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Chapter 4: Problem 8

Carnegie-Mellon multiprocessor C.mmp consists of processors, switches, and memory units. Consider a configuration with 16 processors, \(1664 \mathrm{~K}\) (64-kilobyte) memories, and one switch. At least four processors and four memories are required for a given task. Assume that we have a (constant) failure rate for each processor of \(68.9\) failures \(/ 10^6 \mathrm{~h}, 224\) failures \(/ 10^6 \mathrm{~h}\) for each memory, and a failure rate for the switch of 202 failures \(/ 10^6 \mathrm{~h}\). Compute the reliability function for the system. Also compute the MTTF of the system.

Short Answer

Expert verified
The reliability function is \( R(t) = e^{-374040.4 t} \). MTTF is approximately 2673 hours.

Step by step solution

01

Define the Reliability Function

The reliability function for a component with a constant failure rate is given by \( R(t) = e^{-\lambda t} \), where \( \lambda \) is the failure rate and \( t \) is the time.
02

Calculate Failure Rates

Given failure rates:- Processors: \( 68.9 \) failures per million hours \( (\lambda_p = 68.9 \)- Memories: \( 224 \) failures per million hours \( (\lambda_m = 224 \)- Switch: \( 202 \) failures per million hours \( (\lambda_s = 202 \)
03

Compute System Failure Rate

The system's failure rate \( \lambda_{sys} \) can be found by summing the failure rates of all components: \( \lambda_{sys} = 16\lambda_p + 1664\lambda_m + \lambda_s \)
04

Plug in the Values

Substitute the given values into the failure rate equation: \( \lambda_{sys} = 16(68.9) + 1664(224) + 202 \)Calculate each part:- Processors: \( 16 * 68.9 = 1102.4 \)- Memories: \( 1664 * 224 = 372736 \)- Switch: \( 202 \)
05

Calculate Total System Failure Rate

Sum the calculated values to get the system's failure rate: \( \lambda_{sys} = 1102.4 + 372736 + 202 = 374040.4 \)
06

Compute Mean Time to Failure (MTTF)

The MTTF is the inverse of the system's failure rate: \( MTTF = \frac{1}{\lambda_{sys}} \)Substitute the system failure rate: \( MTTF = \frac{1}{374040.4/(10^6)} \)\( MTTF = \frac{10^6}{374040.4} = 2.673 * 10^3 \) hours

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Failure Rate
The failure rate, denoted as \( \lambda \), is a crucial metric in reliability engineering. It represents the frequency with which an engineered system or component fails within a specific timeframe. In the given problem, the failure rates for different components are provided: \( 68.9 \) failures \/ 10^6 hours for processors, \( 224 \) failures \/ 10^6 hours for memory units, and \( 202 \) failures \/ 10^6 hours for the switch.
To compute the overall system failure rate, sum the failure rates of all individual components:
  • For processors, there are \( 16 \) processors each with a failure rate of \( 68.9 \), giving a total of \( 16 \times 68.9 = 1102.4 \).
  • For memory units, there are \( 1664 \) units each with a failure rate of \( 224 \), giving a total of \( 1664 \times 224 = 372736 \).
  • For the switch, with a single unit, the failure rate is \( 202 \).
The system鈥檚 total failure rate is thus \( \lambda_{sys} = 1102.4 + 372736 + 202 = 374040.4 \/ 10^6 \) hours.
This total failure rate provides insight into the reliability of the entire multiprocessor system.
Mean Time To Failure (MTTF)
Mean Time to Failure (MTTF) is a fundamental measurement used to determine the average time to failure for a system or component. It is the inverse of the failure rate. For a system with a failure rate \( \lambda_{sys} \), the MTTF can be calculated as:
  • \( MTTF = \frac{1}{\lambda_{sys}} \)
For this multiprocessor system, with \( \lambda_{sys} = 374040.4 \/ 10^6 \) hours, the MTTF calculation is:
\( MTTF = \frac{1}{374040.4 \/ (10^6)} = \frac{10^6}{374040.4} = 2.673 \times 10^3 \) hours, or approximately \( 2673 \) hours.
This means, on average, the system is expected to function properly for about 2673 hours before encountering a failure.
System Reliability
System reliability refers to the probability that a system will perform its intended function without failure for a specific period, under stated conditions. It is represented by the reliability function \( R(t) \), usually expressed as:
\( R(t) = e^{-\lambda t} \)
In this formula, \( \lambda \) represents the system failure rate, and \( t \) denotes the time. Using the calculated system failure rate \( \lambda_{sys} = 374040.4 \/ 10^6 \) hours, you can determine the reliability over any desired period. For example, the reliability for 1000 hours is computed as:
\( R(1000) = e^{-(374040.4 \/ (10^6)) \times 1000} \)
By calculating the exponent and taking the exponential, you get the probability of the system not failing within that time frame. Understanding system reliability helps ensure that critical tasks are completed without unexpected interruptions.

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Most popular questions from this chapter

\({ }^{\star}\) Consider a series system of two independent components. The lifetime of the first component is exponentially distributed with parameter \(\lambda\), and the lifetime of the second component is normally distributed with parameters \(\mu\) and \(\sigma^2\). Determine the reliability \(R(t)\) of the system and show that the expected life of the system is $$ \frac{1}{\lambda}\left[1-\exp \left(-\lambda \mu+\frac{\lambda^2 \sigma^2}{2}\right)\right] . $$

\({ }^*\) The problem of dynamic storage allocation in the main memory of a computer system [WOLM 1965] can be simplified by choosing a fixed node size, \(k\), for allocation. Out of the \(k\) units (bytes, say) of storage allocated to a node, only \(k-b\) bytes are available to the user, since \(b\) bytes are required for control information. Let the random variable \(L\) denote the length in bytes of a user request. Thus \(\lceil L /(k-b)\rceil\) nodes must be allocated to satisfy the user request. Thus the total number of bytes allocated is \(X=k\lceil L /(k-b)\rceil\). Find \(E[X]\) as a function of \(k\) and \(E[L]\). Then, by differentiating \(E[X]\) with respect to \(k\), show that the optimal value of \(k\) is approximately \(b+\sqrt{2 b E[L]}\).

Let \(X, Y\), and \(Z\), respectively, denote \(\operatorname{EXP}(1)\), two-stage hyperexponential with \(\alpha_1=.5=\alpha_2, \lambda_1=2\), and \(\lambda_2=\frac{2}{3}\), and two-stage Erlang with parameter 2 random variables. Note that \(E[X]=E[Y]=E[Z]\). Find the mode, the median, the variance, and the coefficient of variation of each random variable. Compare the densities of \(X, Y\), and \(Z\) by plotting on the same graph. Similarly compare the three distribution functions.

Consider a discrete random variable \(X\) with the following pmf: $$ p_X(x)= \begin{cases}\frac{1}{x(x+1)}, & x=1,2, \ldots \\ 0, & \text { otherwise }\end{cases} $$ cant copy graph Figure 4.P.1. An alternative method of computing \(E[X]\) Show that the function defined satisfies the properties of a pmf. Show that the formula (4.1) of expectation does not converge in this case and hence \(E[X]\) is undefined. [Hint: Rewrite \(1 / x(x+1)\) as \(1 / x-1 /(x+1) \cdot]\)

In a Bell System study made in 1961 regarding the dialing of calls between White Plains, New York, and Sacramento, California, the pmf of the number of trunks, \(X\), required for a connection was found to be \begin{tabular}{c|ccccc} \hline\(i\) & 1 & 2 & 3 & 4 & 5 \\ \hline\(p_x(i)\) & \(.50\) & \(.30\) & \(.12\) & \(.07\) & \(.01\) \\ \hline \end{tabular} Determine the distribution function of \(X\). Compute \(E[X], \operatorname{Var}[X]\) and the mode of \(X\). Let \(Y\) denote the number of telephone switching exchanges that this call has to pass through. Then \(Y=X+1\). Determine the pmf, the distribution function, the mean, and the variance of \(Y\).
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