1. Find Interest Rate i = ?
Regular Deposit
(PMT Amount) C = 1000, Time n = 5 Year, Future value FV = 6105.1,
Deposit Frequency = at the end (Ordinary Annuity) of every Year (1/year)
for Future value of Annuity method
Solution:
`C=1000` (Cash flow per year)
`n=5` years (Number of periods)
`FV=6105.1` (Future value)
Now, Future value (Ordinary Annuity) formula is
`FV_("Ordinary Annuity")=C*[((1+i)^n-1)/(i)]`
`:.6105.1=1000*[((1+i)^5-1)/(i)]`
`:.(6105.1)/(1000)=[((1+i)^5-1)/(i)]`
`:.[((1+i)^5-1)/(i)]=6.11`
Now, find one solution using Newton Raphson method
Here `((1+x)^5-1)=6.11x`
`:.(1+x)^5-6.11x-1=0`
Let `f(x) = (1+x)^5-6.11x-1`
`d/(dx)((1+x)^5-6.11x-1)=5*(1+x)^4-6.11`
`d/(dx)((1+x)^5-6.11x-1)`
`=d/(dx)((1+x)^5)-d/(dx)(6.11x)-d/(dx)(1)`
`d/(dx)((1+x)^5)=5*(1+x)^4`
`d/(dx)((1+x)^5)`
`=5*(1+x)^4*d/(dx)(1+x)`
`d/(dx)(1+x)=1`
`d/(dx)(1+x)`
`=d/(dx)(1)+d/(dx)(x)`
`=0+1`
`=1`
`=5*(1+x)^4*1`
`=5*(1+x)^4`
`=5*(1+x)^4-6.11-0`
`=5*(1+x)^4-6.11`
`:. f'(x) = 5*(1+x)^4-6.11`
`x_0 = 0.1`
`1^(st)` iteration :`f(x_0)=f(0.1)=(1+0.1)^5-6.11*0.1-1=-0.00049`
`f'(x_0)=f'(0.1)=5*(1+0.1)^4-6.11=1.2105`
`x_1 = x_0 - f(x_0)/(f'(x_0))`
`x_1=0.1 - (-0.00049)/(1.2105)`
`x_1=0.100405`
Approximate root of the equation `(1+x)^5-6.11x-1=0` using Newton Raphson method is `0.100405` (After 1 iterations)
| `n` | `x_0` | `f(x_0)` | `f'(x_0)` | `x_1` | Update |
| 1 | 0.1 | -0.00049 | 1.2105 | 0.100405 | `x_0 = x_1` |
`:.i=0.100405`
`:.i=10.04 %` per year
2. Find Interest Rate i = ?
Regular Deposit
(PMT Amount) C = 5000, Time n = 3 Year, Future value FV = 16550,
Deposit Frequency = at the end (Ordinary Annuity) of every Year (1/year)
for Future value of Annuity method
Solution:
`C=5000` (Cash flow per year)
`n=3` years (Number of periods)
`FV=16550` (Future value)
Now, Future value (Ordinary Annuity) formula is
`FV_("Ordinary Annuity")=C*[((1+i)^n-1)/(i)]`
`:.16550=5000*[((1+i)^3-1)/(i)]`
`:.(16550)/(5000)=[((1+i)^3-1)/(i)]`
`:.[((1+i)^3-1)/(i)]=3.31`
Now, find one solution using Newton Raphson method
Here `((1+x)^3-1)=3.31x`
`:.(1+x)^3-3.31x-1=0`
Let `f(x) = (1+x)^3-3.31x-1`
`d/(dx)((1+x)^3-3.31x-1)=3*(1+x)^2-3.31`
`d/(dx)((1+x)^3-3.31x-1)`
`=d/(dx)((1+x)^3)-d/(dx)(3.31x)-d/(dx)(1)`
`d/(dx)((1+x)^3)=3*(1+x)^2`
`d/(dx)((1+x)^3)`
`=3*(1+x)^2*d/(dx)(1+x)`
`d/(dx)(1+x)=1`
`d/(dx)(1+x)`
`=d/(dx)(1)+d/(dx)(x)`
`=0+1`
`=1`
`=3*(1+x)^2*1`
`=3*(1+x)^2`
`=3*(1+x)^2-3.31-0`
`=3*(1+x)^2-3.31`
`:. f'(x) = 3*(1+x)^2-3.31`
`x_0 = 0.1`
`1^(st)` iteration :`f(x_0)=f(0.1)=(1+0.1)^3-3.31*0.1-1=0`
`f'(x_0)=f'(0.1)=3*(1+0.1)^2-3.31=0.32`
`x_1 = x_0 - f(x_0)/(f'(x_0))`
`x_1=0.1 - (0)/(0.32)`
`x_1=0.1`
Approximate root of the equation `(1+x)^3-3.31x-1=0` using Newton Raphson method is `0.1` (After 1 iterations)
| `n` | `x_0` | `f(x_0)` | `f'(x_0)` | `x_1` | Update |
| 1 | 0.1 | 0 | 0.32 | 0.1 | `x_0 = x_1` |
`:.i=0.1`
`:.i=10 %` per year
This material is intended as a summary. Use your textbook for detail explanation.
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