Lecture 2IEGR 459: Introduction to Logistics

Management and Supply Chain

James NgeruIndustrial 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 hoursSolution:– Total Operating Time = 75+125+130+325+525 = 3805– Failure rate, λ = 5/3805 = 0.001314

A system operational cycle.

Reliability – Example 2

Failure Rate λ = Number of failures/Total Mission Time

= 6/142 = 0.0422535

Mean Time Between FailureMTBF = 1/ λ

= 23.6667 hours

ReliabilityR(t) = e –t/m

= e –λt

= 0.002479

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.

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 nomograghJoin the points with straight line - 1st point- On MTBF scale = 23.667 - 2nd point - On operating time scale = 10hrsThe 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

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

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

System 1Reliability (Rs) = RA + RB – (RA)(RB)

System 2R(s) = 1- (1- RA) (1-RB )(1-Rc)Þ If n identical components in parallel

Þ R(s) = 1 – (1-R)n

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)