Actuarial Science: Financial Mathematics

STAT 485 Actuaral Scence: Fnancal Mathematcs 1.1.1 Effectve Rates of Interest Defnton Defnton lender. An nterest s money earned by deposted funds. An nterest rate s the rate at whch nterest s pad to the Defnton A compound nterest arses when nterest s added to the ntal nvestment (called prncpal), so that from that moment on, the nterest that has been added tself earns nterest. Example 1.1 The rate of nterest s 9% credted annually. The ntal depost s $1,000. Assumng that there are no transactons on the account other than the annual credtng of nterest, determne the account balance after nterest s credted at the end of three years. Soluton. After one year, the account balance s $1,000(1+0.09)=$1,090. After two years, the account balance s $1,190(1+0.09)=$1,188.10. After three years, the account balance s $1,188.10(1+0.09)=$1,295.03. Let C denote the ntal nvestment (C = $1, 000), and let be the annual nterest rate ( = 0.09). Then after n years(n = 3), the account balance s C(1+)(1+)...(1+) = C(1+) n. In our case, C(1+) n = $1, 000(1+0.09) 3 = $1, 295.03. In practce nterest may be credted monthly or weekly or daly. Each nterest rate may be recalculated nto an equvalent annual rate. For example, a 0.75% monthly nterest rate s equvalent to ((1 + 0.0075) 12 = 1.0938) 9.38% annual rate. Ths annual rate s called an effectve annual rate of nterest. Defnton Two rates of nterest are sad to be equvalent f they result n the same accumulated values at each pont n tme. 1.1.2 Compound Interest Example 1.3 A person deposts $1,000 nto an account on January 1, 2007. The account credts nterest at an effectve annual rate of 5% every December 31. The person wthdraws $200 on January 1, 2009, deposts $100 on January 1, 2010, and wthdraws $250 on January 1, 2012. What s the balance n the account just after nterest n credted on December 31, 2013? Soluton. One approach s to recalculate the balance after every transacton. On December 31, 2008 the balance s $1, 000(1 + 0.05) 2 = $1, 102.50. On January 1, 2009 the balance s $1, 102.50 $200 = $902.50. 1

On December 31, 2009 the balance s $902.50(1 + 0.05) = $947.63. On January 1, 2010 the balance s $947.63 + $100 = $1, 047.63. On December 31, 2011 the balance s $1, 047.63(1 + 0.05) 2 = $1, 155.01. On January 1, 2012 the balance s $1, 155.01 $250 = $905.01. On December 31, 2013 the balance s $905.01(1 + 0.05) 2 = $997.77. An alternatve approach s to accumulate each transacton to the December 31, 2013 date of valuaton and then combne all accumulated values, addng deposts and subtractng wthdrawals. We have ($1, 000)(1 + 0.05) 7 + ($100)(1 + 0.05) 4 ($200)(1 + 0.05) 5 ($250)(1 + 0.05) 2 = $997.77. Defnton The accumulaton factor from tme 0 to tme t s the accumulated value at tme t of an nvestment of $1 made at tme 0. The accumulaton factor wll be denoted by a(t). Defnton The accumulated amount functon, denoted by A(t), s the accumulated amount of an nvestment at tme t, A(t) = A(0)a(t). Example Under compound nterest, the accumulaton factor from 0 to t s a(t) = (1 + ) t where s the effectve annual rate of nterest, and tme t s a postve real number. Example Under compound nterest, the accumulated amount by tme t satsfes the equaton A(t) = A(0)(1 + ) t. Gven any of the three quanttes A(t), A(0),, and t, t s possble to fnd the fourth. The formulas are A(t) = A(0)(1 + ) t, A(0) = A(t)(1 + ) t, t = ln ( A(t)/A(0) ) / ln(1 + ), and = ( A(t)/A(0) ) 1/t 1. For nstance, fnd the tme t takes for an ntal nvestment of $100 to grow to $150 under the compound annual rate of 5%. Soluton: t = ln(150/100)/ ln(1.05) = 8.31. In practce, transactons can take place any tme durng a year. If tme s measured n months, t s common to express t n the form t = m/12 years. If tme s measured n days, use t = d/365 years. 1.1.3 Smple Interest Defnton Smple nterest s an nterest that s pad only on the prncpal amount. The accumulaton functon from tme 0 to tme t at annual smple nterest rate s a(t) = 1 + t where t s measured n years. Comparson of Compound and Smple Interests 2

a(t) (1 + ) t 1 + 1 + t 0 1 t From the pcture, the smple nterest accumulaton factor s larger than that for compound nterest f 0 t 1, and s smaller for t > 1. Interest accumulaton s often based on a combnaton of smple and compound nterests. Compound nterest would be appled over the completed number of years (that s, nterest compoundng perods), and smple nterest would be appled from then to the fractonal pont n the current nterest perod. For example, at annual rate 9% ( = 0.09), over a perod of 4 years and 5 months, an nvestment of $1,000 wll grow to $1, 000(1+) 4 (1+t) where t =5 months= 5/12 years, that s, t wll grow to $1, 000(1.09) 4( 1+(5/12)(0.09) ) = $1, 464.52 wth the accumulaton factor of 1.46452. Defnton A promssory note s a short-term contract requrng the ssuer of the note (the borrower) to pay the holder of the note (the lender) a prncpal amount plus nterest on that prncpal at a specfed annual nterest rate for a specfed length of tme. The length of tme s usually less than one year, and the nterest s calculated on the bass of smple nterest. The nterest rate earned by the lender s sometmes referred to as the yeld rate of the nvestment. Example 1.4 On January 31 Smth borrows $5,000 from Brown and gves Brown a promssory note. The note states that the loan wll be repad on Aprl 30 of the same year, wth annual nterest at 12%. On March 1 Brown sells the promssory note to Jones, who pays Brown amount of money X n return for the rght to collect the payment from Smth on Aprl 30. Jones pays Brown an amount such that Jones yeld from March 1 to the maturty date (Aprl 30) can be stated as an annual rate of nterest of 15%. Assume all calculatons are based on smple nterest and a 365-day year. t 1 = 29/365 t 2 = 60/365 January 31 March 1 Aprl 30 Smth borrows $5,000 from Brown Brown receves X from Jones Jones receves $5,146.30 from Smth 3

(a) Determne the amount Smth was to have pad Brown on Aprl 30. Soluton The payment requred at the maturty date s $5, 000(1+(0.12)(89/365)) = $5, 146.30. (b) Determne X, the amount Jones pad to Brown. Soluton The perod from March 1 to Aprl 30 s t 2 = 60/365. X satsfed the equaton $5, 146.30 = X(1 + (0.15)(60/365)). Thus, X=$5,022.46. (c) Determne the yeld rate Brown earned. Soluton Let t 1 = 29/365 denote the perod from January 31 to March 1, and let j 1 be the annual yeld rate earned by Brown. Then j 1 solves the equaton X = $5, 000(1+j 1 t 1 ) or j 1 = [( $5, 022.46/$5, 000 ) 1 ] /(29/365) = 0.0565 or 5.65%. 1.2 Present Value Defnton Let X be the amount that must be nvested at the start of a tme perod to accumulate $1 at the end of the perod at effectve annual nterest rate. Then X satsfes the equaton X( + ) = 1, or equvalently, X = 1/(1 + ). The quantty v = 1/(1 + ) s called the present value of an amount of $1 due n one tme perod (or present value factor or dscount factor). Example 1.5 A person nvests $X n a rsky fund, whch average annual rate of return s 19.5%. What amount must the person nvest n order to accumulate $1,000,000 n 25 years? Soluton The amount X solves the equaton X(1 + ) 25 = $1, 000, 000. Thus, X = $1, 000, 000 v 25 = $1, 000, 000(1.195) 25 = $11, 635.96. 1.3 Equaton of Value Defnton Any fnancal transacton can be represented n the form of an equaton. The components of the equaton are the nterest rate and the amounts of money dsbursed and receved. These amounts are termed dated cash flows (nflow and outflow). The equaton s called an equaton of value for the transacton, because t balances the accumulated and present values of the payments made at dfferent tme ponts. In order to formulate an equaton of value for a transacton, t s necessary to choose a reference tme pont called valuaton date. At the reference tme pont, the equaton of value balances (1) the accumulated value of all payments already dsbursed plus the present value of all payments yet to be 4

dsbursed, and (2) the accumulated value of all payments already receved plus the present value of all payments yet to be receved. Example 1.7 On February 7,14,21 and 28, a person draws $1,000 credt that charges an effectve weekly nterest rate of 8% on all credt extended. He pays hs debt n four nstallments whch he makes on March 7, 14,21, and 28. He pays $1,100 on each March 7, 14, and 21. How much must he pay on March 28 to repay hs debt completely? Receved Pad Feb 7 Feb 14 Feb 21 Feb 28 Mar 7 Mar 14 Mar 21 Mar 28 $1, 000 $1, 000 $1, 000 $1, 000 $1, 100 $1, 100 $1, 000 X Soluton 1 (straghtforward) Walt s debt on February 7 s $1,000; on February 14, $1, 000(1 + ) + $1, 000; on February 21, ( $1, 000(1 + ) + $1, 000 ) (1 + ) + $1, 000 = $1, 000 ( (1 + ) 2 + (1 + ) + 1 ) ; on February 28, $1, 000 ( (1+) 2 +(1+)+1 ) (1+)+$1, 000 = $1, 000 ( (1+) 3 +(1+) 2 +(1+ ) + 1 ). On March 7, the debt s $, 1000 ( (1 + ) 3 + (1 + ) 2 + (1 + ) + 1 ) (1 + ) $1, 100 = $, 1000 ( (1+) 4 +(1+) 3 +(1+) 2 +(1+) ) ( $1, 100; on March 14, $, 1000 ( (1+) 4 +(1+) 3 +(1+) 2 +(1+) ) ) $1, 100 (1+) $1, 100 = $, 1000 ( (1 + ) 5 + (1 + ) 4 + (1 + ) 3 + (1 + ) 2) $1, 100 ( (1 + ) + 1 ) ( ; on March 21, $, 1000 ( (1 + ) 5 + (1 + ) 4 + (1 + ) 3 + (1 + ) 2) $1, 100 ( (1 + ) + 1 )) (1 + ) $1, 100 = $, 1000 ( (1 + ) 6 + (1 + ) 5 + (1 + ) 4 + (1 + ) 3) $1, 100 ( (1 + ) 2 + (1 + ) + 1 ) ( ; on March 28; X = $, 1000 ( (1 + ) 6 + (1 + ) 5 + (1 + ) 4 + (1 + ) 3) $1, 100 ( (1 + ) 2 + (1 + ) + 1 )) (1 + ) = $, 1000 ( (1 + ) 7 +(1+) 6 +(1+) 5 +(1+) 4) $1, 100 ( (1+) 3 +(1+) 2 +(1+) ) = $2, 273.79. Soluton 2 Choose March 28 as the reference tme pont for valuaton. Then the accumulated value of all payments receved on February 7, 14, 21, 28 are $1, 000(1 + ) 7, $1, 000(1 + ) 6, $1, 000(1 + ) 5, and $1, 000(1 + ) 4, respectvely. The accumulated value of all payments pad on March 7, 14, 21, and 28 are, respectvely, $1, 100(1+) 3, $1, 100(1+) 2, $1, 100(1+), and X. The equaton of value s the dfference between the accumulated values of all the sums receved and all the sums pad, that s, the equaton of value s ] [ ] $1, 000 [(1+) 7 +(1+) 6 +(1+) 5 +(1+) 4 $1, 100 (1+) 3 +(1+) 2 +(1+) X = 0. From here X = $2, 273.79. 5

Soluton 3 Choose February 7 as the reference tme pont. Then the present value of the sums that wll be receved on February 7, 14, 21, 28 are $1, 000, $1, 000/(1 + ) = $1, 000v, $1, 000/(1 + ) 2 = $1, 000v 2, and $1, 000/(1 + ) 3 = $1, 000v 3, respectvely, where v s the present value factor. The present value of the sums that wll be pad on March 7, 14, 21, and 28 are, respectvely, $1, 100/(1 + ) 4 = $1, 100v 4, $1, 100/(1 + ) 5 = $1, 100v 5, $1, 100/(1 + ) 6 = $1, 100v 6, and Xv 7. The equaton of value s the dfference between the present values of all the sums that wll be receved and all the sums that wll be pad, that s, the equaton of value s $1, 000 [ 1 + v + v 2 + v 3] $1, 100 [ v 4 + v 5 + v 6] Xv 7 = 0. From here X = $2, 273.79. Soluton 4 Choose February 21 as the reference tme pont. Then the accumulated value of the payments receved on February 7 and 14 are $1, 000(1+ ) 2 and $1, 000(1 + ), respectvely. The present value of the payments that wll be receved on February 21 and 28 are $1, 000 and $1, 000/(1+), respectvely. The present value of the amounts that wll be pad on March 7, 14, 21, and 28 are, respectvely, $1, 100/(1+) 2, $1, 100/(1+) 3, $1, 100/(1+) 4, and X/(1 + ) 5. The equaton of value s [ $1, 000 (1+) 2 +(1+)+1+ 1 (1 + ) From here X = $2, 273.79. 1.4 Nomnal Rates of Interest ] $1, 100 [ 1 (1 + ) + 1 2 (1 + ) + 1 ] 3 (1 + ) 4 Defnton A nomnal annual nterest rate (m) compounded m tmes per year refers to an nterest rate (m) /m compounded every 1/mth of a year. Example 1.8 A credt card account charges 24% on unpad balances, payable monthly. The quoted rate of 24% s a nomnal annual nterest rate. A person puts $1,000 on hs credt card, and doesn t pay for one year. (a) How much does he owe accordng to hs 13th statement? (b) Compute an effectve annual nterest rate. Soluton (a) The monthly nterest rate s (12) = 24/12 = 2%. Hs 1st statement s for $1,000 and doesn t charge nterest. Hs 13th statement s for $1, 000(1.02) 12 = $1, 268.23. (b) The effectve annual rate of nterest s 26.823%. Defnton A nomnal annual nterest rate s sometmes called an annual percentage rate (APR). The correspondng monthly nterest rate s called 6 X (1 + ) = 0. 5

a perodc rate. Example 1.9 Bank A offers an annual nterest rate of 15.25% compounded semannually, and Bank B offers an annual rate of 15% compounded monthly. Whch bank offers a hgher effectve annual nterest rate? Soluton The perodc rate n Bank A s (2) A = 15.25%/2 = 7.625%. Thus, a depost of $1 n Bank A wll grow to (1 + 0.07625) 2 = 1.158314. The perodc rate n Bank B s (12) B = 15%/12 = 1.25%. A depost of $1 wll grow to (1.0125) 12 = 1.160755. Hence, the effectve annual nterest rates n Bank A and B are 15.8314% and 16.0755%, respectvely. Bank B has a hgher effectve annual nterest rate. The formula that connects the nomnal annual nterest rate (m) and the effectve annual nterest rate s ( (m) ) m 1 + = 1 +. m Example 1.10 Suppose an effectve annual rate of nterest s 12%. Fnd the equvalent nomnal annual rate for m = 1, 12, 365, and. Soluton Usng the formula (m) = m [ (1+) 1/m 1 ], we compute (1) = = 0.12, (12) = 0.1139, (365) = 0.113346. When m, m [ (1 + ) 1/m 1 ] ln(1 + ) = 0.113329. The lmtng case m s called contnuous compoundng. 1.5 Effectve and Nomnal Rates of Dscount Defnton An nterest payable at the end of an nterest perod s called nterest payable n arrears. Defnton An nterest payable at the begnnng of an nterest perod s called nterest payable n advance. Defnton The rate of nterest payable n advance s called the rate of dscount. Example. A person borrows $1,000 for one year at an nterest rate of 10% payable n advance. It means that he receves the loan amount of $1,000 but must mmedately pay the lender $100. Thus, the net loan amount s $900. One year later he must repay $1,000. The effectve annual rate of nterest of ths transacton s 100/900=0.1111 or 11.11%. Defnton The effectve annual rate of dscount from tme 0 to tme 1 s A(1) A(0) d = A(1) 7

where A(t) s the accumulated amount functon. Example In our example, A(1) = 1, 000, A(0) = 900, d = (1000 900)/1000 = 100/1000 = 0.10 or 10%. Recall that the effectve annual rate of nterest satsfes the equalty A(1) = A(0)(1 + ). Therefore, A(1) A(0) =. A(0) From here, the equaltes hold d = 1 + and = d 1 d. Example In our example, = 0.1/(1 0.1) = 1/9 = 0.1111, and d = (1/9)/(1 + 1/9) = 1/10 = 0.10. Defnton A nomnal annual dscount rate d (m) compounded m tmes per year refers to a dscount rate d (m) /m compounded every 1/mth of a year. The relaton between d and d (m) s Proof: ) 1 d = (1 d(m) m. m By defnton, A(0) = A(1)(1 d). Hence, A(0) = A(1/m)(1 d (m) /m) = A(2/m)(1 d (m) /m) 2 =... = A(1)(1 d (m) /m) m, whch proves the formula. Example 1.12 Suppose an effectve annual rate of nterest s 12%. Fnd the equvalent nomnal annual rate of dscount for m = 1, 12, 365, and. Soluton Frst compute the effectve annual rate of dscount d = 1 + = 0.12 1.12 = 0.107143. Usng the formula d (m) = m [ 1 (1 d) 1/m], we compute d (1) = d = 0.107143, d (12) = 0.1128, d (365) = 0.1133. When m, m [ 1 (1 d) 1/m] ln(1 d) = 0.113329 = ln(1 + ). Show that s t always so: ln(1 d) = ln(1 + ). 1.6 The Force of Interest 8

The nterest rate by the nvestment for the 1/mth of a year perod from tme t to tme t + 1/m s A(t + 1/m) A(t). A(t) The nomnal annual rate of nterest compounded m tmes per year s (m) = m A(t + 1/m) A(t) A(t) Lettng m, we get the nomnal annual nterest rate compounded contnuously, ( ) A(t + 1/m) A(t) = lm m m A(t). = A (t) A(t). Example 1.13 From Example 1.10, ( ) = ln(1 + ). Indeed, A(t) = A(0)(1 + ) t, thus, A (t) = A(0)(1 + ) t ln(1 + ) = A(t) ln(1 + ). Hence, ( ) = A (t)/a(t) = ln(1 + ). Defnton The force of nterest at tme t s δ t = A (t) A(t). To express the accumulated amount functon A(t) n terms of the force of nterest, solve the dfferental equaton δ t = A (t) A(t) = d dt ln[a(t)]. The soluton s A(t) = A(0) exp { t 0 δ u du }. Example 1.14 Gven δ t = 0.08 + 0.005t, fnd an accumulated value over fve years of an nvestment of $1,000 made at tme (a) t=0, (b) t=2. Soluton (a) A(0) = $1, 000, and A(5) = A(0) exp { 5 δ 0 u du } = 1000 exp { 5 (0.08+ 0 0.005u)du } = 1000 exp{0.4625} = 1588.04. (b) A(2) = 1000, and A(7) = A(2) exp { 7 2 (0.08+0.005u)du} = 1669.46. If the force of nterest s constant, δ t = δ, then A(t) = A(0) exp { t 0 δ u du } = exp{δ t}. On the other hand, we know that A(t) = A(0)(1+) t for a constant effectve annual nterest rate. Hence, δ = ln(1 + ) or = e δ 1. 9

1.7 Inflaton and the Real Rate of Interest Defnton Inflaton s a rse n the general level of prces of goods and servces n an economy over a perod of tme. Defnton A chef measure of prce nflaton s the nflaton rate, the percentage change n the Consumer Prce Index (CPI) over tme, generally quoted on an annual bass. The change n the CPI measures the effectve annual rate of change n the cost of a specfed basket of consumer goods and servces. Defnton (Explanaton s n Example 1.16). Wth effectve annual nterest rate and annual nflaton rate r, the real rate of nterest for the year s real = value of real return n the end of the year dollars value of nvested amount n theend of the year dollars = r 1 + r. Example 1.16 A person nvests $1,000 for one year at effectve annual nterest rate of 15.5%. At the tme Smth makes the nvestment, the cost of a certan consumer tem s $1. One year later, the nterest s pad and the prncpal returned. Assume that the nflaton s 10%. (a) What s the annual growth rate n purchasng power (the number of tems that can purchased) wth respect to the consumer tem? Soluton (a) At the start of the year, Smth can buy 1,000 tems. At the end of the year, he receves $1, 000(1.155) = $1, 155. Due to nflaton, the cost of the tem at the end of the year s $1.10. Smth s able to buy $1,155.00/$1.10=1,050 tems. Thus, hs purchasng power has grown by (1050-1000)/1000x100%=5%. (b) Compute the real rate of nterest. Soluton real = (0.155 0.1)/(1 + 0.1) = 0.055/1.1 = 0.05 or 5%. Note that the real nterest rate can be defned as the rate of change n purchasng power. Indeed, the general formula for the rate of change n purchasng power s A(0)(1+) P (1+r) A(0) P A(0) P = 1 + 1 + r 1 = r 1 + r. 2.1 Level Payment Annutes Defnton An annuty s a seres of perodc payments. 10

Defnton An annuty-certan s a seres of payments that are defntely made. They are not contngent upon occurrence of any random event. Defnton An annuty-mmedate s a seres of payments made at the end of each perod. Defnton An annuty-due s a seres of payments made at the begnnng of each perod. For now we wll study annuty-mmedate. Example 2.1 A person deposts $30 n a bank account on the last day of each month. The annual nterest rate s 9% compounded monthly, and the nterest s pad nto the account on the last day of each month. Fnd the account balance after the 140th depost. Soluton Let j = (12) = 0.09/12 = 0.0075 or 0.75%. The account balance after the 1st depost s $30; after the 2nd one, $30(1 + j) + $30 = $30 [ (1 + j)+1 ] ; after the 3rd, $30 [ (1+j)+1 ] (1+j)+$30 = $30 [ (1+j) 2 +(1+j)+1 ]. After the 140th depost, the balance s [ ] [ (1 + j) $30 (1+j) 139 +(1+j) 138 140 1 ] +...+(1+j)+1 = $30 = $7, 385.91. (1 + j) 1 Defnton The accumulated value of an n-payment annuty-mmedate of $1 per perod s s n = (1 + )n 1 where s the constant nterest rate per payment perod. Example In our example, s 140 0.0075 = $7, 385.91/30 = $246.197. Example After the n payments are completed, the balance s n, f not wthdrawn, wll contnue to accumulate wth nterest only. After k tme perods, the balance wll be s n (1 + ) k = (1 + )n 1 (1 + ) k = (1 + )n+k (1 + ) k = (1 + )n+k 1 (1 + )k 1 = s n+k s k. Useful formula: s n+k = s n (1 + ) k + s k. Note that n and k are swtched n ths formula n the textbook. Extensons of ths formula prove to be useful. 11

Example 2.4 (Annuty accumulaton wth non-level nterest rate) Note that non-level means non-constant. Suppose n Example 2.1, after 68 months, the nomnal nterest rate changes to 7.5% stll compounded monthly. Fnd the balance after 140 months. Soluton After 68 months, the balance s $30 s 68 0.0075 = $2, 648.50. Durng the next 72 months, the balance accumulates to $2, 648.50(1 + 0.75/12) 72 = $2, 648.50(1.00625) 72 = $4, 147.86. The remanng deposts contnung for the last 72 months [ accumulate to $30 s 72 0.00625 = $2, 717.36. Thus, the total balance s $30 s 68 0.0075 (1.00625) 72 + s 72 0.00625 ] = $6, 865.22. Example 2.5 (Annuty accumulaton wth a changng payment) Suppose that 10 monthly payments of $50 each are followed by 14 monthly payments of $75 each. If the effectve monthly rate s 1%, what s the accumulated value the annuty at the tme of the fnal payment? Soluton At the tme of the fnal payment, 24 months later, the accumulated value of the frst 10 payments s $50 s 10 0.01 (1.01) 14 = $601.30. The value of the fnal 14 payments, also valued at tme 24 months s $75 s 14 0.01 = $1, 121.06. Thus, the total balance s $1,722.36. 2.1.2 Present Value of an Annuty Example 2.6 The 1st wthdrawal from an account wll take place one year from now. There wll be four yearly wthdrawals of $1,000 each. The account has an effectve annual nterest rate of 6%. Calculate an amount of a sngle depost today so that the account balance wll be reduced to 0 after the 4th wthdrawal. Soluton Suppose that the amount of the ntal depost s $X. The account balance after the 1st wthdrawal s $X(1.06) $1, 000. The balance after the 2nd wthdrawal s [ $X(1.06) $1, 000 ] (1.06) $1, 000 = $X(1.06) 2 $1, 000(1.06) $1, 000. The balance after the 3rd wthdrawal s (note a typo n the text) [ $X(1.06) 2 $1, 000(1.06) $1, 000 ] (1.06) $1, 000 = $X(1.06) 3 $1, 000(1.06) 2 $1, 000(1.06) $1, 000. The balance after the 4th wthdrawal s [ $X(1.06) 3 $1, 000(1.06) 2 $1, 000(1.06) $1, 000 ] (1.06) $1, 000 = $X(1.06) 4 $1, 000(1.06) 3 $1, 000(1.06) 2 $1, 000(1.06) $1, 000. Snce the fnal balance s zero, X solves $X(1.06) 4 = $1, 000(1.06) 3 + $1, 000(1.06) 2 + $1, 000(1.06) + $1, 000, or equvalently, $X = $1, 000 1.06 $1, 000 $1, 000 $1, 000 + + + (1.06) 2 (1.06) 3 (1.06) = $1, 000[ v + v 2 + v 3 + v 4] 4 12

= $1, 000 v [ 1 + v + v 2 + v 3] = $1, 000 v 1 v4 1 v = $1, 000 1 1 + 1 v 4 1 1 1+ = $1, 000 1 v4 = $1, 000 1 (1.06) 4 = $3, 465.11. 0.06 Defnton The present value of an n-payment annuty-mmedate of $1 per perod s a n = v + v 2 + + v n = 1 vn where s the constant nterest rate per payment perod. Example 2.7 (Loan repayment) A person takes a $12,000 loan and has to make monthly payments for 3 years, startng one month from now, wth a nomnal nterest rate of 12% compounded monthly. Fnd a monthly payment on ths loan and the total amount of payment. Soluton Let $P be a monthly payment. It solves $12, 000 = $P a 36 0.01. Hence, $12, 000 $12, 000 $P = = a 36 0.01 (1 (1.01) 36 )/0.01 = $398.57 and the total payment s $398.57(36) = $14, 348.52. Defnton A k-perod deferred, n-payment annuty of $1 per perod s a seres of n payments that start k + 1 tme perods from now. Example 2.8 Suppose n the prevous example, the 1st payment starts 9 months from now. Fnd monthly and total payments. Soluton A monthly payment $P satsfes Thus, $P = $12, 000 = $P [ v 9 + v 10 + + v 44] = $P v 8 a 36 0.01. $12, 000 8 $12, 000 = (1.01) = (1.01) 8 ($398.57) = $431.60, v 8 a 36 0.01 a 36 0.01 and the total payment s $431.60(36) = $15, 537.60.. From the relaton v k[ v + v 2 + + v n] = v k a n, we get v k a n = v k 1 vn = vk v n+k = 1 vn+k 1 vk = a n+k a k. A useful formula s a n+k = v k a n + a k. 13

Example A person takes a $12,000 loan and has to make monthly payments for the frst year at the nterest rate of 10% compounded monthly, and for the followng 2 years wth the nterest rate of 12% compounded monthly. The frst payment s due one month from now. Fnd a monthly payment on ths loan and the total amount of payment. Soluton Let $P be a monthly payment. The effectve monthly nterest rate for the 1st year s 1 = 0.10/12 = 0.0083, whle for the next two years, t s 2 = 0.12/12 = 0.01. The amount of monthly payment $P solves the equaton $12, 000 = $P [ ] a 12 1 + (1 + 1 ) 12 a 24 2 [ 1 (1 + 1 ) 12 = $P + (1 + 1 ) 12 1 (1 + 2) 24 ] = (30.6043)$P, 1 2 Thus, $P = $392.10 and the total amount pad s ($392.10)(36) = $14, 115.60. 2.1.2.3. Relatonshp Between a n and s n. Recall that s n s the accumulated value of an annuty-mmedate at the tme of the last nth payment. The quantty a n s the present value of an n-payment annuty-mmedate one perod before the 1st payment. Thus, the valuaton pont of the present value s n perods earler than that for the accumulated value. Therefore, or equvalently, s n = (1 + ) n a n, a n = v n s n. Ths equalty can be verfed algebracally by observng that v n s n = 1 [ (1 + ) n 1 ] = 1 vn (1 + ) n 2.1.2.4 Valuaton of Perpetutes = a n. Note that lm a 1 v n n = lm = 1 n n. Ths expresson can also be derved by summng the nfnte seres of present values of payments v + v 2 + v 3 + + = v(1 + v + v 2 1 + ) = v 1 v = 1 1 + Defnton 1 1 1 1+ = 1. An annuty that has no defnte end s called a perpetuty. Notaton The present value of a perpetuty s denoted by a = 1/. 14

Example 2.10 A person deposts $10,000 n a bank account that offers an effectve annual nterest rate of 8% (e.g., buys a certfed depost). In one year, he wthdraws $800 of nterest. The prncpal amount of $10,000 remans n the account and generates $800 the followng year. If the person keeps wthdrawng only the nterest, the process wll go forever. Ths s what s called lves off the nterest. There s another way to derve the expresson for a. Let $X denote the amount that has to be nvested to generate $1 nterest per tme perod. Then $X solves $X = $1, or X = 1/. 2.1.3 Annuty-Immedate and Annuty-Due Recall n case of an annuty-mmedate payments are made at the end of valuaton perods, whle an annuty-due payment s made at the begnnng of each perod. Defnton The present value of an n-payment annuty-due at the tme of the 1st payment s ä n = 1 + v + v 2 + + v n 1 = 1 vn 1 v = 1 vn d where d = /(1 + ) = 1 1/(1 + ) = 1 v s the rate of dscount. The accumulated value one perod after the fnal payment s [ (1 + ) s n = (1 + ) + (1 + ) 2 + + (1 + ) n n 1 = (1 + ) Note the relatonshps ä n = (1 + ) a n and s n = (1 + ) s n. = (1 + )n 1 d Example 2.11 A person makes monthly deposts of $200 nto a fund earnng 6% annual nterest compounded monthly. The frst depost occurred on January 1, 1995. Compute the accumulated balance on December 31, 2009. Soluton There are 180 deposts. The monthly nterest rate s 0.06/12=0.005 or 0.5%. The accumulated value s. $200 s 180 0.005 = $200(1.005) (1.005)180 1 0.005 = $58, 454.56. 3.1 Amortzaton Method of Loan Payment Defnton A loan repayment nvolves payng down the prncpal amount of the loan as well as the accumulatng nterest on the loan. 15

Defnton An amortzng loan s a loan where the prncpal s pad down over the lfe of the loan, accordng to some amortzaton schedule. The orgn of the word amortze s Latn mort = death. Example Consder a loan of amount $1,000 wth an nterest rate of 10% per year. Suppose there s a payment of $200 at the end of the 1st year, a payment of $500 at the end of the 2nd year, and a fnal payment of $X at the end of the 3rd year. At the end of the 1st year, the amount owed, ncludng nterest, s $1000(1.1)=$1,100. After a payment of $200, the outstandng balance s $900. At the end of the 2nd year, the amount owed s $900(1.1)=$990. The payment of $500 reduces t to $490. At the end of the 3rd year, the amount owed s $490(1.1)=$539. Thus, the amount needed to repay the loan completely s $539. The amortzaton method requres that whenever a payment s made, nterest s pad frst, and any amount remanng s appled toward reducng the prncpal (goes towards prncpal). In partcular, a perodc loan payment cannot be less than accumulated nterest for the tme perod. Example Consder our example from the pont of vew of nterest and prncpal. The amortzaton of the loan can be summarzed n an amortzaton schedule. Tme Payment Interest Prncpal Outstandng (n years) Amount Due Repad Balance 0 $1,000 1 $200 $1,000(0.1)=$100 $200-$100=$100 $1000-$100=$900 2 $500 $900(0.1)=$90 $500=$90=$410 $900-$410=$490 3 $539 $490(0.1)=$49 $539-$49=$490 $490-$490=$0 Note that the outstandng balance can be updated from one year to the next accordng to the equaltes: $1, 000(1.1) $200 = $900, $900(1.1) $500 = $490, $490(1.1) $539 = $0. Now multply the frst equalty by (1.1) 2 and the second one by (1.1) and movng all terms to the left-hand sde, we get $1, 000(1.1) 3 $200(1.1) 2 $900(1.1) 2 = $0, $900(1.1) 2 $500(1.1) $490(1.1) = $0, $490(1.1) $539 = $0. Now addng the equaltes, we obtan $1, 000(1.1) 3 $200(1.1) 2 $500(1.1) $539 = $0, or equvalently, $1, 000 = $200v +$500v 2 +$539v 3. Hence, the orgnal loan amount s equal to the present value of the loan payments (v s the present value factor or dscount factor). 16

Note also that the total amount pad s 200 + $500 + $539 = $1239, whch breaks nto the orgnal loan amount of $1,000 and the nterest of $239. Indeed, as seen from the table, the nterest pad s $100+$90+$49=$239. Defnton An amortzed loan of amount L made at tme 0 wth the dscount factor v to be repad n n payments of amounts K 1, K 2,..., K n s based on the equaton L = K 1 v + K 2 v 2 + + K n v n. Notaton At the tme t, K t s the payment amount, OB t s the outstandng balance after the payment s made, I t s the nterest on outstandng balance snce the prevous payment was made, P R t s the part of the payment K t that s appled toward repayng loan prncpal. The amortzaton schedule n general form s Tme Payment Interest Prncpal Outstandng (n years) Amount Due Repad Balance 0 L = OB 0 1 K 1 I 1 = OB 0 P R 1 = K 1 I 1 OB 1 = OB 0 P R 1 n K n I n = OB n 1 P R n = K n I n OB n = OB n 1 P R n Example 3.1 A loan of amount $1,000 at a nomnal annual nterest rate of 12% compounded monthly s repad by 6 monthly payments, startng one month after the loan s made. The frst three payments are $X each and the fnal three payments are $2X each. Construct the amortzaton schedule for ths loan. Soluton The monthly nterest rate s = 0.12/12 = 0.01 or 1%. The dscount factor s v = 1/(1+) = 1/(1.01). The amount X solves the equaton $1, 000 = $X(v + v 2 + v 3 ) + $2X(v 4 + v 5 + v 6 ) = $Xa 3 0.01 + $2Xv 3 a 3 0.01. From here $X = $115.61 and so $2X = $231.21 to the nearest cent. The 17

amortzaton schedule s as follows: Tme Payment Interest Prncpal Outstandng (n years) Amount Due Repad Balance 0 L = OB 0 = $1, 000 1 K 1 I 1 = OB 0 P R 1 = K 1 I 1 OB 1 = OB 0 P R 1 = $115.61 = $10 = $105.61 = $894.39 2 K 2 I 2 = OB 1 P R 2 = K 2 I 2 = OB 2 = OB 1 P R 2 = = $115.61 = $8.94 $106.67 $787.72 3 K 3 I 3 = OB 2 P R 3 = K 3 I 3 OB 3 = OB 2 P R 3 = $115.61 = $7.88 = $107.73 = $679.99 4 K 4 I 4 = OB 3 P R 4 = K 4 I 4 OB 4 = OB 3 P R 4 = $231.21 = $6.80 = $224.41 = $455.58 5 K 5 I 5 = OB 4 P R 5 = K 5 I 5 OB 4 = OB 4 P R 5 = $231.21 = $4.56 = $226.65 = $228.93 6 K 6 I 6 = OB 5 P R 6 = K 6 I 6 = OB 4 = OB 5 P R 6 = $231.21 = $2.29 $228.92 = $0 The total amount pad on the loan s $1,040.46, of whch $40.47 s the nterest payment and $1,000 s the prncpal payment. 3.2 Amortzaton of a Loan wth Level Payments If a loan s repad wth level payments, the amortzaton schedule has a systematc form. Suppose the payment s of amount 1, that s, K 1 = K 2 = = K = 1. The loan amount s the present value of the payments, L = OB 0 = a n. The outstandng balance just after the tth payment s the orgnal loan amount valued at tme t mnus the accumulated value of the frst t payments OB t = L(1 + ) t s t = a n (1 + ) t s t. Ths s the retrospectve form of the outstandng balance. To obtan a prospectve form, recall that a n = (1 (1 + ) n )/ and s t = ((1 + ) t 1)/. Therefore, OB t = a n (1 + ) t s t = 1 (1 + ) n (1 + ) t (1 + )t 1 = (1 + )t (1 + ) (n t) (1 + )t 1 1 (1 + ) (n t) = = a n t, whch s the present value at tme t of the remanng n t payments. The nterest pad at tme t s I t = OB t 1 = a n t+1 = 1 v n t+1. The prncpal repad at tme t s P R t = K t I t = 1 (1 v n t+1 ) = v n t+1. The outstandng balance at tme t s OB t = OB t 1 P R t = a n t+1 v n t+1 = 1 (1 + ) (n t+1) (1 + ) (n t+1) = 1 (1 + )(1 + ) (n t+1) 18

1 (1 + ) (n t) = The amortzaton schedule s as follows. = a n t. Tme Payment Interest Prncpal Outstandng (n years) Amount Due Repad Balance 0 L = OB 0 = a n 1 K 1 = 1 I 1 = OB 0 P R 1 = K 1 I 1 OB 1 = OB 0 P R 1 = a n = 1 v n = v n = a n v n = a n 1 2 K 2 = 1 I 2 = OB 1 P R 2 = K 2 I 2 OB 2 = OB 1 P R 2 = a n 1 = 1 v n 1 = v n 1 = a n 1 v n 1 = a n 2 n K n = 1 I n = OB n 1 P R n = K n I n OB n = OB n 1 P R n = a 1 = 1 v = v = a 1 v = 0 The total nterest pad s I total = (1 v n )+(1 v n 1 )+ +(1 v) = n a n. The total amount pad s K total = K 1 +K 2 + +K n = n. The total prncpal repad s K total I total = a n = L. Example 3.5 A home buyer borrows $250,000 for 30-year perod wth level monthly payments begnnng one month after the loan s made. The nterest rate s 9% compounded monthly. Fnd the amortzaton schedule. Soluton The monthly nterest rate s 0.09/12=0.0075 or 0.75%. The monthly payment s K that satsfes $Ka 360 0.0075 = $250, 000. Hence, K = $250, 000/ [ (1 (1.0075) 360 )/0.0075 ] = $2, 011.56. The amortzaton schedule s (see page 189, show n Excel) Tme Payment Interest Prncpal Outstandng (n years) Amount Due Repad Balance 0 $250,000 1 $2, 011.56 ($250, 000)(0.0075) $2,011.56-$1,875.00 $250,000-$2,011.56 $1,875.00 =$136.56 $249,863.44 360 $2,011.56 $14.97 $1,996.58 0 3.3 The Snkng-fund Method of Loan Repayment Consder the case of a loan whch calls for perodc payments of nterest only durng the term of the loan, and a repayment of the prncpal amount at the end of the term. For such a loan, the borrower has to make n payments of nterest of amount L, along wth a payment of L at tme n. The borrower may wsh to accumulate the amount L durng the term of the loan by means of n perodc deposts nto an nterest-bearng savngs account called a snkng fund. The method of loan repayment s called the snkng-fund 19

method. In practce t s usually the case that the nterest rate charged by the lender of the loan,, s hgher than that of the snkng fund, j, that s, > j. The borrower s perodc payment to the lender s L and to the snkng fund, L/s n j. Thus, the total perodc outlay s L( + 1/s n j ). Example 3.6 A loan of $100,000 s to be repad by ten annual payments begnnng one year from now. The lender sets the annual rate of 10% and repayment of prncpal after 10 years. The borrower makes 10 annual deposts to a snkng fund earnng 8%. (a) Fnd the borrower s total annual outlay. Soluton The total annual outlay s L(+1/s n j ) = $100, 000(0.1+1/s 10 0.08 ) = $100, 000 ( 0.1 + 0.08/((1.08) 10 1) ) = $16, 902.95. (b) Fnd the level annual payment requred by the amortzaton method at 10%. Soluton K = $100, 000/a 10 0.1 = $16, 274.54. Note that the payment s lower than that for the snkng-fund method. (c) Fnd the rate of nterest for whch the amortzaton method results n the total annual outlay of $16,902.95. Soluton We need to fnd such that $100, 000 = $16, 902.95 a 10, whch results n = 0.1089. 3.3.1 Snkng-Fund Method Schedule The level annual depost nto the snkng fund s L/s n j. The accumulated value n the snkng fund after t deposts s ( L/s n j ) st j. The outstandng balance at tme t s the net amount of the debt, OB t = L ( L/s n j ) st j. The prncpal repad n the tth perod s [ st j s ] t 1 j P R t = OB t 1 OB t = L = s n j L(1 + j)t 1 s n j. The nterest pad every year to the lender s I = L. The nterest earned n the snkng-fund account at tme t s ( L/s n j ) st 1 j j. The net nterest s the dfference I t = L ( ) [ L/s n j st 1 j j = L (1 + j)t 1 1 ]. s n j [ ] Note that I t + P R t = L + 1 s n j s the borrower s total perodc outlay. Example 3.6(contnued...) L = 100, 000, = 0.1, j = 0.08, n = 10. The annual payment of nterest to the lender s $100, 000 (0.1) = $10, 000. The annual level depost to the snkng-fund account s $100, 000/s 10 0.08 = 20

$6, 902.95. The payment schedule s as follows Tme Net Interest Prncpal Outstandng (n years) Due Repad Balance 0 $100, 000 1 $10,000 $6,902.95 $100,000-$6,902.96 =$93,097.05 2 $10,000-$6,902.95(0.08) $6,902.95(1+0.08) $93,097.05-$7,455.19 =$9447.76 =$7,455.19 =$85,641.86 3 $8851.35 $8051.60 $ 77590.26 4 $8207.22 $ 8695.73 $68894.53 5 $7511.56 $9391.39 $59503.15 6 $6760.25 $10142.70 $ 49360.45 7 $5948.84 $10954.11 $ 38406.33 8 $5072.51 $11830.44 $ 26575.89 9 $4126.07 $12776.88 $ 13799.01 10 $3103.92 $13799.03 $ -0.02 4.1 Determnaton of Bond Prces Defnton A bond s a formal contract to repay borrowed money wth nterest at fxed ntervals. Government or a corporaton may ssue bonds to rase funds to cover planned expendtures. Defnton Bonds ssued by corporatons are usually backed by varous corporate assets as collateral. A junk bond s a bond that has lttle or no collateral. For example, a junk bond may be ssued to rase funds to fnance the takeover of another company. Defnton A coupon s the nterest rate that the ssuer pays to the bond holders. Defnton The maturty date s the date on whch the bond ssuer has to repay the prncpal amount. As long as all payments have been made, the ssuer has no more oblgaton to the bond holders after the maturty date. Another type of debt securty s a stock, but the major dfference between a stock and a bond s that stockholders have an equty stake n the company (.e., they are owners), whereas bondholders have a credtor stake n the company (.e., they are lenders). Another dfference s that bonds usually have a defned maturty date, whereas stocks may be outstandng ndefntely. Defnton The straght-term bond s a bond wth regular payments of nterest plus a sngle payment of prncpal at the end of the term. Ths type of bond s smlar to a loan. A bond specfes a face amount (prncpal), a 21

coupon rate (nterest rate), and a maturty date or term to maturty durng whch coupons are to be pad, and the redempton amount that s to be repad on the maturty date. Typcally the face amount and the redempton amount are the same. Let F denote the face value of a bond (also called par value), and let r denote the coupon rate per coupon perod (defned by the bond ssuer). Typcally coupons are pad semannually, so the coupon perod s 6 months (unless stated otherwse). Let C be the redempton amount on the bond. It s typcally equal to F, unless stated otherwse). Denote by n the number of coupon perods untl the term of maturty ( or smply term) of the bond. We wll use j to denote the sx-month yeld rate determned by the market forces (for example, an nterest rate that a person buyng bonds can get somewhere else). Defnton The prce of a bond (or the purchase prce) s the total amount that a bondholder wll receve from the bond ssuer. The present value of the prce P of the bond on the ssue date s calculated as 1 P = C (1 + j) + F r[ 1 n 1 + j + 1 (1 + j) + + 1 ] = C v n + F ra 2 (1 + j) n n j where v = (1 + j) 1 s the present value factor. It s usually the case that C = F, therefore, the bond prce can be expressed as P = F v n + F ra n j. Usng the defnton of the present value of an n-payment annuty-mmedate v n = 1 j a n j, we see that or, alternatvely, P = F (1 ja n j ) + F ra n j = F [ 1 + (r j)a n j ], P = F v n + F r 1 vn j = K + r (F K) j where K = Cv n s the present value of the redempton amount. Ths formula s known as Makeham s Formula. Defnton If r > j, then P > F and the bond s sad to be bought at a premum. If r = j, then P = F and the bond s sad to be bought at par. If r < j, then P < F and the bond s sad to be bought at a dscount. Example 4.1 A 10% bond wth semannual coupons has a face amount of $1,000 and was ssued on June 18, 1994. The frst coupon was pad on December 18, 1994, and the bond has a maturty date on June 18, 2014. 22

(a) Fnd the prce of the bond on ts ssue date usng (2) = 0.05 or 5%. Soluton It s gven that F = C = $1, 000, r = 0.1/2 = 0.05, n = 40 months, and j = 0.05/2 = 0.025. Thus, P = F v n + F ra n j = $1, 000 (1 + 0.025) 40 1 (1 + 0.025) 40 +($1, 000)(0.05) = $1, 627.57. 0.025 Note that r > j, hence P > F. Ths means that the bondholder bought ths bond at a premum. (b) Fnd the prce of the bond on ts ssue date usng (2) = 0.10 or 10%. Soluton Now j = 0.1/2 = 0.05 = r, and thus, P = $1, 000 (1 + 0.05) 40 1 (1 + 0.05) 40 + ($1, 000)(0.05) 0.05 That s, P = F and the bond s bought at par. = $1, 000. (c) Fnd the prce of the bond on ts ssue date usng (2) = 0.15 or 15%. Soluton Now j = 0.15/0.075 > 0.05 = r. P = $1, 000 (1 + 0.075) 40 1 (1 + 0.075) 40 + ($1, 000)(0.05) 0.075 That s, P < F and the bond s bought at dscount. 4.1.3 Bond Prces Between Coupon Dates = $685.14. Suppose we wsh to fnd the purchase prce P t of a bond at tme t, where 0 t 1 coupon perod. The prce of the bond s found as the present value of all future payments (coupons and redempton). Defne P 0 to be the prce of the bond at the begnnng of the consdered coupon perod. Then P t = P 0 (1 + j) t. The purchase prce P t s also called the prce-plus-accrued of the bond. Defnton The market prce of a bond s equal to the prce-plus-accrued mnus the fracton of the coupon accrued to tme t, Market P rce t = P t tf r = P 0 (1 + j) t tf r. Note that the tme t s expressed as fracton of a coupon perod, that s, t = # of days snce last coupon pad # of days n the coupon perod. 23

Example 4.2 (Refer to Example 4.1) A 10% bond wth semannual coupons has a face amount of $1,000 and was ssued on June 18, 2004. The frst coupon was pad on December 18, 2004. Fnd the purchase prce and market prce on August 1, 2014, f (2) = 0.05. Soluton On the last coupon date, June 18, 2014, the purchase prce of the bond was P = P 0 = F v n + F ra n j = $1, 000 (1 + 0.025) 20 1 (1 + 0.025) 20 +($1, 000)(0.05) = $1, 389.73. 0.025 The number of days from June 18 to August 1 s 44, and the number of days n the coupon perod from June 18 to December 18 s 183. Hence, t = 44/183. The purchase prce of the coupon on August 1, 2014, s calculated as P t = P 0 (1 + j) t = $1, 389.73(1 + 0.025) 44/183 = $1, 398.00. The market prce s Market P rce t = $1, 398.00 (44/183)($1, 000)(0.05) = $1, 385.98. 4.2 Amortzaton of a Bond A bond can be vewed as a standard amortzed loan. The bondholder can be thought of as a lender and the ssuer of the bond as the borrower. The purchase prce P s the loan amount, and the coupon and redempton payments are the loan payments made by the borrower. The amortzaton schedule for a bond s Tme Payment Interest Prncpal Outstandng Amount Due Repad Balance 0 P = F [ 1 + (r j)a n j ] 1 F r P j F r P j P F (r j)v n = F [ j + (r j)(1 v n ) ] = F (r j)v n = F + F (r j)a n 1 j 2 F r F [ j + (r j)(1 v n 1 ) ] F (r j)v n 1 F + F (r j)a n 2 j n 1 F r F [ j + (r j)(1 v 2 ) ] F (r j)v 2 F + F (r j)a 1 j n F r + F F [ j + (r j)(1 v) ] F [ 1 + (r j)v ] 0 Example 4.4 A 10% bond wth face amount $10,000 matures 4 years after ssue. Construct the amortzaton schedule for the bond over ts term for nomnal annual yeld rate of 8%. 24

Soluton Gven r = 0.05, j = 0.04, F = 10, 000, and n = 8. The amortzaton schedule s Tme Payment Interest Prncpal Outstandng Amount Due Repad Balance 0 $10,673.27 1 $500 $426.93 $73.07 $10,600.21 2 $500 $424.01 $75.99 $10,524.21 3 $500 $420.97 $79.03 $10,445.18 4 $500 $417.81 $82.19 $10,362.99 5 $500 $414.52 $85.48 $10,277.51 6 $500 $411.10 $88.90 $10,188.61 7 $500 $407.54 $92.46 $10,096.15 8 $10,500 $403.85 $10,096.15 $0 5.1 Internal Rate of Return Defned A general fnancal transacton nvolves a number of cash nflows and outflows at varous ponts n tme. Defnton The nternal rate or return for a transacton s the nterest rate at whch the value of all cash nflows s equal to the value of cash outflows. Suppose that a transacton conssts of a sngle amount L nvested at tme 0, and several future payments K 1, K 2,..., K n to be receved at tmes 1, 2,..., n. Then the equaton of value s 1 L = K 1 1 + + K 1 2 (1 + ) + + K 1 2 n (1 + ), n and there s a unque soluton for, provded that L < n j=1 K j. It s possble to extend the noton of nternal rate of return to more complex transactons. Suppose that at tmes 0, 1,..., n, there are payments receved of amounts A 0, A 1,..., A n and payments dsbursed of amounts B 0, B 1,..., B n. The net amount receved at tme k s C k = A k B k. It can be postve or negatve. If there s no amount pad at tme k, then we assume A k = 0. If there s no dsbursement at tme k, then B k = 0. Defnton Suppose that a transacton has net cashflows of amounts C 0, C 1,..., C n at tmes 0, 1,..., n. The nternal rate of return for ths transacton s any rate of nterest satsfyng the equaton of value n k=0 C 1 k = 0. (1+) k Example A person buys 1,000 shares of stock at $5 per share and pays a commsson of 2%. Sx months later he receves a cash dvdend (a share of 25

profts receved by a stockholder) of $0.20 per share, whch he mmedately renvests commsson-free n shares at a prce of $4 per share. Sx months after that he buys another 500 shares at a $4.50 per share, and pays a commsson of 2%. Sx months later he receves another cash dvdend of $0.25 per share and sells hs exstng shares at $5 per share, agan payng 2% commsson. Fnd the nternal rate of return for the entre transacton n the form (2). Soluton At tme t 0 = 0, A 0 = 0 and B 0 = $5, 100. Sx months later, at tme t 1 = 1, A 1 = $200 and B 1 = $200, snce he receves and mmedately renvests the dvdend of $200 buyng addtonal 50 shares. At tme t 2 = 2, A 2 = 0 and B 2 = $2, 295 (buyng addtonal 500 shares and now ownng a total of 1,550 shares). At t 3 = 3, A 3 = $387.50 + $7, 595 = $7, 982.50 (the dvdend on 1,550 shares plus the proceeds from the sale of the shares after the commsson s pad), and B 3 = 0. The net amount receved are C 0 = $5, 100, C 1 = 0, C 2 = $2, 295, and C 3 = $7, 982.50. The rate of return (2) solves the equaton 1 5, 100 2, 295 (1 + (2) /2) + 7, 982.5 1 2 (1 + (2) /2) = 0. 3 It can be found numercally that (2) = 0.0649 or 6.49%. 5.1.4.1 Proftablty Index Defnton The proftablty ndex s a rato measurng the return per dollar of nvestment. It s I = present value of cash nflows present value of cash outflow. Example An nvestment of $1,000 can be made nto one of two projects. The frst project generates ncome of $250 per year for 5 years startng n one year, and the second project generates ncome of $140 per year for 10 years. The shareholder s requred return rate (called the cost of captal) s 5% per year for both projects. Compare the proftablty ndces for the two projects. Soluton For Project 1, I = $250 a 5 0.05 /$1, 000 = 1.0824. For Project 2, I = $140 a 10 0.05 /$1, 000 = 1.0810. Project 1 has a hgher proftablty ndex and thus should be preferable to Project 2. 5.1.4.2 Payback Perod Defnton Suppose an nvestment conssts of a seres of cash outflows C 0, C 1,..., C t < 0 followed by a seres of cash nflows C t+1, C t+2,..., C n > 0. The payback perod s the number of years requred to recover the orgnal 26

amount nvested, that s, t s the smallest k such that t C s s=0 k r=t+1 Example In our prevous example, for Project 1, C 0 = $1, 000, and C 1 = C 2 = C 3 = C 4 = $250, thus the payback perod s 4 years. For Project 2, the payback perod s just over 7 years, ($140)(7)=$980. C r. 5.2.1 Dollar-Weghted Rate of Return As a rule, the yeld rate (or the return rate) of an nvestment s reported on an annual bass. There are two standard methods for measurng the yeld rate: dollar-weghted and tme-weghted rate of return. Defnton Suppose the followng nformaton s known: () the balance n a fund at the start of a year s A, () for 0 < t 1 < t 2 < < t n < 1, the net depost at tme t k s amount C k, and () the balance n the fund at the end of the year s B. Then the net amount of nterest earned by the fund durng the year s I = B [ A + n C k ], and the dollar-weghed rate of return earned by the fund for the year s the rato [ n ] = I/ A + C k (1 t k ). k=1 The dollar-weghted rate of return solves the equaton of value that equates the fund balance at the end of a year wth all deposts mnus all wthdrawals, both accumulated to the end of the year wth smple nterest: B = A(1 + ) + C 1 (1 + (1 t 1 )) + C2(1 + (1 t 2 )) +... + C n (1 + (1 t n )). Example 5.3 A penson fund began a year wth a balance of $1,000,000. There were contrbutons to the fund of $200,000 at the end of February and agan at the end of August. There was a beneft of $500,000 pad out of the fund at the end of October. The balance remanng n the fund at the start of the next year was $1,100,000. Fnd the dollar-weghted annual rate of return earned by the fund. Soluton Let be the unknown rate of return. The equaton of value for the dollar-weghted return s $1, 000, 000(1 + ) + $200, 000(1 + 10/12 ) + $200, 000(1 + 4/12 ) 27 k=1

Solvng for, we get = $500, 000(1 + 2/12 ) = $1, 100, 000. $1, 100, 000 + $500, 000 $1, 000, 000 $200, 000 $200, 000 $1, 000, 000 + $200, 000(10/12) + $200, 000(4/12) $500, 000(2/12) = 0.1739. Note that f compound nterest were used, then the equaton of value would be $1, 000, 000(1 + ) + $200, 000(1 + ) 10/12 + $200, 000(1 + ) 4/12 $500, 000(1 + ) 2/12 = $1, 100, 000. The numerc soluton s = 0.1740 whch s very close to 0.1739. Note that snce Taylor s expanson of (1+) x s 1+x+o(x), the formula for smple nterest approxmates that for compound one and gves a good approxmaton. 5.2.2 Tme-weghted Rate of Return Defnton Suppose the followng nformaton s known: () the balance n a fund at the start of a year s A, () for 0 < t 1 < t 2 < < t n < 1, the net depost at tme t k s amount C k, () the value of the fund just before the net depost at tme t k s F k, and (v) the balance n the fund at the end of the year s B. Then the tme-weghed rate of return earned by the fund for the year s [ F1 A F 2 F 1 + C 1 F 3 F 2 + C 2 F k F k 1 + C k 1 B ] F k + C k 1. Note that n ths defnton, the tme length t k t k 1 of each pece of the year s rrelevant. The rato F j /(F j 1 + C j 1 ) s the growth factor for the perod from t j 1 to t j. The tme-weghted return s found by compoundng the successve growth factors over the course of the year. Example 5.4 The followng are the penson fund values after every transacton and at the end of the year. Fnd the tme-weghted rate of return. 28