Modelling a Savings Goal with Regular Deposits
A student wants to save $2,400 for a course starting in 18 months. She has $350 saved now and can deposit a fixed amount at the end of each month into an account paying 0.4% interest per month, compounded monthly. She first calculates that she needs (2400 − 350)/18 ≈ $113.89 per month and plans to deposit $115. A friend warns that this ignores interest, but the student thinks interest can only help and that $115 is therefore safe. She also has a fallback: if she falls short, she can borrow the difference at 1.5% per month simple interest for up to 6 months.
1. Set up the recurrence for her balance with monthly deposits and determine the deposit needed to reach $2,400 in 18 months, accounting for interest.
Read the worked answer
Let Bₙ be the balance after n months, with B₀ = 350 and monthly rate r = 0.004. Each month: Bₙ = 1.004·Bₙ₋₁ + d.
After 18 months: B₁₈ = 350(1.004)^18 + d·[(1.004^18 − 1)/0.004].
Compute (1.004)^18 ≈ 1.0744. So the first term is 350 × 1.0744 ≈ 376.05. The annuity factor = (1.0744 − 1)/0.004 = 0.0744/0.004 = 18.60.
Require 376.05 + 18.60d = 2400, so 18.60d = 2023.95, giving d ≈ $108.81.
So about $109 per month suffices, less than the naive $113.89 because interest contributes. Her $115 plan is more than enough; the friend's warning is directionally right but does not threaten her plan.
2. Verify whether the $115 deposit reaches the goal and quantify the surplus or shortfall.
Read the worked answer
With d = 115: B₁₈ = 376.05 + 18.60(115) = 376.05 + 2139 = $2515.05.
Surplus = 2515.05 − 2400 = $115.05, about one extra monthly deposit. So her plan succeeds with room to spare.
Breakdown of the $2,400: $350 initial grows to about $376, her deposits total 18 × 115 = $2,070, and interest on the growing balance adds the remainder, roughly $69. Interest helps but is a modest part of the total; the deposits do most of the work. The naive calculation overstated the required deposit by about $5 per month, or roughly $90 over 18 months, which is why the surplus is close to one deposit.
3. Suppose she deposits only $100 per month. Compute the shortfall at 18 months and the cost of covering it with the fallback loan over 6 months.
Read the worked answer
With d = 100: B₁₈ = 376.05 + 18.60(100) = 376.05 + 1860 = $2236.05.
Shortfall = 2400 − 2236.05 = $163.95.
Loan cost at 1.5% per month simple interest for 6 months: interest = 163.95 × 0.015 × 6 = 163.95 × 0.09 = $14.76. Total repaid ≈ $178.71.
Judgement: depositing $100 saves $15 per month versus $115, or $270 over 18 months, but leaves a $164 gap and $14.76 of loan interest. Raising the deposit to about $109 avoids the loan entirely and costs less than the loan route. The comparison shows that small monthly adjustments beat short-term borrowing here, and the recurrence, not the naive division, is the right tool for setting the target.