1. James Ngeru Industrial and System Engineering
Lecture 2 IEGR 459: Introduction to Logistics Management and Supply Chain
James Ngeru
Industrial and System Engineering

2. 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

3. Reliability Reliability Function R(t) = 1 – F(t)
F(t) is the probability that the system will fail

4. Reliability curve for the exponential distribution

5. Find the failure rate of the component
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 = = 3805
Failure rate, λ = 5/3805 =

6. A system operational cycle.
Reliability – Example 2
A system operational cycle.
Failure Rate
λ = Number of failures/Total Mission Time
= 6/142 =
Mean Time Between Failure
MTBF = 1/ λ
= hours
Reliability
R(t) = e –t/m
= e –λt =

7. 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 =
- 2nd point - On operating time scale = 10hrs
The systems reliability is where the line intersects the reliability scale approx
Check:
RS = e -λt
= e -t/m
= e -10/23.667 =
Reliability nomograph for the exponential failure distribution. Source:NAVAIR /NAVORD OD 41146, Reliability Engineering Handbook, Naval Air Systems Command and Naval Ordnance Systems Command.

8. Typical failure-rate curve relationships.

9. Failure-rate curve with maintenance (software application).

11. 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 , the receiver reliability is and that of power supply is , what is total system reliability if are all connected in series.
Solution:
RS = (0.9712)(0.8521)(0.9357) =
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 – ( )(1000)

12. Reliability - Parallel Network
System 2
R(s) = 1- (1- RA) (1-RB )(1-Rc)
If n identical components in parallel
R(s) = 1 – (1-R)n
System 1
Reliability (Rs) = RA + RB – (RA)(RB)
Example 1:
Consider a system of two identical components configured in parallel series, each with reliability of What is the total system reliability
Solution:
RS = – (0.95)(0.95) =
Example 2:
Consider a system of three identical components configured in parallel, each with reliability of What is the total system reliability
Solution:
RS = 1 – ( )3 =

13. 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)