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Home > Algebra calculators > Present value of Annuity Due example
8. Present value of Annuity Due example ( Enter your problem )
( Enter your problem )
  1. Find Present value (PV) 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, Present value PV = 4169.87,
Deposit Frequency = at the beginning (Annuity Due) of every Year (1/year)
for Present value of Annuity Due method

Solution:
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)

nx_0f(x_0)f'(x_0)x_1Update
10.1-0.000135-6.8618020.09998x_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 method

Solution:
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)

nx_0f(x_0)f'(x_0)x_1Update
10.1-0.004463-2.3290760.098084x_0 = x_1
20.0980840.00001-2.3398430.098088x_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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2. Find Regular Deposit (C) Example
(Previous example)		4. Find Time (n) Example
(Next example)


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