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Learning curve theory question

LLetty4y ago
Tech World is a company which manufactures mobile phone handsets. From its past experiences, Tech World has realised that whenever a new design engineer is employed, there is a learning curve with a 75% learning rate which exists for the first 15 jobs. A new design engineer has just completed their first job in five hours. Note. At the learning rate of 75%, the learning factor (b) is equal to -0.415. How long would it take the design engineer to complete the sixth job? Y = axb b = log 0.75/log 2 = –0.1249/0.3010 = -0.415 Can you please explain me the below calculation: When x = 6, x –0.415 = 1/6 –0.415 = 0.4754 Did they divide it by 6 because we are looking for the time of the sixth jobs? Thank you.
John MoffatJohn MoffatTutor4y ago#1
I can only imagine that you have copied the answer wrongly or that there is a typing error in the answer. If it is either a past exam question or a question in the BPP Revision Kit, then tell me which one and I will check. (The formula is not y = axb, it is y = ax^b ) Have you watched my free lectures on learning curves, which explain exactly how to do questions like this? The time for the 6th job is the time take to produce 6 jobs less the time taken to produce 5 jobs.
LLetty4y ago#2
The formula has been written as such. It is from BPP Revision Kit. I have understood the calculation afterwards, thank you though for your explanation.
John MoffatJohn MoffatTutor4y ago#3
I have looked at the BPP answer and although the answer is correct, they have made one small typing error. You probably remember from school that taking anything to a negative power is the same as taking 1 divided by the same number to the positive power. So......6^(-0.145) is the same as 1/(6^0.415). Depending on your calculator you can do it either way - the answer is the same!! The typing mistake in the BPP answer is that they should have typed it as 1/(6^0.415) (and not as 1/6^-0.415 )
LLetty4y ago#4
Yes indeed. Thank you for your explanation and time Sir. Greatly appreciated!
John MoffatJohn MoffatTutor4y ago#5
You are welcome :-)
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