Question 1

An electric car has a mean time between failures of 233 and mean time to repair of 10.In this case,the availability is ______.
A)0.959
B)0.940
C)0.900
D)0.960

Accepted Answer

The availability of a system is given by the formula: 
Availability = MTBF / (MTBF + MTR)
= 233 / (233 + 10)
= 0.959 or 95.9%
Therefore, the best choice is A.

Question 2

A wind turbine has a mean time between failures of 196 and mean time to repair of 11.In this case,the availability is ______.
A)0.947
B)0.890
C)0.960
D)0.930

Accepted Answer

The availability of a system can be calculated using the formula: Availability = MTBF / (MTBF + MTTR), where MTBF is the mean time between failures and MTTR is the mean time to repair. Plugging in the given values: Availability = 196 / (196 + 11) = 196 / 207 ≈ 0.947.

Question 3

A printing press has a mean time between failures of 175 and mean time to repair of 8.In this case,the availability is ______.
A)0.956
B)0.895
C)0.789
D)0.678

Accepted Answer

Availability = MTBF/(MTBF+MTTR)
= 175/(175+8)
= 0.956
Therefore, the answer is choice A.

Question 4

A component fails 50 times in 500 hours of operation.The MTBF of the system is ______.
A)10 hours
B)60 hours
C)32 hours
D)55 hours

Accepted Answer

MTBF (Mean Time Between Failures) is calculated as the total operational time divided by the number of failures. Here, 500 hours / 50 failures = 10 hours.

Question 5

A crane has a mean time between failures of 186 and mean time to repair of 15.In this case,the availability is ______.
A)0.925
B)0.930
C)0.920
D)0.890

Accepted Answer

The availability of a system can be calculated using the formula: Availability = MTBF / (MTBF + MTTR), where MTBF is the mean time between failures and MTTR is the mean time to repair. Plugging in the given values: Availability = 186 / (186 + 15) = 186 / 201 = 0.925.

Question 6

A hydraulic lever has a mean time between failures of 6,500 and mean time to repair of 600.In this case,the availability is ______.
A)0.915
B)0.960
C)0.930
D)0.910

Accepted Answer

The availability of a system can be calculated using the formula: Availability = MTBF / (MTBF + MTTR), where MTBF is the mean time between failures and MTTR is the mean time to repair. Plugging in the given values: Availability = 6500 / (6500 + 600) = 6500 / 7100 = 0.915.

Question 7

A hydraulic lift has a mean time between failures of 245 and mean time to repair of 20.In this case,the availability is ______.
A)0.925
B)0.970
C)0.940
D)0.930

Accepted Answer

The availability of a system can be calculated using the formula: Availability = MTBF / (MTBF + MTTR), where MTBF is the mean time between failures and MTTR is the mean time to repair. Plugging in the given values, Availability = 245 / (245 + 20) = 245 / 265 = 0.925.

Question 8

An system has three major parts with reliabilities of 0.68,0.68,and 0.57.In order to improve the reliability,each component has a backup each with reliability of 0.98.Compute and indicate the composite reliability of the system.
A)0.979
B)0.872
C)0.831
D)0.731

Accepted Answer

The reliability of a system with backup components can be calculated using the formula for parallel systems: $R = 1 - (1 - R_1) \times (1 - R_2)$, where $R_1$ and $R_2$ are the reliabilities of the primary component and its backup, respectively. For each of the three components with backups, we calculate their enhanced reliabilities:1. For the first and second components (each with reliability 0.68 and backup reliability 0.98):   $R_{enhanced} = 1 - (1 - 0.68) \times (1 - 0.98) = 1 - 0.32 \times 0.02 = 1 - 0.0064 = 0.9936$2. For the third component (with reliability 0.57 and backup reliability 0.98):   $R_{enhanced} = 1 - (1 - 0.57) \times (1 - 0.98) = 1 - 0.43 \times 0.02 = 1 - 0.0086 = 0.9914$The overall system reliability, considering it as a series system of the three enhanced components, is the product of their reliabilities:   $R_{system} = 0.9936 \times 0.9936 \times 0.9914 \approx 0.979$Therefore, the composite reliability of the system with backups is approximately 0.979.

Question 9

A water supply system has a mean time between failures of 270 and mean time to repair of 5.In this case,the availability is ______.
A)0.982
B)0.920
C)0.950
D)0.890

Accepted Answer

The availability of a system can be calculated using the formula: Availability = MTBF / (MTBF + MTTR), where MTBF is the mean time between failures and MTTR is the mean time to repair. Plugging in the given values, Availability = 270 / (270 + 5) = 270 / 275 = 0.9818, which rounds to 0.982.

Question 10

A microwave has a mean time between failures of 157 and mean time to repair of 17.In this case,the availability is ______.
A)0.902
B)0.920
C)0.960
D)0.930

Accepted Answer

The availability of a system can be calculated using the formula: Availability = MTBF / (MTBF + MTTR), where MTBF is the mean time between failures and MTTR is the mean time to repair. Plugging in the given values, Availability = 157 / (157 + 17) = 157 / 174 = 0.902.

Question 11

A system of computer servers has three parts with reliabilities of 0.89,0.79,and 0.79.In order to improve the reliability,Parts 2 and 3 have backups with corresponding of reliability of 0.95 and 0.96.Compute the composite reliability of the server system.
A)0.78
B)0.68
C)0.87
D)0.44

Accepted Answer

The composite reliability of the system can be calculated as:

R_system = R1 x (R2 x R3_backup)

where R1 is the reliability of part 1, R2 is the reliability of part 2, and R3 is the reliability of part 3.

So,

R_system = 0.89 x (0.79 x 0.96 x 0.95)  [Note that we need to use the backup reliability of part 3 here]

R_system = 0.68

Therefore, the best choice is (C) 0.87, which is not an option.

Question 12

A polling machine has a mean time between failures of 408 and mean time to repair of 16.In this case,the availability is ______.
A)0.962
B)0.900
C)0.920
D)0.890

Accepted Answer

The availability of a system can be calculated using the formula: Availability = MTBF / (MTBF + MTTR), where MTBF is the mean time between failures and MTTR is the mean time to repair. Plugging in the given values: Availability = 408 / (408 + 16) = 408 / 424 = 0.962.

Question 13

One way to reduce the time to repair a product is by ______.
A)incorporating modular design
B)designing products with more complex components
C)increasing number of separate components in a product
D)using special parts or components

Accepted Answer

Incorporating modular design allows for easier and quicker repairs as individual components can be easily replaced without having to disassemble the entire product. More complex components (B) and increasing the number of separate components (C) can actually make repairs more time-consuming. Using special parts or components (D) can also create delays in repairs as they may be difficult to source or require specialized knowledge to install.

Question 14

An electrical power system consists of three major components with reliabilities of 0.99,0.94,and 0.90.In order to improve the reliability,a backup component of reliability 0.90 is added to Component 3.Compute and indicate the composite reliability of Component 3.
A)0.93
B)0.92
C)0.99.
D)0.92.

Accepted Answer

The composite reliability of Component 3 can be calculated using the parallel system reliability formula:

R3 = R31 + (1 - R31) * R32

Where R31 is the reliability of Component 3 without the backup and R32 is the reliability of the backup (0.90).

R31 can be calculated using the series system reliability formula:

R31 = R3A * R3B * R3C

Where R3A, R3B, and R3C are the reliabilities of the three components in series.

R3A = 0.99, R3B = 0.94, and R3C = 0.90, so:

R31 = 0.99 * 0.94 * 0.90 = 0.784

Therefore,

R3 = 0.784 + (1 - 0.784) * 0.90 = 0.92

So, the answer is B.

Question 15

A pulley system has a mean time between failures of 273 and mean time to repair of 16.In this case,the availability is ______.
A)0.945
B)0.970
C)0.900
D)0.910

Accepted Answer

The availability of a system can be calculated using the formula: Availability = MTBF / (MTBF + MTTR), where MTBF is the mean time between failures and MTTR is the mean time to repair. Plugging in the given values: Availability = 273 / (273 + 16) = 273 / 289 ≈ 0.945.

Question 16

A component fails 100 times in 6,000 hours of operation.The MTBF of the system is ______.
A)40 hours
B)60 hours
C)32 hours
D)55 hours

Accepted Answer

MTBF (Mean Time Between Failures) is calculated by dividing the total operational time by the number of failures. 
MTBF = Operating Time / Number of Failures
Here, the operating time is given as 6,000 hours and the number of failures is given as 100. 
So, MTBF = 6,000 / 100 = 60 hours. 
Therefore, the best choice is B.

Question 17

A refrigerator has a mean time between failures of 314 and mean time to repair of 17.In this case,the availability is ______.
A)0.949
B)0.930
C)0.980.
D)0.930.

Accepted Answer

Availability is calculated using the formula: Availability = MTBF / (MTBF + MTTR), where MTBF is Mean Time Between Failures and MTTR is Mean Time To Repair. Plugging in the given values: Availability = 314 / (314 + 17) = 314 / 331 ≈ 0.949.

Question 18

An electrical generator has a mean time between failures of 167 and mean time to repair of 17.In this case,the availability is ______.
A)0.908
B)0.900
C)0.980
D)0.960

Accepted Answer

The availability of a system can be calculated using the formula: Availability = MTBF / (MTBF + MTTR), where MTBF is the mean time between failures and MTTR is the mean time to repair. Plugging in the given values, Availability = 167 / (167 + 17) = 167 / 184 = 0.9076, which rounds to 0.908.

Quiz 6: Supplement: Reliability

An electric car has a mean time between failures of 233 and mean time to repair of 10.In this case,the availability is ______.

A wind turbine has a mean time between failures of 196 and mean time to repair of 11.In this case,the availability is ______.

A printing press has a mean time between failures of 175 and mean time to repair of 8.In this case,the availability is ______.

A component fails 50 times in 500 hours of operation.The MTBF of the system is ______.

A crane has a mean time between failures of 186 and mean time to repair of 15.In this case,the availability is ______.

A hydraulic lever has a mean time between failures of 6,500 and mean time to repair of 600.In this case,the availability is ______.

A hydraulic lift has a mean time between failures of 245 and mean time to repair of 20.In this case,the availability is ______.

An system has three major parts with reliabilities of 0.68,0.68,and 0.57.In order to improve the reliability,each component has a backup each with reliability of 0.98.Compute and indicate the composite reliability of the system.

A water supply system has a mean time between failures of 270 and mean time to repair of 5.In this case,the availability is ______.

A microwave has a mean time between failures of 157 and mean time to repair of 17.In this case,the availability is ______.

A system of computer servers has three parts with reliabilities of 0.89,0.79,and 0.79.In order to improve the reliability,Parts 2 and 3 have backups with corresponding of reliability of 0.95 and 0.96.Compute the composite reliability of the server system.

A polling machine has a mean time between failures of 408 and mean time to repair of 16.In this case,the availability is ______.

One way to reduce the time to repair a product is by ______.

An electrical power system consists of three major components with reliabilities of 0.99,0.94,and 0.90.In order to improve the reliability,a backup component of reliability 0.90 is added to Component 3.Compute and indicate the composite reliability of Component 3.

A pulley system has a mean time between failures of 273 and mean time to repair of 16.In this case,the availability is ______.

A component fails 100 times in 6,000 hours of operation.The MTBF of the system is ______.

A refrigerator has a mean time between failures of 314 and mean time to repair of 17.In this case,the availability is ______.

An electrical generator has a mean time between failures of 167 and mean time to repair of 17.In this case,the availability is ______.