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Supplement 7

Supplement 7. Learning Curves. LC estimates (SLOPE):. Every doubling of repetitions results in a constant percentage decrease in the time per repetition Typical decreases range from 10 to 20 percent 10% of Improvement Rate = 90% of Learning Curve Slope = 90% of Learning Percent.

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Supplement 7

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  1. Supplement 7 Learning Curves

  2. LC estimates (SLOPE): Every doubling of repetitions results in a constant percentage decrease in the time per repetition Typical decreases range from 10 to 20 percent 10% of Improvement Rate = 90% of Learning Curve Slope = 90% of Learning Percent

  3. LC estimates (SLOPE): • T2n = Learning Percent xTn • T22 = Learning Percent xT11 • T11 = T22 /Learning Percent • Learning Percent = T22 / T11 Tn = Time for nthunit

  4. Learning

  5. The Learning Effect

  6. Learning Curves: On a Log-Log Graph

  7. The Learning Effect

  8. Learning Illustrated Each time cumulative output doubles, the time per unit for that amount should be approximately equal to the previous time multiplied by the learning percentage. If the first unit of a process took 100 hours and the learning rate is 90%:

  9. Unit Times: Formula Approach

  10. Example: Formula Approach If the learning rate is 90%, and the first unit took 100 hours to complete, how long would it take to complete the 25th unit? Or, Check table 7s.1 (PAGE 255) T25 = T1 x ( ) .90, 25 T25 = 100 x .613 = 61.3 hours

  11. Unit Times: Learning Factor Approach The learning factor approach uses a table that shows two things for selected learning percentages: Unit value for the number of repetitions (unit number) Cumulative value, which enables us to compute the total time required to complete a given number of units.

  12. Example: Learning Factor Approach If the learning rate is 90%, and the first unit took 100 hours to complete, how long would it take to complete the 25th unit? How long would it take to complete the first 25 units?

  13. eg : learning rate is 80%, T1st Jet = 400 Days Estimate the expected number of labor days of direct labor for the 20th jet. T20 = T1 x ( ) .80, 20 T20 = 400 x .381 = 152.4 Days Estimate the expected number of labor days of direct labor for all 20 jets. T1-20 = T1 x ( ) .80, 1-20 T1-20= 400 x 10.485 = 4194 Days Avg Days/Jet = 4194 days / 20jets = 209.7 days/jet

  14. Eg : learning rate is 80, T1st Jet = 400 Days If the company expects a contribution to overhead & profit of $150/day on top of a labor cost $200/day, what should be the Total Price Quote? T1-20 = 4194 Days Charge per day=150+200=350 Total price quote = $350 x 4194days = $1467900 If the company expects a contribution to overhead & profit of $150/day on top of a labor cost $200/day, what should be the Unit Price Quote? Unit price quote = $1467900 / 20jets = $73395

  15. Eg : learning rate is 80, T1st Jet = 400 Days Estimate the expected number of labor days of direct labor for jet 10 through jet 15. T1-15 – T1-9 = T1 x ( ).80, 1-15 – T1 x ( ).80, 1-9 = 400 x 8.511 – 400 x 5.839 = 400 x (8.511 – 5.839) = 1068.8 Days

  16. Eg : learning rate is 80, T1st Jet = 400 Days If the total labor cost of 20 jets is $1467900, what should be the total labor cost for 30 jets and the average cost per unit? T1-20 = 4194 Days $1467900 /4194days = $350/day T1-30 = 400 x 14.020 = 5608 Days Total labor cost =$350x5608days=$1962800 Avg cost/unit =$1962800/30Jets=$65426.70

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