Equations of Value and Time Diagram

Problem 12.1
In return for payments of $5,000 at the end of 3 years and $4,000 at the
end of 9 years, an investor agrees to pay $1500 immediately and to make an additional payment at the end of 2 years. Find the amount of the additional payment if $i^{(4)} = 0.08$.

Solution.
$$5,000(1 + \frac{0.08}{4})^{-3 \times 4} + 4,000(1 + \frac{0.08}{4})^{-9 \times 4} \\ = 1,500 + X(1 + \frac{0.08}{4})^{-2 \times 4} \\ \implies X = 5,159.23627$$

Problem 12.2
At a certain interest rate the present values of the following two payment
patterns are equal:
(i) 200 at the end of 5 years plus 500 at the end of 10 years;
(ii) 400.94 at the end of 5 years.
At the same interest rate 100 invested now plus 120 invested at the end of 5 years will accumulate to P at the end of 10 years. Calculate P.

Solution.
$$\text{Solving for the value of $i$ first, we have} \\ 200(1+i)^{-5} + 500(1+i)^{-10} = 400.94(1+i)^{-5} \\ 500(1+i)^{-10} - 200.94(1+i)^{-5} = 0 \\ 500(1+i)^{-5} = 200.94 \\ \implies i = 19.99\% \; or \; 20\% \\ \text{Solving for the value of $P$, we have} \\ P = 100(1+0.19999)^{10} + 120(1+0.1999)^{5} \\ = 917.7627567$$

Problem 12.3
An investor makes three deposits into a fund, at the end of 1, 3, and 5 years. The amount of the deposit at time $t$ is $100(1.025)^{t}$. Find the size of the fund at the end of 7 years, if the nominal rate of discount convertible quarterly is $\frac{4}{41}$ .

Solution.
$$100(1.025)(1-\frac{\frac{4}{41}}{4})^{-6 \times 4} + 100(1.025)^{3}(1-\frac{\frac{4}{41}}{4})^{-4 \times 4} \\ + 100(1.025)^{-2 \times 4}(1-\frac{\frac{4}{41}}{4})^{-6 \times 4} \\ = 483.1105329$$

Problem 12.4
Brian and Jennifer each take out a loan of $X$. Jennifer will repay her loan
by making one payment of 800 at the end of year 10. Brian will repay his
loan by making one payment of 1,120 at the end of year 10. The nominal
semi-annual rate being charged to Jennifer is exactly one-half the nominal
semi-annual rate being charged to Brian. Calculate $X$.

Solution.
$$800(1+\frac{j}{2})^{-10 \times 2} = 1,120(1+\frac{2j}{2})^{-10 \times 2} \\ \frac{800}{1,120}(1 + \frac{j}{2}) = (1 + \frac{2j}{2}) \\ \text{For simplicity of solving, store the value of $\frac{800}{1,120}$ in your calculator.} \\ \text{But here, we'll just denote the value of $\frac{800}{1,120}$ as $A$. We'll have} \\ A + \frac{Aj}{2} = 1 + j \\ 2A + Aj = 2 + 2j \\ 2A + Aj - 2J = 2 \\ Aj - 2j = 2 - 2A \\ j(A - 2) = 2 - 2A \\ j = \frac{2 - 2A}{A-2} \\ \text{Solving for the value of $x$ using the computer $j$, we have} \\ X(1 + \frac{j}{2}) = 800 \\ \implies X = 568.1479212$$

Problem 12.5
Fund A accumulates at 6% effective, and Fund B accumulates at 8% effective. At the end of 20 years the total of the two funds is 2,000. At the end of 10 years the amount in Fund A is half that in Fund B. What is the total of the two funds at the end of 5 years?

Solution.
$$A(1+0.06)^{20} + B(1+0.08)^{20} = 2,000 \; e.q \; (1) \\ 2A(1+0.06)^{10} = B(1+0.08)^{10} \; e.q \;(2) \\ \text{Multiplying equation 2 by 2 and} \\ \text{and subtracting it to equation 1, we have} \\ A(1+0.06)^{20} + 2A(1+0.06)^{10}(1+0.08)^10 = 2,000 \\ $$

Problem 12.6
Louis has an obligation to pay a sum of $3,000 in four years from now and a sum of $5,000 in six years from now. His creditor permits him to discharge these debts by paying $\$X$ in two years from now, $1000 in three years from now, and a final payment of $\$2X$ in nine years from now. Assuming an annual effective rate of interest of 6%, find $X$.

Solution.
$$3,000(1+0.06)^{-4} + 5,000(1+0.06)^{-6} \\ = X(1+0.06)^{-2} + 1,000(1+0.06)^{-3} + 2X(1+0.06)^{-7} \\ \implies X = 2,440.679235$$

Problem 12.7
Every Friday in February (7, 14, 21,28) Vick places a 1,000 bet, on credit,
with his off-track bookmaking service, which charges an effective weekly interest rate of 8% on all credit extended. Vick looses each bet and agrees to repay his debt in four installments to be made on March 7, 14, 21, and 28. Vick pays 1,100 on March 7, 14, and 21. How much must Vick pay on March 28 to completely repay his debt?

Solution.
$$1,000 + 1,000(1+0.08)^{-1} + 1,000(1+0.08)^{-2} + 1,000(1+0.08)^{-9} \\ = 1,100(1+0.08)^{-4} + 1,100(1+0.08)^{-5} + 1,100(1+0.08)^{-6} \\ + X(1+0.08)^{-7} \\ \implies X = 2,273.792429$$

Problem 12.8
A borrower is repaying a loan by making payments of 1,000 at the end of
each of the next 3 years. The interest rate on the loan is 5% compounded
annually. What payment could the borrower make at the end of the first
year in order to extinguish the loan?

Solution.
$$1,000(1+0.05)^{-1} + 1,000(1+0.05)^{-2} + 1,000(1+0.05)^{-3} \\ = X(1+0.05)^{-1} \\ \implies X = 2,859.410431$$

Problem 12.9
An investor purchases an investment which will pay 2,000 at the end of one year and 5,000 at the end of four years. The investor pays 1,000 now and agrees to pay $X$ at the end of the third year. If the investor uses an interest rate of 7% compounded annually, what is $X$?

Solution.
$$2,000(1+0.07)^{-1} + 5,000(1+0.07)^{-4} = 1,000 + X(1+0.07)^{-3} \\ \implies X = 5,737.654196$$