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Learning Curve Analysis

Learning Curve Analysis. Supplement G. 0.30 – 0.25 – 0.20 – 0.15 – 0.10 – 0.05 – 0 –. Learning curve. Process time per unit (hr). | | | | | | 50 100 150 200 250 300. Cumulative units produced. Learning Curves. 0.30 – 0.25 – 0.20 – 0.15 – 0.10 – 0.05 – 0 –.

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Learning Curve Analysis

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  1. Learning CurveAnalysis Supplement G

  2. 0.30 – 0.25 – 0.20 – 0.15 – 0.10 – 0.05 – 0 – Learning curve Process time per unit (hr) | | | | | | 50 100 150 200 250 300 Cumulative units produced Learning Curves

  3. 0.30 – 0.25 – 0.20 – 0.15 – 0.10 – 0.05 – 0 – Process time per unit (hr) Learning curve Learning period | | | | | | 50 100 150 200 250 300 Cumulative units produced Learning Curves Showing the learning period

  4. Learning Curves 0.30 – 0.25 – 0.20 – 0.15 – 0.10 – 0.05 – 0 – Showing the learning period and the time when standards are calculated Learning curve Process time per unit (hr) Learning period Standard time | | | | | | 50 100 150 200 250 300 Cumulative units produced

  5. Developing Learning Curves • In developing learning curves we make the following assumptions: • The direct labor required to produce the n + 1st unit will always be less than the direct time of labor required for the nth unit. • Direct labor requirements will decrease at a declining rate as cumulative production increases. • The reduction in time will follow an exponential curve. kn = k1nb where k1 = direct labor hours for the 1st unit n = cumulative number of units produced b = log r / log 2 r = learning rate

  6. 80% Conversion Factors for the Cumulative Average Number of Direct Labor Hours per Unit

  7. 90% Conversion Factors for the Cumulative Average Number of Direct Labor Hours per Unit

  8. 50 – 40 – 30 – 20 – 10 – 0 – Manufacturer of diesel locomotives: Labor hours required for first unit = 50,000 Learning rate = 80% Direct labor hours per locomotive (thousands) | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1Developing the 80% Learning Curve

  9. 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% Direct labor hours per locomotive (thousands) | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1Estimating Direct Labor Requirements

  10. 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% Direct labor hours per locomotive (thousands) Labor hours required for 40th unit k40 = 50,000(40)(log 0.8)/(log 2) | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1using the formula

  11. 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% Direct labor hours per locomotive (thousands) k40 = 50,000(40)-0.322 | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1using the formula

  12. 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% Direct labor hours per locomotive (thousands) k40 = 50,000(0.30488) | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1using the formula

  13. 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% Direct labor hours per locomotive (thousands) k40 = 15,244 hours | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1using the formula

  14. 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% Direct labor hours per locomotive (thousands) k40 = 15,244 hours | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1using the formula

  15. 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% Direct labor hours per locomotive (thousands) k40 = 15,244 hours | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1using the formula

  16. 80% Learning Rate (n = cumulative production) 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% n 1 1.00000 2 0.90000 3 0.83403 . . . . . . 38 0.43634 39 0.43304 40 0.42984 64 0.37382 128 0.30269 Direct labor hours per locomotive (thousands) Cumulative average labor hours = | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1using Conversion Factors

  17. 80% Learning Rate (n = cumulative production) 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% n 1 1.00000 2 0.90000 3 0.83403 . . . . . . 38 0.43634 39 0.43304 40 0.42984 64 0.37382 128 0.30269 Direct labor hours per locomotive (thousands) Cumulative average labor hours = 50,000(0.42984) | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1using Conversion Factors

  18. 80% Learning Rate (n = cumulative production) 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% n 1 1.00000 2 0.90000 3 0.83403 . . . . . . 38 0.43634 39 0.43304 40 0.42984 64 0.37382 128 0.30269 Direct labor hours per locomotive (thousands) Cumulative average labor hours = 21,492 hours | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1using Conversion Factors

  19. 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% Direct labor hours per locomotive (thousands) | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1using Unit-doublings

  20. 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% Direct labor hours per locomotive (thousands) | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1 using Unit-doublings

  21. 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% Direct labor hours per locomotive (thousands) Second unit = 50,000(80%) | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1using Unit-doublings

  22. 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% Direct labor hours per locomotive (thousands) Second unit = 40,000 hours | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1using Unit-doublings

  23. 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% Direct labor hours per locomotive (thousands) Fourth unit = 40,000(80%) | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1using Unit-doublings

  24. 50 – 40 – 30 – 20 – 10 – 0 – Labor hours required for first unit = 50,000 Learning rate = 80% Direct labor hours per locomotive (thousands) Fourth unit = 32,000 hours | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced Example G.1using Unit-doublings

  25. 50 – 40 – 30 – 20 – 10 – 0 – Direct labor hours per locomotive (thousands) | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced The 80% Learning Curve forExample G.1

  26. 50 – 40 – 30 – 20 – 10 – 0 – Direct labor hours per locomotive (thousands) | | | | | | | 40 80 120 160 200 240 280 Cumulative units produced The 80% Learning Curve forExample G.1

  27. Application G.1Estimating Direct Labor Requirements The 1st unit of a new product is expected to take 1,000 hours. The learning rate is 80%, how much time should the 50th unit take?

  28. Units per Cumulative Month Month Units 1 2 2 2 3 5 3 5 10 4 8 18 5 12 30 Example G.2Estimating Labor Requirements

  29. 90% Learning Rate (n = cumulative production) Units per Cumulative Month Month Units n 1 1.00000 2 0.95000 3 0.91540 4 0.88905 5 0.86784 . . . . . . 30 0.69090 64 0.62043 128 0.56069 1 2 2 2 3 5 3 5 10 4 8 18 5 12 30 Cumulative Cumulative Average Time Total Hours Month per Unit for All Units Example G.2Estimating Labor Requirements

  30. 90% Learning Rate (n = cumulative production) Units per Cumulative Month Month Units n 1 1.00000 2 0.95000 3 0.91540 4 0.88905 5 0.86784 . . . . . . 30 0.69090 64 0.62043 128 0.56069 1 2 2 2 3 5 3 5 10 4 8 18 5 12 30 Cumulative Cumulative Average Time Total Hours Month per Unit for All Units 1 30,000(0.95000) = 28,500 Example G.2Estimating Labor Requirements

  31. 90% Learning Rate (n = cumulative production) Units per Cumulative Month Month Units n 1 1.00000 2 0.95000 3 0.91540 4 0.88905 5 0.86784 . . . . . . 30 0.69090 64 0.62043 128 0.56069 1 2 2 2 3 5 3 5 10 4 8 18 5 12 30 Cumulative Cumulative Average Time Total Hours Month per Unit for All Units 1 30,000(0.95000) = 28,500 28,500(2) = 57,000 2 30,000(0.86784) = 26,035 Example G.2Estimating Labor Requirements

  32. 90% Learning Rate (n = cumulative production) Units per Cumulative Month Month Units n 1 1.00000 2 0.95000 3 0.91540 4 0.88905 5 0.86784 . . . . . . 30 0.69090 64 0.62043 128 0.56069 1 2 2 2 3 5 3 5 10 4 8 18 5 12 30 Cumulative Cumulative Average Time Total Hours Month per Unit for All Units 1 30,000(0.95000) = 28,500 28,500(2) = 57,000 2 30,000(0.86784) = 26,035 26,035(5) = 130,175 3 30,000(0.79945) = 23,983 23,983(10) = 239,830 4 30,000(0.74080) = 22,224 22,224(18) = 400,032 5 30,000(0.69090) = 20,727 20,727(30) = 621,810 Example G.2Estimating Labor Requirements

  33. Units per Cumulative Month Month Units 1 2 2 2 3 5 3 5 10 4 8 18 5 12 30 Cumulative Cumulative Average Time Total Hours Month per Unit for All Units 1 30,000(0.95000) = 28,500 2 30,000(0.86784) = 26,035 3 30,000(0.79945) = 23,983 4 30,000(0.74080) = 22,224 5 30,000(0.69090) = 20,727 Example G.2Estimating Labor Requirements

  34. Units per Cumulative Month Month Units 1 2 2 2 3 5 3 5 10 4 8 18 5 12 30 Cumulative Cumulative Average Time Total Hours Month per Unit for All Units 1 30,000(0.95000) = 28,500 28,500(2) = 57,000 2 30,000(0.86784) = 26,035 3 30,000(0.79945) = 23,983 4 30,000(0.74080) = 22,224 5 30,000(0.69090) = 20,727 Example G.2Estimating Labor Requirements

  35. Units per Cumulative Month Month Units 1 2 2 2 3 5 3 5 10 4 8 18 5 12 30 Cumulative Cumulative Average Time Total Hours Month per Unit for All Units 1 30,000(0.95000) = 28,500 28,500(2) = 57,000 2 30,000(0.86784) = 26,035 26,035(5) = 130,175 3 30,000(0.79945) = 23,983 4 30,000(0.74080) = 22,224 5 30,000(0.69090) = 20,727 Example G.2Estimating Labor Requirements

  36. Units per Cumulative Month Month Units 1 2 2 2 3 5 3 5 10 4 8 18 5 12 30 Cumulative Cumulative Average Time Total Hours Month per Unit for All Units 1 30,000(0.95000) = 28,500 28,500(2) = 57,000 2 30,000(0.86784) = 26,035 26,035(5) = 130,175 3 30,000(0.79945) = 23,983 23,983(10) = 239,830 4 30,000(0.74080) = 22,224 22,224(18) = 400,032 5 30,000(0.69090) = 20,727 20,727(30) = 621,810 Example G.2Estimating Labor Requirements

  37. Month 1: Units per Cumulative Month Month Units 1 2 2 2 3 5 3 5 10 4 8 18 5 12 30 Cumulative Cumulative Average Time Total Hours Month per Unit for All Units 1 30,000(0.95000) = 28,500 28,500(2) = 57,000 2 30,000(0.86784) = 26,035 26,035(5) = 130,175 3 30,000(0.79945) = 23,983 23,983(10) = 239,830 4 30,000(0.74080) = 22,224 22,224(18) = 400,032 5 30,000(0.69090) = 20,727 20,727(30) = 621,810 Example G.2Estimating Labor Requirements

  38. Month 1: 57,000 – 0 = 57,000 hours Units per Cumulative Month Month Units 1 2 2 2 3 5 3 5 10 4 8 18 5 12 30 Cumulative Cumulative Average Time Total Hours Month per Unit for All Units 1 30,000(0.95000) = 28,500 28,500(2) = 57,000 2 30,000(0.86784) = 26,035 26,035(5) = 130,175 3 30,000(0.79945) = 23,983 23,983(10) = 239,830 4 30,000(0.74080) = 22,224 22,224(18) = 400,032 5 30,000(0.69090) = 20,727 20,727(30) = 621,810 Example G.2Estimating Labor Requirements

  39. Month 1: 57,000 – 0 = 57,000 hours Month 2: 130,175 – 57,000 = 73,175 hours Units per Cumulative Month Month Units 1 2 2 2 3 5 3 5 10 4 8 18 5 12 30 Cumulative Cumulative Average Time Total Hours Month per Unit for All Units 1 30,000(0.95000) = 28,500 28,500(2) = 57,000 2 30,000(0.86784) = 26,035 26,035(5) = 130,175 3 30,000(0.79945) = 23,983 23,983(10) = 239,830 4 30,000(0.74080) = 22,224 22,224(18) = 400,032 5 30,000(0.69090) = 20,727 20,727(30) = 621,810 Example G.2Estimating Labor Requirements

  40. Month 1: 57,000 – 0 = 57,000 hours Month 2: 130,175 – 57,000 = 73,175 hours Month 3: 239,830 – 130,175 = 109,655 hours Month 4: 400,032 – 239,830 = 160,202 hours Month 5: 621,810 – 400,032 = 221,778 hours Units per Cumulative Month Month Units 1 2 2 2 3 5 3 5 10 4 8 18 5 12 30 Cumulative Cumulative Average Time Total Hours Month per Unit for All Units 1 30,000(0.95000) = 28,500 28,500(2) = 57,000 2 30,000(0.86784) = 26,035 26,035(5) = 130,175 3 30,000(0.79945) = 23,983 23,983(10) = 239,830 4 30,000(0.74080) = 22,224 22,224(18) = 400,032 5 30,000(0.69090) = 20,727 20,727(30) = 621,810 Example G.2Estimating Labor Requirements

  41. Month 1: 57,000 – 0 = 57,000 /150 = 380 employees Month 2: 130,175 – 57,000 = 73,175 /150 = 488 employees Month 3: 239,830 – 130,175 = 109,655 /150 = 731 employees Month 4: 400,032 – 239,830 = 160,202 /150 = 1068 employees Month 5: 621,810 – 400,032 = 221,778 /150 = 1479 employees Units per Cumulative Month Month Units 1 2 2 2 3 5 3 5 10 4 8 18 5 12 30 Cumulative Cumulative Average Time Total Hours Month per Unit for All Units 1 30,000(0.95000) = 28,500 28,500(2) = 57,000 2 30,000(0.86784) = 26,035 26,035(5) = 130,175 3 30,000(0.79945) = 23,983 23,983(10) = 239,830 4 30,000(0.74080) = 22,224 22,224(18) = 400,032 5 30,000(0.69090) = 20,727 20,727(30) = 621,810 Example G.2Estimating Labor Requirements

  42. Application G.2Estimating Cumulative Labor Hours An example of using the learning model to test budget constraints: A company has a contract to make a product for the first time. The total budget for the 38-unit job is 15,000 hours. The first unit took 1000 hours, and the rate of learning is expected to be 80 percent. Do you think the 38-unit job can be completed within the 15,000-hour budget? How many additional hours would you need for a second job of an 26 additional units?

  43. Application G.2First 38-unit Job

  44. Application G.2Second additional 26-unit Job

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