1. Find Interest Rate i = ?
Regular Deposit
(PMT Amount) C = 1000, Time n = 5 Year, Present value PV = 3790.79,
Deposit Frequency = at the end (Ordinary Annuity) of every Year (1/year)
for Present value of Annuity methodSolution:`C=1000` (Cash flow per year)
`n=5` years (Number of periods)
`PV=3790.79` (Present value)
Now, Present value (Ordinary Annuity) formula is
`PV_("Ordinary Annuity")=C*[(1-(1+i)^(-n))/(i)]``:.3790.79=1000*[(1-(1+i)^-5)/(i)]`
`:.(3790.79)/(1000)=[(1-(1+i)^-5)/(i)]`
`:.[(1-(1+i)^-5)/(i)]=3.79`
Now, find one solution using Newton Raphson method
Here `((1-(1+x)^-5)/x)=3.79`
`:.(1-1/((1+x)^5))/x-3.79=0`
Let `f(x) = (1-1/((1+x)^5))/x-3.79`
`d/(dx)((1-1/((1+x)^5))/x-3.79)=((5x)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^2)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(30x^3)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^4)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(5x^5)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-1+1/(1+5x+10x^2+10x^3+5x^4+x^5))/(x^2)`
`d/(dx)((1-1/((1+x)^5))/x-3.79)`
`=d/(dx)((1-1/((1+x)^5))/x)-d/(dx)(3.79)`
`d/(dx)((1-1/((1+x)^5))/x)=((5x)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^2)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(30x^3)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^4)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(5x^5)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-1+1/(1+5x+10x^2+10x^3+5x^4+x^5))/(x^2)`
`d/(dx)((1-1/((1+x)^5))/x)`
`=1-1/(1+5x+10x^2+10x^3+5x^4+x^5)`
`=((x) * d/(dx)(1-1/(1+5x+10x^2+10x^3+5x^4+x^5))-(1-1/(1+5x+10x^2+10x^3+5x^4+x^5)) * d/(dx)(x))/(x)^2`
`d/(dx)(1-1/(1+5x+10x^2+10x^3+5x^4+x^5))=5/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(30x^2)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^3)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(5x^4)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)`
`d/(dx)(1-1/(1+5x+10x^2+10x^3+5x^4+x^5))`
`=d/(dx)(1)-d/(dx)(1/(1+5x+10x^2+10x^3+5x^4+x^5))`
`d/(dx)(1/(1+5x+10x^2+10x^3+5x^4+x^5))=-5/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-(20x)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-(30x^2)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-(20x^3)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-(5x^4)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)`
`d/(dx)(1/(1+5x+10x^2+10x^3+5x^4+x^5))`
`=-1/((1+5x+10x^2+10x^3+5x^4+x^5)^2)*d/(dx)(1+5x+10x^2+10x^3+5x^4+x^5)`
`d/(dx)(1+5x+10x^2+10x^3+5x^4+x^5)=5+20x+30x^2+20x^3+5x^4`
`d/(dx)(1+5x+10x^2+10x^3+5x^4+x^5)`
`=d/(dx)(1)+d/(dx)(5x)+d/(dx)(10x^2)+d/(dx)(10x^3)+d/(dx)(5x^4)+d/(dx)(x^5)`
`=0+5+20x+30x^2+20x^3+5x^4`
`=5+20x+30x^2+20x^3+5x^4`
`=-1/((1+5x+10x^2+10x^3+5x^4+x^5)^2)*5+20x+30x^2+20x^3+5x^4`
`=-5/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-(20x)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-(30x^2)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-(20x^3)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-(5x^4)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)`
`=0-(-5/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-(20x)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-(30x^2)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-(20x^3)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-(5x^4)/((1+5x+10x^2+10x^3+5x^4+x^5)^2))`
`=5/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(30x^2)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^3)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(5x^4)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)`
`=((x) * (5/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(30x^2)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^3)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(5x^4)/((1+5x+10x^2+10x^3+5x^4+x^5)^2))-(1-1/(1+5x+10x^2+10x^3+5x^4+x^5)) * (1))/(x^2)`
`=((5x)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^2)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(30x^3)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^4)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(5x^5)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-1+1/(1+5x+10x^2+10x^3+5x^4+x^5))/(x^2)`
`=(((5x)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^2)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(30x^3)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^4)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(5x^5)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-1+1/(1+5x+10x^2+10x^3+5x^4+x^5))/(x^2))-0`
`=((5x)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^2)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(30x^3)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^4)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(5x^5)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-1+1/(1+5x+10x^2+10x^3+5x^4+x^5))/(x^2)-0`
`=((5x)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^2)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(30x^3)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^4)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(5x^5)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-1+1/(1+5x+10x^2+10x^3+5x^4+x^5))/(x^2)`
`:. f'(x) = ((5x)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^2)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(30x^3)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(20x^4)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)+(5x^5)/((1+5x+10x^2+10x^3+5x^4+x^5)^2)-1+1/(1+5x+10x^2+10x^3+5x^4+x^5))/(x^2)`
`x_0 = 0.1`
`1^(st)` iteration :`f(x_0)=f(0.1)=(1-1/((1+0.1)^5))/0.1-3.79=0.000787`
`f'(x_0)=f'(0.1)=((5*0.1)/((1+5*0.1+10*0.1^2+10*0.1^3+5*0.1^4+0.1^5)^2)+(20*0.1^2)/((1+5*0.1+10*0.1^2+10*0.1^3+5*0.1^4+0.1^5)^2)+(30*0.1^3)/((1+5*0.1+10*0.1^2+10*0.1^3+5*0.1^4+0.1^5)^2)+(20*0.1^4)/((1+5*0.1+10*0.1^2+10*0.1^3+5*0.1^4+0.1^5)^2)+(5*0.1^5)/((1+5*0.1+10*0.1^2+10*0.1^3+5*0.1^4+0.1^5)^2)-1+1/(1+5*0.1+10*0.1^2+10*0.1^3+5*0.1^4+0.1^5))/(0.1^2)=-9.684171`
`x_1 = x_0 - f(x_0)/(f'(x_0))`
`x_1=0.1 - (0.000787)/(-9.684171)`
`x_1=0.100081`
`2^(nd)` iteration :`f(x_1)=f(0.100081)=(1-1/((1+0.100081)^5))/0.100081-3.79=0`
`f'(x_1)=f'(0.100081)=((5*0.100081)/((1+5*0.100081+10*0.100081^2+10*0.100081^3+5*0.100081^4+0.100081^5)^2)+(20*0.100081^2)/((1+5*0.100081+10*0.100081^2+10*0.100081^3+5*0.100081^4+0.100081^5)^2)+(30*0.100081^3)/((1+5*0.100081+10*0.100081^2+10*0.100081^3+5*0.100081^4+0.100081^5)^2)+(20*0.100081^4)/((1+5*0.100081+10*0.100081^2+10*0.100081^3+5*0.100081^4+0.100081^5)^2)+(5*0.100081^5)/((1+5*0.100081+10*0.100081^2+10*0.100081^3+5*0.100081^4+0.100081^5)^2)-1+1/(1+5*0.100081+10*0.100081^2+10*0.100081^3+5*0.100081^4+0.100081^5))/(0.100081^2)=-9.680944`
`x_2 = x_1 - f(x_1)/(f'(x_1))`
`x_2=0.100081 - (0)/(-9.680944)`
`x_2=0.100081`
Approximate root of the equation `(1-1/((1+x)^5))/x-3.79=0` using Newton Raphson method is `0.100081` (After 2 iterations)
| `n` | `x_0` | `f(x_0)` | `f'(x_0)` | `x_1` | Update |
| 1 | 0.1 | 0.000787 | -9.684171 | 0.100081 | `x_0 = x_1` |
| 2 | 0.100081 | 0 | -9.680944 | 0.100081 | `x_0 = x_1` |
`:.i=0.100081`
`:.i=10.01 %` per year
2. Find Interest Rate i = ?
Regular Deposit
(PMT Amount) C = 5000, Time n = 3 Year, Present value PV = 12434.26,
Deposit Frequency = at the end (Ordinary Annuity) of every Year (1/year)
for Present value of Annuity methodSolution:`C=5000` (Cash flow per year)
`n=3` years (Number of periods)
`PV=12434.26` (Present value)
Now, Present value (Ordinary Annuity) formula is
`PV_("Ordinary Annuity")=C*[(1-(1+i)^(-n))/(i)]``:.12434.26=5000*[(1-(1+i)^-3)/(i)]`
`:.(12434.26)/(5000)=[(1-(1+i)^-3)/(i)]`
`:.[(1-(1+i)^-3)/(i)]=2.49`
Now, find one solution using Newton Raphson method
Here `((1-(1+x)^-3)/x)=2.49`
`:.(1-1/((1+x)^3))/x-2.49=0`
Let `f(x) = (1-1/((1+x)^3))/x-2.49`
`d/(dx)((1-1/((1+x)^3))/x-2.49)=((3x)/((1+3x+3x^2+x^3)^2)+(6x^2)/((1+3x+3x^2+x^3)^2)+(3x^3)/((1+3x+3x^2+x^3)^2)-1+1/(1+3x+3x^2+x^3))/(x^2)`
`d/(dx)((1-1/((1+x)^3))/x-2.49)`
`=d/(dx)((1-1/((1+x)^3))/x)-d/(dx)(2.49)`
`d/(dx)((1-1/((1+x)^3))/x)=((3x)/((1+3x+3x^2+x^3)^2)+(6x^2)/((1+3x+3x^2+x^3)^2)+(3x^3)/((1+3x+3x^2+x^3)^2)-1+1/(1+3x+3x^2+x^3))/(x^2)`
`d/(dx)((1-1/((1+x)^3))/x)`
`=1-1/(1+3x+3x^2+x^3)`
`=((x) * d/(dx)(1-1/(1+3x+3x^2+x^3))-(1-1/(1+3x+3x^2+x^3)) * d/(dx)(x))/(x)^2`
`d/(dx)(1-1/(1+3x+3x^2+x^3))=3/((1+3x+3x^2+x^3)^2)+(6x)/((1+3x+3x^2+x^3)^2)+(3x^2)/((1+3x+3x^2+x^3)^2)`
`d/(dx)(1-1/(1+3x+3x^2+x^3))`
`=d/(dx)(1)-d/(dx)(1/(1+3x+3x^2+x^3))`
`d/(dx)(1/(1+3x+3x^2+x^3))=-3/((1+3x+3x^2+x^3)^2)-(6x)/((1+3x+3x^2+x^3)^2)-(3x^2)/((1+3x+3x^2+x^3)^2)`
`d/(dx)(1/(1+3x+3x^2+x^3))`
`=-1/((1+3x+3x^2+x^3)^2)*d/(dx)(1+3x+3x^2+x^3)`
`d/(dx)(1+3x+3x^2+x^3)=3+6x+3x^2`
`d/(dx)(1+3x+3x^2+x^3)`
`=d/(dx)(1)+d/(dx)(3x)+d/(dx)(3x^2)+d/(dx)(x^3)`
`=0+3+6x+3x^2`
`=3+6x+3x^2`
`=-1/((1+3x+3x^2+x^3)^2)*3+6x+3x^2`
`=-3/((1+3x+3x^2+x^3)^2)-(6x)/((1+3x+3x^2+x^3)^2)-(3x^2)/((1+3x+3x^2+x^3)^2)`
`=0-(-3/((1+3x+3x^2+x^3)^2)-(6x)/((1+3x+3x^2+x^3)^2)-(3x^2)/((1+3x+3x^2+x^3)^2))`
`=3/((1+3x+3x^2+x^3)^2)+(6x)/((1+3x+3x^2+x^3)^2)+(3x^2)/((1+3x+3x^2+x^3)^2)`
`=((x) * (3/((1+3x+3x^2+x^3)^2)+(6x)/((1+3x+3x^2+x^3)^2)+(3x^2)/((1+3x+3x^2+x^3)^2))-(1-1/(1+3x+3x^2+x^3)) * (1))/(x^2)`
`=((3x)/((1+3x+3x^2+x^3)^2)+(6x^2)/((1+3x+3x^2+x^3)^2)+(3x^3)/((1+3x+3x^2+x^3)^2)-1+1/(1+3x+3x^2+x^3))/(x^2)`
`=(((3x)/((1+3x+3x^2+x^3)^2)+(6x^2)/((1+3x+3x^2+x^3)^2)+(3x^3)/((1+3x+3x^2+x^3)^2)-1+1/(1+3x+3x^2+x^3))/(x^2))-0`
`=((3x)/((1+3x+3x^2+x^3)^2)+(6x^2)/((1+3x+3x^2+x^3)^2)+(3x^3)/((1+3x+3x^2+x^3)^2)-1+1/(1+3x+3x^2+x^3))/(x^2)-0`
`=((3x)/((1+3x+3x^2+x^3)^2)+(6x^2)/((1+3x+3x^2+x^3)^2)+(3x^3)/((1+3x+3x^2+x^3)^2)-1+1/(1+3x+3x^2+x^3))/(x^2)`
`:. f'(x) = ((3x)/((1+3x+3x^2+x^3)^2)+(6x^2)/((1+3x+3x^2+x^3)^2)+(3x^3)/((1+3x+3x^2+x^3)^2)-1+1/(1+3x+3x^2+x^3))/(x^2)`
`x_0 = 0.1`
`1^(st)` iteration :`f(x_0)=f(0.1)=(1-1/((1+0.1)^3))/0.1-2.49=-0.003148`
`f'(x_0)=f'(0.1)=((3*0.1)/((1+3*0.1+3*0.1^2+0.1^3)^2)+(6*0.1^2)/((1+3*0.1+3*0.1^2+0.1^3)^2)+(3*0.1^3)/((1+3*0.1+3*0.1^2+0.1^3)^2)-1+1/(1+3*0.1+3*0.1^2+0.1^3))/(0.1^2)=-4.378116`
`x_1 = x_0 - f(x_0)/(f'(x_0))`
`x_1=0.1 - (-0.003148)/(-4.378116)`
`x_1=0.099281`
`2^(nd)` iteration :`f(x_1)=f(0.099281)=(1-1/((1+0.099281)^3))/0.099281-2.49=0.000003`
`f'(x_1)=f'(0.099281)=((3*0.099281)/((1+3*0.099281+3*0.099281^2+0.099281^3)^2)+(6*0.099281^2)/((1+3*0.099281+3*0.099281^2+0.099281^3)^2)+(3*0.099281^3)/((1+3*0.099281+3*0.099281^2+0.099281^3)^2)-1+1/(1+3*0.099281+3*0.099281^2+0.099281^3))/(0.099281^2)=-4.387515`
`x_2 = x_1 - f(x_1)/(f'(x_1))`
`x_2=0.099281 - (0.000003)/(-4.387515)`
`x_2=0.099282`
Approximate root of the equation `(1-1/((1+x)^3))/x-2.49=0` using Newton Raphson method is `0.099282` (After 2 iterations)
| `n` | `x_0` | `f(x_0)` | `f'(x_0)` | `x_1` | Update |
| 1 | 0.1 | -0.003148 | -4.378116 | 0.099281 | `x_0 = x_1` |
| 2 | 0.099281 | 0.000003 | -4.387515 | 0.099282 | `x_0 = x_1` |
`:.i=0.099282`
`:.i=9.93 %` per year
This material is intended as a summary. Use your textbook for detail explanation.
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