Reliability Probablility

MathsMan1

Ok, I came across this question when studying for my exam and was hoping I could get some explanation here.

A Flow Measurement System for dispensing medications evenly to patients consists of an Orifice Plate (with failure rate of
[FONT=Arial,Arial][FONT=Arial,Arial]λ [/FONT][/FONT]= 0.75), a Differential Pressure Transmitter ([FONT=Arial,Arial][FONT=Arial,Arial]λ [/FONT][/FONT]= 1.0), a Square Root Extractor ([FONT=Arial,Arial][FONT=Arial,Arial]λ [/FONT][/FONT]= 0.1) and a Recorder ([FONT=Arial,Arial][FONT=Arial,Arial]λ [/FONT][/FONT]= 0.1) all connected in series. Calculate the probability of losing the flow measurement after 0.5 years for the following configurations;

1. A single Flow Measurement System as shown; (Just shows a series system of the above compnents in the order they are shown..)

So the solution I have is as follows:
1. Calculate the overall failure rate: 0.75 + 1.0 + 0.1 + 0.1 = 1.95

2. Calculate the probability of a system failure within 0.5 years:
FSyst(t) = 1 - e^-λsyst(t)
= 1- e^1.95(.5)
= 1-e^-0.975
= 1-0.377

Now, I understand everything up until the 0.377 figure. Where does this come out of? The above is the solution from the book (Bentley; An introduction to reliability & quality engineering).
Can anyone explain the .377 to me? Sorry if this is a very basic question...

Michael Collins

e is Euler's Number, a mathematical constant = 2.71828...

e^-0.975 is this number raised to the power of -0.975, which any scientific calculator will tell you is = 0.37719... or 0.377 to 3 decimal places.

click_here!!!

e^(1.95)(0.5)=0.377, See http://www.wolframalpha.com/input/?i=e^((-1.95*0.5))

You probably don't need to know this for that question, but this is based on the Poisson distribution,, where there is a continuous risk of something happening (i.e. the device breaking down).

To get the probability of the device breaking down in the 0.5 years, you get the probability of the device not breaking down in the 0.5 years and take it away from 1. I used the Poisson formula in Wolfram Alpha, which gives you 1-0.377. http://www.wolframalpha.com/input/?i=((e^-((1.95)(0.5)))*((1.95)(0.5))^0)/(0!)

The 0 stands for no failures. Note that anything to the power of 0 is one, and 0! is one also. Multiplying or dividing by 1 is the same as not doing anything.

Hope this helps:).

Fremen

click_here!!! wrote: »

You probably don't need to know this for that question, but this is based on the Poisson distribution,, where there is a continuous risk of something happening (i.e. the device breaking down).

I think you mean the exponential distribution. The Poisson distribution measures the number of events that occur when the probability of a given event is small, but there are a lot of chances for that event to happen - e.g. a soldier in the cavalry getting kicked to death by a horse.

The exponential distribution measures the time of an event with no memory - given that the event hasn't happened at time t, the chances of it happening by time t+s are the same as starting the experiment from scratch and the event happening before time s.

There's a meaningful connection between the Poisson process and the exponential distribution, which is possibly what's confusing you.

click_here!!!

Fremen wrote: »

I think you mean the exponential distribution. The Poisson distribution measures the number of events that occur when the probability of a given event is small, but there are a lot of chances for that event to happen - e.g. a soldier in the cavalry getting kicked to death by a horse.

The exponential distribution measures the time of an event with no memory - given that the event hasn't happened at time t, the chances of it happening by time t+s are the same as starting the experiment from scratch and the event happening before time s.

There's a meaningful connection between the Poisson process and the exponential distribution, which is possibly what's confusing you.

A Poisson process is used to describe events happening over a continuous interval, such as device breakages over time or minefields scattered perfectly randomly over an area. Independence of events is a property of Poisson distributions, as well as the exponential distribution you mentioned previously.

Note that the expected value formula (λt) of the Poisson process is different to the probability distribution function ((e^-(λt)*(λt)^x)/(x!) . Fremen seemed to be referring to the expected value above. I was referring the probability distribution function here. The exponential distribution formula is indeed based on the probability distribution formula of the Poisson distribution where x=0.

Anyways, if you use the Poisson distribution or the exponential distribution, you end up getting the same answer. :P I'm not sure what course the OP was doing, but I thought it would help. I hope I'm not confusing anyone.

1-((e^-((1.95)(0.5)))*((1.95)(0.5))^0)/(0!) = 1-e^((-1.95*0.5))
:P