Question

Consider the data in table below which shows the time required to produce a particular item. Using that data, calculate the learning rate associated with this product.

	Units	Labor Hrs
	5	155
	9	147
	13	137
	17	97
	25	72

Answer 1

Learning equation is Tn = T1*r^ln(n)/ln(2), where Tn represents the time required to produce nth item and r is the learning rate, ln() represents natural logarithm functon

T5 = T1*r^ln(5)/ln(2)

T9 = T1*r^ln(9)/ln(2)

T9/T5 = T1*r^ln(9)/ln(2) / (T1*r^ln(5)/ln(2) )

= r(^{ln(9)/ln(2) - ln(5)/ln(2)}

= r^0.848 ---(1)

T9/T5 = 147/155 = 0.9484 ---(2)

Equating (1) and (2)

r^0.848 = 0.9484

Solving for r, we get, r = e^{ln(0.9484)/0.848} = 0.9394

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Similarly, T13/T9 = r^{(ln(13)/ln(2) - ln(9)/ln(2))} = r^0.5305

T3/T9 = 137/147 = 0.932 or

r^0.5305 = 0.932

Solving for r, we get, e^{(ln(0.932)/0.5305)} = e^-0.1327 = 0.8757

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Similarly, T17/T13 = r^{(ln(17)/ln(2) - ln(13)/ln(2))} = r^0.3870

T17/T13 = 97/137 = 0.708 or

r^0.3870 = 0.708

Solving for r, we get, r = e^{(ln(0.708)/0.3870)} = e^-0.8921 = 0.4098

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Similarly, T25/T17 = r^{(ln(25)/ln(2) - ln(17)/ln(2))} = r^0.5564

T25/T17 = 72/97 = 0.7423

r^0.5564 = 0.7423

Solving for r, we get, r = e^{(ln(0.7423)/0.5564)} = e^-0.5356 =  0.5853

Average learning rate associated with the product is determined as the average of the 4 learning rates as calculated above

Average learning rate = (0.9394+0.8756+0.4098+0.5853)/4 = 0.7025