{"id":4112,"date":"2014-10-29T16:39:05","date_gmt":"2014-10-29T21:39:05","guid":{"rendered":"http:\/\/www.livingreliability.com\/en\/?p=4112"},"modified":"2025-11-06T05:32:04","modified_gmt":"2025-11-06T10:32:04","slug":"random-failure-is-exponential-reliability-decay","status":"publish","type":"post","link":"https:\/\/www.livingreliability.com\/en\/posts\/random-failure-is-exponential-reliability-decay\/","title":{"rendered":"Random failure is exponential reliability decay"},"content":{"rendered":"

When something decays or grows “exponentially” it means that it changes regularly by a constant factor. An\u00a0example of exponential growth is the principle in a compound interest bank account which increases\u00a0at regular intervals by a constant factor<\/em>[1<\/a>]<\/sup>. Assume that you drive your car normally. You replace\u00a0tires whenever the tread depth falls below the manufacturer’s safety recommendation.<\/em>\u00a0[2<\/a>]<\/sup>\u00a0Intuitively, we would agree that you’re no more likely to have a punctured\u00a0tire in any one year than in any other.<\/em>[3<\/a>]<\/sup><\/p>\n

In other words the conditional probability of failure (a flat tire) is constant\u00a0or\u00a0age independent<\/em>. We call this failure pattern “random”.[4<\/a>]<\/sup> \u00a0Assume that the conditional failure probability of a puncture\u00a0in any year is a constant\u00a025%. \u00a0When\u00a0you drive the car off the dealer’s lot for the first time, at that moment the Reliability R1 is 100%. What is the Reliability R2 at the beginning\u00a0of the second\u00a0year? \u00a0In the article here<\/a> we showed the Conditional Probability of Failure to be:<\/p>\n

\n

(1) <\/span>   <\/span>\"\begin{align*}<\/p>\n<\/p>\n

Rearranging and substituting<\/p>\n

\n

(2) <\/span>   <\/span>\"\begin{align*}<\/p>\n<\/p>\n

The reliability at the beginning of year 3 is:<\/p>\n

\n

(3) <\/span>   <\/span>\"\begin{align*}<\/p>\n<\/p>\n

Repeating the calculation in an Excel spreadsheet\u00a0for each subsequent year reveals\u00a0the exponentially decaying Reliability curve:<\/p>\n

\"RandomFailureExcel\"<\/a><\/p>\n

The Reliability graph says that there is a 7.5% chance (row 2 col K) that you will drive for 9\u00a0years without a puncture.[5<\/a>]<\/sup><\/p>\n

Discrete versus continuous reliability analysis<\/h3>\n

In the article\u00a0Random failure and the MTTF <\/a>we showed that the equation describing the exponential decay of\u00a0Reliability (random failure) is:\u00a0\"R\left ( t \right )=e^{-\frac{t}{MTTF}}\". Let us include this equation in our Excel worksheet (in row 27). In rows 20 to 26 we set up the following example: Assuming a sample of 100 cars there will be 75 survivors at the beginning of year two\u00a0at which time\u00a0100 car years will have been logged, 56.25\u00a0survivors at the start of year 3 at which time 175 failure free car years will have been logged, and so on. The MTTF = 4 will be calculated (in row 26) as the total number of car years 369.97 divided by the number of failures 92.49\u00a0in the\u00a0period:<\/p>\n

\"ExponentialDecayRandomFailure5\"<\/a><\/p>\n

Note that the Reliability as calculated by the equation for random failure (row 27) is different from that calculated in row 2\u00a0and suggested by the number of survivors (row 20) at the start of\u00a0each year. That is because the Reliability equation represents a continuous function while the Survivors were calculated in the worksheet at discreet intervals as if all failures occurred on the last day of each year. They would actually have been spread out over the year.<\/p>\n

\"Histogram<\/a>
Histogram of the accumulating failures (from row 21) in each year. The dashed curve represents the discrete calculation. The solid curve drawn through the mid point of each interval\u00a0predicts the accumulated failures from the continuous Cumulative Failure Probability function (the complement of the Reliability function).<\/figcaption><\/figure>\n

It would be more reasonable to use the middle of the year in the histogram\u00a0of cumulative failures (from row 21) as representative of the average<\/em> time of failure in each interval.<\/p>\n

Hence the continuous Reliability Curve\u00a0\"R\left ( t \right )=e^{-\frac{t}{MTTF}}\" of row 27 would show somewhat better\u00a0survival levels at year start than the discrete calculation (row 2). Accordingly row 27 (col K) shows that your chances of driving for nine years without \u00a0a puncture would be closer to 10.5% than to 7.5% as determined from the discrete calculation of row 2. For more information on discrete versus continuous reliability calculation see the articles\u00a0here<\/a>\u00a0and here<\/a>.<\/p>\n

 <\/p>\n\n

    \n\t
  1. [1]<\/sup><\/strong>The factor is 1 plus the interest rate↩<\/a><\/li>\n\t
  2. [2]<\/sup><\/strong>This is an example of applying a Preventive Maintenance strategy in order to gain the desired constant, low conditional failure probability pattern. If we allow the treads to wear beyond the safety depth then tire\u00a0failure would become age related meaning that\u00a0the conditional probability of failure would increase with age\u00a0 conforming to\u00a0Nowlan and Heap’s pattern B<\/a>.↩<\/a><\/li>\n\t
  3. [3]<\/sup><\/strong>Nevertheless the probability of surviving to an\u00a0age t\u00a0decreases with increasing age because, obviously, if you keep driving the car, eventually you will have a flat. This does not contradict the fact (although it seems paradoxical) that the probability of getting a flat in any one year, if you ask the question at the start of the\u00a0year, remains constant.↩<\/a><\/li>\n\t
  4. [4]<\/sup><\/strong>The word “random” when used in the reliability sense differs from the often conjured image of throwing dice. In the latter situation it is impossible to predict the result of the next throw. To the reliability engineer, however, “random” means merely that the conditional failure probability in any interval is independent of the item’s age. Therefore, even when an item’s failure behavior is\u00a0random<\/em>, observed condition data can be used to\u00a0predict its failure.↩<\/a><\/li>\n\t
  5. [5]<\/sup><\/strong>Sometimes engineers might argue that if you replace tires whenever the tread depth reaches the safety limit then the survival curve for a tire will go back up to 100% at that time, so that we no longer have exponential decay. They would be correct if we were dealing with a failure mode subject to wear out. But in this case we are considering only a puncture due to a nail or broken glass on the road. It is going to be exponential decay no matter how often you replace the tires because the probability of getting a puncture is the same in any interval no matter the tire’s age as long as you are above the safety limit. ↩<\/a><\/li><\/ol>\n","protected":false},"excerpt":{"rendered":"

    When something decays or grows “exponentially” it means that it changes regularly by a constant factor. An\u00a0example of exponential growth is the principle in a compound interest bank account which increases\u00a0at regular intervals by a constant factor. Assume that you drive your car normally. You replace\u00a0tires whenever the tread depth falls below the manufacturer’s safety […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[89],"tags":[24],"class_list":["post-4112","post","type-post","status-publish","format-standard","hentry","category-theory-and-definitions","tag-conditional-failure-probability"],"_links":{"self":[{"href":"https:\/\/www.livingreliability.com\/en\/wp-json\/wp\/v2\/posts\/4112","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.livingreliability.com\/en\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.livingreliability.com\/en\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.livingreliability.com\/en\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.livingreliability.com\/en\/wp-json\/wp\/v2\/comments?post=4112"}],"version-history":[{"count":3,"href":"https:\/\/www.livingreliability.com\/en\/wp-json\/wp\/v2\/posts\/4112\/revisions"}],"predecessor-version":[{"id":6675,"href":"https:\/\/www.livingreliability.com\/en\/wp-json\/wp\/v2\/posts\/4112\/revisions\/6675"}],"wp:attachment":[{"href":"https:\/\/www.livingreliability.com\/en\/wp-json\/wp\/v2\/media?parent=4112"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.livingreliability.com\/en\/wp-json\/wp\/v2\/categories?post=4112"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.livingreliability.com\/en\/wp-json\/wp\/v2\/tags?post=4112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}