Reducing-balance depreciation and a sinking fund for a cargo e-bike
A courier co-operative in Porto buys a cargo e-bike for €4,200 and depreciates it at a reducing-balance rate of 18% per year, applied at the end of each year. In the same month it opens a maintenance fund, depositing €150 at the end of every month into an account paying a nominal annual rate of 2.4% compounded monthly; each deposit is made after that month's interest is credited. The co-operative will replace the bike once its book value first falls below €1,500. The replacement it has in mind costs €5,300 today, and the dealer expects prices to rise 2% per year. Two members are arguing about the fund: one expects 'about €5,400, since 36 × 150 = 5,400'; the other insists the interest is 'barely a few euros'. The treasurer wants both claims checked with real calculations, and wants to know how long the deposits actually have to run before the fund can pay for the replacement.
1. Determine the number of complete years after which the e-bike's book value first falls below €1,500. Show the logarithms and a check of the values either side.
Read the worked answer
Book value after n years: V = 4200 × 0.82^n.
Set 4200 × 0.82^n < 1500, so 0.82^n < 1500/4200 = 0.35714.
Take logs: n > ln(0.35714) / ln(0.82) = (−1.0296)/(−0.1985) = 5.19.
n must be a whole number of years, so check both sides of 5.19:
- V(5) = 4200 × 0.37074 = €1,557 (still above €1,500)
- V(6) = 4200 × 0.30401 = €1,277 (below €1,500)
The value first drops below €1,500 at the end of year 6.
The common error is to round 5.19 down to 5; that leaves the bike at €1,557, so the condition is not yet met. The co-operative replaces the bike 6 years after purchase.
2. Calculate the value of the maintenance fund at the end of that period and compare it with the member's €5,400 estimate.
Read the worked answer
6 years = 72 monthly deposits. Monthly rate i = 0.024/12 = 0.002.
FV = 150 × ((1.002^72 − 1)/0.002) 1.002^72 = 1.15475 (1.15475 − 1)/0.002 = 77.375 FV = 150 × 77.375 = €11,606 (to the nearest euro).
Total deposited = 72 × 150 = €10,800, so interest contributes about €806, roughly 7.5% on top of the deposits.
The €5,400 figure is wrong twice over: it assumes only 36 deposits, and it ignores interest completely. Even at 3 years the fund would hold 150 × ((1.002^36 − 1)/0.002) = €5,593, already above €5,400 - so the estimate is not just uninterest-adjusted, it is also short by the wrong number of months.
3. Find the earliest number of monthly deposits needed for the fund to pay the inflated replacement price, and comment on the second member's claim about interest.
Read the worked answer
Replacement price at year 6 = 5300 × 1.02^6 = 5300 × 1.12616 = €5,968.66.
Need 150 × ((1.002^n − 1)/0.002) ≥ 5968.66 (1.002^n − 1) ≥ 0.079582 1.002^n ≥ 1.079582 n ≥ ln(1.079582)/ln(1.002) = 0.076573/0.001998 = 38.3
Check whole months: 38 deposits give €5,916 (short by €53); 39 deposits give €6,078 (surplus €109). 39 monthly deposits is the first month that covers the can.
At that point deposits total 39 × 150 = €5,850 and interest supplies about €228 - over 3.9% of the fund. Interest is not 'a few euros'. The fund is comfortably over-funded if deposits run the full 6 years, so the treasurer could either stop at month 39 or reduce the deposit.