1. REPAIRABLE SYSTEMS 1 AVAILABILITY ENGINEERING

2. A Single Repairable Component With Failure Rate λ and Repair Rate μ The component may exist in one of Two States S O (Working) and S 1 (Failing).It transits from S O to S 1 by Failure and back from S 1 to S O by Repair SOSO S1S1 λ μ The Concept of AVAILABILITY STATE GRAPH After time interval Δt At Present SOSO S1S1 SOSO 1 - λ Δtλ Δt S1S1 μ Δtμ Δt1 - μ Δt TRANSITION MATRIX It is required to evaluate Probabilities P O and P 1 Of the component being in states S O and S 1

3. TRANSITION EQUATIONS After time interval Δt At Present SOSO S1S1 SOSO 1 - λ Δtλ Δt S1S1 μ Δtμ Δt1 - μ Δt Multiply both sides by

4. Then At Steady state as t tends to infinity, the component(system) will be AVAILABLE with probability

5. After time interval Δt At Present SOSO S1S1 SOSO 1 - λ Δtλ Δt S1S1 μ Δtμ Δt1 - μ Δt As t tends to infinity, Po = Availability = const and dPo / d t =0, then λ=1/MTBFμ=1/MTTR Therefore Another Approach to AVAILABILITY

6. Availability of A System with One Unit Standby

7. This system could exist in one of three states S o, S 1 and S 2: S o state of system operating with probability P o S 1 state of system with the main component failed and switched to the standby unit to operate ( probability of being in this state P 1, S 2 State of having both units main and standby failed with probability P 2 SOSO S1S2 λ λ μ 2 μ SOS1S2 So1 - λ Δtλ Δtλ Δt0 S1μ Δtμ Δt1 - λ Δt - μ Δtλ Δtλ Δt S202 μ Δt1 – 2 μ Δt Two Repair gangs are available STATE GRAPH TRANSITION MATRIX

8. TRANSITION EQUATIONS SOS1S2 SO1 - λ Δtλ Δtλ Δt0 S1μ Δtμ Δt1 - λ Δt - μ Δtλ Δtλ Δt S202 μ Δt1 – 2 μ Δt Taking the Limit as the time tends to infinity, we find

9. Substitute (4) and (5) in (1), we find This Result could be obtained directly by a method called HARMONIC BALANCE Apply Harmonic Balance to state So SOSO Expected IN [ µ P1 ] = Expected out [ λ Po ] μ S1 λ Similarly, apply Harmonic Balance to state S2 S1S2 λ 2μ2μ

10. The System is AVAILABLE in case of being in either states So and S1

11. In a metal cutting workshop, there are two different machines: Machine A: a mechanical cutting with capacity of 10 tons per hour. The machine has a constant failure rate of 3 failures per month and mean repair time of 5 days Machine B: a gas cutting machine with capacity of 20 tons per hour. The machine has a constant failure rate of 4 failures per month and mean repair time of 3 days. Evaluate the EFFECTIVE CAPACITY of the workshop. _____________________________________________________________________ The Workshop could exist in one of the following states So Both machines are working. S1 Mechanical Cutting failed and gas Cutting works S2 Gas Cutting failed and Mechanical Cutting works S3 Both failed S1S1 S2S2 S3S3 λ m μ m λ g Required to find the FOUR probabilities, Po, P1, P2, P3 The first equation: Po + P1 + P2 + P3 = 1 ………………………….(1) The other THREE equation will be obtained by applying HARMONIC BALANCE to the three states So, S1, S2 as follows: SOSO μ m λ m λ g μ g

12. Apply to So Apply to S1 Apply to S2 Omitting P3 from the two equations (3) and (4), we get Consider (2) and (5) to omit P2, we find Substitute in (2) Substitute in (3)

13. Effective Capacity = (20+10)*Po + 20*P1 + 10 * P2 Substitute in (1), we getPo + P1 + P2 + P3 =1……….(1)

14. Evaluation of the RELIABILITY of the above considered REPAIRABLE SYSTEM Consider the following three equations in the three unknowns: Apply Laplace Transform on the above three equations

15. Where, Is the Laplace Transform of Pm. Therefore, Solving the above system of algebraic equations

16. At t = 0, P O =1 : Therefore, A is the AVAILABILITY λ=1/MTTFμ=1/MTTR Therefore