###### Software-Engineering

October 5, 2023###### Software-Engineering

October 5, 2023# Software-Engineering

Question 15 |

The following table shoes the time between failures for a software system

The reliability of the system for one hour of operation assuming an exponential model is

0.45 | |

0.63 | |

0.84 | |

0.95 |

Question 15 Explanation:

MIBF = ∑(Start of downtime – Start of uptime)/No. of failures

MIBF = (6+4+8+5+6)/5 = 29/5

The probability or reliability that the product will work for a defined period of time without failure is given by

R(T) = exp(-T/MTBF); T = 1 hour

R(1) = e

MIBF = (6+4+8+5+6)/5 = 29/5

The probability or reliability that the product will work for a defined period of time without failure is given by

R(T) = exp(-T/MTBF); T = 1 hour

R(1) = e

^{(-1/(29/5))}= e^{(-5/29)}= 0.84Correct Answer: C

Question 15 Explanation:

MIBF = ∑(Start of downtime – Start of uptime)/No. of failures

MIBF = (6+4+8+5+6)/5 = 29/5

The probability or reliability that the product will work for a defined period of time without failure is given by

R(T) = exp(-T/MTBF); T = 1 hour

R(1) = e

MIBF = (6+4+8+5+6)/5 = 29/5

The probability or reliability that the product will work for a defined period of time without failure is given by

R(T) = exp(-T/MTBF); T = 1 hour

R(1) = e

^{(-1/(29/5))}= e^{(-5/29)}= 0.84 Subscribe

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