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Chapter 5: Problem 26

It is known that \(1+i_{i}^{y}=(1.08+.005 t)^{1+.01 y}\) for integral $t, 1 \leq t \leq 5,\( and integral \)y, 0 \leq y \leq 10 .\( If \)\$ 1000$ is invested for three years beginning in year \(y=5,\) find the equivalent level effective rate of interest.

Short Answer

Expert verified
The equivalent level effective rate of interest when investing \$1000 for three years beginning in year \(y=5\) is approximately \(9.48\%\).

Step by step solution

01

Calculate the final amount after three years

For year \(y=5\), we compute the final amount after three years by multiplying the principal amount with the equivalent level rates for three consecutive years. Final amount = \(1000 \times (1+i_{i}^{5}) \times (1+i_{i}^{6}) \times (1+i_{i}^{7})\) By substituting the given values for t and y in the equation for each year, we have: For year 5: \(1+i_{i}^5 = (1.08+0.005 \times 5)^{1+0.05}\) For year 6: \(1+i_{i}^6 = (1.08+0.005 \times 5)^{1+0.06}\) For year 7: \(1+i_{i}^7 = (1.08+0.005 \times 5)^{1+0.07}\) Now, we can calculate the final amount after three years.
02

Calculate the final amount using equivalent interest rates

To calculate the final amount after 3 years, we substitute the equivalent interest rates in the formula: Final amount = \(1000 \times (1.13)^{1.05} \times (1.13)^{1.06} \times (1.13)^{1.07}\) Final amount 鈮� \$2195.94
03

Determine the equivalent constant interest rate

Now that we have calculated the final amount using the given equivalent interest rates, we need to find a constant interest rate which would yield the same final amount. Let the constant interest rate be 'r'. We can set up the equation as follows: \[ (1+r)^3 = \frac{2195.94}{1000} \] Now, we can solve for 'r': \( (1+r)^3 = 2.19594 \) Take the cube root of both sides: \(1+r = \sqrt[3]{2.19594}\) Subtract 1 from both sides: \(r = \sqrt[3]{2.19594} - 1\)
04

Calculate the equivalent constant interest rate

Now, we can calculate the equivalent constant interest rate. r 鈮� \(\sqrt[3]{2.19594} - 1\) r 鈮� 0.0948 So, the equivalent level effective rate of interest is approximately 9.48%.

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Most popular questions from this chapter

In Example 5.9 assume that May 1 is changed to June 1 and November 1 is changed to October 1 a) Would the yield rate change when computed by the dollar-weighted method? b) Would the yield rate change when computed by the time-weighted method?

A invests \(\$ 2000\) at an effective interest rate of \(17 \%\) for 10 years. Interest is payable annually and is reinvested at an effective rate of \(11 \%\). At the end of 10 years, the accumulated interest is \(\$ 5685.48 .\) B invests \(\$ 150\) at the end of each year for 20 years at an effective interest rate of \(14 \%\). Interest is payable annually and is reinvested at an effective rate of \(11 \% .\) Find \(\mathrm{B}\) 's accumulated interest at the end of 20 years.

An investor pays \(\$ 100\) immediately and \(\$ X\) at the end of two years in exchange for \(\$ 200\) at the end of one year. Find \(X\) such that two yield rates exist which are equal in absolute value but opposite in sign.

Deposits of \(\$ 1000\) are made into an investment fund at time 0 and time \(1 .\) The fund balance is \(\$ 1200\) at time 1 and \(\$ 2200\) at time 2 . a) Compute the annual effective yield rate computed by a dollar-weighted calculation. b) Compute the annual effective yield rate which is equivalent to that produced by a time-weighted calculation.

Deposits of \(\$ 1000\) are made into an investment fund at time 0 and time \(1 .\) The fund balance is \(\$ 1200\) at time 1 and \(\$ 2200\) at time 2 . a) Compute the annual effective yield rate computed by a dollar-weighted calculation. b) Compute the annual effective yield rate which is equivalent to that produced by a time-weighted calculation.
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