IntroductionDeferred means postponed, put off until later.There are situations in which the buyer only starts paying after a grace period: "buy today and start paying in three months". In this lesson we will calculate the present value of these deferred sequences and then the future value (accumulated amount) of a uniform sequence.Deferred uniform sequencesA uniform sequence is deferred when the first payment does not occur on date 1 but on date m + 1. On the timeline: dates 1, 2, …, m have no payment (this is the grace period), and the n equal installments R fall due on dates m + 1, m + 2, …, m + n.To calculate the present value (PV) of this sequence on date 0, we work in two steps:1. We calculate the value of the sequence on date m (Vm), that is, one period before the first payment. On that date the sequence is an ordinary uniform sequence, and the formula from the previous lesson applies:\[ V_m = R \left[\frac{(1 + i)^n - 1}{(1 + i)^n \cdot i}\right] \]2. We bring Vm from date m back to date 0, discounting it for m periods:\[ PV = \frac{V_m}{(1+i)^m} \]Be careful with m: it is the number of periods until the first payment, minus one. If the first installment is due three months after the purchase, m = 2, not 3.ExampleA plot of land is sold for R$ 50,000 in cash or in six equal monthly installments, the first due three months after the purchase. If the financing interest rate is 2% per month, what is the value of each installment?m = 2 (the first installment is due on date 3)n = 6i = 0.02PV = 50,000R = ?First we carry the cash price forward to date m = 2:\[ 50{,}000 = \frac{V_2}{(1+0.02)^{2}} \Rightarrow V_2 = 52{,}020 \]On date 2, the six installments form an ordinary uniform sequence:\[ V_2 = R \left[\frac{(1 + 0.02)^{6} - 1}{(1 + 0.02)^{6} \cdot 0.02}\right] \]\[ 52{,}020 = R \cdot 5.601431 \]\[ R = 9{,}286.91 \]Each installment is R$ 9,286.91.A store advertises: buy today and pay in 5 equal monthly installments, the first one 4 months from now. When using the deferred present value formula, what is m?Answer: m = 3 — the uniform sequence formula gives the value one period before the first payment; if that payment is on date 4, the value sits on date 3 and must be brought back 3 periods to date 0. Using m = 4 is the most common mistake: it discounts one month too many.An item costs R$ 10,000.00 in cash or 5 equal monthly installments, the first 4 months after the purchase, at 2% per month. What is the value of each installment?Answer: R$ 2,251.44 — with m = 3, V3 = 10,000 × 1.02³ = 10,612.08, and R = 10,612.08 / 4.713460 = 2,251.44. R$ 2,296.47 is what you get with m = 4, and R$ 2,121.58 is the installment with no grace period.Future value of a uniform sequenceWe have seen the present value of a uniform sequence. Now we will look at its future value, that is, the accumulated amount.The future value of the sequence on date n is the sum of the accumulated values of each installment R, invested from the date it was made until date n. With deposits on dates 1, 2, …, n, the last one (made on date n itself) earns nothing, the second-to-last earns one period, and so on, up to the first, which earns n − 1 periods.Denoting the future value by M:\[ M = R + R \cdot (1+i)^{1} + R \cdot (1+i)^{2} + ... + R \cdot (1+i)^{n-1} \]This is again a geometric progression, now with n terms, first term R and ratio (1 + i). Using the formula for the sum of a GP:\[ S = \frac{a_1 \cdot (q^n - 1)}{q - 1} \]we get:\[ M = \frac{R \cdot ((1+i)^n - 1)}{1 + i - 1} \]\[ M = R \left[\frac{(1+i)^n - 1}{i}\right] \]This is the future value at the moment of the last deposit.ExamplesAn investor deposits R$ 2,000.00 every month in an investment fund that pays compound interest of 1% per month. If the investor makes seven deposits, what is the accumulated amount at the moment of the last deposit?R = 2,000i = 1% per monthn = 7\[ M = 2{,}000 \left[\frac{(1+0.01)^{7} - 1}{0.01}\right] = 14{,}427.07 \]The accumulated amount is R$ 14,427.07.An executive, thinking about future retirement, decides to make 240 monthly deposits of R$ 700.00 each in a fund that earns 0.4% per month (real rate). The goal is to build up an amount that will allow withdrawals of x reais per month for 360 months, until the balance runs out. Find x, assuming the first withdrawal is made one month after the last deposit. Assume all amounts are in real terms relative to the date the deposits begin.Accumulated amount right after the 240th deposit:\[ M = 700 \left[\frac{(1+0.004)^{240} - 1}{0.004}\right] = 281{,}172.52 \]This amount is the present value of the withdrawals, which start one month later. Therefore:\[ 281{,}172.52 = x \left[\frac{(1+0.004)^{360} - 1}{(1+0.004)^{360} \cdot 0.004}\right] \Rightarrow x = 1{,}475.21 \]The executive will be able to withdraw R$ 1,475.21 per month, in today's money.Check the spreadsheet below and make a copy for your own retirement experiments.https://docs.google.com/spreadsheets/d/1xYi51tznE1N46asp-zQE7YbpWJmR6Fv5dCEwaDbTju8/edit?usp=sharingYou deposit R$ 500.00 per month for 24 months in an investment that earns 1% per month. What is the accumulated amount right after the last deposit?Answer: R$ 13,486.73 — M = 500 × (1.01^24 − 1) / 0.01 = 500 × 26.973465. R$ 12,000.00 is just the sum of the deposits, with no interest; R$ 14,121.60 would be the balance one month after the last deposit (or with 25 deposits).In the future value formula, why is the last term of the sum R × (1 + i) raised to n − 1, and not to n?Answer: Because the first deposit earns n − 1 periods until date n, and the last one earns nothing — with deposits on dates 1 to n, the one made on date 1 stays invested for n − 1 periods and the one made on date n counts at face value. That is why the sum has n terms, with exponents from 0 to n − 1.Exercises1) A person deposits R$ 3,500.00 every month, for seven months, in a fund that pays 0.9% per month. What is the accumulated amount right after the last deposit?2) How much must a person deposit every month, for 15 months, in an investment fund that earns 1.8% per month, to have R$ 60,000.00 right after the last deposit?3) A company must pay a bill of R$ 50,000.00 one year from now. How much must it invest every month, starting today, if the deposits are equal and earn 0.85% per month, so that one month after the last deposit the balance is enough to pay the bill?4) How much should I deposit every month, in an investment fund that pays 0.8% per month, to have R$ 2,000,000.00 at the time of the 18th deposit?5) An executive, planning to supplement retirement income, decides to make 180 monthly deposits of R reais each on dates 1, 2, 3, …, 180, in order to withdraw R$ 1,500.00 per month on dates 181, 182, …, 420. Find R, assuming the money earns 0.5% per month.6) Suppose you retire at 70 with savings of R$ 800,000.00 invested in a fund. How much can you withdraw from this fund and spend each year, over the next 25 years, assuming constant spending in real terms? Assume a real return of 4% per year and the first withdrawal one year after retirement.7) Continuing the previous exercise: how much would you need to invest each year (a constant deposit in real terms) in this fund, from age 35 to 70 (36 deposits), to build savings of R$ 800,000.00?8) A sofa set is sold for R$ 6,000.00 in cash or in four equal monthly installments, the first due three months after the purchase. What is the value of each installment if the financing interest rate is 2.8% per month?9) A motorcycle is advertised in ten equal monthly installments of R$ 2,000.00 each, the first due two months after the purchase. What is the cash price if the financing rate is 3.5% per month?10) A sound system is sold for R$ 2,300.00 in cash. On credit, it is sold in six equal monthly installments with a two-month grace period, that is, the first installment is only due three months after the purchase. The store charges 2% per month. Find the value of each installment.Answers1) R$ 25,171.512) R$ 3,519.953) R$ 3,941.94 (12 deposits, on dates 0 to 11; the bill is due on date 12)4) R$ 103,746.145) R$ 719.946) R$ 51,209.577) R$ 10,309.508) R$ 1,697.679) R$ 16,070.7310) R$ 427.20
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