1. Equipment Replacement Four Really Great Models 1. Deterministic Age Replacement 2. Minimal Repair Model 3. Repair versus Replace 4. Block Replacement

2. Applications Continuous operating equipment Transportation Systems Radar sites Power generating equipment Computers and communications systems Intermittent operating equipment Vehicles Aircraft Appliances and entertainment components Lighting Systems

3. Deterministic Age Replacement Section 12.6 with extensions When to replace?

4. Some Very Good Assumptions Equipment is used continuously or time is measured in operating units Negligible downtime for repair and maintenance Infinite planning horizon Identical replacement equipment Only maintenance and replacement costs are considered Objective is to minimize long-run costs Time value of money not addressed

5. A Simple Model Let t = the decision variable, age of equipment at replacement time (length of the replacement cycle) K = replacement cost C(u) = maintenance cost rate ($/unit time) at age u Cost per replacement cycle of length t:

6. Solving a simple model: Average cost per unit time: To find the value of t that minimizes G(t):

7. The necessary first example Maintenance cost on the first year of a power generator was $1200. Replacement cost is $64,000.

8. A more general replacement model S(u) = salvage value at age u

9. Still a more general model

10. A Specific Case Rearranging terms:

11. The necessary second example Replacement cost of an automobile is $10,000. The car loses 15% of its value each year.

12. More of the necessary second example First year maintenance cost was $200 and is increasing at the rate of 40 percent per year.

13. Bringing it home… 49778.426 4.19812.65 4.29851.775 4.39896.092 4.49945.903 4.510001.52 4.610063.27 4.710131.48 4.810206.51 4.910288.73 510378.5

14. Revised maintenance cost First year maintenance cost was $200 and is increasing at the rate of 20 percent per year. Let M = rate of increase in maintenance cost I 0 = first year maintenance cost D = yearly rate of depreciation as a fraction Then b = ln(1 + M) = ln(1 +.2) =.18232 a = I 0 b / M = 200 (.18232) /.2 = 182.322 c = K = 10,000 d = -ln(1 – D) = -ln(1-.15) =.1625

15. Our new solution 8.59572.033 8.69650.86 8.79734.041 8.89821.688 8.99913.913 910010.83 9.110112.57 9.210219.23 9.310330.96 9.410447.86 9.510570.08

16. Yet another general case Then the average cost per unit time is given by:

17. Minimize G(t) Directly The Die Forge Company operates several die forges die forging pounds or presses metal between two dies (called tooling) that contain a precut profile of the desired part. Parts from a few ounces to 60,000 lbs. can be made using this process. Each unit cost $150,000 with C(t) = 200t 1.5 and S(t) = 500t -.15

18. The Solution Using Solver t* = 17.31128 yr. G(t) 1514625.37 15.514537.92 1614474.38 16.514432.87 1714411.72 17.3114408.18 1814424.73 18.514456.38 1914503.34 19.514564.66 2014639.47

19. Let’s take derivatives… tG'(t) 15-200.206 15.5-150.323 16-104.456 16.5-62.1349 17-22.9577 17.31128-0.00028 1847.30427 18.578.94795 19108.5795 19.5136.3968 20162.5735

20. A MINIMAL REPAIR MODEL Replacement Based upon random failures

21. Minimal repair model Equipment is restored upon failure Restoration results in minimal repair Equipment continues to deteriorate over time Define  (t) = failure rate at time t (an increasing function) N(t) = number of failures in time t

22. Power Law Process (t) = a b t b-1, a,b > 0 A six year old regional transit bus experiences minimal repair upon failure. It was found to have an intensity function given by  (t) =.5 t 1.5 with t measured in years. The expected number of failures during the first two years is given by The expected number of failures during the first 5 years is given by

23. Replacement Model C u = unit cost, C f = cost of a failure t = replacement time (decision variable)

24. Replacement Model – Power Law process what if b <= 1?

25. A Really Good Example A repairable machine has a NHPP with an intensity function of  (t) = 2 x 10 -6 t with t measured in operating hours. If the cost of a failure (repair) is $500 and the unit cost is $21000, then Replace at 6,481 hr or 810 days if used 8 hr/day

26. REPAIR VERSUS REPLACE Decisions…Decisions

27. Repair vs. Replacement f = the number of part failures over the life of the system, c = unit cost of the part, a r = fixed cost of repair a d = fixed cost of discarding where a d < a r b r = cost to repair a failure b d = cost to remove and replace a part where b d < b r, k = condemnation fraction A Cost Trade-off Model repair cost = a r + b r f + c k f discard cost = a d + (c + b d ) f

28. Repair vs. Replacement a d + (c + b d ) f <= a r + b r f + c k f If Then discard c f discard repair b r -b d 1-k

29. Example - circuit board a r = $ 20,000 (primarily test equipment and facilities), a d = $ 1200 (warehouse overhead for the spares), b r = $ 768 / failure ($48 / hr labor x 8 hr MTTR x 2 crew members), b d = $ 24 / failure ( $48 / hr labor x.5 hr R&R x 1 crew member), k =.05 c = 783.2 + 19789.5 / f.

30. BLOCK REPLACEMENT Block replacement for a group of items (pages 736-738) When is it more economical to replace a group of items that will eventually fail at the same time rather than one at a time?

31. Block Replacement Let p k = fraction of items that fail in period k n 0 = number of items placed in service at time 0 PeriodNumber of failures 1n 1 = n 0 p 1 2n 2 = n 0 p 2 + n 1 p 1 3n 3 = n 0 p 3 + n 1 p 2 + n 2 p 1 kn k = n 0 p k + n 1 p k-1 + …+ n k-1 p 1...

32. Group Replacement Continues Let a 1 = cost of an individual replacement a 2 = cost of replacing all n 0 items number periods for block replacementaverage cost 1a 2 + a 1 n 1 2[a 2 + a 1 (n 1 + n 2 )]/2 k n j is the expected number of failures in period j

33. Expected Cost without Block Replacement The expected lifetime of a single item: Therefore the expected (steady-state) cost for the entire block of items: Example: 200 items cost $5 each to replace and have an expected lifetime of 25 months. Then (200)(5)/25 = $40 per month (on the average, there will be 8 failures per month).

34. The mandatory example A large sign us lit by 8,000 bulbs. The bulbs cost $2 each to replace as they fail but can be replaced for 30 cents each when they are replaced all at once. Based upon past history: monthsfraction failing 10.02 20.03 3 40.05 50.08 60.09 70.07 80.1 90.11 100.13 110.15 120.14 1 n 0 = 8,000 n 1 = n o p 1 = (8,000) (.02) = 160 n 2 = n o p 2 + n 1 p 1 = (8,000) (.03) + (160)(.02) = 243 etc. Use Excel a1 =2 a2 =2400 n0 =8000

35. From Excel… nbr mo. fraction failing123456789101112njSum njG(k) 10.02160160.0 2720.0 20.032403.2243.2403.21603.2 30.032404.84.9249.7652.91235.2 40.054004.87.35.0417.11070.01135.0 50.0864087.37.58.3671.11741.11176.4 60.0972012.812.27.512.513.4778.42519.51239.8 70.0756014.419.512.5 20.115.6654.63174.01249.7 80.180011.221.920.020.920.123.413.1930.54104.51326.1 90.118801617.022.533.433.623.419.618.61064.05168.51415.2 100.13104017.624.317.537.553.738.919.627.921.31298.46466.91533.4 110.15120020.826.825.029.260.462.332.727.931.926.01542.98009.81674.5 120.1411202431.627.541.747.070.152.446.531.939.030.91562.49572.31795.4 sums18000

36. This is the End