t10.interest rates&present value

t,&"?

Then you ought to have investedmy money with the bankers,and at my coming I should havereceived what was mine with nterest.-tvtarrnew l5: l /

Jhis chapter treats some basic topics in the mathematics of f inance. The main concern is howI the values of investments and loans at different times are affected by interest rates.

Sections 1.2 and 4.9 have already discussed some elementary calculations involving interestrates. This chapter goes a step further and considers different interest periods. lt also discusseseffective rates of interest, continuously compounded interest, present values of future claims,annuities, mortgages, and the internal rate of return on investment projects. The calculationsinvolve the summation formula for qeometric series. which we therefore derive.

10.1 Interest Periods and Effective RatesIn advertisements that offer bank loans or savings accounts, interest is usually quoted as anannual rate, also called a nominal rate, even if the actual interest period is different. Thisinterest period is the time which elapses between successive dates when interest is addedto the account. For some bank accounts the interest period is one year, but recently it hasbecome increasingly common for financial institutions to offer other interest schemes. Forinstance, many U.S. bank accounts now add interest daily, some others at least monthly. If abank offers 9Eo annloalrate of interest with interest payments each month, then (l ll2)97o :O.'lSVo of the capital accrues at the end of each month. The annual rate must be divided bythe number of interest periods to get the periodic rate-that is, the interest per period.

Suppose aprincipal (orcapital) ofSo yields interestattherate p%o perperiod (forexampleone year). As explained in Section 1.2, after / periods it will have increased to the amount

: So(l *r) ' where : pll00

Each period the principal increases by the factor 1 + r. Note thaf p%a means p/100, andwe say that "the interest is p%o" or that "the interest rate is r".

t

350 CHAPTER 1O / INTEREST RATES AND PRESENT VALUES

The formula requires the interest to be added to the principal at the end of each period-Suppose that the annual interest is p%o, but that we are offered biannual interest paymeilrat the rate p 127o. Then the principal after half a year will have increased to

o/2 / r \so + so too

: so\l + t,)Each halfyear the principal increases by the factor | * r 12. After 2 periods (= one year, fowill have increased to Ss(l -f r l2)2. and after / years it will have increased to

- / , \2 'to( . ' * t /

Note that abiannual interest payment at th erate |r is betterfor a lenderthan an annual interespayment at the rate r. This follows from the fact that (l * r l2)2 : I -f r + 12 4 . | + r-

More generall suppose that interest at the rate p lnTo is added to the principal at adifferent times distributed more or less evenly over the year. For example, : 4 if interesis added quarterly, n : 12 if it is added monthly, etc. Then the principal will be multipliedby afactor (l + rln)' eachyear. Affer t years, the principal will have increased to

/ r \ ' tt ' ( t * ; , ) ( r 'The greater is zl, the faster interest accrues to the lender. (See Problem 10.2.6.)

A deposit of f5000 is put into an account earning interest at the annual rate of 9Vo, wtinterest paid quarterly. How much will there be in the account after 8 years?

Solution: The periodic rate r f n is 0.09 14 : 0.0225 and the number of periods nr is4 . 8 : 32, so formula (1) gives:

5000(1 +0.022132 x rc1-90.52

ff iHowlongwi l l i t takeforthef,5000inExample1(withannual interestrate9voainterest paid quarterly) to increase to f,15 000?

Solution: After r quarterly periods the account will increase to 5000(l * 0.0225). So

5000( l +0.022rt : 15000 or l .O225t :3

To find f we take the natural losarithm of each side:

t LnI.0225 : ln3 (because lnap - plna)ln3

, : ;Tf f ix49.37

Thus it takes approximately 49.37 quarterly periods, that is approximately l2years and foumonths- before rhe accounr has increased to f15000. I

mh peritr:t p&\ fneni-.

One \ea.rr

nl interes:- l> l - r .ncipai at r4 if intere:: muhiplie;dto

rt 9'r. s irL

rrods l is

I

E 9Vc and

':rsl ' . so

rs and fourI

SECTION 10.1 / INTEREST PERIODS AND EFFECTIVE RATES 351

Effective Rate of InterestA consumer who needs a loan may be faced with several offers from competing financialinstitutions. It is therefore important to know how to compare various offers. The conceptof effective interest rate is often used in making such comparisons.

Consider a loan which implies an annual interest rate of 97o with interest at the rate0.7 57a added 12 times a year. If no interest is paid in the meantime, after one year an initialprincipalof56willhavegrowntoadebtof Ss(l+0.09112)t2 ry So.l.0g4.Infact,aslongasno interest is paid, the debt will grow at a constant proportional rate that is (approximately)9.4 7o pet year. For this reason, we call 9.4Vo the effective yearly rate. More generally:

: r r.Erts.'.:+t+hlt vt:dEWhen interest is added n times during the year at the rate r f n per period, thenthe effective yearlv rate R is defined as

n:(r+

The effective yearly rate is independent of the amount 56. For a given value of r > 0, it isincreasing in n. (See Problem 10.2.5.)

E What is the effective yearly rate R corresponding to an annual interest rate of 97o withinterest compounded: (i) each quarter; (ii) each month?

Solution:(i) Applying formula (2) with r : 0.09 and n :4, rhe effective rate is

R:(1 +0.0914)4 - I : ( l +0.02214-Ir0.0931 or 9.3rvo

(ii) In this case / : 0.09 and n -- 12, so the effective rate is

R:(1 +0.0gln)12 -1:( l+0.0075)12- I =0.0938 or 9.3lva *

A typical case in which we can use the effective rate of interest to compare different financialoffers is the followine.

, When investing in a savings account, which of the following offers are better: 5.9Vo withinterest paid quarterly or 67o with interest paid twice a year?

Solution: According to (2), the effective rates for the two offers are

R: ( t +0.05g1q4 - t =0.0603. R: ( t + 0.06/2)2 - t :0.0609The second offer is therefore better for the saver. I

(2); ) ' - ,

352 CHAPTER . IO / INTEREST RATES AND PRESENT VALUES

NOTE I In many countries there is an official legal definition of effective interest rate whichtakes into account different forms of fixed or "closing" costs incurred when initiating a loan.The effective rate of interest is then defined as the rate which implies that the combinedpresent value of all the costs is equal to the size of the loan. This is the internal rate of return.as defined in Section 10.7. lPresent values are discussed in Section 10.3.)

rFR B'!.E:M.SI:F08

1.

SECTION 10.1

(a) What will be the size of an account after 5 years if $8000 is invested at an annualinterest rate of 57o compounded (i) monthly; (ii) daily (with 365 days in a year)?

(b) How long does it take for the $8000 to double with monthly compounding?

2. (a) Anamount$5000earnsinterest af3Eoperyear. Whatwil lthisamounthavegrownto after 10 vears?

t

3.

4.

5.

6.

(b) How long does it take for the $5000 to triple?

What annual percentage rate of growth is needed for a100 times as large after 100 years? ( '0t 00 x 1.041.)

(a) An amount of 2000 euros is invested at77o per year.account after (i) 2years; (ii) 10 years?

country's GDP to become

What is the balance in the

(b) How long does it take (approximately) for the balance to reach 6000 euros?

Calculate the effective yearly interest if the nominal rate is lTVo and interest is added(i) biannually; (ii) quarterly; (iii) monthly.

Which is preferable for a borrower: (i) to take out a loan with an annual interest rate of2l.5Vo, with interest paid yearly; or (ii) to take out a loan with an annual interest rateof 207o, with interest paid quarterly?

7. (a) The sum of $ 12 000 is invested at 4Vo anwal interest. What will this amount har egrown to after 15 years?

(b) How much should you have deposited in a bank account 5 years ago in order to har e$50 000 today, given that the interest rate has been 57o per year over the period?

(c) A credit card is offered with interest on the outstanding balance charged at2%c gmonth. What is the effective annual rate of interest?

8. What is the nominal yearly interest rate if the effective yearly rate js 28Vo and intereris compounded quarterly?

t

trate whichIting a loan.) combinedte of return,

rt an annualin a year)?Cing?

have grown

' to becomet

ance in the

uros?

ast is added

Erest rate ofinterest rate

mount have

rder to havee period?

d at27o pet

nd interest

lf

SECTION 10.2 / CONTINUOUS COMPOUNDING 353

"10.2 Cont inuous CompoundingWe saw in the previous section that if interest at the nte r f n is added to the princip al Ss at ndifferent times during the year, the principal will be multiplied by a factor (l * r ln)" eachyear. After / years, the principal will have increased to .ls( I * r f n)nt . In practice, there is alimit to how frequently interest can be added to an account. However, let us examine whathappens to the expression as the annual frequency tends to infinity. We put rf n : lf m.Then:mrarrdso

/ r \ ' ' / l \ " t l l l \ ' l ' rso(\r+; / :so(r+-/ : ro l ( ' * ; ) ) (1)As -+ oo (with r fixed), so m : nf r -> oo, and according to Example 7 .11.2, we haveQ * llm)^ -> . Hence, the expression in (l) approaches Sse'/ as r tends to infinity,implying that interest is compounded more and more frequently. In the limit, we talk aboutcontinuous comp ounding of interest:

c orN,TJ,t!1"[J 0 u 5'],c,G M P o B N 0,1 l\t G,+St lNlt R E 51:

The formulaS(r) - 50"'r

shows how much a principal of Ss will have increased to after r years, if theannual interest is r, and there is continuous compounding ofinterest.

f NM9l'i:1'., , Suppose the sum of 5000 euros is invested in an account earning interest at an annualrate of 97o. What is the balance after 8 years if interest is compounded continuously?

.

Solution: Using formula (2) with r : 9 /700 : 0.09, we see that the balance is

50000'0e'8 : 5000e0'72. x l0 27 2.77

(Compare with the result in Example I of the previous section.)

If S(/) : Soe't, then S'(l) - Ssre't : rS(/) (according to formula (2) in Section 6.10),and so S'(l)/S(t) : r. Using the terminology introduced in Section 6.4:

.

WithcontinuouScompoundingofintereStattherater,theprincipalincreasesatthe constant relative rate r, so that S'(/)/S(t) : r.

From (2), we infejr that S(l) : Soer, so that the principal increases by the factor e' duringthe first year. In general, S(r + l) - S0r(t+1) : Sye't e' : S(t)e' . Hence:Wth contirutolts cotllpounding of interest, the principal increases each year by a fixedfactor e'.

(2)

354 CHAPTER 1o / INTEREST RATES AND PRESENT VALUES

Comparing Different Interest PeriodsGiven any fixed interest rate of p%o (: 100r) per year, continuous compounding of interesris best for the lender. (See Problem 6.) For comparatively low interest rates, however, thedifference between annual and continuous compounding of interest is quite small, when thenumber of years of compounding is relatively small.

Find the amount K by which $1 increases in the course of a year when the interest rateisSVo per year and interest is added: (a) yearly; (b) biannually; (c) continuously.

Solution: In this case r : 8/100 : 0.08, and we obtainK:(1+0.08):1.08K:(1 +0.081D2:1.0816K-eo'osxl .0832g

If we increase either the interest rate or the number of years over which interest accumulates.then the difference between yearly and continuous compounding of interest increases.

In the previous section the effective yearly interest was defined by the formula(I * r/n)'- 1, when interest is compounded times a year with rate rf n per period.Letting n approach infinity in this formula, we see that the expression approaches

e'-7

This is called the effective interest rate with continuous compounding at the annual rate r

i:;. .:, i,,i::i:1r...:r',, sl ::l:*rlii,1l.*i;ii;rF'r;' lir'i:jffi1 . (a) How much does $8000 grow to after 5 years if the annual interest rate is 5%, with

continuous compounding?

(b) How long does it take before the $8000 has doubled?

2. An amount $1000 earns interest at 57o per year. What will this amount have grou'nto after (a) 10 years, and (b) 50 years, when interest is compounded (i) yearly, or(ii) monthly, or (iii) continuously?

3. (a) Find the effective rate corresponding to an annual rate of 107o compounded con-tinuously.

(b) What is the maximum amount of comDound interest that can be eamed at an annualrate of lOVo?

4. The value r' of a new car depreciates continuously at the annual rate of I}%o-that is.rr : ',,e-it u'here :0. I istherate of depreciation. Howmany years does ittakeor *e !-ar to lo! 90-r of its orieinal value?.

(al(b)(c)

(-') |

rg of intereslo\\'ever. therll. rvhen the

rnterest ratet .

I

rumulates.reases.

r formula

ryr period.gs

(3)

nual rate r'.

J.,5or. with

e -grown

barlv, or

ded con-

n annual

-that is.s it take

SECTION 10.3 / PRESENT VALUE 355

5. The value of a machine depreciates continuously at the annual rate of 6Vo. How manyyears will it take for the value of the machine to halve?

HARDER PROBLEMS

6. weshowedthat( l r r ln)n --> e 'asn --> @. Foreachf ixedr > 0weclaimthat(l * r /n)' is strictly increasing in . In particular, this implies that

( I*r /n)" 0 for all > 0. What conclusion can you draw?

,' , =;:,:.',,= .. ,lr;ijipiii,;:lril'r*fuii1;'t,!irfu;;ili1r,', ''..:.' .i:,,,-.,.

10.3 Present ValueThe sum of $1000 in your hand today is worth more than $1000 to be received at somefuture date. One important reason is that you can invest the $1000 and hope to earn someinterest or other positive return.l If the interest rate is IlVo per year, then after I year theoriginal $1000 will have grown ro rhe amount 1000(1 + l1/100) : l l l0, and after 6 years,it wil l have grown to 1000( 1 + 11/ 100)6 : 1000 . (l. l l)6 ry 1g70. This shows rhar, ar rheinterest rate ll%o per year, $1000 now has the same value as $1 I 10 next year, or $1g70 in 6years time. Accordingly, if the amount $ I I 10 is due for payment I year from now and theinterest rate is ll%o per year, then the present value of this amount is $1000. Because $ 1000is less than $1110, we often speak of $1000 as the present discounted value (or PDV) of$l1l0nextyear. Therat io$1000/$1110: 1l( t+ i l /100) ry 0.9009iscal ledthe(annual)discount factor, whose reciprocal 1.11 is one plus the discount rate, makjngthe discountrate equal to the interestrate of llVo.

Similarly, if the interest rare is I 7Vo per year, then the pDV of $ l g70 due 6 years fromnow is $1000. Again, the ratio $1000/$1870

-

0.53 is called the discountfactor.Suppose that an amount K is due for payment / years after the present date. What is

the present value when the interest rate is pvo per year? Equivalently, how much must bedeposited today earning pqa anil:l interest in order to have the amount K affer / years?

r l fpr icesareexpectedtoincrease,anotherreasonforpreferr ing$l000todayisinf lat ion,because$1000 to be paid at some furure date will buy less then than $1000 does today.

(*)

s'(x) :sr" l [ r ' (r


If interest is paid annually, an amount A will have increased to A(1 * p ll}O)t after tyeafs,sothatweneedA(1

- l p/ l})r : K. Thus, A: K( l t p/100)- t : K( l *r)- , ,wherer : p/100. Heretheannualdiscountfactoris (1+r)-r, and (1f r)- ' isthediscountfactor appropriate for I years.

If interest is compounded continuously, then the amount A will have increased to Ae'tafter I years. Hence, Ae't : K, or A : Ke-,t. Here e-rt is the discount factor. Tosummarize:

:If theinterestordiscount rateis pToperyearandr: p/I00,anamount K lhat :is payable in t years has the present value (or present discounted value, or PDV): ,,

K(l * r)-t , with annual interest paymentsK e-" , with continuous compounding of interest

Find the present value of g 100 000 which is due for payment after 15 years if the interestrate is 6Vo per year, compounded (i) annually, or (ii) continuously.

Solution:(i) According to (l), the present value is 100 000(1 * 0.OO-ts x 4fi 26.5I.

(ii) According to (l), the present value is l6ggggu-0.00'15 : 100000e-0.e x 40656.9j.As expected, the present value with continuous compounding is the smaller, becausecapital increases most rapidly with continuously compounding of interest. t

flVhen to Harvest a Thee?) Consider a tree that is planted at time f : 0, and let p (t)be its current market value at time /, where P() is differentiable. Assume that the interestrafeis lOOrTo per year, and assume continuous compounding ofinterest.(a) When should this tree be cut down in order to maximize its present discounted value?(b) The optimal cutting time f* depends on the interest tate r. Find an expression for

dt* ldr.

Solution: (a) The present value is f (t) : P (t)e-,t , whose derivative is

f ' (t) : p' (t)e-'t r p(t)(-r)e-rt - r-rtfpt Q) - r p 6l (*)

A necessary condition for /* > 0 to maximize /(l) is that ft(t*): 0. We see that thisoccurs when

P'( t*) - rP(t*) (**)The tree. therefore, should be cut down at a time /* when e relative rate of increase in thevalue of e tree is precisely equal to the interest rate. Of course, some conditions have tobe placed on / in order for r* to be a maximum point. It suffices to have pt(t)/p(t) > rt 'orr < t - and P'$l lP(t \ < r for f > f* .

(1)

f r

SECTION 10.3 / PRESENT VALUE 357

f aer l - r t - t .

rdisr-txnr

d to -{s-'

iptor. To

b interes

t

M56.97.r, because

I

d let P( lb iterest

tsd value?

ession for

(*)

r that this

(x*)ease in thens have tolP(t)>r

(b) Differentiating (x*) w.r.t r yields

. . dt*P" (r*) '

:

Solving for dt* ldr,

. dt*P(f ' t +rP'( t" )

*

P(t*)

?RO

dt*dr:

NOTE 1 Differentiating (*) w.r.t. r yields

Pt '7t* - r Pt( t*) (x*x)

f" (t) : P" (t)e-'t - r P' (t)e-'t - P'(t)re-'t + r2 P 1te-'t

Using (**) we see that the second-order condition f " (t*) < 0 is satisfied if and only if

e- ' t f Pt ' ( t*) - 2r P'( t " ) + 12 P(t*) l : e- ' t lPt ' ( t*) - rPl( t*) l < 0

in which case dt* f dr < 0. Thus the optimal growing time shortens as r increases (whichmakes the foresters more impatient). In particular, given any r > 0, the optimal /* is lessthan the time i that maximizes current market value P(r), which is optimal only if r : 0.

NOTE 2 In the example, we did not consider how the ground the tree grows on may beused after cutting-for example, by planting a new tree. This generalization is studied inProblem 10.4.8.

Find the present value of 350 000 which is due after 10 years if the interest rcte is 8Voper year (i) compounded annually, or (ii) compounded continuously.

Find the present value of 50000 which is due after 5 years when the interest rate is5 .7 57o per year, paid (i) annually, or (ii) continuously.

With reference to the tree-cutting problem of Example 2, consider the case where

f( t ) : ( t 5zr-oost r>0

(a) Find the value of r that maximizes /(). (Study the sign variation of f '(t).)(b) Find lim*- f (t) anddraw the graph of /.

1.

2.

3.

3s8 CHAPTER 1O / INTEREST RATES AND PRESENT VALUES

10.4 Geometric SeriesThis section studies finite and infinite geometric series. These have many applications ineconomics and finance. Here we shall use them to calculate annuities and mortgage pay-ments.

Finite Geometric Series

This year a firm has an annual revenue of $100 million that it expects to increase by 16%per year throughout the next decade. How large is its expected revenue in the tenth year.and what is the total revenue expected over the whole period?

Solution: The expected revenue in the second year (in millions of dollars) amounts to100(1 + 16/100) : 100 . 1.16, and in the third year, it is 100 . (l.lq2.In the tenth year.the expected revenue is I 00 . ( I . I 6)e. The total revenue expected during the decade is thus

100 + 100. 1.16 + 100. (1.16)2 +. . .+ 100' (1.16)e

If we used a calculator to add the ten different numbers, we would find that the sumapproximatel y $2132 million.

Finding the sum in Example I by adding 10 different numbers on a calculator would bevery tedious. When there are infinitely many terms, it is obviously impossible. There is aneasier method, as we now explain.

Consider the z numberc a, ak, ak2, . . . , ek'-r . Each term is obtained by multiplyingthe previous one by a constant k. We wish to find the sum

sn : a * ak * ak2 + . . . I ak"-z r ak"- l

of these numbers. We call this sum a (finite) geometric series with quotient ft. The sum inExample 1 occurs in the case when a : 100, k = 7.76,and n : 10.

To find the sum sn of the series, we use a trick. First multiply both sides of (1) by ft toobtain

ksn : s l< ak2 + ak3 + . . . { ak"-1 q l

SECTION 10.4 / GEOMETRIC SERIES 3s9

In conclusion:

lions iEe pa)-

b1 16"lh 1'ear-

lnl-s l\a

lh lear- is us

|l,r.n i'I

puld befE is an

lipll ing

r l '

sum in

lbr I ttr

r l r

Erms inI - l ts . .

a + ak + ak2 + . ' . - f akn- l : o '=n - , t

k-r (k+r) (3)

For the sum in Example I we have a : 100, k : 1.16, andn : 10. Hence, (3) yields

' ' , ,1n

100 + 100. 1.16 + . . .+ 100. (1.16)e : roo (1,19) ' ' ; I1.16 - INow it takes many fewer operations on the calculator than in Example I to show that thesumis about2l32. I

lnf inite Geometric SeriesConsider the infinite sequence of numbers

l l1 l lt ' t ' 4 ' 8 ' G' n 'Each term in the sequence is formed by halving its predecessor, so that fhe nth term is| 12"-r . The sum of the n first terms is a finite geometric series with quotient k : I 12 andthe first term a : l. Hence, (3) gives

, r 1 I l - (+) ' ^

II r 1+ /2-r . . . * r ,_r- : t _ I :2- 2*,

We now ask what is meant by the "infinite sum"

(*)

l l1 1I

-L - - . , t - - -L - -L. . .'

- , - Z,

- t

- r " ' - r 2n-t- r

" ' (*x)

Because all the terms are positive, and there are infinitely many terms, you might be inclinedto think that the sum must be infinitely large. However, if we look at formula (*), we seethat the sum of the first terms is equal lo 2 - | l2'- I . This number is never larger than 2,irrespective of our choice of 2. As n increases, the term I12"-t comes closer and closer to0, and the sum in (x) tends to 2 as limit. This makes it natural to define the infinite sum in(**) as the number 2.

In general, we ask what meaning can be given to the "infinite sum"

a * ak I akz . , .+ ak"-r +. . . (4)


I We use the same idea as in (**), and consider the sum sn of the n first terms in (4). Accordingi to (3),

l -k"sn:a

l -k &+l)

What happens to this expression as tends to infinity? The answer evidently depends onft", because only this term depends on 2. In fact, k" tends to 0 if -1 < k < l, whereas /cndoesnottendtoaaylimitif fr > 1orft < -1. (If youarenotyetconvincedthatthisclaimis true, study the cases ft : -2, k : -1, k : -l/2, k : l/2,and k = 2.) It follows thatif ll < 1, then the sum sn of the z first terms in (4) will tend to the limit a/(l - k) as ntends to infinity. In this case, we let (4) define the sum, and say that the infinite series in (4)converges. To summarize:

a * ak + ak2 +. . .* ak ' - t * . . . : & i f l f t l l. When fr : 1, then sn : na, which tends to +oo if a > 0or to -oo if a < 0. When /c : -1, then sn is a when n is odd, but 0 when r is even; againthere is no limit as n --> x (if a y' 0).

A rough estimate of the total oil and gas reserves under the Norwegian continental shelfat the beginning of 1999 was 13 bitlion (13 . 10e) tons (ofoil equivalent). Output that yearwas approximately 250 million (250 . 106) tons.

When will the reserves be exhausted if output is kept at the same constant level?Suppose that output is reduced each year by 2Vo per leatbeginning in 1999. How longwill the reserves last in this case?

Solution:(a) The number of years for which the reserves will last is given by

(a)(b)

13. lOe- -

^dccording

bpends onrhereas ft"tthis claimollows thar[- f r )asnries in (4)

SECTION 10.4 / GEOIVFTRIC SERIES 361

(b) In 1999, ourpurwas a :250.106. In2000, itwouldbea -2a/100 : a.0.9g.rn2}ol,it becomes a '0.982, and so on. If this continues foreveq the total amount extracted willbe

a I a.0.98 +a. (0.9g)2 +. . . +a. (0.98),-1 +. . .This geometric series has quotient ft : 0.98. According to (5), the sum is

5----o -5goI _ 0.98

since a : 250' 106, s : 50' 250. 106 : r2.5 . r}e,whichis less than 13 . 10e. Theextraction, therefore, may be continued indefinitely, leaving 500 million (: 0.5 . l0e)tons which will never be extracted.

(5t

t

1 . Find the sum sn of the following finite geometric series

r+l+**.JJ'

When n approaches infinity, what is the limit of s,? Find the sum In, #.

2. Find the sums of the following geometric series:

(a) { + r{r '+ (+)3 + { t )4 +. . .(b) 0.1 + (0.1)2 + (0.1)3 + (0.1)4 +. . .(c) 517 +517(1.1)-1 +5nQ.D-2 *517(1.1)-3+.. .(d) a- la( I - la)-1 - ta( I*a)-2 +a.( I+a)-3 *a(I*a)-4 +. . . , (a > 0)

5.3 5.32 5 .3n-r(e)5* . , +

, +. . .+- ; r +. . .

3. Determine whether the following series are geometric, and find the sums of thosegeometric series that do converge.

(a) 8*r+r l8+r/64+.. . (b) _2+6_18 +s4_.. .@) 2t/3 * l 2- t /z +2-2r +. . . (d) 1- t /2+t/3-t /4+.. .

4. Examine the convergence of the following geometric series, and find their sums when

I+_' 3n-1

(6r

rite ) sum-o. i fa>0rcn: again

tsl shelfl -lat vea

pll

tou'long

they exist:

l1(a) -*-

pp'- lI* r* . . .\/x

I+_+

p- (b) x+Ji+ @) D,',"n: l


5. Find the sum I

r(t * #)--,p. o.Total world consumption of iron was approximatel y 794 . 106 tons in 197 1 . If consump-tion had increased by 57o eachyear and the resources available for mining in 1971 were249 -lO9 tons, how much longer would the world's iron resources have lasted?

The world's total consumption of natural gas was 1824 million tons oil equivalent(mtoe) in 1994. The reserves at the end of that year were estimated to be 128 300 mtoe.If consumption had increasedby 2Vo in each of the coming years, and no new sourceswere ever discovered, how much longer would these reserves have lasted?

Consider Example 10.3.2. Assume that immediately after one tree is felled, a new treeof the same type is planted. If we assume that a new tree is planted at times t, 2t, 3r

-

etc., then the present value of all the trees will be

f ( t ) : P(t)e- ' t + P(t)e-z ' t " '

Find the sum of this infinite geometric series.

Prove that it f (r) has a maximum for some f * > 0, then

P'( t*) rPf f i :1--*

(c) Examine the limit of P'(t.) I P (/*) as r -+ 0.

;E:Eiq* i.r,*:iqi:;" i,i.fiii;,;ll.: iti.l

10.5 Total Present ValueSuppose that thee successive annual payments are to be made, with the amount $100[tfalling due after 1 year, then $1500 after 2 years, and $2000 after 3 years. How muchmust be deposited in an account today in order to have enough savings to cover these threepayments, given that the interest rcteis lIVo per year? We call this amount the present valueof the three payments.

In order to have $1000 after I year, we must deposit an amount -r1 today, where

6.

7.

#

8.

(a)(b)

xr(1 *0.11) : 1000, sothar 1000 1000^t - l +o. l l - 1 l lIn order to have $ 1500 after 2 years, we must deposit an amount x2 today, where

-\;( I r 0.1 l.l : 15gg. so rhat 15oo 15oo12: J :nP: (r . t ,F

SECTION 10.5 / TOTAL PRESENT VALUE

Finally, to have $2000 after 3 years, we must deposit an amount x3 today, where

353

MrIl rrct

lrltflllm,m-tllrw

E! IIEE

, lr- f,r-

[t(mrebiihee'w,ll

xz\ -l0.11)3 : 2000, so that 2000 200013: a +oJTF: { , . t '

So the total present value of the three payments, which is the total amount A that must bedeposited today in order to cover all three payments, is given by

1000 1500 2000^: t r f - { t r t - ( t .ur ,

The total is approximarely A x 900.90 + 1217.43 + 1462.39: 3590.71.Suppose, in general, that successive payments ar, . . . , an aretobe made, with a1 being

paid after L year, a2 after 2 years, and so on. How much must be deposited into an accounttoday in order to have enough savings to cover all these future payments, given that theannual interest is r? In other words, what is the present value of all. these payments?

In order to have al after l yea4 we must deposit al/Q f r) today, tohave a2 after 2yearswe must deposit a2/Q -f r)2 today, and so on. The total amount Pn thatmust be depositedtoday in order to cover all n payments is therefore

Here P, is the present value of the n instalments.An annuity is a sequence of equal payments made at fixed periods of time over some

time span. If a1 - e2 : .' . : an : a in (1), then (l) represents the present value of anannuity. In this case the sum in (1) is a finite geometric series with r? terms. The first term isa / (l I r) and the quotient is 1 / ( 1 * r). According to the summation formula for a geometricseries, (3) in the previous section, with fr : (1 * r)-1, the sum is

pn - --L _Jl:llt.t-l] =!lt- -L l( l * r ) t l - ( l * r ) - t j r l ' ( l * r )"J

(The second equality holds because the denominator of the middle expression reduces to r.)Hence, we have the following.

The present value of an annuity of a per payment period for n periods at therate of interest r per period, where each payment is at the end of the period, isgiven by

pn--! -+. . .+ - : - : ! l t - - f I wherer:olr0o' " I l r ( l - l r ) ' r l (1*r)nl - . - r '

(1)

(2)


This sum is illustrated below:

a

1+/

o +;P:

0+rY

Formula (2) gives the present value of n future claims, each of a (say) dollars. If we wanrto find how much is accumulated in the account after n periods, immediately after the lastdeposit, then the future value F, ofthe annuity is given by:

Fn : a * a( l * r ) * a( l + r)z +. . . * a( l + r)n- lThis different sum is illustrated below:

a,a

a( l I r )

a( I * r )z

:

a( l + r)"-r

The summation formula for a geometric series yields:

-

a l l - ( l * r ) ' l a - . -h' :

'T i t +; ; : - [ ( l * r ) ' - t lWe can also find the future value directly by calculating the interest earned on the presenrvalue P, in n periods: Fn: P,(l i r)n - t( l + r), - l l . So:

FUTURE VAIUE O.F AN ANNUITY

An amount a is deposited in an account each period forn periods, earning interestat r per period. The future (total) value of the account, immediately after the lastdeposit, is

4,.Fr:- [ ( l * r )"-1]

r

ExAMPLE 1 Compute the present and the future values of a deposit of $1000 in each of the comine8 reas ifthe annual interest rateis6Vc.

SECTION 10.5 / TOTAL PRESENT VALUE 365

Solution: To find the present value, we apply formula (2) with a : 1000, : 8 andr : 61100: 0.06. This gives

1000/ I \o06f ' - , , ro/x62oe'7e

The future value is found by applying formula (3), which gives

or: #[ t1.06)s - r ]ogsst .+tAlternatively, Fs : Ps(1.06)s * eZOg.l9(1.06)8 x 9891.47.

rs. If we wantI' after the last

rn the presenl

Estlst

f the coming

If r > 0 and we let n approach infinity in (2), then (l * r)' approaches infinity and Pnapproaches alr. So in the limit,

aa_+_+.. .l+r ( l i r )2

a

r( r>o) (4)

This corresponds to the case where an investment pays 4 per period in perpetuity when theinterest rate is r.

EXAMPLE 2 Compute the present value of a series of deposits of $1000 at the end of each year inperpetuity when the annual interest rate is 147o.

Solution: According to formula (4), we obtain1000 1000 1000

*._- l - . . . : -x7142.86l+0.14 (r+0.t4f 0.14

l? l

Present Value of a Continuous Future Income StreamWe have discussed the present value of a series of future payments made at specific discretemoments in time. It is often more natural to consider revenue as accruing continuouslv. likethe timber yield from a large growing forest.

Suppose that income is to be received continuously from time f : 0 to time / : T attherate of /(r) dollars per year at time /. We assume that interest is compounded continuouslyatrate r per year.

Let P (t) denote the present discounted value (PDV) of all payments made over the timeinterval [0, ]. This means that P(I) represents the amount of money you would have todeposit at time / : 0 in order to match what results from (continuously) depositing theincome stream /(r) over the time interval [0, Z]. If Ar is any number, the present value ofthe income received during the interval lt,t + Ltl is P(/ * Lt) - P(t).If A/ is a smallnumber, the income received during this interval is approximately f (t) Lt, and the PDV ofthis amount is approximately f (t)e-'t A. Thus, P(t + Lt) - P(t) x f (t)e-'t A/ and so

lP(t + Lt) - P(t) l /6 x f ( t )e- ' t

CHAPTER 1 O / INfEREST RATES AND PRESENT VALUES

This approximation gets better the smaller is Ar, and in the limit as A/ -+ 0, we haveP'( t ) : f ( t )e-" . Bythedef in i r ionof thedef in i te integral , P(T)-P(0) : { ; f ( t )e- ' t dt .Because P(0) : 0, we have the following:

PRESENT VATUE OF: A CONTINUCUS INCOME..STRE:AM

The present (discounted) value (at time 0) of a continuous income stream at therate of /(l) dollars per year over the time interval [0, Z], with continuouslycompounded interest atrate r per year, is given by (5)

pDV: [ ' f , , ) r - , , 0,,lo

Equation (5) gives the value at time 0 of income stream /(r) received during time interval[0, Z]. The value of this amount at time Z, with continuously compounded interest atrate r.is e'r ld f G)e-" d/. Because the number e'r is aconstant, we can rewrite the integral a-sIl f Or"' -t) dt. This is called the future discounted value (FDV) of the income srream:

FUTURE VALUE OF A CONTINUOUS.INCOME SIREAM

The future (discounted) value (at time Z) of a continuous income stream at therate of /(r) dollars per year over the time interval [0, Z], with continuouslycompounded interest atrate r per year, is given by

An easy modification of (5) will give us the discounted value (DV) at any time s e [0. f ,of an income stream /(l) received during time interval [s, Z]. In fact, the DV at time s ofincome /(l) received in the small time interval [t, t + dtl is f (t)-rtt-') d/. So we halethe followine:

DISCOUNIED VALUE OF A CONTINUOUS lN:COM.E,STREAM

The discounted value at any time s of a continuous income stream at the rate of/ (r) dollars per year over the time interval [s, Z], with continuously compoundedinterest at rate r per year, is given by

FDV: lo' f , , rr , ,r . , o,

",

: I:,f

(t)e-,(t-s) t

SECTION 10.5 / TOTAL PRESENT VALUE 367

f XAtMirU!] Find the PDV and the FDV of a constant income stream of $1000 per year over rhe next10 years, assuming an interest rate of r : 87o :0.08 annually, compounded continuously.

r'. e halel te d .

: inten raI rate r.fiegral - 0. Then( l )hasauniquesolut ionsatisfying 1 * r* > 0, that is a unique internal rate of return rx > -1. Also, the internalrate of return is positive if I:o a > 0. You are asked to prove these results in Problem 3.

3.

4.

5.

6.

1 . An investment project has an initial outlay of $50 000 and at the end of each of the nexttwo years has returns of $30 000. Find the associated internal rate of return r.

2. Suppose that in (1) we have as < 0 and ai : a > 0 for i : I, 2, . . .. Find an expressionfor the internal rate of return in the limit as n -+ oo.

Consider an investment project with an initial loss, so that ao < O, and thereafter nolosses. Suppose also that the sum of the later profits is larger than the initial loss. Pror-ethe two claims in Note l. (Hint: Define f (r) as the expression on the left side of (l r.Then study the signs of flr) and f'{r) on the interval (0, oo).)

An investment in a certain machine is expected to eafii a profit of $400 000 each year-What is the maximum price that should be paid for the machine if it has a lifetime of -years, the interest rate is l7 .5%, and the annual profit is earned at the end of each year.'

HARDER PROBLEMS

An investment project has an initial outlay of $100000, and at the end ofeach ofthenext 20 years has a return of $10 000. Show that there is a unique positive internal rateof return, and find its approximate value. (Hint: lJse s : (1 + r)-l as a new variable.Prove that the equation you obtain for s has a unique positive solution. Verify thars : 0.928 is an approximate root.)

A is obliged to pay B $1000 yearly for 5 years, the first payment in I year's time. Isells this claim to C for $4340 in cash. Find an equation that determines the rate ofreturn p that C obtains from this investment. Prove at it is a little less than 5%.

e;{,+Xtffi;1. (a) An amount $5000 earns interest at3Eo peryear. What will this amount have grorm

to after 10 years?

(b) How long does it take for the $5000 to double?

2. An amount of 8000 euros is invested at 5Vo per year

la) What is the balance in the account after 3 years?tbl \\'hat is e balance after 13 years?t c r Hor' long does it take (approximately) for the balance to reach 32 000 euros ?

runique solutiontlso. the internalIts in Problem

_3.

l .='each

of the nerteturn r.

nd an expression

nd thereafter notial loss. Pror,eleft side of (l ).

)000 each year.s a lifetime of 7I of each year?

I of each of theire internal ratea new variable.on. Verify that

lear's time. Bnes the rate ofl.an 5Vo.

' l t

nt have grown

3.

4.

RIVITW PROBLEM5 FOR CHAPTER 1O 5I>

Which is preferable for a borrower: (i) to borrow at the annual interest rate of lTVowith interest paid yearly; or (ii) to borrow at annual interest rate 10Vo with interest paidmonthly?

Suppose the sum off,l5 000 is invested in an account earning interest at an annual rateof 7 7o . What is the balance after 12 years if interest is compounded continuously?

5. (a) How much has $8000 increased to after 3 years if the annual interest rafe s 6Vo,with continuous compounding?

(b) How long does it take before the $8000 has doubled?

6. Find the sums of the followins infinite series:

(a) 44 * 44 .0.56 + 44 . (0.56)2+ . . .

3.2 3.22 3.2n-r(c)3* -

*

-

* . . .*

-

* . . .) ) -

o) i ^(i)'@i,*

F

7. (a) Find the present discounted value (PDV) of a constant income stream of a dollarsper year over the next Z years, assuming an interest rate of r annuall compoundedcontinuously.

(b) What is the limit of the PDV as Z -> oo? Compare this result with ( 1 0.5.4).

8. (a) At the beginning of a year $5000 is deposited in an account earning 4Vo annlalinterest. What is the balance after 4 years?

(b) At the end ofeach year for four years, $5000 is deposited in an account eaning4%oannual interest. What is the balance immediately after the fourth deposit?

(c) Suppose you had $10000 in your account on lst January 1996. You agreed todeposit a fixed amount K each year for 8 years, the first deposit on lst January1999. What choice of the fixed amount K will imply that you have a balance of$70 000 immediately after the last deposit? The annual interest rafe is 4%o.

9. A business borrows 500000 euros from a bank at the beginning of one year, and issupposed to pay it off in 10 equal instalments at the end of each year, with interest at'7 Eo compotrnding annually.

(a) Find the annual payment. What is the total amount paid to the bank?

(b) What is the total amount if the business has to pay twice ayear?I) euros?


10. Lucy is offered the choice between the following three options:

(a) She gets $3200 each year for 10 years. First payment due after I year.(b) She gets $7000 today, and thereafter $3000 each year for 5 years. First payment

after I year.

(c) She gets $4000 each year for 10 years. First payment only due after 5 years.The annual interest rute is 8Vo. Calculate the present values of the three options. Whatwould you advise Lucy to choose?

1'l . The revenue produced by a new oil well is $1 million per year initially (r : 0), andit is assumed that it will rise uniformly to $5 million per year after 10 years. If wemeasure time in years and let f (t) denote the revenue (in millions of dollars) per unitof time at time , it follows that f (t) - 1+ O.4t.If F(l) denotes rhe rotal revenuewhich accumulates over the time interval [0, ], then Ft(t) : f (t).(a) Calculate the total revenue earned during the 10 year period (i.e. F'(10)).(b) Find the present value of the revenue stream over the time interval [0, 10], if we

assume continuously compounded interest at the rate r : 0.05 per year.

F. l t r rC

t10.interest rates&present value

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