Part 3: Product, Process and Service Design Flashcards
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
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
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
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.
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
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
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
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.
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
C - Geometric tolerance
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
A - Primary Datum
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
A - Installation qualification
The goal is to make sure the equipment involved is installed properly.
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
C - Concept phase
In the concept phase, all possible solutions are explored. Non-feasible solutions are eliminated.
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
B - Quality function deployment (QFD)
QFD which uses the “house of quality” is helpful in turning customer needs into product characteristics or design concepts
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
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.
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?
B - Is the failure rate acceptably low?
Reliability includes metrics such as failure rates for the design under consideration.
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
C - 7.95 and 8.05
This bilateral tolerance with allowable variation in both directions.
For a process, confirming that specified requirements have been fulfilled is known as:
A) validation
B) verification
C) qualification
D) performance
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).
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
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)
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
D - Design for six sigma
Design for Six Sigma is an approach used to eliminate problems before they occur
True/False. The reliability of a product can be measured at its release time.
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.
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
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.
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
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
Which of the following probability distributions is the only one to have a constant failure rate?
A) Uniform
B) Normal
C) Exponential
D) Weibull
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.
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
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
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
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
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
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
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
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
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
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.
True/False: In a parallel system, if one subsystem fails, the entire system will fail.
A) True
B) False
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.
True/False: The reliability of a series system is lower than its weakest subsystem.
A) True
B) False
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.
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
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
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
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
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
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
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
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
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
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
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%
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
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
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
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
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
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%
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%
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
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
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
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
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
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.
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
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.
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
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
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
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.
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
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.
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.
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.
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
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.
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
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
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
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
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.
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.
