OpenTuition | ACCA | CIMA
Free ACCA and CIMA on line courses | Free ACCA, CIMA, FIA Notes, Lectures, Tests and Forums
Spread the word
If you have benefited from our materials, please spread the word so more students can benefit.
To help us keep materials up to do and add new content you can also donate
August 22, 2018 at 5:00 am
Hi John It’s interesting how the learning rate is calculated. Thanks a lot
John Moffat says
August 22, 2018 at 6:42 am
Thank you fo r your comment 🙂
August 15, 2018 at 12:49 pm
Sir, I just wanted to say you are a wonderful lecturer and your lectures are helping me understand the paper F5!
August 15, 2018 at 7:05 pm
Thank you very much for your comment 🙂
July 6, 2018 at 6:28 pm
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??
July 6, 2018 at 6:31 pm
Sorry sir, My whole comment is wrong 🙁 Forgive me…..
July 6, 2018 at 7:13 pm
No problem 🙂
June 4, 2018 at 6:21 pm
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
June 5, 2018 at 5:28 am
You must ask this sort of question in the Ask the Tutor Forum, and not as a comment on a lecture.
January 20, 2018 at 10:15 am
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?
January 20, 2018 at 11:38 am
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.
December 1, 2017 at 5:32 pm
sorry its a mistake should be 14 units****
December 1, 2017 at 5:30 pm
or the first unit take 100 hrs and the average time for next 14 hours is 60 what is the learning rate?
December 2, 2017 at 8:48 am
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.
December 1, 2017 at 5:26 pm
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 ?
December 2, 2017 at 8:50 am
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%
November 19, 2017 at 10:05 pm
At the end of this lecture – is the learning rate 37.5%?
First unit – 100
Fourth Unit – 50
Average time – 100+50 /4 = 37.5
37.5/100 = 37.5% learning curve effect
November 20, 2017 at 8:11 am
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.
November 20, 2017 at 8:55 pm
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?
November 21, 2017 at 1:16 pm
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.
November 21, 2017 at 11:42 pm
Hi John, Here the last bit calculation is without second unit: So, r2 = 50/100 r=70.71% Is it the correct answer.
November 21, 2017 at 11:50 pm
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.
November 22, 2017 at 8:49 am
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.
November 22, 2017 at 11:22 am
Ok. Thank you for the clarification of my mistake sir.
November 22, 2017 at 3:31 pm
You are welcome 🙂
October 19, 2017 at 5:36 pm
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)?
October 20, 2017 at 8:40 am
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??
October 20, 2017 at 11:08 am
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 🙂 🙂
October 20, 2017 at 3:38 pm
You must be logged in to post a comment.