Question 4 of 6: Interest and Investment Costs

Question 4: Interest and Investment Costs (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Simple versus compound interest

Simple interest is charged only on the original principal for the whole term: the interest each period is $P\,i$ and never itself earns interest, so the debt grows linearly, $F=P(1+iN)$. Compound interest is charged on the running balance — principal plus all interest accrued so far — so interest earns interest and the debt grows geometrically, $F=P(1+i)^N$. Over a single period the two are identical; the gap widens with time and with the interest rate, and it is always in the lender's favour. Because it reflects the real time-value of money, compound interest is the basis of all engineering-economy calculations.

(b) Amount owed after 15 years (simple then compound)

Given. Principal $P=\$10{,}000$; first phase $i_1=4\%$ simple for $N_1=5$ years; second phase $i_2=6\%$ effective compound for $N_2=10$ more years; no payments made at any time.

Find. The total amount owed at the end of 15 years.

Approach. Grow the debt in two stages: apply simple interest over the first 5 years to get the balance carried forward, then compound that balance at 6 % for the remaining 10 years.

1. Phase 1 — 5 years of simple interest. Simple interest accrues on the original principal only: $$F_5 = P\,(1 + i_1 N_1) = \$10{,}000\,[\,1 + (0.04)(5)\,] = \$10{,}000(1.20) = \boxed{\$12{,}000}$$ This $\$12{,}000$ is the unpaid balance carried into the extension.
2. Phase 2 — 10 years of compound interest. The carried balance now compounds annually at 6 %: $$F_{15} = F_5\,(1 + i_2)^{N_2} = \$12{,}000\,(1.06)^{10} = \$12{,}000(1.79085) = \boxed{\$21{,}490}$$ The total owed at the end of 15 years is therefore about $\$21{,}490$.