JAYANN WALTERS ASSIGNMENT #1
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UNIVERSITY OF TECHNOLOGY, JAMAICA
FACULTY OF ENGINEERING AND COMPUTING
Risk and Reliability Engineering
Title: Assignment #1
Submitted by: Jayann Walters
ID No: 0804251
Lecturer: Professor Yolanda Silvera
JAYANN WALTERS ASSIGNMENT #1
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QUESTION 1
SEE FAULT TREE ANALYSIS ATTACHED
QUESTION 2
PART A
The system reliability for a 100-hour mission can be calculated from the formula:
Where t= time
λ- failure rate
λ= 0.0045/ hr
t= 100 hr
R(100)= e-(0.0045*100) (1+(0.0045*100))
R(100)= e-0.45 (1.45)
R(100)= 0.637628(1.45)
ANS: R(100)= 0.92456 or 92.46%
The mean time to failure of the standby unit remains as good as new in its standby mode and
failure detection and unit replacement mechanisms are 100% reliable can be calculated from the
following formula:
MTTF= (K+1)/ λ, where K=1
MTTF= (1+1)/0.0045
MTTF= 2/0.0045
ANS: MTTF= 444.44 hours
JAYANN WALTERS ASSIGNMENT #1
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𝑛
1
𝑀𝑇𝐹𝐹 = ∑ 1⁄𝑖
𝜆
𝑖=𝑘
Where n=2, k=1, λ=0.0045/ hr
1
= 0.0045 * (1/1 +1/2)
1
1
+
(0.0045) 2(0.0045)
222.22+111.11
ANS: MTTF=333.33 hours
PART B
The system mean time to failure if the unit failure rate is 0.0035 failures per hour can be given by
the formula:
𝑛
1
𝑀𝑇𝐹𝐹 = ∑ 1⁄𝑖
𝜆
𝑖=𝑘
Where n=4, k=3, λ=0.0035/ hr
1
= 0.0035 * (1/3 +1/4)
1
1
+
3(0.0035) 4(0.0035)
95.238095+71.42857
ANS: MTTF=166.67 hours
JAYANN WALTERS ASSIGNMENT #1
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QUESTION 3
PART A
Let UWI= 1, UTECH= 2, CMU= 3
1
𝟏 0.80
P= 𝟐 0.30
𝟑 0.20
2
3
0.20 0.00
0.40 0.30
0.10 0.70
Probability that the grandson of a man that went to UWI went to UWI:
P2= P.P
0.80 0.20
P = 0.30 0.40
0.20 0.10
2
0.00
0.80 0.20 0.00
0.30 × 0.30 0.40 0.30
0.70
0.20 0.10 0.70
0.70 0.24 0.06
P2≈ 0.42 0.25 0.33
0.33 0.15 0.52
1
2
𝟏 0.70 0.24
P2≈ 𝟐 0.42 0.25
𝟑 0.33 0.15
3
0.06
0.33
0.52
The numbers in the product have been rounded to the same number of decimal places as in matrix
P. The entry in row 1, column 1 gives the probability that a man that went to UWI will have a
grandchild that goes to UWI. This number 0.70 or 70%.
JAYANN WALTERS ASSIGNMENT #1
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PART B
Let UWI= 1, UTECH= 2, CMU= 3
1
𝟏 1.00
P= 𝟐 0.30
𝟑 0.20
2
3
0.00 0.00
0.40 0.30
0.10 0.70
Probability that the grandson of a man that went to UWI went to UWI:
P2= P.P
1.00 0.00
P2= 0.30 0.40
0.20 0.10
0.00
1.00 0.00 0.00
0.30 × 0.30 0.40 0.30
0.70
0.20 0.10 0.70
1.00 0.00 0.00
P ≈ 0.48 0.19 0.33
0.37 0.11 0.52
2
1
2
𝟏 1.00 0.00
P2≈ 𝟐 0.48 0.19
𝟑 0.37 0.11
3
0.00
0.33
0.52
The numbers in the product have been rounded to the same number of decimal places as in matrix
P. The entry in row 1, column 1 gives the probability that a man that went to UWI will have a
grandchild that goes to UWI. This number 1.00 or 100%.
QUESTION 4
To find Reliability of the component, we use formula:
Where t=1500 hrs, β=4, θ=3000, ϒ= 2000
R (1500)= e-(
1500−2000 4
)
3000
ANS: R (1500)= 0.9992 or 99.92%
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To find hazard rate for an operating time of 1500 hours we use formula:
Where t=1500 hrs, β=4, θ=3000, ϒ= 2000
h (1500) =
h (1500) =
h (1500) =
4(1500−2000)4−1
30004
4(1500−2000)3
30004
4(−500)3
30004
ANS: h (1500) = −𝟔. 𝟏𝟕𝟑 ∗ 𝟏𝟎 − 𝟔 failures/hr or 0 failures/hour
Therefore, no failure is expected per hour for an operating time of 1500 hours.