Question

A club would like to know what the learning rate is for a certain welded assembly. They conducted a study on a total of four assembles. However, the data on samples 2 and 3 were corrupted. The records for sample 1 and sample 4 were 6 hours and 4.86 hours respectively. Determine the learning rate for this assembly using (a) doubling method and (b) formula method.

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Answer 1

a)

Doubling Method -

Time Required for nth unit = T*L^x

Where,

T = Time Required for First unit

L = learning Rate

x = Number of time T will be doubled

Here, x will be equal to 2 as there will be twice doubling, 1 to 2 to 4

Time Required for 4th Sample = 6*(L)^2

4.86 = 6*L^2

L^2 = (4.86/6)

L = (4.86/6)^(1/2)

L = 0.9

Learning Rate = 90%

b)

Tn = T1*N^b

Where

T1 = Time required for first unit

N = Unit Number

b = ln(learning rate)/ln(2)

4.86 = 6*4^b

4^b = (4.86/6)

4^b = 0.81

b = ln(0.81)/ln(4)

b = -0.152

ln(Learning Rate)/ln(2) = -0.152

ln(Learning Rate) = -0.152*ln(2)

ln(Learning Rate) = -0.105358

Learning rate = e^-0.105358

Learning Rate = 0.90 or 90%

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