If it takes 40 hours to complete the first unit and the company knows from experience that the learning rate should be 0.75.
The hours required to produce the nth unit may be computed as follows:
TN=T1(Nb)
where TN = time for the Nth unit T1 = hours to produce the first unit
b = (log of the learning rate)/(log 2) = slope of the learning curve
Here, T1 = 40 and b = 0.75
Using the above formula, we get
T2 = 40*2^(LOG(0.75)/LOG(2)) = 30
T3 = 40*3^(LOG(0.75)/LOG(2)) = 25.35
T4 = 40*4^(LOG(0.75)/LOG(2)) = 22.50
T5 = 40*5^(LOG(0.75)/LOG(2)) = 20.51
T6 = 40*6^(LOG(0.75)/LOG(2)) = 19.02
T7 = 40*7^(LOG(0.75)/LOG(2)) = 17.84
T8 = 40*8^(LOG(0.75)/LOG(2)) = 16.88
T9 = 40*9^(LOG(0.75)/LOG(2)) = 16.07
T10 = 40*10^(LOG(0.75)/LOG(2)) = 15.38
So, Total time = T1 + T2 + ...+ T9 + T10 = 40 + 30 +...+ 16.07 + 15.38
Total time to complete all 10 units = 223.54
The range of values to don't forget for the mastering rate is much less than 1. zero and greater than 10^-6. A conventional default price for the getting-to-know-the-charge is 0.1 or zero.01, and this will constitute a good starting pay as the mastering fee critical
Typically, a huge learning rate permits the version to analyze faster, at the value of arriving at a sub-most reliable final set of weights. A smaller getting to know the price may also permit the model to analyze a more optimal or maybe globally most beneficial set of weights however might also take considerably longer to train.
Point to your hassle. getting to know the rate offers the fee of velocity wherein the gradient moves at some point of gradient descent. setting it too high would make your direction unstable, and too low could make convergence slow. placed it to zero manners your model isn't always gaining knowledge of whatever from the gradients.
Learn more about the learning rate here:
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