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How to Calculate System Reliability 2
- Thread starter jay165
- Start date
PUMPDESIGNER
Mechanical
Reliabiity electrically, hydraulically, off line, cost, what?
Very simplistic and casual analyses would be 3 times the pumps = 3 times the problems, plus you must add in the added complexity of co-ordinating the 3 pumps electrically and hydraulically, so the final result could be 4-6 times less reliable if the engineer knew what he/shee was doing, 100 times less reliable if the engineer does not know much about multiple pumps, and I am not exaggerating on that.
Soooo,
You start at 85% reliability for all three pumps as a whole, and with a good engineer on board that probably is reduced to 70% - 80% actual for the whole system, and with an unknowledgeable designer on board you can get down to almost zero %, and do not laugh because I have seen it.
PUMPDESIGNER
Very simplistic and casual analyses would be 3 times the pumps = 3 times the problems, plus you must add in the added complexity of co-ordinating the 3 pumps electrically and hydraulically, so the final result could be 4-6 times less reliable if the engineer knew what he/shee was doing, 100 times less reliable if the engineer does not know much about multiple pumps, and I am not exaggerating on that.
Soooo,
You start at 85% reliability for all three pumps as a whole, and with a good engineer on board that probably is reduced to 70% - 80% actual for the whole system, and with an unknowledgeable designer on board you can get down to almost zero %, and do not laugh because I have seen it.
PUMPDESIGNER
-
2
- #4
Risk assessment means using probabilities.
When dealing with units in parallel the failure of the system requires both pumps to fail, since there is no other convenient way to combine failure rates. Thus, if each pump has a failure rate of 0.05, their individual reliability R would be = e[sup]-0.05t[/sup] = 0.95. Their failure probability would be P = 1-R = 0.05.
When dealing with parallel units their failure probabilities are multiplied. Thus, the total reliability for 2 pumps in paralel, R[sub]T[/sub]=1-(0.05*0.05)=0.9975.
If the three pumps were working in parallel:
R[sub]T[/sub]=1-(0.05*0.05*0.05)=0.9998
Now, assuming pumpdesigner's experience take a failure rate of 0.7, The individual reliability would be
e[sup]-0.7t[/sup]=0.5. The failure probability: 1-0.5=0.5.
The total reliability of 2 units in parallel: 1-(0.5*0.5)=0.75, and for 3 pumps in parallel: 1-(0.5*0.5*0.5)=0.875.
When dealing with units in parallel the failure of the system requires both pumps to fail, since there is no other convenient way to combine failure rates. Thus, if each pump has a failure rate of 0.05, their individual reliability R would be = e[sup]-0.05t[/sup] = 0.95. Their failure probability would be P = 1-R = 0.05.
When dealing with parallel units their failure probabilities are multiplied. Thus, the total reliability for 2 pumps in paralel, R[sub]T[/sub]=1-(0.05*0.05)=0.9975.
If the three pumps were working in parallel:
R[sub]T[/sub]=1-(0.05*0.05*0.05)=0.9998
Now, assuming pumpdesigner's experience take a failure rate of 0.7, The individual reliability would be
e[sup]-0.7t[/sup]=0.5. The failure probability: 1-0.5=0.5.
The total reliability of 2 units in parallel: 1-(0.5*0.5)=0.75, and for 3 pumps in parallel: 1-(0.5*0.5*0.5)=0.875.
- Thread starter
- #5
25362 is correct, how ever to add some more details for the system mentioned by you i.e. three pumps in parallel, with two pumps running and one as stand by, the reliability of the system is as guven below
R = (R1*R2*R3)+(R1*R2*P3)+(R1*P2*R3)+(P1*R2*R3)
Where R1, R2, R3 are the reliability of individual pumps and P1, P2, P3 are the failure probalility
Hence System reliability
R = 0.95*0.95*0.95 + 0.95*0.95*0.05 + 0.95*0.05*0.95 + 0.05*0.95*0.95
R = 0.99275
R = 99.275 %
V.Venkat
R = (R1*R2*R3)+(R1*R2*P3)+(R1*P2*R3)+(P1*R2*R3)
Where R1, R2, R3 are the reliability of individual pumps and P1, P2, P3 are the failure probalility
Hence System reliability
R = 0.95*0.95*0.95 + 0.95*0.95*0.05 + 0.95*0.05*0.95 + 0.05*0.95*0.95
R = 0.99275
R = 99.275 %
V.Venkat
to j165,
It was another (deleted) thread and forum where you mentioned the probability of getting a head in the fifth throw of a coin after getting 4 consecutive tails.
Allow me to tell you my thoughts, although I'm not an expert in probabilistics, on why one is generally in error when assuming always an a priori probability of 1/2 for all cases.
When looking for a probability in a sequence of events we apply Bayes' theorem of conditional probability, which states that the probability of A happening, given C as a prior event, is P(A/C) = P(A&C)/p(C)
In an unknown family with two children, the four possible time sequences are GG, GB, BB, BG to which we attached the same probability. Knowing that at least one child is a girl (G), P(C)=3/4. To find the probability P(A) of whether the other child is also a girl (GG), we estimate it from P(A&C)=(1/4). Then P(A)=(1/4)/(3/4)=1/3.
Easily seen w/o any theoretical help just from counting the cases.
In short, to estimate the correct probability of an event one has to see the right context, or, in technical jargon, build the right model.
If one knows beforehand that four throws resulted in tails, you are absolutely right in that the chances for the next being heads is 1/2. The same would happen with the family above, had we said the first born child was a girl. In this case we are left with two options: GB, and GG, and the chances are again 1/2.
If one knows more about the family or about genetical or hereditary influences, or about the coin's metallurgy or geometry, or of any external influences, barring the chance of its standing on the edge, the conclusion isn't right simply because the model used isn't right.
If one asked what is the probability of exactly one head in five throws of a coin the probability is 1/5. The model would then be composed of: HTTTT, THTTT, TTHTT, TTTHT, TTTTH.
If the question were about the probability of getting "at least" one H, then again the model changes and so does the probability. Can I say: Quod erat demonstrandum ?![[pipe]](/data/assets/smilies/pipe.gif "[pipe] [pipe]")
It was another (deleted) thread and forum where you mentioned the probability of getting a head in the fifth throw of a coin after getting 4 consecutive tails.
Allow me to tell you my thoughts, although I'm not an expert in probabilistics, on why one is generally in error when assuming always an a priori probability of 1/2 for all cases.
When looking for a probability in a sequence of events we apply Bayes' theorem of conditional probability, which states that the probability of A happening, given C as a prior event, is P(A/C) = P(A&C)/p(C)
In an unknown family with two children, the four possible time sequences are GG, GB, BB, BG to which we attached the same probability. Knowing that at least one child is a girl (G), P(C)=3/4. To find the probability P(A) of whether the other child is also a girl (GG), we estimate it from P(A&C)=(1/4). Then P(A)=(1/4)/(3/4)=1/3.
Easily seen w/o any theoretical help just from counting the cases.
In short, to estimate the correct probability of an event one has to see the right context, or, in technical jargon, build the right model.
If one knows beforehand that four throws resulted in tails, you are absolutely right in that the chances for the next being heads is 1/2. The same would happen with the family above, had we said the first born child was a girl. In this case we are left with two options: GB, and GG, and the chances are again 1/2.
If one knows more about the family or about genetical or hereditary influences, or about the coin's metallurgy or geometry, or of any external influences, barring the chance of its standing on the edge, the conclusion isn't right simply because the model used isn't right.
If one asked what is the probability of exactly one head in five throws of a coin the probability is 1/5. The model would then be composed of: HTTTT, THTTT, TTHTT, TTTHT, TTTTH.
If the question were about the probability of getting "at least" one H, then again the model changes and so does the probability. Can I say: Quod erat demonstrandum ?
- Thread starter
- #8
to j165, now that you mention cops, it comes to mind the so-called prosecutor's fallacy, when using biological markers such as DNA to identify a criminal.
The accused person may be placed in two different contexts (models):
1. What is the probability that an individual's DNA will match the crime sample, given that he or she is innocent ?
2. What is the probability that the suspect is innocent, given a DNA match ?
While in the first case, the individual is conceptually placed in a large population for scientific convenience, in the 2nd case, he or she is placed in a less well defined but more relevant group -those who might have reasonably committed the crime.
The prosecutor's fallacy refers to the confusion between the two probabilities. The second question should be the one concerning the court. Bayes' theorem could come to the rescue. The "a priori" probability would be that based on guilt as assessed from the evidence obtained.
Sorry for the digression. I thought it may support the points we raised above.
The accused person may be placed in two different contexts (models):
1. What is the probability that an individual's DNA will match the crime sample, given that he or she is innocent ?
2. What is the probability that the suspect is innocent, given a DNA match ?
While in the first case, the individual is conceptually placed in a large population for scientific convenience, in the 2nd case, he or she is placed in a less well defined but more relevant group -those who might have reasonably committed the crime.
The prosecutor's fallacy refers to the confusion between the two probabilities. The second question should be the one concerning the court. Bayes' theorem could come to the rescue. The "a priori" probability would be that based on guilt as assessed from the evidence obtained.
Sorry for the digression. I thought it may support the points we raised above.
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