How Does Exponential Decay Model the Lifespan of Computer Chips?

Eleni
Messages
14
Reaction score
0

Homework Statement


The probability that a particular computer chip fails after a hours of operation is given by
0.00005∫e^(-0.00005t)dt on the interval [a, ∝]
i. find the probability that the computer chip fails after 15,000 hours of operation
ii. of the chips that are still operating after 15,000 hours, what fraction of these will operate for at least another 15,000 hours?
iii. Evaluate 0.00005∫e^(-0.00005t)dt [0, ∝]t and interpret its meaning in the context of the situation.

Homework Equations


0.00005∫e^(-0.00005t)dt [a, ∝]
a= hours of operation

The Attempt at a Solution


i. do I evaluate P(0≤ x≤15000)= 0.00005∫e^(-0.00005t)dt on the interval [0, 15000]
ii. do I evaluate P(15000≤ x≤ 30000)= 0.00005∫e^(-0.00005t)dt on the interval [15000, 30000]
iii. would the meaning of this integral evaluation be the total operating life expectancy of the computer chip (hrs)?

please accept my apologies if this appears quite messy. I am not very good at using latex or typing equations and would much rather the old school pen and paper.
Please note: -0.00005t is the supercript for e in the formula

THANK YOU.
 
Physics news on Phys.org
Assuming that by ∝ you mean ∞:
Eleni said:
i. do I evaluate P(0≤ x≤15000)= 0.00005∫e^(-0.00005t)dt on the interval [0, 15000]
Yes, but why? What is the formula for ##P(X \leq 15000)##?
(Here I interpret "fails after 15,000 hours of operation" as you did, meaning "fails after at most 15,000 hours of operation". It could also mean "fails after at least 15,000 hours of operation", though.)
Eleni said:
ii. do I evaluate P(15000≤ x≤ 30000)= 0.00005∫e^(-0.00005t)dt on the interval [15000, 30000]
##P(15000≤ X≤ 30000)## is the probability that a chip fails between 15000 and 30000 hours. That is not what you is asked for.
You need to evaluate ##P(X \geq 30000|X \geq15000)##, a conditional probability.
Eleni said:
iii. would the meaning of this integral evaluation be the total operating life expectancy of the computer chip (hrs)?
No, you compute the probability that the chip will fail before ...
 
Last edited:
Eleni said:

Homework Statement


The probability that a particular computer chip fails after a hours of operation is given by
0.00005∫e^(-0.00005t)dt on the interval [a, ∝]
i. find the probability that the computer chip fails after 15,000 hours of operation
ii. of the chips that are still operating after 15,000 hours, what fraction of these will operate for at least another 15,000 hours?
iii. Evaluate 0.00005∫e^(-0.00005t)dt [0, ∝]t and interpret its meaning in the context of the situation.

Homework Equations


0.00005∫e^(-0.00005t)dt [a, ∝]
a= hours of operation

The Attempt at a Solution


i. do I evaluate P(0≤ x≤15000)= 0.00005∫e^(-0.00005t)dt on the interval [0, 15000]
ii. do I evaluate P(15000≤ x≤ 30000)= 0.00005∫e^(-0.00005t)dt on the interval [15000, 30000]
iii. would the meaning of this integral evaluation be the total operating life expectancy of the computer chip (hrs)?

please accept my apologies if this appears quite messy. I am not very good at using latex or typing equations and would much rather the old school pen and paper.
Please note: -0.00005t is the supercript for e in the formula

THANK YOU.

First hint (for problem solving in general): simplify notation!

You have that the time to failure, ##X##, satisfies
P(X > a) = \int_a^{\infty} r e^{-rt} \, dt
where ##r = 0.00005 = .5 \times 10^{-4} = 1/20000## (##\text{hr}^{-1}##). Even if you don't know LaTeX (or care to learn it), the expression "int{e^(-rt) dt, t=a..infinity}" is still a lot easier to type than what you gave, and is easier to read as well. Isn't it a lot better to type "r" than "0.00005"? Anyway, if ##t## is in 10s of thousands of hours, writing 0.00005t as t/20000 is more "revealing": when t = 15000 the product is rt = 15/20 = 0.75, and when t = 30000 it is rt = 30/20 = 1.5.

Second hint: do the integral first, before doing anything else.

As for your questions: (i) and (ii) what do YOU think, and why? Can you write the answer to (ii) in terms of ##P(X > 30000)## and ##P(C > 15000)##? For question (iii): do not try to answer it until you have actually done the integral, as I suggested.

Note that PF rules forbid us from being more helpful until you have done some work on the problem: give us your answers first, then we can talk.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
1
Views
2K
Replies
1
Views
3K
Replies
13
Views
3K
4
Replies
175
Views
25K
Back
Top