1. Find Interest Rate i = ?
Regular Deposit
(PMT Amount) C = 1000, Time n = 5 Year, Future value FV = 6715.61,
Deposit Frequency = at the beginning (Annuity Due) of every Year (1/year)
for Future value of Annuity Due method

Solution:
`C=1000` (Cash flow per year)

`n=5` years (Number of periods)

`FV=6715.61` (Future value)

Now, Future value (Annuity Due) formula is
`FV_("Annuity Due")=C*[((1+i)^n-1)/(i)]*(1+i)`

`:.6715.61=1000*[((1+i)^5-1)/(i)]*(1+i)`

`:.(6715.61)/(1000)=[((1+i)^5-1)/(i)]*(1+i)`

`:.[((1+i)^5-1)/(i)]*(1+i)=6.72`

Now, find one solution using Newton Raphson method

Here `((1+x)^5-1)*(1+x)=6.72x`

`:.((1+x)^5-1)(1+x)-6.72x=0`

Let `f(x) = ((1+x)^5-1)(1+x)-6.72x`

`d/(dx)(((1+x)^5-1)(1+x)-6.72x)=5*(1+x)^5+((1+x)^5-1)-6.72`

`d/(dx)(((1+x)^5-1)(1+x)-6.72x)`

`=d/(dx)(((1+x)^5-1)(1+x))-d/(dx)(6.72x)`

`d/(dx)(((1+x)^5-1)(1+x))=5*(1+x)^5+((1+x)^5-1)`

`d/(dx)(((1+x)^5-1)(1+x))`

`=(d/(dx)((1+x)^5-1))(1+x)+((1+x)^5-1)(d/(dx)(1+x))`

`d/(dx)((1+x)^5-1)=5*(1+x)^4`

`d/(dx)((1+x)^5-1)`

`=d/(dx)((1+x)^5)-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)^4-0`

`=5*(1+x)^4`

`d/(dx)(1+x)=1`

`d/(dx)(1+x)`

`=d/(dx)(1)+d/(dx)(x)`

`=0+1`

`=1`

`=(5*(1+x)^4)(1+x)+((1+x)^5-1)*1`

`=5*(1+x)^5+((1+x)^5-1)`

`=(5*(1+x)^5+((1+x)^5-1))-6.72`

`=5*(1+x)^5+((1+x)^5-1)-6.72`

`:. f'(x) = 5*(1+x)^5+((1+x)^5-1)-6.72`

`x_0 = 0.1`


`1^(st)` iteration :

`f(x_0)=f(0.1)=((1+0.1)^5-1)(1+0.1)-6.72*0.1=-0.000439`

`f'(x_0)=f'(0.1)=5*(1+0.1)^5+((1+0.1)^5-1)-6.72=1.94306`

`x_1 = x_0 - f(x_0)/(f'(x_0))`

`x_1=0.1 - (-0.000439)/(1.94306)`

`x_1=0.100226`


Approximate root of the equation `((1+x)^5-1)(1+x)-6.72x=0` using Newton Raphson method is `0.100226` (After 1 iterations)

`n`	`x_0`	`f(x_0)`	`f'(x_0)`	`x_1`	Update
1	0.1	-0.000439	1.94306	0.100226	`x_0 = x_1`

`:.i=0.100226`

`:.i=10.02 %` per year

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2. Find Interest Rate i = ?
Regular Deposit
(PMT Amount) C = 5000, Time n = 3 Year, Future value FV = 18205,
Deposit Frequency = at the beginning (Annuity Due) of every Year (1/year)
for Future value of Annuity Due method

Solution:
`C=5000` (Cash flow per year)

`n=3` years (Number of periods)

`FV=18205` (Future value)

Now, Future value (Annuity Due) formula is
`FV_("Annuity Due")=C*[((1+i)^n-1)/(i)]*(1+i)`

`:.18205=5000*[((1+i)^3-1)/(i)]*(1+i)`

`:.(18205)/(5000)=[((1+i)^3-1)/(i)]*(1+i)`

`:.[((1+i)^3-1)/(i)]*(1+i)=3.64`

Now, find one solution using Newton Raphson method

Here `((1+x)^3-1)*(1+x)=3.64x`

`:.((1+x)^3-1)(1+x)-3.64x=0`

Let `f(x) = ((1+x)^3-1)(1+x)-3.64x`

`d/(dx)(((1+x)^3-1)(1+x)-3.64x)=3*(1+x)^3+((1+x)^3-1)-3.64`

`d/(dx)(((1+x)^3-1)(1+x)-3.64x)`

`=d/(dx)(((1+x)^3-1)(1+x))-d/(dx)(3.64x)`

`d/(dx)(((1+x)^3-1)(1+x))=3*(1+x)^3+((1+x)^3-1)`

`d/(dx)(((1+x)^3-1)(1+x))`

`=(d/(dx)((1+x)^3-1))(1+x)+((1+x)^3-1)(d/(dx)(1+x))`

`d/(dx)((1+x)^3-1)=3*(1+x)^2`

`d/(dx)((1+x)^3-1)`

`=d/(dx)((1+x)^3)-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)^2-0`

`=3*(1+x)^2`

`d/(dx)(1+x)=1`

`d/(dx)(1+x)`

`=d/(dx)(1)+d/(dx)(x)`

`=0+1`

`=1`

`=(3*(1+x)^2)(1+x)+((1+x)^3-1)*1`

`=3*(1+x)^3+((1+x)^3-1)`

`=(3*(1+x)^3+((1+x)^3-1))-3.64`

`=3*(1+x)^3+((1+x)^3-1)-3.64`

`:. f'(x) = 3*(1+x)^3+((1+x)^3-1)-3.64`

`x_0 = 0.1`


`1^(st)` iteration :

`f(x_0)=f(0.1)=((1+0.1)^3-1)(1+0.1)-3.64*0.1=0.0001`

`f'(x_0)=f'(0.1)=3*(1+0.1)^3+((1+0.1)^3-1)-3.64=0.684`

`x_1 = x_0 - f(x_0)/(f'(x_0))`

`x_1=0.1 - (0.0001)/(0.684)`

`x_1=0.099854`


Approximate root of the equation `((1+x)^3-1)(1+x)-3.64x=0` using Newton Raphson method is `0.099854` (After 1 iterations)

`n`	`x_0`	`f(x_0)`	`f'(x_0)`	`x_1`	Update
1	0.1	0.0001	0.684	0.099854	`x_0 = x_1`

`:.i=0.099854`

`:.i=9.99 %` per year

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