23 Yield Rate and Reinvestment Rate

23.1 Yield Rate

Suppose that a single amount \(L\) is invested for an \(n\)-year period and the value of the investment at the end of \(n\) years is \(M\). A reasonable definition of the annual yield rate earned by the investment over the \(n\) year period is the rate \(i\) that satisfies the equation \(L \times (1+i)^n =M\).

Contrary to the internal rate of return, which is simply the interest rate at which a financial transaction (e.g. a loan) is made, the yield rate does not assume that the amount of interest received is reinvested at the same loan rate, but also take into account the reinvestment rate \(j\).

23.2 Reinvestment Rate

23.2.1 Single Deposit

Assume that a single deposit of \(1.00\) is made into a fund crediting interest at rate \(i\). If the interest is withdrawn annually from the first fund and reinvested in a second fund crediting interest at rate \(i'\), the amount in both funds combined at the end of \(n\) years is:

\[AV_n = 1 + i\times s_{\overline{n|}i'}\] ### Annual Deposit

Assume that a single deposit of \(1.00\) is made into a fund crediting interest at rate \(i\) at the beginning of each year in \(n\) years. If the interest is withdrawn annually from the first fund and reinvested in the second fund crediting interest at rate \(i'\), the amount in both funds combined at the end of \(n\) years is: \[AV_n = n + i\times (Is)_{\overline{n|}i'}\]

23.3 Examples

Example: Smith owns a \(10,000\) savings bond that pays interest monthly at \(i^{(12)} = 0.6\). Upon receipt of a bond interest payment, he immediately deposits it into an account earning interest, payable monthly, at a rate of \(i^{(12)} = .12\). Find the accumulated value of this account just after the 12th, 24th and 36th deposit, and find the annual yield rate \(i^{(12)}\) for each case based on the initial investment of \(10,000\). Assume that the savings bond may be cashed in at any time for \(10,000\).

Solution: The savings bond pays interest of \(10,000 \times 0.6 / 12 = 50\) per month, so the accumulated values in the account are:

\(50\times s_{\overline{12|}0.01} = 634.13\)
\(50\times s_{\overline{24|}0.01} = 1348.67\)
\(50\times s_{\overline{36|}0.01} = 2153.84\)

Let \(j\) be the average monthly yield rate earned on the initial \(10,000\) investment, then if Smith was to cash in his bond at the end of

12 months: \(j = (10,634.13 / 10,000)^{1/12} = 0.00514 \implies i^{(12)} = 0.0616\)
24 months: \(j = (11,348.67 / 10,000)^{1/24} = 0.00529\implies i^{(12)} = 0.0634\)
36 months: \(j = (12,153.84 / 10,000)^{1/36} = 0.00543 \implies i^{(12)} = 0.0652\)

Example: Suppose that Smith, on a payroll savings plan, buys a savings bond for \(100\) at the end of every month, with the bond paying monthly interest at \(i^{(12)} = .06\). The interest payments generated are reinvested in an account earning \(i^{(12)} = .12\). Find the accumulated value in the deposit account just after the 12th, 24th and 36th deposit, and find the annual yield rate \(i^{(12)}\) that Smith realizes over each of these time periods on his investment.

Solution: At the end of 12 months Smith will have bought \(1,200\) in bonds. The first \(100\) bond was bought at the end of the first month, so Smith received interest of \(0.50\) at the end of the second month, then \(1.00\) at the end of the third month, then \(1.50\) at the end of the fourth month, and so on. Therefore, the cashflows into the second account make an increasing annuity-immediate.

\(0.50\times (Is)_{\overline{11|}0.01} = 34.13\)
\(0.50\times (Is)_{\overline{23|}0.01} = 148.67\)
\(0.50\times (Is)_{\overline{35|}0.01} = 353.84\)

Let \(j\) be the average monthly yield rate earned. Since he buys the saving bonds monthly, hence \(j\) would be the solution of the equations \(100\times s_{\overline{n|}j} = AV_n(\text{both funds})\). Thus, for each case of \(n\) equals to:

12 months: \(100\times s_{\overline{12|}j} = 1,234.13 \implies j = .00508 \implies i^{(12)} = 0.061\)
24 months: \(100\times s_{\overline{24|}j} = 2,548.67 \implies j = .00518 \implies i^{(12)} = 0.0622\)
36 months: \(100\times s_{\overline{36|}j} = 3,953.84 \implies j = .00529 \implies i^{(12)} = 0.0634\)