27/3/2014 CGMS401, Module4- Learning Objectives
After completing this module, you should be able to:
Perform simple reliability computations
Explain the purpose of redundancy in a system
Reliability: The ability of a product, part or system to perform its intended function under a prescribed set of
Reliability is measured as the PROBABILITY that an item will function as intended.
Failure is when the product, part or system does not function.
Each time you turn the start mechanism on your car you expect it to start. Usually it does but not
always. Similarly when you press the “ON” button on your cell phone, you expect it to turn on.
If an item is switched on 100 times and it functions 95 times but fails to turn on 5 times, then the item’s
reliability is .95 or 95%, i.e., it turns on 95% of the time. The same item’s failure rate is .05 or 5%, i.e., it fails
to turn on 5% of the time.
We calculate reliability based on the rules of probability that you would have learned in the CQMS102 course
at Ryerson University in probability and statistics or an equivalent course taken at another post-secondary
college or university.
There are two considerations for reliability:
Will the item or system function at a given POINT in time when activated?
Will the item or system function for a given LENGTH of time?
Probability of Functioning When Activated
Do you recall the concept of INDEPENDENT events?
A common example is the toss of the unbiased and fair coin. If the coin were to land on heads 10 times in a
row, the 11th toss still has a probability of .50 heads and .50 tails. The 11th toss is an “independent random
trial”. Its outcome is totally independent of the previous 10 tosses. Another way of saying it is that the coin has
no memory. This basic concept of probability is the basis for calculating reliability.
“If two or more events are independent and SUCCESS is defined as that probability that all of the events will
occur then the probability of success is equal to the product of the probabilities of the events.
Suppose a room has two lamps, but to have adequate light, both lamps MUST work. One lamp has a
probability of working of .90 and the other of .80.
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The probability of both working is .90*.80 = .72.
This also means that the probability of failure (here defined that EITHER or BOTH lamps fail to function) is the
This means that even though the individual components of a system might have high reliabilities, the system
as a whole will have a lower reliability.
Further, the more components in such a system, the lower the overall reliability will be.
For example: A system of 4 components with reliabilities of .95, .90, .80, and .75 will have a reliability
of .95*.90*.80*.75 = .513
Figure 4S1 on page 3 of the supplement demonstrates this effect.
How can the reliability of an item or system be improved?
There are a number of possibilities:
Improve the quality of materials used in the manufacture of the product.
Improve manufacturing by the use of newer better machines more able to hold closer tolerances.
Train and motivate workers to do a better job.
Redesign the product to include “redundant” or backup components that will improve the reliability of
The aircraft industry uses many backups in the aircraft they design for those critical components that must
function in order to keep the aircraft in the sky. Businesses use backup power systems to ensure they have
power to operate their computers. The following link will take you to the website of a manufacturer of backup
power supplies: http://www.upspower.ca/?gclid=CIm56NXm_qsCFY3KKgodCnn-mg
As another example, I have a backup sump pump system in my basement to avoid my basement being
flooded. I increased the probability of success by designing it such that if the main pump system fails the
backup system automatically switches on.
It is this backup principle that is covered by Rule 2.
If two or more events are INDEPENDENT and SUCCESS is defined as the probability that AT LEAST ONE of
the events will occur, then the probability of success Ps is equal to 1 – probability that none of the events will
occur, i.e., 1 - (1 - P1)(1 - P2)(1 - P3). . . .
Simplifying, Ps = P1 + (1 - P1) P2 + (1 - P1) (1 - P2) P3 +……
For a three component system this can be expressed as
Ps = P1 + (1 - P1) x P2 + (1 - P1) x (1 - P2) x P3
P[#1 operates] + [ #1 fails and #2 operates] + [ #1 fails and #2 fails and 3 operates]
Examples in the textbook show how to calculate the probability under these conditions.
Probability of Functioning for a Given Length of Time
How Long Will an Item or System Continue to Function?
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Probabilities are determined relative to a specified length of time. This approach is normally used for product
warranties which pertain to a specified length of time after the purchase of a product. TVs and home
electronics equipment tend to provide a one-year warranty. Motor vehicles are typically under warranty for 5
years or 80,000 kilometers—whichever comes first.
Product failure rates are usually modeled as either:
A negative exponential distribution.
A normal distribution.
Which of these two it is would be determined by similar products that have been studied and data collected to
determine the appropriate model.
Negative Exponential Distribution:
A typical product failure pattern can frequently be modeled as a “bathtub” curve - see Figure 4S-2 on page 5
of the text. As you can see there are 3 “zones”:
Infant Mortality: things that are truly defective will tend to fail sooner rather than later. The other major problem
in the early life of a product is misuse. Users fail to read the owner’s manual - especially for common office
products such as photocopiers - and have the attitude “if you can use one copier you can use them all”.
Clearly this is not the case as the history of new photocopiers in offices is that there is a high initial failure rate
due to misuse.
Normal Lifetime: very few failures.
Wear Out Failures: simply due to age and wear out of the equipment.
MTTF or Mean Time To Failure
Mean Time To Failure (MTTF) is defined as the average length of time before failure of a product or
component. Information on the distribution of failure rate and the length of each phase requires collection of
historical data and the