An Introduction to Financial Mathematics, 2nd Edition

Appendix D

Solutions to Odd-Numbered Problems

Chapter 1

1.  If interest is compounded quarterly, then 2A(0) = A(0)(1 + r/4)^6.4, hence r = 4(2^1/24 − 1) ≈ .1172. If interest is compounded continuously, then 2A(0) = A(0)e^6r, hence r = (ln2)/6 ≈ 0.1155.

3.  By (1.4), A₅ = $29,391 and A₁₀ = $73,178.

5.  By (1.4), the number of payments n must satisfy

$400\frac{{\left(1.005\right)}^n-1}{.005}\ge 30,000.$

Using a spreadsheet, one determines that the smallest value of n satisfying this inequality is 63 months.

7.  Let

$x=\frac{ln\left(P\right)-ln\left(P-i{A}_0\right)}{ln\left(1+i\right)}.$

By (1.6), N = ⌊x⌋ + 1 unless x is an integer, in which case x = N. The number for the second part is ⌊138.976⌋ + 1 = 139. (The 138th withdrawal leaves $1,941.85 in the account.)

9.  The number of withdrawals n must satisfy

$A_n=\frac{{\left(1.005\right)}^n[\mathrm{300\; ...}}{}$