(a) NOTE: Assumptions of the solution are:
- Current time is t=0 or beginning of the first month
- First deposit comes in at the end of the first month
Savings Tenure = 30 years or 360 months, Compounding Frequency = Monthly, APR = 6 % or (6/12) = 0.5 % per month
Target Value of Savings = $ 450000
Let the monthly deposits be $ S
Therefore, S x (1.005)^(359) + S x (1.005)^(358) + ..........+ S x (1.005) + S = 450000
S x [{(1.005)^(360) - 1}/{1.005-1}] = 450000
S x 1004.515 = 450000
S = 450000 / 1004.515 ~ $ 447.98
NOTE: If the deposits are made at the beginning of each month, then the first deposit comes in now at t=0 and the last one comes in at t=359 (end of the 359th month or beginning of the 360th month). In that case the required deposit(S) would be calculated as:
S x (1.005)^(360) + S x (1.005)^(359) + ..........+ S x(1.005)^(2) + S x (1.005)^(1) = 450000
S x [{(1.005)^(360)-1}/{1.005-1}] x (1.005) = 450000
S x 1009.538 = 450000
S = 450000 / 1009.538 ~ $ 445.75
As the question does not mention anything about beginning or end of month deposits both answers are correct depending on the solver's assumption about the timing of each deposit.
(b) Assumption 1: End of Month Deposits (first deposit is at t=1)
Let the required number of months be T1
Therefore, 550 x (1.005)^(T1-1) + 550 x (1.005)^(T1-2) +.......+ 550 x (1.005)^(1) + 550 = 450000
550 x [{(1.005)^(T1) - 1}/{1.005-1}] = 450000
{(1.005)^(T1) -1} = (450000 x 0.005) / 550 = 45 /11
(1.005)^(T1) = 56 / 11
= T1
T1 = 326.3 months ~ 326 months
Assumption 2: Beginning of Month Deposits (first deposit is at t=0)
Let the required number of months be T2
Therefore, 550 x (1.005)^(T2) + 550 x (1.005)^(T2-1) +.......+ 550 x (1.005)^(2) + 550 x (1.005) = 450000
550 x [{(1.005)^(T2) - 1}/{1.005-1}] x 1.005 = 450000
{(1.005)^(T2) -1} = (450000 x 0.005) / (550 x 1.005) = 3000/737
(1.005)^(T2) = 3737 / 737
= T2
T2 = 325.5 months ~ 326 months
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