Lesson One level more complicated non-standard problems on ordinary annuity plans
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<H2>One level more complicated non-standard problems on ordinary annuity plans</H2> <H3>Problem 1</H3>Find the future value of the ordinary annuity saving plan in 10 years from now, given that a person makes 10 deposits of $10,000 each at the end of each of 10 years. The interest rate in the bank is 8% per year compounded quarterly. <B>Solution</B> <pre> It is a non-traditional accumulative saving plan. The complication is that the deposits are made annually, while compounding are made quarterly. So, we should construct an equivalent model, which will smoothly combine/treat these features. An account with 8% annual interest rate compounded quarterly works as the effective quarterly rate r = 0.08/4; then the effective annual growth coefficient is t = {{{(1 + 0.08/4)^4}}} = {{{1.02^4}}} = 1.08243216, or an effective annual rate q = 0.08243216. Now this given saving plan as an equivalent to the ordinary annuity with annual deposits of $10,000 and with effective annual rate q = 0.071859031 compounding yearly. Therefore, we can apply the standard formula for future value of such ordinary annuity FV = {{{10000*((1.08243216^10-1)/0.08243216)}}} = 146549.56. (rounded). <U>ANSWER</U>. The future value is $146549.56. </pre> We equivalently transformed the given saving plan into another saving plan, where deposits are synchronized with compounding. <H3>Problem 2</H3>Mrs. Cook makes deposits of $950 at the end of every 6 months for 15 years. If interest is 3% compounded monthly, how much will Mrs. Cook accumulate in 15 years? <B>Solution</B> <pre> Again, it is a non-traditional accumulative saving plan with $950 deposited semi-annually and compounded monthly at 3% per annum. It means that the monthly effective growth factor is (1+0.03/12) = 1.0025. Then the semi-annual effective growth factor is {{{(1+0.03/12)^6}}} = {{{1.0025^6}}} = 1.01509406308652. So, we can now write the formula for the future value of the ordinary annuity in 15 years with the semi-annual deposits of $950 at the end of every 6 months with the found effective rate FV = {{{950*((1.01509406308652^(2*15)-1)/0.01509406308652)}}} = 35713.39 (rounded). <U>ANSWER</U>. In 15 years, the accumulated amount will be about 35713.39 dollars. </pre> We equivalently transformed the given saving plan into another saving plan, where deposits are synchronized with compounding. <H3>Problem 3</H3>Keys Company has a target of establishing a fund of $50,000. If $10,000 is deposited at the end of every six months, and the fund earns interest at 4% compounded quarterly, how long will it take to reach the target? <B>Solution</B> <pre> It is a non-traditional accumulative saving plan with $10,000 deposited semi-annually and compounded quarterly at 4% per annum. It means that the quarterly effective rate is 0.04/4 = 0.01 and the equivalent semi-annual effective rate is {{{(1+0.01)^2}}} = {{{1.01^2}}} = 1.0201. +-------------------------------------------------------------------+ | So, this non-traditional accumulative saving plan is equivalent | | to the ordinary annuity with semi-annual deposits of $10,000 | | and the semi-annual effective rate of compounding r = 1.0201. | +-------------------------------------------------------------------+ Now use the general formula for a classic Ordinary Annuity saving plan FV = {{{P*(((1+r)^n-1)/r)}}}, (1) where FV is the future value of the annuity; P is the semi-annual deposit; r is the semi-annual effective percentage yield presented as a decimal; n is the number of deposits. Under the given conditions, P = 10000; r = 0.0201. So, according to (1), the formula for the future value is FV = {{{10000*(((1+0.0201)^n-1)/0.0201)}}}. So, we should find n, the number of deposits (or the number of semi-annual periods) from this equation 50000 = {{{10000*((1.0201^n-1)/0.0201)}}}. Simplify it by dividing both sides by 10000 {{{50000/10000}}} = {{{((1.0201^n-1)/0.0201)}}}, or 5 = {{{((1.0201^n-1)/0.0201)}}}. Simplify it further, step by step 5*0.0201 = {{{1.0201^n-1}}}, 0.1005 = {{{1.0201^n-1}}}, 0.1005 + 1 = {{{1.0201^n}}}, 1.1005 = {{{1.0201^n}}}. Take logarithm base 10 of both sides log(1.1005) = n*log(1.0201) and find n n = {{{log((1.1005))/log((1.0201))}}} = 4.81 (approximately). Finally, round the decimal value of 4.81 to the closest GREATER integer value of 5 in order for the bank be in position to complete the last semi-annual compounding. At this point, the solution is complete. The <U>ANSWER</U> is: 5 semi-annual periods are needed. </pre> We equivalently transformed the given saving plan into another saving plan, where deposits are synchronized with compounding. <H3>Problem 4</H3>What payment made at the end of each year for 18 years will amount to $16,OOO at 4.2% compounded monthly? <B>Solution</B> <pre> As it is given, this annuity is not standard: the payments are made at the end of each year, while compounding is made at the end of each month. Analytic formulas exist only for coinciding schedules of payments and compounding. But we can transform to an equivalent standard scheme, considering payments at the end of each year and compounding at the end of each year with the effective annual multiplicative growth rate 1+r = {{{(1+0.042/12)^12}}} = 1.042818007. (1) Now we can use a well known formula for such ordinary annuity FV = {{{P*(((1+r)^18-1)/r)}}}. (2) In this formula, FV is the future value in 18 years; P is the annual payment, the unknown value which we should find. We calculate the factor in the formula (2) first {{{((1+r)^18-1)/r}}} = {{{((1.042818007)^18-1)/0.042818007}}} = 26.31908947. Then from formula (2) we find P = {{{FV/26.31908947}}} = {{{16000/26.31908947}}} = 607.93 dollars. Thus we found out the annual payment value. It is $607.94. <U>ANSWER</U> </pre> We equivalently transformed the given saving plan into another saving plan, where deposits are synchronized with compounding. 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