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Chapter 4S

• Reliability– The ability of a product, part, or system to perform its

intended function under a prescribed set of conditions– Reliability is expressed as a probability:

• The probability that the product or system will function when activated

• The probability that the product or system will function for a given length of time

Independent event– Events whose occurrence or non-occurrence do not

influence one another

Rule 1

If two or more events are independent and success is defined as the probability that all of the events occur, then the probability of success is equal to the product of the probabilities of the events

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Example – Rule 1

• A machine has two buttons. In order for the machine to function, both buttons must work. One button has a probability of working of .95, and the second button has a probability of working of .88.

.836

.88.95

Works) 2 Button(Works) 1 Button( Works) Machine(

PPP

Button 2.88

Button 1.95

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• Enhance reliability by utilizing redundancy.i.e., The use of backup components to increase reliability

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Rule 2 (see example)

– If two events are independent and success is defined as the probability that at least one of the events will occur, the probability of success is equal to the probability of either one plus 1.00 minus that probability multiplied by the other probability

Example– Rule 2

• A restaurant located in area that has frequent power outages has a generator to run its refrigeration equipment in case of a power failure. The local power company has a reliability of .97, and the generator has a reliability of .90. The probability that the restaurant will have power is

Generator.90

Power Co..97

Scenario 1Generator

.90

Power Co..97

If the Power Co. works, then whole system works

Generator in this scenario is the “redundancy”, i.e., generator works or not does not affect power.

Probability of Scenario 1 = Probability of Power Co. works = P (Power Co.)

= 0.97

Scenario 2Generator

.90

Power Co..97

Probability of Power Co. fails = 1 - Probability of Power Co. works= 1 – P(Power Co.)= 1 -.97 = 0.03

Probability of Scenario 2 = Probability of Power Co. fails X Probability of Generator works= (1 – P(Power Co)) X P (Generator)= 0.03 X 0.9=0.027

If the Power Co. fails, but the Generator works, then the whole system has power too.

.997

.97)(.90)-(1.97

Generator)(Co.))Power (-(1Co.)Power (

2Pr1Pr

Power)(

PPP

ScenarioofobabilityScenarioofobability

P

Generator.90

Power Co..97

Scenario 1 Scenario 2

Generator.90

Power Co..97

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Example 3

• A student takes three calculators (with reliabilities of .85, .80, and .75) to her exam. Only one of them needs to function for her to be able to finish the exam. What is the probability that she will have a functioning calculator to use when taking her exam?

Calc. 2.80

Calc. 1.85

Calc. 3.75

Calc. 2.80

Calc. 1.85

Calc. 3.75

Subsystem 1

Subsystem 2

Calc. 2.80

Calc. 3.75

Subsystem 1

Refer to example 2.

Probability of Subsystem 1 works = P (calc 3 works) + P (calc 3 fails but calc2 works)= P (calc 3 works) + P (calc3 fails) X P(calc 2 works)= .75 + (1- .75) X .8= .95

Calc. 1.85 Subsystem 2

Probability or subsystem 2 works = Probability of Calc 1 works= .85

Subsystem 1.95

Subsystem 2.85

Refer to example 2.

Probability of whole system works = P (subsystem 1 works) + P (subsystem 1 fails but subsystem 2 works)= P (subsystem 1 works) + P (subsystem 1 fails) X P(subsystem 2 works)= .95 + (1- .95) X .85= .9925

In class exercise:

• Example of two lamps, at the bottom part of page 171

• example of three lamps, on the top part of page 172

• optional: example 4s – 1, on page 172 ( “optional” means it is not required)