Answer to Question #161171 in Financial Math for Bdll

Question

Darcey would like to accumulate $80 000 in savings before she retires

A 20 years from now. She intends to make the same deposit at the end of

each month in an RRSP that pays 6.3%/a compounded monthly.

a) Draw a timeline to represent the annuity.

b) What regular payment will let Darcey reach her goal?

c) Suppose Darcey decides to wait 5 years before starting her

deposits. What regular payment would she have to make to reach the same goal?

Answer 1

The annuity is as P_n = $d*[\frac{(1+\frac{r}{k})^{n*k}-1}{\frac{r}{k}}]$

where,

P_n = balance in the account after n years

d = regular deposit (each year, each month etc.)

r = annual interest rate per annum in decimal form = 6.3/100 = 0.063

k = number of compounding periods in one year = 12

n = 20 years

for the first year, P₁ = $d*[\frac{(1+\frac{0.063}{12})^{1*12}-1}{\frac{0.063}{12}}] = (12.4)d$

for 2nd year, P₂ = $d*[\frac{(1+\frac{0.063}{12})^{2*12}-1}{\frac{0.063}{12}}] = (25.5)d$

for 3rd year, P₃ = $d*[\frac{(1+\frac{0.063}{12})^{3*12}-1}{\frac{0.063}{12}}] = (39.5)d$

similarly, we can find the annuity for all years upto 20th year as :

for 4th year, P₄ = (54.4)d

for 5th year, P₅ = (70.3)d

for 6th year, P₆ = (87.2)d

for 7th year, P₇ = (105.2)d

for 8th year, P₈ = (124.4)d

for 9th year, P₉ = (144.8)d

for 10th year, P₁₀ = (166.6)d

for 11th year, P₁₁ = (189.7)d

for 12th year, P₁₂ = (214.4)d

for 13th year, P₁₃ = (240.6)d

for 14th year, P₁₄ = (268.6)d

for 15th year, P₁₅ = (298.4)d

for 16th year, P₁₆ = (330.1)d

for 17th year, P₁₇ = (363.8)d

for 18th year, P₁₈ = (399.8)d

for 19th year, P₁₉ = (438.1)d

for 20th year, P₂₀ = (478.8)d

where, each block of y axis equal to 20d where, d = monthly deposit

b).

The regular payment will let Darcey reach her goal is:

n = 20 years

80,000 = $d*[\frac{(1+\frac{0.063}{12})^{20*12}-1}{\frac{0.063}{12}}]$

80,000 = d(478.8)

d = $167.1 monthly deposit

c).

If Darcey decides to wait 5 years before starting her deposits, the regular payment would she have to make to reach the same goal is:

n = 15 years

80,000 = $d*[\frac{(1+\frac{0.063}{12})^{15*12}-1}{\frac{0.063}{12}}]$

80,000 = d(298.4)

d = $268.1 monthly deposit