Part 3: Product, Process and Service Design

Question 1

Which of the following quality characteristics is an attribute?

A) Body temperature
B) Number of defects
C) Length of a bolt
D) Width of a board

Answer 1

B - Number of Defects

a defect is an attribute. Either defect is present or it is not. A defect has not real measurement on a continuous scale

Question 2

A part will be inspected using a limit gage. An acceptable part is one that:

A) mates with both ends of the limit gage (both “go” and “no-go”)
B) mates only with the “no-go” end of the gage
C) mates with neither end of the limit gage
D) mates only with the “go” end of the limit gage

Answer 2

D - mates only with the “go” end of the limit gage

An acceptable part should only fit within the “go” end of the limit gage. If it passes on both ends, it is an unacceptable part.

Question 3

A system has two components in parallel, each with a reliability of 0.95. The system reliability is approximately:

A) 0.9999
B) 0.9500
C) 0.9975
D) 0.0975

Answer 3

C - 0.9975

For components in parallel, each with reliability of p, the system reliability is given by
Rs(t) = 1 - (1 - 0.95)^2 = 0.9975

Question 4

Consider the product/process development life cycle. The phase in which the product or process is released to the customer for use is referred to as the:

A) prototype phase
B) production phase
C) distribution phase
D) normal use phase

Answer 4

D - normal use phase

The normal use phase of the development life cycle includes release of the product/process to the customer for its intended use or function. Normal wear and maintenance occur in this phase as well.

Question 5

Consider the following control frame.
//     (L)     0.01     A     B     C

0.01 represents the:
A) primary datum
B) geometric feature or characteristic
C) geometric tolerance
D) modifier

Answer 5

C - Geometric tolerance

Question 6

Consider the following control frame.
//     (L)     0.01     A     B     C

“A” represents the:

A) Primary Datum
B) Secondary Datum
C) Tertiary Datum
D) modifier

Answer 6

A - Primary Datum

Question 7

Which of the following involves demonstrating that all equipment being used in a process has been installed properly (as per manufacturer’s specifications)?

A) Installation qualification
B) Operational qualification
C) Performance qualification
D) Software qualification

Answer 7

A - Installation qualification

The goal is to make sure the equipment involved is installed properly.

Question 8

Consider the product/process development life cycle. The phase in which all passible solutions are explored is referred to as the:

A) specification phase
B) prototype phase
C) concept phase
D) distribution phase

Answer 8

C - Concept phase

In the concept phase, all possible solutions are explored. Non-feasible solutions are eliminated.

Question 9

A tool that is useful for converting customer needs into design concepts is:

A) process validation
B) quality function deployment
C) statistical process control
D) acceptance sampling

Answer 9

B - Quality function deployment (QFD)

QFD which uses the “house of quality” is helpful in turning customer needs into product characteristics or design concepts

Question 10

Design reviews should be conducted by:

A) top management only
B) a quality engineer
C) a cross-functional team including quality, manufacturing, and suppliers, and so on
D) Customers only

Answer 10

C - a cross functional team including quality, manufacturing, suppliers, etc.

Because of the complexity of design review, a cross functional team consisting of many different personnel should conduct the review. The ream should include, but not be limited to, quality personnel, manufacturing, suppliers, customers, etc.

Question 11

For each design review conducted, there are many different points of view from which the design should be considered. One of these viewpoints is from a reliability standpoint. Which of the following would be a question from a reliability view point?

A) Can the design be sufficiently tested?
B) Is the failure rate acceptably low?
C) Can the product be shipped without damage?
D) Can the parts needed for the design be obtained at an acceptable cost?

Answer 11

B - Is the failure rate acceptably low?

Reliability includes metrics such as failure rates for the design under consideration.

Question 12

A specification is given as 8 (+.05/-.05). The limits for this specification are:

A) 8.00 and 8.05
B) 7.95 and 8.00
C) 7.95 and 8.05
D) 8. 5

Answer 12

C - 7.95 and 8.05

This bilateral tolerance with allowable variation in both directions.

Question 13

For a process, confirming that specified requirements have been fulfilled is known as:

A) validation
B) verification
C) qualification
D) performance

Answer 13

B - Verification

Verification deals with determining whether the product or process satisfies specified requirements, that is, it meets all specifications or requirements. But this does not guarantee that the product or process specifications meet or exceed predetermined requirements (that is, it may not work like it is supposed to).

Question 14

Validation of a product, process, or service is normally carried out:

A) during production or development
B) when customer requirements are determined
C) after completion of the product, process or service
D) None of the above

Answer 14

C - After completion of the product, process, or service

Validation is used to determine whether predetermined requirements are being met (product/process/service is working as intended)

Question 15

The Design for X concept that focuses on eliminating problems before they occur is known as:

A) Design for reliability
B) Design for cost
C) Design for manufacturability
D) Design for six sigma

Answer 15

D - Design for six sigma

Design for Six Sigma is an approach used to eliminate problems before they occur

Question 16

True/False. The reliability of a product can be measured at its release time.

Answer 16

False

Reliability can not be measured at the release time of the product; it can only be predicted. This is because the reliability of a product is dependent on many factors, including the system design, the reliability of its components, the operating environment, the environmental factor, san manufacturing defects.

Question 17

The cumulative distribution function of failure at time t for a type of air conditioner is defined as F(t) = 4/5000. Which of the following statements is/are true

A) The probability that a randomly selected air conditioner fails by time t is 4/5000
B) The probability that a randomly selected air conditioner will instantly fail by time t is 4/5000
C) the fraction of all units in the lot of air conditioners that fail by time t is 4/5000
D) A and C
E) All of the above

Answer 17

D - A and C

The cumulative distribution function (cdf) of failure F(t) has two interpretations;

1) the probability that a randomly selected unit from the population will fail be time t
2) the fraction of all units in the population that fail by time t

The hazardous rate function h(t) defines the instantaneous failure rate at time t.

Question 18

The cumulative distribution function of failure at time t for a computer model is 0.992. What fraction of this computer model fails by time t?

A) 0.992
B) 0.008
C) 1
D) None of the above

Answer 18

A - 0.992

The cumulative distribution function (cdf) of failure can be interpreted as the fraction of all units in the population that fail by time t. Since F(t) = 0.992 for this computer model, the fraction of this model that fail by time t is 0.992

Question 19

Which of the following probability distributions is the only one to have a constant failure rate?

A) Uniform
B) Normal
C) Exponential
D) Weibull

Answer 19

C - Exponential

The exponential distribution is the only probability distribution to have a constant failure rate. For an exponential distribution with parameter λ, the hazard rate function h(t) = λ.

Question 20

450 electrical components for a machine were subjected to a reliability test. The observed failures during 100-hour intervals are shown in the following table.

Upper bound (hrs)     Number of failures, x
0                                  0
100                              7
200                             16
300                             19
400                             27

What is the estimated reliability at time t=200?

A) 0.0156
B) 0.9489
C) 0.0511
D) 0.0004

Answer 20

B - 0.9489

Upper bound (hrs) Number of failures, x Number of survivors
0 0 450
100 7 443
200 16 427
300 19 408
400 27 381

The estimated reliability at time t = 200 is
R(t)= n2/N= 427/450 = 0.9489

Question 21

450 electrical components for a machine were subjected to a reliability test. The observed failures during 100-hour intervals are shown in the following table.

Upper bound (hrs)     Number of failures, x
0                                  0
100                              7
200                             16
300                             19
400                             27

What is the estimated failure density for time 300 < t < 400?

A) 0.0006
B) 0.9067
C) 0.0933
D) 0.8467

Answer 21

A - 0.0006

Upper bound (hrs) Number of failures, x Number of survivors
0 0 450
100 7 443
200 16 427
300 19 408
400 27 381

The failure density is the probability density function defined as
f(t) = (n3-n4) / ((t3-t4) × N) = (408-381) / ((400-300) × 450) = 0.0006

Question 22

450 electrical components for a machine were subjected to a reliability test. The observed failures during 100-hour intervals are shown in the following table.

Upper bound (hrs)     Number of failures, x
0                                  0
100                              7
200                             16
300                             19
400                             27

What is the instantaneous rate of failure for time 300 < t < 400

A) 0.90667
B) 0.09333
C) 0.15333
D) 0.00066

Answer 22

D - 0.00066

Upper bound (hrs) Number of failures, x Number of survivors
0 0 450
100 7 443
200 16 427
300 19 408
400 27 381

The instantaneous rate of failure is the hazard rate function h(t) where:

h(t) = f(t) / h(t) = (n3-n4) / ((t3-t4) × n3) = (408 - 381) / ((400 - 300) × 408) = 0.00066

Question 23

A series system is composed of 5 subsystems, each with reliability 0.995. What is the reliability of the system?

A) 0.9950
B) 0.9999
C) 0.9752
D) None of the above

Answer 23

C - 0.9752

A series system has n components connected end to end. If one component fails. The reliability of a series system made up of n components is:
Rs(t) = R1(t) × R2(t) × … ×Rn(t)

For this series system, the system reliability, therefore is:
Rs(t) = 0.995^5 = 0.9752

Question 24

A parallel system is composed of 5 subsystems, each with reliability 0.995. The reliability of this system is closest to which of the following values?

A) 0.9950
B) 1.0000
C) .9752
D) None of the above

Answer 24

B - 1.0000

A parallel system is made up of components so that if one or more paths fail, the remaining path(s) are still able to perform properly. Therefore, the system fails when all units fail. The reliability of a parallels system is:
Rs(t) = 1 - F1(t) x F2(t) x … x Fn(t)

For this series system, the system reliability therefore is:
Rs(t) = 1 - ((1-0.995)^5) = 0.999999

The reliability of the system is essentially 1.

Question 25

True/False: In a parallel system, if one subsystem fails, the entire system will fail.

A) True
B) False

Answer 25

B - False

A parallel system is made of n components connected in parallel. Therefore, the system will fail only if all the components or subsystems fail.

Question 26

True/False: The reliability of a series system is lower than its weakest subsystem.

A) True
B) False

Answer 26

A - True

A series system is mad up of n components connected end to end. Therefore, if one of the components fails, the entire system fails. The system reliability, therefore, is the product of the reliability of each sub system. This means that the system reliability will be lower than the reliability of the subsystem with the lowest reliability.

Question 27

A machine has six subsystems and will fail when four or more of these subsystems fail. Each subsystem has a reliability of .088. What is the reliability of the machine?

A) 0.5997
B) 0.4644
C) 0.9998
D) 0.9975

Answer 27

D - 0.9975

The machine has a k-out-of-n system design where k = 3 and n = 6. The system reliability of a k-out-of-n system is:
Rs(t) = ∑[^n_(i=k)] = (n i) x p^i x (1-p)^(n-i)
where p = 0.88

Therefore the system reliability is
= (6 3)(0.88^3)(0.12^3) + (6 4)(0.88^4)(0.12^2) + (6 5)(0.88^5)(0.12^1) + (6 6)(0.88^6)(0.12^0)
= (20)(0.88^3)(0.12^3) + (15)(0.88^4)(0.12^2) + (6)(0.88^5)(0.12^1) + (1)(0.88^6)(0.12^0)
= 0.02355 + 0.1295 + 0.3799 + 0.4644
= .9975

Question 28

A machine has a standby system design with four identical components in standby mode. Each component has a constant rate of failure λ = 0.08. What is the reliability of the machine at 100 hours of continuous operation?

A) 0.9999
B) 0.0001
C) 0.0424
D) 0.7164

Answer 28

C - 0.0424

The reliability of a standby system with n standby components is
Rs(t) = e^(-λt) ∑ [^n _(i=0)] (λt)^i / i!

In this case, the machine is a standby system where n = 4, λ = 0.08, and t = 100.

Therefore the system reliability is
Rs(t) = e^(-.08)(100) x ∑ [ ^4 _i=0] (0.08x100)^i / i!
Rs(t) = e^(-8) x [(8^0)/0! + (8^1)/1! + (8^2)/2! + (8^3)/3!
Rs(t) = 0.0424

Question 29

750 machines were tested for 5 hours. 28 failures occurred. What is the failure rate for these machines?

A) 133
B) 0.037
C) 27
D) 0.0075

Answer 29

D - 0.0075

The failure rate, λ, is the number of failures divided by the total test time.
Therefore,
λ = (28) / (750 x 5) = 0.0075

Question 30

The reliability of a component at t = 400 hours is 0.645. Assuming that the reliability of this component is exponential, which of the following is closest to the mean time to failure (MTTF)?

A) 0.0011
B) 910
C) 0.355
D) 3

Answer 30

B = 910

We are given that R(400) = 0.645, Recall that MTTF = 1/failure rate = 1 - λ. 
Since the reliability of the component is exponential, we have
R(t) = e^(-λt)
e^(-λ)(400) = 0.645
ln(e^(-λ)(400)) = ln(0.645)
-400λ = ln(0.645)
λ = ln(0.645) / (-400)
λ = 0.0011

Therefore,
MTTF = 1 / λ
= 1 / 1.0011
= 909.09 ≈ 910 hours

Question 31

The MTTF for a model of dishwasher is 115 months. What is the reliability that a randomly selected dishwasher of this model type survives past 115 months? Assume that the reliability function is an exponential distribution.

A) 0.368
B) 2.718
C) 0.009
D) 0.632

Answer 31

A = 0.368

MTTF = 115 - (1 / λ) = 1 / Failure Rate
This means that the failure rate, λ, is λ = 1/115. Since the reliability of the dishwasher has an exponential distribution, the probability that a randomly selected dishwasher survives past t months is R(t) = e^-λt. Therefore, the probability that a randomly selected dishwasher survives past 115 months is

R(115) = e^(-115/115) = e^-1 = 0.3697

Question 32

The MTTF for a machine is 525 hours. What fraction of machines survives past 1000 hours? Assume that the reliability of the machine has an exponential distribution

A) 85.12%
B) 14.88%
C) 0.19%
D) 99.81%

Answer 32

B - 14.88%

Another interpretation of reliability is the faction of all machines that survives past a time t. We can find this fraction using the reliability function R(t) = e^-λt.

Since MTTF = 525,
λ = 1 / MTTF
λ = 1 / 525

R(1000) = e^(-1000/525)
R(1000) = 0.1488

Question 33

What is the probability a randomly selected rechargeable battery fails by six hours is the mean time between failures (MTBF) is five hours? Assume that the reliability of the component is exponentially distributed.

A) 0.3012
B) 0.1667
C) 0.8333
D) 0.6988

Answer 33

D - 0.6988

The rechargeable battery is a repairable system (via recharging). Since the mean time between failures (MTBF) is five hours, the failure rate λ = 1/5 = 0.2.
The probability that a battery fails by time t describes the failure function:
F(t) = 1 - e^(-6/5)
F(t) = 1 - 0.3012 = 0.6988

Question 34

If the MTBF of a component is 900 hours, what is the reliability of this component at 500 hours? Assume that the reliability of the component is exponentially distributed.

A) 0.1653
B) 0.8347
C) 0.4262
D) 0.5738

Answer 34

D - 0.5738

The mean time between failures (MTBF) is 900 hours. Therefore, the failure rate λ = 1/900. Assuming that the reliability of the component is exponential, the reliability at time t = 500 is defined as

R(500) = e^-λt
R(500) = e^(-500/900)
R(500) = 0.5738

Question 35

The failure rate of a machine is 0.0008. Assuming the life distribution of this type of machine is exponential, what fraction of machines in the population survives longer than the MTTF?

A) 36.79%
B) 63.21%
C) 1250
D) 99.92%

Answer 35

A - 36.79%

The reliability can be interpreted as the fraction of all components that survive past some time t.

Therefore, R(t) = e^-λt
and R(MTTF) = e^-λ(MTTF)

Since the failure for this component is λ = 0.0008
MTTF = 1/λ = 1/0.0008 = 1250
R(MTTF) = R(1250) = e^(-.0008)(1250) = 0.3679 = 36.79%

Question 36

The failure rate of a machine is 0.0005. Assuming the life distribution of this type of machine is exponential and the MTTF is five hours, what is the steady-state availability of this machine?

A) 0.004
B) 1250
C) 0.996
D) 1255

Answer 36

C - 0.996

The failure rate for this machine is λ=0.0008, which means that the mean time between failures (MTBF) is
1/λ = 1/0.0008 = 1250 hours

Since MTTR = 5 hours, the steady-state availability of the machine is

A = MTBF/(MTBF + MTTR) = 1250/(1250+5) = 0.996

Question 37

In a test of a repairable component of a machine, 600 components were tested for 25 hours and 515 components failed. If the mean time to repair (MTTR) these components is eight hours, what is the steady-state availability of this component?

A) 29.13
B) 0.2155
C) 0.7845
D) 0.0343

Answer 37

C = 0.7845

λ = Failure rate = #failures / Total test time
λ = 515 / (600 x 25) = 0.0343

Therefore,
MTBF = 1/λ = 1/0.0343 = 29.1262

Since the mean time to repair the components is eight hours, the steady-state availability of the component is

A = MTBF / (MTBF + MTTR)
A = 29.1262 / (29.1262 + 8) = 0.7845

Question 38

Which type of maintenance policy allows for the maximum run time between repairs?

A) Preventative maintenance policy
B) Corrective maintenance policy
C) Predictive maintenance policy
D) None of the above

Answer 38

B - Corrective maintenance policy

A corrective maintenance policy does not require any repairs or other preventative maintenance until a failure occurs. Therefore, this type of policy allows for the maximum run time between repairs.

Question 39

A preventative maintenance policy is recommended if:

A) The cost to repair the system after its failure is less than the cost of maintaining the system before its failure.
B) the failure rate of the function of the system is monotonically increasing.
C) The cost to repair the system after its failure is greater than the cost of maintaining the system before its failure
D) Both B and C
E) None of the above

Answer 39

D - Both B and C

A preventative maintenance policy is recommended when the cost to repair the system after its failure is greater than the cost of maintaining the system before its failure and the failure rate is monotonically increasing. If the failure rate is decreasing, then the system is more likely to improve over time, and preventative actions would be a waste of resources.

Question 40

In the second region identified in the general failure rate model (the bathtub curve), what is the most common cause for failures?

A) The manufacturing process
B) Fatigue
C) Random Failures
D) All of the above

Answer 40

C - Random failures

The second region of the bathtub curve is the constant failure rate region. This region is characterized by the inherent failure rate of the components of a product

Question 41

Which of the following probability distributions is often used to model the second region of the bathtub curve?

A) Uniform
B) Weibull
C) Exponential
D) Normal

Answer 41

C - Exponential

The second region of a bathtub curve is characterized by a constant failure rate. The exponential distribution is the only probability distribution that has a constant failure rate; therefore, the exponential distribution is often used to model the second region of the bathtub curve.

Question 42

Which of the following probability distributions is often used to model the third regain of the bathtub curve?

A) Uniform
B) Weibull
C) Exponential
D) Normal

Answer 42

B - Weibull

The third region of the bathtub curve, the wear-out region, is characterized by an increasing failure rate over time. This time-dependent failure rate is often modeled using the Weibull distribution.

Question 43

What is the main goal of FMEA?

A) To help reduce the negative effects of potential failures before they occur.
B) To organize the workplace environment.
C) To communicate hierarchical relationships between events.
D) To reduce the number of defects produced in a process.

Answer 43

A - To help reduce the negative effects of potential failures before they occur.

The main goal of an FMEA is to help a team identify and eliminate the potential negative effects of a potential failure before it occurs.

Question 44

A hospital is preparing to do an FMEA for a surgical procedure. Which of the following groups of employees should make up the FMEA team members?

A) One doctor that performs this surgical procedure
B) A nurse and a doctor
C) A doctor, nurse, technician, engineer, business office employee, and manager
D) None of the above

Answer 44

C - A doctor, nurse, technician, engineer, business office employee, and manager

An FMEA team should be composed of a cross-section of employees across several fields. At a minimum, the FMEA team should have a doctor, nurse, technician, engineer and manger. An FMEA team should have members that are both technical and nontechnical to have many viewpoints represented during the process.

Question 45

Which of the following are components of a FMEA?

A) Potential failure modes
B) The risk associated with potential failure modes
C) Corrective action plans
D) All of the above

Answer 45

D = All of the above

An FMEA has several components:
Potential failure modes
Risk associated with each failure mode
Corrective action plans (for most significant failure modes)
Documentation on the system

Question 46

An FMEA team has identified a potential failure and assessed the following score for severity, occurrence, and detection for this failure mode: 8, 2, 6. What is the RPN for this failure mode?

A) 16
B) 8
C) 96
D) None of the above

Answer 46

C - 96

The Risk Priority Number (RPN) is determined by multiplying the three components of risk together: severity, occurrence, and detection. Therefore,
RPN = 2 x 8 x 6 = 96

Question 47

What is the main goal of an FMECA?

A) To reduce the number of defects produced in a process
B) To develop a prioritization scheme for corrective action plans.
C) To organize the workplace environment
D) To communication hierarchical relationships between events.

Answer 47

B - To develop a prioritization scheme for corrective action plans.

An FMECA is an assessment of risk that provides a prioritization of corrective action based on severity and occurrence of failure modes.