Thank you sir for such great lectures But the way I did was like more complicated and time consuming but still I got 0.833333 I did the reverse of the doubling method by making the learning effect as x and total time as y and for the second unit y+25. So 80x/100(average time) = y(total time for first unit) 55x/100(average time) = y+25(total time for second unit)

And then I substituted y with 80x/100 and found out x and got 0.8333

Hi, I’m a little confused with this question from the specimen CBE.

The first phase of production has now been completed for the new car seat. The first unit actually took 12.5 hours to make and the total time for the first eight units was 34.3 hours, at which point the learning effect came to an end. Chair Co are planning on adjusting the price to reflect the actual time it took to complete the eighth unit.

What was the actual rate of learning which occurred (to two decimal places)?

They have shown the workings but I can’t calculate how they arrived at the answer. The equation throws me off when I get to 0.343 = r3

You could not be asked this in the exam because it would mean using the formula ‘backwards’ which is not in the syllabus. You can only be asked this sort of question if it involves doubling.

I’ve tried countless ways of doing this but can’t see how one could work their way back to get the learning rate.

You’re saying the first unit takes 100 hours, and the third and fourth have an average rate of 50 hours (combined). This makes an unknown value to get to the second unit’s average time by the use of simple arithmetic, yes?

But the second unit does not have ‘an average time’!!

Suppose the first unit take 100 hours and the second unit takes 50 hours. Then the average time per unit when they make two units is (100 + 50) / 2 = 75 hours.

The learning rate applies to the average time per unit, so in this case the learning rate would be 75/011 = 0.75 or 75%

I do suggest that you watch both of the lectures again.

Please watch the lecture again. I do not write that the time for the 4th unit is 50. I write that the average time if they make 4 units is 50 per unit. (This would be the total time for all 4 units divided by 4 to get the average time per unit).

67.5 is certainly not the learning rate – it is the average time per unit when they make two units. The learning rate is 67.5/80, using the doubling rule. Have you watched part (a) of this lecture??

Actually i was a bit confused but now i can clearly understand about the example you have shown in this video and differentiate the average time per unit and the learning rate

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addisanopacourage says

Hi John

It’s interesting how the learning rate is calculated. Thanks a lot

John Moffat says

Thank you fo r your comment 馃檪

fani92 says

Sir, I just wanted to say you are a wonderful lecturer and your lectures are helping me understand the paper F5!

John Moffat says

Thank you very much for your comment 馃檪

swaroop12 says

Thank you sir for such great lectures

But the way I did was like more complicated and time consuming but still I got 0.833333

I did the reverse of the doubling method by making the learning effect as x and total time as y and for the second unit y+25.

So 80x/100(average time) = y(total time for first unit)

55x/100(average time) = y+25(total time for second unit)

And then I substituted y with 80x/100 and found out x and got 0.8333

Can this be a possibility??

swaroop12 says

Sorry sir,

My whole comment is wrong 馃檨

Forgive me…..

John Moffat says

No problem 馃檪

tanya18 says

Hi, I’m a little confused with this question from the specimen CBE.

The first phase of production has now been completed for the new car seat. The first unit actually took 12.5 hours to make and the total time for the first eight units was 34.3 hours, at which point the learning effect came to an end. Chair Co are planning on adjusting the price to reflect the actual time it took to complete the eighth unit.

What was the actual rate of learning which occurred (to two decimal places)?

They have shown the workings but I can’t calculate how they arrived at the answer. The equation throws me off when I get to 0.343 = r3

John Moffat says

You must ask this sort of question in the Ask the Tutor Forum, and not as a comment on a lecture.

maqmukul says

Average time for the 1st unit 100

Average time for the 4th unit 50

So, learnig rate will be 50/100=50%

Is the correct sir?

John Moffat says

No it is not correct.

If the average time per unit when they make 4 units is 50, then the learning rate will be the square root of 50/100.

Ahmed says

sorry its a mistake should be 14 units****

Ahmed says

or the first unit take 100 hrs and the average time for next 14 hours is 60 what is the learning rate?

John Moffat says

You could not be asked this in the exam because it would mean using the formula ‘backwards’ which is not in the syllabus. You can only be asked this sort of question if it involves doubling.

Ahmed says

Sir,

I still dit not get the last bit tried a lot i know its silly but im stuck,,could you please help me?

if first unit take 100 hours and the average time for 4 unit is 50 what is the learning rate?

could you please show the calculation? 100

.

.

50 ?

John Moffat says

4 units involves doubling twice (1 to 2; 2 to 4)

Therefore the learning rate is the square root of 50/100 = 0.71 or 71%

colinjsweetman says

At the end of this lecture – is the learning rate 37.5%?

IE

First unit – 100

Fourth Unit – 50

Average time – 100+50 /4 = 37.5

37.5/100 = 37.5% learning curve effect

thanks!

John Moffat says

No. If you want the average then you need the times for the second and third as well!!

Anyway, 50 is the average time if we make 4 units, not the time for the 4th unit.

colinjsweetman says

HI John

Thank you for the quick reply.

I’ve tried countless ways of doing this but can’t see how one could work their way back to get the learning rate.

You’re saying the first unit takes 100 hours, and the third and fourth have an average rate of 50 hours (combined). This makes an unknown value to get to the second unit’s average time by the use of simple arithmetic, yes?

John Moffat says

But the second unit does not have ‘an average time’!!

Suppose the first unit take 100 hours and the second unit takes 50 hours. Then the average time per unit when they make two units is (100 + 50) / 2 = 75 hours.

The learning rate applies to the average time per unit, so in this case the learning rate would be 75/011 = 0.75 or 75%

I do suggest that you watch both of the lectures again.

skaneesh says

Hi John,

Here the last bit calculation is without second unit:

So, r2 = 50/100

r=70.71%

Is it the correct answer.

skaneesh says

I think my early comment is wrong.

r2 = [(50+100)/4] / 100 = 0.375

r = 61.24%

Is this the right way. I’m getting bit confused, please help.

John Moffat says

Please watch the lecture again. I do not write that the time for the 4th unit is 50. I write that the average time if they make 4 units is 50 per unit. (This would be the total time for all 4 units divided by 4 to get the average time per unit).

Then r^2 will be equal to 50/100.

skaneesh says

Ok. Thank you for the clarification of my mistake sir.

John Moffat says

You are welcome 馃檪

sattar786 says

hi sir,

i wanted to know why did you divide the 67.5 with the 80?

isn’t the 67.5 the learning rate?

if it is not then what is it called (the 67.5)?

John Moffat says

67.5 is certainly not the learning rate – it is the average time per unit when they make two units.

The learning rate is 67.5/80, using the doubling rule.

Have you watched part (a) of this lecture??

sattar786 says

hi sir,

yes i have watched the part (a)

Actually i was a bit confused but now i can clearly understand about the example you have shown in this video and differentiate the average time per unit and the learning rate

thank you so much sir for guiding me 馃檪 馃檪

John Moffat says

You are welcome 馃檪