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Using the formula

$\dfrac {\log \left(1+\frac {Br}{P(1+r)} \right)}{\log (1+r)} = n$

from https://math.stackexchange.com/a/2101796/164710,

with values

B = 10000
r = 5% (0.00416667 monthly)
P = 250

I get 36.93 periods (months), but when I calculate on an amortization table, I get 43 months:

Period  Balance    Interest Paid   Principal Paid

0       $10,000.00 $41.67          $208.33
1       $9,791.67  $40.80          $209.20
2       $9,582.47  $39.93          $210.07
3       $9,372.39  $39.05          $210.95
4       $9,161.44  $38.17          $211.83
5       $8,949.62  $37.29          $212.71
6       $8,736.91  $36.40          $213.60
7       $8,523.31  $35.51          $214.49
8       $8,308.82  $34.62          $215.38
9       $8,093.44  $33.72          $216.28
10      $7,877.17  $32.82          $217.18
11      $7,659.99  $31.92          $218.08
12      $7,441.91  $31.01          $218.99
13      $7,222.91  $30.10          $219.90
14      $7,003.01  $29.18          $220.82
15      $6,782.19  $28.26          $221.74
16      $6,560.45  $27.34          $222.66
17      $6,337.78  $26.41          $223.59
18      $6,114.19  $25.48          $224.52
19      $5,889.67  $24.54          $225.46
20      $5,664.21  $23.60          $226.40
21      $5,437.81  $22.66          $227.34
22      $5,210.46  $21.71          $228.29
23      $4,982.17  $20.76          $229.24
24      $4,752.93  $19.80          $230.20
25      $4,522.74  $18.84          $231.16
26      $4,291.58  $17.88          $232.12
27      $4,059.46  $16.91          $233.09
28      $3,826.38  $15.94          $234.06
29      $3,592.32  $14.97          $235.03
30      $3,357.29  $13.99          $236.01
31      $3,121.28  $13.01          $236.99
32      $2,884.28  $12.02          $237.98
33      $2,646.30  $11.03          $238.97
34      $2,407.33  $10.03          $239.97
35      $2,167.36  $9.03           $240.97
36      $1,926.39  $8.03           $241.97
37      $1,684.42  $7.02           $242.98
38      $1,441.43  $6.01           $243.99
39      $1,197.44  $4.99           $245.01
40      $952.43    $3.97           $246.03
41      $706.40    $2.94           $247.06
42      $459.34    $1.91           $248.09
43      $211.25    $0.88           $249.12

What am I doing wrong?

1 Answers 1

1

The formula is $B\cdot q^n=P\cdot \frac{1-q^n}{1-q}$, with $q=1+r$

Multiplying both sides of the by $1-q$

$B\cdot q^n-B\cdot q^n\cdot q=P-P\cdot q^n$

$B\cdot q^n-B\cdot q^n\cdot q+P\cdot q^n=P$

$q^n(B-Bq+P)=P$

$q^n(B(1-q)+P)=P$

$1+r=q \Rightarrow 1-q=-r$

$q^n(B(-r)+P)=P$

$q^n(P-Br)=P$

$q^n=\frac{P}{P-Br}$

taking logs

$\log(q^n)=\log\left(\frac{P}{P-Br} \right)$

$n\cdot \log(q)=\log\left(\frac{P}{P-Br} \right)$

$$n=\frac{\log\left(\frac{P}{P-Br} \right)}{\log(q)}$$

Calculating $n$ with the given numbers

$$n=\frac{\log\left(\frac{250}{250-10,000\cdot \frac{0.05}{12}} \right)}{\log(1+\frac{0.05}{12})}=43.85\approx 44$$

  • 0
    That is actually the formula I got when I solved for `n` in `P = B(r(1 + r)^n) / ((1 + r)^n - 1)`. I should have used mine. :-) – 2017-01-17
  • 0
    @Ralph Yeah, this is the same formula like mine, only solved for P. – 2017-01-17