Question

An operator has taken 70 minutes to produce 5^th unit and has taken 48.6 mins to produce the 20^th unit.

         (a) What is the learning percentage.

            (b) What is the learning curve equation of the operator on this process

Answer 1

a)

Tn = T1*n^(ln(l)/ln(2)) [ Tn = Time taken to complete nth job and l = learning rate]

So, for n = 5,

70 = T1*5^(ln(l)/ln(2)) -----(1)

For n = 20,

48.6 = T1*20^(ln(l)/ln(2)) -----(2)

by dividing 1 by 2 we get,

70/48.6 = 5^(ln(l)/ln(2))/20^(ln(l)/ln(2))

or, 1.440329218 = 0.25^(ln(l)/ln(2))

or, (ln(l)/ln(2)) = ln(1.440329218)/ln(0.25) = -0.263199304

or, ln(l) = -0.182435856

or, l = 83%
b)

Substituting l in equation 1,

T1 = 70/5^(ln(0.83)/ln(2)) = 107.8929061

So, learning curve equation of the operator on this process = 107.8929061*n^(ln(0.83)/ln(2)) where Tn = Time taken to complete nth job