Lesson Annuity Due saving plans and geometric progressions
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<H2>Annuity Due saving plans and geometric progressions</H2> Geometric progressions are widely used in <B>financial mathematics</B>. Particularly, they are working tools in calculating and analyzing <B>saving plans</B>. You can learn from this lesson how to apply geometric progressions to the <B>'Annuity Due'</B> type saving plans. An <B>annuity</B> is a sequence of equal periodic deposits. The periodic deposits may be made annually, quarterly, monthly, or daily. If the deposit is made at the beginning of each payment period and the interest is compounded at the end of the payment period, the annuity is called <B>'annuity due'</B>. Another type of an annuity is <B>'ordinary annuity'</B> saving plan, when the deposit is made at the end of each payment period and the interest is credited at the same time. This type saving plan is considered in the lesson <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Ordinary-Annuity-saving-plans-and-geometric-progressions.lesson>Ordinary Annuity saving plans and geometric progressions</A> under the current topic in this site. <H3>Problem 1 ('Annuity Due' saving plan)</H3>Ann has so called <B>annuity due</B> bank saving plan. According to the plan Ann deposits $100 in an account at the beginning of each month for 2 years. The account pays annual interest rate of 4%, compounded monthly at the end of each month. What is Ann's balance at the end of 2 years? <B>Solution</B> This <B>annuity due</B> saving plan works in such a way that the first deposit earns interest at the monthly rate of {{{4/12}}}% = {{{0.3333}}}% for full 24 months. The second deposit earns interest for 23 months, the third deposit earns interest for 22 months, and so on. Using the formula for compound interest, you can write the total of all 24 deposits as <B>Total</B> = {{{100*1.003333^1}}} + {{{100*1.003333^2}}} + {{{100*1.003333^3}}} + . . . + {{{100*1.003333^24}}}. This is the sum of the 24 terms of the geometric progression with the first term 100*1.003333 and the common ratio 1.003333. For the formula for the sum of a geometric progression see the lesson <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Geometric-progressions.lesson>Geometric progressions</A> in this site. So, Ann's balance at the end of the second year is <B>Total</B> = 100*1.003333*{{{((1.003333^24-1)/(1.003333-1))}}} = 2502.60. <B>Answer</B>. Ann's balance at the end of the second year is $2502.60. Below is one more problem on the annuity due saving plan. <H3>Problem 2 ('Annuity Due' saving plan)</H3>It was just described in the previous <B>Problem 1</B> how an <B>annuity due</B> saving plan works. During the time period of <B>t</B> years, an individual deposits a fixed amount of <B>P</B> dollars to the account at the beginning of each month. The account is compounded monthly at the end of each month at the monthly interest rate <B>i</B> = {{{r/12}}}%, where <B>r</B> is per annum interest rate. Derive a general formula for the balance at the end of the annuity due saving plan and find the balance if a) P = $100, t = 10 years, r = 4%; b) P = $500, t = 30 years, r = 4%. <B>Solution</B> The saving plan spans the time period of <B>n</B>=12*t months. The first deposit will earn interest at the monthly rate of <B>i</B>={{{r/12}}}% for full <B>n</B>=12*t months. This means that the amount of money generated by the first deposit will increase to the value of {{{P*(1+i)}}} at the end of the first month; to the value of {{{P*(1+i)^2}}} at the end of the second month; to the value of {{{P*(1+i)^3}}} at the end of the third month, and so on. The amount of money generated by the first deposit will increase to the value of {{{P*(1+i)^n}}} at the end of the last, n-th month. Similarly, the second deposit will earn interest at the monthly rate of <B>i</B>={{{r/12}}}% for <B>n-1</B>=12*t-1 months. This means that the amount of money generated by the second deposit will reach the value of {{{P*(1+i)^(n-1)}}} at the end of the saving plan. The third deposit will earn interest at the monthly rate of <B>i</B>={{{r/12}}}% for <B>n-2</B> months. It will generate the amount of money {{{P*(1+i)^(n-2)}}} at the end of the saving plan. The last, n-th deposit, will earn interest at the monthly rate of <B>i</B>={{{r/12}}}% for only 1 months. It will generate the amount of money {{{P*(1+i)}}} at the end of the saving plan. Thus, at the end of the saving plan the total amount of money in the account will be {{{P*(1+i)}}} + {{{P*(1+i)^2}}} + {{{P*(1+i)^3}}} + . . . + {{{P*(1+i)^(n-1)}}} + {{{P*(1+i)^n}}}. This is the sum of the geometric progression with the first term {{{P*(1+i)}}} , the common difference {{{1+i)}}} and the number of terms equal to <B>n</B>=12*t. Applying the formula for the sum of a geometric progression from the lesson <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Geometric-progressions.lesson>Geometric progressions</A> you get the general formula for the balance of the annuity due saving plan Total = {{{P*(1+i)}}}*{{{((1+i)^n-1)/i)}}} = {{{P*(1+i)}}}*{{{((1+i)^(12*t)-1)/i)}}} = {{{P*(1+(r/12))}}}*{{{ ( (1+(r/12))^(12*t)-1)/(r/12) ) }}}. Now, let us apply this formula to cases a) and b). a) There are <B>n</B> = 12*t = 12*10 = 120 monthly payment periods in this case. The monthly rate is {{{i}}} = {{{r/12}}} = {{{0.04/12}}} = {{{0.003333}}}. Total = {{{100*(1+0.003333)}}}*{{{(1.003333^120-1)/0.003333}}} = {{{100.333333}}}*{{{0.490832682/0.003333}}} = 14,774.06. <B>Answer</B>. The balance is equal to $14,774.06. b) There are <B>n</B> = 12*t = 12*30 = 360 monthly payment periods in this case. The monthly rate is {{{i}}} = {{{r/12}}} = {{{0.04/12}}} = {{{0.003333}}}. Total = {{{500*(1+0.003333)}}}*{{{(1.003333^360-1)/0.003333}}} = {{{501.6667}}}*{{{2.313498/0.003333}}} = 348,181.45. <B>Answer</B>. The balance is equal to $348,181.45. You can compare these results with the similar calculations for the <B>ordinary annuity</B> saving plans of the lesson <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Ordinary-annuity-saving-plans-and-geometric-progressions.lesson>Ordinary Annuity saving plans and geometric progressions</A> in this site to see the difference. <H3>Problem 3</H3>Jane deposits $1,300 in an account at the beginning of each 3-month period for 12 years. If the account pays interest at the rate of 4%, compounded quarterly, how much will she have in her account after 12 years? <B>Solution</B> <pre> It is a classic Annuity Due saving plan. The general formula is FV = {{{P*(1+r)*(((1+r)^n-1)/r)}}}, (1) where FV is the future value of the account; P is the quarterly payment (deposit) at the beginning of each payment period; r is the quarterly percentage yield presented as a decimal; n is the number of deposits (= the number of years multiplied by 4, in this case). Under the given conditions, P = 1300; r = 0.04/4 = 0.01; n = 12*4 = 48. So, according to the formula (1), Jane will get at the end of the 4-th year FV = {{{1300*(1+0.01)*(((1+0.01)^48-1)/0.01)}}} = {{{1300*1.01*((1.01^48-1)/0.01)}}} = $80385.28 (rounded). Note that Jane will deposit only 12*4*$1300 = $62400 in 12 years. The rest is what the account earns/accumulates in 12 years. </pre> <H3>Problem 4</H3>Find the quarterly payment for the annuity due whose future value should achieve $20,000 in 17 years. Assume that the compounding period is the same as the payment period; interest rate is 5%. <B>Solution</B> <pre> It is a classic Annuity Due saving plan. The general formula is FV = {{{P*(1+r)*(((1+r)^n-1)/r)}}}, (1) where FV is the future value of the account; P is your quarterly payment (deposit), which is made at the beginning of each quarter; r is the effective quarterly percentage yield presented as a decimal; n is the number of deposits (= the number of years multiplied by 4, in this case). Under the given conditions, P = 20000; r = 0.05/4; n = 17*4 = 68. So, according to the formula (1), the quarterly payment should be P = {{{20000/((1+0.05/4)*(((1+0.05/4)^(17*4)-1)/((0.05/4))))}}} = $186.02. <U>ANSWER</U> Note that you deposit only 17*4*$186.02 = $12,649.36. The rest is what the account earns/accumulates in 17 years. </pre> My other lessons in this site associated with annuity saving plans and retirement plans are - <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Geometric-progressions.lesson>Geometric progressions</A> - <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/The-proofs-of-the-formulas-for-geometric-progressions.lesson>The proofs of the formulas for geometric progressions</A> - <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Ordinary-Annuity-saving-plans-and-geometric-progressions.lesson>Ordinary Annuity saving plans and geometric progressions</A> - <A HREF=https://www.algebra.com/algebra/homework/Sequences-and-series/Solved-problem-on-Ordinary-Annuity-saving-plans.lesson>Solved problems on Ordinary Annuity saving plans</A> - <A HREF=https://www.algebra.com/algebra/homework/Sequences-and-series/Finding-present-value-of-an-annuity.lesson>Finding present value of an annuity, or an equivalent amount in today's dollars</A> - <A HREF=https://www.algebra.com/algebra/homework/Sequences-and-series/Withdrawing-a-certain-amount-of-money-periodically-from-a-compounded-saving-account.lesson>Withdrawing a certain amount of money periodically from a compounded saving account</A> - <A HREF=https://www.algebra.com/algebra/homework/Sequences-and-series/Solved-problems-on-Annuity-saving-plans.lesson>Miscellaneous problems on retirement plans</A> OVERVIEW of my lessons on geometric progressions with short annotations is in the lesson <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Review-of-lessons-on-geometric-progressions.lesson>OVERVIEW of lessons on geometric progressions</A>. Use this file/link <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-II.
