Reliability of CD-ROMs. In Reliability Ques (March 2004), the exponential distribution was used to model the lengths of life of CD-ROM drives in a two-drive system. The two CD-ROM drives operate independently, and at least one drive must be operating for the system to operate successfully. Both drives have a mean length of life of 25,000 hours.

a. The reliability R(t) of a single CD-ROM drive is the probability that the life of the drive exceeds t hours. Give a formula for R(t).

b. Use the result from part a to find the probability that the life of the single CD-ROM drive exceeds 8,760 hours (the number of hours of operation in a year).

c. The Reliability S(t) of the two-drive/CD-ROM system is the probability that the life of at least one drive exceeds thours. Give a formula for S(t). [Hint: Use the rule of complements and the fact that the two drives operate independently.]

d. Use the result from part c to find the probability that the two-drive CD-ROM system has a life whose length exceeds 8,760 hours.

e. Compare the probabilities you found in parts b and d.

The two CD-ROM drives operate independently in a two-drive system. The length of life of CD-ROM drives in a two-drive system is exponentially distributed with a mean of 25000 hours.

Let $T_1$and $T_2$represents the length of life of two CD-ROMs, respectively.

The probability density function of each is:

$f\left(t\right)=\frac{1}{25000}{e}^{\frac{-t}{25000}};t>0$

.

a.

The formula of a single CD-ROM drive is obtained as follows:

$\begin{array}{rcl}R\left(t\right)& =& P\left(T>t\right)\\ & =& e^{\frac{-t}{25000}}\\ & & \end{array}$.

Thus the formula for Reliability is $R\left(t\right)=e^{\frac{-t}{25000}}$.

b.

The probability that the life of the single CD-ROM drive exceeds 8760 hours is obtained as:

$\begin{array}{rcl}P\left(T>8760\right)& =& e^{\frac{-8760}{25000}}\\ & =& e^{-0.3504}\\ & =& 0.7044\\ & & \end{array}$.

Therefore, the required probability is 0.7044.

c.

The system's Reliability is that the life of at least one drive exceeds t hours.

That is, $1-P\left(Nodriveexceesthours\right)$.

Where,

$P\left(Nodriveexceesthours\right)=P\left(T_1⩽tand{T}_2⩽t\right)$.

Since the two drives operate independently,

$\begin{array}{rcl}P\left(Nodriveexceesthours\right)& =& P\left(T_1⩽t\right)P\left(T_2⩽t\right)\\ & =& \left[1-P\left(T_1>t\right)\right]\times \left[1-P\left(T_2>t\right)\right]\\ & =& \left[1-e^{\frac{-t}{25000}}\right]\left[1-e^{\frac{-t}{25000}}\right]\\ & =& 1-2e^{\frac{-t}{25000}}+-e^{\frac{-2t}{25000}}\end{array}$

Therefore,

$\begin{array}{rcl}S\left(t\right)& =& 1-P\left(Nodriveexceesthours\right)\\ & =& 1-\left[1-2e^{\frac{-t}{25000}}+e^{\frac{-2t}{25000}}\right]\\ & =& 2e^{\frac{-t}{25000}}-e^{\frac{-2t}{25000}}\\ & & \end{array}$.

Thus, the formula for the Reliability of the CD-ROM system is $S\left(t\right)=2e^{\frac{-t}{25000}}-e^{\frac{-2t}{25000}}$.

d.

The probability that the two-drive CD-ROM system has a life whose length exceeds 8,760 hours is obtained as:

$\begin{array}{rcl}S\left(8760\right)& =& 2e^{\frac{-8760}{25000}}-e^{\frac{-2\times 8760}{25000}}\\ & =& 2e^{-0.3504}-e^{-0.7008}\\ & =& 2\times 0.7044-0.4962\\ & =& 1.4088-0.4962\\ & =& 0.9126\end{array}$.

Thus the required probability is 0.9126.

e.

The Reliability of the single CD-ROM drive is 0.7044 while that of a two-drive CD-ROM system is 0.9126, which means the system with two drives CD-ROM is more reliable than the single CD-ROM drive.