8.4, #2. The future value of the installment plan 3 years from now when the final payment is made is

5000 + 5000e^{0.08} + 5000e^{2(0.08)} + 5000e^{3(0.08)} = 22640.24

The future value of $25000 paid 3 years from now is $25000. Thus the lump payment is preferable. Alternatively, one can calculate the present values of the two options to see the lump payment is preferable.

8.4, #3. The present value of the income stream is

∫_{0}^{15} 3000e^{−0.06t} dt = 3000(\frac{e^{−0.06t}}{−0.06} )|_{0}^{15}
= 29671.52.

The future value of the income stream is

∫_{0}^{15} 3000e^{0.06(15−t)} dt = 3000e^{0.9} ∫_{0}^{15} e^{−0.06t} dt

= 3000e^{0.9}(\frac{e^{−0.06t}}{−0.06} )|_{0}^{15}

= 72980.16

8.4, #5. If P dollars is deposited at a continuous each year into the account, the future value of the stream in 10 years time is

∫_{0}^{10} P e^{0.1(10−t)} dt = P e ∫_{0}^{10} e^{−0.1t} dt = −P e\frac{e^{−0.1t}}{0.1} |_{0}^{10} = 10P e(1 − e^{−1}).

This is $100000 if P = 10000/(e − 1) = 5819.77. The future value at the end of 10 years of a lump sum of L dollars deposited now is Le^{(0.1)×10} = Le. This is $100000 if L = 100000/e = 36787.94.

8.4, #7. In (a) the future value of the deposit stream at time T is

∫_{0}^{T} 1000e^{0.05(T −t)} dt = −20000e^{0.05(T −t)}|_{0}^{T} = 20000(e^{0.05T} − 1).

20000(e^{0.05T} − 1) = 10000 when e^{0.05T} = 1.5, that is , when 0.05T = ln 1.5 and T = 8.11 years. In (b) the future value of the deposit stream at time T is

2000e^{0.05T} + 20000(e^{0.05T} − 1) = 22000e^{0.05T} − 20000.

22000e^{0.05T} − 20000 = 10000 when e^{0.05} = 32/22 = 1.3636, that is, when 0.05T = 0.3102 and T = 6.20 years.