# Power System Reliability

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POWER SYSTEMS RELIABILITY
By:
Dioscoro M. Loquinte

Subject             : Probability and Statistics
Target Audience : High School Level (Grade 12)
Time Requirements: This module is expected to require approximately 4 45-minute periods
(or 3 hours of class time)
General Objective : To teach the students the concept of reliability as an application of
probability
Specific Objectives : After completing this module, the student will be able to:
1) define reliability
2) define a system
3) identify the different system configurations
4) give examples of the three system models
5) calculate system reliabilities of the three system models
6) find the effects of system reliabilities when the number of
components is increased
7) find the “best” system configuration for a given specific problem
a) by finding an optimum reliability
b) by minimizing the cost of operation (monetary and non-
monetary resources, such as time)
c) by finding the least number of components in the system for
maximum reliability
8) describe the lifetime of a population of many products, as well as
humans, using the bathtub curve.

Connections to State/National Standards : Data Analysis and Probability
STATE GOAL 10: Collect, organize and analyze data using statistical methods; predict
results; and interpret uncertainty using concepts of probability.

Assumed Pre-requisite Knowledge: basic concepts of statistics and probability, experimental
probability, theoretical probability, conditional probability
2
Activities/Lessons and Supporting Materials

Educational Module Background
     Introduction

Things fail. Probably, some people have experienced a lawn-mower casing crack, a washing machine
fail, a car battery go dead, a toaster oven electrical plug burn, a water-heater leak, a floppy disk drive go bad, a
TV remote control quit functioning, an automobile engine starter fail, and a house roof leak.

The cracked lawn-mower casing might be a result of its aluminum construction having insufficient
strength to withstand the stresses placed on it. The car battery, the engine starter, and the washing-machine
motor may have experienced wearout after a “normal” life. The toaster oven plug burn because of poor design,
considering the amount of current passing through it. Corrosion might be the cause that the hot water tank
caused it to leak. The corrosion might be partly attributed to the lack of preventive maintenance, which
required periodical draining of the bottom of the tank. The failure of the disk drive might be a result of an
unknown (premature) mechanical failure, and the TV remote control’s failure was caused by a “random”
electronic component failure. Poor construction resulted in the house roof leaking adjacent to the dormers.
Some of these failures caused much inconvenience in addition to their economic impact.

       Educational Module Overview

In this module, the students will apply their knowledge of probability and statistics to solve real-life
problems involving reliability.

In the first activity, Activity 1: “What’s Your Buying Pattern”, the students will be required to answer
a quick survey which describes how they choose to buy their Walkman. The purpose of this survey is for the
students to have a first experience in collecting actual data. They will then be required to construct a frequency
distribution, draw and interpret their graphs.

Activities 2 and 3 are concerned with series system configurations. For activity 2, Activity 2: Finding
A Series System’s Reliability, the students will be required to use the formula for finding the reliability of the
system in series given the components’ reliabilities. After computing the reliability of the system, they will be
asked to describe the general effect of component reliability in a series system. For activity 3, Activity 3:
Effect of the Number of Components in a Series System, the students will be required to solve for the
reliability of the system in series when the number of components increases and are asked to make a
generalization based from the graph of system reliability vs number of components in series.

Activities 4 and 5 deal with parallel systems. Activity 4 requires the students to find the reliability of
the system in parallel and make a description as to the general effect of component reliability in a parallel
system. After completing Activity 5, the students will be able to describe the effect of increasing the number
of components in a parallel system.

Activity 6 asks the students to find the reliability of a combination of series and parallel structures
and in Activity 7, the students will describe different real-life applications of the bathtub curve.
3

        Project Description

Phase 1: Project Design Competition
1. The participants/students will be asked to form 10 groups of 3 members
2. Each group will be given n components with corresponding reliabilities and cost of operation.
3. Each group will design a system with the best configuration (using series, parallel and/or
combination)
4. Each group will present/explain their “best” configuration

Phase 2: Application/Simulation

Using their “best” configuration, the students will be required to do a Monte Carlo simulation on one of
the following situations:
a) an online ordering (subscription) business
 total number of orders per day
 costs (price/volume served)
 number of operators

b) repair facility business
 number of items handled per day
 time of the repair and the average queue length
 cost of operation
 number of operators

c) emergency service
 number of operators
 volume served/demands
 cost of operation

I. Warm-Up

Teacher Notes

Objective:
To give the students their first experience in collecting actual data. The students will then be required
to construct a frequency distribution, draw and interpret their graphs.

Materials:
Activity 1 worksheet (one copy per student), graphing calculator, ruler, protractor

Suggested Teaching Strategies:
Give each student a copy of Activity 1 worksheet. Each student will answer items 1 – 5 of this
worksheet for 5 minutes. For the remaining 15 minutes, a student representative will be chosen to act as a
facilitator to make the frequency distribution of the whole class. Each student will then construct pie diagrams
for items 2 and 3, and be required to interpret their graphs.
4

1. Below is a list of some of the product attributes that would help you determine in buying a Walkman
CD player. Rank them on a scale from 1 (most important) to 15 (least important).

Performance _____           lasts a long time (reliability)  ______      service           _____
Service           _____     easily repaired (maintainability) ______     warranty          _____
Easy to use       _____     appearance                        ______     brand name        _____
Packaging display _____     latest model                      ______     price              _____
Availability      _____     recommended                      _______    other (pls specify) _____
2. How frequently do you use this product?
______ Every day ______ Few times a week            _____ about once a week ____ other

3.     What is the brand name of your walkman?
_____ Aiwa      ______ Casio    _______ Denon     ______ Kenwood ____ Sony _____ Other

4.  Would you recommend this product to someone else?
_____ Yes          _____ No
5. Would you buy this product again under the same circumstances?
_____ Yes          _____ No
Tally Sheet 1: Product Attributes vs Rank
Rank
Product attributes 1    2    3      4   5     6    7    8    9    10       11    12      13     14     15
performance
service
easy to use
packaging display
availability
lasts a long time
(reliability)
easily repaired
(maintainability)
appearance
latest model
recommended
service
warranty
brand name
price
other (pls specify)
5

Frequency of        frequency     %         no. of
Usage                           of total    degrees
every day

few times a week

about once a week

other

total

Brand name          frequency     %         no. of
of total   degrees
Aiwa

Casio

Denon

Kenwood

Sony

Other

total
6
II. Reliability; System (10 minutes)

Teacher Notes

        What is reliability?
Reliability is a broad term that focuses on the ability of a product to perform its intended function.
Mathematically speaking, assuming that an item is performing its intended function at time equals zero,
reliability can be defined as the probability that an item will continue to perform its intended function
without failure for a specified period of time under stated conditions. Please take note that the product defined
here could be an electronic or mechanical hardware product, a software product, a manufacturing process or
even a service.
So, I if I say that the reliability of my walkman is 0.98 (or 98%), I mean that my walkman is working
98% of the time and failing 2% of the time.

Thus, the term reliability refers to the proper functioning of equipment and systems and thus
encompasses hardware, software, human, and environmental factors.

        Why is Reliability Important?
There are a number of reasons why reliability is an important product attribute, including:
 Reputation. A company’s reputation is very closely related to the reliability of their products.
The more reliable a product is, the more likely the company is to have a favorable reputation.

    Customer Satisfaction. While a reliable product may not dramatically affect customer
satisfaction in a positive manner, an unreliable product will negatively affect customer
satisfaction severely. Today’s consumer is more intelligent and product-aware than the
consumer of years past. This consumer will no longer tolerate products that do not perform in a
reliable fashion, or as promised and advertised. Customer dissatisfaction with a product’s
reliability can have disastrous financial consequences to the manufacturer. Statistics show that
when a customer is satisfied with a product they might tell 8 other people; however, a
dissatisfied customer will tell 22 people, on average.

    Warranty Costs. If a product fails to perform its function within the warranty period, the
replacement and repair costs will negatively affect profits, as well as gain unwanted negative
attention. Introducing reliability analyses is an important step in taking corrective action,
ultimately leading to a product that is more reliable.

    Repeat Business. A concentrated effort towards improved reliability shows existing customers
that a manufacturer is serious about their product, and committed to customer satisfaction. This
type of attitude has a positive impact on future business.

    Cost Analysis. Manufacturers may take reliability data and combine it with other cost
information to illustrate the cost-effectiveness of their products. This life cycle cost analysis
can prove that although the initial cost of their product might be higher, the overall lifetime cost
is lower than a competitor’s because their product requires fewer repairs or less maintenance.

    Customer Requirements. Many customers in today’s market demand that their suppliers have
an effective reliability program. These customers have learned the benefits of reliability
analysis from experience.
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   Competitive Advantage. Many companies will publish their predicted reliability numbers to
help gain an advantage over their competition who either does not publish their numbers or has
lower numbers.

      What is the Difference Between Quality and Reliability?
Although the terms reliability and quality are often used interchangeably, there is a difference between
these two disciplines. While reliability is concerned with the performance of a product over its entire lifetime,
quality control is concerned with the performance of a product at one point in time, usually during the
manufacturing process. As stated in the definition, reliability assures that components, equipment and systems
function without failure for desired periods during their whole design life, from conception (birth) to junking
(death). Quality control is a single, albeit vital, link in the reliability process. Quality control assures
conformance to specifications. This reduces manufacturing variance, which can degrade reliability. Quality
control also checks that the incoming parts and components meet specifications, that products are inspected
and tested correctly and that the shipped products have a quality level equal to or greater than that specified.
The specified quality level should be one that is acceptable to the users, the consumer, and the public. No
product can perform reliably without the inputs of quality control, because quality parts and components are
needed to go into the product so that its reliability is assured.

Even though a product has a reliable design, when the product is manufactured and used in the field, its
reliability may be unsatisfactory. The reason for this low reliability may be that the product was poorly
manufactured. So, even though the product has a reliable design, it is effectively unreliable when fielded
which is actually the result of a substandard manufacturing process.
As an example, cold solder joints could pass initial testing at the manufacturer, but fail in the field as a
result of thermal cycling or vibration. This type of failure did not occur because of an improper design, but
rather it is the result of an inferior manufacturing process. So while this product may have a reliable design, its
quality is unacceptable because of the manufacturing process.
Just like a chain is only as strong as its weakest link, a highly reliable product is only as good as the
inherent reliability of the product and the quality of the manufacturing process.

       What is a system?
A system is a collection of components, subsystems and/or assemblies arranged to a specific design in
order to achieve desired functions with acceptable performance and reliability. The types of components, their
quantities, their qualities and the manner in which they are arranged within the system have a direct effect on
the system’s reliability. A simplified computer system with a redundant fan configuration is shown below:

FAN

POWER                                  PROCESSOR                     HARD
SUPPLY                                                               DRIVE

FAN

A simple reliability block diagram
8
III. Models of A System (15)
Many systems can be modeled using series structures, parallel structures or combinations of both.
So first here are a few pictures and explanations to explain each of the above configurations.

 Series Structures:
The following are examples of series structures:

1              2

For this system to work, both components 1 and 2 must work.

1              2            ∙∙∙               n

There may be many components in a series. In this case all n components must work in order for the whole
system to work.

For example, a personal computer may consist of four basic subsystems: the power supply, the motherboard,
the processor and the hard drive.

POWER             MOTHERBOARD             PROCESSOR              HARD
SUPPLY                                                           DRIVE

A simple series system

 Parallel Structures
The following are examples of parallel structures:

1                                                 1

2                                                2

∙
∙
∙

n

In a parallel system, the system will work as long as at least one component works.
9

 Combination of Series and Parallel Structures
A system may also combine both series and parallel structures:

1               2

3

IV.     Finding a System’s Reliability (65 minutes)

A) A Series System

1               2               ∙∙∙            n

For a pure series system, the system reliability is equal to the product of the reliabilities of its constituent
components. Or :

Rs = (R1)(R2)…(Rn)
where
Rs = reliability of the system,
Ri = reliability of the ith component, i = 1, 2, …, n

Example:
Three components are connected in series and make up a system. Component 1 has a reliability of 0.995,
component 2 has a reliability of 0.987 and component 3 has a reliability of 0.973 for a mission of 100 hours.
Find the overall reliability of the system for a 100-hour mission.
Solution:
Rs = (R1)(R2)(R3)
= (0.995)(0.987)(0.973)
= 0.955

Effect of Component Reliability in a Series System

In a series configuration , the component with the smallest reliability has the biggest effect on the
system’s reliability. There is a saying that “a chain is only as strong as its weakest link.” This is a good
example of the effect of a component in a series system. In a chain, all the rings are in series and if any of the
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rings break, the system fails. In addition, the weakest link in the chain is the one that will break first. The
weakest link dictates the strength of the chain in the same way that the weakest component dictates the
reliability of a series system. As a result, the reliability of a series system is always less than the reliability of
the least reliable component.

What is the effect of the number of components in a series system?

Activity 2: A Series System’s Reliability
A. Computing Series System Reliability
In each of the following systems, the reliability of each component is given. Find the reliability of
the system.
1)
0.8             0.9                              Rs = ________________________

2)
0.7              0.6                                Rs = _______________________

Based on your answers to the above problems,
a) what can you say about the reliability of the system compared to the least reliable component?
_____________________________________________________________________________________

b) describe in general the effect of component reliability in a series system.
_________________________________________________________________________
B. Effect of the Number of Components in a Series System
Consider a system that consists of a single component. Assume that the reliability of the component is
0.95 ( or 95%), thus the reliability of the system is also 0.95 (or 95%). What would the reliability of the
system be if there were more than one component (with the same individual reliability) in series?

Sketch the graph of system reliability vs. number of components.
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1.0

0.9
s
y    0.8
s
t
e    0.7
m
0.6
r
e    0.5
l
I
a    0.4
b
i    0.3
l
I    0.2
t
y    0.1

0.0
Number       of   Components

Sketch of System Reliability vs. Number of Components (in series)

Conclusion/Generalization:
_____________________________________________________________________________

Challenge Problem: An example of a long series system is seen in long natural gas pipelines which lack
enroute storage capabilities. A long gas pipeline can have approximately 125 welded connections per mile and
the pipelines can easily be 1500 miles long. This represents 187,500 welded connections in series! If the
system needs an overall reliability of 90%, what must the reliability of the individual connections be
(assuming all reliability values to be the same)?
12
B) A Parallel System
Now we’ll consider a simple parallel system. For this system to work, we need at least one of the two
components to work.

1

2

∙
∙
∙

n

For a pure parallel system, the overall system unreliability is equal to the product of the component
unreliabilities.
Note: If R1 is the reliability of component 1, then its unreliability is 1 – R1.

Thus, the reliability of the parallel system is then given by:

Rs  1  1  R1 1  R2  ... 1  Rn 
Example:
Three components are connected in parallel and make up a system. Component 1 has a reliability of
0.995, component 2 has a reliability of 0.987 and component 3 has a reliability of 0.973 for a mission of 100
hours. Find the overall reliability of the system for a 100-hour mission.
Solution:
Rs  1  1  R1 1  R2 1  R3 
= 1  1  0.995 1  0.987 1  0.973 
= 0.999998245
= 99.998%
_________________________________________________________________________________

Effect of Component Reliability in a Parallel System
When we examined a system of components in series, we found that the least reliable component has
the biggest effect on the reliability of the system. However, the component with the highest reliability in a
parallel configuration has the biggest effect on the system’s reliability, since the most reliable component is
the one that will most likely fail last. This is a very important property of the parallel configuration,
specifically in the design and improvement of systems.

What is the effect of the number of components in a parallel system?
13
Activity 3: A Parallel System’s Reliability

A) Computing Parallel System’s Reliability
In each of the following systems, the reliability of each component is given. Find the reliability of the
system.
1)                                                 2)

0.9                                                0.7

0.8                                                0.8

Rs = ____________________                         Rs = ____________________

Based on your answers to the above problems,
c) what can you say about the reliability of the system compared to the highest reliable component?

_____________________________________________________________________________________

d) describe in general the effect of component reliability in a parallel system.

_____________________________________________________________________________________

B) Effect of the Number of Components in a Parallel System
Consider a system that consists of a single component. The reliability of the component is 60% (or
0.60), thus the reliability of the system is 60%. What would the reliability of the system be if there were more
than one component (with the same individual reliability) in parallel?
Using the formula Rs = 1 – [(1- R1)(1- R2)…(1- Rn)], we have the following table that gives the
reliability of the system for the different number of components.

Number of components                         System Reliability
1                                       0.600000
2                                       0.840000
3                                       0.936000
4                                       0.974400
5                                       0.989760
6                                       0.995904
7                                       0.998362
8                                       0.999345
9                                      0.9997379
10                                      0.9998951

Draw the graph of system reliability vs. number of components.
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1.0

0.9
s
y    0.8
s
t
e    0.7
m
0.6
r
e    0.5
l
I
a    0.4
b
i    0.3
l
I    0.2
t
y
0.1

0.0
1     2       3    4         5      6   7           8       9       10
Number       of    Components
Graph of System Reliability vs. Number of Components (in parallel)

Conclusion/Generalization:
________________________________________________________________________________________

Challenge Problem: Consider a system with three components connected reliability-wise parallel. If the
system needs an overall reliability of 99%, what must the reliability of the individual component be (assuming
all reliability values to be the same)?

Note: Clearly, the reliability of the system can be improved by adding redundancy. (Units in parallel are also
referred to as redundant units). However, it must be noted that doing so is usually costly in terms of additional
components, additional weight, volume, etc. Redundancy is widely used in the aerospace industry. Other
applications include the computer hard drive systems, brake systems and support cables in bridges.
The cost as a function of the reliability for each component must be quantified before attempting to
improve the reliability. Otherwise, the design changes may result in a system that is needlessly expensive or
overdesigned. Thus, by associating cost values to the reliabilities of the system’s components, we can find an
optimum design that will provide the required reliability at a minimum cost.
For example, consider a system that initially consists of a single unit. The cost of that unit, including
all associated mounting and hardware costs, is one dollar. The reliability of this unit for a given mission time
is 30%. It has been determined that this is inadequate and that a second component is to be added in parallel to
boost the reliability. The reliability for the two-unit parallel system is Rs = 1 – (1 – 0.3)2 = 0.51, or 51%. So,
the reliability has increased by a value of 21%, and the cost has increased by one dollar. In similar fashion, we
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can continue to add additional units in parallel, thus increasing the reliability and cost (see table and graph
below).

Number of components     System Reliability         System Cost, \$
1                   0.300                       1.00
2                   0.510                       2.00
3                   0.657                       3.00
4                   0.7599                      4.00
5                   0.83193                     5.00
6                   0.882351                    6.00
7                   0.9176457                   7.00
8                   0.94235199                  8.00
9                   0.959646393                 9.00
10                   0.9717524751               10.00

\$10

\$9
s
y    \$8
s
t
e    \$7
m
\$6
c
o    \$5
s
t
\$4

\$3

\$2

\$1

\$0
10%     20%   30%   40%      50%  60%    70%        80%   90%    100%
System       Reliability
Graph of System Reliability vs. Cost
16
C) Combination of Series and Parallel Structures
While many smaller systems can be accurately represented by either a simple series or parallel
configuration, there may be larger systems that involve both series and parallel configuration in the overall
system. Such systems can be analyzed by calculating the probabilities for the individual series and parallel
sections and then combining them in the appropriate manner. Such a methodology is illustrated in the
following example.

Example:
Consider a system with three components. Units 1 and 2 are connected in series and Unit 3 is
connected in parallel with the first two, as shown in the figure below.

1               2

3

What is the reliability of the system if R1 = 99.5%, R2 = 98.7%, and R3 = 97.3% at 100 hours?
Solution:
First the reliability of the series segment consisting of Units 1 and 2 is calculated:
R1,2 = (R1)(R2)
= (0.995)(0.987)
= 0.982065 or 98.2065%

The reliability of the overall system is then calculated by treating Units 1 and 2 as one with a reliability of
98.2065% connected in parallel with Unit 3. Therefore:

R1,2

3

Rs = 1 - [(1 – 0.982065)(1 – 0.973000)]
= 0.999515755 or 99.95%
17
Activity 4: Combination of Series and Parallel Structures

In each of the following systems, the reliability of each component is given. Find the reliability of
the system.

0.6
0.7

0.5
18
V. The Bathtub Curve (25 minutes)
Reliability specialists often describe the lifetime of a population of products using a graphical
representation called the bathtub curve. It characteristically describes the life of many products (as well as
humans).

f
a      EARLY LIFE                USEFUL LIFE                               WEAROUT LIFE
i                                 (or normal life)
l
u       (burn-in or
r       break-in or
e         infant mortality
period)
r
a
t
e
(failure rate                                                              (failure rate
decreases with time)                                                  increases with time)
(failure rate approx. constant)

time (hours, miles, cycles, etc)

The bathtub curve consists of three periods:
a) early life (burn-in or break-in or infant mortality period)
b) useful life or normal life period
c) wearout period

Systems having this graph experience decreasing failure rates early in their life cycle (early life),
followed by a nearly constant failure rate (useful life), followed by an increasing failure rate (wearout).

Some individual units will fail relatively early (infant mortality failures), others (we hope most) will
last until wear-out, and some will fail during the relatively long period typically called normal life. Failures
during infant mortality are highly undesirable and are always caused by defects and blunders: material defects,
design blunders, errors in assembly, etc. Normal life failures are normally considered to be random cases of
“stress exceeding strength”. However, as statistics show, many failures often considered normal life failures
are actually infant mortality failures. Wear-out is a fact of life due to fatigue or depletion of materials (such as
lubrication depletion in bearings). A product’s useful life is limited by its shortest-lived component. A product
manufacturer must assure that all specified materials are adequate to function through the intended product
life.
We are interested in the characteristics illustrated by the entire bathtub curve. The infant mortality
period is a time when the failure rate is dropping, but is undesirable because a significant number of failures
occur in a short time, causing early customer dissatisfaction and warranty expense. Infant mortality does not
mean “products that fail within 90 days” or any other defined time period.
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Theoretically, failures during normal life occur at random but with a relatively constant rate when
measured over a long period of time. Because these failures may incur warranty expense or create service
support costs, we want the bottom of the bathtub as low as possible.
The wear-out portion of the curve represents those parts that begin to show the limits of their
endurance. Most parts have some kind of finite life and sooner or later will slowly deteriorate from prolonged
use. From a design point of view, this wear-out feature has to be out far enough in its life cycle so that the
customer at least feels he/she bought a reasonable bargain. This is especially true if the product was properly
maintained and not abused. Competition, reputation, cost of the product, and other factors determine what
kind of bathtub curve a company can live with. A product that has little or no infant mortality, a very low
constant failure rate, and a wear-out portion that is far out in its life cycle will certainly sell much better than a
product that does not have these features, sometimes even if the price is higher. Reliability then becomes a
trade-off between cost, competition, and the desire to satisfy the customer.

The table below summarizes some of the distinguishing features of the bathtub curve.

Period                 Caused by                                       Reduced by

Early life             Manufacturing defects:                          Burn-in testing
welding flaws, cracks,                        Screening
defective parts, poor quality                 Quality control
control, contamination, poor                  Acceptance testing
workmanship

Useful life            Environment                                     Redundancy
Random loads                                    Excess strength
Human error
“Acts of God”
Chance events

Wear-out               Fatigue                                         Derating
Corrosion                                       Preventive maintenance
Aging                                           Parts replacement
Friction                                        Technology

Some Applications of the Bathtub Curve
The bathtub curve is readily evident in everyday life.
9) A new car comes with a warranty to cover problems during its break-in phase.
When money is borrowed from a bank to buy a car, the loan term is typically
three or four years --- timed to be paid off before the car enters the wear-out
phase.

10) New shoes may be uncomfortable until they are worn in and then remain
comfortable until worn out.

11) At the infant mortality period, infants are vulnerable to numerous illnesses until
they grow stronger and build up immunities. The useful life period represents
the peak health years for living things. Accidents still occur but at a lower rate.
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Finally, due to aging, older people are more prone to breaking bones in a fall
than younger people.

12) Similarly, products may fail soon after being put to use due to manufacturing
defects, material imperfections, or poor workmanship. During the useful life,
accidents and random events still occur, but at a lower rate. And due to aging, it
takes less stress to cause failure of these products.
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VI.    Project (45 minutes)

Phase 1: Project Design Competition
1. The participants/students will be asked to form 10 groups of 3 members
2. Each group will be given n components with corresponding reliabilities and cost of operation.
3. Each group will design a system with the best configuration (using series, parallel and/or
combination)
4. Each group will present/explain their “best” configuration

Phase 2: Application/Simulation

Using their “best” configuration, the students will be required to do a Monte Carlo simulation on one of the
following situations:
d) an online ordering (subscription) business
 total number of orders per day
 costs (price/volume served)
 number of operators

e) repair facility business
 number of items handled per day
 time of the repair and the average queue length
 cost of operation
 number of operators

f) emergency service
 number of operators
 volume served/demands
 cost of operation
22

Module Assessment
I.   Pre-Assessment in Probability/Systems

Name: ________________________________________ Period: ___ Date: ____________
---------------------------------------------------------------------------------------------------------------------
A. Directions: Read each question below. Write the letter of your answer on the blank provided before
each number.

______ 1. Which of the following is an experiment?

a) tossing a coin                              b) rolling a single die
c) choosing a marble from a jar                d) all of the above

______ 2. Which of the following is an outcome?

a) rolling a pair of dice                      b) landing on red
c) choosing 2 marbles from a jar               d) none of the above

______ 3. Which of the following experiment does NOT have equally likely outcomes?

a) choose a number at random from 1 to 7                       b) toss a coin
c) choose a letter at random from the word SCHOOL              d) none of the above

______ 4. What is the probability of choosing a vowel from the alphabet?

a) 21/26               b) 5/26                 c) 1/21                   d) none of the above

______ 5. A number from 1 to 11 is chosen at random. What is the probability of choosing an odd number.

a) 1/11                b) 5/11                 c) 6/11                          d) none of the above

______ 6. Which of the following is the sample space when 2 coins are tossed?

a) {H, T, H, T}        b) {H, T}               c) {HH, HT, TH, TT}              d) none of the above

______ 7. At South Shore High School, 3 out of 5 students make honor roll. What is the probability that a
student does not make honor roll?

a) 65 %                b) 40 %                 c) 60 %                   d) none of the above

______ 8. A large basket of fruit contains 3 oranges, 2 apples, and 5 bananas. If a piece of fruit is chosen at
random, what is the probability of getting an orange or a banana?

a) 4/5                 b) ½                    c) 7/10                   d) none of the above
23

Pre-assessment (cont’d)

______ 9. A pair of dice is rolled. What is the probability of getting a sum of 2?

a) 1/6                 b) 1/3                 c) 1/36                 d) none of the above

______ 10. In a class of 30 students, there are 17 girls and 13 boys. A total of 5 students made an A grade on
their report card, three of which were girls. If a student is chosen at random, what is the probability of getting
a girl or an A student?

a) 19/30               b) 11/15               c) 17/180               d) none of the above

______ 11. In the United States, 43% of people wear a seat belt while driving. If two people are chosen at
random, what is the probability that both of them wear a seat belt?

a) 86 %                b) 18 %                c) 57 %                 d) none of the above

______ 12. Three cards are chosen at random from a deck. What is the probability of getting an ace, a king
and a queen in order?

8                     1                       6
a)                     b)                     c)                      d) none of the above
16575                  2197                   35152

______ 13. A city survey found that 47% of teenagers have a part time job. Of those teenagers, 78% also
plan to attend college. If a teenager is chosen at random, what is the probability that he/she has a part time job
and plans to attend college?

a) 60 %                b) 63 %                c) 37 %                 d) none of the above

______ 14. In a school, 14% of students take drama and computers, and 67% take computer class. What is
the probability that a student taking computers also takes drama?

a) 81 %                b) 21 %                c) 53 %                 d) none of the above

______ 15. In a shipment of 100 televisions, 6 are defective. If a person buys two televisions, what is the
probability that both are defective?

3                      9                      1
a)                     b)                     c)                      d) none of the above
100                    2500                   330
24

B.    Answer the following questions to the best of your knowledge.

1) Define in your own words what a system is.

2) How are systems modeled? Draw a schematic diagram of each of the models of a system.

3) Describe the advantages and disadvantages of each of the models.

-------------------------------------------------------------------------------------------------------------------------------------

Answers: 1) d              2) b         3) c            4) b            5) c        6) c               7) b              8) a
9) c              10) a       11) b           12) a            13) c 14) b                15) c
25

II.   Objective Assessment of the Whole Concept

Name: ________________________________________ Period: ___ Date: ____________

In each of the following systems, the reliability of each component is given. Find the reliability of
the system.

1)
0.75           0.83                                Rs = ________________________

2)
0.85          0.94        ∙∙∙   0.77              Rs = ________________________
n

3)                                                4)

0.8                                                0.6

0.7                                                0.8

0.7

Rs = ____________________                         Rs = ____________________

5)

0.7             0.9

Rs = ____________________

0.5

6) Discuss the advantages and disadvantages of the series, and parallel connections.

7) Give some examples of systems that are connected in a) series       b) parallel     c) series-parallel
26

8) Describe at least two real-life applications of the bathtub curve.
III. Self Assessment (Rubric)

Name: ________________________________________ Period: ___ Date: ____________

4                        3                     2                     1           Score

I contributed each        I contributed most      I said things some-     I was there
Participation    time my group             of the time.            times. I worked          wasn’t I?
met.

I listened to all        I listened. If people   I usually got my       I don’t care
suggestions. I made      disagreed, I let them   way. I didn’t like      what they
Cooperation      suggestions. I helped    solve it. Sometimes     people telling me      decide. If that’s
my group reach           I changed my choice     what to do. Some-       what they want,
decisions without        because the group       times I did what       they can do it.
telling them what to wanted something everyone wanted.
do.                  else.

I did what my group I did my work. I I did some of my When’s
expected of me. I finished work that work. Other       time to go?
Work Load         did not take work other people       people had to
away from other     weren’t doing.   finish my work.
people.

I showed                      I gave some      Sometimes, I
appreciations to              appreciations. I showed
Appreciation my group each time            made a couple of appreciations.                 Shut up.
we met. I never               put-downs.       Most of the
gave put-downs.                                time I forgot
or I made put-
downs.

Total Score

Teacher’s Signature: ____________________________ Date: ______________

** Adapted from: http://www.globalclassroom.org/assessment.htm
27

IV. Team Assessment (Rubric)

Name: ________________________________________ Period: ___ Date: ____________

Complete the following questions as a team.

Low              High

1. Did all of the members of our group contribute ideas?               1    2    3     4       5

2. Did all of the members of our group listen carefully to             1    2    3     4       5
the ideas of other members?

3. Did all of the members of our group encourage other                 1     2   3     4       5
members to contribute their thoughts and opinions?

4. Three ways that we helped each other learn the material were:
 ______________________________________________

   ______________________________________________

   ______________________________________________

5. a) One difficulty our group had was (explain fully):

b) To resolve the situation, we could:

Group signatures:

_______________________________________             ________________________________________

_______________________________________             ________________________________________
28
** Adapted from: http://www.curriculumfutures.org/assessment/a03-10.html

V. Rubric for Reliability Project

Name: ________________________________________ Period: ___ Date: ____________

Strong Satisfactory Progressing   Needs Improvement   Unsatisfactory    Score
5          4           3                 2                  1

Logic of Design

Quality of diagrams

Correctness of
reliabilities’ set-up

Correctness of
mathematical
computations

Quality/clarity of
explanations

Appropriateness
of report (tone,
persuasion)

Project was on time

Over-all score

Teacher’s Signature: __________________________________ Date: _______________
29

Teacher Resources
Below are the MATLAB codes for the simulation:
function [co, sys_cop, sys_ccop] = two_tier_cap_out (ico_table);
% capacity outage probability tables for two tier systems
%
% [co, sys_cop, sys_ccop] = two_tier_cap_out (icop_table);
%
% icop_table is a matrix of concatenated individual capacity outage
%    probability tables of the form [cap. out., ind. prob.] where
%    each new generator table begins with the zero cap. out. state:
%    icop_table = [
%       g_1_cap_out_1, g_1_prob_out_1;
%             .             .
%       g_1_cap_out_n, g_1_prob_out_n;
%             .             .
%       g_m_cap_out_1, g_m_prob_out_1;
%             .             .
%       g_m_cap_out_n, g_m_prob_out_n;
% co is the system capacity outage vector
% sys_cop is the system capacity outage probability table
% sys_ccop is the system capacity cumulative outage probability table

% copyright August 2004: Alexander J. Flueck, Illinois Institute of Technology

%       -----         -----
%    ---| A |---------| C |---
%    | -----          ----- |
% ---|                       |---
%    | -----          ----- |
%    ---| B |---------| D |---
%       -----         -----

%   max cap: min(Acap,Ccap) + min(Bcap,Dcap)
%   ACcap = min(Acap,Ccap);
%   ACfor = 1 - (1-Afor)*(1-Cfor);
%   icopAC = [0 1-ACfor; ACcap ACfor];
%   BDcap = min(Bcap,Dcap);
%   BDfor = 1 - (1-Bfor)*(1-Dfor);
%   icopBD = [0 1-BDfor; BDcap BDfor];
%   [sys1_co, sys1_cop, sys1_ccop] = cap_out ([icopAC;icopBD]);
%   maxcap = ACcap + BDcap;
%   sys1_cap = maxcap-sys1_co;

%       -----         -----
%    ---| A |---   ---| C |---
%    | ----- |     | ----- |
% ---|         |---|         |---
%    | ----- |     | ----- |
%    ---| B |---   ---| D |---
%       -----         -----

%   max cap: min(Acap+Bcap,Ccap+Dcap)
%   icopA = [0 1-Afor; Acap Afor];
%   icopB = [0 1-Bfor; Bcap Bfor];
%   [ABco, ABsys_cop, ABsys_ccop] = cap_out ([icopA;icopB]);
%   ABcap = Acap+Bcap;
30
%   icopC = [0 1-Cfor; Ccap Cfor];
%   icopD = [0 1-Dfor; Dcap Dfor];
%   [CDco, CDsys_cop, CDsys_ccop] = cap_out ([icopC;icopD]);
%   CDcap = Ccap+Dcap;

%   ACfor = 1 - (1-Afor)*(1-Cfor);
%   icopAC = [0 1-ACfor; ACcap ACfor];
%   maxcap = min(ABcap,CDcap)
%   sys2_cap = maxcap-sys2_co;

nrow=length(ico_table(:,1));
gndx=find(ico_table(:,1)==0);
ngen=length(gndx);
pndx=find(ico_table(:,1)>0);
ival=min(gcd(ico_table(:,1),min(ico_table(pndx,1))));
sys_cop=[1];
for i=1:ngen-1,
g_cap_out = [ico_table(gndx(i),2);zeros(ico_table(gndx(i+1)-1,1)/ival,1)];
for j=1:gndx(i+1)-1-gndx(i),
k = ico_table(j+gndx(i),1)/ival;
g_cap_out(k+1) = ico_table(j+gndx(i),2);
end
sys_cop = conv(sys_cop, g_cap_out);
end
g_cap_out = [ico_table(gndx(ngen),2);zeros(ico_table(nrow,1)/ival,1)];
for j=1:nrow-gndx(ngen),
k = ico_table(j+gndx(ngen),1)/ival;
g_cap_out(k+1) = ico_table(j+gndx(ngen),2);
end
sys_cop = conv(sys_cop, g_cap_out);
sys_ccop = [1; zeros(length(sys_cop)-1,1)];
for i=2:length(sys_cop),
sys_ccop(i) = sys_ccop(i-1) - sys_cop(i-1);
end
co = [0:ival:ival*(length(sys_cop)-1)]';
%
% plot results versus system capacity outage
%
figure(1)
stem(co,sys_cop,'k*-')
%axis([1,ntrials, 0,1])
xlabel('capacity outage (MW)')
ylabel('probability')
grid off
title_text = sprintf('system capacity outage probability table');
title(title_text)
figure(2)
stairs([-ival;co],[sys_ccop;0],'k*-')
%stem(cap,sys_ccop,'k*-')
%axis([1,ntrials, 0,1])
xlabel('capacity outage (MW)')
ylabel('cumulative probability')
grid off
title_text = sprintf('system cumulative capacity outage probability table');
title(title_text)
31

function two_tier_monte_carlo;

%   Monte Carlo reliability analysis for two tier systems
%
%   [co, sys_cop, sys_ccop] = two_tier_monte_carlo (icop_table);
%
%   icop_table is a matrix of concatenated individual capacity outage
%      probability tables of the form [cap. out., ind. prob.] where
%      each new generator table begins with the zero cap. out. state:
%      icop_table = [
%         g_1_cap_out_1, g_1_prob_out_1;
%               .             .
%         g_1_cap_out_n, g_1_prob_out_n;
%               .             .
%         g_m_cap_out_1, g_m_prob_out_1;
%               .             .
%         g_m_cap_out_n, g_m_prob_out_n;
%   co is the system capacity outage vector
%   sys_cop is the system capacity outage probability table
%   sys_ccop is the system capacity cumulative outage probability table

% copyright August 2004: Alexander J. Flueck, Illinois Institute of Technology

comp_cap = [ 12 10 14     5];
comp_for = [.10 .05 .10 .05];
Acap = comp_cap(1);
Afor = comp_for(1);
Bcap = comp_cap(2);
Bfor = comp_for(2);
Ccap = comp_cap(3);
Cfor = comp_for(3);
Dcap = comp_cap(4);
Dfor = comp_for(4);

%% analytical derivation of sys1 reliability (probabilistic capacity)
%% max cap: min(Acap,Ccap) + min(Bcap,Dcap)
%       -----         -----
%    ---| A |---------| C |---
%    | -----          ----- |
% ---|                       |---
%    | -----          ----- |
%    ---| B |---------| D |---
%       -----         -----
ACcap = min(Acap,Ccap);
ACfor = 1 - (1-Afor)*(1-Cfor);
icopAC = [0 1-ACfor; ACcap ACfor];
BDcap = min(Bcap,Dcap);
BDfor = 1 - (1-Bfor)*(1-Dfor);
icopBD = [0 1-BDfor; BDcap BDfor];
[sys1_co, sys1_cop, sys1_ccop] = cap_out ([icopAC;icopBD]);
maxcap = ACcap + BDcap;
sys1_capa = maxcap-sys1_co;
32

%% analytical derivation of sys2 reliability (probabilistic capacity)
%% max cap: min(Acap+Bcap,Ccap+Dcap)
%       -----         -----
%    ---| A |---   ---| C |---
%    | ----- |     | ----- |
% ---|         |---|         |---
%    | ----- |     | ----- |
%    ---| B |---   ---| D |---
%       -----         -----
icopA = [0 1-Afor; Acap Afor];
icopB = [0 1-Bfor; Bcap Bfor];
[ABco, ABsys_cop, ABsys_ccop] = cap_out ([icopA;icopB]);
ABcap = Acap+Bcap;
icopC = [0 1-Cfor; Ccap Cfor];
icopD = [0 1-Dfor; Dcap Dfor];
[CDco, CDsys_cop, CDsys_ccop] = cap_out ([icopC;icopD]);
CDcap = Ccap+Dcap;
maxcap = min(ABcap,CDcap);

for i=1:length(ABco),
for j=1:length(CDco),
sys2_capa(i,j) = min(ABcap-ABco(i),CDcap-CDco(j));
end
end
%sum(sum(sys2_prob))
sys2_co=[];
sys2_cop=[];
for i=1:length(ABco),
for j=1:length(CDco),
co = maxcap-sys2_capa(i,j);
ind = find(co==sys2_co);
if length(ind)>0, % co already found?
sys2_cop(ind) = sys2_cop(ind) + ABsys_cop(i)*CDsys_cop(j); % add indpt probs
else % insert new co value and prob value
ind = find(co<sys2_co);
if length(ind)>0,
ind = min(ind); % insert values by shifting some to the end
sys2_co = [sys2_co(1:ind-1);co;sys2_co(ind:end)];
sys2_cop = [sys2_cop(1:ind-1);ABsys_cop(i)*CDsys_cop(j);sys2_cop(ind:end)];
else % append values to the end
sys2_co = [sys2_co;co];
sys2_cop = [sys2_cop;ABsys_cop(i)*CDsys_cop(j)];
end
end
end
end
sys2_capa = maxcap-sys2_co;

% eliminate storage of zero probability events
ind = find(sys1_cop);
sys1_capa = sys1_capa(ind);
sys1_cop = sys1_cop(ind);
fprintf('cap   prob_1\n')
fprintf('%3d   %6.4f\n',[sys1_capa,sys1_cop]')
ind = find(sys2_cop);
sys2_capa = sys2_capa(ind);
sys2_cop = sys2_cop(ind);
fprintf('cap   prob_2\n')
fprintf('%3d   %6.4f\n',[sys2_capa,sys2_cop]')
33

% create load/minute vector for one day (60 min/hr x 24 hr/day = 1440 min/day)
hourly_avg = [10 5 1 0.5 0 1 5 5 10 5 10 15 10 10 5 10 10 5 10 15 20 25 20 15];
for i = 1:24,
for j = 1:60,
end
end
fprintf('Total requests for today = %d, Average requests per minute = %3.1f\n',...

%% Monte Carlo simulation/determination of reliability (probabilistic capacity)
for i=1:1440,
Arand = rand;
Brand = rand;
Crand = rand;
Drand = rand;
sys1_cap(i) = min(Acap*(Arand<1-Afor),Ccap*(Crand<1-Cfor)) + ...
min(Bcap*(Brand<1-Bfor),Dcap*(Drand<1-Dfor));
sys2_cap(i) = min(Acap*(Arand<1-Afor)+Bcap*(Brand<1-Bfor),...
Ccap*(Crand<1-Cfor)+Dcap*(Drand<1-Dfor));
end

%% Calculation of load served (orders processed), given capacity and load
%% at each of 1440 samples
fprintf('System 1: cumulative avail today = %d, orders processed today = %d\n',...
sum(sys1_cap), sum(sys1_ord))
fprintf('System 2: cumulative avail today = %d, orders processed today = %d\n',...
sum(sys2_cap), sum(sys2_ord))

figure(1)
grid
xlabel('Time (min) [60 min/hr x 24 hr/day = 1440 min/day]')
ylabel('Load (# of order requests)')
figure(2)
xlabel('Requests per minute (# of potential orders)')
ylabel('Histogram of order requests (# of minutes at each load level)')

input('Press <return> to plot system availability vs time')

figure(2);
plot(sys1_cap,'+')
grid
xlabel('Time (min) [60 min/hr   x 24 hr/day = 1440 min/day]')
ylabel('System 1 Availability   (# of potential orders)')
figure(3);
plot(sys2_cap,'+')
grid
xlabel('Time (min) [60 min/hr   x 24 hr/day = 1440 min/day]')
ylabel('System 2 Availability   (# of potential orders)')

input('Press <return> to plot capability histograms')

figure(2);
hist(sys1_cap,sys1_capa(end:-1:1))
p1 = hist(sys1_cap,sys1_capa(end:-1:1))/1440;
34
%ind = find(n);
%ind = ind(end:-1:1);
%fprintf('%3d   %6.4f\n',[ind;n(ind)])
%fprintf('cap   prob\n')
%fprintf('%3d   %6.4f\n',[sys1_capa';n(end:-1:1)])
grid
xlabel('System 1 Availability per minute (# of potential orders)')
ylabel('Histogram of potential orders (# of minutes)')
figure(3);
hist(sys2_cap,sys2_capa(end:-1:1))
p2 = hist(sys2_cap,sys2_capa(end:-1:1))/1440;
%ind = find(n);
%ind = ind(end:-1:1);
%fprintf('%3d   %6.4f\n',[ind;n(ind)])
%fprintf('cap   prob\n')
%fprintf('%3d   %6.4f\n',[sys2_capa';n(end:-1:1)])
grid
xlabel('System 2 Availability per minute (# of potential orders)')
ylabel('Histogram of potential orders (# of minutes)')

%fprintf('cap    prob_1   cap   prob_2\n')
%fprintf('%3d    %6.4f   %3d   %6.4f\n',[sys1_capa';n1(end:-1:1);sys2_capa';n2(end:-
1:1)])
%ind = find(n1);
%n1 = n1(ind);
fprintf('cap   prob_1\n')
%fprintf('%3d    %6.4f\n',[sys1_capa(ind)';n1(end:-1:1)])
fprintf('%3d   %6.4f\n',[sys1_capa';p1(end:-1:1)])
%ind = find(n2);
%n2 = n2(ind);
fprintf('cap   prob_2\n')
%fprintf('%3d    %6.4f\n',[sys2_capa(ind)';n2(end:-1:1)])
fprintf('%3d   %6.4f\n',[sys2_capa';p2(end:-1:1)])
35

Appendices

   Activity Sheets for Teacher-Participants
o Activity 1: Pre-Assessment
o Activity 2: A Series System’s Reliability
o Activity 3: A Parallel System’s Reliability
o Activity 4: Combination of Series and Parallel Structures
o Project

 Activity Sheets for Students
o Activity 1: What’s Your Buying Pattern?
o Activity 2: Finding A Series System’s Reliability
o Activity 3: Effect of the Number of Components in a Series System
o Activity 4: Finding A Parallel System’s Reliability
o Activity 5: Effect of the Number of Components in a Parallel System
o Activity 6: Combination of Series and Parallel Structures
o Activity 7: Real-life Applications of the Bathtub Curve
o Project
 Module Assessment Sheets
o Pre-Assessment in Probability/Systems
o Objective Assessment of the Whole Concept
o Self Assessment (Rubric)
o Team Assessment
o Rubric for Reliability Project

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