average percentages

[sciencer]

If a quantity of 100 in year 1 changes to 11 in year 10, the change would be -89 %. But when this change is converted to an annual value (-89/10) and then applied to each year, year 2 would calculate as 91.1. This is repeated such that according to the annual value, year 10 quantity is 39, which is a big difference from 11! What is the correct way to apply an annual change, please?

[CaptainPanic]

If a quantity of 100 in year 1 changes to 11 in year 10, the change would be -89 %. But when this change is converted to an annual value (-89/10) and then applied to each year, year 2 would calculate as 91.1. This is repeated such that according to the annual value, year 10 quantity is 39, which is a big difference from 11! What is the correct way to apply an annual change, please?

A quantity can decrease in multiple ways in a 10 year period.

 

It can decrease linearly. That means you lose 89/10 (or 8.9) in year 1. And again 8.9 in year 2. And again 8.9 in year 3.

 

It can also be an exponential decay. You can lose for example 50% every year. Starting with 100, you lose 50 in year 1. You have 50 left, and therefore you lose only 25 in year 2. And only 12.5 in year 3.

 

What you just did is that you assumed that the decrease (-89/10) is the linear value. But then you applied it to the exponential decay.

 

Since this sounds an awful lot like homework, I will not give you the answer.

[sciencer]

>

> A quantity can decrease in multiple ways in a 10 year period.

>

> It can decrease linearly. That means you lose 89/10 (or 8.9) in year 1. And again 8.9 in year 2. And again 8.9 in year 3.

>

> It can also be an exponential decay. You can lose for example 50% every year. Starting with 100, you lose 50 in year 1. You have 50 left, and therefore you lose only 25 in year 2. And only 12.5 in year 3.

>

> What you just did is that you assumed that the decrease (-89/10) is the linear value. But then you applied it to the exponential decay.

>

A linear change is the requested assumption. Thanks for the explanation of the exponential application, which was not understood and hence presumably the source of the error. If the change of -8.9% is applied to the original quantity, the result in year 10 becomes 34, still wrong. If anyone knows the relevant areas of maths that should be researched (i.e. don't know what search criterion to use!), it would be appreciated to know; thank you.

>

> Since this sounds an awful lot like homework, I will not give you the answer.

>

>

Not sure what constitutes homework so cannot comment.

Edited March 17, 2011 by sciencer

[CaptainPanic]

A linear change is the requested assumption. Thanks for the explanation of the exponential application, which was not understood and hence presumably the source of the error. If the change of -8.9% is applied to the original quantity, the result in year 10 becomes 34, still wrong. If anyone knows the relevant areas of maths that should be researched (i.e. don't know what search criterion to use!), it would be appreciated to know; thank you.

A linear change would mean: subtracting 8.9 every time.

 

100 - 8.9 = 91.1

91.1 - 8.9 = 82.2

82.2 - 8.9 = 73.3

 

Do that 10 times, and you will get to 11.

[sciencer]

A linear change would mean: subtracting 8.9 every time.

 

100 - 8.9 = 91.1

91.1 - 8.9 = 82.2

82.2 - 8.9 = 73.3

 

Do that 10 times, and you will get to 11.

 

Understood, thank you.

 

Is the correct terminology 'linear change', whilst the other phenomenon is 'exponential change'? Seems strange because the inference of the term 'exponential' is that the same end point is obtained by rapidly increasing/decreasing rate of change. A polynomial factor would achieve the exponential change, but no polynomial was applied in this example?

[CaptainPanic]

There was a polynomial. It was a 1st degree polynomial, also known as a linear function.

 

Exponential functions look and behave different.

Edited March 17, 2011 by CaptainPanic