3 # items in service Repair shop repaired units stock time Failures failed units
4 Criteria for Decision Making 1.Instant reliability 2.Interval reliability 3.Cost minimization 4.(Process) Availability
5 Scenario Plant has 62 electric motors on their conveyor systems (Mining company) MTBReplacements of motors is 3000 days (8 years) Planning horizon is 1825 days (5 years) Cost of spare motor is 15,000 $ Value of unused spare is 10,000 $ Cost of emergency spare is 75,000 $ MTTRepair a motor is 80 days Cost of plant downtime for a single motor is 1000 $ per day Holding cost of a spare is 4.11 $ per day (10% of value of part/annum) QUESTION: HOW MANY SPARE PARTS TO STOCK?
6 Results: Repairable Parts Electric motors Interval reliability: 95% reliability requires 7 spares Instant reliability: 95% reliability requires 4 spares Cost minimization: requires 6 spares. Associated plant availability is % Availability of 95%: requires 0 spares. Associated electric motor availability is 97.4% [Note: If availability of 99% was required (rather than the specified 95%) then spares required would be 2]
7 Reference Case Population100 transformers Failure Rate0.005 failures/transformer/yr Repair Time1 yr Replacement Time0.001yr Interval1 yr
9 Case studies: 1.Fume fan shaft, blast furnace in a steel operation: non- repairable part, decision support 2.Power train component, haul trucks: repairable parts, multiple criteria 3.Frigate control system: supportability interval Additional Cases
10 1. Fume fan shaft – steel mill Spares provisioning optimization project Part: fume fan shaft used in a Blast Furnace Decision: should there be 0 or 1 spares? Complication: Part has long lifespan (25-40 years). Long lead time (22 weeks). If part fails, results are catastrophic (loss of almost $6 million per week). Inventories are trying to be minimized. SMS was used to quantify the risk involved in not having a spare Decision support
11 How many spares – Fume fan shaft?
12 2. Repairable components – haul trucks Open pit mining operation in South America Haul truck power train component: COMPONENT X
13 6,600 operating hours per truck per year (average fleet utilization) Preventive replacement policy at 9,000 operating hours in place Repairable components - Data / 1 General Two parameter Weibull distribution, fitted using Weibull events: 86 failures, 85 suspensions (preventive replacements) Beta = Eta = 14,650 operating hours Mean time to replacement = 6,420.3 operating hours Time to replacement
14 Based on estimations provided by maintenance personnel Estimated at MTTR = 452 operating hours Repairable components - Data / 2 Time to repair Downtime: estimated using operational indicators (value of lost production, $2,173.3 / op. hour) Holding: 25% of the value of the part per annum ($1.51/ op. hour) Cost of downtime and holding costs
15 Repairable components - Data Summary SMS can perform the optimization based on four criteria ParameterValue Number of components in operation78 MTBReplacements (μ) (op. hours) Planning horizon (T)6600 (op. hours) MTTRepair (μ R )452 (op. hours) Holding cost for one spare$1.51 per op. hour (25% of value of part/annum) Cost of plant downtime for a single component$ per op. hour
17 4. Frigate control system – supportability intervals (S.I.) Determination of supportability intervals Several electronic components – control system Parts no longer available - discontinued Decision: how long can we support the operation of the system using only the current stock? (achieving the desired reliability of the stock)
18 Case Study – Supportability Interval (S.I.) Summary of Data Supportability interval (RUL of current stock) can be rapidly calculated using SMS – NEWLY ADDED FEATURE
19 Case Study (S.I.) / 2 Results Supportability for the system is influenced by the (shortest) supportability for any of its critical parts
20 Case Study (S.I.) / 3 Informed decisions for: Adequate timing for replacement of current system Placement of final orders for discontinued parts Maximum supportability interval for current stock – procurement planning