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

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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

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