1. Find Interest Rate i = ?
Regular Deposit
(PMT Amount) C = 1000, Time n = 5 Year, Present value PV = 4169.87,
Deposit Frequency = at the beginning (Annuity Due) of every Year (1/year)
for Present value of Annuity Due methodSolution:`C=1000` (Cash flow per year)
`n=5` years (Number of periods)
`PV=4169.87` (Present value)
Now, Present value (Annuity Due) formula is
`PV_("Annuity Due")=C*[(1-(1+i)^(-n))/(i)]*(1+i)``:.4169.87=1000*[(1-(1+i)^-5)/(i)]*(1+i)`
`:.(4169.87)/(1000)=[(1-(1+i)^-5)/(i)]*(1+i)`
`:.[(1-(1+i)^-5)/(i)]*(1+i)=4.17`
Now, find one solution using Newton Raphson method
Here `((1-(1+x)^-5)/x)*(1+x)=4.17`
`:.((1-1/((1+x)^5))/x)(1+x)-4.17=0`
Let `f(x) = ((1-1/((1+x)^5))/x)(1+x)-4.17`
`d/(dx)(((1-1/((1+x)^5))/x)(1+x)-4.17)=((5x)/((1+x)^5)+x(1-1/((1+x)^5))-(1-1/((1+x)^5))(1+x))/(x^2)`
`d/(dx)(((1-1/((1+x)^5))(1+x))/x-4.17)`
`=d/(dx)(((1-1/((1+x)^5))(1+x))/x)-d/(dx)(4.17)`
`d/(dx)(((1-1/((1+x)^5))(1+x))/x)=((5x)/((1+x)^5)+x(1-1/((1+x)^5))-(1-1/((1+x)^5))(1+x))/(x^2)`
`d/(dx)(((1-1/((1+x)^5))(1+x))/x)`
`=((x) * d/(dx)((1-1/((1+x)^5))(1+x))-((1-1/((1+x)^5))(1+x)) * d/(dx)(x))/(x)^2`
`d/(dx)((1-1/((1+x)^5))(1+x))=5/((1+x)^5)+(1-1/((1+x)^5))`
`d/(dx)((1-1/((1+x)^5))(1+x))`
`=(d/(dx)(1-1/((1+x)^5)))(1+x)+(1-1/((1+x)^5))(d/(dx)(1+x))`
`d/(dx)(1-1/((1+x)^5))=5/((1+x)^6)`
`d/(dx)(1-1/((1+x)^5))`
`=d/(dx)(1)-d/(dx)(1/((1+x)^5))`
`d/(dx)(1/((1+x)^5))=-5/((1+x)^6)`
`d/(dx)(1/((1+x)^5))`
`=-5/((1+x)^6)*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)^6)*1`
`=-5/((1+x)^6)`
`=0-(-5/((1+x)^6))`
`=5/((1+x)^6)`
`d/(dx)(1+x)=1`
`d/(dx)(1+x)`
`=d/(dx)(1)+d/(dx)(x)`
`=0+1`
`=1`
`=(5/((1+x)^6))(1+x)+(1-1/((1+x)^5))*1`
`=5/((1+x)^5)+(1-1/((1+x)^5))`
`=((x) * (5/((1+x)^5)+(1-1/((1+x)^5)))-((1-1/((1+x)^5))(1+x)) * (1))/(x^2)`
`=((5x)/((1+x)^5)+x(1-1/((1+x)^5))-(1-1/((1+x)^5))(1+x))/(x^2)`
`=(((5x)/((1+x)^5)+x(1-1/((1+x)^5))-(1-1/((1+x)^5))(1+x))/(x^2))-0`
`=((5x)/((1+x)^5)+x(1-1/((1+x)^5))-(1-1/((1+x)^5))(1+x))/(x^2)-0`
`=((5x)/((1+x)^5)+x(1-1/((1+x)^5))-(1-1/((1+x)^5))(1+x))/(x^2)`
`:. f'(x) = ((5x)/((1+x)^5)+x(1-1/((1+x)^5))-(1-1/((1+x)^5))(1+x))/(x^2)`
`x_0 = 0.1`
`1^(st)` iteration :`f(x_0)=f(0.1)=((1-1/((1+0.1)^5))/0.1)(1+0.1)-4.17=-0.000135`
`f'(x_0)=f'(0.1)=((5*0.1)/((1+0.1)^5)+0.1(1-1/((1+0.1)^5))-(1-1/((1+0.1)^5))(1+0.1))/(0.1^2)=-6.861802`
`x_1 = x_0 - f(x_0)/(f'(x_0))`
`x_1=0.1 - (-0.000135)/(-6.861802)`
`x_1=0.09998`
Approximate root of the equation `((1-1/((1+x)^5))/x)(1+x)-4.17=0` using Newton Raphson method is `0.09998` (After 1 iterations)
| `n` | `x_0` | `f(x_0)` | `f'(x_0)` | `x_1` | Update |
| 1 | 0.1 | -0.000135 | -6.861802 | 0.09998 | `x_0 = x_1` |
`:.i=0.09998`
`:.i=10 %` per year
2. Find Interest Rate i = ?
Regular Deposit
(PMT Amount) C = 5000, Time n = 3 Year, Present value PV = 13677.69,
Deposit Frequency = at the beginning (Annuity Due) of every Year (1/year)
for Present value of Annuity Due methodSolution:`C=5000` (Cash flow per year)
`n=3` years (Number of periods)
`PV=13677.69` (Present value)
Now, Present value (Annuity Due) formula is
`PV_("Annuity Due")=C*[(1-(1+i)^(-n))/(i)]*(1+i)``:.13677.69=5000*[(1-(1+i)^-3)/(i)]*(1+i)`
`:.(13677.69)/(5000)=[(1-(1+i)^-3)/(i)]*(1+i)`
`:.[(1-(1+i)^-3)/(i)]*(1+i)=2.74`
Now, find one solution using Newton Raphson method
Here `((1-(1+x)^-3)/x)*(1+x)=2.74`
`:.((1-1/((1+x)^3))/x)(1+x)-2.74=0`
Let `f(x) = ((1-1/((1+x)^3))/x)(1+x)-2.74`
`d/(dx)(((1-1/((1+x)^3))/x)(1+x)-2.74)=((3x)/((1+x)^3)+x(1-1/((1+x)^3))-(1-1/((1+x)^3))(1+x))/(x^2)`
`d/(dx)(((1-1/((1+x)^3))(1+x))/x-2.74)`
`=d/(dx)(((1-1/((1+x)^3))(1+x))/x)-d/(dx)(2.74)`
`d/(dx)(((1-1/((1+x)^3))(1+x))/x)=((3x)/((1+x)^3)+x(1-1/((1+x)^3))-(1-1/((1+x)^3))(1+x))/(x^2)`
`d/(dx)(((1-1/((1+x)^3))(1+x))/x)`
`=((x) * d/(dx)((1-1/((1+x)^3))(1+x))-((1-1/((1+x)^3))(1+x)) * d/(dx)(x))/(x)^2`
`d/(dx)((1-1/((1+x)^3))(1+x))=3/((1+x)^3)+(1-1/((1+x)^3))`
`d/(dx)((1-1/((1+x)^3))(1+x))`
`=(d/(dx)(1-1/((1+x)^3)))(1+x)+(1-1/((1+x)^3))(d/(dx)(1+x))`
`d/(dx)(1-1/((1+x)^3))=3/((1+x)^4)`
`d/(dx)(1-1/((1+x)^3))`
`=d/(dx)(1)-d/(dx)(1/((1+x)^3))`
`d/(dx)(1/((1+x)^3))=-3/((1+x)^4)`
`d/(dx)(1/((1+x)^3))`
`=-3/((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`
`=-3/((1+x)^4)*1`
`=-3/((1+x)^4)`
`=0-(-3/((1+x)^4))`
`=3/((1+x)^4)`
`d/(dx)(1+x)=1`
`d/(dx)(1+x)`
`=d/(dx)(1)+d/(dx)(x)`
`=0+1`
`=1`
`=(3/((1+x)^4))(1+x)+(1-1/((1+x)^3))*1`
`=3/((1+x)^3)+(1-1/((1+x)^3))`
`=((x) * (3/((1+x)^3)+(1-1/((1+x)^3)))-((1-1/((1+x)^3))(1+x)) * (1))/(x^2)`
`=((3x)/((1+x)^3)+x(1-1/((1+x)^3))-(1-1/((1+x)^3))(1+x))/(x^2)`
`=(((3x)/((1+x)^3)+x(1-1/((1+x)^3))-(1-1/((1+x)^3))(1+x))/(x^2))-0`
`=((3x)/((1+x)^3)+x(1-1/((1+x)^3))-(1-1/((1+x)^3))(1+x))/(x^2)-0`
`=((3x)/((1+x)^3)+x(1-1/((1+x)^3))-(1-1/((1+x)^3))(1+x))/(x^2)`
`:. f'(x) = ((3x)/((1+x)^3)+x(1-1/((1+x)^3))-(1-1/((1+x)^3))(1+x))/(x^2)`
`x_0 = 0.1`
`1^(st)` iteration :`f(x_0)=f(0.1)=((1-1/((1+0.1)^3))/0.1)(1+0.1)-2.74=-0.004463`
`f'(x_0)=f'(0.1)=((3*0.1)/((1+0.1)^3)+0.1(1-1/((1+0.1)^3))-(1-1/((1+0.1)^3))(1+0.1))/(0.1^2)=-2.329076`
`x_1 = x_0 - f(x_0)/(f'(x_0))`
`x_1=0.1 - (-0.004463)/(-2.329076)`
`x_1=0.098084`
`2^(nd)` iteration :`f(x_1)=f(0.098084)=((1-1/((1+0.098084)^3))/0.098084)(1+0.098084)-2.74=0.00001`
`f'(x_1)=f'(0.098084)=((3*0.098084)/((1+0.098084)^3)+0.098084(1-1/((1+0.098084)^3))-(1-1/((1+0.098084)^3))(1+0.098084))/(0.098084^2)=-2.339843`
`x_2 = x_1 - f(x_1)/(f'(x_1))`
`x_2=0.098084 - (0.00001)/(-2.339843)`
`x_2=0.098088`
Approximate root of the equation `((1-1/((1+x)^3))/x)(1+x)-2.74=0` using Newton Raphson method is `0.098088` (After 2 iterations)
| `n` | `x_0` | `f(x_0)` | `f'(x_0)` | `x_1` | Update |
| 1 | 0.1 | -0.004463 | -2.329076 | 0.098084 | `x_0 = x_1` |
| 2 | 0.098084 | 0.00001 | -2.339843 | 0.098088 | `x_0 = x_1` |
`:.i=0.098088`
`:.i=9.81 %` per year
This material is intended as a summary. Use your textbook for detail explanation.
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