Question

The answers previously provided are all over the place. Without Excel Please.

- -

Your sister has estimated that the maintenance costs on the “slightly used” car she just purchased will be $500 in the first year. She anticipates that this cost will increase by $150 each year. She wants to set aside enough money today to cover all anticipated maintenance costs for the 10 years that she plans to own the car. How much should she deposit today in this account if the account earns 5% interest per year?

Answer 1

First, we will estimate the annual equivalent maintenance cost

A = A1 + G[((1 + i)ⁿ – in – 1))/i(1 + i)ⁿ – i]

A = A1 + G(A/G, i, n)

Where (A/G, i, n) is called uniform gradient series factor

A = Annual equivalent amount

A1 = Amount at the end of the first year = $500

G = Equal increment amount = $150

n = Number of interest periods = 10

i = Interest rate = 5% or 0.05

A = 500 + 150[((1 + 0.05)¹⁰ – (0.05×10) – 1))/0.05(1 + 0.05)¹⁰ – 0.05]

A = 500 + 150(A/G, 5%, 10)

A = 500 + 150[((1.05)¹⁰ – (0.05×10) – 1))/0.05(1.05)¹⁰ – 0.05]

A = 500 + (150 × 4.099)

A = 500 + 614.85

A = 1,114.85

This is equivalent to receiving an equivalent amount of $1,114.85 at the end of every year for the next 10 years. The present worth of this revised series is following

P = A[((1 + i)ⁿ – 1)/i(1 + i)ⁿ]

P = A(P/A, i, n)

(P/A, i, n) = Equal payment series present worth factor

A = Annual equivalent payment = 1,114.85

n = Number of interest periods = 10

i = Interest rate = 5% or 0.05

P = Present worth

P = 1,114.85[((1 + 0.05)¹⁰ – 1)/0.05(1 + 0.05)¹⁰]

P = 1,114.85(P/A, 5%, 10)

P = 1,114.85[((1.05)¹⁰ – 1)/0.05(1.05)¹⁰]

P = 1,114.85 × 7.722

P = 8,608.87

So, she should deposit $8,608.87 today in the account.