Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let's say I have a number, 10, and over a period of a year that number has increased by 100% to 20. I would like to figure out what the monthly increase should be for 10 to reach 20 over 12 months.

If I convert 100% to a coefficient I get 2. Now by trying several numbers manually I figured the monthly coefficient should be around 1.065:

M.  NUMBER          COEFF
1   10              1.065
2   10.65           1.065
3   11.34225        1.065
4   12.07949625     1.065
5   12.8646635063   1.065
6   13.7008666342   1.065
7   14.5914229654   1.065
8   15.5398654581   1.065
9   16.5499567129   1.065
10  17.6257038992   1.065
11  18.7713746527   1.065
12  19.9915140051   1.065

Q: How can I obtain the monthly coefficient (something around 1.065) from 2 and 12?

share|improve this question

4 Answers 4

Actually, your exponent is off by one: if the table represents the number at the end of each month, it should start at 0. Otherwise, if it represents the number at the beginning of the month, it should go to 13.

Since you're arriving at 20 by multiplying repeatedly by $x_{\mathrm{month}}$, we have $$x_{\mathrm{year}}=(x_{\mathrm{month}})^{12}$$ so $$x_{\mathrm{month}}=(x_{\mathrm{year}})^{1/12}=\sqrt[12]{x_{\mathrm{year}}}$$

Numerically, we arrive at ca. $1.059$, or $5.9\%$.

share|improve this answer
    
you had the idea but we need to use 11 instead of 12 for the root calculation (see my answer). –  user359650 Feb 1 '12 at 11:27
    
@user359650 Whilst yours is the correct answer to the question that is actually asked, I suspect that this is the correct answer to the question that was intended to be asked... –  Chris Taylor Feb 1 '12 at 11:34
1  
@user359650 In order to reproduce your table, yes, 11 is the correct exponent. But that does not give you monthly growth from yearly growth: Pin down your numbers to a day rather than just a month, and you will notice that your table is too short. Say, M=1 means 1st January, then M=12 will be 1st December; but the year goes on to 31st December, so you have to add a month. –  Ansgar Esztermann Feb 2 '12 at 16:12

Ansgar had the solution, although with an imperfection.

If you use 12 months for the root calculation you get this

COEFF = (1 + 100%/100) ^ (1/12)
M.  NUMBER          COEFF
1   10  
2   10.5946309436   1.0594630944
3   11.2246204831   1.0594630944
4   11.89207115     1.0594630944
5   12.5992104989   1.0594630944
6   13.3483985417   1.0594630944
7   14.1421356237   1.0594630944
8   14.9830707688   1.0594630944
9   15.8740105197   1.0594630944
10  16.8179283051   1.0594630944
11  17.8179743628   1.0594630944
12  18.8774862536   1.0594630944

The good way of doing it is use 11, because there are only 11 increases.

COEFF = (1 + 100%/100) ^ (1 / (12-1) )
M.  NUMBER          COEFF
1   10  
2   10.6504108944   1.0650410894
3   11.343125222    1.0650410894
4   12.080894444    1.0650410894
5   12.8666489801   1.0650410894
6   13.7035098472   1.0650410894
7   14.5948010568   1.0650410894
8   15.5440628177   1.0650410894
9   16.5550655977   1.0650410894
10  17.6318250999   1.0650410894
11  18.7786182132   1.0650410894
12  20              1.0650410894
share|improve this answer

Please see @Graphth's post for the math. This is just to clarify the crucial point @user359650 made, but I fear a bit unclearly.

If there are $12$ monthly compounding periods, then there are $13$ time points, since each period is an interval with an adjacent starting and ending time point. It is customary to call the first left endpoint $t=0$. (How many tick marks to you see on the $x$ axis below, and how many intervals between tick marks?)

plot(10*2^(x/12),(x,0,12)).show(figsize=(4,1),ymin=0,ymax=20)

yearly doubling compound interest

In sage, we calculate:

def grow(ratio, periods):
    r = ratio^(1/periods); print 'growth per period =',r.n()
    print '%2s %-12s' % ('Period', 'Ending Balance')
    for n in range(periods+1):
        print '%5d %12.6f' % (n, 10*r^n)
grow(2,12)

to get (values commensurate with the graph above):

growth ratio = 1.05946309435930
Month Ending Balance
    0    10.000000
    1    10.594631
    2    11.224620
    3    11.892071
    4    12.599210
    5    13.348399
    6    14.142136
    7    14.983071
    8    15.874011
    9    16.817928
   10    17.817974
   11    18.877486
   12    20.000000

or (erroneously, if you want a period of one year with a monthly-based calendar system):

grow(2,11)

to get:

growth per period = 1.06504108943996
Period Ending Balance
    0    10.000000
    1    10.650411
    2    11.343125
    3    12.080894
    4    12.866649
    5    13.703510
    6    14.594801
    7    15.544063
    8    16.555066
    9    17.631825
   10    18.778618
   11    20.000000
share|improve this answer

$$10 (1 + j)^{12} = 20$$

is equivalent to

$$(1 + j)^{12} = 2,$$

which is equivalent to

$$(1 + j) = \sqrt[12]{2}.$$

share|improve this answer
    
This is a good way to formulate the solution. $j$ is the monthly interest rate. One should also point out that this is called compound interest and there is a wealth of reference literature. $V(t)=10(1+j)^t$ is the value at the end of month $t$, starting at $t=0$ which is more natural. You then would have a table of values of $V(t)$ for $t\in\{0,1,\dots,12\}$, having a unit interval between integral values of $t$ for the span of each month. –  bgins Apr 1 '12 at 20:04

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.