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
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
This material is intended as a summary. Use your textbook for detail explanation.
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