number of defects
per unit
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P-Chart
Example 4:- an inspection counted the number of defective monthly
billing statements of a company telephone in each of 20 samples. Using
The following information, construct a control chart that will describe
99.74 percent of the chance variation in the process when the process
Is in control. Each sample contained 100 statements.(z=3)
Sample
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Number of defectives
4
10
12
3
9
11
10
22
13
10
8
12
9
10
21
10
8
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18
19
20
12
10
16
∑220
Solution:-
̅
(
̂
√
̅(
̅)
Control limit are
̅
̅
( ̂)
(
( ̂)
(
)
)
)
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C-chart
Example 4:- roll of coiled wire are monitored using c-chart eighteen rolls
have been examined and the number of defects per roll has been recorded
in the Following table. Is the process in control? Using three Standard
deviation control limits.z=3
Sample
1
2
3
4
5
6
7
8
9
10
11
12
13
Number
of defects
3
2
4
5
1
2
4
1
2
1
3
4
2
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14
15
16
17
18
4
2
1
3
1
∑45
̅
̅
√̅
√
=7.24
̅
=-2.24
√̅
√
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Employee Relationship Management
Employees are the major assets of an organization. It is essential that the employees
perform together as a collective unit and contribute equally towards the realization of a
common goal. No task can be accomplished if the individuals are engaged in constant
conflicts and misunderstandings. It has been observed that targets are achieved at a much
faster rate if the employees work together and share a warm relationship with each other.
.Employees must be comfortable with each other to deliver their best and enjoy their work
What is employee relationship management?
Employee relationship management refers to managing the relation between the various
employees in an organization. The relationship can be between employee and the employer
.as well as between employees at the same level
What is Management?
Management is nothing but a technique which brings people together on a common
platform and guides them so that they achieve their desired targets without fighting with
each other. In a layman’s language, management is nothing but managing things effectively
so that tasks are accomplished without any hassles and confusions. Management is required
.everywhere
Every individual goes for shopping. The moment you enter in an outlet, a sales person would
come to you and assist you in your shopping. He would try his level best to convince you and
guide you in selecting an outfit according to your taste as well as budget. The moment you
finalize something, you automatically would be directed to the billing section for the
monetary transactions. Your shopping basket in no time would reach the packing area where
the officials would nicely put the outfits in a smart carry bag flaunting the logo of the store.
Finally there would be a supervisor who would recheck your bill and thank you for your
.valuable time
How do you think such a smooth coordination is possible? Not a single moment, there was
any confusion. All this is possible through management. Every thing was well managed and
organized effectively to avoid confusions and meet the ultimate objective of the store ie
.selling the product as well as making the customer happy
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Employee relationship management is an art which effectively monitors and manages the
relation between individuals either of the same team or from different teams. Employee
relationship management activity helps in strengthening the bond among the employees
.and ensures that each one is contented and enjoys a healthy relation with each other
Employee relationship management includes various activities undertaken by the superiors
or the management to develop a healthy relation among the employees and extract the best
.out of each team member
Simplex Method
When decision variables are more than 2, we always use
Simplex Method
Slack Variable: Variable added to a  constraint to
convert it to an equation (=).
A slack variable represents unused resources
A slack variable contributes nothing to the objective
function value.
Surplus Variable: Variable subtracted from a 
constraint to convert it to an equation (=).
A surplus variable represents an excess above a
constraint requirement level.
Surplus variables contribute nothing to the calculated
value of the objective function.
Cont….
Basic Solution(BS) : This solution is obtained by setting
any n variables (among m+n variables) equal to zero and
solving for remaining m variables, provided the
determinant of the coefficients of these variables is nonzero. Such m variables are called basic variables and
remaining n zero valued variables are called non basic
variables.
Basic Feasible Solution(BFS) : It is a basic solution
which also satisfies the non negativity restrictions.
Cont…..
BFS are of two types:
Degenerate BFS: If one or more basic
variables are zero.
Non-Degenerate BFS: All basic variables are
non-zero.
Optimal BFS: BFS which optimizes the
objective function.
Example
Max. Z = 13x1+11x2
Subject to constraints:
4x1+5x2 < 1500
5x1+3x2 < 1575
x1+2x2 < 420
x1, x2 > 0
Solution :
Step 1: Convert all the inequality constraints into equalities
by the use of slack variables.
Let S1, S2 , S3 be three slack variables.
Introducing these slack variables into the inequality constraints
and rewriting the objective function such that all variables are on
the left-hand side of the equation. Model can rewritten as:
Z - 13x1 -11x2 = 0
Subject to constraints:
4x1+5x2 + S1 = 1500
5x1+3x2 +S2= 1575
x1+2x2 +S3 = 420
x1, x2, S1, S2, S3 > 0
Cont…
Step II: Find the Initial BFS.
One Feasible solution that satisfies all the
constraints is: x1= 0, x2= 0, S1= 1500,
S2= 1575, S3= 420 and Z=0.
Now, S1, S2, S3 are Basic variables.
Step III: Set up an initial table as:
Cont…
Row
NO.
Basic
Variable
Coefficients of:
Sol.
Rati
o
Z
x1
x2
S1
S2
S3
A1
Z
1
-13
-11
0
0
0
0
B1
S1
0
4
5
1
0
0
1500
375
C1
S2
0
5
3
0
1
0
1575
315
D1
S3
0
1
2
0
0
1
420
420
Step IV: a) Choose the most negative number from row A1(i.e Z row). Therefore,
x1 is a entering variable.
b) Calculate Ratio = Sol col. / x1 col. (x1 > 0)
c) Choose minimum Ratio. That variable(i.e S2) is a departing
variable.
Cont….
Step V: x1 becomes basic variable and S2 becomes non basic
variable. New table is:
Row
NO.
Basic
Varia
ble
Coefficients of:
Sol.
Ratio
Z
x1
x2
S1
S2
S3
A1
Z
1
0
-16/5
0
13/5
0
4095
B1
S1
0
0
13/5
1
-4/5
0
240
92.3
C1
x1
0
1
3/5
0
1/5
0
315
525
D1
S3
0
0
7/5
0
-1/5
1
105
75
Cont…
Next Table is :
Row
NO.
Basic
Variab Z
le
Coefficients of:
Sol.
x1
x2
S1
S2
S3
A1
Z
1
0
0
0
15/7
16/7
4335
B1
S1
0
0
0
1
-3/7
-13/7
45
C1
x1
0
1
0
0
2/7
-3/7
270
D1
x2
0
0
1
0
-1/7
5/7
75
Optimal Solution is : x1= 270, x2= 75, Z= 4335
Example
Max. Z = 3x1+5x2+4x3
Subject to constraints:
2x1+3x2 < 8
2x2+5x3 < 10
3x1+2x2+4x3 < 15
x1, x2, x3 > 0
Cont…
Let S1, S2, S3 be the three slack variables.
Modified form is:
Z - 3x1-5x2-4x3 =0
2x1+3x2 +S1= 8
2x2+5x3 +S2= 10
3x1+2x2+4x3+S3= 15
x 1 , x 2 , x 3 , S1 , S2 , S3 > 0
Initial BFS is : x1= 0, x2= 0, x3=0, S1= 8,
S2= 10, S3= 15 and Z=0.
Cont…
Basic
Variable Z
Coefficients of:
Sol.
Ratio
x1
x2
x3
S1
S2
S3
Z
1
-3
-5
-4
0
0
0
0
S1
0
2
3
0
1
0
0
8
8/3
S2
0
0
2
5
0
1
0
10
5
S3
0
3
2
4
0
0
1
15
15/2
Therefore, x2 is the entering variable and S1 is the
departing variable.
Cont…
Basic
Variable Z
Coefficients of:
Sol.
Ratio
x1
x2
x3
S1
S2 S3
Z
1
1/3
0
-4
5/3
0
0
40/3
x2
0
2/3
1
0
1/3
0
0
8/3
-
S2
0
-4/3
0
5
-2/3 1
0
14/3
14/15
S3
0
5/3
0
4
-2/3 0
1
29/3
29/12
Therefore, x3 is the entering variable and S2 is the
departing variable.
Cont…
Basic
Variable Z
Coefficients of:
Sol.
x1
x2
x3
S1
Z
1
-11/15
0
0
17/15 4/5
0
256/15
x2
0
2/3
1
0
1/3
0
8/3
4
x3
0
-4/15
0
1
-2/15 1/5
0
14/15
-
S3
0
41/15
0
0
2/15
1
89/15
89/41
S2
0
-4/5
Ratio
S3
Therefore, x1 is the entering variable and S3 is the
departing variable.
Cont…
Basic
Variable Z
Coefficients of:
Sol.
x1
x2
x3
S1
S2
S3
Z
1
0
0
0
45/41
24/41
11/41
765/41
x2
0
0
1
0
15/41
8/41
-10/41
50/41
x3
0
0
0
1
-6/41
5/41
4/41
62/41
x1
0
1
0
0
-2/41
-12/41 15/41
89/41
Optimal Solution is : x1= 89/41, x2= 50/41, x3=62/41, Z= 765/41
Example
Min.. Z = x1 - 3x2 + 2x3
Subject to constraints:
3x1 - x2 + 3x3 < 7
-2x1 + 4x2 < 12
-4x1 + 3x2 + 8x3 < 10
x1, x2, x3 > 0
Cont…
Convert the problem into maximization problem
Max.. Z’ = -x1 + 3x2 - 2x3 where Z’= -Z
Subject to constraints:
3x1 - x2 + 3x3 < 7
-2x1 + 4x2 < 12
-4x1 + 3x2 + 8x3 < 10
x1, x2, x3 > 0
Cont…
Let S1, S2 and S3 be three slack variables.
Modified form is:
Z’ + x1 - 3x2 + 2x3 = 0
3x1 - x2 + 3x3 +S1 = 7
-2x1 + 4x2 + S2 = 12
-4x1 + 3x2 + 8x3 +S3 = 10
x1, x2, x3 > 0
Initial BFS is : x1= 0, x2= 0, x3=0, S1= 7, S2= 12, S3 = 10
and Z=0.
Cont…
Basic
Variable Z’
Coefficients of:
Sol.
Ratio
x1
x2
x3
S1
S2
S3
Z’
1
1
-3
2
0
0
0
0
S1
0
3
-1
3
1
0
0
7
-
S2
0
-2
4
0
0
1
0
12
3
S3
0
-4
3
8
0
0
1
10
10/3
Therefore, x2 is the entering variable and S2 is the
departing variable.
Cont…
Basic
Variable Z’
Coefficients of:
Sol.
Ratio
x1
x2
x3
S1
S2
S3
Z’
1
-1/2
0
2
0
3/4
0
9
S1
0
5/2
0
3
1
1/4
0
10
4
x2
0
-1/2
1
0
0
1/4
0
3
-
S3
0
-5/2
0
8
0
-3/4
1
1
-
Therefore, x1 is the entering variable and S1 is the
departing variable.
Cont…
Basic
Variable Z’
Coefficients of:
Sol.
x1
x2
x3
S1
S2
S3
Z’
1
0
0
13/5
1/5
8/10
0
11
x1
0
1
0
6/5
2/5
1/10
0
4
x2
0
0
1
3/5
1/5
3/10
0
5
S3
0
0
0
11
1
-1/2
1
11
Optimal Solution is : x1= 4, x2= 5, x3= 0,
Z’ = 11
Z = -11
Example
Max.. Z = 3x1 + 4x2
Subject to constraints:
x1 - x2 < 1
-x1 + x2 < 2
x1, x2 > 0
Cont…
Let S1 and S2 be two slack variables
.
Modified form is:
Z -3x1 - 4x2 = 0
x1 - x2 +S1 = 1
-x1 + x2 +S2 = 2
x1 , x 2 , S1 , S2 > 0
Initial BFS is : x1= 0, x2= 0, S1= 1, S2= 2
and Z=0.
Cont…
Basic
Variable Z
Coefficients of:
Sol.
Ratio
x1
x2
S1
S2
Z
1
-3
-4
0
0
0
S1
0
1
-1
1
0
1
-
S2
0
-1
1
0
1
2
2
Therefore, x2 is the entering variable and S2 is the
departing variable.
Cont…
Basic
Variable Z
Coefficients of:
Sol.
Ratio
x1
x2
S1
S2
Z
1
-7
0
0
4
8
S1
0
0
0
1
1
3
-
x2
0
-1
1
0
1
2
-
x1 is the entering variable, but as in x1 column every no. is less
than equal to zero, ratio cannot be calculated. Therefore given
problem is having a unbounded solution.
The Break-Even Point
The break-even point is that quantity
of output where total revenues equals
total costs : that is, where the
operating income is zero.
Abbreviations
COST-VOLUME-PROFIT ANALYSIS
USP = Unit selling price
UVC = Unit variable costs
UCM = Unit contribution margin (USP-UVC)
CM% = Contribution margin percentage (UCM/USP)
FC = Fixed costs
Q = Quantity of output units sold (and manufactured)
OI = Operating Income
TOI = Target operating income
TNI = Target net income
CHAPTER 3
COST-VOLUME-PROFIT ANALYSIS
Objective 3
methods for determining break- even point
Cost Accounting: A managerial emphasis
1. Equation method:
Revenues – Variable costs – Fixed Costs = Operating Income
(USP*Q) – (UVC*Q) – FC = OI
Provides the most general and easy-to-remember
approach
Example:
Variable cost=120$/unit
Sell price =200$/unit
Fixed cost=2000$
$200Q - $120Q -$2000= $0
Q= $2000/$80
Q= 25 units
$80Q= $2000
CHAPTER 3
COST-VOLUME-PROFIT ANALYSIS
2.Contribution margin method:
It uses the concept of contribution margin to rework the
equation method.
(USP*Q) – (UVC*Q) – FC = OI
By rewriting,
(USP – UVC) * Q = FC + OI
UCM*Q = FC + OI
Q = (FC +OI)/UCM
At break-even, OI=0, therefore:
Q = FC /UCM
Break-even no. of units = Fixed costs/Unit contribution
margin
CHAPTER 3
COST-VOLUME-PROFIT ANALYSIS
Objective 3
Cost Accounting: A managerial emphasis
Substituting,
Break-even no. of units = Fixed costs/Unit contribution
margin
Break-even no. of units = $2000/$80 per unit = 25 units
Calculating break-even revenues,
Break-even in revenue dollars = Break-even no. of units X USP
= (FC*USP)/UCM
= FC/(UCM/USP)
Since, CM% = UCM/USP
= $80/$200 = 40%
= FC/CM%
= $2000/40%
Break-even in revenue dollars = $5000
TR
Revenue and Costs
Y
3. Graph method:
Profits
Break Even Point
TC
TVC
TFC
Loss
0
Q
Output
X
Y
Revenue and Costs
TR
Break Even Point
Profits
TC
TFC
Loss
0
Q
Output
Diagram of Break Even Point
X
CHAPTER 3
COST-VOLUME-PROFIT ANALYSIS
Objective 3
Cost-Volume-Profit Graph for Do-All Software
Operating
Income area
$12,000.00
Break-even point
$10,000.00
Operating Income
Cost Accounting: A managerial emphasis
$14,000.00
$8,000.00
$6,000.00
Total revenues
Total costs
$4,000.00
$2,000.00
Operating Loss
area
$0.00
0
5
10
15
20
25
30
35
Units Sold
40
45
50
55
60
EXAMPLE
• John sells a product for $10 and it
cost $5 to produce (UVC) and has
fixed cost (FC) of $25,000 per year
• How much will he need to sell to
break-even?
• How much will he need to sell to
make $1000?
1. Equation method:
Revenues – Variable cost – Fixed cost = OI
(USP x Q) – (UVC x Q) – FC = OI
$10Q - $5Q – $25,000 = $ 0.00
$5Q = $25,000
Q = 5,000
What quantity demand will earn $1,000?
$10Q - $5Q - $25,000 = $ 1,000
$5Q = $26,000
Q = 5,200
2.Contribution Margin Method
(USP – UVC) x Q = FC + OI
Q= FC + OI
UMC
Q= $25,000 + 0
$5
Q= 5,000
What quantity needs sold to make $1,000?
Q = $25,000 + $1,000
$5
Q = 5,200
3.Graphical Method:
Dollars
70,000
Total Cost
60,000
Line
50,000
40,000
30,000
20,000
Total Revenue
Break-even point
10,000
Line
0
1000 2000 3000 4000 5000 6000
Quantity
Graphical Method :Cont.
Dollars
70,000
Total Cost
60,000
Line
50,000
40,000
30,000
20,000
Total Revenue
Break-even point
10,000
Line
0
1000 2000 3000 4000 5000 6000
Quantity
Break-even Analysis:
Comparing different variables
• Company XYZ has to choose
between two machines to purchase.
The selling price is $10 per unit.
• Machine A: annual cost of $3000 with
per unit cost (VC) of $5.
• Machine B: annual cost of $8000 with
per unit cost (VC) of $2.
Break-even analysis:
Comparative analysis Part 1
• Determine break-even point for
Machine A and Machine B.
• Where: V =
FC
SP - VC
Break-even analysis:
Part 1, Cont.
Machine A:
v = $3,000
$10 - $5
= 600 units
Machine B:
v = $8,000
$10 - $2
= 1000 units
Part 1: Comparison
• Compare the two results to
determine minimum quantity sold.
• Part 1 shows:
– 600 units are the minimum.
– Demand of 600 you would choose
Machine A.
Part 2: Comparison
Finding point of indifference between
Machine A and Machine B will give
the quantity demand required to
select Machine B over Machine A.
Machine A
FC + VC
$3,000 + $5 Q
$3Q
Q
=
Machine B
=
FC + VC
= $8,000 + $2Q
= $5,000
= 1667
Part 2: Comparison
Cont.
• Knowing the point of indifference we
will choose:
• Machine A when quantity demanded
is between 600 and 1667.
• Machine B when quantity demanded
exceeds 1667.
Part 2: Comparison
Graphically displayed
Dollars
21,000
18,000
Machine A
15,000
12,000
9,000
Machine B
6,000
3,000
0
500 1000 1500 2000 2500 3000
Quantity
Part 2: Comparison
Graphically displayed Cont.
Dollars
21,000
18,000
Machine A
15,000
12,000
9,000
Machine B
6,000
Point of indifference
3,000
0
500 1000 1500 2000 2500 3000
Quantity
Exercise 1:
• Company ABC sell widgets for $30 a
unit.
• Their fixed cost is$100,000
• Their variable cost is $10 per unit.
• What is the break-even point using
the basic algebraic approach?
Exercise 1:
Answer
Revenues – Variable cost - Fixed cost = OI
(USP x Q) – (UVC x Q) – FC
$30Q - $10Q – $100,00
$20Q
Q
=
=
=
=
OI
$ 0.00
$100,000
5,000
Exercise 2:
• Company DEF has a choice of two
machines to purchase. They both
make the same product which sells
for $10.
• Machine A has FC of $5,000 and a per
unit cost of $5.
• Machine B has FC of $15,000 and a
per unit cost of $1.
• Under what conditions would you
select Machine A?
Exercise 2:
Answer
Step 1: Break-even analysis on both
options.
Machine A:
v = $5,000
$10 - $5
= 1000 units
Machine B:
v = $15,000
$10 - $1
= 1667 units
Exercise 2:
Answer Cont.
Machine A
FC + VC
$5,000 + $5 Q
$4Q
Q
=
Machine B
=
FC + VC
= $15,000 + $1Q
= $10,000
= 2500
• Machine A should be purchased if
expected demand is between 1000
and 2500 units per year.
Summary:
• Break-even analysis can be an
effective tool in determining the cost
effectiveness of a product.
• Required quantities to avoid loss.
• Use as a comparison tool for making
a decision.