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Home > Algebra calculators > Future value of Annuity Due example
4. Future value of Annuity Due example ( Enter your problem )
( Enter your problem )
  1. Find Future value (FV) Example
  2. Find Regular Deposit (C) Example
  3. Find Interest Rate (i) Example
  4. Find Time (n) Example
 		Other related methods
  1. Future value using Simple Interest
  2. Future value using Compound Interest
  3. Future value of Annuity
  4. Future value of Annuity Due
  5. Present value using Simple Interest
  6. Present value using Compound Interest
  7. Present value of Annuity
  8. Present value of Annuity Due
  9. Contineous Compounding
2. Find Regular Deposit (C) Example
(Previous example)		4. Find Time (n) Example
(Next example)

3. Find Interest Rate (i) Example


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)

nx_0f(x_0)f'(x_0)x_1Update
10.1-0.0004391.943060.100226x_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)

nx_0f(x_0)f'(x_0)x_1Update
10.10.00010.6840.099854x_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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2. Find Regular Deposit (C) Example
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