The doubling rule applies to the average time per unit. If the first takes 18 hours and the second takes 10 hours, then the average time per unit is (18 + 10)/2 = 14 hours.

Using the doubling rule, if the learning rate is r, then the average time per unit if 2 are made = 18 x r = 14. So r = 0.78 (or 78%)

If the learning rate was 56% then the average time per unit if 2 were made would be 18 x 0.56 = 10 hours. This is NOT the time for the second unit, but the average time per unit is we make 2. So the total time for 2 units is 2 x 10 = 20 hours. Since the first unit took 18 hours, the time for the second = 20 – 18 = 2 hours.

I do suggest that you watch the free lectures on this.

Sir,I am confused as well by Q2 If you need to calculate the time for 7 units you need to take the time for 8 and then subtract the hrs spent for the first right?

So

1 unit 42 hrs 2 units average 33.6 4 units average 26.88 6 units average 21.504 8 units average 17.20

I thought I should calculate average 17.20*8units= 137.26 hrs and then take off 42 hrs = 95.62 hrs Please can you explain again exactly why you multiplied by ^3

Your table is wrong. When using the doubling rule, it is 1 unit, then 2 units, then 4 units, then 8 units. You cannot get 6 units by doubling.

Because it is doubling 3 times (1 to 2; 2 to 4; 4 to 8) you can set up a table as you have done, or (faster) take 42 to the power of 3. (^3 is the standard way of typing ‘to the power of 3’)

Did you not watch the free lectures before attempting the test?

Activity level 800 units 1200 unit Total cost $16400 $23600

The fixed cost of the business step up by 40 % at 900 units

What is variable cost per unit

A : $8.00 B : $18 c : $19.67 d: $20:00

sir this is the question about high low method from kaplan practice question i calculate variable cost $18 but answer saying its $8. is there anything i missing ?

sir for q1, the average time per unit for 2 units = (18 + 10) / 2 = 14 hours so what i am getting it that for 2nd unit we require 14 hours that is average time and if there are 3 units so we divide it by 3 for average and the answer would give the average time required specifically for making 3rd unit..is i am right

That’s the tabular approach in in learning curve use only when production doubles. To solve the problem of calculating time for the 3rd or 5th unit u need to use the formula – y = axb (y is average time per unit to produce x units, a is time taken for the first unit, x is the total number of units and be is to the power: log the learning curve over log 2) of course u can use the formula on even number of units. I have tried it ?Hope this helps.

Sir the cost for the lowest production shouldn’t be the lowest cost overall. Yes I have use the lowest and highest production levels but by right automatically the cost suppose to be the lowest and highest for the levels of production selected. However the lowest cost there is not under the lowest units produced. Rather it’s under the level of production of 20,160

You should watch my free lectures on the high-low method. We use the costs for the highest and lowest production levels (regardless of whether or not they are the highest and lowest costs).

Because 8 involves doubling 3 times. I do suggest that you watch the free lecture, which explains the doubling rule and works through an almost identical example. (You should not attempt the tests unless you have watched the relevant lecture first 🙂 )

What you are calculating is the time for the first 7, which is not what is asked for.

The have already made 1, and so another 7 means a total of 8. The only way you can do it is to calculate the time for the first 8 and subtract the time for the first. By all means use the formula, but it is faster to use the doubling rule (especially since you are not given a value for b – it is always given in the exam if you need to use the formula), and, of course, you need to be able to use the doubling rule anyway because you can be specifically tested on it.

I do suggest that you watch our free lectures on learning curves where all of the above is explained, with examples. (Our lectures are a complete course for Paper F5 and cover everything you need to be able to pass the exam well)

With regard to Q1, the second unit takes 10 hrs which means 18 * r^1. That gives r = 56%. Please correct me if I am wrong. Also, when we deduct 56% from 18, it gives 10 hours which is what it takes for the second unit.

The current software does not allow for the calculations to be shown (which is why you should ask here if you have a problem and then I will explain). When we can afford to upgrade the software then the calculations will appear, but given that we provide all this free of charge there is obviously a limit as to what we can afford 🙂 )

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briandean2002 says

Q1 is very confusing, I got 56% as the answer.

Allow me to rewrite the question to illustrate my point.

A company is intending to produce a new product.

They have produced one test unit which took 18 hours. The learning curve is 56%

How many hours will the second product take to product (to the nearest hour)?

A: 5 hours

B: 3 hours

C: 14 hours

D: 10 hours (correct answer)

John Moffat says

What you have written is wrong!

The doubling rule applies to the average time per unit.

If the first takes 18 hours and the second takes 10 hours, then the average time per unit is (18 + 10)/2 = 14 hours.

Using the doubling rule, if the learning rate is r, then the average time per unit if 2 are made = 18 x r = 14. So r = 0.78 (or 78%)

If the learning rate was 56% then the average time per unit if 2 were made would be 18 x 0.56 = 10 hours. This is NOT the time for the second unit, but the average time per unit is we make 2. So the total time for 2 units is 2 x 10 = 20 hours.

Since the first unit took 18 hours, the time for the second = 20 – 18 = 2 hours.

I do suggest that you watch the free lectures on this.

briandean2002 says

Thank you!

I understand it now.

hitsui says

Sir, so if we are asked about more than 2 units for example 4 or 8, we’ll do it the same way you just did right?

John Moffat says

You are welcome 🙂

blgl says

Q2?sir?I don’t understand the0.8^3

why is it 3?

thanks!

John Moffat says

Because to make 8 in total means doubling 3 times.

Have you not watched the free lectures on this?

marghe says

Sir,I am confused as well by Q2

If you need to calculate the time for 7 units you need to take the time for 8 and then subtract the hrs spent for the first right?

So

1 unit 42 hrs

2 units average 33.6

4 units average 26.88

6 units average 21.504

8 units average 17.20

I thought I should calculate average 17.20*8units= 137.26 hrs and then take off 42 hrs = 95.62 hrs

Please can you explain again exactly why you multiplied by ^3

Thanks

John Moffat says

Your table is wrong. When using the doubling rule, it is 1 unit, then 2 units, then 4 units, then 8 units. You cannot get 6 units by doubling.

Because it is doubling 3 times (1 to 2; 2 to 4; 4 to 8) you can set up a table as you have done, or (faster) take 42 to the power of 3. (^3 is the standard way of typing ‘to the power of 3’)

Did you not watch the free lectures before attempting the test?

marghe says

Yes I watched them but still I was multiplying by 2 instead of doubling ! My bad, dumb mistake thanks for taking time to explain 🙂

John Moffat says

No problem 🙂

ali says

Activity level 800 units 1200 unit

Total cost $16400 $23600

The fixed cost of the business step up by 40 % at 900 units

What is variable cost per unit

A : $8.00

B : $18

c : $19.67

d: $20:00

sir this is the question about high low method from kaplan practice question i calculate variable cost $18 but answer saying its $8.

is there anything i missing ?

John Moffat says

Kaplan’s answer is correct, but ask in the Ask the Tutor Forum and I will explain (not as a comment on a lecture).

raheelislam says

sir for q1, the average time per unit for 2 units = (18 + 10) / 2 = 14 hours so what i am getting it that for 2nd unit we require 14 hours that is average time and if there are 3 units so we divide it by 3 for average and the answer would give the average time required specifically for making 3rd unit..is i am right

John Moffat says

No, you are not right.

You need to watch my lectures on learning curves.

preetierc says

That’s the tabular approach in in learning curve use only when production doubles. To solve the problem of calculating time for the 3rd or 5th unit u need to use the formula – y = axb (y is average time per unit to produce x units, a is time taken for the first unit, x is the total number of units and be is to the power: log the learning curve over log 2) of course u can use the formula on even number of units. I have tried it ?Hope this helps.

preetierc says

*That’s the tabular approach use in learning curve only when production doubles

John Moffat says

Which is as I explain in my free lectures. It is important to watch the lectures before attempting the tests.

rcbudhu says

Referring to Question 3 in this practice questions,

Using the highlow method sir I thought it was supposed to be 960000-883200 not 885120 that you use

John Moffat says

No – you use the highest and lowest levels of production.

preetierc says

Sir the cost for the lowest production shouldn’t be the lowest cost overall. Yes I have use the lowest and highest production levels but by right automatically the cost suppose to be the lowest and highest for the levels of production selected. However the lowest cost there is not under the lowest units produced. Rather it’s under the level of production of 20,160

John Moffat says

You should watch my free lectures on the high-low method. We use the costs for the highest and lowest production levels (regardless of whether or not they are the highest and lowest costs).

preetierc says

Okay sir thank u. U are really doing a good job.

jonathanforstudying says

Hi Sir, i don’t think question 4 is covered in the lecture right? i don’t remember learning it.

jonathanforstudying says

ok i get it, it is from chapter 11 instead.

John Moffat says

I am glad you have got it 🙂

jonathanforstudying says

Thank you Sir

John Moffat says

You are welcome 🙂

ogunseye says

Using the highlow method sir I thought it was supposed to be 960000-883200 not 885120 that you used.

Or am I getting this wrong sir?

John Moffat says

Which question are you referring to?

khadijaali says

Q3 sir I believe the lowest vc is $ 883,200 not $ 885,120

Please explain further

Thanks

Khadija

John Moffat says

We use the highest and lowest of the independent variable, which is units in order to decide which months to use.

The answer is correct, and do watch the free lecture on this.

aamir2111 says

isn’t the question 3 in practical questions of quantitative analysis of budgeting actually from CVP?

John Moffat says

No – it is testing the high-low method (which is covered in our lectures on quantitative analysis).

aamir2111 says

Thanks.

aamir2111 says

And what about question 4? Multi product profit volume charts?

ckinnrossli says

sir, why 0.8^3? i didn’t get where is 3 comes from.

John Moffat says

Because 8 involves doubling 3 times.

I do suggest that you watch the free lecture, which explains the doubling rule and works through an almost identical example.

(You should not attempt the tests unless you have watched the relevant lecture first 🙂 )

ali imran says

In Q.2 by solving Y = axb formula we get :42*7^-.321928=22.448. But this is wrong. plz identify my mistake.

Thank you.

John Moffat says

What you are calculating is the time for the first 7, which is not what is asked for.

The have already made 1, and so another 7 means a total of 8. The only way you can do it is to calculate the time for the first 8 and subtract the time for the first.

By all means use the formula, but it is faster to use the doubling rule (especially since you are not given a value for b – it is always given in the exam if you need to use the formula), and, of course, you need to be able to use the doubling rule anyway because you can be specifically tested on it.

I do suggest that you watch our free lectures on learning curves where all of the above is explained, with examples.

(Our lectures are a complete course for Paper F5 and cover everything you need to be able to pass the exam well)

ali imran says

Ok,

got it,

Thanks.

John Moffat says

Great 🙂

Renee says

Question 2 how was the decimal 0.83 arrived at?

John Moffat says

Oops – it is a typing error.

The average time per unit for 8 units = 42 x 0.80^3 = 21.504

Everything else in the solution is fine.

I will have it corrected.

jasmine says

Hi Sir, the first took 18 hour and second took 10 hour , how calculate the learning rate?

John Moffat says

The average time per unit for 2 units = (18 + 10) / 2 = 14 hours

Therefore the learning rate = 14/18 = 0.78 (or 78%)

triplec says

Mr.John,

With regard to Q1, the second unit takes 10 hrs which means 18 * r^1. That gives r = 56%. Please correct me if I am wrong. Also, when we deduct 56% from 18, it gives 10 hours which is what it takes for the second unit.

Thanks,

Sathish

mohsin1525 says

Sir kindly also update the calculations workings for arriving at a particular solution especially for the numerical

John Moffat says

The current software does not allow for the calculations to be shown (which is why you should ask here if you have a problem and then I will explain).

When we can afford to upgrade the software then the calculations will appear, but given that we provide all this free of charge there is obviously a limit as to what we can afford 🙂 )

Steven says

Hi Sir, not sure I understood the question here – using high low wouldn’t you end up with a vc of 1.81 hence having a formula tc = fc + vc

960000= Fc + 22080(1.81)?

John Moffat says

I don’t know where you got 1.81 from.

Using high low, the variable cost = (960,000 – 885,120) / (22,080 – 19,200) = 26 per unit.

(The Paper F2 free lectures on high-low will help you)

maamaan says

but the lowest of tatal overheads is 883200 not 885120 please recheck sir