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Chapter 7: Problem 33

Each time a certain machine breaks down it is replaced by a new one of the same type. In the long run, what percentage of time is the machine in use less than one year old if the life distribution of a machine is (a) uniformly distributed over \((0,2) ?\) (b) exponentially distributed with mean \(1 ?\)

Short Answer

Expert verified
In the long run, the machine is in use less than one year old for: (a) 50% of the time when its life distribution is uniformly distributed over \((0, 2)\). (b) Approximately 63.21% of the time when its life distribution is exponentially distributed with mean \(1\).

Step by step solution

01

(a) Uniform distribution on \((0, 2)\)

: Find the probability that the machine is less than one year old. To find the probability that the machine is less than one year old for a uniform distribution defined over the interval \((0, 2)\), we first need to find the probability density function (PDF). Since the area under the curve of the PDF over the entire interval must equal \(1\), we know that the PDF for a uniform distribution is simply the constant \(1/(b-a)=1/(2-0)=1/2\). Next, we need to integrate the PDF over the interval \((0, 1)\) to find the probability that the machine is less than one year old: \[\int_{0}^{1} \frac{1}{2} dx\]
02

Integrate the PDF:

: Evaluate the integral to find the probability. Evaluating the integral: \[\int_{0}^{1} \frac{1}{2} dx = \left[ \frac{1}{2}x \right]_0^1 = \frac{1}{2}(1) - \frac{1}{2}(0) = \frac{1}{2}\] So, 50% of the time, the machine is less than one year old when it follows a uniform distribution.
03

(b) Exponential distribution with mean \(1\)

: Find the probability that the machine is less than one year old. For the exponential distribution with mean \(1\), we first need to find the probability density function (PDF). The PDF for an exponential distribution with mean \(\mu\) is given by: \[f(x) = \frac{1}{\mu} e^{-x/\mu}\] Since the mean is \(1\), the PDF becomes: \[f(x) = e^{-x}\] Now, we need to integrate this PDF over the interval \((0, 1)\) to find the probability that the machine is less than one year old: \[\int_{0}^{1} e^{-x} dx\]
04

Integrate the PDF:

: Evaluate the integral to find the probability. Evaluating the integral: \[\int_{0}^{1} e^{-x} dx = \left[ -e^{-x} \right]_0^1 = -(e^{-1} - e^0) = 1 - e^{-1}\] So, approximately 63.21% of the time, the machine is less than one year old when it follows an exponential distribution with mean \(1\) (rounded to two decimal places). In conclusion, the long-term percentage of time the machine is in use less than one year old is: (a) 50% for a uniform distribution on \((0, 2)\). (b) Approximately 63.21% for an exponential distribution with mean \(1\).

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Most popular questions from this chapter

For a renewal process, let \(A(t)\) be the age at time \(t\). Prove that if \(\mu<\infty\), then with probability 1 $$ \frac{A(t)}{t} \rightarrow 0 \quad \text { as } t \rightarrow \infty $$

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Consider a single-server queueing system in which customers arrive in accordance with a renewal process. Each customer brings in a random amount of work, chosen independently according to the distribution \(G\). The server serves one customer at a time. However, the server processes work at rate \(i\) per unit time whenever there are \(i\) customers in the system. For instance, if a customer with workload 8 enters service when there are 3 other customers waiting in line, then if no one else arrives that customer will spend 2 units of time in service, If another customer arrives after 1 unit of time, then our customer will spend a total of \(1.8\) units of time in service provided no one else arrives. Let \(W_{i}\) denote the amount of time customer \(/\) spends in the system. Also, define \(E[W]\) by $$ E[W]=\lim _{n \rightarrow \infty}\left(W_{1}+\cdots+W_{n}\right) / n $$ and so \(E[W]\) is the average amount of time a customer spends in the system. Let \(N\) denote the number of customers that arrive in a busy period. (a) Argue that $$ E[W]=E\left[W_{1}+\cdots+W_{N}\right] / E[N] $$ Let \(L_{1}\) denote the amount of work customer \(i\) brings into the system; and so the \(L_{i}, i \geq 1\), are independent random variables having distribution \(G\). (b) Argue that at any time \(t\), the sum of the times spent in the system by all arrivals prior to \(t\) is equal to the total amount of work processed by time \(t\). Hint: Consider the rate at which the server processes work. (c) Argue that $$ \sum_{i=1}^{N} W_{i}=\sum_{i=1}^{N} L_{i} $$ (d) Use Wald's equation (see Exercise 12) to conclude that $$ E[W]=\mu $$ where \(\mu\) is the mean of the distribution \(G\). That is, the average time that customers spend in the system is equal to the average work they bring to the system.
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