SoftwareEngineering
October 5, 2023SoftwareEngineering
October 5, 2023SoftwareEngineering
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^{(1/(29/5))} = e^{(5/29)} = 0.84
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
Correct 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^{(1/(29/5))} = e^{(5/29)} = 0.84
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
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