The reliability of the computer system is made to the time t a probability of normal operation this time, it indicates the start operation (t = 0) by R (t).
Failure rate is the ratio of the total number of elements as the number of failures per unit time, he expressed in λ, when λ is constant, the relationship between reliability and failure rate are: R (t) = е ^ (- λt).
| System Type | reliability | Failure Rate |
| Series system | R=R1×R2×...×Rn | l = l 1 + l 2 + ... + k n |
| Parallel System | R=1-(1-R1)×(1-R2)×...×(1-Rn) |  |
| TMR system |  |
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1) serial system: Assuming a system of n sub-systems, if and only if all the subsystems have to work properly, the system to work properly, this system is called a tandem system
Each subsystem reliability system provided respectively R1, R2, R3 ......, Rn denotes the reliability of the system is
R = R1 × R2 × R3 × ...... × Rn
if the failure rate of each sub-system are represented by λ1 , λ2, λ3 ......, λn represented, the failure rate of the system is
λ = λ1 + λ2 + λ3 + ...... + λn
the system MTBF is:
the MTBF =. 1 / [lambda]
is assumed that the present problem three subsystems are connected in series, n is = 3, R1 = R2 = R3 = 0.9, λ1 = λ2 = λ3 = 1/10000 = 0.0001, then:
reliability R = R1 × R2 × R3 = 0.9 × 0.9 × 0.9 = 0.729
system failure rate λ = λ1 + λ2 + λ3 = 0.0001 + 0.0001 + 0.0001 = 0.0003
system MTBF =. 1 the MTBF / 0.0003 = 3333
(2) parallel system: if a system is composed of n sub-systems, as long as there is a sub-system to work properly, the system can normal work.
Each subsystem reliability system provided respectively R1, R2, R3 ......, Rn denotes the reliability of the system
R = 1- (1-R1) × (1-R2) × (1-R3) × ...... × (1-Rn)
if the failure rate of each sub-system are λ, the failure rate of the system is [mu] is
the mean time between failures of the system:
the MTBF =. 1 / [mu]
according to the present title understood meaning of the questions, n = 3, R1 = R2 = R3 = 0.9, λ1 = λ2 = λ3 = 1/10000 = 0.0001, then:
reliability R = (1-R1) × (1-R2) × (1-R3) = 1- (1-0.9 ) × (1-0.9) × (1-0.9 ) = 0.999
The system failure rate μ = 1 / ((1 / 0.0001) * (1/1 + 1/2 + 1/3)) = 6 / (10000 * 11)
System MTBF MTBF = 10000 * 11/6 = 18333
X-nines reliability software
X 9 represents a 1-year period during use of the system software, system uptime to the total time (1 year) ratio, usually 3 to 5.
- 3 9: (99.9%) * 365 * 24 = 8.76 hours, indicating that the software system in continuous operation time up to one year duration of service interruption may be 8.76 hours.
- 4 9: (1-99.99%) * 365 * 24 = 0.876 hours = 52.6 minutes, indicating that the software system in continuous operation time up to one year duration of service interruption may be 52.6 minutes.
- 5 9: (1-99.999%) * 365 * 24 * 60 = 5.26 minutes, indicating that the software system in continuous operation time up to one year duration of service interruption may be 5.26 minutes.
Then the 9-X where X represents the number 3 to 5, why not 1-2, and no more than 6 it? We went down to calculate:
- 9 1: (1-90%) * 365 = 36.5 days
- 2 9: (99%) * 365 = 3.65 days
- 6 9: (1-99.9999%) * 365 * 24 * 60 * 60 = 31 seconds
And you can see a 9, 2 9 respectively, within a year of service may be interrupted time is 36.5 days, 3.65 days, perhaps not this level of reliability with the use of "reliability" of the word; and 6 9 said the service interruption time in a year at most 31 seconds, then the level of reliability is not impossible to achieve, but to achieve increased from 9 reliability 5 "6 9, then the latter than the former to pay several times costs, so we all talk about in the enterprise (3-5) nines.