### pptx file

```Lecture 2
IEGR 459: Introduction to Logistics
Management and Supply Chain
James Ngeru
Industrial and System Engineering
Reliability
• Definition:
– The probability that a system or a product will preform in a
satisfactory manner for a given period of time when used
under specified operating conditions.
• Note that the definition emphasizes on the following
elements:
–
–
–
–
Probability
Satisfactory performance
Time
Operating conditions
– Reliability is important as the following are dependent on
it:
• Availability
– Maintainability
Reliability
• Reliability Function
– R(t) = 1 – F(t)
F(t) is the probability that the system will fail
Reliability curve for the exponential distribution
Reliability – Example
• Find the failure rate of the component
– Component 1 failed after 75 hours
– Component 2 failed after 125 hours
– Component 3 failed after 130 hours
– Component 4 failed after 325 hours
– Component 5 failed after 525 hours
Solution:
– Total Operating Time = 75+125+130+325+525 = 3805
– Failure rate, λ = 5/3805 = 0.001314
Reliability – Example 2
A system operational cycle.
Failure Rate
λ = Number of failures/Total Mission Time
= 6/142
= 0.0422535
Mean Time Between Failure
MTBF = 1/ λ
= 23.6667 hours
Reliability
R(t) = e –t/m
= e –λt
= 0.002479
Reliability Nomograph.
Example:
If for the example in previous slide, what would
be reliability of the system if it is expected to run
for 10hrs? Find using the reliability nomogragh
Join the points with straight line
- 1st point- On MTBF scale = 23.667
- 2nd point - On operating time scale = 10hrs
The systems reliability is where the line
intersects the reliability scale approx.. 0.65
Check:
RS = e -λt
= e -t/m
= e -10/23.667
= 0.65538
Reliability nomograph for the exponential failure distribution. Source:NAVAIR 00-65-502/NAVORD OD 41146, Reliability Engineering
Handbook, Naval Air Systems Command and Naval Ordnance Systems Command.
Typical failure-rate curve relationships.
Failure-rate curve with maintenance (software application).
Reliability - Series Network
Reliability (RS) = (RA)(RB)(RC)
• R(t) = e-λt
• R(s) = e-(λ1 + λ2 + λ3 + …… + λn )t
Example 1:
Consider a system that includes transmitter (A), receiver (B) and power supply (C). If the transmitter reliability is
0.9712, the receiver reliability is 0.8521 and that of power supply is 0.9357, what is total system reliability if are all
connected in series.
Solution:
RS = (0.9712)(0.8521)(0.9357) = 0.7743
Example 2:
A system with four (A, B, C and D) subsystems configured in series consists is expected to run for 1000 hours. The
MTBF for A = 6000hrs, B = 4500 hrs, C = 10,500 hrs and D = 3200 hrs. Determine the overall reliability of the series
network.
Solution:
RS = e – (0.000797)(1000)
Reliability - Parallel Network
System 1
Reliability (Rs) = RA + RB – (RA)(RB)
System 2
R(s) = 1- (1- RA) (1-RB )(1-Rc)
 If n identical components in parallel
 R(s) = 1 – (1-R)n
Example 1:
Consider a system of two identical components configured in parallel series, each with reliability of 0.95. What is the
total system reliability
Solution:
RS = 0.95 +0.95 – (0.95)(0.95) = 0.9975
Example 2:
Consider a system of three identical components configured in parallel, each with reliability of 0.95. What is the total
system reliability
Solution:
RS = 1 – (1 - 0.95)3 = 0.999875
Some combined series–parallel networks.
(Rs) = RA (RB + RC – RBRC)
(Rs) = (RA + RB – RARB) (RC + RD – RCRD)
Note In the book:
Rs = [(1-(1-RA)(1-RB)][1-(1-RC)(1-RD)]
= [1-(1-RA-RD+RARB)][1-(1-RC-RD +RCRD)]
= (1-1+RA+RB-RARB)(1-1+RC+RD-RCRD)
=(RA+RB-RARB)(RC+RD-RCRD)
(Rs) = [1-(1-RA)(1-RB)(1-RC)]*RD*[(RE + RF-RERF)
```