FINANCIAL MATHEMATICS

3 LESSON FINANCIAL MATHEMATICS Annutes What s an annuty? The term annuty s used n fnancal mathematcs to refer to any termnatng sequence of regular fxed payments over a specfed perod of tme. Loans are usually pad off by an annuty. If payments are not at regular (rregular) perods, we are not workng wth an annuty. We get two types of annutes: The ordnary annuty Ths s an annuty whose payments are made at the end of each perod. (At the end of each week, month, half year, year, etc.) Payng back a car loan, a home loan The annuty due Ths s an annuty whose payments are made at the begnnng of each perod. Deposts n savngs, rent payments, and nsurance premums are examples of annutes due. On the tme lne: If we look at the tmelne, t s clear to see that f we are lookng at nvestng money nto an account, then we wll be workng wth the future value of these payments. Ths s so because we are savng up money for some use one day n the future. If we want to consder the present value of a seres of payments, then we wll be lookng at a scenaro where a loan s beng pad off. Ths s so because we get the money today, and pay that money wth nterest back to the fnancng company some tme n the future. Present Value works backwards Future Value works forwards T 4 T 5 T 6 T 7 T 8 x x x x x x x x Regular perodc payments x Future value of an annuty: When we calculate the future value of an annuty, t s mportant to realze that each of the regular payments s a present value that wll collect nterest durng the term / perod of the nvestment. The present values then become a sequence of future values when we move them forward towards the end of the tme lne. Collectvely, ther sum results nto a future value of the nvestment. The future value of the annuty thus conssts of the sum of each payment s future value, and forms a geometrc sequence of whch we determne the sum of the payments. Page 32

T 4 T 5 x x x x x 1 2 3 4 5 The x values are all present values whch need to go forward to become future values when the annuty matures. Ths gong forward occurs by addng nterest for the perod. Lets us see how ths works: In the dagram alongsde we see that 5 payments have been made nto an account at regular yearly ntervals. Each payment nto ths account happened at a dfferent tme, but they are all spaced at once a year. Each payment must therefore move to tmelne T 5 to become a future value. That means that when we pad n the Rx at the begnnng of each year, that Rx was a present value at the tme whch became a future value as t moved forwards on the tmelne. The sum of all these separate future values gves us the future value of the annuty. Note that the above scenaro was an annuty due as each payment was made at the begnnng of the perod. The formula: Let us nvest Rx monthly nto a savng account that pays r % nterest per annum compounded annually, for a total of n payments. The frst payment s made at the end of the month and the last payment at the end of n months. So ths s an ordnary annuty. The tmelne: T 4 T 5 x x x x x FV1 2 3 4 5 At : At : At : We pay n Rx at the end of the frst year Ths Rx has accrued one year s nterest to become a future value at compound nterest. So we get that here usng the compound ncrease formula = P v (1 + ) n : at = x(1 + ) 1. At the end of the second month we add another payment of Rx. So now the money n the bank has ncreased by addng nterest to the frst payment and then addng another Rx to the account to become F v at = x(1 + ) 1 + x. The moneys accumulated at now moves forward to by addng nterest to the whole amount: = at = [x(1 + ) 1 + x(1 + ) = x(1 + ) 2 + x(1 + ) We fnally add another payment so that ths amount becomes at = x(1 + ) 2 + x(1 + ) + x. Note that the earler two payments have nterest that s accumulatng as t moves forward. 2009 Page 1 Lesson 1 Algebra Page 33

At T 4 : At T 5 : The money from moves forward one perod and we then add the new payment to get: at T 4 = [x(1 + ) 2 + x(1 + ) + x(1 + ) + x = x(1 + ) 3 + x(1 + ) 2 + x(1 + ) + x The money from T 4 moves forward one perod and we then add the new payment to get: at T 5 = [x(1 + ) 3 + x(1 + ) 2 + x(1 + ) + x(1 + ) + x = x(1 + ) 4 + x(1 + ) 3 + x(1 + ) 2 + x(1 + ) + x If we want to do ths for 120 payments the process wll be very lengthy, so we need a shorter method that does the same thng. If we look at the sequence of terms we see that: = x + x(1 + ) + x(1 + ) 2 + x(1 + ) 3 + x(1 + ) 4 Ths s a geometrc sequence of values wth frst term x and constant rato (1 + ). Let s see how ths works: S n = _ a(rn 1) : a = x, r = (1 + ) and n = 5 r 1 \ S 5 = x((1 + )5 1) (1 + ) 1 = x [ (1 + )5 1 So the future value of these fve payments wll be: (1 + )5 1. x x x x x There are two mportant thngs to menton here. We are movng forward on the tmelne so the powers of the (1 + ) rato wll be postve as we are addng nterest to every payment. Secondly there s no perod open wth payments. So f we wsh to generalze ths formula, we can say that for the followng ordnary annuty for n perodc payments nto an account that pays an effectve r% per perod, the tmelne wll look as follows: T n 2 T n-1 T n And the future value wll be: (1 + )n 1 x x x x x x Page 34

T 4 T 5 x x x x x If we work wth an annuty due where the frst payment s made mmedately and the last payment s made at the begnnng of the last perod, then the stuaton looks rather dfferent snce the perod at the end nfluences the future value calculaton. The scenaro changes to the followng: At : At : At : At T 4 : At T 5 : at = x(1 + ) 1 + x at = x(1 + ) 2 + x(1 + ) + x at = x(1 + ) 3 + x(1 + ) 2 + x(1 + ) + x at T 4 = x(1 + ) 4 + x(1 + ) 3 + x(1 + ) 2 + x(1 + ) + x at T 5 = x(1 + ) 5 + x(1 + ) 4 + x(1 + ) 3 + x(1 + ) 2 + x(1 +) If we look at the sequence of terms we see that: = x(1 + ) + x(1 + ) 2 + x(1 + ) 3 + x(1 + ) 4 + x(1 + ) 5 We can remove a common factor of (1 + ) to get: = (1 + ) [x + x(1 + ) + x(1 + ) 2 + x(1 + ) 3 + x(1 + ) 4 T 4 T 5 F v x x x x x Future value for 5 payments One perod of nterest to take the payments to T 5 Now we already know that for the square brackets we have a formula whch says that S 5 (1 + )5 1. So we use ths to get (1 + )5 1 (1 + ). Ths would be the same as movng all the payments to a future value at T 4,, startng at, and then addng one more perod of nterest to take ths accumulated future value to the end of the fve year perod. Ths s very mportant to remember, because ths s exactly how the future value formula works. If you understand fully how to adjust the standard formula, then you won t have major dffcultes wth the future value formulae for fnance. Let us summarze the future value of an annuty by lookng at dfferent tmelnes and showng how the standard formula adjusts to accommodate the dfferent payment perods: For the ordnary annuty for n perodc payments nto an account that pays an effectve r% per perod, 2009 Page 1 Lesson 1 Algebra Page 35

the tmelne wll look as follows: T n 2 T n-1 T n The formula: (1 + )n 1 x x x x x x For the annuty due for n perodc payments nto an account that pays an effectve r% per perod, the tmelne wll look as follows: The formula: T n 2 T n-1 T n (1 + )n 1 (1 + ) x x x x x x Some worked examples: Example Example 1: An nvestment of R300 per month, wth the frst, of 30 payments, made n one month s tme, matures to Rx after three years. Interest s pad at a rate of 18% per annum, compounded monthly. Determne x. Effectve monthly nterest rate: _ 0,18 or 18_ 12 1 200 = 0,015. So: (1 + )n 1 \ = 300 ( (1,015)36 1 0,015 ) \ = 300(47,27596921) \ = R14 182,79 Example Example 2 How much money must be nvested monthly nto an ordnary annuty to realse R1 000 000 n 20 year s tme f the current rate of nvestment s calculated at an effectve 9% per annum compounded annually? (The payments stretch over the 20 year perod) 40 1 000 000 Page 36 x x x x We need an effectve monthly rate: (1 + _ 12) 12 = (1 + ) \ _ 12 12 = 12 1,09 _ 1 = 0,00720732331 = x ( (1 + )n 1 ) \ 1 000 000 = x ( (1,00720732331)240 1 0,00720732331 )

\ 1 000 000 = x(638,8517021) \ = R1 565, 31 Example 3 How long should an nvestor contnue to make monthly nvestments of R1 200 at a rate of 12% per annum compounded monthly f he wshes to have at least R200 000 n order to buy a car cash? Assume that hs frst payment s mmedately and that hs last payment s made on the day the nvestment matures. Example T n 200 000 1 200 1 200 1 200 1 200 1 200 Effectve monthly nterest rate: _ 0,12 or _ 12 12 1 200 = 0,01. Notce that the payments start mmedately and end on the last day. So the tmelne shows payments from to T n whch s n + 1 payments: (1 + )n + 1 1 \ 200 000 = 1 200 ( (1,01)n + 1 1 0,01 ) 200 000 \ (_ 1 200 0,01 ) + 1 = (1,01) n + 1 \ _ 8 3 = (1,01)n + 1 log 8 3 \ n + 1 = _ log 1,01 = 98,5725 \ n = 97,5725 \ n = 98 months Example 4 Manuel decdes to save a monthly amount of R250 for the next ten years. Hs bank offers hm an nterest rate of 8% p.a. compounded monthly for ths perod. After the ten years the bank wll offer hm 12% p.a. compounded monthly f he does not wthdraw the money for the followng three years. How much wll he have n the bank at the end of the 13 years? Example 20 56 250 250 250 250 F v Effectve monthly nterest rate: 1 = _ 0,08 or 8_ 12 1 200 = 0,00 6 and 2 = _ 0,12 or 12_ 12 1 200 = 0,01. Notce that the payments stop after ten years and the money then remans n the account to accumulate nterest for a further three years: = x ( (1 + 1 )120 1 )(1 + 2 ) 36 (1 + _ 0,08 12 \ = 250 [ ) 120 1 0,08 (1,01)36 12 \ = 250(182,9460352)(1,01) 36 \ = 45736,5088(1,01) 36 2009 Page 1 Lesson 1 Algebra Page 37

\ = R65 438, 37 Notce that the annuty matures the moment the payments end, and then ths value smply accumulated more nterest at a normal compound nterest rate. Actvty Actvty 1 1. Peter nvests R300 per month nto an account that pays 14% p.a. compounded monthly for a perod of fve years. Hs frst payment s mmedately, and hs last payment s on the day the nvestment matures. How much does he have n ths account at the end of 5 years? 2. Bongwe wants to save up some money so that she can gve her unborn daughter R16 000 on her 16 th brthday. On the day that her daughter s born, she starts makng equal monthly payments nto an account that pays 8% p.a. compounded monthly. Her last payment nto ths account s due one month before her daughter turns 16. 2.1 Calculate the sze of the monthly payment. Page 38

2.2 How much should Bongwe nvest monthly f she wants to gve her daughter R21 000 on her 21 st brthday, nstead of the R16 000 on her 16 th brthday? Her last payment nto ths account s due one month before her daugher turns 21. 3. Whch nvestment wll be the better opton: monthly deposts of R100 nto an account payng 12%p.a. compounded monthly or quarterly nvestments of R300 nto an account payng 4% per quarter. Both nvestments run for one year. 4. Sam makes monthly deposts of R200 nto an ordnary annuty for a perod of four years and earns nterest at 6%p.a. compounded monthly. At the end of each year, he deposts an addtonal R1 000 nto ths account. How much money wll Sam have n ths account at the end of four years? 2009 Page 1 Lesson 1 Algebra Page 39

Snkng Funds What s a snkng fund? A snkng fund s an nvestment that s made to replace expensve equpment / tems n a few years tme. It s used as a savngs account that wll accumulate funds over a perod of tme, whch wll enable the nvestor to purchase expensve tems or to fund expensve captal outlays n a few years tme. The workngs of the snkng fund problems are best explaned by lookng at a few worked problems. Example Soluton Example 1 Machnery s purchased at a cost of R550 000 and s expected to rse n cost at 15% per annum, compound nterest, and deprecate n value at a rate of 8% per annum compounded annually. A snkng fund s started to make provson for replacng the old machne. The snkng fund pays 16% per annum compounded monthly, and you make monthly payments nto ths account for 10 years, startng mmedately and endng one month before the purchase of the new machne. Determne: 1. the replacement cost of a new machne ten years from now 2. the scrap value of the machne n ten years tme. 3. the monthly payment nto the snkng fund that wll make provson for the replacement of the new machne. Soluton We need to separate the nformaton that s gven so that we do not mx up some rates and some perods. So: Current Machne : = R550 000 Apprecaton rate: r = 15%p.a. compounded annually Deprecaton rate: r = 8% p.a. compounded annually Term: n = 10 years Snkng Fund : Interest rate: r = 16% p.a. compounded annually Payment = Rx per month Term: n = 10 years Page 40 1. Snce the cost of a new machne apprecates at an effectve yearly rate of 15% p.a : = P(1 + ) n = 550 000(1,15) 10 F Ths future value s known as the V = R2 225 056,76 REPLACEMENT COST.

2. Snce the old machne deprecates at an effectve yearly rate of 8% p.a : = P(1 + ) n = 550 000(1 0,08) 10 = 550 000(0,92) 10 = R238 913,65 Ths deprecated value s also called the SCRAALUE. 3. The scrap value s always used as a part payment on the new machne. So f we take the replacement cost and we subtract the scrap value, we obtan the value of the snkng fund. So: Snkng fund = R2 225 056,76 R238 913,65 = R1 986 143,11 The Tmelne: 20 x x x x x 19 The effectve monthly rate: _ 0,16 12 = x ( (1 + )n 1 )(1 + ) or (1 + _ 0,16 12 1986143,11 = x [ ) 120 1 0,16 (1 + _ 0,16 12 ) 12 1986143,11 = x(296,4715095) x = R6 699,27 1 986 143,11 _ 16 1200 = 0,01 3 It s clear to see that there are a few steps that wll be standard n each problem that nvolves a snkng fund. Frstly we need to always calculate the replacement cost, and the scrap value of the machne. To fnd the value of the snkng fund, we need to determne the dfference between the replacement value and the scrap value as the old machne s always sold to defray costs of purchasng a new machne. We then set up a future value annuty to determne the amount of the monthly nstalments. Example 2 A compact dsc press s purchased for R 1,2 mllon and s expected to rse n cost at a rate of 9 % per annum compounded annually, whlst t wll deprecate at a rate of 7,5% per annum compounded annually. A snkng fund s set up to make provson for the replacement of the machne n ten years tme, and pays nterest at a rate of 9,2% per annum compounded monthly. 1. Determne the monthly amount that has to be nvested nto the snkng fund to realze enough money for a replacement machne n ten years tme. Payments start mmedately and end on the day that the replacement machne s purchased. 2. After fve years new technology n Compact Dscs are ntroduced to the market. Ths machne wll cost R 2 mllon. If you decde to replace your current machne mmedately, how much money wll you have to borrow to purchase the new equpment, f you use the snkng fund and the sales of the old machne toward payng for ths new machne? Example 2009 Page 1 Lesson 1 Algebra Page 41

Soluton Soluton Current Machne : = R1 200 000 Apprecaton rate: r = 9% p.a.c.a Deprecaton rate: r = 7,5% p.a.c.a Term: n = 10 years 1. Replacement cost: Scrap Value: = (1 + ) n = (1 ) n = 1 200 000(1,09) 10 = 1 200 000(1 ) n = R2 840 836,41 = R550 298,81 So the value of the snkng fund: R2 840 836,41 R550 298,81 = R2 290 537,60 The tmelne: 19 20 2 290 537,60 x x x x x x Notce that there are 121 payments ths tme round. So:. The effectve monthly rate: _ 0,092 9,2 or _ 12 1200 = 0,007 6 = x ( (1 + )n 1 ) 2290537,60 = x (. (1,007 6 ). 121 1) 0,0076 2290537,60 = x(198,21807) x = R11 555,64 2. We now need to move everythng to the fve year pont nstead of the ten year pont. Scrap Value: = (1 ) n =1 200 000(1 0,075) 5 = R812 624,50 The snkng fund at the end of fve years: = x ( (1 + )n 1 ) = 11 555,64 (. (1,007 6 ). 61 1) 0,0076 = 11 555,64(77,40269292) = R894 437,65 So we need to borrow R2 000 000 R894 437,65 R 812 624,50 = R292 937,85. Page 42

Actvty 2 1. A company purchases a bus for a prce of R x. It s expected to have a useful lfe of n years and a scrap value of 18% of ts orgnal purchase prce. If the annual deprecaton rate s estmated to be 21% p.a. on the reducng balance, determne the value of n. Actvty 2. A machne whch costs R120 000 s estmated to have a useful lfe of 10 years and then a scrap value of R45 000. Determne the annual deprecaton rate on: 2.1 the reducng balance bass. 2.2 on the straght lne bass. 3. A hydraulc lfter costs R550 000 and s expected to have a useful lfetme of 8 years. It deprecates at 10% p.a. on the reducng balance bass. The cost of a replacement lfter s expected to escalate at 18% p.a. effectve. A snkng fund s set up to fnance the replacement hydraulc lfter n 8 years tme. Fnd, at the tme of purchase of the new hydraulc lfter: 3.1 The scrap value of the old hydraulc lfter. 2009 Page 1 Lesson 1 Algebra Page 43

3.2 The expected cost of a new hydraulc lfter. 3.3 The value that the snkng fund must attan, f the scrap value of the old hydraulc lfter s used to defray expenses. 3.4 The value of the monthly nstallments that are made nto the snkng fund f payments start mmedately and end on the day of the replacement and the snkng fund earns nterest of 12% p.a. compounded monthly. 4. A prntng press currently costs R850 000. The value of the machne s expected to drop at a rate of 7% per annum smple nterest, whlst the cost of a replacement machne escalates at a rate of 14% p.a. compounded annually. The press s expected to have a useful lfetme of 8 years. 4.1 Calculate the scrap value of the old machne. Page 44

4.2 Calculate the cost of the replacement machne. 4.3 Calculate the amount needed to replace the old machne, f the scrap value s used as part of the payment for the new machne. 4.4 A snkng fund s set up to provde for ths balance, payng nterest at 15% p.a. compounded monthly. Determne the monthly amount that should be pad nto the snkng fund to realze ths. Payments start mmedately and end 6 months before replacement. 5. Mchelle s the proud owner of an nteror decoratng busness and needs to purchase machnery to restore antque furnture. The cost of ths equpment s R 1,2 mllon and t s estmated that the machnery wll deprecate n value at a rate of 9% per annum compounded annually. She decdes to nvest money nto a snkng fund, to make provson for the replacement of the equpment n fve year s tme. The snkng fund pays 14% per annum compounded quarterly. It s estmated that the replacement cost of the new equpment rses at a rate of 6% per annum compounded annually. 5.1 Determne the scrap value of the current equpment n fve years tme. 2009 Page 1 Lesson 1 Algebra Page 45

5.2 Determne the replacement value of the new machnery n fve years tme. 5.3 If the scrap value of the machne s used toward payng for the new machnery, determne the quarterly amount that must be pad nto the snkng fund to make provson for the replacement of the old machnery n fve years tme. The frst payment s made mmedately and the last payment at the end of fve years. 5.4 Three years after purchasng the current machne, new technology becomes avalable and Mchelle wants to replace the machnery mmedately. How much money wll she have avalable to do so f at ths pont n tme she has made 13 payments nto the snkng fund? Page 46

Present Value Annutes When we calculate the present value of an annuty, each of the regular payments are future (accumulated) values that are made up of captal and nterest added. These future values then become a sequence of present values when we move them backwards towards the begnnng of the tme lne. Collectvely, ther sum results nto a present value of the nvestment or loan- that s each payment wthout nterest. Each future value (payment) has thus moved backwards on the tmelne to become a present (prncpal) value at the begnnng of the tme lne. The present value of the annuty thus conssts of the sum of each payment s present value, and forms a geometrc sequence of whch we determne the sum of the payments. We say that an nterest bearng loan / debt s amortsed f both the prncpal and the nterest are pad by means of an annuty. T 4 T 5 1 2 3 4 5 x x x x x The x values are all future values whch need to go backwards to become present values at the begnnng of the tme lne. By gong backwards we are removng nterest for the perod. Lets us see how ths works: In the dagram above we see that 5 payments have been made nto an account at regular yearly ntervals. Each payment nto ths account happened at a dfferent tme, but they are all spaced at once a year. Each payment must therefore move to tmelne to become a present value. That means that when we pad n the Rx at the end of each perod, Rx was a future value at the tme whch became the equvalent value as t moved backwards on the tmelne. The sum of all these separate present values gves us the present value of the annuty. Note that the above scenaro was an ordnary annuty as each payment was made at the end of the perod. The formula: Let us pay back a loan by makng yearly down payments of R x nto ths account. Interest s charged at r % per annum compounded annually, for a total of n payments. The frst payment s made at the end of the month and the last payment at the end of n months. So ths s an ordnary annuty. The tmelne: T 4 T 5 1 2 3 4 5 x x x x x 2009 Page 1 Lesson 1 Algebra Page 47

We move each of the payments back to. So: Snce: = (1 + ) n F \ = _ V (1 + ) n \ = (1 + ) n From : From : From : From T 4 : From T 5 : 1 = x(1 + ) 1 2 = x(1 + ) 2 3 = x(1 + ) 3 4 = x(1 + ) 4 5 = x(1 + ) 5 So now: a = x(1 + ) 1 ; r = (1 + ) 1 ; n = 5 S 5 = x(1 + ) 1 [1 (1 + ) 5 1 (1 + ) 1 = x(1 + ) 1 [1 (1 + ) 5 1 _ (1 + )1 1 (1 + ) (1 + ) 1 [1 (1 + ) = x 5 (1 + ) 1 1 (1 + ) 5 If we now add all these present values together we get: = x(1 + ) 1 + x(1 + ) 2 + x(1 + ) 3 + x(1 + ) 4 + x(1 + ) 5 So for the fve payments made: 1 (1 + ) 5 In general If the frst of n payments on a loan starts at the end of the frst perod then the standard tmelne wll be: And the formula that apples wll be T n 1 (1 + ) n P v Notce that the powers are negatve x x x x because we are movng backward Notce that there s one perod at on the tmelne towards the present the begnnng of the tmelne where value. nothng happens as the frst payment s made at the end of the frst perod. (Ordnary annuty) Example Example 1 A loan of R100 000 s repad by means of 10 sem-annual payments of Rx each. If nterest on the loan s charged at 16% per annum compounded semannually, 1. determne x f the frst payment s made at the end of the frst half year. 2. determne the sem-annual payment f the fst payment s n sx months tme and f a depost of R15 000 was gven. Page 48

Soluton 1. Frst payment s made at the end of the frst sem annum: We need an effectve sem annual rate, so = _ 0,16 or 16_ 2 200 = 0,08; P = 100 000; n = 10 V 0 Soluton 100 000 x x x x 1 (1 + ) n \ 100 000 1 (1,08) 10 0,08 \ 100 000 = x(6,710081399) \ x = R14 902,95 2. If we gve a depost, ths money s taken off the value of the loan. So the loan amount wll now be R100 000 R15 000 = R85 000. So: 1 (1 + ) n \ 85 000 1 (1,08) 10 0,08 \ 85 000 = x(6,710081399) \ x = R12 667,51 It s very mportant to understand how the formula adjusts for dfferent scenaros. Once you grasp ths, you wll not have a problem wth the present value annutes. Example 2 A loan of R1 000 s pad off by equal monthly payments of R88,85 per month at a rate of 12% p.a. compounded monthly. How long does t take to amortse the loan, f the frst payment s made at the end of the frst perod. Ths s an ordnary annuty, wth tmelne: Example 1 100 000 x x x x Notce that the perod of ths loan s unknown and that we wll have to solve for n. Furthermore we also need to use an effectve monthly rate so we have: = _ 0,12 or 12_ 12 1 200 = 0,01 Now the formula apples: 1 (1 + ) n \ 1 000 = 88,85 [ 1 (1,01) n 0,01 \ (_ 1 88,85) 000 0,01 = 1 (1,01) n \ (1,01) n = 1 (_ 1 88,85) 000 0,01 \ (1,01) n = 0,8874507597 log 0,8874507597 \ n = or log log 1,01 1,01 0,8874507597 2009 Page 1 Lesson 1 Algebra Page 49

\ = 11,999826 \ n = 12 months Example Example 3 A loan s amortsed by 24 monthly payments of R2 000 each made nto an ordnary annuty wth nterest charged at 14% per annum compounded monthly. Determne the value of the loan. Notce that we do not have the loan amount, but we do have the future value of each of the payments. We need to remove the nterest from these payments by brngng all 24 back to : 4 2 000 2 000 2 000 2 000 The effectve monthly rate wll be: = _ 0,14 or _ 14 12 1 200 = 0,011 6 1 (1 + ) n \ = 2 000 [ 1 ( 14 1200 ) 24 14_ 1200 \ = 2 000(20,82774314) \ = R41 655,49 If = 0,019 : RHS = 1 (1,019) 12 = 10,6405815 ¹ LHS (We are movng further 0,019 away) Actvty Actvty 3 1. A man plans to buy a house on a 24 year mortgage and can only afford to pay R2 700 per month. If the nterest rate s currently 22% per annum compounded monthly, determne the sze of the mortgage he can take, f he starts payng one month after the mortgage was approved. Page 50

2. A loan of R150 000 s amortzed by equal quarterly payments for a perod of 10 years at a rate of 14% p.a. compounded quarterly. Determne the sze of each quarterly payment. 3. Lous buys a car at a purchase prce of R75 000. He can repay ths car n 24 equal monthly nstalments, but has a choce of 3.1 14% p.a. smple nterest 3.2 16% p.a. compounded monthly. 3.3 Whch one should he choose and why? 4. Determne the amount that must be nvested now to realse equal monthly wthdrawals of R3 500 for the next 15 years f nterest s at 16% p.a. compounded monthly. The frst wthdrawal wll be n one months tme. 2009 Page 1 Lesson 1 Algebra Page 51

5. Sandra buys a slmmng machne, and pays a depost of R2 000 on the purchase. The balance s pad off by 36 equal monthly nstalments of R 1800 each. Interest s calculated at 23% p.a. compounded monthly. 5.1 Calculate the purchase prce of the slmmng machne f the frst payment s made at the end of the frst month. 5.2 What amount would she have saved f she made the purchase cash? 5.3 If she took a loan for the balance and pad ths loan back at 23% p.a. compounded quarterly wth equal quarterly payments, would she have saved on the loan repayments? 6. A competton makes a startlng clam that you can wn a prze of 1 mllon rand. The small prnt nforms us that the prze wll be pad out n equal annual nstalments of R50 000 over the next 20 years, startng now. Assumng an average nflaton rate of 14% p.a. over the next twenty years, show that the present value of the prze s sgnfcantly less than the clamed 1 mllon rand. Page 52

7. When consderng the purchase of a house, Mr Pllay has to take the followng nto account : The house s on the market for R570 000 He has R80 000 avalable as a depost The Bank s condton for grantng a mortgage bond (loan) for the balance s that hs monthly repayments may not exceed _ 1 3 of hs monthly salary. Hs salary s R17 500 per month The bank s offerng mortgage bonds at 16,25% p.a compounded monthly, repayable n equal monthly nstallments over 20 years. 7.1 Show that Mr Pllay does not meet the thrd requrement above. Set your argument out clearly. 7.2 As he s determned to purchase the house, he decdes on a two-pronged strategy : to put n an offer to purchase whch s R50 000 less than the askng prce; to ask the bank to let hm repay the bond over a longer perod Calculate how many years he wll need to pay off the loan. Outstandng balance on a loan Often a person wth debt decdes to settle ther debts when they come across some money by wnnng a lotto game or by nhertng some money from a frend or a relatve. We also hear qute often that the governor of the reserve bank announces a change n the nterest rate. In both these scenaros the banks have to fnd an outstandng balance on your loan, to calculate what amount of the orgnal captal amount of the loan, s stll owed to the bank. Let s see how the captal amount gets smaller wth each payment: 2009 Page 1 Lesson 1 Algebra Page 53

A loan of R180 000 s repad by means of 10 sem-annual payments of R x each. Interest on the loan s charged at 14% per annum compounded sem-annually. If the frst payment was made at the end of the frst perod (an ordnary annuty) the sem-annual payment wll be R25 627,95: The effectve sem anuual rate wll be = _ 0,14 or 14_ 2 200 = 0,07 The tmelne: 0 180 000 25 627,95 25 627,95 25 627,95 25 627,95 1 (1 + ) n 180 000 1 (1,07) 10 0,07 180 000 0,07 x = 1 (1,07) 10 x = R25 627,95 Let us dscuss the outcomes of the example above: For the ordnary annuty, the frst payment s made at the end of sx months (sem-annum). Consder the table below: Tmelne Balance at Tmelne Balance Outstandng after payment Regular Payments Interest 0 180000,00 1 192600,00 166972,05 25627,95 12600,00 2 178660,09 153032,14 25627,95 11688,04 3 163744,39 138116,44 25627,95 10712,25 4 147784,59 122156,64 25627,95 9668,15 5 130707,61 105079,66 25627,95 8550,96 6 112435,24 86807,29 25627,95 7355,58 7 92883,80 67255,85 25627,95 6076,51 8 71963,76 46335,81 25627,95 4707,91 9 49579,31 23951,36 25627,95 3243,51 10 25627,95 0 25627,95 1676,59 10 payments Owng R0 Total pad: R256279,50 Total pad n nterest: R76279,50 If we want to manually calculate the porton of nterest on the frst payment we need to do the followng: 180 000(1,07) 180 000(0,07) = 180 000(0,07)=R12 600 the frst sem annum nterest s added to the value of the loan So we could have sad R180 000 0,07 = R12 600. To then calculate the captal amount pad we take the regular payment and subtract the nterest: R25 627,95 R12 600 = R13 027,95. Page 54

The tmelne below llustrates ths clearly. If we wsh to calculate the balance outstandng after the ffth payment, we can do ths n one of two ways. These methods are based on the fact that you must move to the same pont on the tmelne to be able to compare thngs. Method 1: Our frst method moves forward on the tmelne as s shown n the dagram below. We need to move the orgnal amount of the loan forward wth nterest, and also the regular payments that were made. Balance of the loan just after the payment s made at T5: (loan + nterest) (nstalments + nterest) 180 000(1,07) 5 25 627,95 [ (1,07)5 1 0,07 = 105 079,66. Method 2: Ths method just focuses on the payments that stll have to be made and works backwards on the tmelne by calculatng the present value of these payments at T5. T 4 T 5 180 000 25 627,95 25 627,95 25 627,95 25 627,95 25 627,95 T 5 T 6 T 7 T 8 T 9 0 25627,95 25627,95 25627,95 25627,95 25627,95 We need to move each payment that has not been made back to tmelne T5. Ths would mean removng nterest from each payment up to T5. So we are thus workng wth the present value of an annuty: OB T 5 = 25 627,95 [ 1 (1,07) 5 0,07 = 25 627,95(4,100197436) = R105 079,66 It s clear that method 2 wll be more tme effcent for you to apply. 2009 Page 1 Lesson 1 Algebra Page 55

At ths pont t s mportant to draw your attenton to some mportant facts about outstandng balance: Outstandng balance on any loan s always calculated drectly after the last payment s made. Method two suggests that the outstandng balance s the present value of all payments yet to be made. Example Example 1 Lets solve some problems where the outstandng balance s needed. Trevor purchases a house for R780 000 on a mortgage for 20 years wth nterest at 17% p.a. compounded monthly. The mortgage payments are made at the end of each month. 1. Calculate the monthly nstalment 2. Calculate the outstandng balance after 10 years 3. At ths pont, how much has been pad nto the bond account and how much captal was pad off on ths loan? 4. Calculate the new monthly payment f the nterest rate changes to 19% p.a. compounded monthly after ten years. 5. If Trevor rather asked the bank for a 25 year mortgage, assumng that the rate remans fxed at 17% p.a. compounded monthly for the duraton of the loan, how much less wll he pay every month? Soluton 1. 40 780 000 x x x x The effectve monthly rate s: = _ 0,17 or _ 17 12 1200 = 0,014166666 = 0,0141 6 So: 1 (1 + ) n 780 000 = x [ 1 (1,0141 6 ) 240 0,01416 780 000 = x(68,17559487) x = R11 441,04 2. After 10 years, the number of payments that were made s 120. There s thus stll 120 payments left. So: OB 20 = 11 441,04 [ 1 (1,0141 6 ) 120 0,01416 = 11 441,04(57,53817667) = R658 296,58 3. After ten years, Trevor has pad nto the bond account: 120 R11 441,04 = R1 372 924,80 The captal pad off on ths loan after payng for ten years: R780 000 R658 296,58 Page 56

= R121 703,42. 4. The new effectve monthly rate wll be: = _ 0,19 or _ 19 12 1200 = 0,015833333 = 0,0158 3 The outstandng balance at ths pont s gven as R658 296,58, so: 658 296,58 = x [ 1 (1,0158 3 ) 120 0,01583 658 296,58 = x(53,56979602) x = R12 288,58 So Trevor wll now have to pay R12 288,58 on hs bond wth the nterest rate ncrease. 5. Askng the bank for a 25 year mortgage mght save money monthly, but n the long run you pay more for the house: 1 (1 + ) n 780 000 = x [ 1 (1,0141 6 ) 300 0,01416 780 000 = (69,55086772) x = R11 214,81 He wll only save R226,23 per month. Actvty 4 Actvty 1. Mark wants to purchase a bg screen televson set at a cash prce of R 39 000. He wshes to obtan a loan from the bank at an nterest rate of 18% p.a. compounded monthly. Mark can only afford to pay R1 200 per month for a perod of 36 months. 1.1 Wll the bank gve hm a loan based on ths nformaton? 1.2 If the bank agrees to accept R1 200 per month, how long wll t take Mark to pay back the loan? 2009 Page 1 Lesson 1 Algebra Page 57

2. A loan of R120 000 s repad by 72 equal monthly payments nto an ordnary annuty of Rx each. The nterest rate charged s 16% per annum compounded monthly. 2.1 Determne the amount of the monthly payment. 2.2 Determne the balance outstandng after the 24th payment. 2.3 If the nterest rate s changed to 18% per annum compounded monthly, drectly after the 24 th payment, determne the new monthly payment to settle the loan over the same tme perod. Page 58

3. Brad purchased a car for R72 000. He agrees to a loan of 54 months at a rate of 19,25% p.a. compounded monthly, wth hs frst payment due at the end of the frst month. After the 36 th payment, the bank agrees that he can settle the account. How much must he pay the bank? 4. A house that was bought 8 years ago for R50 000 s now worth R100 000. Orgnally the house was fnanced by payng 20% depost wth the rest fnanced through a 20 year mortgage at 10,5% nterest per annum compounded monthly. The owner, after makng 96 equal monthly payments, s n need of cash, and would lke to refnance the house. The fnance company s wllng to loan 80% of the new value of the house, less any amount stll owng. How much cash wll the owner be able to borrow? 2009 Page 1 Lesson 1 Algebra Page 59

5. Mathew bought a house for R825 000 n Wapadrand and took a mortgage bond at a rate of 16,25% p.a. compounded monthly. Ths bond has to be pad back over a perod of 20 years by makng monthly deposts nto the mortgage account. If Mathew put down a depost of R84 000, 5.1 Determne the monthly payment on hs bond account. 5.2 After the 200th payment, the bank rate changes to 17,5% p.a. compounded monthly. Determne the outstandng balance on the loan at ths tme 5.3 Determne the new monthly payment that wll amortze the loan. Page 60

5.4 How much dd Mathew actually pay for hs house? Solutons to Actvtes Actvty 1 1. T 60 F v 300 300 300 300 300 The effectve monthly rate s _ 0,14 or 14_ 12 1 200 = 0,011 6 Notce that the payments start mmedately and end on the last day, so 61 payments are made. = x ( (1 + )n 1 ) \ = 300 ( (1,011666 )61 1 0,01166666 ) \ = 300(88,20073489) \ = R26 460,22 2.1 91 92 x x x x x Effectve monthly rate: _ 0,08 or 8_ 12 1 200 = 0,00 6 Note ths s an annuty due wth 16 12 = 192 payments made nto the annuty. = x ( (1 + )n 1 )(1 + ) \ 16 000 = x ( (1,00666 )192 1 0,00666 )(1,00666 ) \ 16 000 = x(387,2091494)(1,00666 ) F v 2009 Page 1 Lesson 1 Algebra Page 61

\ 16 000 = x(389,7905437) \ x = R41,05 2.2 51 52 x x x x x Effectve monthly rate: _ 0,08 or 8_ 12 1 200 = 0,00 6 Note that the perod s changng to accommodate 21 12=252 payments = x ( (1 + )n 1 )(1 + ) \ 21 000 = x ( (1,00666 )252 1 0,00666 )(1,00666 ) \ 21 000 = x(650,3587456)(1,00666 ) \ 21 000 = x(654,6944706) \ x = R32,08 3. For 12% p.a. c.m. effectve _ 0,12 or 12 100 ( (1,01)12 1 0,01 ) = 1 268,25 F v _ 12 1 200 = 0,01: For 4% p.q. c.q. ths s an effectve rate so here = 0,04: 300 ( (1,04)4 1 0,04 ) = 1 273,94 So clearly the quarterly nvestment s better at these rates that were gven. 4. 2 4 6 T 48 F v 200 200 200 200 200 200 200 1 000 1 000 1 000 1 000 Effectve monthly rate s needed for the monthly deposts: _ 0,06 or 6_ 12 1 200 = 0,005 The effectve annual rate s needed for the annual deposts: (1 + ) = (1 + _ 12) 12 \ = (1 + _ 0,06 12 ) 12 1 \ = 0,06167781186 We treat ths scenaro as two separate annutes that are runnng at the same tme: = 200 ( (1,005)48 1 0,005 ) + 1 000 ( (1,06167781186)4 1 0,06167781186 ) the frst annuty at R200p.m. the second annuty at R1 000 p.a. \ = 200(54,09783222) + 1 000(4,385518113) \ = R15 205,08 Page 62 Actvty 2 1. Scrap Value = 0,18x Deprecaton: = (1 ) n 0,18x = x(1 0,21) n

0,18 = (0,79) n log 0,18 n = _ log 0,79 n = 7,274654049 n = 7,27 years 2.1 = (1 ) n 45 000 = 120 000(1 ) 10 0,375 = (1 ) 10 1 = 10 0,375 _ = 0,09342627726 = 0,0934 r = 9,34% p.a.c.a 2.2 = (1 n) 45 000 = 120 000(1 10) 0,375 = (1 10) 10 = 1 0,375 10 = 0,625 r = 6,25%SI 3.1 = (1 ) n = 550 000(1 0,1) 8 = R236 756,97 3.2 = (1 + )n = 550 000(1,18) 8 = R2 067 372,56 3.3 R2 067 372,56 R236 756,97 = R 1 830 615,59. 3.4 = x ( (1 + )n 1 ) 1 830 615,59 = x ( (1,01)97 1 0,01 ) 1 830 615,59 = x(162,5265655) = R11 263,49 4.1 = (1 n) = 850 000(1 0,07 8) = 850 000(0,44) F v = R374 000 4.2 = (1 + ) n = 850 000(1,14) 8 = R2 424 698,46 4.3 Snkng fund = R 2 050 698,46 4.4 = x ( (1,0125)90 1 0,0125 )(1,0125) 6 2 050 698,46 = x ( (1,0125)90 1 0,0125 )(1,0125) 6 2009 Page 1 Lesson 1 Algebra Page 63

2 050 698,46 = x(164,7050076)(1,0125) 6 x = R11 556,46 5.1 = (1 ) n = 1 200 000(1 0,09) 5 = R748 838,57 5.2 = (1 + ) n = 1 200 000(1,06) 5 = R1 605 870,69 5.3 Snkng fund = R 857 032,12 = x ( (1 + )n 1 ) 857 032,12 = x ( (1,035)21 1 0,035 ) 857 032,12 = x(30,26947068) x = R28 313,42 5.4 Scrap Value of current equpment: = (1 ) n = 1200 000(1 0,09) 3 = R904 285,20 In the snkng fund: = x ( (1 + )n 1 ) = 28 313,42 ( (1,035)13 1 0,035 ) = 28 313,42(16,1130303) = R456 214,99 She wll have R1 360 500,19 avalable. Actvty 3 1. 88 2 700 2 700 2 700 2 700 the effectve monthly rate wll be: = _ 0,22 or 22_ 12 1 200 = 0,018 3 2. T 40 150 000 x x x x - the effectve quarterly rate wll be: = _ 0,14 or _ 14 4 400 = 0,035 1 (1 + ) n \ = 2 700 [ 1 (1,018333333) 288 0,0183333333 \ = 2 700(54,254081) \ = R146 486,02 1 (1 + ) n \ 150 000 1 (1,035) 40 0,035 \ 150 000 = x(21,35507234) \ = R7 024,09 Page 64

3. 4 75 000 x x x x 3.1 The rate s a smple nterest rate, so ths s a hre purchase agreement: Now: x = _ 24 = P (1 +.n) V 24 75 000(1 + 0,14 2) \ x = 24 \ x = R4 000 So the nstalment s R4 000 per month at a smple nterest rate. 3.2 the effectve monthly rate wll be: = _ 0,16 or _ 16 12 1 200 = 0,01 3 1 (1 + ) n \ 75 000 = x [ 1 (1,01 3 ) 24 0,013 \ 75 000 = x(20,42353906) \ x = R3 672,23 3.3 Lous should opt for the compound nterest agreement as he wll save: 4. 24 (4 000 3 672,23) = R7 866,48 on the deal. 80 1 (1 + ) n \ = 3 500 [ 1 (1,01333333) 180 0,013333333 3 500 3 500 3 500 3 500 \ = 3 500(68,08738987) \ P The effectve monthly rate: = _ 0,16 16 or _ 12 1 200 = 0,01 3 V = R238 305,86 5. 6 2 000 1 800 1 800 1 800 1 800 The effectve monthly rate: = _ 0,23 or _ 23 12 1 200 = 0,0191 6 1 (1 + ) n 5.1 2 000 \ 2 000 = 1 800 [ 1 (1,0191666 ) 36 0,0191666 \ 2 000 = 1 800(25,83330388) \ = R46 499,95 + R2 000 \ = R48 499,95 5.2 She pad 36 1 800 + 2 000 = R66 800 If she purchased t cash she would have pad R48 499,95. Ths means she would have saved R18 300,05. 5.3 The effectve quarterly rate: = _ 0,23 or 23_ 4 400 = 0,0575 2009 Page 1 Lesson 1 Algebra Page 65

The perod wll reman 3 years but the number of payments wll be 12 quarterly payments. 1 (1 + ) n \ 46 499,95 = x [ 1 (1,0575) 12 0,0575 \ 46 499,95 = x(8,499955647) \ x = R5 470,61 She now pays 12 R5 470,61+R2 000 = R67 647,32. So she would not have saved n fact she pays R847,32 more f the payments happen quarterly. 6. The effectve yearly rate wll be: = _ 14 100 = 0,14. The tmelne: 0 50 000 50 000 50 000 50 000 1 (1 + ) n 0,14 = x [ \ = 50 000 [ 1 (1,14) 20 \ = 50 000(6,623130552) \ = R331 156,53 Ths s R668 843,47 short of R1 mllon 7.1 He gves a depost of R80 000. So the loan amount wll be R490 000. Hs monthly salary s R17 500 and a thrd of ths s R5 833,33. The effectve monthly rate s: = _ 0,1625 or _ 16,25 12 1 200 = 0,013541 6 1 (1 + ) n \ 490 000 = x [ 1 (1,013541 6 ) 240 0,0135416 \ 490 000 = x(70,91969686) \ x = R6 909,22 Mr Pllay does not qualfy by R1 075,89. 7.2 If he pays a depost of R80 000 and offers R50 000 less, the bond amount wll be R440 000. He can only afford to pay R5 833,33 per month. So: 1 (1 + ) n \ 440 000 = 5 833,33 [ 1 (1,013541 6 ) n 0,0135416 440 000 \ 5 833,33 0,01354166 = 1 (1,013541 6 ) n \ (1,0135416 ) n = 0,021429155 log 0,021429155 \ n = log 1,0135416 \ n = 285,7081773 (ths answer s n months, so _. by 12) \ n = 23 years and 10 months Page 66

Actvty 4 1.1 We need to fnd the present value of the 36 payments of R1500 each. The effectve monthly rate wll be: = _ 0,18 or _ 18 12 1200 = 0,015. Now: 1 (1 + ) n = 1 200 [ 1 (1,015) 36 0,015 = 1 200(27,66068431) x = R33 192,82 Payng R1200 per month for 36 months does not settle the loan whch s R39000. He stll shorts R5807,18. 1.2 If the bank accepts the R1200 per month, then the perod wll be longer than 36 months. We need to fnd n for the present value of R39000: 1 (1 + ) n 39 000 = 1 200 [ 1 (1,015) n 0,015 39 000 _ 0,015 = 1 (1,015) n 1 200 (1,015) n = 1 _ 39000 1200 0,015 (1,015) n = 0,5125 log 0,5125 n = or log log 1,015 1,015 (0,5125)= 44,8970 n = 45 months 2.1 The effectve monthly rate wll be: = _ 0,16 or _ 16 12 1 200 = 0,01 3. 1 (1 + ) n 120 000 = x [ 1 (1,01 3 ) 72 0,013 120 000 = x(46,10028344) x = R2 603,02 2.2 After the 24 th payment, there are stll 48 payments left to be made. SO: OB 4 = 2 603,02 [ 1 (1,01 3 ) 48 0,013 = 2 603,02(35,28546548) = R91 848,77 2.3 For the balance of R91 848,77 the new effectve monthly rate wll be: = _ 0,18 or _ 18 12 1 200 = 0,015 1 (1 + ) n 91 848,77 1 (1,015) 48 0,015 91 848,77 = x(34,04255365) x = R2 698,06 3. We frst need to fnd how much Brad must pay the bank monthly. The effectve monthly rate s = _ 0,1925 or _ 19,25 12 1 200 = 0,016041 6. 2009 Page 1 Lesson 1 Algebra Page 67

1 (1 + ) n 72 000 = x [ 1 (1,016041 6 ) 54 0,0160416 72 000 = x(35,94224527) x = R2 003,21 To settle after the 36 th payment, we obtan the present value of the last 18 payments on ths day: 1 (1 + ) 18 OB 36 = 2 003,21 [ 1 (1,016041 6 ) 18 0,0160416 = 2 003,21(15,52717265) = R31 104,19 He thus has to pay R31 104,19. 4. We frstly need to work out what the monthly payments were on the orgnal loan of R50 000 depost: Loan amount = R50 000 20% of R50 000 = R50 000 R10 000 = R40 000 The nterest rate: = _ 0,105 or _ 10,5 12 1 200 = 0,00875. So: 1 (1 + ) n 40 000 1 (1,00875) 240 0,00875 40 000 = x(100,1622742) x = R399,35 After 96 months, the house s worth R100 000, and the bank wll loan R80000 to the homeowner. We now need the outstandng balance after 8 years that s the present value of the last 144 payments that must stll be made: 1 (1 + ) 144 OB 96 = 399,35 [ 1 (1,00875) 144 0,00875 = 399,35(81,68995711) = R32 622,88 So he wll receve: R80 000 R32 622,88 = R47 377,12. 5.1 Effectve monthly rate: = _ 0,1625 or _ 16,25 12 1200 = 0,013541 6 1 (1 + ) n depost 825 000 84 000 = x [ 1 (1,013541 6 ) 240 0,0135416 741 000 = x(70,91969686) Page 68

x = R10 448,44 5.2 There are 40 payments left: OB 200 = 10 448,44 [ 1 (1,013541 6 ) 40 0,0135416 = 10 448,44(30,72765908) = R321 056,10 5.3 The new monthly payment for the remanng R321 056,10 at an effectve rate of = _ 0,175 or _ 17,5 12 1 200 = 0,01458 3 : 1 (1 + ) n 321 056,10 = x [ 1 (1,01458 3 ) 40 0,014583 321 056,10 = x(30,14462647) x = R10 650,53 5.4 Mathew pad a total of: 200 R10 448,44 + 40 R10 650,53 + R84 000 = R2 599 709,20 for hs house. 2009 Page 1 Lesson 1 Algebra Page 69